02.

STRENGTH OF MATERIALS
1. Simple Stress and Strain
1. Poisson's ratio is expressed as
(a) Lateral stress/lateral strain
(b) Longitudinal stress/longitudinal strain
(c) Lateral strain/longitudinal strain
(d) Lateral stress by longitudinal stress
VIZAG MT, 14.12.2020
TSPSC Manager (Engg.) HMWSSB 12.11.2020
RPSC Vice Principal ITI 2018
RPSC AE 2018 σ ∝ ∈; σ = E∈
HPPSC AE 2018 So, Hooke's law holds good upto proportionality limit.
TRB Polytechnic Lecturer 2017 E is proportionality constant known as Young's
CGPSC Polytechnic Lecturer 2017 Modulus of elasticity.
HPPSC Asstt. Prof. 20.11.2017 Note-E (Elastic constant) is a material property.
GPSC Lect. (Auto) 16.10.2016 3. The relation between modulus of elasticity (E).
Rajasthan Nagar Nigam AE 2016, Shift-I modulus of rigidity (G) and bulk modulus (K)
CGPSC AE 16.10.2016 is given by:
Mizoram PSC AE/SDO Paper-II 2014 K+G 3KG
TNPSC AE 2013 (a) (b)
APPSC AEE 2012 3K + G 3K + G
UKPSC AE 2012 Paper-I K+G 9KG
(c) (d)
APPSC IOF, 2009 9K + G 3K + G
Ans. (c) : Poisson ratio (µ)-The ratio of the transverse MECON MT 2019
contraction of a material to the longitudinal extension Oil India Limited Sr. Engineer (Drilling) 30.11.2019
strain in the direction of the stretching force is the OPSC AEE 2019 Paper-I
Poisson's Ratio for a material. RPSC AE 2018
This Poisson's Ratio for most of the materials is in the TNPSC AE 2013
range of 0 to 0.5. UJVNL AE 2016
When the Poisson's Ratio is 0 there is no reduction in APPSC AE 2012
the diameter or one can even say there is no laterally UKPSC AE 2012, 2013 Paper-I
TRB Polytechnic Lecturer 2017
contraction happening when you are elongating the GPSC Engineer, Class-II Pre-19.01.2020
material but the density would reduce. The value 0.5 CSE Pre-1995
indicates the volume of the material or object will ESE 2009
remain the same or constant during the elongation Ans. (d) : Relationship between young modulus (E),
process or when the diameter decrease of material when Rigidity modulus (G), Bulk modulus (K), and poisson
the material is elastomeric. ratio (µ)-
Rubber (µ) = 0.5, Cork (µ) = 0. E = 2G (1 + µ), E = 3K (1 – 2µ)
2. Hooke's law holds good upto 9 KG
E=
(a) Yield point 3K + G
(b) Limit of proportionality 4. In terms of Poisson's ratio (v), the ratio of
(c) Ductile limit Young's modulus (E) to Shear modulus (G) of
(d) Breaking point plastic material is :
Haryana PSC AE (PHED) 05.09.2020, Paper-II (a) 2 (1+v) (b) 2 (1 – v)
OPSC AEE 2019 Paper-I (c) (1 + v)/2 (d) (1 – v)/2
Rajasthan AE (Nagar Nigam) 2016 Shift-3 PPSC Asstt. Municipal Engg. 15.06.2021
Assam PSC CCE Pre 2015 Oil India Limited Sr. Engineer (Production) 30.11.2019
UPRVUNL AE 2014 GPSC ARTO Pre 30.12.2018
Mizoram PSC AE/SDO Paper-II 2014 Nagaland PSC (CTSE) Paper-I 2018
SJVN ET 2013 GPSC Executive Engg. 23.12.2018
TNPSC ACF 2012 TSPSC AEE 2017
VIZAG Steel MT 2011 HPPSC Asstt. Prof. 18.11.2016
UKPSC AE 2007 Paper -I CGPSC AE 26.04.2015 Shift-I
VIZAG Steel MT 2011
Ans. (b) : In this stress-strain diagram. Straight line O- GATE-2004
A is called proportionality limit.
Strength of Materials 157 YCT
Ans. (a) : The relation is, Ans. (b): Elongation due to self weight
E = 2G (1 + v) = 3K (1 – 2v) ρgL2 γL2
E δ self = = [∵ γ = ρg ]
= 2 (1 + v) 2E 2E
G 7. The relation between modulus of elasticity E,
5. The stress induced in a body, when suddenly modulus of rigidity G, bulk modulus K and
loaded, is _________ the stress induced when Poisson's ratio µ is
the same load is applied gradually. (a) E = G ( µ + 1) (b) E = 2G ( µ + 1)
(a) equal to (b) one-half
(c) twice (d) four times (c) E = 4G ( 2µ + 1) (d) E = 2G ( µ − 1)
GPSC DEE, Class-2 (GWSSB) 04.07.2021 Haryana PSC AE (PHED) 05.09.2020, Paper-II
Sikkim PSC (Under Secretary), 2017 WBPSC AE, 2017
WBPSC AE, 2017 ISRO Scientist/Engineer 17.12.2017
APPSC AE Subordinate Service Civil/Mech. 2016 (CGPSC Polytechnic Lecturer 2017)
HPPSC Asstt. Prof. 29.10.2016 APPSC AEE Screening Test 2016
MPPSC AE 08.11.2015 Mizoram PSC AE/SDO Paper-II 2014
Haryana PSC Civil Services Pre, 2014 ISRO Scientist/Engineer 2009
APPSC AEE 2012
APPSC AE 04.12.2012 Ans. (b) : E = 2G ( µ + 1)
Ans. (c) : 8. The relation between modulus of elasticity (E)
(1) Gradually applied load- & bulk modulus (k) is
(a) E= 3k(1 – 2*Poisson's ratio)
(b) E= 3k(1 + Poisson's ratio)
(c) E= 3k(1 + 2 Poisson's ratio)
(d) E= 3k(1 – Poisson's ratio)
Rajasthan Nagar Nigam AE 2016, Shift-II
Rajasthan AE (Nagar Nigam) 2016 Shift-3
J & K PSC Screening, 2006
UPRVUNL AE 21.08.2016
OPSC Civil Services Pre 2006
W GATE-2002
σgradual =
A Ans. (a) : The relation between modulus of elasticity
(2) Suddenly applied load (E) & bulk modulus (k)
E = 3 k (1–2µ)
E = 2G(1 + µ)
9KG
E=
3K + G
3K − 2G
µ =
6K + 2G
Where µ – Poisson's ratio
9. The value of Poisson's ratio for steel varies
2W
σsudden = from:
A (a) 0.20 to 0.25 (b) 0.25 to 0.35
σsudden = 2σ gradual (c) 0.35 to 0.40 (d) 0.40 to 0.55
GPSC DEE, Class-2 (GWSSB) 04.07.2021
(3) Impact Load- HPPSC AE 2018
W 2h  Nagaland PSC (CTSE) Paper-I 2018
σimpact = 1 + 1 +  Vizag Steel (MT) 2017
A δ 
GMB AAE 25.06.2017
6. A solid metal bar of uniform diameter D and Rajasthan Nagar Nigam AE 2016, Shift-II
length L is hung vertically. If ρ is density and
E is the Young's Modulus, then the total Ans. (b) : Poisson's ratio (µ) – the ratio of the
elongation due to self weight is transverse contraction of a material to the longitudinal
(a) ρLg/2E (b) ρL2g/2E extension strain in the direction of stretching force. is
(c) ρEg/2L 2
(d) ρL2g/E the Poisson's ratio of material.
SJVN ET 2019 The value of Poisson's ratio for
ISRO Scientist/Engineer 22.04.2018 Steel (µ) = 0.25 to 0.33
Nagaland PSC (CTSE) Paper-I 2018 Rubber (µ) = 0.50
TSPSC AEE 28.08.2017 (Civil/Mech.)
Cork (µ) = 0
GPSC Poly. Lect. 07.12.2014
UPSC JWM Adv. No.-16/2009 Aluminium (µ) = 0.32
UKPSC AE 2007 Paper -I Concrete (µ) = 0.20
Strength of Materials 158 YCT
10. The deformation of a bar under its own weight Assam PSC AE (IWT) 14.03.2021
as compared to that when subjected to a direct VIZAG Steel MT 24.01.2021
axial load equal to its own weight will be APPSC AEE SCREENING 17.02.2019
(a) half (b) double SJVN ET 2019
(c) one fourth (d) the same Rajasthan Nagar Nigam AE Shift-II 2016
RPSC Lect. (Tech. Edu. Dept.) 13.03.2021, Paper-I TNPSC AE (Industries) 09.06.2013
Nagaland PSC (CTSE) Paper-I 2018 Ans. (d) : Shear modulus or modulus of rigidity is
UKPSC AE-2013, Paper-I defined as the ratio of shear stress to the shear strain.
APPSC AEE 2012, ESE 1998
Shear stress (τ)
Ans. (a) : Deflection due to load P Modulus of rigidity (G) =
Shear strain ( γ )
PL
δ1 = 14. A bar uniformly tapering from a diameter d1
AE at one end to a diameter d2 at the other end, is
Deflection due to self load subjected to an axial tensile load P. The length
PL of the bar is ℓ and the young's modulus is E.
δ2 = The extension of the bar is
2AE
4Pℓ PℓE
 PL  (a) (b)
πEd d 4 π d1d 2
δ2  2AE  1 1 2
= = Pℓ 4Pℓd1d 2
δ1  PL  2 (c) (d)
 AE  4πEd1d 2 πE
11. Which of the following materials is most GPSC Engg. Class-2 Pre 19.01.2020
elastic. GPSC Executive Engineer 23.12.2018
(a) glass (b) brass APPSC AE Subordinate Service Civil/Mech. 2016
(c) plastic (d) steel APPSC IOF, 2009
GWSSB AAE, 27.12.2015 CSE Pre-1996
APPSC AEE 2012
ESE 1995
APPSC AE 04.12.2012
J&K PSC Civil Services Pre, 2010 Ans. (a) :
CSE Pre-1995
Ans. (d) : Elasticity is the property of solid materials to
return to their original shape and size after the forces
deforming them have been removed.
Steel is more elastic than any other thing because steel
comes back to its original shapes faster than rubber
when the deforming forces are removed.
12. A steel bar of 40 mm x 40 mm square cross-
section is subjected to an axial compressive Consider a small strip of thickness 'dx' at a distance 'x'
load of 200 kN. If the length of the bar is 2m from smaller dia side.
and E = 200 GPa, the elongation of the bar will Let the dia of the bar be d' at distance x
be:  d −d 
(a) 1.25 mm (b) 2.70 mm d' = d1 –  1 2  x
(c) 4.05 mm (d) 5.40 mm  ℓ 
ISRO Scientist/Engineer (RAC) 22.04.2018  d − d2 
RPSC Vice Principal ITI 2018 Let,  1 =K
RPSC ADE 2016  ℓ 
UPRVUNL AE 2014 ∴ d ' = d − Kx
1
VIZAG Steel MT 18.06.2013
VIZAG Steel MT 10.06.2012 ∴ cross sectional area at distance x from the smaller
GATE-2006 end = A' = π d12 = π ( d − kx )2
1
Pl 4 4
Ans. (a) : Elongation or shortenig δl = Intensity of stress on the section = σ'
AE
3 P 4P
200 × 10 × 2 = =
= π ( d1 − kx )
2
A '
( 0.040 × 0.040 ) × 200 ×109
=1.25×10–3 m σ' 4P
∴ Strain = e' = =
=1.25 mm E πΕ ( d1 − kx )2
13. Modulus of rigidity is defined as the ratio of
(a) longitudinal stress to longitudinal strain ∴ Extension of the length dx = e'dx
(b) volumetric stress to volumetric strain 4P
(c) lateral stress to lateral strain = .dx
πE ( d1 − kx )
2
(d) shear stress to shear strain
Strength of Materials 159 YCT
 ℓ Ans. (a) : If the value of Poisson's ratio is zero then it
4P dx
∴ Total extension of the bar = δ =
πE ∫ (d
0 1 − kx )
2
means that the material is rigid.
17. The unit of elastic modulus is the same as those
ℓ of
4P  1  (a) Stress, strain and pressure
δ=−   (b) Strain, shear modulus and pressure
πEK  d1 − kx 0
(c) Shear modulus, stress and force per unit area
4P  1 1 (d) Strain, shear modulus and force
δ=  −  GPSC Executive Engg. 23.12.2018
πEK  d1 − kℓ d1 
HPPSC Asstt. Prof. 18.09.2017
d − d2 APPSC AE 04.12.2012
put, K = 1
ℓ UKPSC AE 2012 Paper-I
 APPSC AEE 2012
4Pℓ 1 1
δ=  −  Ans. (c) : Unit of elastic modulus (E) = MPa or Pa or
πE ( d1 − d 2 )  d1 − d1 + d 2 d1  N/m2
4Pℓ 1 1 (i) Shear modulus (G) = MPa or Pa or N/m2
δ=  −  (ii) Stress (σ) = MPa or Pa or N/m2
πE ( d1 − d 2 )  d 2 d1 
F
4Pℓ d − d2 (iii) Force per unit area = N / m2
δ= . 1 A
πE ( d1 − d 2 ) d1d 2 18. For a linearly elastic, isometric and
4Pℓ homogeneous material , the number of elastic
δ= constants required to relate stress and strain
πEd1d 2
are :
For the particular case when the rod is of uniform dia. (a) Four (b) Two
d1 = d2 = d (c) Three (d) Six
4Pℓ UPRVUNL AE 05.07.2021
and for this case, δ =
πEd 2 RPSC Lect. (Tech. Edu. Dept.) 13.03.2021, Paper-I
15. If Poisson's ratio is 0.25, then the modulus of OPSC AEE 2015 Paper-I
rigidity (G) and the modulus of elasticity (E) TSPSC AEE 2015
are related as APPSC AEE 2012
(a) G = 0.33 E (b) G = 0.5 E ESE 1998
(c) G = 0.4 E (d) G = 0.25 E CSE Pre-1990
CGPSC AE 15.01.2021 Ans : (b) For a linearly elastic, isometric and
RPSC 2016 homogeneous material, the number of elastic constant
APPSC AEE Screening Test 2016 required to relate stress and strain are two.
Kerala LBS Centre For Sci. & Tech. Asstt. Prof.
2014 19. If the radius of wire stretched by a load is
APPSC Poly. Lect. 2013 doubled, then its Young's Modulus will be
TNPSC AE, 2008 (a) Doubled
Ans. (c) Given, µ = 0.25 (b) Halved
We know that, E = 2G(1+µ) (c) Become four times
E = 2G(1+.25) (d) Become one fourth
E = 2G × 1.25 (e) None of the above
E = 2.50G TNPSC AE 2017
E = 2G(1+µ) Rajasthan Nagar Nigam AE 2016, Shift-II
10 CGPSC Poly. Lect. 22.05.2016
G= E
25 ISRO Scientist/Engineer 2006
2 Ans. (e) : The Young's Modulus of any materials is
G= E equal to the ratio of stress and the strain.
5
G = 0.4E Its formula is given by –
σ
16. If the value of Poisson's ratio is zero then it E=
means that : ε
(a) The material is rigid or
(b) The material is perfectly plastic F/ A
(c) There is no longitudinal strain in the material E=
(d) The longitudinal strain in the material is ∆L / L
infinite. The Young's Modulus of any material is inherited
OPSC AEE 2019 Paper-I property. Its only depends on the temperature and the
GPSC ARTO Pre 30.12.2018 pressure.
Kerala PSC IOF 19.04.2016 So, if the radius of a wire stretched by a load is
KPSC ADF 2015 doubled then its Young's Modulus will remain
ESE 1994 unaffected.
Strength of Materials 160 YCT
20. The number of elastic constants for a Example–Wood, Aluminium, Copper, Steel and Gold.
completely anisotropic elastic material are :
(a) 3 (b) 4 Isotropic Material–The material which exhibit the
(c) 21 (d) 25 same elastic properties in all direction are called
Kerala PSC Poly. Lect., 2017 isotropic.
APPSC AEE 2012 Example–Steel, Brass.
APPSC AE 04.12.2012
ESE 1999 Homogeneous and Isotropic–A material is said to be
Ans. (c) : homogeneous and isotropic when it exhibits same
Material Number of independent elastic elastic property in any direction at any point.
constant Example–Steel, Copper, Gold, Aluminum.
Homogeneous 2
& Isotropic Orthotropic Material–A material is said to be
Orthotropic 9 orthotropic when it exhibits different elastic property in
Anisotropic 21 orthogonal direction at a given point.
21. A steel bar of 5 mm is heated from 15 °C to Example–Any layered material like plywood, graphite.
40°C and it is free to expand. The bar will
induce Anisotropic Material–A material which exhibits
(a) No stress (b) Shear Stress direction dependent elastic properties is known as
(c) Tensile Stress (d) Compressive stress anisotropic material.
Nagaland PSC (CTSE) Paper-I 2018 Example–Composite material.
RPSC ADE 2016 24. For a given material. Young's Modulus is
APPSC AEE Mains (Civil Mechanical) 2016 200 GN/m2 and Modulus of Rigidity is
Haryana PSC Civil Services Pre, 2014 80 GN/m2. Its Poisson Ratio will be:
Ans. (a) : A bar subjected to heating and allow to free (a) 0.15 (b) 0.20
expansion will not experience any stress. The stress (c) 0.25 (d) 0.35
produced due to oppose of expansion or contraction. SJVN ET 2013
UPSC JWM Adv. No-16/2009
22. A solid uniform metal bar of cross-section area UKPSC AE 2007 Paper -I
A and length L is hanging vertically from its CSE Pre-2007
upper end. If w is the total weight of bar and E
the Young's modulus of elasticity, the Ans. (c) : Given, E = 200 GN/m2
elongation of bar due to self-weight would be G = 80 GN/m2
(a) WL/2AE (b) WL/4AE E = 2G (1 + µ)
(c) WL/AE (d) 2WL/AE 200 = 2 × 80 (1 + µ)
APPSC AEE SCREENING 17.02.2019 µ = 0.25
WBPSC AE, 2017
25. Which one of the following is rupture stress?
APPSC AEE Mains (Civil Mechanical) 2016 (a) Breaking stress
DRDO Scientist 2009 (b) Maximum load/original cross-sectional area
Ans. (a) : Specific weight, γ = Weight/Volume (A)
Axγ (c) Load breaking point /Area (A)
σ= = xγ
A (d) Load at breaking point/neck area
σ xγ Assam PSC AE (PHED) 18.10.2020
∈= = TSPSC AEE 28.08.2017 (Civil/Mechanical)
E E CSE Pre-2006
Total elongation of bar due to self weight,
L L
TSPSC (Telangana)
xγ Ans. (a&d) :
= ∫ ∈ dx = ∫ dx
0 0
E
γL2 WL
= =
2E 2AE
23. The material which exhibit the same elastic
properties in all direction are called
(a) homogeneous (b) inelastic
(c) isotropic (d) anisotropic
B = Rupture stress or Breaking stress.
APPSC AEE SCREENING 17.02.2019
26. If a material expands freely due to heating it
Mizoram PSC AE/SDO 2014, Paper-II will develop
APPSC AEE 2012 (a) Thermal stresses (b) Tensile stress
APPSC AE 04.12.2012 (c) No stress (d) Bending
Ans. (c) : Homogeneous Material–A material is said Assam PSC AE (PHED), 18.10.2020
to be homogeneous when it exhibits same elastic (CGPSC Polytechnic Lecturer 2017)
properties at any point in a given direction i.e. UPRVUNL AE 2014
properties are independent of point. ISRO Scientist/Engineer 2007
Strength of Materials 161 YCT
Ans. (c) : If a material expands freely due to heating it Ans. (a) :
will develop no stress i.e. zero stress.
27. Young's modulus of elasticity and Poisson's
ratio of a material are 1.25 × 105 MPa and 0.34
respectively. The modulus of rigidity of the
material is
(a) 0.4025 ×105 MPa (b) 0.4664 ×105 MPa If temperature is increased by ∆T then
5
(c) 0.8375 ×10 MPa (d) 0.9469 ×105 MPa Stress developed in the rod is zero. (σth = 0) because
APGCL AM, 2021 there is no any restriction.
APPSC AEE 2012 α∆T × L
APPSC AE 04.12.2012 Strain developed in the rod ε =
CSE Pre-1994 L
Ans. (b) : ε = α∆T
Given:- 30. A steel rod 2m long is heated through a
E = 1.25 × 105 MPa temperature of 100°C. The coefficient of linear
µ = 0.34 expansion is 6.5 × 10–6/°C and Young's
E = 2G (1 + µ) Modulus is 2 × 106 N/m2, the stress induced in
the bar will be :
1.25 × 105 (a) 1000 N/m2 (b) 1200 N/m2
=G
2 (1 + 0.34 )
2
(c) 1300 N/m (d) 1400 N/m2
G = 46641.79 MPa GPSC ARTO 01.05.2016
5
= 0.4664 × 10 MPa MPPSC AE 08.11.2015
MPPSC State Forest Service Exam, 2014
28. The bulk modulus of a body is equal to
Ans. (c) : Given, α = 6.5 × 10–6/°C, L0 = 2m
mE mE E = 2 × 106 N/m2 ∆t = 100°C
(a) (b)
3( m – 2) 3( m + 2) ∆L = L0 α ∆T
mE mE Thermal stress = E α ∆T
(c) (d) = 2 × 106 × 6.5 × 10–6 × 100
2 ( m – 2) 2 ( m + 2) = 1300 N/m2
1 31. Bars of copper and steel form a composite
(where = Poisson's Ratio and E = Young's
m system. They are heated to a temperature of
Modulus of Elasticity) 400C. What type of stress is induced in the
RPSC IOF, 2020 copper bar?
Nagaland PSC (CTSE) Paper-I 2018 (a) Tensile
(b) Compressive
TNPSC ACF 2012
(c) Both tensile and compressive
Ans. (a) : The bulk modulus of a body (d) Shear.
mE GPSC Poly. Lect. 07.12.2014
K=
3( m – 2) ESE 2013
TRB Asstt. Prof., 2012
Where, K = bulk modulus Ans. (b) : αcopper = 16×10–6/°C
E = modulus of elasticity αsteel = 12×10–6/°C
1 Thermal expansion of copper material is more as
= µ = Poissons ratio compare to steel hence as temperature increases the
m
copper material elongate more than steel hence copper
29. A circular rod of length L and area of cross- material experiences compression while steel material
section A has a modulus of elasticity E and experiences tension.
coefficient of thermal expansion α. One end of 32. The maximum stress produced in a bar of
the rod is fixed and the other end is free. If the tapering section is at
temperature of the rod is increased by ∆T, (a) smaller end (b) larger end
then (c) middle (d) anywhere
(a) Stress developed in the rod is zero and strain Haryana PSC Civil Services Pre, 2014
developed in the rod is α∆T VIZAG STEEL MT 2011
(b) Both stress and strain developed in the rod Ans. (a) : The maximum stress produced in a bar of
are zero tapering section is at smaller end.
33. The Young's modulus of elasticity of a material
(c) Stress developed in the rod is Eα∆T and is 2.5 times its modulus of rigidity. The
strain developed in the rod is α∆T poission's ratio for the material will be:
(d) Stress developed in the rod is Eα∆T and (a) 0.50 (b) 0.75
strain developed in the rod is zero (c) 0.33 (d) 0.25
Assam Engg. College AP/Lect. 18.01.2021 HPPSC Asstt. Prof. 18.09.2017
ISRO Scientist/Engineer 07.05.2017 UPRVUNL AE 21.08.2016
TSPSC AEE 2015, GATE 2014 CSE Pre-1997
Strength of Materials 162 YCT
Ans. : (d) Relation between Young's Modulus and 37. Deformation per unit length in the direction of
modulus of rigidity force is known as
E = 2G (1 + µ) (a) Lateral strain (b) Linear strain
E = 2.5 G (c) Tensile stress (d) Linear stress
2.5 G = 2G (1+µ) HPPSC Poly. Lect. 05.07.2021
2.5 = 2 + 2µ Rajasthan Nagar Nigam AE 2016, Shift-I
2µ = 0.5 UPRVUNL AE 2014
µ = 0.25 Ans. (b) : Deformation per unit length in the direction
34. The ratio of modulus of rigidity to modulus of of force is known as linear strain.
elasticity for a Poisson's ratio of 0.25 would be
(a) 0.5 (b) 0.3
(c) 0.4 (d) 0.2
(e) None of the above
CGPSC AE 16.10.2016
GPSC Poly. Lect. 07.12.2014
ESE 2007
Ans. (c) : As we know,
Deformation ( δℓ ) per unit length in the direction of
E
G= force/load is know as linear/longitudinal strain.
2(1 + µ) δℓ ℓ f − ℓ o
E G 1 ∴linear or longitudinal strain ( (∈long ) = = .
So, = 2(1 + µ) or = ℓo ℓo
G E 2(1 + µ) 38. A steel rod 10 mm in diameter and 1 m long is
G 1 heated from 20 ºC to 120 ºC,E = 200 GPa and
= = 0.4
E 2(1 + 0.25) α = 12 × 10–6 per ºC. If the rod is not free to
35. A rod of length L and diameter D is subjected expand, the thermal stress developed is
to a tensile load P. Which of the following is (a) 120 MPa (tensile) (b) 240 MPa (tensile)
sufficient to calculate the resulting change in (c) 120 MPa (comp. (d) 240 MPa (comp.)
diameter ? RPSC Vice Principal ITI 2018
(a) Young's modulus TSPSC AEE 28.08.2017 (Civil/Mechanical)
(b) Shear modulus UPRVUNL AE 21.08.2016, CSE Pre 2003
(c) Poisson's ratio Ans. (d) : Thermal strain = e = α∆T
(d) Both Young's modulus and Shear modulus
GPSC ARTO Pre 30.12.2018 Thermal stress = σ = E.e = E α ∆T
BPSC Asstt. Prof. 29.11.2015 = (200 × 109) × (12 × 10–6) × (120 – 20)
GATE-2008 = 200 × 109 × 12 × 10–6 × 100
Ans. (d) : We can find longitudinal elongation using = 240 MPa (compressive)
Young's modulus– Since, body tries to expand, but it's expansion is
PL restricted.
∆L =
AE ∴ Compressive stress will be induced.
Now to find decrease in diameter of rod, Poisson's ratio
(µ) should be known – 39. In a simple tensile test, Hooke's law is valid
upto the
Lateralstrain or diametralstrain
µ=− (a) elastic limit
Longitudinalstrain (b) limit of proportionality
Hence we need both E and µ but by knowing E and G (c) ultimate stress
we can find µ, E = 2G(1 + µ) (d) breaking point
So, both Young's modulus and shear modulus.
36. If poisson's ratio for a material is 0.5, then the TSPSC AEE 28.08.2017 (Civil/Mechanical)
elastic modulus for the material is- TNPSC AE (Industries) 09.06.2013
(a) 3 times its shear modulus APPSC AE 04.12.2012, CSE Pre 1998
(b) 4 times its shear modulus Ans. (b) : In a simple tensile test, Hooke's law is valid
(c) Equal to its shear modulus upto the limit of proportionality.
(d) Indeterminate
APPSC IOF, 2009 40. The independent elastic constants for a
ISRO Scientist/Engineer 2006 homogeneous and isotropic material are
ESE 1995 (a) E, G, K, ν (b) E, G, K
Ans. (a) : Given– (c) E, G, ν (d) E, G
Poisson's ratio (µ) = 0.5 APPSC AEE 2012
E = 2G (1 + µ) CSE Pre-1995
µ = 0.5, Ans. (d) : Isotropic materials have the same properties
E = 2G (1 + 0.5) in all directions. The number of independent elastic
E constants for such materials is 2. Out of E, G, K and µ if
= G(1.5)
2 any two constants are known for any linear elastic and
E = 3G isotropic material then rest two can be derived.
Strength of Materials 163 YCT
41. The true strain ∈t and engineering strain ∈ Ans : (c) If a material expands or contract freely due to
relationship is heating or cooling. Then no stress will develop in
(a) ∈t = ln(1− ∈) (b) ∈t = ln(1+ ∈) material but if this expansion and contraction is
prevented than internal resisting forces are developed in
1 the material and because of these internal in the
(c) ∈t = ln(1 − 2 ∈) (d) ∈t = ln
(1+ ∈) material.
UPPSC AE 12.04.2016 Paper-I 44. Two tapering bars of the material are
GPSC ARTO 01.05.2016 subjected to a tensile load P. The length of
GATE-2014 both the bars are the same. The larger
Ans : (b) True strain:- diameter of each of the bars is D. The diameter
of the bar A at its smaller end is D/2 and that
Lf δℓ L 
∈T = ∫ = [ l n ]Lof = l n  f 
L
of the bar B is D/3. What is the ratio of
Lo ℓ L
 o elongation of the bar A to that of the bar B?
(a) 3 : 2 (b) 2 : 3
 L o + ∆L  (c) 4 : 9 (d) 1 : 3
∈T = l n  
 L o  OPSC AEE 2019 Paper-I
TSPSC Managers, 2015
∈T = l n (1+ ∈) ESE 2006
Ans : (b) :

Elongation of bar A
4 PL
σT = σ0 (1 + ε ) (δ A ) =
π Ed1d 2
42. Which of the relationship between bulk 4 PL 8PL
modulus (K), modulus of elasticity (E) and = =
modulus of rigidity (G) is correct. π E ( D )( D / 2 ) π ED 2
9KE 9KE
(a) G = (b) G =
K + 3E E + 3K
3KE 9 3 1
(c) G = (d) = +
E + 9K E G K Elongation of bar B, if d1 = D, d2 = D/3
ISRO Scientist/Engineer 22.04.2018
4 PL 4 PL 12 PL
RPSC LECTURER 16.01.2016 (δ B ) = = =
ESE 2008 π Ed1d 2 π E ( D )( D / 3) π ED 2
Ans. (d) : δA 8 2
9 KG = =
E= δ B 12 3
3K + G 45. For a ductile material, toughness is a measure
1 of
E= (a) resistance to scratching
 3 K   G 
 9 KG  +  9 KG  (b) ability to absorb energy up to fracture
(c) ability to absorb energy till elastic limit
1 (d) resistance to indentation
E= Gujarat PSC AE 2019
 1   1 
 3G  +  9 K  GPSC Executive Engineer 23.12.2018, GATE 2013
Ans. (b) : Toughness of a material is defined by energy
1 1 1 absorbed up to fracture. It is equal to area load
= +
E 3G 9 K deformation diagram. Area under stress strain diagram
up to fracture represents energy absorbed per unit
9 3 1
= + volume which is known as modulus of toughness, it is
E G K measure of toughness of material.
43. A steel rod of diameter 1 cm and 1m long is 46. If a block of material of length 25 cm, breadth
heated from 200C its α = 12×10–6/K and E = 200 10 cm and height 5 cm undergoes a volumetric
GN/m2. If the rod is free to expand, the thermal
stress developed in it is– strain of 1/5000, then change in volume will be:
(a) 12×104 N/m2 (b) 240 KN/m2 (a) 0.50 cm3 (b) 0.25 cm3
3
(c) Zero (d) Infinity (c) 0.20 cm (d) 0.75 cm3
TSPSC AEE 28.08.2017 (Civil/Mechanical) HPPSC Asstt. Prof. 18.09.2017
RPSC 2016 TSPSC AEE 28.08.2017 (Civil/Mechanical)
CSE Pre-2002 CSE Pre-1997
Strength of Materials 164 YCT
 ∆V Ans. (b) : The ratio of volumetric stress to the
Ans. (b) : Volumetric strain = ev = corresponding volumetric strain in a body is always
V
Initial volume V = (25 × 10 × 5) cm3 constant is known as bulk modulus of elasticity.
1 51. The ratio between the change in volume and
ev = ( given ) original volume of the body is called
5000 (a) tensile strain (b) compressive strain
∆V = ev × V (c) volumetric strain (d) shear strain
1
= × ( 25 ×10 × 5 ) UP Jal Nigam AE 2016
5000 3 APPSC AEE 2012
= 0.25 cm Ans : (c)
47. A material has a young's modulus of 1.25 × 105 changein volume
N/mm2 and a poisson's ratio of 0.25. The bulk Volumetric strain = original volume
modulus of the material will be
(a) 83 × 105 N/mm2 (b) 0.83× 105 N/mm2 ∆V
3
(c) 8.3× 10 N/mm 2
(d) 8.3× 105 N/mm2 εv =
V
ISRO Scientist/Engineer (RAC) 22.04.2018
Mizoram PSC AE/SDO Paper-II 2014 52. A rod of copper originally 305 mm long is
pulled in tension with a stress of 276 MPa. If
Ans. (b) : Given, the modulus of elasticity is 110 GPa and the
E = 1.25 × 10 N/mm , µ = 0.25
5 2
deformation is entirely elastic, the resultant
then K=? elongation will be nearly
E = 3K (1-2 µ) (a) 1.0 mm (b) 0.8 mm
5
1.25 × 10 = 3 × K × (1 - 0.5) (c) 0.6 mm (d) 0.4 mm
2.50 ESE 2020
K= × 10 5

3 ESE 2019
4
K = 8.33 × 10 Ans. (b)
K = 0.833 × 105 N/mm2 Elongation,
48. A free bar of length 1 is uniformly heated from Pℓ σℓ 276 × 305
δℓ = = = = 0.765 mm ≃ 0.8 mm
0ºC to a temperature tºC. α is the coefficient of AE E 110 × 103
linear expansion and E is the modulus of
53. Temperature stress are set up in a material when
elasticity. The stress in the bar is
(a) α TE (b) E/αT (a) It is free to expand or contract
(c) Zero (d) None of the above (b) It is first heated then cooled
ISRO Scientist/Engineer (RAC) 07.05.2017 (c) It is first cooled and then heated
Karnataka PSC Lect., 27.05.2017, GATE 1995 (d) its expansion and contraction is restrained
Nagaland PSC CTSE 2017, 2016 Paper-I
Ans. (c) : The stress in the heated bar will be zero
because the bar is free to expand so therefore no stress Ans. (d) : Whenever there is some increase or decrease
produced in heated bar. in the temperature of a body, it causes the body to
expand or contract, there is no stresses are induced in
49. Robert Hooke's discovered experimentally that the body. But, if the deformation of the body is
within elastic limit prevented, some stresses are induced in the body, such
(a) Stress = strain stresses are known as thermal stresses or temperature
(b) Stress/strain = a constant stresses.
(c) Stress x strain = 1 54. A 200 × 100 × 50 mm steel block is subjected to
(d) None of these a hydrostatic pressure of 15 MPa. The Young's
ISRO Scientist/Engineer 2011 modulus and Poisson's ratio of the material
ESE 2020 are 200 GPa and 0.3 respectively. The change
Ans. (b) : Hooke's law—According to Hook's law the in the volume of the block is
stress is directly proportional to strain within elastic (a) 100 mm3 (b) 110 mm3
3
limit i.e. normal stress (σ) ∝ Normal strain (ε) (c) 85 mm (d) 90 mm3
3
σ = Eε (e) 80 mm
And shear stress (τ) ∝ shearing strain (γ) CGPSC AE 26.04.2015 Shift-I
or τ = Gγ GATE 2007
50. The ratio of volumetric stress to the Ans. (d) : Given, 6 3
corresponding volumetric strain in a body is Volume (V) = 200 × 100 × 50 = 10 mm
always constant is known as : Hydrostratic pressure (σ) = 15 MPa = 15 N/mm2
(a) Surface tension Young's modulus (E) = 200 GPa = 200 × 103 N/mm2
(b) Bulk modulus of elasticity Poisson's ratio (µ) = 0.3
(c) Compressibility [ ∵ For Hydrostratic pressure, σx = σy = σz = σ]
(d) Viscosity
3σ
Karnatka PSC AE (WRD) 31.07.2021 Volumetric strain (ev) = (1 − 2µ )
MECON MT 2019 E
Strength of Materials 165 YCT
 ∆V 3σ
= (1 − 2µ )
V E
3σ V
∆V = (1 − 2 µ )
E
3 × 15 × 106 (1 − 2 × 0.3) 3. Stress-strain curve for elastic - Perfectly plastic
= = 90 mm3
200 × 103 material.
55. ______ is the capacity of material to absorb
energy when it is elastically deformed and then
upon unloading, to have this energy recovered.
(a) Toughness (b) Tensile strength
(c) Plasticity (d) Resilience
CIL MT 26.03.2017 4. Stress-strain curve for an ideal elastic material with
ESE 2016 strain hardening material.
Ans. (d) : Resilience- It is energy absorbed by a
member in elastic region. It denotes the capacity of
material to absorb energy when it is elastically
deformed and then upon unloading, to release this
energy.
Toughness- It is energy absorbed by member 5. stress-strain curve for rigid - Linear hardening
just before its fracture. material
56. The stress-strain curve of an ideal elastic
material with strain hardening will be as-
(a)

57. The ability of the material to absorb energy

before fracture is known as:
(a) Toughness (b) Ductility
(b) (c) Cold shortness (d) Hardness
UPRVUNL AE 07.10.2016
SJVN ET 2013
Ans. (a) : The ability of the material to absorb energy
before fracture is known as toughness.

(c)

1
58. The equation for relationship between , C
(d) m
& K is,
1 3K − 2C 1 2C − 3K
(a) = (b) =
m 6 K + 2C m 2C + 6 K
1 2 K − 3C 1 3K + 2C
(c) = (d) =
m 2C + 6 K m 6 K − 2C
RPSC AE 2018 TNPSC AE 2013
CSE Pre-1998 BPSC AE 2012 Paper - VI
Ans. (d) : 1. The stress-strain curve for an ideal elastic Ans. (a) : We know that relation between poisson ratio
material.  1
µ or m  , modulus of rigidity (C) and Bulk Module
 
(K) is given as
2C (1 + µ ) = 3K (1 − 2µ )
2C + 2Cµ = 3K − 6Kµ
2. The stress-strain curve for rigid - Perfectly plastic 1 3K − 2C
µ= =
material m 6K + 2C

Strength of Materials 166 YCT

59. In a body, thermal stress is induced because of (c) Resist fracture due to high impact.
the existence of : (d) Retain deformation produced under load
(a) Latent heat (b) Total heat permanently.
(c) Temperature gradient (d) Specific heat UKPSC AE 2012 Paper-I
OPSC AEE 2019 Paper-I VIZAG STEEL MT 2011
ESE 2013 Ans. (a) : Regain the original shape after the removal of
Ans : (c) : Thermal stress- If the material is restrained applied force.
from expanding or contracting while the temperature 65. The relation between E (modulus of elasticity)
change, then stress builds within the part. and k (bulk modulus of elasticity) is
60. True strain for a steel bar which is doubled its  2  2
length by tension is : (a) E = k 1 −  (b) E = 2k 1 − 
(a) 0.307 (b) 0.5  m  m
(c) 0.693 (d) 1.0  2   2
OPSC AEE 2019 Paper-I (c) E = 3k 1 −  (d) E = 4k 1 − 
 m  m
ISRO Scientist/Engineer 2012 UPRVUNL AE 07.10.2016
Ans : (c) : UKPSC AE 2007 Paper -I
Initial length of bar, ℓi = ℓ
 2
Final length of bar, ℓf = 2ℓ Ans. (c) : E = 3k 1 − 
 m
∆L 2ℓ i − ℓ i 66. Which one of the following is connect respect
Engineering strain, ∈= = =1 of poisson's ratio (µ) limits for an isotropic
ℓ ℓi
elastic solid:
True strain (∈true ) = ln (1+ ∈) = ln (1 + 1) (a) 0 < µ < 1 (b) –1 < µ < 1
∈true = ln ( 2 ) (c) –1 < µ < 0.5 (d) 0 < µ < 0.5
= 0.693 PTCUL AE 25.06.2017
ESE 2004
61. Resilience of material should be considered
when it is subjected to Ans. (c) : The Poisson's ratio (µ) for any material lies in
(a) shock load (b) electroplating −1 < µ < 0.5
(c) chemical coating (d) polishing 67. Whenever some external system of forces act
RPSC AE 2016 on a body, it undergoes deformation. As the
GWSSB DEE 07.07.2016 body goes under deformation, it sets up some
RPSC ACF-2011 resistance to the deformation. What is this
Ans : (a) Resilience of material should be considered resistance per unit area of deformation called
when it is subjected to shock loading. as?
Resilience:- It is the property of a materials to absorb (a) Strain
energy and to resist shock and impact loads. It is (b) Stress
measured by the amount of energy absorbed per unit (c) Pressure
volume within elastic limit this property is essential for (d) Modulus of elasticity
spring materials. HPPSC Poly. Lect. 05.07.2021
62. Strain in direction at right angle to the VIZAG STEEL MT 2011
direction of applied force is known as:- Ans. (b) : Resistance force per unit area of deformation
(a) Lateral strain is called stress.
(b) Shear strain
68. Which one of the following information cannot
(c) Volumetric strain be obtained from the static tensile test of a
(d) None of the above mild steel specimen?
UKPSC AE-2013, Paper-I (a) Modulus of elasticity
APPSC AEE 2012 (b) Qualitative determination of toughness
Ans. (a) : Strain in direction at right angle to the (c) Ductility
direction of applied force is known as lateral strain. (d) Weldability
63. Stress and Strain are tensor of TSPSC AEE 28.08.2017 (Civil/Mechanical)
(a) zero-order (b) first order CSE Pre-2006
(c) second order (d) None of the above
HPPSC Asstt. Prof. 2014 Ans. (d) : Weldability cannot be obtained from static
UKPSC AE 2012 Paper-I tensile test.
Ans. (c) : Stress in second order tensor. Magnitude, 69. A straight bimetallic strip of copper and steel
direction and plane are required to describe stress that’s is heated. It is free at ends. The strip will
why it is known as second order tension. (a) Expand and remain straight
64. What does the elasticity of material enables it (b) Not expand but bend
to do ? (c) Expand and bend also
(a) Regain the original shape after the removal (d) First only
of applied force. TSPSC AEE 28.08.2017 (Civil/Mechanical)
(b) Draw into wires by the application of force. CSE Pre 2002
Strength of Materials 167 YCT
Ans. (c) : If heated and free to expand K 1
since αcopper = 18 × 10–6 /ºc =
E  2
αsteel = 11 × 10–6 /ºc 3 1 − 
Copper expands more than steel and hence the  4
bimetallic strip will bend with outer fiber as copper. 4
=
70. The work done per unit volume in elongating a 3× 2
body by a uniaxial force is :
(a) stress/strain (b) stress × strain K
= 2/3
1 E
(c) stress × strain (d) None of the above
2 75. If modulus of rigidity of a material is 80 GPa
APPSC AE 04.12.2012 and modulus of elasticity is 200 GPa, what will
APPSC AEE 2012 be the bulk modulus of material?
Ans. (c) : (a) 133.3 GPa (b) 153.3 GPa
1 1 (c) 163.3 GPa (d) 173.3 GPa
Workdone/volume = stress × strain = σ.ε APPSC Poly Lect. 13.03.2020
2 2 ISRO Scientist/Engineer (RAC) 07.05.2017
71. The Poisson's ratio a material which has Ans. (a) : Given data : G = 80 GPa, E = 200 GPa
Young's modulus of 120 GPa and shear K=?
modulus of 50 GPa, is ∵ E = 2G(1+µ) ⇒ 200 = 2×80 (1+µ)
(a) 0.1 (b) 0.2
200 10 1
(c) 0.3 (d) 0.4 ⇒ 1+ µ = ⇒ µ = − 1 = = 0.25
RPSC Lect. (Tech. Edu. Dept.) 13.03.2021, Paper-I 2 × 80 8 4
UKPSC AE-2013, Paper-I Now,
Ans. (b) : As, we know, the relation between E, µ & G E = 3K (1–2µ)
E = 2G [1+µ] E 200 200
Bulk modulus (K) = = =
120 = 50 × 2 [1+µ] 3 (1 − 2µ ) 3 (1 − 0.50 ) 3 × 0.50
µ = 0.2
⇒ K = 133.33 GPa
72. Thermal Strain, developed in a rod of length 'l'
and temperature rise '∆T', is given by- 76. A uniform slender cylindrical rod is made of a
(Where α = Coefficient of Thermal Expansion; homogeneous and isotropic material. The rod
E = Young's modulus) rests on a frictionless surface. The rod is
(a) α∆T (b) Eα∆Tl heated uniformly. If the radial and
(c) Eα∆T (d) Zero longitudinal thermal stresses are represented
RPSC Lect. (Tech. Edu. Dept.) 13.03.2021, Paper-I by σr and σz respectively, then
APPSC AEE 2012 (a) σr = 0, σz = 0 (b) σr ≠ 0, σz = 0
Ans. (a) : As we know, (c) σr = 0, σz ≠ 0 (d) σr ≠ 0, σz ≠ 0
∆L = α∆TL [∴ General equation] TSPSC AEE 2017
GATE-2005
∆L
Strain, = α∆T Ans. (a) : When any material is heated or cooled and
L allowed to expand or contract without any restriction
73. Which one of the following properties is more then they will not generate any stress, this is known as
sensitive to increase in strain rate? free expansion or contraction.
(a) Yield strength (b) Proportional Limit Hence, σr = 0 and σz = 0
(c) Elastic Limit (d) Tensile Strength 77. The Young's modulus of a material is 125 GPa
APPSC AEE 2012 and Poisson's ratio is 0.25. The modulus of
RPSC Lect. (Tech. Edu. Dept.) 2011 rigidity of the material is
Ans. (a) : Yield strength is more sensitive to increase in (a) 30 GPa (b) 50 GPa
strain rate. (c) 80 GPa (d) 100 GPa
Note : No answer is given by commission. Assam PSC CCE Pre 2015
74. The ratio of Bulk Modulus to Young's JPSC AE 2013, Paper-V
Modulus of a Poisson's ratio of 0.25 will be : Ans. (b) : Given as,
(a) 1/3 (b) 2/3 E = 125 GPa
(c) 1 (d) 3/2 µ = 0.25
APPSC AE Subordinate Service Civil/Mech. 2016 E = 2G (1+µ)
RPSC AE GWD, 2011 125 = 2G (1+0.25)
Ans. (b) : Given, Poisson's ratio (µ) = 0.25 125 = 2G (1.25)
G = 50 GPa
E = 3K(1 − 2µ)
78. When a wire is stretched to double its length,
K 1 the longitudinal strain produced in it is
=
E 3(1 − 2µ) (a) 0.5 (b) 1.0
(c) 1.5 (d) 2.0
K 1
= Assam PSC CCE Pre 2015
E 3(1 − 2 × 0.25) UKPSC AE-2013, Paper-I
Strength of Materials 168 YCT
 Ans. (b) : Given, Initial length, ℓi = ℓ 81. A rod of length 20 mm is stretched to a length
of 40 mm. Subsequently, it is compressed to
Final length, ℓf = 2ℓ make a rod of final length 10 mm. Consider
Change in Length ( δl ) the longitudinal tensile strain as positive and
Longitudinal strain = compressive strain as negative. What is the
Original Length ( lo ) total true longitudinal strain in the rod?
If lo be the original length and if stretched to double of (a) –0.5 (b) –1.0
its length, change in length (δl) = (ℓf – ℓi) = (2 × l – l ) (c) –0.75 (d) –0.69
= l Assam Engg. College AP/Lect. 18.01.2021
GATE-2017
δl l Ans. (d) : Total true longitudinal strain
So, strain (∈)= = =1
lo l (εT) = (εT)1 + (εT)2
79. Two threaded bolts A and B of the same   l     l 
material and length are subjected to identical = l n  f   +  l n  f  
tensile load. If the elastic strain energy stored   li  1   li   2
in bolt A is 4 times that of bolt B and the mean  40   10 
diameter of bolt A is 12 mm, the mean = l n   + l n   = –0.693
diameter of bolt B in mm is:  20   40 
(a) 16 (b) 24 82. The elastic stress-strain behavior of rubber is -
(c) 36 (d) 48 (a) Linear
NPCIL ET, 2015 (b) Non-linear
GATE-2013 (c) Plastic
Ans. (b) : Strain energy is given by (d) No fixed relationship
P2L TNPSC AE 2018
U= GMB AAE 25.06.2017
2AE
Given that- Ans. (b) : The most common example of this kind of
UA = 4UB material is rubber whose stress-strain relationship can
2 2 be defined as non-linearly Elastic, Isotropic, in
PL PL
=4 compressible and generally Independent of strain role.
2A A E 2A B E Hyper elasticity provides a means of modeling the
Since bolts are of some material, length and subjected to stress-strain behavior of such material.
identical tensile load.
1 1
=4
AA AB
1 1
2
=4 2
DA DB
DB = 2DA
DB = 2 × 12 83. For the material loaded within the elastic limit,
DB = 24 mm the Young's modulus is defined as the ratio of
80. A solid cube of side 1 m is kept at a room (a) longitudinal strain in lateral strain
temperature of 32ºC. The coefficient of linear (b) volumetric stress to volumetric strain
thermal expansion of the cube material is (c) longitudinal stress to longitudinal strain
1×10–5 /ºC and the bulk modulus is 200 GPa. If (d) shear stress to shear strain
the cube is constrained all around and heated Haryana PSC AE (PHED) 05.09.2020, Paper-II
uniformly to 42ºC, then the magnitude of GPSC Lect. 23.10.2016
volumetric (mean) stress (in MPa) induced due
to heating is______ SJVN ET 2013
(a) 180 (b) 240 VIZAG Steel MT 2011
(c) 30 (d) 60 Ans. (c) :
Assam Engg. College AP/Lect. 18.01.2021 Longitudinalstress(σ)
GATE-2019 Young's modulus (E) =
Ans. (d) : Given, Longitudinalstrain (ε)
α = 1×10 /ºC, K = 200 GPa, T1 = 32ºC, T2 = 42ºC
–5
84. When a body is subjected to three mutually
Volumetric strain εv = 3 × ε perpendicular stresses of equal intensity, the
= 3 × α∆T ratio of direct stress to the expanding
= 3 × 1×10–5 × (42 – 32) volumetric strain is known as :
= 3 × 10–4 (a) Young's modulus
σv Direct stress (b) Modulus of rigidity
Bulk modulus K = =
ε v Volumetricstrain (c) Bulk modulus
σv = K × ε v (d) Poisson's ratio
= 200 × 103 × 3 × 10–4 ISRO Scientist/Engineer (RAC), 10.03.2019
Volumetric stress σv = 60 MPa Punjab PSC SDE 12.02.2017
Strength of Materials 169 YCT
Ans. (c) :
σ σ
Bulk modulus (K) = =
 ∆V  Volumetricstrain
 
 V 
85. A copper rod 400 mm long is pulled in tension
to a length of 401.2 mm by applying a tensile
load of 330 MPa. If the deformation is entirely
elastic, the Young's modulus of copper is
(a) 110 GPa (b) 110 MPa
(c) 11 GPa (d) 11 MPa
GPSC EE Pre, 28.01.2017
ESE 2012 (d) None of the above
Ans. (a) : Given- BPSC Asstt. Prof. 29.11.2015
CSE Pre-1996
l1 = 400 mm
Ans. (b) : (i) Linear Elastic Materials-
l2 = 401.2 mm
Elastic of modulus = (E)
l –l
Strain (ε) = 2 1
l1
401.2 – 400
= (ii) Rigid materials-
400
1.2
Strain (ε) = = 0.003
400
stress
Modulus of elasticity (E) =
strain (iii) Perfectly Plastic (Non-Strain hardening)-
330 × 106

(E) =
0.003
330 × 106 ×103
=
3
(E) = 110 × 109 = 110 GPa (iv) Rigid hardening –
86. A composite bar made of copper and steel is
heated to 120 °C from room temperature. If αc
> αs, the stress induced in copper bar is:
(a) No stress (b) Compressive stress
(c) Shear stress (d) Tensile stress
CIL MT 27.02.2020 (v) Elasto Plastic Material-
GPSC EE Pre, 28.01.2017
Ans. (b) : Given, T = 120 °C
αcopper > αsteel

88. stress-strain relationship best describes the

behavior of brittle materials such as ceramics
and thermosetting plastics,
(σ = stress and ε = strain)?

As the coefficient of expansion of copper is more as

compared to that of steel.
The final expansion is less than the free expansion of
copper due to temperature rise and thus compressive
stress will be developed in copper bar.
87. The stress-strain relationship for the perfectly
rigid material is :
Strength of Materials 170 YCT
 both in tension and compression. The ratio Et /
Ec will be approximately equal to
(a) 0.5 (b) 0.75
(c) 1.0 (d) 1.25
HPPSC Poly. Lect. 05.07.2021
Ans. (b) : 0.75
94. The modulus of rigidity of a material whose
Young's modulus is 200 Nmm-2 and poisson's
ratio is 0.5 is
(a) 133.33 Nmm-2 (b) 80.67 Nmm-2
(e) None of the above (c) 66.67 Nmm-2 (d) 242.67 Nmm-2
CGPSC Poly. Lect. 22.05.2016 GPSC DEE, Class-2 (GWSSB) 04.07.2021
GATE-2015 Ans. (c) : Given,
Ans. (d) : In stress-strain relationship best describes E = 200 N/mm2
describes the behavior of brittle materials such as µ = 0.5
ceramics and thermosetting plastics We know,
(σ = stress and strain = ε) E = 2G(1+µ)
200 = 2 × G (1+0.5)
200 = 2G (1.5)
Modulus of Rigidity, G = 66.67 N/mm2
95. Resilence is the
(a) energy stored in a body when strained within
the elastic limit
(b) energy stored in a body when strained up to
breaking point
89. Normal stresses are divided into (c) max strain energy stored
(a) tensile and torsion (d) none of the above
(b) compressive and thermal GPSC DEE, Class-2 (GWSSB) 04.07.2021
(c) shear and bending Ans. (a) : Resilence is the energy stored in a body when
(d) tensile and compressive strained within the elastic limit.
UPRVUNL AE 05.07.2021
96. In a bolt and tube assembly, due to the
Ans. (d) : Normal stresses are divided into tensile and tightening of the nut on the bolt, _____ stress is
compressive. developed in the bolt and ______ stress is
90. Eccentric loading applies to ______ stress developed in the tube.
(a) Bending (b) torsional (a) compressive, compressive
(c) tensile (d) thermal (b) tensile, compressive
UPRVUNL AE 05.07.2021 (c) compressive, tensile
Ans. (a) : Eccentric loading applies to bending stress. (d) tensile, tensile
91. True stress at fracture ______ during tensile VIZAG Steel MT 24.01.2021, Shift-I
testing. Ans. (b) : In a bolt and tube assembly, due to the
(a) becomes zero tightening of the nut on the bolt tensile stress is
(b) is the same as ultimate stress developed in the bolt and compressive stress is
(c) is higher than ultimate stress developed in the tube.
(d) is lower than ultimate stress 97. A copper bar of rectangular section of 20 mm
UPRVUNL AE 05.07.2021 × 30 mm and a length of 500 mm is subjected
Ans. (c) : True stress at fracture is higher than ultimate to a compressive load of 60 kN. The
stress during tensile testing. compressive stress acting on the bar is _____
92. Factor of safety is equal to _______ N/mm2.
Allowable stress (a) 150 (b) 50
(a) (c) 200 (d) 100
Failure stress VIZAG Steel MT 24.01.2021, Shift-I
Allowable load Ans. (d) : Given,
(b)
Failure load Area, A = 20 mm × 30 mm
(c) Allowable stress × failure stress = 600 mm2
Compressive Load, P = 60 kN, length, l = 500 mm
Failure stress
(d) P 60 × 103
Allowable stress Compressive stress, σc = =
UPRVUNL AE 05.07.2021 A 600
σc = 100 N/mm2
Failure stress
Ans. (d) : Factor of safety = 98. A rod of length 2 m and diameter 50 mm is
Allowable stress elongated by 5 mm when an axial force of 400
93. Measurements have been made for the Young's kN is applied. The modulus of elasticity of the
modulus of elasticity for mild steel specimen material of rod will be nearly
Strength of Materials 171 YCT
 (a) 66 GPa (b) 72 GPa Ans. (*) :
(c) 82 GPa (d) 96 GPa dL dL
APGCL AM, 2021 True strain at any instant (ε T ) = ∫ =∫
L L o (1 + t 2 )
Ans. (c) :
Given:- As, L = Lo(1 + t2)
dL = Lo2tdt
L = 2m,
dL 2tdt
εT = ∫
Lo (1 + t 2 ) ∫ (1 + t 2 )
P = 400kN = = ln[1 + t 2 ]
δL = 5mm
d = 50mm
 Let,1 + t 2 = x, 
π  
Area = ( 50 )
2
 dx = 2tdt 
4
True strain at the end of 3 minutes will be
Load 400 × 103 εT = ln[1 + (3)2] = ln10 = 2.30
σ= =
Area π Note- There is no option in the given question.
× (50) 2
4 102. The relationship between conventional strain,
σ = 203.82 MPa ε0 and true strain, εT is:
δL 5 (a) ε0 = {eεT} (b) ε0 = {eεT} – 1
Strain, ε = = = 2.5×10 -3
(c) ε0 = {e } –εT
(d) ε0 = {eεT} + 1
L 2000 AAI Jr. Executive 26.03.2021
Now stress, σ = Eε Ans. (b) : Conventional strain/Engg. strain = ε0
203.82 = E×2.5×10-3 True strain = εT
E =
203.82
= 81.5 GPa εT = ln(1 + ε0)
2.5 × 10−3 1 + ε0 = eεT
99. A tensile test was conducted on mild steel bar. ε0 = eεT – 1
The load at elastic limit was 250 kN and the ε0 = {eεT } − 1
diameter of steel bar was 3 cm. The value of
103. If a deformation δ is produced in the rod BC of
stress is cross-sectional area A of the figure below by a
4 2
(a) 35368 × 10 N/m load P, what load is required to cause the
4 2
(b) 32463 × 10 N/m same deformation in the rod B 'C' of the same
4 2
(c) 35625 × 10 N/m length L, but of cross-sectional area 2A?
4 2
(d) 37562 × 10 N/m
APGCL AM, 2021
Ans. (a) : Given,
P=250 kN = 250 × 103 N
d=3cm=3 × 10-2m
P 250 × 103
Stress=σ = = = 353677.65 × 103 N/m2
A π × 3 × 10−2 2
4
( )
≈35368 × 104 N/m2 (a) 6P (b) 4P
100. An annealed copper strip, 9 inch (228mm) (c) 2P (d) P
wide and 1 inch (25mm) is rolled to a thickness CGPSC AE 15.01.2021
of 0.80 inch (20mm) in one pass. Ans. (c) : Given,
The absolute value of true strain that the strip
undergoes is :
(a) 0.20 (b) 0.25
(c) 0.223 (d) 0.225
JPSC AE 10.04.2021, Paper-II
Ans. (c) : Given, ti = 25mm, tf = 20mm
t 
True strain, εT = ln (1 + ε) = ln  f 
 ti 
 25  PBC L
= ln  
 20  δ BC =
AE
ε T = 0.223 P L
101. Aluminium rod with original length Lo is δ B 'C ' = B 'C '
2AE
strained. The length of the rod at any instant is Acc to question, δ BC = δB 'C '
given by L(t) = Lo(1+t2), where t is in minutes.
The true strain at the end of 3 minutes will be: PBC L PB 'C ' L
(a) 6 (b) 9 =
AE 2AE
(c) 0.6 (d) 0.9 PB 'C ' = 2PBC
AAI Jr. Executive 26.03.2021
Strength of Materials 172 YCT
104. A steel rod of 30 mm diameter and 5 m long is Ans. (d) : As we know, thermal stress occur only when
connected to two grips and the rod is there is resistance to its deformation. Since here its free
maintained at 95°C. Determine the stress to deform hence no thermal stress will be developed.
induced in the rod when it is cooled to 30°C, if 107. According to Hooke's law, when a material is
E = 2 × 105 MN/m2; α = 12 × 10–6 /°C . loaded within its elastic limit, the stress is
(a) 156 N/mm2 (b) 65 N/mm2 (a) directly proportional to the strain
2 2
(c) 165 N/mm (d) 185 N/mm (b) inversely proportional to the strain
VIZAG Steel MT 24.01.2021 (c) proportional to the square of the strain
Ans. (a) : Given, (d) proportional to the inverse of the square of
Diameter (d) = 30 mm the strain
Length (ℓ) = 5 m Assam PSC AE (IWT) 14.03.2021
T1 = 95°C, T2 = 30°C Ans. (a) : According to Hooke's law, when a material is
E = 2 × 105 MN/m2, α = 12 × 10–6/°C loaded within its elastic limit the stress is directly
∴ Thermal stress (σT) = α∆TE proportional to the strain.
= 12 × 10–6 × (95 – 30) × 2 × 105 × 106 σ∝ε
= 12 × 65 × 2 × 105 σ = εE
= 1560 × 105 N/m2 108. The Poisson's ratio of an elastic material is 0.3,
= 156 N/mm2 then the ratio of its young's modulus and bulk
105. A bar, having 1000 mm2 cross-sectional area, modulus is ______.
is subjected to axial loads as shown in the (a) 1.2 (b) 2.1
figure. What is the value of stress in QR (c) 1.4 (d) 1.6
section? VIZAG MT, 14.12.2020
Ans. (a) : µ = 0.3
E
=?
K
(a) 30 MPa Compressive E = 3K(1 – 2µ)
(b) 30 MPa Tensile E
(c) 50 MPa Compressive = 3(1 − 2µ)
(d) 50 MPa Tensile K
RPSC Lect. (Tech. Edu. Dept.) 13.03.2021, Paper-I = 3(1 – 2 × 0.3)
Ans. (a) : E
= 1.2
K
109. Maximum theoretical value of Poisson's ratio
is
(a) infinity (b) 1
(c) 0.707 (d) 0.5
Haryana PSC AE (PHED) 05.09.2020, Paper-II
Ans. (d) : The Poisson's ratio of a stable, isotropic,
linear elastic material must be between –1.0 to +0.5
because of the requirement for young's modulus, the
shear modulus and bulk modulus to have positive values
most material have poisson's ratio values ranging
between 0.0 and 0.5.
110. A bar of steel of length L = 0.3m, fixed at one
end and free at the other end as shown in the
Stress in QR section figure is uniformly heated by increasing its
P temperature by 400C. If the coefficient of
σ=
A thermal expansion is 12 × 10–6/0C and modulus
30 × 103 of elasticity is 210 GPa, the thermal stress
= = 30 MPa (Compressive)
1000 induced in the bar is
106. A steel rod, with all its surfaces free to deform,
has young's modulus 'E', Poisson's ratio 'ν',
coefficient of thermal expansion 'α'. If the rod
is uniformly heated with '∆T' temperature
rise, what is the pressure developed? (a) -100.8 MPa (b) 100.8 GPa
2
(a) -α∆T/(1 - 2ν) (b) α∆T/(1 - 2ν) (c) 100.8 N/m (d) 0
(c) Eα∆T/(1 - 2ν) (d) 0 Haryana PSC AE (PHED) 05.09.2020, Paper-II
RPSC Lect. (Tech. Edu. Dept.) 13.03.2021, Paper-I, Ans. (d) : Since the bar is free to expand. There are no
GATE 2014 stresses developed.

Strength of Materials 173 YCT

111. A 8 m long round bar made of aluminium (c) limit of proportionality
(Modulus of elasticity 70 GPa) carries a tensile (d) lower yield point
load of 720 kN. If the maximum allowable UPRVUNL AE 05.07.2021
elongation is 10 mm, the minimum required Ans. (b) : For a stress-strain diagram for ductile
diameter of the bar is material, the work hardening starts after the upper yield
(a) 103 mm (b) 30 mm point.
(c) 79 mm (d) 72 mm 116. During plastic deformation, induced stress in
Haryana PSC AE (PHED) 05.09.2020, Paper-II the material is:
Ans. (a) : Given, (a) greater than its proportional limit but lesser
P = 720 × 103 N than lower yield strength
l =8m (b) greater than its lower yield strength but lesser
than upper yield strength
δl = 10 × 10-3 m (c) greater than its ultimate strength but lesser
E = 70 GPa = 70 ×109 m than its fracture strength
Pl (d) greater than its yield strength but lesser than
Elongation, δl = its fracture strength
AE
NLCIL GET 17.11.2020, Shift-II
720 ×103 × 8 Ans. (b) : Plastic deformation is the permanent
10 × 10-3 =
π 2 distortion that occurs when a material is subjected to
D × 70 × 109 tensile, compressive bending or torsion stresses that
4
D = 0.1023 m exceed its yield strength and cause it to elongate
D = 102.3 mm ≈ 103 mm compress buckle, bend or twist.
Induced stress is greater than its ultimate strength but
112. Failure in the case of ductile materials in static lesser than its fracture strength.
tensile loading is due to
(a) Shear stress (b) Tensile stress 117. The stress-strain curve of brittle material has:
(a) only upper yield point
(c) Compressor stress (d) Contact stress (b) only lower yield point
Haryana PSC AE (PHED) 05.09.2020, Paper-II (c) no clear yield point
Ans. (a) : Failure in case of ductile materials in static (d) both lower and upper yield point
tensile loading is due to shear stress. NLCIL GET 17.11.2020, Shift-II
113. The value of Poisson's ratio depends upon: Ans. (c) : A stress strain curve for brittle material will
(a) Material of the test specimen be linear and no clear yield point for some materials
(b) Nature of load, tensile or compressive such as concrete tensile strength is negligible compared
(c) Dimensions of the test specimen to the compressive strength.
(d) Magnitude of load 118. The temperature stress is a function of, 1. Co-
CIL MT 27.02.2020 efficient of linear expansion 2. Temperature
Ans. (a) : The value of Poisson's ratio depends upon rise 3. Modulus of elasticity. The correct
material of the test specimen. answer is
114. A square bar of side 5 mm and 200 mm long, (a) 1 and 2 only (b) 1 and 3 only
experience an extension of 0.01 mm upon a (c) 2 and 3 only (d) 1, 2 and 3
TSPSC Manager (Engg.) HMWSSB 12.11.2020
load of 100 N. Then Young's Modulus of the
material is: Ans. (d) : Thermal stress = α∆TE
(a) 800 MPa (b) 200 GPa Where, α = Coefficient of thermal expansion
(c) 400 MPa (d) 80 GPa E = Young's modulus
CIL MT 27.02.2020 ∆T = Temperature rise
119. Simple stress is often called
Ans. (d) : Given, ℓ = 200 mm (a) direct stress (b) transverse stress
Side = 5 mm (c) total stress (d) complete stress
In square bar (A) = (Side)2 TSPSC Manager (Engg.) HMWSSB 12.11.2020
A = 25 mm2 Ans. (a) : Simple stress is often called direct stress.
δℓ = 0.01 mm 120. A rod of length 2 m and diameter 50 mm is
P = 100 N elongated by 5 mm when an axial force of 400
Young modulus of material, kN is applied. The modulus of elasticity of the
material of the rod will be nearly
Pℓ (a) 66 GPa (b) 72 GPa
E=
Aδℓ (c) 82 GPa (d) 96 GPa
100 × 200 ESE 2020
E=
25 × 0.01 Pℓ
E = 80 GPa Ans. (c) : Elongation, δ =
115. For a stress-strain diagram for ductile AE
material, the work hardening starts after the 400 × 103 × 2000
5=
______. π
× 502 E
(a) elastic limit 4
(b) upper yield point E = 81487.3 MPa = 81.5 GPa ≈ 82 GPa
Strength of Materials 174 YCT
121. A rod of 2m long at a temperature of 10°C is (a) 0.25 (b) 0.29
heated to 80°C. If the coefficient of thermal (c) 0.33 (d) 0.38
expansion of the rod is 0.000012/°C and E is APPSC Poly Lect. 13.03.2020
100 GPa, the thermal stress under the Ans. (c) : Given data;
complete prevention of expansion is : di = 12 mm
(a) 96MPa (b) 84 MPa df = 11.98 mm
(c) 10 MPa (d) 12 MPa Li = 200 mm, Lf = 201 mm
NLCIL GET 17.11.2020, Shift-II L − Li
Ans. (b) : Given, rod length (L) = 2m Longitudinal strain (∈long) = f
∆t = (80°C – 10°C) Li
∆t =70°C 201 − 200 1
= = = 5 × 10−3
The coefficient of thermal expansion (α) = 0.000012°C 200 200
Young modulus (E) = 100 GPa d − di
(E) = 100 × 103 MPa Lateral strain (∈lat) = f
Thermal stress (σth) = αE∆t di
= 0.000012 × 100 × 10 × 70
3 11.98 − 12
= = = −1.666 × 10−3
σth = 84 MPa 12
122. During a tensile test on a specimen of 1 cm2 Lateral strain
cross-section, maximum load observed was 8 ∴ Poisson's ratio (µ) =
tonnes and area of cross-section at neck was Longitudinal strain
0.5 cm2. Ultimate tensile strength of specimen 1.666 × 10−3
is =
(a) 4 tonnes/cm2 (b) 8 tonnes/cm2 5 × 10−3
2 2
(c) 16 tonnes/cm (d) 22 tonnes/cm ⇒ µ = 0.333 ⇒ µ = 0.33
RPSC IOF, 2020
Ans. (b) : Ultimate tensile strength is the maximum 125. The stiffness of an axially loaded bar is defined
load upon original area. as ________. [E = Modulus of elasticity, A =
Area of cross-section of bar, L = Length of
F
σ= bar]
A EA L
8 (a) (b)
σ= L EA
1 E AL
σ = 8 tonnes/cm2 (c) (d)
AL E
123. The elongation of a freely hanging uniform APPSC Poly Lect. 13.03.2020
steel rope. If its length is doubled will increase Ans. (a) : We know that stiffness is the ratio of applied
in the ratio of load to per unit elongation or deflection.
(a) 2 : 1 (b) 1 : 1
(c) 4 : 1 (d) 8 : 1 Stiffness for axially loaded bar (s)
RPSC IOF, 2020 Load ( P )
=
Ans. (c) : The elongation of the freely hanging bar elongation ( δ)
ρg l 2 P
δl1 = =s=
2E PL / AE
δl ∝ l2
AE
If its length is doubled ⇒ stiffness ( s ) =
ρg(2l ) 2 L
δl2 = 126. A 1.25 cm diameter steel bar is subjected to a
2E
load of 2500 kg. The stress induced in the bar
δl l 2
Ratio of the elongation ⇒ 1 = will be
δl2 (2l )2 (a) 200 MPa (b) 210 MPa
(c) 220 MPa (d) 230 MPa
δl1 1
= ESE 2020
δl2 4
P 2500 × 9.81
δl2 4 Ans. (a) : σ = = = 199.4 = 200 MPa
= A π
δl1 1 (12.5 ) 2

4
124. A specimen of an alloy is made into a 127. A copper rod 3mm in diameter when subjected
cylindrical bar of 12.00 mm diameter and to a pull of 495 N extends by 0.07 mm over a
200mm length. This bar is loaded in axial gauge length of 100 mm. The Young's Modulus
tension up to the proportional limit when the
load is 40 kN. At this load the length was for copper will be :
measured as 201 mm and diameter was (a) 1 × 105 N/mm2 (b) 1 × 106 N/mm2
measured as 11.98 mm. What will be the (c) 7 × 10 N/mm
5 2
(d) 1 × 107 N/mm2
Poisson's ratio? ISRO Scientist/Engineer (RAC), 10.03.2019
Strength of Materials 175 YCT
Ans. (a) : Given, 131. Failure of a material is called fatigue when it
l = 100 mm falls
d = 3mm (a) at the elastic limit
P = 495 N (b) below the elastic limit
δ = 0.07 mm (c) at the yield point
PL (d) below the yield point
Deflection, (δ) = Gujarat PSC AE 2019
AE
Ans : (d) : Fatigue- When a material is subjected to
495 × 100
0.07 = repeated stress, it fails at stress below the yield point
π 2 stress. Such type of failure of a material is known as
×3 × E
4 fatigue.
495 × 100 132. Poisson's ratio of perfectly linear elastic
E= = 100040.2499 material is
π
× 9 × 0.07 (a) 0 (b) 1
4 (c) 0.3 (d) 0.5
E = 1× 105 N / mm 2 Gujarat PSC AE 2019
128. Engineering stress-strain curve and true Ans : (d) : Volumetric strain for liner elastic material is
stress-strain curve are equal up to : given by
(a) Proportional limit ∆V
(b) Elastic limit ∈V =
(c) Yield point V
(d) Tensile strength point
OPSC AEE 2019 Paper-I =
( )
(1 − 2µ ) σ x + σ y + σ z
Ans : (c) : Engineering stress-strain curve and true E
stress curve are equal upto yield point. ∆V = 0,
129. Strain is defined as the ratio of : 1 − 2µ = 0;
(a) Change in volume to original volume µ = 0.5
(b) Change in length to original length
(c) Changes in cross-sectional area to original 133. A 10 mm diameter aluminium alloy test bar is
cross-sectional area subjected to a load of 500 N. If the diameter of
(d) Any one of these the bar at this load is 8 mm, the true strain is
OPSC AEE 2019 Paper-I (a) 0.2 (b) 0.25
Ans : (d) : Strain is defined as the ratio of- (c) 0.22 (d) 0.1
Change in volume to original volume Gujarat PSC AE 2019
Change in length to original length. Ans : (*) : True strain is given as
Change in cross-sectional area to original cross
sectional area. A 
∈ t = ln  0 
130. The stress-strain curve of an rigid-plastic
material will be as :  Af 
d 
= 2 ln  o 
 df 
(a) (b) Since, d0 = 10 mm
df = 8 mm
Therefore, we get,
 10 
∈ t = 2ln  
8
(c) (d) = 0.446
134. The following diagram is a stress-strain
diagram of any material. Which kind of
OPSC AEE 2019 Paper-I material is it?
Ans : (b)

(a) Plastic (b) Linear Elastic

(c) Non-linear Elastic (d) Visco-elastic
APPSC AEE SCREENING 17.02.2019
Strength of Materials 176 YCT
Ans. (d) : Purely elastic material (a) 10 MPa (b) 1 MPa
(c) 0.1 MPa (d) 100 MPa
BHEL ET 2019
Ans. (b) : Given - Dimension = 100 × 100 × 10

Loading and unloading to energy loss.

Viscoelastic material

A = 100 × 100 mm2

P = 10 × 103 N
Direct shear stress
P 10 ×10 3
Loading and unloading the area inside the curve, τ= = = 1N / mm 2
hysteresis loop is energy loss. A 100 × 100
135. A metal sphere of diameter D is subjected to a = 1 MPa
suniform increase in temperature ∆T. E, ν and 139. A copper rod with initial length lo is pulled by
α are the Young's modulus, Poisson's ratio and a force. The instantaneous length of the rod is
Coefficient of thermal expansion respectively. given by l = lo ( 1 + 2e4t), where t represents
If the ball is free to expand, the hydrostatic time. True strain rate at t = 0 is :
stress developed within the ball due to 1 8
temperature change is (a) (b)
α∆TE 3 3
(a) 0 (b) 4 2
1 − 2ν (c) (d)
α∆TE α∆TE 3 3
(c) − (d) BHEL ET 2019
1 − 2ν 3(1 − 2ν )
APPSC AEE SCREENING 17.02.2019 Ans. (b) : Given - Initial length = lo
Ans. (a) : The hydrostatic stress developed within the Instantaneous length = l = lo (1 + 2e4t)
ball due to temperature change is zero as the ball is free at, t = 0
to expand. l = lo (1 + 2eo)
136. Which mechanical property gets affected in an l = lo (1 + 2) = 3 lo
alloy, when it is over-aged condition : dl
(a) lower hardness True strain ∈T = ∫
l
(b) low strain hardening rate
(c) higher yield strength change in length
∈T =
(d) higher tensile strength Instantaneous length
BHEL ET 2019
= l o ( 8e 4t )
dl
Ans. (a) : Change in length
• The ageing phenomena are associated with aluminium dt
alloy mainly. at, t = 0
• Age hardening curve of the alloy basically contain of dl
= lo × 8
initial hardness it maximum hardness subsequently dt
decrease hardness with ageing time which called the
under ageing. 8l 8
True strain ∈T = o ∈T =
137. After which point of the Stress-Strain Diagram 3 lo 3
does metal cutting start?
(a) Proportional point (b) Ultimate point 140. True stress experienced by a material is
(c) Fracture point (d) Yield point ______ than the engineering stress at a given
BHEL ET 2019 load.
Ans. (c) : After fracture point of stress-strain diagram, (a) lower
metal cutting start. (b) higher
138. A block is dimensions of upper surface 100 (c) equal
mm × 100 mm. The height of the block is 10 (d) higher or lower
mm. A tangential force of 10 kN is applied at BHEL ET 2019
the centre of the upper surface. The block is Ans. (b) : True stress experienced by a material is
displaced by 1 mm with respect to lower face.
Direct shear stress in the element is : higher than the engineering stress at a given load.
Strength of Materials 177 YCT
 Ans. (c) : Data given-
d = 30 mm, ∆d = 0.0039 mm
l = 200 mm, ∆l = 0.09 mm
P = 60 kN
we know that
Poisson's ratio
 ∆d   0.0039 
 d   30 
µ=   = 
 ∆l   0.09 
 l   200 
µ = 0.2888
141. A 1m long rod is fixed at one end. There is a 144. The loads acting on a 3 mm diameter bar at
rigid wall at a distance 1 mm from the free end different points are as shown in the figure:
of the rod as depicted in the figure. What is the
thermal stress generated in the rod if its
temperature is increased by 100ºC?
Take E = 200 GPa and α = 12 × 10-6/ºC

If E = 205 GPa, the total elongation of the bar

will be nearly.
(a) 29.7 mm (b) 25.6 mm
(c) 21.5 mm (d) 17.4 mm
ESE 2019
Ans. (a) :
(a) 40 MPa (b) 80 MPa
(c) 120 MPa (d) 240 MPa
APPSC AEE SCREENING 17.02.2019
Ans. (a) : Free expansion = l α ∆T Total elongation is equal to sum of elongation of each
σ bar
ℓα∆t − ∆ =
E δ = δAB + δBC + δCA
ℓα∆t = (1000) (12 × 10-6) (100) = 1.2 mm pℓ pℓ pℓ
= 1 1 + 2 2 + 3 3
Expansion prevented = 1.2 - 1 = 0.2 mm A1E1 A 2 E 2 A 3E 3
Pl A1E1 = A2E2 = A3E3 = AE
0.2 =
AE 10 × 103 × 2000 8 × 103 × 1000 5 ×103 × 3000
= + +
 P AE AE AE
∵ σ =  43 × 106 43 × 106
 A
= = = 29.68 mm
2 (σ )(1000) AE π 2
= × 3 × 205 × 103
10 200 × 103 4
145. Rails are laid such that there will be no stress
σ = 40 MPa in them at 24°C. If the rails are 32 m long with
142. If a material is heated up, its Elastic modulus an expansion allowance of 8 mm per rail,
(a) decreases (b) increases coefficient of linear expansion α =11×10–6/°C
(c) remains constant (d) None of the above and E = 205 GPa, the stress in the rails at 80°c
APPSC AEE SCREENING 17.02.2019 will be nearly.
Ans. (a) : As the material is heated up, it becomes soft. (a) 68 MPa (b) 75 MPa
(c) 83 MPa (d) 90 MPa
It undergoes more strain for a given stress
ESE 2019
 σ
∵ E =  The modulus of elasticity
Ans. (b) :
 ∈
decreases.
143. A bar of 30 mm diameter is subjected to a pull
of 60 kN. The measured extension on gauge Given,
length of 200 mm is 0.09 mm and change in ℓ = 32 m ∆ = 8 mm
diameter is 0.0039 mm. Find its Poisson's ratio. α = 11 × 10−6 /°C
(a) 0.309 (b) 0.299 E = 205 GPa = 205 × 103 N/mm2
(c) 0.289 (d) 0.279 ∆T = 80 − 24 = 56°C
TNPSC AE 2019 if ∆ is the permissible expansion allowance then thermal
Strength of Materials 178 YCT
 Lα∆t − ∆ Ans. (a) : Given,
stress σ = Initial diameter (di) = 12.8 mm
L
σℓ Final diameter (df) = 10.7 mm
ℓ α ∆T − ∆ = Engineering fracture strength (σf) = 460 MPa
E True stress at fracture
σ × 32 × 103 A
32 × 103 × 11 × 10−6 × 56 − 8 = (σt) = σf i
205 × 103 Af
11.712 = 0.156 σ 2
σ = 75.03 N/mm2  d2   12.8 
= 460  i2  = 460  
σ = 75.03 MPa  df   10.7 
146. When a load of 20 kN is gradually applied at a = 658.27 MPa ≈ 660 MPa
particular point in a beam, it produces a 148. If for a given material, E = 2G (E is modulus of
maximum bending stress of 20 MPa and a elasticity, G is the shear modulus), then the
deflection of 10 mm. What will be the height bulk modulus K will be
from which a load of 5 kN should fall onto the
beam at the same point if the maximum E E
(a) (b)
bending stress is 40 MPa? 2 3
(a) 80 mm (b) 70 mm E
(c) 60 mm (d) 50 mm (c) E (d)
ESE 2019 4
APPSC AEE SCREENING 17.02.2019
Ans. (c) : For 20 kN static load (P1 = 20 kN) Ans. (b) : Given, E = 2G
δ1 = 10 mm We have, E = 2G (1 + µ)
(σ1)static = 20 MPa 2G = 2G (1 + µ)
For 5 kN impact load (P2 = 5 kN)
µ=0
σmax = 40 MPa
and E = 3K (1 - 2µ)
(σ2)static = ? δ2 = ? E = 3K [1 - 2(0)]
From equation E = 3K
P
σ= E
A ⇒ K=
σ∝P 3
149. Consider the state of stress at any point as σxx
( σ2 )static P2 = 250 MPa, σzz = 250 MPa. The Young's
=
( σ1 )static P1 modulus and Poisson's ratio of the material is
P 5 considered as 2 GPa and 0.18 respectively.
(σ2)static = ( σ1 )static × 2 = 20 × = 5 MPa Determine the εzz at the point.
P1 20 (a) -0.125 (b) 0.103
deflection (δ) ∝ P (c) -0.103 (d) 0.125
δ2 P2 APPSC AEE SCREENING 17.02.2019
=
δ1 P1 Ans. (b) : σxx = 250 MPa
P 5 σyy = 0
δ2 = δ1 2 = 10 × = 2.5 mm σzz = 250 MPa
P1 20 E = 2 × 109 Pa
As we know for impact load µ = 0.18
 2h  σ zz σ yy σ xx
σmax = σstatic 1 + 1 +  ε zz = −µ −µ
 δstatic  E E E
σ zz σ xx σ xx
 2h  = −µ = (1 − µ )
40 = 5 1 + 1 +  E E E
 2.5  (∵ σ xx = σ zz )
h = 60 mm
250 × 106
147. A cylindrical specimen of steel having an = (1 − 0.18)
2 × 109
original diameter of 12.8 mm is tensile tested = 0.1025 = 0.103
to fracture and found to have engineering
150. Find out the Lame constants (λ and µ) for an
fracture strength σf of 460 MPa. If its cross isotropic material having modulus of elasticity
sectional diameter at fracture is 10.7 mm, the (E) and Poisson's ratio (v) as 200 GPa and 0.2,
true stress at fracture will be respectively.
(a) 660 MPa (a) 80 GPa, 80 GPa
(b) 645 MPa (b) 35.71 GPa, 166.6 GPa
(c) 630 MPa (c) 55.55 GPa, 83.33 GPa
(d) 615 MPa (d) 73.33 GPa, 66.66 GPa
ESE 2019 APPSC AEE SCREENING 17.02.2019
Strength of Materials 179 YCT
Ans. (c) : E = 200 GPa
Poisson's ratio ν = 0.2
Eν
Lame constant (λ ) =
(1 + ν )(1 − 2ν )
200 × 0.2
λ= = 55.55GPa
(1 + 0.2)(1 − 2(0.2))
2E ν 2 × 200 × 0.2
µ= = =83.33 GPa
(1 + ν )(1 −ν ) (1 + 0.2)(1 − 0.2) 154. The value of yield stress obtained by offset
151. The deformation per unit length is called : method is known as :
(a) Tensile stress (b) Poissons ratio (a) True stress (b) Elastic stress
(c) Youngs modulus (d) Strain (c) Proof stress (d) Breaking stress
MECON MT 2019 Oil India Limited Sr. Engineer (Drilling) 30.11.2019
Ans. (d) : The deformation per unit length is called Ans. (c) : The value of yield stress obtained by offset
strain. method is known as proof stress.
152. A bar of 20 mm diameter is subjected to pull of
50 kN. If the measured extension over the
gauge length of 20 cm is 0.1 mm and the
change in diameter is 0.0035 mm, Poisson's
ratio and young's modulus are:
(a) 0.18, 200 GPa (b) 0.1, 108 GPa
(c) 0.25, 180 GPa (d) 0.35, 318 GPa
Oil India Ltd. Sr. Engg. (Production) 30.11.2019
Ans. (d) : Given,
Diameter of bar (d) = 20 mm 155. The unit of strain is
Pull load (F) = 50 kN (a) N-mm (b) N/mm
Gauge length (lg) = 20 cm = 200 mm (c) mm (d) no unit
Extension (δl) = 0.1 mm Nagaland PSC (CTSE) 2018, Paper-I
Change in diameter (δd) = 0.0035 mm Ans. (d) : As we know, strain is the ratio of change in
∈lateral δd / d 0.0035/ 20 length to the total length so it has no. dimension (unit).
µ= = = ∆L
∈longitudinal δl / lg 0.1/ 200 ε=
µ = 0.35 L
156. The modulus of elasticity for mild steel; is
approximately equal to
σ F/ A F × lg 50 ×103 × 200 (a) 10 kN/mm2 (b) 80 kN/mm2
E= = = = 2
∈longitudinal δℓ / ℓ g A × δl π 20 2 × 0.1 (c) 100 kN/mm (d) 210 kN/mm2
( ) Nagaland PSC (CTSE) 2018, Paper-I
4
= 318.31GPa Ans. (d) : As we know, the modulus of elasticity for
mild steel is approximately equal to 210 GPa or 210
153. The statement ''Total effect of several loads kN/mm2.
applied on a body is the sum of effects of
individual loads applied separately'' is known 157. Two bars of different materials and same size
as : are subjected to the same tensile force. If the
(a) Generalized Hooke's law bars have unit elongation in the ratio of 2 : 5,
(b) St. Venant's principle then the ratio of modules of elasticity of the
(c) Principle of superposition two materials will be
(d) Uniqueness theorem (a) 2:5 (b) 5:2
(c) 4:3 (d) 3:4
Oil India Limited Sr. Engineer (Drilling) 30.11.2019
Nagaland PSC (CTSE) 2018, Paper-I
Ans. (c) : Principle of superposition- Simply states Ans. (b) : As we know,
that on a linear elastic structure, the combined effect of PL ∆L 2
several loads acting simultaneously is equal to the E= ∴ 1 =
algebraic sum of the effects of each load acting A ∆ L ∆ L 2 5
individually. 1
So, E1 ∝
In other words, for a linearly elastic ∆L1
member/structure the effect of several loads acting on a E 1 ∆L 2 5
member is equal to the summation of the loads acting = =
E 2 ∆L1 2
separately.
158. When a bar is cooled to –50°C, it will develop
(a) no stress (b) shear stress
(c) tensile stress (d) compressive stress
Nagaland PSC (CTSE) 2018, Paper-I
Strength of Materials 180 YCT
Ans. (c) : When a bar is cooled to –5 temperature, it 163. Allotropic metal,
must goes to tensile stress, as the bar will resists to (a) exists in more than one type of lattice
expand, so tensile stress will developed. structure depending upon temperature
159. If the modules of elasticity for a given material (b) has equal stresses in all directions
is twice its modules of rigidity, then bulk (c) has only one lattice structure of all
modules is equal to temperature
(a) 2C (b) 3C (d) gives equal strain in all direction
(c) 2C/3 (d) 3C/2 TNPSC AE 2018
Nagaland PSC (CTSE) 2018, Paper-I Ans. (a) : Allotropic metal, exists in more than one type
9KC of lattice structure depending upon the temperature.
Ans. (c) : We know, E = 164. Which of the following is correct relation
C + 3K among elastic constants E (modulus of
Given, E = 2C
elasticity), G (modulus of rigidity), ν (Poisson's
9KC
⇒ 2C = ratio) and K (bulk modulus)?
3K + C (a) E = 3K (1 − ν ) = 2G (1 + ν )
⇒ 2C(3K + C) = 9 KC
⇒ 2
6KC + 2C = 9 KC (b) E = 2G (1 − ν ) = 3K (1 + ν )
⇒ 2C2 = 3 KC (c) E = 3K (1 − 2ν ) = 2G (1 + ν )
2C = 3 K
2C (d) E = 2K (1 − 2ν ) = 3G (1 + ν )
∴ K=
3 (e) E = 3K (1 + 2ν ) = 2G (1 − ν )
160. Consider a bar of length 'L', breadth 'b' and CGPSC AE 25.02.2018
thickness 't' subjected to an axial pull or Ans. (c) : We know that
tension 'P'. The resulting volumetric strain will E = 3K [1 – 2ν)
be equal to E = 2G [1 + ν)
(a) ∈ (1 − 2µ ) (b) 2 ∈ (1 − 2µ ) 165. The area of under the stress-strain diagram up
(c) ∈ (1 + 2µ ) (d) 3 to the rupture point is known as
(a) Proof resilience
GPSC Executive Engineer 23.12.2018 (b) Modulus of toughness
 σ x + σ y + σz  (c) Modulus of elasticity
Ans. (a) : ∈v =∈x + ∈y + ∈z =   (1 − 2µ ) (d) Modulus of resilience
 E  HPPSC AE 2018
∈v =∈ (1 − 2µ ) Ans. (b) :
161. If Poisson's ratio of a material is 0.75, then the
elastic modulus for the material is
(a) 3.5 times its shear modulus
(b) 4 times shear modulus
(c) equal to its shear modulus
(d) 5 times its shear modulus
GPSC Executive Engineer 23.12.2018
Ans. (a) : Given,
µ = 0.75
E = 2G(1+µ)
= 2G (1+0.75)
Point A – Proportionality limit
= 2 × G × 1.75 Point B – Elastic limit
=3.50 G Point C – Upper yield Point
E = 3.5G Point D – Lower yield Point
162. Lateral strain (∈) can be expressed as DE – Yielding Region
EF – Strain hardening region
δl l
(a) (b) F – Ultimate point
l δl FG – Necking region
(c) γ ∈ (d) −γ ∈ G – Breaking [Rupture point]
TNPSC AE 2018 Modulus of Toughness [M.O.T.]–Modulus of
Ans. (d) : Lateral strain (∈ ') Toughness is defined as energy observed by a
component per unit volume just before its rupture.
= −γ × longitudinal strain M.O.T.–Total area of stress vs strain curve per unit
= −γ×∈ volume.
where γ → poisson ratio 166. Modulus of Rigidity is related to-
lateral strain (a) Length (b) Shape
= (c) Size (d) Volume
longitudnal strain RPSC AE 2018
Strength of Materials 181 YCT
Ans. (b) : Modulus of Rigidity—The modulus of
rigidity is the elastic coefficient when a shear force is
applied resulting in lateral deformation. It gives us a
measure of how rigid a body is

(a) 4060 N (b) 6780 N

(c) 12240 N (d) 2412 N
ISRO Scientist/Engineer 22.04.2018
Ans. (a): After Increasing the temperature of composite
bar. Total deflection due to temperature rise is given by
δTotal = (αtl ) s + (αtl ) Al
F = 12 × 10−6 × 10 × 3000 + 11 × 10 −6 × 10 × 2000
τ xy  A  δTotal = 0.58mm.......(1)
G= = Another way to finding total elongation
γ xy  ∆x  PLs PLA
  δTotal = +
 l  ( AE ) s ( AE ) Al
F ×l
= P × 3000 P × 2000
A ∆x δTotal = +
200 × 210 × 103 400 × 70 ×103
where P
F δTotal = .......(2)
• τ xy = is shear stress. 7000
A P = 7000 × δTotal
• F is the force acting on the object. P = 7000 × 0.5 = 4060 N
∆x
• γ xy = is the shear strain. 170. A round uniformly tapered bar of length L and
l Young's modulus E has diameter of 'd' at one
• ∆x is the transverse displacement. end and '2d' at the other end. If the bar is
167. Detrimental property of a material for shock pulled by an axial force F, the extension
load application is- produced will be
(a) High density (b) Low toughness 2 FL 4 FL
(c) High strength (d) Low hardness (a) (b)
π Ed 2
π Ed 2
RPSC AE 2018
8 FL 16 FL
Ans. (b) : Detrimental property of a material for shock (c) (d)
load application is low toughness. π Ed 2
π Ed 2
168. A metallic rod of length 2.5 m and cross- ISRO Scientist/Engineer 22.04.2018
section of 20 mm2 is elongated by a load of 500 Ans. (a): Extension produced in tapered bar
kN. If Poisson ratio is 0.25, then the increase in 4 FL
δ=
volume is: (assume E = 2 × 105 N/mm2) πd × 2d × E
(a) 3000 mm3 (b) 2125 mm3 2 FL
(c) 2000 mm 3
(d) 3125 mm3 δ=
AAI Jr. Executive 29.11.2018 πEd 2
Ans. (d) : Given, 171. A rigid beam of negligible weight is supported
L = 2.5 m in a horizontal position by two rods of steel
A = 20 mm2 and aluminium, 2 m and 1 m long, having
P = 500 kN values of cross sectional areas 100 mm2 and
µ = 0.25 200 mm2, and young’s modulus of 200 GPa
E = 2 × 105 N/mm2 and 100 GPa, respectively. A load P is applied
as shown in the figure below:
∈v =
(1 − 2µ )( σ x ) = ∆V
3E V
(1 − 0.25 × 2 ) × 500 ×103 = ∆V = ∆V
3 × 2 ×105 20 V 20 × 2500
∆V = 3125mm3
169. A composite bar made of aluminum and steel
is fixed rigidly at supports P and Q as shown in
figure. When the temperature of the bar is
If the rigid beam is to remain horizontal then
increased by 10ºC, the reactions at the (a) the force P must be applied at the centre of
supports are (given Eal = 70 GPa, Es = 210 the beam
GPa, αal = 11E-6 mm/mmºC and αs = 12 E-6 (b) the force on the steel rod should be twice the
mm/mmºC) force on the aluminium rod
Strength of Materials 182 YCT
 (c) the force on the aluminium rod should be
twice the force on the steel rod
(d) the forces on both the rods should be equal
ESE 2018
Ans. (c) :

W = x × A × L (x- specific weight N/m3)

W = (δh) × A × L
δgA × L × L
∴ Total elongation =
2AE
δgL2
=
2E
174. A round bar length l, elastic modulus E and
Let P1 = Force in steel Poisson's ratio is subjected to an axial pull µ, is
P2 = Force in aluminium subjected to an axial pull 'P'. What would be
From the given condition that the rigid beam to remain the change in volume of the bar?
horizontal. Pl Pl (1 − 2µ )
A1 = 100 mm2 (a) (b)
A2 = 200
(1 − 2 µ ) E E
L1 = 2 m = 2000 mm Pl µ Pl µ
(c) (d)
L2 = 1 m = 100 mm E (1 − µ ) E
E1 = 200 GP TSPSC AEE 28.08.2017 (Civil/Mechanical)
E2 = 100 GP
Ans. (b) : Volumetric strain for cylinder
δ1 = δ2
ev = eℓ + 2ed
P1L1  P2 L 2 
=  P/A
A1E1  A 2 E 2  eℓ =
E
P1 × 2000 P2 × 1000
= P/A
100 × 200 100 × 200 ed = –µ
2P1 = P2 E
P
172. The percentage elongation of a material as ∴ ev = (1 − 2µ )
AE
obtained from static tension test, depends on
∆V P P
(a) diameter of the test specimen ev = = = (1 − 2µ )
(b) gauge length of the specimen V AE AE
(c) nature of end grips of the testing machines P
(d) geometry of the test specimen
⇒ ∆V = (A × ℓ) (1 − 2µ )
AE
TSPSC AEE 28.08.2017 (Civil/Mechanical) Pℓ
Ans. (b) : The percentage elongation of a material as ∆V = (1 − 2µ )
E
obtained from static tension test, depends on gauge
175. A 2m long rod of diameter 2 mm is subjected
∆ℓ
length of the specimen (% Elongation = × 100 ). to an axial pull of 12 kN. The rod is extended
ℓ by 0.5 cm. What is the approximate value of
173. A heavy uniform rod of length 'L' and the modulus of elasticity of the material of the
material density δ is hung vertically with its rod?
top end rigidly fixed. How is the total (a) 12 GPa (b) 180 GPa
elongation of the bar under its own weight (c) 125 GPa (d) 200 GPa
expressed? TSPSC AEE 28.08.2017 (Civil/Mechanical)
2δL2 g δL2 g Ans. (*) : Given data is wrong
(a) (b) Length of rod 'ℓ' = 2 m
E E
δL g
2
δL2 g Diameter 'd' = 2 mm = 2 × 10–3 m
(c) (d) Axial pull 'P' = 12 kN
2E 2E
Changed in length '∆ℓ' = 0.5 cm = 0.5 × 10–2 M
TSPSC AEE 28.08.2017 (Civil/Mechanical)
Pℓ
Ans. (d) : Given : ∆ℓ =
Length of uniform rod = 'L' AE
Material density = δ
E=
Pℓ
=
( )
12 × 103 × 2
Let weight of rod be w A∆ℓ π 2 × 10−3 2 × 0.5 × 10−2
Total elongation of rod under its own weight ( 4
) ( )
WL = 1520 × 109 N/m2
=
2AE = 1520 GPa
Strength of Materials 183 YCT
176. If a material has numerically the same value From similar triangle
for its modulus of rigidity and bulk modulus, δA 1
then what is Poisson's ratio? =
δC 0.6
(a) 0.25 (b) 0.2
(c) 0.15 (d) 0.125  FA ⋅ L 
TSPSC AEE 28.08.2017 (Civil/Mechanical)  
 AE  = 1
Ans. (d) : Given :  FC ⋅ L  0.6
Modulus of rigidity = Bulk modulus  
G=K  AE 
Also FA 1
=
2G (1 + µ) = 3K (1 – 2µ) FC 0.6
2 + 2µ = 3 – 6µ FC = 0.6 FA .....(1)
8µ = 1 Now ∑Mo = 0
1 (FA × 1) + (FC × 0.6) = 20 × 0.8
µ = = 0.125 FA + 0.6 FC = 16 .....(2)
8
177. What is the maximum possible value of On solving equation (1) & (2)
Poisson's ratio for a non-dilatants material? F A = 11.76 kN
(a) 0.67 (b) 0.50 FC = 7.05 kN
(c) 0.33 (d) 0.25 180. A steel rod, 2 m long and 20 mm × 20 mm in
TSPSC AEE 28.08.2017 (Civil/Mechanical) cross section, is subjected to a tensile force of
Ans. (c) : For non dilatants material, maximum value of 40 kN. What will be elongation of the rod when
poisson's ratio is 0.33. the modulus of elasticity is 200 × 103 N/mm2?
178. Modulus of elasticity and Poisson's ratio of a (a) 0.5 mm (b) 1.0 mm
material are 2.1 × 105 Nmm2 and 0.25 (c) 1.5 mm (d) 2.0 mm
respectively. What is the value of modulus of (e) 2.5 mm
5
rigidity of the same material in 10 N/mm ? 2 (CGPSC Polytechnic Lecturer 2017)
(a) 0.84 (b) 0.70 Ans. (b) : Data given;
(c) 1.40 (d) 0.50 L = 2 m = 2 × 103 mm
TSPSC AEE 28.08.2017 (Civil/Mechanical) A = 20 × 20 mm2
Ans. (a) : E = 2G (1 + µ) F = 40 kN
2.1 × 105 = 2G (1 + 0.25) E = 200 × 103 N/mm2
2.1× 105 δℓ = ?
G= = 0.84 N / mm 2 We know that,
2 × 1.25
G = 0.84 × 105 N/mm2 F× L 40 × 103 × 2 ×103
Elongation, δℓ = =
179. A rigid bar ACO as shown is hinged at O and A × E 20 × 20 × 200 × 103
is held in a horizontal position by two identical 181. For copper, the yield stress σy and the brittle
vertical steel wires AB and CD. A point load of fracture stress σf are related as:
20 kN is hung at the position shown. The (a) σf > σy (b) σy > σf
tensions in wires AB and CD are (c) σy = σf (d) σf << σy
TRB Polytechnic Lecturer 2017
Ans. (a) : For copper
σf > σy

(a) 15.2 kN and 7.1 kN (b) 11.8 kN and 7.1 kN

(c) 15.2 kN and 5.0 kN (d) 11.8 kN and 5.0 kN
ESE 2017
Ans. (b) :
182. Proof stress–
(a) Is the safe stress
(b) Cause a specified permanent deformation in a
material usually 0.1% or less
(c) Is used in connection with acceptance tests
for materials
(d) Does not exist
Nagaland PSC CTSE 2017, Paper-I
Strength of Materials 184 YCT
Ans. (b) : When material such as aluminium does not Ans : (a) : Because necking begins to occur, where by
have an obvious yield point and yet undergoes large engineering stress becomes less than the true stress.
strain after the proportional limit is exceeded, an
arbitrary yield stress may be determined by the offset of
0.1 or 0.2% of strain.

183. The stress strain curve for glass rod during 186. Two dimensionally equal blocks made of
tensile test would exhibit– material M1 and M2 are placed on a flat
(a) A straight line (b) A parabola horizontal surface as shown in figure below.
(c) A sudden break (d) None of the above
Young's modulus of the materials are E1 & E2
Nagaland PSC CTSE 2017, Paper-I
resp. A uniform pressure 'p' is applied over the
Ans. (c) : Stress-strain curve for a glass rod during
blocks through a thick plate (AB)
tensile test would exhibit, a sudden break. point Occur
symmetrically placed over the blocks. For E1 >
(due to a glass becomes a Brittle material).
E2, consider the statements below.

184. A bar of mild steel 200 mm long and 50 mm ×

50 mm in cross section is subjected to an axial
load of 200 kN. If E is 200 GPa, the elongation
(a) Stress in both blocks are same
of the bar will be
(b) Strain in both blocks are same
(a) 0.16 mm (b) 0.08 mm
(c) Plate AB tilts and point B moves downward
(c) 0.04 mm (d) 0.02 mm
JWM 2017 (d) Plate AB tilts and point A moves downward
Ans. (b) : Length of bar, L = 200 mm ISRO Scientist/Engineer 17.12.2017
Area of bar, A = 50 × 50 mm = 250 mm 2 Pℓ
3
Ans. (c) : Deflection (∆) =
Axial load, P = 200 × 10 N AE
E = 200 × 103 N/mm2 1
∆∝
Elongation of bar, E
PL 200 × 10 3 × 200 as E1 > E2
δ= =
AE 50 × 50 × 200 ×10 3 ∆1 < ∆2
δ = 0.08mm Hence AB tilts and B moves downward
P
185. The stress-strain plot for ductile materials Stress = (Independent of E)
exhibits peak at ultimate strength A
(a) because necking begins to occur where by As two blocks are dimensionally equal having equal
engineering stress becomes less than the true stresses.
stress 187. A horizontal bar AB assumed to be rigid, is
(b) because the material starts becoming weaker supported by two wires CE & DF, having
at microstructural level lengths L and 2L respectively and pinned at A.
(c) due to strain softening of the material If each wire has cross sectional area A, the
(d) None of the above tensile stress σ1 and σ2 in wires CE and DF are
BPSC AE Mains 2017 Paper - VI respectively.

Strength of Materials 185 YCT

 1 PL P 2 L  π 2
= ×P× = ∵ A = 4 d 
2 AE 2AE
1 PA2 L A 1 PB2 L B
⇒ = × 16
2 AA EA 2 A BEB
1 1
= × 16
AA AB
AB = 16AA
π 2 π
d B = × 16 × d 2A
4 4
d 2B = 16d 2A
dB = 16 × (13)2
3P 6P 2P 4P dB = 4 × 13
(a) and (b) and = 52 mm
5A 5A 5A 5A
3P 6P P 189. The figure shows an Aluminium rod of 25 mm2
(c) and (d) each cross sectional area. It is loaded at four points,
A A A K, L, M and N. Assume E = 67 GPa for
ISRO Scientist/Engineer 17.12.2017 Aluminium. The total change in length of the
Ans. (d) : rod due to loading as shown in close to

(a) 30 µm (b) −10 µm

(c) −30 µm (d) 10 µm
ISRO Scientist/Engineer 07.05.2017
Ans. (c) :
Taking moment about point A
P × 3b – P2 × 2b – P1 × b = 0
3P = P1 + 2P2 .....(1)
∆2 = 2∆1
P2 × 2L P ×L
=2 1
AE AE
P2 = P1 .....(2)
From equation (1) and (2)
P = P1 = P2
Tensile stresses in CE and DF
P
σ1 =
A
P
σ2 =
A
188. Two bolts A and B of same material and length
are subjected to identical tensile load. If the
elastic strain energy stored in bolt A is 16
times that of the bolt B and the mean diameter
of bolt A is 13 mm, the mean diameter of bolt
B (mm) is
(a) 52 (b) 36 Let, change in length of the 'rod' is ∆L
(c) 24 (d) 16  PL 
So, ∆LKL =  
ISRO Scientist/Engineer 07.05.2017  AE  KL
Ans. (a) : Given, 100 × 500
=
PA = PB 25 × 67 ×103
UA = 16 UB 50
⇒ (Strain energy)A = 16 (Strain energy)B =
25 × 67
1  PL  50
SE = Pδ ∵δ = (∆L)KL = = 0.0298 (Tension) ≈ 0.03
2  AE  1675
Strength of Materials 186 YCT
  PL  60 × 10−5
(∆L)LM =   ∈ℓ = = 200 × 10−5
 AE LM 0.3
150 × 800 σ = ∈E
= 300
25 × 67 ×103 E= = 150 GPa
120 200 × 10−5
= = 0.07 (Compression) 192. The modulus of rigidity of an elastic material
1675 is found to be 38.5% of the value of its Young’s
 PL  modulus. The Poisson’s ratio µ of the material
(∆L)MN =  
 AE MN is nearly
50 × 400 (a) 0.28 (b) 0.30
= (c) 0.33 (d) 0.35
25 × 67 ×103 ESE 2017
20 Ans. (b) : Given, G = 0.385 E
= = 0.01 (Tension)
1675 We know that E = 2G (1 + µ)
Total change in length (∆L) = (∆L)KL + (∆L)LM E 1
+ (∆L)MN 1+µ= =
2G 2 × 0.385
= 0.03 – 0.07 + 0.01 1 + µ = 1.297
= –0.03 mm µ = 0.297
(∆L) = –30 µm 193. Material Stress-strain relationship expressed
190. A 10 mm diameter bar of mild steel of elastic by σ = Kεn; where in K & n are :
modulus 200×109 Pa is subjected to a tensile (a) K is strength coefficient & n is yield strain
load of 50000N, taking it just beyond its yield (b) K is yield stress & n is work hardening
point. The elastic recovery of strain that would exponent
occur upon removal of tensile load will be (c) K is strength coefficient & n is work
(a) 1.38×10–3 (b) 2.68×10–3 hardening exponent
(c) 3.18×10–3 (d) 4.62×10–3 (d) None of the above
ESE 2017 Punjab PSC SDE 12.02.2017
Ans. (c) : Given, Ans. (c) : σ = K∈n
d = 10 mm Where,
E = 200 × 109 Pa = 200 × 103 MPa K is strength coefficient
P = 50000 N n is work hardening exponent
P P 4P 194. An aluminum bar (E = 72 GPa, v = 0.33) of
stress (σ) = = = 2 diameter 50mm cannot exceed a diameter of
π
A d 2 πd 50.1mm when compressed by axial force P.
4 The maximum acceptable compressive load P
is approximately.
(a) 190 kN (b) 200 kN
(c) 470 kN (d) 860 kN
HPPSC Asstt. Prof. 20.11.2017
4 × 50000 Ans. (d) : Given,
=
π×102 E = 72 GPa, Poisson ratio (v) = 0.33, d0 = 50mm,
= 636.94 N/mm2 df=50.1mm
= 636.94 MPa δd d f − d 0 σ P
σ = E∈ ∈lat. = = =v =v
d0 d0 E A.E
636.94
∈= df − d0 P
200 × 103 =v
= 3.18 × 10 −3
d0 π 2
d 0 .E
191. A bar produces a lateral strain of magnitude 4
60×10−5 m/m when subjected to a tensile stress (d − d 0 ) π d 02 × E
of magnitude 300 MPa along the axial P= f × ×
d0 4 v
direction. What is the elastic modulus of the
material if the Poisson’s ratio is 0.3? 0.1 π (50) × 72 × 103
2
P= × × = 856.79kN
(a) 200 GPa (b) 150 GPa 50 4 0.33
(c) 125 GPa (d) 100 GPa ≈ 860 kN
ESE 2017 195. A copper tube with wall thickness of 8mm
Ans. (b) : Given, must carry an axial tensile force of 175 kN.
Lateral strain = 60 × 10−5 The allowable tensile stress is 90 MPa. The
σ = 300 MPa minimum required outer diameter is
µ = 0.3 approximately.
We know that (a) 60 mm (b) 72 mm
(c) 85 mm (d) 93 mm
Lateral strain = µ × longitudinal strain = µ × ∈ℓ HPPSC Asstt. Prof. 20.11.2017
Strength of Materials 187 YCT
Ans. (c) : Stress
P P Ans. (a) : Young's modulus (E) = (N/m2)
Stress (σ) = = D 2 = D1 + 2t Strain
A π (D 2 − D 2 ) 200. True stress is
2 1
4 (a) average load/original area
4× P (b) average load/minimum area
D 22 − (D 2 − 2t) 2 = (c) average load/ average area
π× σ
(d) instantaneous load/instantaneous area
4 × 175 × 103
D 22 − (D 2 − 2 × 8) 2 = WBPSC AE, 2017
π× 90 Ans. (d) : • True stress represents the ratio of
D 2 ≈ 85mm instantaneous load to instantaneous area.
W
196. A nylon bar (E = 2.1 GPa) with diameter σT = i
12mm, length 4.5 m, and weight 5.6 N hangs Ai
vertically under its own weight. The elongation • Engineering stress represents the ratio of average
of the bar at its free and is approximately? load to original cross-section of the specimen.
(a) 0.05 mm (b) 0.07 mm W
(c) 0.11 mm (d) 0.17 mm σ Engineering =
Ao
HPPSC Asstt. Prof. 20.11.2017
Ans. (a) : Given, E = 2.1 × 109 Pa, d = 12mm, L = 201. A load of 10000 kg is applied to copper
4.5m, W = 5.6 N. cylinder 20cm long and 5cm in diameter causes
the length to increase 0.4mm and diameter to
WL decreases 0.04 mm. The Poisson ratio for
Elongation due to self weight (δSelf-Weight) =
2AE copper would be
5.6 × 4.5 (a) 0.23 (b) 0.25
δSelf -Weight = (c) 0.33 (d) 0.4
π
2 × × (12) 2 × 2.1× 109 WBPSC AE, 2017
4
Lateral strain
25.2 Ans. (d) : Poisson ratio (µ) = −
δSelf-Weight = Longitudinal strain
1899.072 ×103
δSelf-Weight = 5.305 × 10–5 m = 0.053mm 0.04 / 50 0.04
µ= = = 0.4
197. A copper bar (d = 10 mm, E = 110 GPa) is 0.4 / 200 0.1
loaded by tensile load P = 11.5 kN. The 202. A reinforced concrete beam is considered as
maximum shear stress in the bar is made of
approximately. (a) Homogeneous material
(a) 73 MPa (b) 87 MPa (b) Heterogeneous material
(c) 145 MPa (d) 150 MPa (c) Isotropic material
HPPSC Asstt. Prof. 20.11.2017 (d) Clad material
Ans. (a) : Given, P = 11.5kN, d = 10mm WBPSC AE, 2017
P 11.5 × 103 Ans. (b) : A reinforced concrete beam is considered as
Normal stress (σ t ) = = made of heterogeneous material.
A π × (10) 2
203. The compressive lateral strain when a cube is
4
subjected to mutually perpendicular tensile
= 146.42MPa stresses of equal magnitude is given by
σ 146.42 1 E 1 m
Maximum shear stress (τmax) = t = (a) × (b) ×
2 2 σ m σ E
= 73.21MPa 1 σ
(c) × (d) None of the above
198. The 'Modular ratio' is the ratio of : m E
(a) Stress and strain Where, σ = stress, E = Young's modulus,
(b) Strain and stress 1
(c) Elasticity of two materials and = Poisson's ration
m
(d) Change in length to original length Karnataka PSC AE, 10.09.2017
GMB AAE 25.06.2017
Ans. (c) :
Ans. (c) : Modular ratio is defined as the ratio between
lateral strain
modulus of elasticity of steel and modulus of elasticity µ=
of concrete it is used to finding losses in pre-stress Longitudinal strain
because of the influence of creep. ∆L 
199. What is the unit of Young's Modulus ? Lateral strain = µ  
 L 
(a) N/m2 (b) N.m
µσ σ
(c) N/m3 (d) N/m4 = =
GMB AAE 25.06.2017 E mE
Strength of Materials 188 YCT
204. If the modulus of elasticity of a material is 207. A compound bar having two members X and
twice to its modulus of rigidity, then the Y of length L when subjected to tensile force P
Poisson's ratio of the material is equal would have elongation equal to (in usual
to______ notations)
(a) One (b) Zero PL PL PL
(c) 0.25 (d) 0.5 (a) (b) +
AX EX + AY EY AXEX AYEY
GPSC EE Pre, 28.01.2017
Ans. (b) : Given- PL PL PL
(c) (d) −
Modulus of elasticity (E) = 2 × modulus of rigidity (G) AX EX − AY EY AXEX AYEY
∴ E = 2G APPSC AEE Mains 2016 (Civil Mechanical)
From, E = 2G (1 + µ) Ans. (a) : We know,
2G = 2G (1 + µ) PL
1+µ=1 ε= , ∴ [P & L Remains constant]
Poisson's ratio, µ = 0 AE
So for X & Y member
205. A steel bar of 500 mm length is under tensile PL
stress of 100 N/mm2. If the modulus of elasticity ε = EX + EY =
is 2×1011 N/m2, then the total elongation of the AXEX + AYEY
bar will be 208. A vertical prismatic bar fixed at the top end
(a) 0.25 mm (b) 2.50 mm loaded with P at the bottom free end, is having
(c) 5.00 mm (d) 0.50 mm a unit weight of W and its length is L. If σ is
APPSC AEE Screening Test 2016 the working stress, the safe cross sectional area
Ans. (a) for the when P and self weight are considered
Given, is given by
length (ℓ) = 500 mm P P
(a) (b)
σ = 100 N/mm2 σ + WL σ − WL
E = 2 × 1011 N/m2 = 2 × 1011 × 10–6 N/mm2 σ + WL σ − WL
(c) (d)
= 2 × 105 N/mm2 P P
We know that, APPSC AEE Mains 2016 (Civil Mechanical)
PL P
elongation (δℓ) = Ans. (b) : We Know, stress =
AE A
σL 100 × 500 So, Total stress = σ – WL
= =
E 2 ×105 P
σ – WL =
5 1 A
= =
20 4 P
= 0.25 mm A=
σ − WL
206. An axial load P is applied on circular section of 209. If temperature changes from t0 to t in a simple
diameter D. If the same load is applied on a bar with both ends constrained and having
hollow circular shaft with inner diameter as modulus of elasticity E and thermal coefficient
D/2, the ratio of stress in two cases would be
(a) 4/3 (b) 9/8 α, the thermal stress due to the temperature
(c) 1 (d) 8/3 rise is
Nagaland PSC CTSE 2016 Paper-I (t − t 0 ) (t − t0 )
(a) E (b) α
Ans. (a) : Axial load (P) on circular section & diameter α E
(D) if the same load is applied on a hollow circular shaft Eα
with inner diameter as (D/2), then ratio of stress will be, (c) Eα( t − t 0 ) (d)
(t − t0 )
d = D/2
APPSC AEE Mains 2016 (Civil Mechanical)
P P
σ1 = σ1 = Ans. (c) : We know,
A1 π 2 t – t0 = α ∆ T, 'α' thermal expansion
D
4 F
P P t = t0 (1 + α ∆ T) ⇒ thermal stress = = t × strain
σ2 = σ2 = A
π( 2 = E α (t – t0)
A2 D − d2 ) So,
4 210. Volumetric strain of a sphere of diameter
P increasing from d1 to d2 is equal to
π 2 d −d (d − d )
(D ) (a) 2 1 (b) 3 2 1
σ1 4 D2 4D 2 4 d1 d1
= = = =
σ2 P D
2
3D 2
3 d + d1 d +d
π (4D 2 − D2 ) D2 −   (c) 3 2 (d) 2 1
2 2d1 2d1
4 4 APPSC AEE Mains 2016 (Civil Mechanical)
Strength of Materials 189 YCT
Ans. (b) : As, We know, 214. A brass tube has enclosed a steel bar and they
Change in volume equal cross-sectional area. The Young's
Volumetric strain = modulus of elasticity is 200 GPa and 100 GPa
Original volume
for steel and brass respectively. Then the ratio
Change in diameter
'εv' for sphere = 3 × of stress developed in the steel bar to that in
Diameter the brass tube compression is
(d 2 − d1 ) (a) 0.5 (b) 1
= 3×
d1 (c) 1.5 (d) 2
211. If a circular rod having Poisson's ratio µ is APPSC AEE Mains 2016 (Civil Mechanical)
subjected to an axial pull and if the strain is e, Ans. (d) : As we know,
then the lateral strain is equal to As = Ab,
(a) –µe (b) µe Es = 200 GPa
(c) –e/µ (d) e/µ Eb = 100 GPa
APPSC AEE Mains 2016 (Civil Mechanical) As we know,
σb A b E b 100
Ans. (a) : For a circular section, = =
Lateralstrain (L) σs A s E s 200
Poisson's ratio = −
Longitudinalstrain (e) σs
So, =2
Lateral strain (L) = –µ × e σb
∵ For circular sec tion,  215. Two bars of same area and length but of
 d − d0 
different materials are subjected to same
 Lateralstrain = f  tensile force. If the bars have their axial
 d 0  elongation in the ratio of 4:6, then the ratio of
modulus of elasticity of the two materials
212. If a timber beam 8 cm wide and 16 cm deep is would be
to be converted into an equivalent steel section (a) 4 : 6 (b) 6 : 4
of the same depth for analysis purpose, then
(c) 2 : 6 (d) 6 : 2
the width of the equivalent section for a APPSC AEE Mains 2016 (Civil Mechanical)
modular ratio of 20 will be Ans. (b) : As we know,
(a) 160 cm (b) 2.5 cm
PL
(c) 0.4 cm (d) 12 cm Elongation, δL1 = 1 1
APPSC AEE Mains 2016 (Civil Mechanical) A1E1
Ans. (c) : We know, Elasticity  L1 = L 2 
∴  A1 = A 2 
Modulus of Elasticity of steel PL
Modular Ratio = So for δL 2 = 2 2
Modulus of Elasticity timber A2E 2
 P1 = P2 
Es Elongation ratio
20 = ⇒ 20 Et = Es
Et δL1 P1L1 A 2 E 2
= ×
So, We know, δL 2 A1E1 P2 L 2
As × Es = Et × At δL1 E 2
=
E 16 × 8 δl2 E1
bs = s × = 0.4cm
20 16 × Es E1 δL 2 6
= =
213. The value of Poisson's ratio for which bulk E 2 δL1 4
modulus of a material will be equal to its
E1 6
Young's modulus =
(a) 0.33 (b) 0.15 E2 4
(c) 0.45 (d) 0.25 216. The bulk modulus of elasticity of a materials is
APPSC AEE Mains 2016 (Civil Mechanical) twice its modulus of rigidity. The poisson's
Ans. (a) : We know, ratio of the material is
K = Bulk Modulus (a) 1/7 (b) 2/7
E = Young's Modulus (c) 3/7 (d) 4/7
E = 3K (1–2µ) APPSC AEE Mains 2016 (Civil Mechanical)
So, E = K, substitute Above eq. Ans. (b) : We know,
We get E = 2G(1 + µ) & Ε = 3Κ(1− 2µ)
1 3Κ − 2G
= 1 − 2µ µ= Given : ⇒ K = 2G
3 2G + 6K
1 6G − 2G 4G 2
µ = = 0.33 So, µ= = =
3 2G +12G 14G 7
Strength of Materials 190 YCT
217. The diameter of a tapering rod varies from 'D' 219. Let the strain produced in length and diameter
to 'D/2' in length of 'L' m. If it is subjected to of cylindrical rod be 0.02 and 0.005
an axial tension of 'P' the change in length is respectively. Then the volumetric strain is
(a) 4 PL / (πED2) given by
(b) 8 PL / (πED2) (a) 0.03 (b) 0.025
(c) 2 PL / (πED2) (c) 0.015 (d) 0.01
(d) None of the given answers APPSC AEE Mains 2016 (Civil Mechanical)
APPSC AEE Mains 2016 (Civil Mechanical) Ans. (a) : Volumetric strain ε v = εl + 2εd
Ans. (b) : We know, for prismatic bar = 0.02 + 2 × 0.005 = 0.03
PL 220. A material with large deformation at failure is
Elongation, δℓ =
AE termed
(a) Brittle (b) Elastic
(c) Ductile (d) Elasto-plastic
APPSC AEE Mains 2016 (Civil Mechanical)
Ans. (c) : The material in which large deformation in
possible before the absolute failure or rupture is termed
as ductile.
221. A system of two prismatic bars of equal length
(L) and equal cross-section (A) carries a
For taper bar, vertical load P as shown in the figure given
PL PL below. If their modulus of elasticity is E, the
δℓ = = vertical displacement of the hinge 1 is
π π D
d1d 2 E D× ×E
4 4 2
[Given, d2 = D, d1 = D/2]
PL
Elongation, δℓ =
π D2 

 × E
4 2 
PL 2 PL
8PL (a) (b)
δℓ = AE AE
πD 2 E 3PL 2.5PL
218. The ratio of load shared by parts 'AB' and (c) (d)
AE AE
'BC' of the bar shown below is
APPSC AEE Mains 2016 (Civil Mechanical)
Ans. (b) : As we know,
PL
Elongation, δℓ =
AE
Here, P at 1 is twice the load
(a) 1 : 1 (b) 2 : 1 PL
(c) 3 : 1 (d) 1 : 2 So, δℓ = 2 ×
APPSC AEE Mains 2016 (Civil Mechanical) AE
Ans. (d) : We know, 222. Which of the following is true (µ = Poisson's
δℓ = δℓ1 + δℓ 2 = 0 ratio):
(a) 0 < µ < 1/2 (b) 1 < µ < -1
(c) 1 < µ < 0 (d) ∞ < µ < -∞
UJVNL AE 2016
Ans : (a)

P1L1
So, δℓ1 =
AE
P2 L 2
δℓ 2 =
AE
P1 L 2 L/3 1
= = =
P2 L1  2L  2
 
 3  Total strain in x-x direction
Strength of Materials 191 YCT
 Impact loading:-
σx σy σ
ex = − µ. −µ z
E E E
σ  σ y σz 
ex = x − µ  + 
E  E E 
Where σ x = σ y = σ z = σ
σ σ σ
ex = − µ + 
E E E
σ σ
ex = − 2µ  
E E
W 2h 
σ σimpact = 1 + 1 + 
e x = ( 1 − 2µ ) A δ 
E
ex + e y + ez = ev 225. A steel bar 10mm×10mm cross section is
subjected to an axial tensile load of 20 kN. If
3σ the length of bar is 1 m and E = 200 GPa, then
ev = ( 1 − 2µ ) elongation of the bar is :
E
(a) 1 mm (b) 0.5 mm
δv 3σ
= ( 1 − 2µ ) (c) 0.75 mm (d) 1.5 mm
Kerala PSC IOF 19.04.2016
v E
This limits 2µ to maximum of 1 or the poisson's ratio lie Ans. (a) Given,
to 0.5. No material is known to have a higher Value for Area = 10 × 10 = 100 mm2
poisson's ratio although µ for materials like rubber Length = 1 m = 1000 mm
approaches this value. Load, P = 20 kN
Poisson's ratio PL
Elongation (δ) =
0 < µ < 1 /2 AE
223. A steel rod of 100 cm long and 1 sq cm cross 20 × 103 × 1000
sectional area has a young's modulus of = = 1mm
100 × 200 × 103
elasticity 2 × 106 kgf/cm2. It is subjected to an 226. The modulus of rigidity and Poisson's ratio of
axial pull of 2000 kgf. The elongation of the rod a material are 80GPa and 0.3 respectively, Its
will be. Young's Modulus will be:
(a) 0.05 cm (b) 0.1cm (a) 160 GPa (b) 208 GPa
(c) 0.15cm (d) 0.20cm (c) 120 GPa (d) 104 GPa
UJVNL AE 2016 Kerala PSC IOF 19.04.2016
Ans. (b) : Given, G = 80 GPa, µ = 0.3
Ans : (b) We know, E = 2G(1 + µ)
E = 2 × 80 (1 + 0.3)
Modulus of Elasticity, E = 208 GPa
227. Yield point in fatigue loading as compared to
static loading is .......... .
l = 100 cm =1m, A = 1 × 10-4m2 (a) same
P = 2 × 104 N. (b) higher
E = 2 × 106kgf/cm2 = 2×1011N/m2 (c) lower
Pl (d) depends on other factors
δl = GWSSB DEE 07.07.2016
AE
Ans. (c) : Yield point in fatigue loading as compared to
2 × 10 4 × 1 static loading is lower.
δl =
1× 10-4 × 2 × 1011 228. Residual stress in the materials ........... .
(a) acts when external load is applied
δl = 0.1cm.
(b) becomes zero when external load is removed
224. A test used to determine the behavior of (c) is independent of external
materials when subjected to high rates of (d) is always be harmful
loading is known as : GWSSB DEE 07.07.2016
(a) Hardness test (b) Impact test Ans. (c) : Residual stressess are locked in stresses
(c) Fatigue test (d) Torsion test within a metal object, even though the object is free of
HPPSC W.S. Poly. 2016 external forces.
These stresses are the result of one region of the metal
Ans : (b) A test used to determine the behavior of
materials when subjected to high rate of loading for being constrained by adjacenet regions from expanding,
small time is known as impact test. contracting, or releasing stresses can be tensile or
compressive.
Strength of Materials 192 YCT
229. Ability of material to resist fracture due to Ans : (b)
high impact load is called ......... .
stress 200 ×106
(a) toughness (b) stiffness E= = = 2 × 1011 Pa
(c) plasticity (d) hardness strain 0.001
GWSSB DEE 07.07.2016 µ = 0.3 ⇒
Ans. (a) : Ability of material to resist fracture due to E = 2G(1 + µ)
high impact load is called toughness.
2 × 1011 = 2G (1 + 0.3)
230. Which of the following statement is correct?
(a) The stress is the pressure per unit area G = 7.69 × 1010 Pa
(b) The strain is expressed in mm G = 76.62GPa
(c) Hook's law holds good up to the breaking
point 235. When a body is subjected to stress in all the
(d) Stress is directly proportional to strain within directions, the body is said to be under.........
elastic limit strain.
UP Jal Nigam AE 2016 (a) compressive (b) tensile
Ans. (d) : Stress is directly proportional to strain within (c) shear (d) volumetric
elastic limit. (HPPSC LECT. 2016)
231. Area under the stress-strain curve when load Ans : (d)
is gradually applied in tension represents the
(a) Strain energy
(b) Strain energy density
(c) Strain energy per unit weight
(d) Strain energy per unit area
RPSC LECTURER 16.01.2016
Ans. (b) : Area under the stress-strain curve when load When a body is subjected to stress in all the direction,
is gradually applied in tension represents the strain the body is said to be under volumetric strain.
energy density. 236. Hooke's law is applicable:
232. For ductile materials, the largest value of (a) Plastic range, strain is proportional to stress
tensile stress that can be sustained by material (b) Elastic range, strain is proportional to stress
before breaking is known as: (c) In both elastic and plastic range, strain is
(a) Modulus of elasticity proportional to stress
(b) Ultimate tensile strength (d) None of the above
(c) Yield strength (HPPSC LECT. 2016)
(d) Toughness KPCL AE 2016
UPRVUNL AE 07.10.2016 Ans : (b) Hooke's law is applicable up to elastic range,
Ans. (b) : For ductile materials, the largest value of strain is proportional to stress.
tensile stress that can be sustained by material before Hooke's law:- The slope of this line is the ratio of stress
breaking is known as ultimate tensile strength. to strain and in constant for a material. In this range, the
233. When the temperature of a solid metal material also remains elastic. When a material behaves
increases- elastically and also exhibits a linear relationship
(a) strength of the metal decreases but ductility between stress and strain, it is called linearly elastic.
increases The slope of stress- strain curve is called the modulus of
Elasticity
(b) both strength and ductility decrease
(c) both strength and ductility increase 237. The ratio of modulus of rigidity to bulk
modulus for a Poisson's ratio of 0.25 would be:
(d) strength of the metal increases but ductility (a) 2/3 (b) 2/5
decreases (c) 3/5 (d) 1.0
RPSC 2016 HPPSC W.S. Poly. 2016
Ans : (a) Strength of the metal decreases but ductility Ans : (c) E = 3K (1-2µ) .......…….. (i)
increases. When the temperature of a solid metal E = 2G (1+µ) …………..(ii)
increases, then its intermolecular bonds breaks and 3K (1-2µ) = 2G (1+µ)
G 3 (1 − 2µ )
strength of solid metal decreases. Due to decreases its
strength, the elongation of the metal increases, when we =
apply the load i.e. ductility increases. K 2 (1 + µ )
234. A rod is subjected to a uniaxial load with in G = Modulus of rigidity
linear elastic limit. When the change in the K = Bulk modulus
E = Modulus of Elasticity
stress is 200 MPa, the change in strain is 0.001. µ = Poission's ratio
If the Poisson's ratio of the rod is 0.3, the
modulus of rigidity (in GPa) is– G 3 (1 − 2 × 0.25 )
=
(a) 75.31 (b) 76.92 K 2 (1 + 0.25 )
(c) 77.23 (d) 76.11 G K=3 5
RPSC 2016
Strength of Materials 193 YCT
238. Two steel rails each of 12 m length are laid Ans. (a) : Given,
with a gap of 1.5 mm at ends at a temperature Inside diameter (ID) = d
of 24°C. The thermal stress produced at a Outside diameter (OD) = D
temperature of 40°C is (take E = 2 × 105 Hoop stress (σh)
N/mm2, coefficient of thermal expansion = 12 × (σ h ) = E × strain
10–6 /°C)
2 2
(a) 10.5 N/mm (b) 12.5 N/mm D–d
(c) 13.4 N/mm2 (d) 15.5 N/mm2 So, strain =  
ISRO Scientist/Engineer 03.07.2016  d 
Ans. (c) : Given, L = 12 m = 12 × 103 mm  D−d 
λ = gap between rails σh = E ×  
 d 
∆T = (40 – 24) = 16°C
E = 2 × 105 N/mm2 241. A tensile force of P is applied on a compound
α = 12 × 10–6/°C bar having two members X and Y. Then the
load shared by the member X is
( αTL − λ )
Then, σ thermal = E (a)
PA X E X
(b)
PA Y E Y
L AX EX + AY E Y AX EX + AY E Y
(12 × 10−6 × 16 × 12 ×103 − 1.5 ) P PE X
σth = × 2 × 105 (c) (d)
12 × 103 2 EY
σth = 13.4 N/mm2
APPSC AEE Screening Test 2016
239. An aluminum tensile test specimen has a
diameter, do = 25 mm and a gauge length of Lo Ans. (a) : Let us consider a compound bar having two
= 250 mm. If a force of 175 kN elongates the member x and y and their corresponding area Ax and
gauge length by 1.25 mm, the modulus of Ay respectively
elasticity of the material is nearly
(a) 71 GPa (b) 71 MPa
(c) 142 GPa (d) 142 MPa
ISRO Scientist/Engineer 03.07.2016
Ans. (a) : Given,
L0 = 250 mm,
P = 175 kN
= 175 × 103 N
∆ℓ = 1.25 mm
d0 = 25 mm Let the load shared by x - section be Px and similarly
for y - section be Py.
π π
A = ( d 0 ) = × ( 25 ) ∴ Px + Py = total load i.e. P ----------- (1)
2 2

4 4 If the compound bars are connected in parallel, the

π axial deflection in both the sections x and y will be the
A = × ( 625 ) same.
4
∴ Deflection in x - section = deflection in y -
PL
Elongation ∆ℓ = section
AE Px L Py L
=
175 × 103 × 250 A x .E x A y E y
1.25 =
π
× 625 × E Px .A y .E y
4 Therefore, Py = ----------- (2)
4 × 7 ×103 × 10 A x .E x
1.25 = From (1) and (2)
π× E
Px + Py = P
28 × 10 4
E= Px A y E y
1.25 × π P = Px +
E = 71301.41 N/mm2 A x .E x
E = 71.301 × 103 N/mm2 Px A x E x + Px A y E y
E = 71 GPa P=
A x .E x
240. Hoop stress for a wooden with steel flat tyre in Px [AxEx+AyEy] = PAxEx
terms of outside diameter of the wheel D, inside
P Ax Ex
diameter of the tyre d and Young's modulus E Px =
is Ax Ex + AyEy
(a) E(D-d)/d (b) E(D/d) 242. Stress concentration may be caused by
(c) E(D/(D-d)) (d) E(d/D) (a) Change in cross sectional area
APPSC AEE Screening Test 2016 (b) Change in shape
Strength of Materials 194 YCT
 (c) Change in dimension 246. It is desired to shrink a thin steel tyre on to a
(d) A hole or a notch in the body rigid wheel of 3600 mm diameter. What is the
APPSC AEE Screening Test 2016 internal diameter of the tyre, if after fitting the
Ans. (d) : Stress concentrations occur when there are hoop stress in the tyre is 95 N/mm2. Take
irregularities in the geometry or material of a structural E=2×105 MPa
component that cause on interruption to the flow of (a) 3598.29 mm (b) 3592.46 mm
stress. This arises from such details as hole, grooves, (c) 3614.27 mm (d) 3618.38 mm
notches and fillets. Stress concentration may also occur APPSC AE Subordinate Service Civil/Mech. 2016
from accidental damage such as nicks and scratches. Ans. (a) : Given, D = 3600 mm, σH = 95 N/m2
E = 2 × 105 MPa
According to Hooke's law
σ = Eε
σ 95
ε= =
E 2 × 105
ε = 4.75 × 10−4
δD
= ε = 4.75 × 10−4
D
δD = 4.75 × 10–4 × 3600
δD = 1.71 mm
D – d = 1.71 mm
d = D–1.71
d = 3600 – 1.71
243. What is the amount of stress induced in a 5 d = 3598.29 mm
mm steel bar of length 2 m when heated from 247. A bar 36 mm × 36 mm × 260 mm long is
20º C to 140º C while it is free to expand? subjected to a pull of 100 kN, in the direction
(a) 600 Pa (b) 1200 Pa of length. The decrease in each lateral
(c) 48 Pa (d) None of these dimension is 0.004 mm. The lateral strain is
KPCL AE 2016 (a) 1.11 × 10–4 (b) 1.25 × 10–4
Ans. (d) : Thermal stress = α∆TE (c) 1.32 × 10 –4
(d) 1.45 × 10–4
∵ For free to expand a thermal stress always zero. APPSC AE Subordinate Service Civil/Mech. 2016
244. Value of Poisson's ratio for ionic solids is in the Ans. (a) :
range of:
(a) 0.1 (b) 0.3
(c) 0.4 (d) 0.2
UPRVUNL AE 21.08.2016
Ans. : (d) Value of poission's ratio for ionic solids is in
the range of 0.2.
245. Find the modulus of elasticity for a rod which Decrease in lateral dimension = 0.004 m (Given)
tapers uniformly from 30 mm to 15mm Lateral strain = change in length of lateral dimension
diameter in a length of 350mm. The rod is Original length of lateral dimension
subjected to an axial load of 5.5 kN and
0.004
extension of the rod is 0.025 mm εy = = 1.11× 10−4
(a) 2.78 × 10 N/mm5 2
(b) 2.18 × 10 N/mm
5 2 36
(c) 3.18 × 105 N/mm2 (d) 3.78 × 105 N/mm2 248. The shear strength of a metal is usually
UPRVUNL AE 21.08.2016 (a) greater than its tensile strength
Ans. : (b) Given, (b) less than its tensile strength
F = 5.5 × 103 N (c) equal to its tensile strength
l = 350 mm (d) none of these
GPSC Lect. 23.10.2016
d1 = 350 mm
Ans. (b) : Shear strength is the load acting along a plane
d2 = 15 mm that is parallel to the direction of the force. The ratio of
∆l = 0.025 mm shear stress to the corresponding shear strain for shear
uniformly tapers bars– stresses below the proportional limit of the material. So,
4Fl shear strength of a metal is usually less than its tensile
∆l = strength.
πEd1d 2
249. During tensile test on a specimen of 1 cm2
4Fl cross-section, maximum load observed was 80
E=
π× d1d 2 ×∆l kN and area of cross-section at neck was 0.5
4 × 5.5 × 10 × 350
3 cm2. U.T.S. of the specimen is:
E= (a) 400 MPa (b) 800 MPa
π × 350 × 15 × 0.025
(c) 1600 MPa (d) 2200 MPa
E = 2.18 × 10 N/mm
5 2
RPSC VPITI 14.02.2016
Strength of Materials 195 YCT
Ans. (b) : Given, 253. Yield strength of a material represents the
Maximum load = 80 × 103 N stress below which the deformation is :
Original cross-section area = 100 mm2 (a) almost elastic (b) plastic
(c) zero (d) none of the above
Maximum load 80 × 10 3
U.T.S. = = N / mm 2 GPSC Lect. (Auto) 16.10.2016
Cross − section area 1× 100 Ans. (a) : Yield strength of a material represents the
UTS = 800 MPa stress below which the deformation is almost elastic.
250. Two wires of different materials but of same
diameter are connected end to end and a force
is applied which stretches them by 1 cm. The
two wires will have the
(a) Same stress and strain
(b) Same stress but different strain
(c) Same strain but different stress
(d) Different stress and strain
RPSC VPITI 14.02.2016 254. A specimen of steel, 20 mm diameter with a
Ans. (b) : gauge length of 200 mm is tested to
destruction. It has an extension of 0.25 mm
under a load of 80 kN and the load at elastic
limit is 102 kN. The modulus of elasticity is
Stress is depend on resisting area and not a material. (a) 203718 N/mm2 (b) 259740 N/mm2
2
P (c) 209740 N/mm (d) 253718 N/mm2
σ1 = σ 2 = (e) 222718 N/mm 2
A CGPSC AE 26.04.2015 Shift-I
∆L Stress
Strain = = Ans. (a) : diameter (d) = 20 mm
L E length (l) = 200 mm
So, strain depends on modulus of elasticity which is extension (δ) = 0.25 mm
property of material. load (P) = 80 kN = 80,000 N
So strain will be different in different wires. load at elastic limit = 102 kN
251. For the bar of composite section We know that
(a) The extension in different materials is Pl
different δ=
(b) The total external load is equal to the total AE
sum of the loads carried by different Pl Pl 80000 × 200
E= = =
materials Aδ π d 2 × δ π
× (20) 2 × 0.25
(c) The compression in different materials is 4 4
different E = 203718.32 N/mm2
(d) Strain in all materials is unequal 255. A circular road of 25 mm diameter and 500
Rajasthan Nagar Nigam AE 2016, Shift-II mm long is subjected to a tensile force of 60
Ans. (b) : For the bar of composite section the total kN. Determine modulus of rigidity and bulk
external load is equal to the total sum of loads carried modulus if Poisson's ratio = 0.3 and Young's
by different materials. modulus E = 2 × 105 N/mm2
(a) 0.7692 × 105 N/mm2 and 1.667 × 105 N/mm2
252. The maximum stress a material can stand (b) 0.667 × 105 N/mm2 and 1.857 × 105 N/mm2
before it breaks is called the (c) 0.1852× 105 N/mm2 and 1.6567 × 105 N/mm2
(a) Working stress (d) 0.4692× 105 N/mm2 and 1.545 × 105 N/mm2
(b) Ultimate tensile stress (e) 1.7562× 105 N/mm2 and 1.117 × 105 N/mm2
(c) Compressive stress CGPSC AE 26.04.2015 Shift-I
(d) Transverse stress Ans. (a) : Data given as
Rajasthan Nagar Nigam AE 2016, Shift-II d = 25 mm, l = 500 mm
Ans. (b) : Ultimate tensile stress i.e. defined as the F = 60 kN, E = 2 × 105 N/mm2
maximum stress that material can withstand when a We know that
force is applied. When the material are pushed beyond E = 2 G (1 + µ)
UTS they experience the cracking. 2 ×105
G= = 0.7692 ×105 N / mm 2
2 × (1 + 0.3)

G = 0.7692 × 105 N / mm 2
E = 3 K (1–2µ)
2 × 105
K=
3 × (1 − 2 × 0.3)
K = 1.667 ×105 N / mm 2

Strength of Materials 196 YCT

256. The steel bar AB varies linearly in diameter (a) δl = AE/ Pl (b) δl = Pl/AE
from 25 mm to 50 mm in a length 500 mm. It is (c) δl = PE/Al (d) δl= P/AlE
held between two unyielding supports at room TSPSC AEE 2015
temperature. What is the stress induced in the Ans : (b)
bar, if temperature rises by 25ºC? Take E = 2
× 105 N/mm2 and α = 1.667× 10-6/ºC
(a) 110 N/mm2 (b) 140 N/mm2
(c) 120 N/mm 2
(d) 150 N/mm2 Hooke's low:-
(e) 170 N/mm 2 Stress ∝ strain
CGPSC AE 26.04.2015 Shift-I σ = Eε
Ans. (c) : Thermal stresses in bars of tapering section P
σ=
A
p
e=
AE
δℓ p
=
ℓ AE
pℓ
Given δl =
length of bar (l) = 500 mm AE
dia of smaller end of bar (d1) = 25 mm 260. A steel bar 100 mm long is subjected to a
dia of bigger end of bar (d2) = 50 mm tensile stress σ. If change in length of bar is
change in temperate (∆t) = 25ºC 0.05 mm what is the value of σ?
Co-efficient of thermal expansion (α) = 12 × 10-6/ºC E = 200 GPa
Young's Modulus (E) = 2 × 105 N/mm2 (a) 200 MPa (b) 100 MPa
d2 (c) 80 MPa (d) 150 MPa
σ = Eα∆t TANGEDCO AE 2015
d1
Ans. (b) : σ = ∈E
50
= 2 × 105 × 12 ×10−6 × 25 × = 120 N/mm2 δL
25 σ= ⋅E
257. For a circular cross section beam is subjected L
to a shearing force F, the maximum shear σ = 0.05mm × 200 GPa
stress induced will be (where d = diameter) 100mm
(a) F/πd2 (b) 4F/πd2 = 5 × 10–4 × 200 × 109 Pa
(c) 2F/πd2 (d) F/4d2 = 1000 × 105 Pa
TSPSC AEE 2015 σ = 100 MPa
shear force 261.The value of modulus of elasticity for steel is
Ans. (b) : τmax = (a) 70 GPa (b) 100 GPa
shear area (c) 125 GPa (d) 200 GPa
F Assam PSC CCE Pre 2015
= Ans. (d)
π 2
4 d Material Modulus of Elasticity (E)
  ● Steel 200 to 220 GPa
4F ● Wrought iron 190 to 200 GPa
τmax = 2 ● Cast iron 100 to 160 GPa
πd
● Copper 90 to 110 GPa
258. Two identical circular rods made of cast iron ● Brass 80 to 90 GPa
and mild steel are subjected to same magnitude
of axial force. The stress developed is within ● Aluminium 60 to 80 GPa
proportional limit. Which of the following 262.Bulk modulus is measured in terms2 of
observation is correct ? (a) N/m (b) N/m
(a) Both roads elongate by same amount (c) N-m/s (d) N-s/m2
(b) MS rod elongates more Assam PSC CCE Pre 2015
(c) Cl rod elongates more Ans. (b) : We know that,
(d) Both stress and strain are equal in both roads
OPSC AEE 2015 Paper-I Direct stress ( σ )
Bulk Modulus (K) =
Ans : (c) Two identical circular rods made of cast iron Volumetric strain (∈v )
and mild steel are subjected to same magnitude of axial Bulk modulus is measured in terms of N/m2.
force. The stress developed is within proportional limit.
CI rod elongates more, because in proportional limit 263. Which of the following materials has the least
cast iron elasticity is less than mild steel elasticity. elasticity?
259. The change in length due to tensile or (a) Mild steel (b) Aluminium
compressive force acting on a body is given by (c) Rubber (d) Zinc
(with usual notations) KPSC ADF 2015
Strength of Materials 197 YCT
Ans. (c) : The word elastic means to regain its original P P+0
position after load is removed, the highest load that can Ans. (d) : σ gradual = , and Load
bear is steel and it is most elastic. While rubber has the A Z
least elasticity and rubber has least elastic. 2 P
σ Sudden =
264. Two wires A & B are of the same materials. A
Their lengths are in the ratio 1 : 2 and the 2P
diameters are in the ratio 2 : 1. If they are σimpact =
A
pulled by the same force, their increase in
length will be in the ratio of : P  h × AE 
(a) 2 : 1 (b) 1 : 4 = 1 + 1 + 
A Pl 
(c) 1 : 8 (d) 8 : 1
KPSC ADF 2015 267. Bulk modulus for a material is 200 GPa and
Ans. (c) : the Poisson's ratio is 0.3. Young's modulus for
Given data : l1 : l2 = 1 : 2 that material will be
Diameter; d1 : d2 = 2 : 1 (a) 120 GPa (b) 160 GPa
For same material, E1 = E2 = E (c) 210 GPa (d) 240 GPa
TSGENCO AE 14.11.2015
Ans. (d) : Bulk modulus (K) = 200 GPa
Poisson's ratio (µ) = 0.3
Young's modulus (E) = 3K (1–2µ)
= 3 × 200 (1–2 ×0.3)
= 600 (1–0.6)
= 600 × 0.4 = 240 GPa
l1 1 268. A hydraulic press exerts a total load of 3.5
∵ = ⇒ l2 = 2l1 MN. This load is carried by two steel rods,
l2 2 supporting the upper head of the : press. The
And
d1 2
= ⇒ d1 = 2d 2 safe stress is 85 MPa and E = 210 kN/mm2. The
d2 1 diameter of the rod will be
(a) 160 mm (b) 161 mm
Fl1 Fl2  π 2 (c) 162 mm (d) 165 mm
∴ δ1 = and δ 2 = ∵ A = 4 d 
A1E A2E   MPPSC AE 08.11.2015
π 2 Ans. (c) : Given data :
Fl1 d1 E 2
δ1 4 l d
∴ = = 1 2
δ 2 Fl π d 2 E l2 d12
2 2
4
2
δ1 ld 1
= 1 2 2 =
δ 2 2l1 × 4d 2 8
⇒ δ1 : δ2 = 1: 8
265. Specific modulus of a composite is given by :
(or)
The specific stiffness is defined as a ratio of :
(a) Square root of Young's modulus/density safe stress (σ) = 85 MPa
(b) Young's modulus/density = 85 × 106 N/m2
(c) Ultimate strength/density Total load (P) = 3.5 MN = 3.5 × 106 N
(d) Square root of strength/density Let, P1 and P2 load distributed equally in the steel rods
APPSC Poly Lect. 13.03.2020
P 3.5 × 106 N
KPSC ADF 2015 i.e., P1 = P2 = = = 1.75 × 106 N
Ans. (b) : Specific modulus is a materials property 2 2
consisting of the elastic modulus per mass density. It is Let, d = diameter of each steel rods
also known as the stiffness to weight ratio or specific load
stiffness. ∴ stress(σ) =
cross − section area of steel rod
Young's Modulus E
Specific modulus = =
Density ρ P1 1.75 ×106 N
⇒ σ= ⇒ 85 × 106 N / m 2 =
266. A rigid body is very slowly dropped on another π 2 π d2
d 4
body and a deflection δst occurs in the second 4
body. If the body be placed suddenly, the value 4 ×1.75 2
of the impact factor will be : ⇒ d2 = m = 0.0262
85π
(a) 0 (b) 1.0
⇒ d = 0.1619m ≈ 0.162m
(c) ∞ (d) 2.0
BPSC Asstt. Prof. 29.11.2015 ⇒ d = 162 mm

Strength of Materials 198 YCT

269. If all the dimension of a prismatic bar be 273. The elongation of a Conical bar under its own
increased in the ratio of k:1, then maximum weight is_____that of prismatic bar of the
stress produced in the bar due to its own same length.
weight will increase in the following ratio (a) Equal to (b) Half
(a) 1:k (b) k2:1 (c) One-third (d) Two-third
(c) k3:1 (d) k:1 TSPSC Managers, 2015
ISRO Scientist/Engineer (RAC) 29.11.2015 Ans. (c) : Elongation of conical bar under self weight is
Ans : (d) Max. stress is given as 1
σ = ρgx of the elongation of identical prismatic bar under its
3
for max x = λ
self weight
( σmax )1 = ρgℓ γL2 1 γL2 ∵ γ = Specific weight 
max. stress ( σmax ) 2 when λ2 = kλ ( δ L ) = ⇒ ×  γ = ρg 
6E 3 2E  
( σmax )2 = ρgkℓ γL2
Where, = Prismaticelongation
(σ max )2 2E
= k :1
(σ max )1 274. The Poisson's ratio for cast iron varies from:
270. In a tensile testing experiment on a specimen of (a) 0.23 to 0.27 (b) 0.25 to 0.33
1 cm2 area, the maximum load observed was 5 (c) 0.31 to 0.34 (d) 0.32 to 0.42
2
tones and neck area 0.25 cm . The ultimate TSPSC Managers, 2015
tensile strength of specimen is- Ans. (a) : The Poisson's ratio
(a) 2.5 tones/cm2 (b) 10 tones/cm2 For cast iron is 0.23 to 0.27
(c) 5 tones/cm 2
(d) 20 tones/cm 2 For steel is 0.25 to 0.33
ISRO Scientist/Engineer (RAC) 29.11.2015 For concrete is 0.2
Ans : (c) The ultimate tensile strength of specimen 275. The use of compound tubes subjected to
internal pressure are made to :
max .load
σ= (a) even out the stresses
Initial Area (b) increases the thickness
5 (c) increases the diameter of the tube
σ= (d) increase the strength
1 HPPSC AE 2014
σ =5 tones/cm2
Ans : (a) The use of compound tubes subjected to
Always take initial area to calculate the value of stress
internal pressure are made to even out the stresses.
except value of true stress.
• So to obtain a more uniform hoop stress distributing
271. For a given material assume, Young's modulus compounding tubes are used.
2
E= 300 GN/m and modulus of rigidity G= 150
GN/m2. Its bulk modulus K will be- 276. When a bar is subjected to a change of
(a) 120 GN/m 2
(b) 100 GN/m 2 temperature and its deformation is prevented,
(c) 200 GN/m2 (d) 250 GN/m2 the stress induced in the bar is called
ISRO Scientist/Engineer (RAC) 29.11.2015 (a) tensile stress (b) compressive stress
(c) shear stress (d) thermal stress
9KG
Ans : (b) E = Mizoram PSC AE/SDO 2014, Paper-II
3K + G Ans. (d) : When the temperature of the bar is raised and
or the bar tries to expand and exerts axial pressure on wall.
EG 300 ×109 × 150 × 109 At the some time wall puts equal and opposite pressure
K= =
3 [3G − E ] 3 [3 ×150 − 300] ×10 9 on the bar which will develop compressive stress in bar.
2 There is no mention of heating or cooling of bar. So
K = 100 GN/m thermal stress is correct option.
272. A steel wire hangs vertically under its own 277. In a tensile test of a specimen, the ratio of
weight. If its density is 10000 kg/m3 and maximum load to the original cross sectional
allowable stress is 3000 kg/cm2 then how much
length it can sustain. area of the test piece is called
(a) 2500 m (b) 1250 m (a) ultimate stress (b) safe stress
(c) 3000 m (d) 5000 m (c) breaking stress (d) permissible stress
ISRO Scientist/Engineer (RAC) 29.11.2015 Mizoram PSC AE/SDO 2014, Paper-II
Ans : (c) Given, density (ρ) = 10000 kg/m allowable Ans. (a) : In a tensile test of a specimen, the ratio of
3

stress = 3000 kg/cm2 maximum load to the original cross-sectional areas of

We know that stress in a member hangs vertically under the test piece is called ultimate stress.
its own weight 278. The shear stress required to cause plastic
σ = ρ× x deformation of solid metal is called
3000 × 10 4 (a) proof stress (b) flow stress
For length x = = 3000 m (c) ultimate stress (d) none of these
10000 Mizoram PSC AE/SDO 2014, Paper-II
Strength of Materials 199 YCT
Ans. (b) : The shear stress required to cause plastic
deformation of solid metal is called flow stress.
279. Consider the following statements :
Assertion (A) : An isotropic material is always
homogeneous.
Reason (R) : An isotropic material is one in
which all the properties are
same in all the directions at
every point. 157 175
of these statements, (a) mm (b) mm
168 168
(a) both (A) and (R) are true and (R) is the
175 157
correct explanation of (A) (c) mm (d) mm
(b) both (A) and (R) are true but (R) is not a 186 186
correct explanation of (A) ISRO Scientist/Engineer 24.05.2014
Ans. (b) : Length of Aluminium bar
(c) (A) is true but (R) is false
lAl = 0.75 m
(d) (A) is false but (R) is true AAl = 25 × 10-4m2
TNPSC AE 2014 Length of steel bar
Ans. (d) : lst = 0.50 m
280. What would be the value of Young’s modulus Ast = 15 × 10-4m2
Esteel = 200 GPa
(Modulus of rigidity), for Poisson’s ratio to be EAl = 70 GPa
0.4? Axial compressive force = 175 kN.
(a) 5/12 (b) 11/5
(c) 14/5 (d) 1/12
UPRVUNL AE 2014
Ans : (c) E = 2 G (1+ µ)
E = 2 G (1 + 0.4)
E 28 14
= 2 × 1.4 = 2.8 = =
G 10 5
281. A uniform rod of cross sectional area 2 mm2 is
heated from 00C to 400C. What would be the Pst Lst
δ st =
value of energy stored per unit volume if Ast Est
Young’s modulus is 1011 N/m2 and linear 175 ×103 × 0.5
=
expansion coefficient is 12 × 10–6 per0C? 15 ×10−4 × 200 × 109
(a) 10250 J/m3 (b) 12500 J/m3 = 2.9167 × 10-4 m (compression)
3
(c) 11520 J/m (d) 55120 J/m3 P L
UPRVUNL AE 2014 δ Al = Al Al
AAl E Al
Ans : (c) Given 175 × 103 × 0.75
A = 2mm2 ∆t = 40ºC, E= 1011N/m2, α = 12×10-6/ºC =
25 × 10−4 × 70 ×109
Thermal stress ( (σ th ) = α ∆tE = 7.5 × 10-4 m (compression)
σ th = 12 × 10-6 × 40 × 1011 δ = δ st + δ Al
= (2.9167 + 7.5) × 10-4 m
σ th = 48 × 106 = 10.4167 × 10-4 m
( σ th )
2
175
Energy store per unit volume (U) = = 1.04167 mm = mm
2E 168
283. Member CD of the assembly shown is an
(48 × 106 ) 2
= = 11520 J / m 3
aluminium bar 0.80 meter long. Cross-
2 × 1011 sectional dimension of the bar is such that, its
282. A 0.75 meter aluminium bar 25 × 10-4 m2 in width is three times its thickness. For
cross-sectional area is attached to a 0.50 meter functional design of the bar CD, the allowable
steel bar 15 × 10-4 m2 in cross-sectional area, as axial stress is 70 MPa, and the total elongation
shown in the figure. Take E (Young's modulus) is not to exceed 0.72 mm. E value for
value of 200 GPa for steel & 70 GPa for aluminium is 70 GPa. Assume the connection
aluminium. Total shortening due to an axial at C does not reduce the net area of the bar.
compressive force of 175 kN is The thickness of the bar in metre is
Strength of Materials 200 YCT
 17 7
(a) GPa (b) GPa
102 120
17 7
(c) GPa (d) GPa
120 102
ISRO Scientist/Engineer 24.05.2014
Ans. (d) : Given, P = 20 kN
2 5 LAl = 2m, Lst = 2m
(a) m (b) m
21× 10 5
21× 105 EAl = 70 GPa, Est = 200 GPa
2 3 AAl = 120 × 10-6m2
(c) m (d) m Ast = 60 × 10-6m2
21× 103 21× 103
ISRO Scientist/Engineer 24.05.2014
Ans. (c) :

Its parallel connected-

So, δ Al = δ st
PAL LAL P .L
= st st
Taking moment at B. AAL E AL Ast .Est
∑MB = 0 PAL L A E
6000 × 0.6 = P × 0.1 or = st × AL × AL
Pst LAL Ast Est
P = 36000 N
PL 120 × 10−6 70
δ CD = = 1× ×
AE 60 × 10−6 200
PL 7
or A= =
E δ CD 10
For in Parallel,
36000 × 0.8 × 103 4 PAL + Pst = P
= =
70 ×109 × 0.72 7 ×103 PAL + Pst = 20
4 140
3t 2 = or PAL = kN
7 × 103 17
4 P 140 × 103
t2 = σ AL = AL =
21×10 3
AAL 17 ×120 ×10−6
2 14
t= m = × 109 N / m 2
21 × 103 12 × 17
284. A 20 kN weight is suspended by two wires as 7 7
shown in the figure. The length of each wire is σ AL = GPa = GPa
6 ×17 102
2 meter. The steel wire (E value 200GPa) has 285.A bar of bronze and a bar of steel are placed
an cross-sectional area of 60 × 10-6 m2 and the between two immovable supports, A and B as
aluminium wire (E value 70GPa) has an cross- shown in the figure. The physical constants
-6 2
sectional area of 120 × 10 m . The stress in are: (where L= length of bars, A = cross-
aluminium wire is
sectional area, α = linear co-efficient of
thermal expansion, E = Young's modulus)

Steel Bronze
L = 0.5m L=0.75m
A = 6 × 10-4 m2 A = 12 × 10-4 m2
α = 12 × 10 /ºC
-6
α = 20 × 10-6/ºC
E = 200 GPa E = 100 GPa
The stresses in the steel & bronze, when
temperature drops by 50ºC is
Strength of Materials 201 YCT
 (a) Tension (b) Compression
(c) Shear (d) None of these By Geometry,
ISRO Scientist/Engineer 24.05.2014 In ∆ ABE and ∆ ACD
Ans. (a) :
δ1 δ 2
Steel Bronze =
L = 0.5m L = 0.75m 2 3
A = 6 × 10-4m2 A = 12× 10-4m2 OR 3 δ1 = 2δ2
α = 12 × 10 /ºC
-6
α = 20 × 10-6/ºC
 P1L1
E = 200 GPa E = 100 GPa ∵ δ1 = A E
δ1 2 
=
1 1

δ2 3  P2 L 2
 δ1 = A E
 2 2

P1L1
A1E1 2
In series, =
P2 L 2 3
∆ = ∆b + ∆ s
A2E 2
= α b Lb ∆t + α s Ls ∆t
= 20 × 10 −6 × 0.75 × 50 + 12 × 10 −6 × 0.5 × 50 P1 2  A1  E1   L 2 
= ×    
= 1.05m (shortening) < (0.75 + 0.5 = 1.25) P2 3  A 2  E 2   L1 
Note:-Due to rigid support, the shortening is prevented
and tensile stresses are developed in each bars. P1 2  3 × 10−4   200   2 
= × × × 
As temperatures the bar tends to gets shortens, to P2 3  2 ×10−4   200   1.5 
present shortening tensile stress will develop.
P1 2 3 2 4
286. A rigid bar with wires at B & C is shown in the = × × = -----------(1)
figure below. The cross sectional area of the P2 3 2 1.5 3
wire at B is 3 × 10-4 m2 at C is 2 × 10-4 m2. The Taking a moment about A. ∑ M A = 0
wires are elasto-plastic with strength 250 MPa
P × 2 = P1 × 2 + P2 × 3
and E value 200 GPa. The ultimate load P that
Or 2P1 + 3 P2 = 2P ------------(2)
can be applied to the rigid bar, as shown is
P1 (Based on strength criterion)
= σ ultimate × A1
= 250 × 106 × 3 ×10−4
P1 = 75 kN
P2 (Based on strength criterion)
= σ ultimate × A 2
= 250 × 106 × 2 ×10−4
= 500 ×102
P2 = 50000 = 50 kN
According to Equation (1) -
P1 4
=
P2 3
(a) 50 kN (b) 75 kN 4 4
(c) 100 kN (d) 150 kN P1 = × P2 = × 50
3 3
ISRO Scientist/Engineer 24.05.2014
= 66.67 kN
Ans. (d) : Given, Cross-section Area of wire 'B' = From Equation (2)-
3 × 10−4 m 2 2P = 2P1 + 3 P2
Cross-section Area of wire 'C' = 2 × 10 −4 m 2 2P = 2 × 66.67 + 3 × 50
σ ultimate = 250 MPa = 283.44
E = 200 GPa P = 141.67 kN
The closest answer is option (d) 150 kN.

Strength of Materials 202 YCT

287. Stress and strain are linearly related by Ans. (d) : Material Poisson’s ratio
Hooke's law in the ____ region. Cork 0
(a) plastic Rubber 0.5
(b) elastic Mild steel 0.3
(c) both elastic and plastic Cast Iron 0.25
(d) Neither elastic nor plastic 293. If D is the diameter of a sphere, then
MPSC HOD (Govt. Poly. Colleges) 04.10.2014 volumetric strain is equal to
Ans. (b) : Stress and strain are linearly related by (a) 2 times the strain of D
Hooke's law in the elastic region, but Hooke's law holds (b) 1.5 times the strain of D
good upto proportionality limit only. (c) 3 times the strain of D
288. The thermal stress in terms of coefficient of (d) strain of D
linear expansion (α), rise in temperature (∆T) APPSC Poly. Lect. 2013
and the modulus of elasticity (E) is given by Ans. (c) : If D is the diameter of a sphere, then the
(a) ∆T (b) 1/(Eα∆T) volumetric strain is equal to three times of strain of D.
(c) E∆T/α (d) Eα∆T
δV 3δD 4
Kerala LBS Centre For Sci. & Tech. Asstt. Prof. 2014 Volumetric strain = = V = πr 3
APPSC Poly. Lect. 2013 V D 3
Ans. (d) : Thermal stress (σT) = Eα∆T π D
Thermal strain (εT) = α∆T V = D3 ......(1) r =
6 2
289. Strain energy is the δV 3δD
(a) energy stored in a body when strained within ∈v = =
elastic limits V D
(b) energy stored in a body when strained upto 294. For a material with the Poisson's ratio v, the
the breaking of a specimen modulus of elasticity (E) and the bulk modulus
(c) maximum strain energy which can be stored of elasticity (K) are same. Which of the
in a body following is correct?
(d) proof resilience per unit volume of a material (a) The material has v = 0
Haryana PSC Civil Services Pre, 2014 (b) The material has v = 1/2
Ans. (a) : Strain energy– When a body is subjected to (c) The material has v = 1/3
external load it undergoes some deformation and stores (d) The material has v = 3/4
some amount of energy. This stored energy will release RPSC AEN Pre-2013
when load is removed is known as strain energy. Ans. (c) : Given, Poisson's ratio = v
Energy stored by body when, it is strained within elastic Modulus of elasticity = E
limit is known as strain energy. Bulk modulus = K
290. For a given material, Young's modulus is 1 × E = K (given)
106 N/m2 and modulus of rigidity is 0.4 × 106 According to,
N/m2 The bulk modulus of elasticity will be. E =3K(1 – 2 v)
(a) 5 × 106 N/m2 (b) 0.67 × 106 N/m2 E =3E(1 – 2 v)
(c) 67 × 10 N/m
6 2
(d) 6.7 × 106 N/m2 1
MPPSC State Forest Service Exam, 2014 1 − 2ν =
3
Ans. (b) : As we know,
1
G = 0.4 × 106 N/m2 1 − = 2ν
9 × KG 3
E= 2
3K + G6 2ν =
E = 1 × 10 N/m2 3
9 × K × 0.4 ×106 1
1×106 = v=
3 × K + 0.4 × 106 3
K = 0.67 × 106 N / m 2 295. A solid cube faces similar equal normal force
291. The no. of elastic constants in Triclinic on all faces. Ratio of volumetric strain to linear
material is: strain on any of three axes will be:
(a) 36 (b) 9 (a) 1 (b) 2
(c) 2 (d) 21 (c) 3 (d) 3
HPPSC Asstt. Prof. 2014 SJVN ET 2013
Ans. (d) : No. of elastic constant Material Ans. (c) :
Isotropic 2 δV 3σ
Anisotropic (Triclinic) 21 ∈v = = (1 − 2µ )
V E
Orthotropic 9
 σ
292. The Poisson's ratio (µ) of bottle cork is: ∈v = 3∈ (1 – 2µ) ∵ ∈= E 
(a) 0.5 (b) 0.3
(c) 0.25 (d) 0 • It is 3 times because cube is subjected to 3 mutually
HPPSC Asstt. Prof. 2014 perpendicular stress.
Strength of Materials 203 YCT
296. When a bar is subjected to a push of P, its Ans. (a) : Given, A = 1 cm2, L = 1m
(a) length, width and thickness increase E = 2×106 kg/cm2, P = 2000 kgf
(b) length, width and thickness decrease PL 2000 × 1
(c) length increases, width and thickness decrease Elongation δ = = = 10−3
(d) length decreases, width and thickness increase AE 1× 2 × 106
SJVN ET 2013 δ = 1mm = 0.1cm
Ans. (d) : When a bar is subjected to a push of P, its 301. If a material has a modulus of elasticity of 2 ×
length decreases width and thickness increases. 106 kg/cm2 and a modulus of rigidity of 0.8 ×
106 kg/cm2, then the approximate value of the
Poisson's ratio of the material will be
(a) 0.26 (b) 0.31
(c) 0.45 (d) 0.2
APPSC AEE 2012
297. In a tensile test, when a material is stressed Ans. (a) : E = 2G(1 + µ)
beyond elastic limit, the tensile strain ___ as 2×106 = 2×0.8×106 (1 + µ)
compared to the stress.
(a) decreases slowly ( 2 ×106 ) = 1.25
(1 + µ ) =
(b) increases slowly
(c) decreases more quickly ( 2 × 0.8 ×106 )
(d) increases more quickly µ = 0.25
JPSC AE 2013, Paper-V
Ans : (d) : In a tensile test, when a material is stressed 302. The ratio of the deformation of a bar due to its
beyond elastic limit the tensile strain increases more own weight, to the deformation due to axial
quickly as compared to the stress. load equal to its weight, is :
(a) 1 (b) 1/2
298. A uniform taper rod of diameter 30 mm to 15 (c) 2 (d) 4
mm, length of 314 mm is subjected to 4500 N.
The Young's modulus of the material is 2 × 10 5 APPSC AE 04.12.2012
2
N/mm . Extension of the bar is Ans. (b) : We know that -
(a) 0.05 mm (b) 0.5 mm WL
(c) 0.25 mm (d) 0.005 mm Deformation of a bar due to its own weight (∆1) =
2AE
TNPSC AE 2013
Deformation due to axial load equal to its weight (∆2)
Ans. (*) : Data given-
D = 30 mm d = 15 mm WL
=
L = 314 mm P = 4500 N 2AE
E = 2 × 105 N/mm2 WL
We know that extension of the taper bar. ∆ 2AE = 1
∴ 1 =
4 P.L ∆ 2
WL 2
δ= AE
π Dd.E 303. To measure _______ strain, strain rosettes are
4 × 4500 × 314 used :
δ=
3.14 × 30 × 15 × 2 ×10 5 (a) Linear (b) Shear
δ = 0.02mm (c) Volumetric (d) Any of the above
299. A load of 4000 N has to be placed at the end of APPSC AE 04.12.2012
a vertical rod with stress induced is 80 N/mm2. Ans. (a) : • Strain rosettes method is used to measure
The diameter of the rod is : Linear strain
(a) 79.79 mm (b) 7.979 mm 304. A localised compressive stress at the area of
(c) 18.181 mm (d) 81.81 mm contact between two members is known as :
TNPSC AE (Industries) 09.06.2013 (a) Shear (b) Crushing
Ans. (b) : Load, P = 4000 N (c) Bending (d) Tensile
Stress, σ = 80 N/mm 2
APPSC AE 04.12.2012
P 4P
σ= = 2 Ans. (b) : A localised compressive stress at the area of
A πd contact between two member is known as crushing.
4 × 4000
80 = ⇒ d = 7.979 mm 305. A square section with side 'X' of a beam is
πd 2 subjected to a shear force 'S'. The magnitude
300. A steel rod of 1 sq. cm cross-sectional area is 1 of shear stress at the top edge of the square is :
m long and has a Young's modulus of elasticity 1.5S S
2 × 106 kg/cm2. It is subjected to an axial pull (a) 2
(b)
X X2
of 2000 kgf. The elongation of the rod will be
0.5S
(a) 0.1 cm (b) 0.05 cm (c) (d) Zero
(c) 0.20 cm (d) 0.15 cm X2
APPSC AEE 2012 APPSC AE 04.12.2012
Strength of Materials 204 YCT
Ans. (d) : At the top edge of the square, the magnitude 1
of the shear stress should be zero. Because shear stress (a) (b) 1
is a surface phenomena. 2
306. Stress-strain analysis is conducted to know 1
(c) 2 (d)
which of the following properties of material? 4
(a) Physical properties APPSC AEE 2012
(b) Optical properties
(c) Mechanical properties Ans : (d)
(d) Magnetic properties Bar 'A ' Bar 'B'
BPSC AE 2012 Paper - VI ℓ1 = ℓ ℓ2 = ℓ
Ans : (c) : Mechanical properties.
A1 = A A2 = A/2
307. The ratio between tensile stress and tensile
strain or compressive stress and compressive E1 = 2E E2 = E
strain is termed as P1 = P P2 = P
(a) modulus of rigidity Pℓ Pℓ
(b) modulus of elasticity δℓ1 = 1 1 δℓ 2 = 2 2
A1E1 A 2 E2
(c) bulk modulus
(d) modulus of subgrade reaction Pℓ 2Pℓ
APPSC AEE 2012 δℓ1 = δℓ 2 =
2AE AE
Ans : (b) According to Hooke's law δℓ1 1
stress ∝ strain (upto proportionality limit) =
σ∝ε δℓ 2 4
σ 312. The elongation of beam of length 'l' and cross-
=E
ε sectional area 'A' subjected to a load 'P' is δl.
E = Modulus of Elasticity. If the modulus of elasticity is halved, the new
where stress and strain both are tensile or compressive elongation will be
nature.
δℓ
308. A prismatic bar has (a) (b) 2(δℓ)
(a) maximum ultimate strength 2
(b) maximum yield strength (c) δℓ (d) 2δℓ
(c) varying cross-section
(d) uniform cross-section APPSC AEE 2012
BPSC AE 2012 Paper - VI Ans : (b) Case 1st :-
Ans : (d) : A prismatic bar or beam is a straight Elongation of beam
structural piece that has the same cross section through Pℓ
its length. δℓ1 =
AE
309. The failure criterion for ductile materials is
based on Case 2nd :-
(a) yield strength (b) ultimate strength E
Modulus of elasticity is halved E′ =
(c) shear strength (d) limit of proportionality 2
BPSC AE 2012 Paper - VI 2Pℓ
Ans : (a) : The failure criterion for ductile materials is δℓ 2 =
based on yield strength. AE
310. The elongation of a conical bar due to its self δℓ 2 = 2 × δℓ1
weight is
313. A 16mm diameter central hole is bored out of a
γℓ 2 γℓ 2
(a) (b) steel rod of 40mm diameter and length 1.6m.
6E 2E The tensile strength because of this operation
γℓ 2
γℓ 2 (a) increases (b) remains constant
(c) (d)
2E E (c) decreases (d) None of these
Where γ = unit weight of the material. APPSC AEE 2012
APPSC AEE 2012 Ans : (c) A 16mm diameter central hole is bored out of
Ans : (a) Elongation of conical bar due to self weight. a steel rod of 40mm diameter and length 1.6m. The
γℓ 2 1 tensile strength because of this operation decreases.
∆= = × Deflection of prismatic bar
6E 3 314. The relationship between Young's modulus
γ = Specific weight 1
and shear modulus when = 0, is
l = length of bar m
E = Modulus of Elasticity. (a) E = 2G
311. Two bars A and B are of equal length but B (b) E = 3G
has an area half that of A and bar A has
young's modulus double that of B. When a load (c) E = 2G+1
'P' is applied to the two bars, the ratio of (d) G = 2E
deformation between A and B is APPSC AEE 2012
Strength of Materials 205 YCT
Ans : (a) We know 320. Select the proper sequence for the following :
1. Proportional limit
 1
E = 2G 1 +  .....(i) 2. Elastic limit
 m  3. Yield point
 2 4. Fracture/failure point
E = 3K 1 −  .....(ii) (a) 1-2-3-4 (b) 2-1-3-4
 m (c) 1-2-4-3 (d) 2-1-4-3
1 UKPSC AE 2012 Paper-I
If = 0 then
m Ans. (a) : 1-2-3-4
E = 2G 321. A test specimen is stressed slightly beyond
E = 3K yield point then unloaded. Its yield strength
315. If a rigidly connected bar of steel and copper is will
heated, the copper bar will be subjected to (a) decreases
(a) compression (b) shear (b) increase
(c) tension (d) None of these (c) remain same
APPSC AEE 2012 (d) become equal to ultimate tensile strength
Ans : (a) If a rigidly connected bar of steel and copper TRB Poly. Lect., 2012
is heated, the copper bar will be subjected to Ans. (b) : A test specimen is stressed slightly beyond
compression. the yield point and then unloaded its yield strength will
• Because αcopper > αsteel. increase because the specimen is strain hardened due to
316. Which of the following has no unit ? plastic deformation.
(a) Kinematic viscosity 322. A mild steel bar 1 m long, 100 mm in diameter
(b) Strain is subjected to an axial tensile load, so that
(c) Surface Tension change in its length is 0.1 mm. If Poisson's
(d) Bulk Modulus ratio of mild steel is 0.3, what is the change in
UKPSC AE 2012 Paper-I its diameter :
δL (a) 1 micron (b) 0.3 micron
Ans. (b) : Strain = (c) 0.03 micron (d) None of these
L PSPCL AE, 2012
317. In a static tension tests of a low carbon steel
sample, the gauge length affects δd
(a) yield stress δd /100
Ans. (d) : Poisson ratio 'µ' = d or 0.3 =
(b) ultimate tensile stress δl 0.1/1000
(c) percentage elongation l
(d) percentage reduction in cross-sectional area 0.3 × 100 × 0.1
UKPSC AE 2012 Paper-I or δd =
Ans. (c) : In a static tension test of a law carbon steel 1000
= 3 × 10–3 = 3 micron
sample, the gauge length affects the percentage
elongation. 323. A steel cube of volume 8 × 106 mm3 is
subjected to all round stress of 135 N/mm2.
318. One end of a metallic rod is fixed rigidly and
The bulk modulus of the material is 1.35 × 105
its temperature is raised. It will experience N/mm2. The volumetric change is :
(a) zero stress (a) 8000 mm3 (b) 800 mm3
(b) tensile stress (c) 80 mm 3
(d) 8 mm3
(c) compressive stress APGENCO AE 2012
(d) None of the above Ans. (a) : Given, Volume (V) = 8 × 106mm3
UKPSC AE 2012 Paper-I σ = 135 N/mm2
Ans. (a) : zero stress Bulk modulus of material (K) = 1.35 × 105 N/mm2
σ
Bulk modulus (K) =
dV
V
319. A metallic cube is subjected to equal pressure σV
dV =
(P) on its all the six faces. If ∈v is volumetric K
P 135 × 8 × 106
strain produced, the ratio is called dV =
∈v 1.35 × 105
3
(a) Elastic modulus d V = 8000mm
(b) Shear modulus 324. A steel bar 2 m length is fixed at both ends at
(c) Bulk modulus 20°C. The coefficient of thermal expansion is
(d) Strain-Energy per unit volume 12 × 10–6/°C. The modulus of elasticity is 2 ×
UKPSC AE 2012 Paper-I 105N/mm2 If the temperature is reduced to
Ans. (c) : Bulk modulus 18°C, the bar will experience a stress of :
Strength of Materials 206 YCT
 (a) 2.4 MPa (compressive) Force in LM section
(b) 2.4 MPa (tensile)
(c) 4.8 MPa (tensile)
(d) 4.8 MPa (compressive)
APGENCO AE 2012
Ans. (c) : Given, length (l) = 2m Force in MN section
Temperature reduce (∆T) ↓ from 20°C to 18°C
So, (∆T)↓ – It will tensile stress
α = 12 × 10–6 /°C, E = 2 × 105 N/mm2
Thermal stress (σT) = α∆TE
= 12 × 10–6 × (20–18) ×2×105 PL
Total change in length = ∑
= 12 × 10–6 × 2 × 2 ×105 AE
48 × 1 "Its applied when cross-section area of each section and
= = 4.8 MPa (Tensile) 'E' is same."
10
100 × 500 − 150 × 800 + 50 × 400
325. A bar of 40mm diameter is subjected to an =
axial load of 4kN. The extension of the bar 25 × 200 × 103
over a gauge length of 200 mm is 0.3 mm. The −50000
=
decrease in diameter is 0.018mm. The 25 × 200 × 103
Poisson's ratio is : = –10 µm
(a) 0.33 (b) 0.35 327. A circular rod of 100 mm diameter and 500
(c) 0.30 (d) 0.25 mm length is subjected to a tensile force of
APGENCO AE 2012 1000 kN. Determine the modulus of rigidity
δd 0.018 (G) if E = 2 × 105 N/mm2 and Poisson's = 0.3
Ans. (c) : Given, = (a) 0.335 ×105 N/mm2
d 40 (b) 0.521 ×105 N/mm2
= 4.5 × 10−4 (c) 0.7692 ×105 N/mm2
δl 0.3 (d) 0.2256 ×105 N/mm2
= = 0.0015 = 15 × 10−4 ISRO Scientist/Engineer 2011
l 200
Ans. (c) : E = 2 × 105 N/mm2
δd Poisson's ratio µ = 0.3
4.5 × 10−4 E = 2G (1 + µ)
∵ Poisson's ratio (µ) = d = 0.30
δl 15 × 10−4 2 × 105 = 2G (1 + 0.3)
l 1× 105
326. The figure below shows a steel rod of 25 mm 2 G =
1.3
cross sectional area. It is loaded at four points, G = 0.7692 × 105 N/mm2
K,L,M,N. Assume E steel = 200 GPa. The total 328. What is the stress which varies from a
change in length of the rod due to loading is minimum value to a maximum value of the
same nature called as ?
(a) Repeated stress (b) Yield stress
(c) Fluctuating stress (d) Alternating stress
VIZAG STEEL MT 2011
Ans. (c) : The stress which varies from a minimum
value to a maximum value of the same nature called as
fluctuating stress.

(a) 1µm (b) -10µm

(c) 10µm (d) -20µm
ISRO Scientist/Engineer 2012
GATE-2004 Fluctuating fatigue load
Ans. (b) : Given- (i.e. only magnitude of load change)
Cross-section Area of K,L,M,N section each is 25mm2 329. The highest stress that a material can
Assume Esteel = 200 GPa. withstand for a specific length of time without
excessive deformation is called :
(a) Fatigue strength (b) Endurance strength
(c) Creep rupture (d) Creep strength
J&K PSC Civil Services Pre, 2010
Force in KL section Ans. (d) : Creep Strength – Creep strength is the
maximum stress that the material can withstand for a
specific length of time without excessive deformation.
Creep Rupture Strength– Creep rupture strength is the
maximum stress that the material can withstand for a
specific length of time without rupture.
Strength of Materials 207 YCT
330. A cast iron specimen when tested in Ans. (a) : E = 200 GPa v ≅ 0.3, Volume V = 1 m3
compression just fails at a compressive stress E
of 320 N/mm2. The shear strength of cast iron K=
is : 3 (1 − 2v )
(a) 320 N/mm2 (b) 160 N/mm2 200
(c) 140 N/mm 2
(d) 80 N/mm2 = = 166.66 GPa
UPSC JWM Advt. No.-52/2010 3 (1 − 0.6 )
Ans. (b) : Solution : Shear strength (τ) PV
K= −
CompressiveStrength of cast Iron ∆V
=
2 1
166.66 × 10 =
9
(σ c )ult ∆V
=
2 ∆V = –6×10-6 m3
320 334. A steel rod of length 300 mm is held between
= = 160 N / mm 2 two fixed supports so that the rod cannot
2
elongate or contract in the axial direction. If
331. A bar of length L, area A and Young's the temperature of the rod is raised by 20ºC,
modulus E is subjected to a pulling force P. the axial stress induced in the rod due this rise
The strain energy stored in the bar is
in temperature is (for steel, E = 200 GPa and α
PL2 PL2 = 11.5 × 10–6/ºC
(a) (b)
AE 2 EI (a) 46 MPa tension
P2 L P2 L (b) 46 MPa compression
(c) (d) (c) 23 MPa tension
2 AE AE (d) 23 MPa compression
ISRO Scientist/Engineer 2009 DRDO Scientists 2009
Ans. (c) : Given,
Length = L Ans. (b) : ∆L = ( α∆T ) L
Area = A ∆L
Young modulus = E = α∆T
L
Applied force = P
σ
1 E=
Strain energy = × P × δ α∆T
2 σ th = Eα∆T
1 PL  PL 
= 200×103×11.5×10–6×20
= × P× ∵ δ = AE 
2 AE   = 46 MPa compression
P2 L 335. Consider the following statements :
= State of stress at a point when completely
2 AE
specified enables one to determine the :
332. A bar of 20 mm dia is tested in tension. It is (I) principal stresses at the point
observed that when a load of 38 kN is applied (II) maximum shearing stress at the point
the extension measured over a gauge length of (III) stress components on any plane containing
200 mm is 0.12 mm and contraction in the point
diameter is 0.0036 mm. Find the Poisson's
Of these statements :
ratio
(a) 0.2 (b) 0.3 (a) I, II and III are correct
(c) 0.25 (d) 0.33 (b) I and III are correct
ISRO Scientist/Engineer 2009 (c) II and III are correct
Ans. (b) : Given, ∆l = 0.12 mm (d) I and II are correct
l = 200 mm APPSC IOF, 2009, ESE-1996
∆d = 0.0036, d = 20 mm Ans. (a) : The state of stress at a point when completely
Poisson's ratio specified enables one to determine, principal stresses at
the point, maximum shearing stress at the point and
 ∆d   ∆l  0.0036 / 20
= /  = stress components on any arbitrary plane containing that
 d   l  0.12 / 200 point.
= 0.3 336. Proof resilience is the greatest energy at
333. A steel cube of 1 m × 1 m × 1 m is subjected to (a) limit of proportionally (b)elastic limit
hydorstatic pressure of 1 MPa. If for steel (c) plastic limit (d) none of these
Young's modulus is 200 GPa and Poisson's TNPSC AE, 2008
ratio is 0.3, the approximate change in volume
of the cube due to the hydrostatic pressure will Ans. (b) : Proof resilience is the maximum energy
be stored at elastic limit.
(a) –6 × 10–6 m3 (b) –2 × 10–6 m3 337. The ultimate tensile strength and yield
–6 3
(c) –1 × 10 m (d) –0.5 × 10–6 m3 strength of most of the metals, when
DRDO Scientists 2009 temperature falls from 0 to -150ºC will
Strength of Materials 208 YCT
 (a) Increase (c) true stress in the rod depends on the type of
(b) Decrease material of the rod, but engineering stress in
(c) Remain same the rod does not depend on the type of
(d) First increase and then decrease material of the rod
ISRO Scientist/Engineer 2008 (d) both engineering stress and true stress in the
Ans. (a) : Ultimate tensile stress—It is the maximum rod does not depend on the type of material
nominal stress attained by a test specimen during a of the rod
simple tensile test the ultimate strength of the materials DRDO Scientists 2008
increase with decrease in temperature. P
Stainless steel has high strength and mostly preferred at Ans. (c) : σ E = A ← Independent of material property
Cryogenic applications. Engineering stress σE, True stress σT,
Yield stress—It is the stress at which the strain of a Engineering strain ∈E
material shows a rapid increase in stress, when σ = σE (1+ ∈E ) 
subjected to a simple tensile test. 
σ  ← Dependent on material property
338. At a point in a body the normal stresses are σx ∈E = E 
= σ and σy = σ. E is the Young's modulus and v E
is the Poisson's ratio of the material of the 342. The extension of a mild steel bar 4m long, 2000
2
body. Assuming the material to be linearly mm cross section under the action of an axial
load of 20kN, if E = 2×105 N/mm2, is
elastic and isotropic, for plane stress condition (a) 2mm (b) 0.2mm
the ratio of σx to ϵx is (c) 0.5mm (d) 0.05mm
E ISRO Scientist/Engineer 2007
(a) (b) E
1− v Ans. (b) : Given data
E E l = 4m = 4000 mm
(c) (d) A = 2000 mm2
v 1+ v P = 20 kN
DRDO Scientists 2008 E = 2 ×105 N/mm2
σ x − ν(σ y ) PL
Ans. (a) : ∈x = ∆L =
E AE
(1 − ν) σ x 20 × 103 × 4000
∈x = = = 0.2 mm
E 2000 × 2 × 105
σx E 343. The elongation produced in a bar due to its
= self-weight is given by
∈x 1 − v
9.81ρl 2 9.81ρl 2
339. The linear relation between the stress and (a) (b)
strain of a material is valid until E 2E
(a) fracture stress (b) elastic limit 9.81ρl 9.81ρ2l
(c) ultimate stress (d) proportional limit (c) (d)
E 2E
DRDO Scientists 2008 UKPSC AE 2007 Paper -I
Ans. (d) : The linear relation between the stress and
strain of a material is valid until proportional limit. 9.81ρl 2
Ans. (b) :
340. A material has a Poisson's ratio of 0.5. If a 2E
body is made of this material and subjected to 344. Which of the following is not the characteristic
external forces (within the elastic limit) then of stress-strain curve for mild steel?
the final volume of the body is (a) The stress is proportional to the strain up to
(a) thrice that of the initial volume of the body the proportional limit
(b) twice that of the initial volume of the body (b) Percentage reduction in area may be as high
(c) zero as 60-70%.
(d) equal to the initial volume of the body (c) A neck is formed due to high stress level
(d) During plastic stage no strain hardening takes
DRDO Scientists 2008 place
Ans. (d) : Bulk hardness of rigidity UKPSC AE 2007 Paper -I
E = 3 K(1 – 2µ) Ans. (d) : During plastic stage no strain hardening takes
K→∞ for µ = 0.5 where µ is Poisson's ratio place.
It means material is rigid so no volume change occurs. 345. When a body is in a state of equilibrium under
341. When a uniaxial tensile load is applied to a rod the action of any force system, the normal
fixed at one end. stress at a point within the body depends upon
(a) both engineering stress and true stress in the (a) elementary area ∆A surrounding the point
rod depend on the type of material of the rod (b) elemental force ∆F acting normal to ∆A
(b) engineering stress in the rod depends on the (c) the plane orientation containing the point
type of material of the rod, but true stress in (d) all the above three
the rod does not depend on the type of UKPSC AE 2007 Paper -I
material of the rod Ans. (d) : All the above three

Strength of Materials 209 YCT

346. A steel bolt passes centrally through a copper Ans. (c) :
tube. At the ends nuts and washers are
provided. The area of cross-section of both
bolt and tube is same. If the young's modulus
of steel is twice that of copper and if the
assembly is tightened by rotating a nut
through 600 on the thread of pitch 3.6 mm,
then the contraction of the tube is
(a) 0.6 mm (b) 0.4 mm For max strength h = 2 b
(c) 0.2 mm (d) 0.1 mm b2 + h2 = D2
WBPSC AE, 2007 b2 + 2b2 = D2
Ans. (b) : Given, D
As = Ac = A or b=
Es = 2Ec Pitch, P = 3.6 mm 3
The nut is rotated by 600 2
h= × D = 2b
θ 3
The number of turns, n =
3600 30 10 × 3 × 3
60 1 b= = = 10 3
= = 3 3
360 6 h = 2 × b = 10 6
Total tensile force in bolt
= Total compressive force in copper tube. 349. Stiffness of a material is expressed in terms of :
σsAs = σcAc (a) mass density
(b) hardness number
σs = σ c (c) modulus of elasticity
Total axial movement of nut is equals to the elongation (d) impact strength
of steel bolt + contraction of copper tube. J & K PSC Screening, 2006
σs σc Ans. (c) : modulus of elasticity–
np = L+ L AE
Es Ec stiffness (k) =
ℓ
1 σ σ Where, AE = Axial rigidity
× 3.6 = c L + c L
6 αE c Ec 350. If the area, length and the stress to which a bar
3σc L is subjected be all doubled, then the elastic
0.6 = strain energy of the bar will become
2E c (a) doubled (b) halved
σc .L 2 × 0.6 (c) eight times (d) four times
= = 0.4mm WBPSC AE 2003
Ec 3
Ans. (d) : The strain energy formula (U)
347. The area under the stress-strain curve
1  PL 
represents: U = Pδ ∵ δ = AE 
(a) breaking strength of material 2
(b) toughness of material 1 PL
(c) hardness of material = P×
2 AE
(d) energy required to cause failure 2
1 P L
J & K PSC Screening, 2006 = ×
Ans. (d) : • The area under the stress strain curve 2 AE
represents energy required to cause failure. 1 σ 2 × AL
= ×
• Area under stress strain curve upto fracture is known 2 E
as modulus of toughness. σ2
• Area under stress strain curve upto elastic limit is the U= × (Volume)
2E
modulus of resilience. Hence, If the area, length and the stress to which a bar
348. The width and depth of the strongest beam of is subjected be all doubled, then the elastic strain energy
rectangular section that can be cut out of a of the bar will become four times.
cylindrical log of wood whose diameter is 30 351. The iron bar which is fixed at both ends are
cm are heated to higher temperature. Then it is cooled
(a) 10 3 cm and 10 cm slowly. During cooling, what kind of thermal
stress is produced in the rod ?
(b) 10 cm and 10 2 cm
(a) Tensile stress (b) Shear stress
(c) 10 3 cm and 10 6 cm (c) Compressive stress (d) Yield stress
(d) 10 cm and 10 3 cm TNPSC AE (Industries) 09.06.2013
OPSC Civil Services Pre 2006 Ans. (a) : Tensile stress

Strength of Materials 210 YCT

352. A steel bar of 40 mm × 40 mm square cross-
section is subjected to an axial compressive 2. Principle Stress and Strain
load of 200 kN. If the length of the bar is 2 m
355. On principal plane the shear stress is..........
and E = 200 GPa, the contraction in mm of the
(a) zero
bar will be :
(b) unity
(a) 1.25 (b) 2.7 (c) double the value of principal stress
(c) 4.05 (d) 5.4 (d) half the value of principal stress
HPPSC Asstt. Prof. 18.11.2016 RPSC Lect. (Tech. Edu. Dept.) 13.03.2021, Paper-I
PL JPSC AE 10.04.2021, Paper-II
Ans. (a) : δ L =
AE GMB AAE 25.06.2017
Where, P = 200 km ⇒ 200 × 10 N 3 (HPPSC LECT. 2016)
L = 2 m = 2000 mm HPPSC Asstt. Prof. 29.10.2016
RPSC ADE 2016
E = 200 GPa = 200 × 10 MPa 3
GPSC Lect. 23.10.2016
A = 40 × 40 mm2 BPSC AE 2012 Paper - VI
200 × 103 × 2000 APPSC AEE 2012
δL =
40 × 40 × 200 × 103 APPSC AEE Mains 2016 (Civil/Mech.)
δ L = 1.25mm APPSC AE 04.12.2012
Ans : (a) on principal plane the shear stress is zero.
353. Property of absorbing large amount of energy Principal Stresses and Strains:-
before fracture is known as:- It has been observed that at any point in a strained
(a) Ductility material, there are three planes, mutually perpendicular
(b) Toughness to each other, which carry direct stresses only, and no
(c) Elasticity shear stress. A little consideration will show that out of
(d) Hardness these three direct stresses; one will be maximum, the
UKPSC AE-2013, Paper-I other minimum and the third an intermediate between
Ans. (b) : Property of absorbing large amount of energy the two. These particular planes, which have no shear
before fracture is known as Toughness. stress, are known as principal planes. The magnitude of
direct stress, across a principal plane, is known as
354. A 200 × 100 × 50 mm steel block is subjected to principal stress.
a hydrostatic pressure of 15 MPa. The Young's
356. Maximum shear stress in Mohr's circle is
modulus and Poisson's ratio of the material are equal to
200 GPa and 0.3 respectively. The change in (a) Radius of circle
the volume of the block in mm3 is (b) Diameter of circle
(a) 85 (b) 90 (c) Center of circle from y-axis
(c) 100 (d) 110 (d) Chord of circle
VIZAG Steel MT 18.08.2013 Gujarat PSC AE 2019
VIZAG Steel MT 10.06.2012 HPPSC AE 2018
Gate 2007 APGENCO AE, 2017
Ans. (b) : σ1 = σx = σy = σz = –15 × 106 Pa GPSC Lect. 23.10.2016
(Compression) APPSC AEE Screening Test 2016
Now, Longitudinal strain- ISRO Scientist/Engineer (RAC) 29.11.2015
σ µσ µσ Mizoram PSC AE/SDO 2014, Paper-II
∴ εx = x − y − z (µ = Poisson's ratio) UKPSC AE-2013, Paper-I
E E E
ISRO Scientist/Engineer 2010, 2007
σ
= 1 (1 − 2µ ) ESE 2014, 2008
E Ans. (a) : Maximum shear stress
15 × 106 σ −σ
= (1 − 2 × 0.3) τ max = 1 2
200 ×109 2
εx =
6
= 3 × 10 −5 where σ 1 and σ 2 are principal stresses.
200 ×103 σ1 − σ 2
τ max = = Radius of circle
∆V 2
Volumetric strain, ε v = = 3ε x
V 357. A Mohr's circle reduces to a point when the
∆V = 3 × 3 × 10 −5 × 200 × 100 × 50 body is subjected to
(a) Pure shear
∆V = 90 mm 3
(b) Uniaxial stress only
Strength of Materials 211 YCT
 (c) Equal and opposite axial stress on two Ans : (c) Given,
mutually perpendicular planes, the plaines σ1 = 100 MPa
being free of shear. σ2 = 40 MPa
(d) Equal axial stress on two mutually
perpendicular planes, the planes being free of
shear.
MPPSC AE 08.11.2015
ISRO Scientist/Engineer 24.05.2014
ISRO Scientist/Engineer 2012
TRB Poly. Lect., 2012
ISRO Scientist/Engineer 2009
Ans. (d) : Two equal mutually perpendicular plane-
Mohr's circle—
Let two equal stress σ1 and σ2 having mutually Maximum shear stress
perpendicular i.e. σ − σ 2 100 − 40
τ max = 1 =
σ1 = σ2 = σ 2 2
τmax = 30MPa
360. For a general two dimensional stress system,
what are the co-ordinates of the centre of
Mohr's circle?
σx − σ y σx + σ y
(a) ,0 (b) 0,
2 2
σx + σy σx − σ y
(c) ,0 (d) 0,
2 2
Here, τ = 0 i.e. planes free of shear. UJVNL AE 2016
Only two equal Axial stress σ and σ UKPSC AE-2013, Paper-I
358. σx, σy and τxy are the rectangular stress APPSC AEE 2012
components of a point. The radius of Mohr's ESE 2007
circle is Ans : (c)
2  σx + σ y 
 σ x −σ y  Centre of Mohr ' s circle =  ,0 
(a) σ x − σ y + τ xy
2 2 2
(b)   + τ xy
2
2
 2   
2
( σ y −σ x )  σx − σ y 
2

( )
2
(c) σ y − σ x + τ xy
2 2 2
(d) + τ xy2 Radius =   + τ xy
2  2 
CGPSC AE 25.02.2018 361. In Mohr's circle drawn for a general case of
UPRVUNL AE 07.10.2016 biaxial stresses, the distance of centre of circle
MPSC HOD (Govt. Poly. Colleges) 04.10.2014 from the Y-axis is
ISRO Scientist/Engineer 2008 σx − σ y σx + σy
ESE 2006 (a) (b)
GATE 1993 2 2
Ans. (b) : Radius of Mohr's Circle σx + σy σx − σ y
2 (c) (d)
 σ x −σ y  (σ x − σ y ) 2 + 4τ xy2 2 2
R=   + τ xy =
2
APGCL AM, 2021
 2  4
WBPSC AE, 2017
1
R= (σ x − σ y ) 2 + 4τ xy2 Nagaland PSC CTSE 2016, Paper-I
2 Ans. (b) :
359. If the principle stresses in a plane stress
problem are σ1= 100 MPa, σ2= 40 MPa the
magnitude of the maximum shear stress (in
MPa) will be
(a) 60 (b) 50
(c) 30 (d) 20
GPSC Executive Engg. 23.12.2018
GPSC ARTO Pre 30.12.2018
RPSC 2016
TNPSC AE 2014 362. When two mutually perpendicular principal
UPRVUNL AE 2014 stresses are unequal but alike, the maximum
OPSC Civil Services Pre. 2011
shear stress is represented by
GATE 2009
Strength of Materials 212 YCT
 (a) the diameter or Mohr's circle σx + σy
(b) half the diameter of Mohr's circle i.e., = 0 or σ x + σ y = 0
(c) one-third the diameter of Mohr's circle 2
(d) one-fourth the diameter of Mohr's circle σx + σy = 0
TSPSC Managers, 2015
TRB Asstt. Prof., 2012 Principle stresses are σx and –σy. Hence they are equal
APPSC IOF, 2009, ESE 1994 in magnitude but unlike in direction.
2
366. When a thick plate is subjected to external
 σ − σy  loads:
Ans. (b) : Radius = τmaximum =  x  + τxy
2
1. State of plane stress occurs at the surface
 2  2. State of plane strain occurs at the surface
 σx + σ y  3. State of plane stress occurs in the interior
Centre of Mohr's circle =  ,0  part of the plate
 2 
4. State of plane strain occurs in the interior
part of the plate
Which of these statements are correct?
(a) 1 and 3 (b) 2 and 4
(c) 1 and 4 (d) 2 and 3
Gujarat PSC AE 2019
ESE 2013
Ans : (c) : For a plain strain case for a given load the
• Maximum shear stress is represented by half the strain in the thickness direction is negligible because
diameter of Mohr's circle. more material is available in thickness direction which
363. If σ1 and σ 2 be the major and minor tensile will resist any deformation in that direction (due to
stresses, then maximum value of tangential Poisson's effect) so strain in thickness direction in thick
stress is equal to- plate is assumed to be zero.
If you thick a cracked body (part through crack) loaded
(a) σ 2 (b) σ1 - σ 2
in tension, the crack front in the interior will have plane
σ1 -σ2 strain (as the mid section is surrounded by sufficient
(c) σ1 + σ 2 (d) volume of material thus making it analogous to thick
2
Assam Engg. College AP/Lect. 18.01.2021 section) whereas crack front at the surface will have
CIL MT 26.03.2017 plane stress.
ISRO Scientist/Engineer (RAC) 29.11.2015 367. A rectangular plate in plane stress is subjected
GATE 2018 to normal stresses σx = 35 MPa, σy = 26 MPa,
PPSC Asstt. Municipal Engineer 15.06.2021 and shear stress τxy = 14 MPa. The ratio of the
Ans : (d) Maximum value of tangential stress– magnitudes of the principal stresses (σ1/σ2) is
σ − σ2 approximately :
τ max. = 1 (a) 0.8 (b) 1.5
2 (c) 2.1 (d) 2.9
364. The minimum number of strain gauges in a RPSC Vice Principal ITI 2018
strain rosette is HPPSC Asstt. Prof. 20.11.2017
(a) One (b) Two Ans. (d) : Given,
(c) Three (d) Four
HPPSC AE 2018 σ x = 35MPa
HPPSC Asstt. Prof. 2014 σ y = 26 MPa
MPPSC State Forest Service Exam, 2014 τxy = 14 MPa
APPSC AEE 2012 2
Ans. (c) : The minimum number of strain gauge in a σx + σy  σ − σy 
strain rosette is Three. σ= ±  x  + τ xy
2

2  2 
Strain Gauge Rosette–A strain gauge rosette is a term
for an arrangement of two or more strain gauge that are σx + σy  σ − σy 
2

positioned closely to measure strains along different σ1 = +  x  + τ xy

2

365. If the centre of Mohr's stress circle coincides 2  2 

with the origin on the σ - τ coordinates, then 2
35 + 26  35 − 26 
(a) σx + σy = 0 (b) σx - σy = 0 = +   + 14
2

(c) σx + σy = 1/2 (d) σx - σy = 1/2 2  2 

Gujarat PSC AE 2019 2
61 9
TNPSC AE 2019 = +   + (14)2
BPSC Main 2017 Paper-VI 2 2
Ans : (a) : = 45.205 MPa
σx + σy 2
By centre of Mohr's circle lie at σx + σy  σ − σy 
2 σ2 = −  x  + τ xy
2

It's coincide with origin 2  2 

Strength of Materials 213 YCT

 2
61 9
σ2 = −   + (14)2
2 2
= 15.79 MPa
The ratio of the magnitude of principal stress,
σ1 45.205 (a)
= = 2.86
σ 2 15.790
≈ 2.9
368. For an element in pure shear (+τ xy ) , the
principal stresses will be given as : (b)
(a) σ1,2 = ±τ xy (b) σ1,2 = ±τ xy / 2
(c) σ1,2 = ±τ xy 2 (d) σ1,2 = 0 + τ xy
CIL MT 26.03.2017
UPRVUNL AE 07.10.2016
Ans. (a) : We know that (c)
For an element in pure shear (±τxy), then principal
stresses.

(d)
APPSC AEE SCREENING 17.02.2019
ISRO Scientist/Engineer 2011
BPSC Asstt. Prof. 29.11.2015
σx = σy = 0 (Pure shear) ESE 1996
Then principal stresses Ans. (d) : State of stress
σx +σ y 1
σ 1,2 = ± (σ x − σ y ) 2 + 4τ xy2
2 2
σ 1,2 = ± τ xy
369. The normal stresses at a point are σx = 10
MPa, σy = 2 MPa, and the shear stress at the at "Pure shear"
this point is 3 MPa. The maximum principal Mohr's circle diagram
stress at this point would be
(a) 15 MPa (b) 13 MPa
(c) 11 MPa (d) 09 MPa
JWM 2017
J&K PSC Civil Services Pre, 2010
ESE 1998 371. If a Mohr's circle is drawn for a fluid element
Ans. (c) : Maximum principle stress, inside a fluid body at rest, it would be :
2
(a) a circle not touching the origin
σx + σy  σx − σy  2 (b) a circle touching the origin
σ1 = ±   + τ xy
2 2 (c) a point on the normal stress axis
 
(d) a point on the shear stress axis
σ x = 10 MPa
RPSC ACF & FRO, 26.02.2021
σ y = 2 MPa CGPSC AE 26.04.2015 Shift-I
τ xy = 3 MPa TSPSC AEE 28-08-2014 (Civil/Mech.)
Ans. (c) : Mohr's circle for a fluid element inside a fluid
2
10 + 2  10 − 2  2 body at rest it would be a point on the normal stress
σ1 = +   +3
2  2  axis.
Mohr's circle reduces to a point in case of
= 6 + 16 + 9 = 6 + 5 = 11 MPa hydrostatic loading.
σ 1 = 11 MPa
370. Which one of the following figures is the
correct sketch of Mohr's circle of the given
state of stress

Strength of Materials 214 YCT

372. If the principal stresses corresponding to a 374. Under uniaxial strain, the ratio of maximum
two-dimensional state of stress are σ1 and σ2. shearing strain to uniaxial strain is–
When σ1 is greater than σ2 and both are (a) 2.0 (b) 0.5
tensile, then which of the following would be (c) 1.0 (d) 1.5
correct criterion for failure by yielding, GPSC AE (Part-B Civil) 2018
according to maximum shear stress theory? Nagaland PSC CTSE 2017, Paper-I
σ – σ2 σ yp σ σ yp Ans. (c) :
(a) 1 =± (b) 1 = ± 2
2 2 2 2 σx + σy  σ − σy 
σ yp Stress, σ1 / σ 2 = +  x  + τxy
2
σ2 2  2 
(c) =± (d) σ1 = ±2σ yp
2 2 2
RPSC IOF, 2020  σ − σy 
Punjab PSC (Lect.) 06.08.2017 τmax =  x  + τxy
2

 2 
Ans. (b) : According to maximum shear stress theory
2 2
for plane condition, ∈x + ∈y  ∈ − ∈y   φxy 
Strain, ∈1 / ∈2 = ±  x  + 
σ − σ 2 σ1 σ 2 
Absolute, τmax = Larger of  1 , ,  2  2   2 
 2 2 2 2 2
(1) If, σ1 and σ2 are alike in nature and σ1 > σ2 φmax  ∈ − ∈y   φxy 
=  x  + 
σ 2  2   2 
Then, absolute τmax = 1 = Ssy
2 In question ∈x = ∈, ∈y =φxy =0
(2) If, σ1 and σ2 are unlike in nature, ∈ φ ∈
σ − σ2 So, ∈1 = x and max = x or φmax =∈x
Then, absolute τmax = 1 = Ssy 2 2 2
2 ∈x = uniaxial strain
According to maximum shear stress theory, φmax = max shearing strain
σ yp σ yp φ
= 2 ⇒ Ssy = Ratio = max = 1
Ssy 2 ∈x
σ As, ∈x = φmax
σ1
So, = ± yp (Because σ1 and σ2 are alike 375. A body is subjected to a tensile stress of 1200
2 2 MPa on one plane and another tensile stress of
in nature) 600 MPa on a plane at right angles to the
373. A body is subjected to a direct tensile stress of former. It is also subjected to a shear stress of
300 MPa in one plane accompanied by a 400 MPa on the same planes. What would be
simple shear stress of 200 MPa. What would the maximum normal stress?
be the maximum normal stress? (a) 400 MPa (b) 500 MPa
(a) –100 MPa (b) 250 MPa (c) 900 MPa (d) 1400 MPa
(c) 300 MPa (d) 400 MPa TNCSC AE 2020
GPSC DEE, Class-2 (GWSSB) 04.07.2021 JPSC AE 2013, Paper- V
RPSC IOF 2020 VIZAG STEEL MT 2011
VIZAG STEEL MT 2011 Ans. (d) : Given, σx = 1200 MPa, σy = 600 MPa
ESE 2016 τxy = 400 MPa
Ans. (d) : Maximum normal stress,
2
 σ + σy   σx − σ y 
σ1 =  x +   + τxy
2

 2   2 
2
 1200 + 600   1200 − 600 
= +   + (400)
2

 2   2 
Given, σx = 300 MPa
σy = 0 = 900 + (300) 2 + (400)2
τxy = 200 MPa σ1 = 900 + 500
Maximum normal stress, σ1 = 1400 MPa
2
σx + σy  σ − σy  376. A plane stressed element is subjected to the
σ1 = +  x  + τxy
2
state of stress given by σx = τxy = 100kgf/cm2
2  2  and σ = 0. Maximum shear stress in the
 300 + 0   300 − 0  (
2 element is equal to
 + 200 )
2
σ1 =  +  (a) 50 3 kgf / cm 2 (b) 100 kgf/cm2
 2   2 
(c) 50 5 kgf / cm 2 (d) 150 kgf/cm2
= 150 + (150)2 + (200)2
APPSC AEE 2012
= 400 MPa APPSC AE 04.12.2012, ESE 1997
Strength of Materials 215 YCT
Ans. (c) : σx = τxy = 100 kgf/cm2, σy = 0 Ans : (c)
2 2
 σx – σ y  2  100  2
τ max =   + τ xy =   + 100
 2   2 
2 2
= 50 + 100 = 50 1 + 4
τ max = 50 5 kgf/cm
2

377. Principal stresses at a point in plane stressed BC –A plane which is inclined at an angle (90–θ) with
element are σx = σy = 500 N/mm2. Normal the line of action of applied un-axial stress σx
stress on the plane inclined at 45º angle to x- P = σ × AB
axis will be x x [ t = 1 assume]
(a) 0 N/mm 2
(b) 700 N/mm 2 Pn = Px × cos θ = σ x × AB × cos θ
(c) 500 N/mm2 (d) 1000 N/mm2 The normal stress at inclined plan BC
UKPSC AE 2013 Paper-I
APPSC AEE 2012, ESE 1993 σ n = Pn = σ x × AB × cos θ cos θ =
AB AB 
, BC =
σx + σy σx − σ y BC  AB   BC cos θ 
Ans. (c) : σθ = + cos 2θ  cos θ 
2 2
500 + 500 500 − 500 σ n = σ x cos θ2

= + cos90
2 2 381. For a plane stress case, σ1 = 50 MPa,
σθ = 500N / mm 2
σ 2 = −100 MPa, τ12 = 40 MPa, the maximum and
378. If a body carries two unlike principal stresses, minimum principal stresses are, respectively,
the maximum shear stress is given by (a) 60 MPa, − 110 MPa,
(a) sum of the principal stresses (b) 50 MPa, − 110 MPa
(b) difference of the principal stresses (c) 40 MPa, − 120 MPa
(c) half the difference of the principal stresses (d) 70 MPa, − 130 MPa,
(d) half the sum of the principal stresses PPSC Asstt. Municipal Engineer 15.06.2021
CGPSC AE 15.01.2021 Ans. (a) : σ1= 50 MPa σ2=− 100 MPa σ12= 40 MPa
TSPSC AEE 2015 Max and Min principal stresses
APPSC AEE 2012 2
Ans. (c) If a body carries two unlike principal stresses, σ σ1 + σ 2  σ − σ2 
max, min= ±  1  +τ12
2
the maximum shear stress is given by half the difference 2  2 
of the principal stresses.
 50 − ( −100 ) 
2
Or 50 − 100
If a body carries two unlike principal stresses the = ±   + 40
2

maximum shear stress is given by half the sum of the 2  2 

principal stresses. = − 25 ± 85
379. Mohr's circle can be used to determine σmax, min = −25+85 & −25−85
following stress on inclined surface σmax, min = 60, −110
(a) Normal stress (b) Principal stress σmax = −110MPa, σ min = 60MPa
(c) Tangential stress (d) All of the above
UJVNL AE 2016 382. The tensile stresses at a point across two
UKPSC AE-2013, Paper-I mutually perpendicular planes are 120 MPa
Ans : (d) Mohr's circle can be used to determine and 60 MPa. The normal stress on a plane
Normal stress, principal stress, and tangential stress. inclined at 30º to the axis of the minor axis is
Mohr's Circle of Stresses:- ______.
The Mohr's circle is a graphical method of finding the (a) 110 Mpa (b) 60 Mpa
normal, tangential and resultant stresses on an inclined (c) 45 Mpa (d) 105 Mpa
plane. It is drawn for the following two cases:- VIZAG Steel MT 24.01.2021, Shift-I
(i) When the two mutually perpendicular principal Ans. (d) : Given,
stresses are unequal and like. σx = 120 MPa
(ii) When the two mutually perpendicular principal σy = 60 MPa
stresses are unequal and unlike. θ = 30º
380. Normal stress on a plane, the normal........ Figure
Which is inclined at an angle θ with the line of
Normal stress, ( σ n )θ = ( σ x + σ y ) + ( σ x − σ y ) cos 2θ
1 1
action of applied un axial stress σ x is given by 2 2
(a) σ x / sin θ 2
(b) σ x / cos θ
2 1 1
= (120 + 60 ) + (120 − 60 ) cos 60º
(c) σ x cos θ 2
(d) σ x sin 2 θ 2 2
1 1
GPSC Engg. Class-II Pre-19.01.2020 = 90 + × 60 × = 90 + 15 = 105 MPa
UPPSC AE 12.04.2016 Paper-I 2 2
Strength of Materials 216 YCT
383. The stresses on two perpendicular planes Ans. (b) :
through a point in a body are 100 MPa and 40
MPa both tensile along with a shear stress of
40 MPa. The maximum failure stress as per
maximum normal stress theory is __________.
(a) 180 MPa (b) 110 MPa
(c) 70 MPa (d) 120 MPa
VIZAG Steel MT 24.01.2021
Ans. (d) : Given,
σx = 100 MPa
σy = 40 MPa
τxy = 40 MPa Given,
P
σx + σ y  σ − σy 
2
σx =
σ1,2 = ±  x  + τxy
2
A
2  2  σy = 0
2
100 + 40 100 − 40 
= ±   + (40)
2 τ xy = 0
2  2  2
 σ − σy 
= 70 ± (30) 2 + (40) 2 Maximum shear stress (τmax) =  x  + τ xy
2

σ1,2 = 70 ± 50  2 
Maximum (σ1) = 70 + 50 = 120 MPa 2
P 
 A −0
Minimum (σ2) = 70 – 50 = 20 MPa
Hence, the maximum failure stress as per maximum =   +0
2

normal stresses (σ1 = 120 MPa)  2 

384. If a body is subjected to stresses in xy plane  
with stresses of 100 N/mm2 and 20 N/mm2 P
acting along x and y axes respectively. Also τmax =
the shear stress acting is 30 N/mm2. Find the 2A
2 386. Mohr's circle for membrane stresses in a thin
maximum amount of shear stress in N/mm to
which the body is subjected. walled spherical pressure vessel with radius r
and wall thickness t, under internal pressure p
(a) 30 5 (b) 50
will be:
(c) 10 5 (d) Zero
APGCL AM, 2021
Ans. (b) :
Given
σ x = 100N / mm 2
σ y = 20N / mm 2
τxy = 30N / mm 2
Maximum Shear Stress = Radius of mohr circle.
2
 σx − σ y 
τmax =  + τxy
2

 2 
2
 100 - 20 
 + ( 30 )
2
= 
 2 
( 40 ) + ( 30 )
2 2
=
τmax = 50N/mm2
385. If a cylindrical rod of cross section A is applied
with uniaxial tension P, the maximum shear JPSC AE 10.04.2021, Paper-II
stress in the rod will be : Ans. (c) : For thin spherical pressure vessel
1P Pd
(a) 0 (b) σ h = σL =
2A 4t
P 2P
(c) (d) Pd
A A σ1 = σ 2 =
JPSC AE 10.04.2021, Paper-II 4t

Strength of Materials 217 YCT

 (b) mutually perpendicular planes which are
parallel to the planes of pure shear
(c) mutually perpendicular planes which are
perpendicular to the planes of pure shear
(d) mutually perpendicular planes which are at
For pure shear – 450 to the planes of pure shear
GPSC DEE, Class-2 (GWSSB) 04.07.2021
Ans. (d) : When an element is in a state of simple shear,
maximum direct stresses are induced on mutually
perpendicular planes which are at 450 to the planes of
pure shear.
389. The state of stress at a point P in a two
dimensional loading is such that the Mohr's
For Biaxial Tension – circle is a point located at 145 MPa on the
positive normal stress axis. The maximum and
minimum principal stresses respectively from
the Mohr's circle are :
(a) 0, 0
(b) 0, –145 MPa
(c) + 145 MPa, + 145 MPa
For Uniaxial Tension – (d) + 145 MPa, – 145 MPa
RPSC ACF & FRO, 26.02.2021
Ans. (c) :

387. Maximum shear stress for the shown stress

condition will be : Maximum principle stress = 145 MPa
Minimum principle stress = 145 MPa
Maximum /minimum shear stress = 0 MPa
390. Principal strains occur on three planes which
are
(a) inclined to one another at 45°
(b) mutually perpendicular to each other
(c) inclined to one another at 22½°
(a) σ/2 (b) σ (d) None of the above
(c) 2σ (d) 0 Assam PSC AE (IWT) 14.03.2021
JPSC AE 10.04.2021, Paper-II Ans. (b) : For the three-dimensional case it is now
Ans. (d) : Given, demonstrated that three planes of zero shear stress exist,
σx = σ that these planes are mutually perpendicular and that on
these planes the normal stresses have maximum or
σy = σ
minimum values.
τxy = 0
391. A spherical container made of steel has a 5m
2
 σ − σy  outside diameter and a 10mm uniform wall
Maximum shear stress =  x  + τ xy
2
thickness. If the internal pressure is 400 kPa,
 2  the maximum normal stress and maximum
2 shear stress in the container are :
σ−σ (a) σmax = 77.4 MPa, τmax = 38.7 MPa
=   +0
2

 2  (b) σmax = 89.0 MPa, τmax = 44.5 MPa

τmax = 0 (c) σmax = 49.8 MPa, τmax = 2.91 MPa
(d) σmax = 38.7 MPa, τmax = 51.9 MPa
CGPSC AE 15.01.2021
Ans. (*) : Given, diameter of spherical container = 5m
thickness (t) = 10 mm
Internal pressure (p) = 400 kPa
Mohr's circle will be point.
Pd
388. When an element is in a state of simple shear, Max. Normal stress, σC =
maximum direct stresses are induced on 4t
(a) mutually parallel planes which are at 450 to 400 × 5
=
the planes of pure shear 4 ×10 × 10−3
Strength of Materials 218 YCT
 = 50 × 103 kPa The co-ordinates of the centre of Mohr's circle
= 50 MPa are
(a) (0, 0) (b) (100, 200)
σC 50 (c) (200, 100) (d) (50, 0)
Max. shear stress, τmax = =
2 2 UPPSC AE 13.12.2020, Paper-I
≃ 25 MPa Ans. (d) : Co-ordinate of the centre of Mohr's circle is
Note- Official answer is (c).  σ x + σ y  
392. Consider a linear elastic rectangular thin sheet   ,0
 2  
of metal, subjected to uniform uniaxial tensile
200 − 100 
stress of 100 MPa along the length direction. Centre =  ,0 = (50,0)
Assume plane stress condition in the plane  2 
normal to the thickness. Young's modulus E = 396. When a member is subjected to a direct stress
200 MPa and Poisson's ratio v = 0.3. The (σ) in one plane, then the tangential stress on
principal strains in the plane of sheet are_____ an oblique plane which is inclined at an angle
(a) (0.5, – 0.15) (b) (0.5, – 0.5) Φ with the normal cross section is _______ .
(c) (0.5, 0.0) (d) (0.35, –0.15) σ(sin 2φ)
Assam Engg. College AP/Lect. 18.01.2021 (a) σ(sin2φ) (b)
Ans. (a) : (0.5, – 0.15) 2
σ(cos 2φ) σ(cos 2 φ)
(c) (d)
2 2
VIZAG MT, 14.12.2020
Ans. (b) : Tangential stress along inclined plane
σx = 100
σy = 0
v = 0.3
σ σy σy σ
εx = x – v εy = –v x
E E E E
100 100
= – 0.3 × 0 = 0 – 0.3 ×
200 200  σ − σy 
εx = 0.5 εy = – 0.15 τφ = −  x  sin 2θ + τ xy cos 2θ
 2 
393. A strain gauge rosette is used when : τxy = 0
(a) direction of strain is known σx = σ θ = (90 + 90 – φ)
(b) only magnitude of stress is known σy = 0 = (180 – φ)
(c) both magnitude and direction of strains are
σ −
τφ = −   sin 2(180 − φ)
unknown 0

(d) magnitude is known but the direction is  2 
unknown σ
APPSC Poly Lect. Automobile Engg., 2020 τφ = sin 2φ
Ans. (c) : A strain gauge rosette is a term for an 2
arrangement of two or more strain gauges that are 397. If a element is subjected to uniform equal
positioned closely to measure strains along different compressive stresses of 1 MPa in all directions
direction of the component under evaluation. Single so that σx = σy = σz . The Mohr's circle for this
strain gauge can only measure strain effectively in one state of stress is a
direction so the use of multiple strain gauges enables (a) Circle with infinite radius
more measurements to be taken. Providing a more (b) Circle with zero radius
(c) Circle with unit radius
precise evaluation of strain on the surface being (d) None of the above
measured. Haryana PSC AE (PHED) 05.09.2020, Paper-II
394. The gauge factor of a semiconductor gauge Ans. (b) : σx = σy = σz the radius of Mohr's circle will
compared to a wire strain gauge is : be zero.
(a) high (b) the same
(c) infinite (d) low  σ − σy 
R=  x  + (τ xy )
2
APPSC Poly Lect. Automobile Engg., 2020  2 
Ans. (a) : The strain gauge should have a high value of R = 0+0 =0
gauge factor. As high gauge factor indicates a large
change in resistance that leads to high sensitivity. The 398. For an element in plane stress, if the larger of
the principal stresses is 3 times that of the
gauge resistance should be high so as to minimize the smaller one, the value of the maximum shear
effect of undesirable variation of resistance in stress is equal to
measurement circuit. (a) the smaller principal stress
395. Consider a two dimensional state of stress for (b) the larger principal stress
an element (c) half of the higher principal stress
Where, σx = 200 MPa (d) sum of the two principal stress
σy = –100 MPa Haryana PSC AE (PHED) 05.09.2020, Paper-II
Strength of Materials 219 YCT
Ans. (c) : σ1 = 3σ 404. Mohr’s circle for the state of stress defined by
σ2 = σ  20 0 
σ3 = 0  0 20 is a circle with :
σ1 − σ3  
Shear stress ( τ ) = (a) Centre at (0, 0) and radius 20 MPa
2 (b) Centre at (0, 0) and radius 40 MPa
3σ − 0 1 (c) Centre at (20, 0) and radius 20 MPa
τ= = σ1
2 2 (d) Centre at (20, 0) and radius zero radius
1 OPSC AEE 2019 Paper-I
τ = higher principal stress Ans : (d) :
2
399. For an element in plane stress, the planes of Data given as-
algebraically maximum shear stress occur at σ x = 20 MPa
(a) At 900 to the principal planes σ y = 20 MPa
(b) At the principal planes
(c) At 450 to the principal planes τ xy = 0 MPa
(d) None of the above Center of Mohr's circle
Haryana PSC AE (PHED) 05.09.2020, Paper-II  σ x + σ y   20 + 20  
TSPSC AEE 28.08.2017 (Civil/Mech.) = ,0  =   ,0 = ( 20,0 )
Ans. (c) : Plane of max shear stress is at 450 to the  2   2  
principal plane. and we know that
400. A strain gauge oriented in a direction at an 2
 σx − σ y 
2
angle θ with x-axis, measures:  20 − 20 
τ =   + τ 2
=   +0 =0
2

 2   2 
max xy
(a) εxsin2 θ + εycos2 θ + τxycosθsinθ
(b) εxcos2 θ + εysin2 θ + τxycosθsinθ 405. The state of stress at a point in a 2-D loading is
(c) εxcos2 θ + εysin2 θ – τxycosθsinθ such that the Mohr's circle is a point located at
(d) εxcos θ + εysin θ + 2τxycosθsinθ
2 2
175 MPa on the positive normal stress axis.
CIL MT 27.02.2020 The maximum and minimum principle stress,
Ans. (b) : A strain gauge oriented in a direction at an respectively, from Mohr's circle are :
angle θ with x-axis measures (a) 0; 0 MPa
= εxcos2 θ + εysin2 θ + τxycosθ.sinθ (b) + 175 MPa; + 175 MPa
401. A wheat stone bridge is used for electrical (c) + 175 MPa; – 175 MPa
strain gauges, because it has: (d) + 175 MPa; 0 MPa
(a) high sensitivity (b) zero sensitivity BHEL ET 2019
(c) low sensitivity (d) infinite sensitivity GATE 2003
CIL MT 27.02.2020
Ans. (b) : σ x = 175 MPa ( Tensile )
Ans. (a) : A wheat stone bridge circuit has high
sensivity for detecting very small variation in the values σ y = 175 MPa ( Tensile )
of resistance.
We can connect the strain gauge in one of the arms of a σx + σy  σx − σy 
2

 + ( τ xy )
2
wheat stone bridge and measure the strain in term of σ= ± 
variation in resistance. 2  2 
402. If normal stress on a normal plane is σ, the
175 + 175  175 − 175 
maximum shear stress on a plane inclined 45 = ±  +0
degree to the normal plane will be 2  2 
(a) σ/2 (b) σ σ maximum = 175 MPa
(c) σ sin θ (d) σ cos θ
TSPSC Manager (Engg.) HMWSSB 12.11.2020 σ minimum = 175 MPa
Ans. (a) : σ/2 406. A point in a structural member is subjected to
403. At a point in a strained material, tensile stress normal stresses (σx and σy) along two mutually
of 100 MPa and compressive stress of 60 MPa perpendicular directions and shear (τxy) in xy
are found to be principal stresses. Maximum plane. The orientation of first principal plane
shear stress at that point is : (θp1) is expressed as :
(a) 20 MPa (b) 40 MPa
(c) 80 MPa (d) 60 MPa  2τ xy  1  2τ xy 
(a) tan −1   (b) tan −1 
NLCIL GET 17.11.2020, Shift-II σ −σ  2  σ − σ 
Ans. (c) : Given,  x y   x y 

σ1 = +100MPa ( Tensile )  σx − σ y  1  σx − σ y 
(c) 2 tan −1   (d) tan −1  
σ 2 = −60MPa ( Compressive )  2τ  2  2τ
 xy   xy 
σ − σ2 Oil India Limited Sr. Engineer (Drilling) 30.11.2019
Maximum shear stress τmax = 1
2 Ans. (b) : Given,
100 − (−60) By using force equilibrium along an a perpendicular
= = 80 MPa
2 to the inclined plane

Strength of Materials 220 YCT

We get,
 σx + σ y   σx − σ y 
σn =  +  cos 2θ + τ xy sin 2θ
 2   2 

(a) 0º, 90º, τ and –τ

(b) 30º, 120º, τ and –τ
(c) 45º, 135º, τ and –τ
τ τ
(d) 45º, 135º, and −
2 2
APPSC AEE SCREENING 17.02.2019
Ans. (c) :

 σx − σ y 
τn = −   sin 2θ + τ xy cos 2θ
 2 
We know that on the principal planes shear stress
always zero.
 σx − σ y 
∴ −  sin 2θ + τ xy cos 2θ = 0
 2  Due to pure shear diagonal tension (σ1 = +τ) and
 σx − σ y  diagonal compression (σ2 = –τ) develops.
  sin 2θ = τ xy cos 2θ The angle between principal planes is 90º.
 2  ∴ θ1 = 45º, θ2 = 135º, σ1 = τ, σ2 = –τ
2τ xy 409. Where does principal stress occur in a
tan 2θ =
( σx − σ y ) component?
(a) Along the plane
1 −1
 2τ xy  (b) Perpendicular to the plane
θ = tan 
2  σ − σ  (c) On mutually perpendicular planes
 x y  (d) Along the direction of load
So, the orientation of first principal planes (θp1 ) is TNPSC AE 2019
expresses as : Ans. (c) : On mutually perpendicular planes principal
stress does occur in a component.
• Shear stress τ = 0
410. A body is subjected to a direct tensile stress of
300 MPa in one plane accompanied by a
simple shear stress of 200 MPa. The maximum
shear stress will be
(a) 150 MPa (b) 200 MPa
(c) 250 MPa (d) 300 MPa
JPSC AE PRE 2019
1  2 τ  Ans. (c) : Given,
θ p1 = tan −1  xy

2  σ − σ  σx = 300 MPa
 x y 
σy = 0
407. What is the number of non-zero strain
components for a plane stress problem? τxy = 200 MPa
2
(a) 6 (b) 4 σ +σ y  σ −σ y 
(c) 3 (d) 2 σ1 = x +  x  + τ xy
2

APPSC AEE SCREENING 17.02.2019 2  2 

Ans. (c) : In a plane stress (2D) problem the number of 2
non-zero strain components are ∈x , ∈y and φxy σ +σ y  σ −σ y 
σ2 = x −  x  + τ xy
2

Total three number. 2  2 

∈
 x φ xy  300 + 0
2
The 2D tensor is    300 − 0 
φ ∈ σ1 = +   + (200)
2

 yx y  2  2 
408. The state stress at a point is shown below. θ
= 150 + (150 ) + (200)2
2
represents the principal plane corresponding
to principal stress σ1 and σ2 (σ1 > σ2). Values = 150 + 250
of θ, σ1, and σ2 are = 400 MPa
Strength of Materials 221 YCT
 2 Ans. (b) : Radius of Mohr's circle is represented by
300 + 0  300 − 0 
σ2 = −   + (200)
2
σ p1 − σ p2
2  2  R=
2
= 150 − (150 ) + (200) 2 414. If principle stress σp1 = 100 N/mm2 (tensile) σp2
2

= 150 – 250 = 40 N/mm2 (compressive), then maximum

= –100 MPa shear stress will be:
(a) 70 N/mm2 (b) 50 N/mm2
σ1 − σ 2 (c) 30 N/mm2 (d) 10 N/mm2
τ max = 2
2 (e) 5 N/mm
400 − (−100) CGPSC AE 25.02.2018
=
2 Ans. (a) : Data given-
= 250 MPa σp1 = 100 N/mm2, σp2 = – 40 N/mm2
411. At a point in a stressed body there are normal Then max, shear stress is given as
stresses of 1N/mm2 (tensile) on a vertical plane σ p1 − σ p2
and 0.5 N/mm2 (tensile) on a horizontal plane. τ max = Radius of Mohr's circle =
2
The shearing stresses on these planes are zero.
What will be the normal stress on a plane 100 − [ −40] 2
τ max = = 70N / mm
making an angle 50o with vertical plane? 2
[Given, (cos 50o)2 = 0.413] 415. In a 3-D state of stress, the independent stress
(a) 0.6015 N/mm2 (b) 0.4139 N/mm2 components required to define state-of-stress
2
(c) 0.5312 N/mm (d) 0.7065 N/mm2 at a point are -
SJVN ET 2019 (a) 3 (b) 6
Ans. (d) : Given as, (c) 12 (d) 9
σ x = 1Pa, σ y = 0.5Pa, τ xy = 0 RPSC AE 2018
Ans. (b) :
θ = 50º
σx + σ y σx − σ y σ xx τ xy τ xz 
We know that, [ σn ]θ=50º = + cos 2θ  
2 2 τ yx σ yy τ yz 
1 + 0.5   1 − 0.5   τ zx τ zy σ zz 
[ σn ]θ=50º =  +  cos ( 2 × 50º ) then,
 
 2   2 
= 0.7065Pa σ x − x , σ y − y ,σ z − z
412. Assertion (A): A plane state of stress always τ xy = τ yx
results in a plane state of strain. τ xz = τ zx
Reason (R) : A uniaxial state of stress results
in a three-dimensional state of strain. τ yz = τ zy
(a) Both A and R are individually true and R is
the correct explanation of A.
(b) Both A and R are individually true and R is
not the correct explanation of A.
(c) A is true but R is false
(d) A is false but R is true
Gujarat PSC AE 2019
ESE 2010
Ans : (d) :
In a plane stress condition, the stress in the In a 3-D state of stress, the independent stress
perpendicular direction to the plane is zero. But strain in components required to define state of stress at a point
that direction need not to be zero. Thus assertion is is six (three normal stress and three shear stress).
wrong. 416. A state of plane stress consists of a uni-axial
tensile stress of magnitude 8 kPa, exerted on
413. Radius of Mohr's circle is represented as : vertical surface and of unknown shearing
[where σp1 > 0 and σp2 > 0 are the major and stresses. If the largest stress is 10 kPa, then the
minor value of principle stresses] magnitude of the unknown shear stress will be
σ p1 − σ p2 (a) 6.47 kPa (b) 5.47 kPa
(a) σ p1 − σ p2 (b) (c) 4.47 kPa (d) 3.47 kPa
2 UPPSC AE 13.12.2020 Paper-I
σ p1 + σ p2 ESE 2018
(c) σ p1 + σ p2 (d)
2 Ans. (c) : Given,
σ p1 × σ p2 σx = 8 kPa, σ1 = 10 kPa
(e) σy = 0
2 Maximum principal stress is given by
CGPSC AE 25.02.2018
Strength of Materials 222 YCT
 2 (a) 70 MPa (b) 60 MPa
σx + σ y  σx − σ y  (c) 50 MPa (d) 10 MPa
σ1 = +   + τ xy
2
GPSC EE Pre, 28.01.2017
2  2 
2
ESE 2012
8+0 8−0 Ans. (c) : Given-
10 = +   + τ xy
2
σx = 70 MPa
2  2 
σy = 10 MPa
10 = 4 + 42 + τ2xy τxy = 0
θ = 45º
τxy = 20 Normal stress (σx)
τxy = 4.47 kPa
σx + σ y σ x − σ y
417. In a bi-axial stress problem, the stresses in x σx = + cos 2θ + τ xy sin 2θ
and y direction are σx = 100 MPa, σy = 100 2 2
MPa respectively. The maximum principal 70 + 10 70 – 10
σx = + × cos90º +0 × sin 90º
stress in MPa, is : 2 2
(a) 50 (b) 100 80
(c) 150 (d) 200 (σx) = = 40 MPa
Vadodara Muncipal Corp. DEE, 2018 2
σx – σ y
Ans. (b) : Given, Shear stress (τ) = sin 2θ – τ xy cos 2θ
σx = 100MPa 2
σy = 100MPa 70 – 10
= sin ( 90º ) – 0 × cos(90º )
τxy = 0 2
60
τ= = 30 MPa
2
( σx ) + ( τ)
2 2
Resultant stress (σr) =

= ( 40 ) + ( 30 ) = 1600 + 900
2 2

= 50 MPa
maximum principal stress 420. A horizontal beam under bending has a
2 maximum bending stress of 100 MPa and
σx + σ y  σx − σy  maximum shear stress of 20 MPa. What is the
σ1 = +   + τ xy
2

2 2 maximum principal stress in the beam?

  (a) 20 (b) 50
2
100 + 100  100 − 100  (c) 50 + 2900 (d) 100
= +   +0
2
HPPSC Asstt. Prof. 18.09.2017
2  2 
CSE Pre-2004
σ1 = 100 MPa Ans. (c) : Given,
418. Consider the following statements : σb = 100 MPa
1 . Two-dimensional stresses applied to a thin τmax = 20 MPa
plate in its own plane represent the plane stress σ
2
σ 
condition. Max. Principle stress, σ = b +  b  + τ 2xy
2. Under plane stress condition, the strain in the 2  2 
direction perpendicular to the plane is zero. 2
 100   100 
 + ( 20 )
2
3. Normal and shear stresses may occur =  + 
simultaneously on a plane  2   2 
Which of the above statements is /are correct?
= 50 + ( 50 ) + ( 20 )
2 2
(a) 1 only (b) 1 and 2
(c) 2 and 3 (d) 1 and 3 σ = 50 + 2900 MPa
GPSC ARTO Pre 30.12.2018
ESE 2009 421. What is the normal stress on a plane inclined
at 45º to the axis of a square rod of side a
Ans. (d) : The correct statements are – subjected to an axial tensile force of T
(i) Two –dimensional stress applied to a thin plate in its (a) T/a2 (b) T/2a2
own plane represents the plane stress condition. (c) T/4a2 (d) T/8a2
(ii) Under plane stress condition, the strain in the TSPSC AEE 28.08.2017 (Civil/Mechanical)
direction perpendicular to the plane is not zero. Ans. (b) :
(iii) Normal and shear stress may occurs simultaneously
on a plane.  σ x + σ y   σx − σ y 
σθ =  +  cos 2θ + τ xy sin 2θ
419. A piece of material is subjected, to two  2   2 
perpendicular tensile stresses of 70 MPa and T
10 MPa. The magnitude of the resultant stress σx = 2
on a plane in which the maximum shear stress a
occurs is σy = 0, τxy = 0
Strength of Materials 223 YCT
 σx σx Ans. (a) : Maximum shear stress on any plane
∴ σθ = + cos 2θ
2 2 σ − σ2
In plane τmax = 1
T T 2
σ 45º = 2 + 2 cos90º
2a 2a τ max σ1 − σ 2
= =1:2
σ 45º = 2
T σ1 − σ2 2 ( σ1 − σ 2 )
2a 425. A specimen is subjected to pure shear, the
422. Principal stresses at a point in a stressed solid shear stress being q. Tensile and compressive
are 400 MPa and 300 MPa respectively. The stresses of intensity Q occur on planes inclined
normal stress on plane inclined at ± 45º to the at 45º to the shear stress. What is the value of
principal planes will be the ratio Q/q?
(a) 200 MPa and 5 MPa (a) 2 (b) 1.5
(b) 350 MPa and both planes (c) 1.25 (d) 1
(c) 100 MPa and 600 MPa TSPSC AEE 28.08.2017 (Civil/Mechanical)
(d) 1500 Mpa and 500 MPa Ans. (d) : Case of pure shear
TSPSC AEE 28.08.2017 (Civil/Mechanical), CSE Pre 2000
Ans. (b) :

Q
∴ =1
q
426. Following figure shows the state of stress at a
certain point in a stressed body. The
magnitudes of normal stresses in the x and y

σθ =
(σ x + σy )
+
σx + σy
cos 2θ + τ xy sin 2θ
directions are 100 MPa and 20 MPa
respectively. The radius of Mohr's stress circle
2 2 representing this state of stress is
τxy = 0
Normal stress on plane inclined at 45º
400 + 300 400 + 300
σθ = + × cos90º
2 2
= 350 MPa on both planes.
423. When the two principal stresses are equal and
like, the resultant stress on any plane is
(a) equal to principal stress
(b) zero (a) 120 (b) 80
(c) one half of the principal stress (c) 60 (d) 40
(d) one third of the principal stress TSPSC AEE 2017
TSPSC AEE 28.08.2017 (Civil/Mechanical) Ans. (c) :
Ans. (a) : When two principal stresses are equal and
like

σx − σy
100 − (−20)
r= = = 60 MPa
2 2
So, Radius of Mohr's circle = 60 MPa
Resultant stress on any plane 427. Under which loading condition Mohr's circle
σ R = σθ2 + τθ2 touches the origin of shear stress axis and
having one positive principal stress?
σθ = σ (irrespective of any plane) (a) Uniaxial compression
τθ = 0 (b) Uniaxial tension
σR = σ (Magnitude of principal stress) (c) Pure torsion
424. The ratio of the maximum shear stress to the (d) None of the above
Punjab PSC SDE 12.02.2017
difference of the two principal stresses is
(a) 1/2 (b) 1/3 Ans. (b) : For Uniaxial Tension Mohr's circle touches
(c) 1/4 (d) 1.6 the origin of shear stress axis and having one positive
TSPSC AEE 28.08.2017 (Civil/Mechanical) principal stress.
Strength of Materials 224 YCT
 Critical stress, σ y = 240 MPa
According to maximum shear stress theory-
 σ1 − σ 2 σ − σ3 σ − σ1  σy
 or 2 or 3  =
 2 2 2  max 2 × N
80 240
=
428. Mohr's circle is used to determine the stresses 2 2× N
on an oblique section of a body subjected to : Factor of safety N = 3
(a) Direct tensile stress in one plane 431. The principle stresses σ1, σ2 and σ3 at a point
accompanied by a shear stress respectively are 80 MPa, 30 MPa and – 40
(b) Direct tensile stress in two mutually MPa. The maximum shear stress is:
perpendicular directions (a) 60 MPa (b) 55 MPa
(c) Direct tensile stress in two mutually (c) 35 MPa (d) 25 MPa
perpendicular directions accompanied by a TRB Polytechnic Lecturer 2017
simple shear stress Ans. (a) : Given as,
(d) All of the above σ1 = 80 MPa
Punjab PSC SDE 12.02.2017 σ2 = 30 MPa
Ans. (d) : Mohr's circle is used to determine the stresses σ3 = –40 MPa
on an oblique section of a body subjected to. We know that maximum shear stress will be,
(i) Direct tensile stress in one plane accompanied by a τmax. = Maximum of
shear stress.
(ii) Direct tensile stress in two mutually perpendicular  σ1 − σ 2   σ 2 − σ3   σ3 − σ1 
 , , 
directions.  2   2   2 
(iii) Direct tensile stress in two mutually perpendicular  80 − 30   30 − (−40)   −40 − 80 
directions accompanied by a simple shear stress. = Max. of  ,  , 
429. A shaft with a circular cross-section is  2   2 2 
subjected to pure twisting moment. The ratio τmax. = 60 MPa
of the maximum shear stress to the largest 432. Consider a plane stress case, where σx = 3 Pa,
principal stress is σy = 1 Pa and τxy = 1 Pa. One of the principal
(a) 2.0 (b) 1.0 directions w.r.t. x-axis would be
(c) 0.5 (d) 0 (a) 0º (b) 15º
GPSC Asstt. Director of Transport 05.03.2017 (c) 22.5º (d) 45º
Ans. (b) : BPSC AE Mains 2017 Paper - VI
Ans : (c) : Given,
σx = 3 Pa
σy = 1 Pa
τxy = 1 Pa
2τ xy 2 ×1
tan2θ = =
σx − σy 3 −1
=1
tan2θ = tan45°
θ = 22.5º
Pure twisting means pure shear, we have 433. An elastic body is subjected to a tensile stress
σ1 = τmax X in a particular direction and a compressive
Where, σ1 is largest principle stress. stress Y in its perpendicular direction. X and
So, Ratio is equal to 1. Y are unequal in magnitude. On the plane of
430. At a point in a bi-axially loaded member, the maximum shear stress in the body there will
be:
principal stresses are found to be 60 MPa and (a) No normal stress
80 MPa. If the critical stress of the material is (b) Also the maximum normal stress
240 MPa, what could be the factor of safety (c) The minimum normal stress
according to the maximum shear stress (d) Both normal and shear stress
theory? HPPSC Asstt. Prof. 18.11.2016
(a) 2 (b) 3
Ans. (d) : Maximum shear stress plane
(c) 4 (d) 5
Normal stress on maximum shear stress plane
JWM 2017
σ + σ2 σ x + σ y
Ans. (b) : Principle stress, σ1 = 80 MPa = 1 =
2 2
σ 2 = 60 MPa Shear stress on maximum shear stress plane
σ 3 = 0 MPa (τmax)in plane = ± Radius of Mohr's circle.

Strength of Materials 225 YCT

434. The maximum tangential stress acting on a
plane inclined to the direction of the axial px + py p x2 + p 2y − 2p x p y + 4p x p y
stress (P) in a tensile bar is equal to = ±
2 4
(a) P (b) 2P
2
(c) P/2 (d) 3P/2 p + py  p + py 
APPSC AEE Mains 2016 (Civil Mechanical) = x ±  x 
Ans. (c) : We know that maximum tangential stress 2  2 
occur at 45º, So the Axial Stress P/2→ Rmax for 45º. px + py px + py
435. If an element in a body is in equilibrium under σ1 = + = px + py
shearing stresses only, such a state of stress is 2 2
called px + py px + py
(a) Pure rotation σ2 = − =0
(b) Pure shear 2 2
(c) Pure bending 438. The maximum shearing stress produced by
(d) None of the given answers shrinkfit between two cylinders (mounted one
APPSC AEE Mains 2016 (Civil Mechanical) inside the other) of E=2×106 kg(f)/m2 and
Ans. (b) : When only two Normal stresses are present shrinkage factor = 0.002, is equal to
the state of stress is called Biaxial. In pure shear, the (a) 2000 kg(f)/m2 (b) 2500 kg(f)/m2
2
element is subjected to plane shearing stresses only. (c) 4000 kg(f)/m (d) 1000 kg(f)/m2
APPSC AEE Mains 2016 (Civil Mechanical)
Ans. (a) : j
439. The three dimensional state of stress at a point
is given by:
 60 20 −20
[σ] =  20 0 40  MN/m 2
 −20 40 0 
What is the shear stress in the xy-plane at the
436. An element is subjected to σx = 35 (Mpa same point?
(tensile) and σy = 20 MPa (tensile) and shear (a) –20 MN/m2 (b) –40 MN/m2
2
stress τ = 7.5 MPa. Then the direction of (c) 20 MN/m (d) 40 MN/m2
2
principal stresses is (e) 300 MN/m
(a) 45º (b) 22.5º CGPSC Asstt. Workshop Supt., 17.07.2016
(c) 30º (d) 15º Ans. (c) : As, we know,
APPSC AEE Mains 2016 (Civil Mechanical) Three dimensional state of stress, of a point is given by–
Ans. (b) : We know that,
σ x τxy τ xz   60 20 −20 
2τ τ 
tan 2θ = σy τyx  = 20 0 40 
σ1 − σ 2  yx 
2 × 7.5  τ zx τ xy σ z   −20 40 0 
tan 2θ = So, τ = τ = 20 MN/m 2
35 − 20 yx xy
tan 2θ = 1 440. When the principal stresses are unlike, in
2θ = 45º ⇒ θ = 22.5º Mohr's circle, the normal resultant stresses
437. If normal stresses of same nature px and py and will be
shear stress q are acting on two perpendicular (a) Negative for points on the circle to the left of
plans and q = (px py)1/2, then the major and the vertical axis through the origin
minor principal stresses respectively are (b) Negative for points on the circle to the right
(a) px + py and px – py of the vertical axis through the origin
(b) px and px – py (c) Positive for points on the circle to the left of
(c) 0.5 (px + py) and 0.5(px – py) the vertical axis through the origin
(d) px + py and zero (d) Either negative or positive depends on the
APPSC AEE Mains 2016 (Civil Mechanical) shear stress and radius of the Mohr's circle.
Ans. (d) : APPSC AE Subordinate Service Civil/Mech. 2016
Ans. (a) : Mohr's circle for unlike principle stresses–
σ1  p x + p y (p x − p y ) 2
= ± +q 2

σ2  2 4

Strength of Materials 226 YCT

When the principal stresses are unlike in Mohr's circle, σx = 55 N/mm2, σy = – 45 N/mm2, τxy = 50 N/mm2
the normal and resultant stresses will be negative for
55 + ( −45 )  55 − ( −45 ) 
2
points on the circle to the left of the vertical axis 2
through the origin. σ minor = −   + 50
441. Complementary shear stresses of intensity q
2  2 
are induced at a point on a body, the plane σ minor = 5 – 70.71 = –65.71 N/mm 2

carrying the maximum shear stress are at the σminor = 65.71 N/mm (compressive)
2

following angle with the principle planes 443. A body is subjected to tensile stress of 1200
(a) 450 (b) 300 MPa on one plane and another tensile stress of
(c) 600 (d) 900 600 MPa on a plane at right angles to the
APPSC AE Subordinate Service Civil/Mech. 2016 former. It is also subjected to a shear stress of
HPPSC Workshop Suptd. 08.07.2021 400 MPa on the same planes. The maximum
HPPSC Lect. 2016 shear stress will be
Ans. (a) : (a) 500 MPa (b) 1400 MPa
(c) 900 MPa (d) 400 MPa
APPSC AE Subordinate Service Civil/Mech. 2016
Ans. (a) :

From mohr circle for max shear

2 θ = 900
θ = 450
Alternatively by analytical method
σ − σ2
τxy = 1 × sin 2θ
2
For τxy to be max,
sin 2θ = 1 (Numerically)
2θ = 900 or 2700
θ = 450 or 1350 σ1 − σ2
442. The normal stresses of 55 N/mm2 tensile and τmax =
45 N/mm2 compressive on two mutually 2
perpendicular planes are acting at a point in a 2
σx + σy  σx − σy 
piece of elastic material. These planes also σ1 = σmajor = +  2
carry shear stresses of 50 N/mm2. The minor 2 2  + τ xy
 
principal stress will be
(a) 65.71 N/mm2 compressive 1200 + 600  1200 − 600 
2
2
(b) 75.71 N/mm2 tensile = +   + 400
(c) 65.71 N/mm2 tensile 2  2 
(d) 75.71 N/mm2 compressive = 900 + 500 = 1400 MPa
APPSC AE Subordinate Service Civil/Mech. 2016
σx + σy  σx − σ y  2
Ans. (a) : σ2 = σminor = −   − τ xy
2  2 
= 900 – 500 = 400 MPa
1400 − 400
τmax = = 500 MPa
2
444. For an inclined plane in a rectangular block
subjected to two mutually perpendicular
normal stresses 100 MPa and 400 MPa and
shear stresses 400 MPa, the maximum normal
stress will be
(a) 1200 MPa (b) 700 MPa
Sign conversion, Tensile = + ve (c) 600 MPa (d) 200 MPa
Comp. = –ve APPSC AEE Screening Test 2016
2
Ans. (*) : Given,
σx + σy  σx − σy  2 σx = 100 MPa
σmajor/minor = ±   + τ xy
2  2  σy = 400 MPa
Considering positive sign for major principle stress and τxy = 400 MPa
negative sign for minor principle stress. Maximum normal stress (σ1)max
Strength of Materials 227 YCT
 2 Ans : (c) Draw Mohr's circle like equal normal stress
σx + σ y  σx − σ y  2 without shear stress :-
σ1,2 = ±   + τ xy
2  2 

 (100 − 400 ) 
2
100 + 400
 + ( 400 )
2
= ± 
2  2 

= 250 ± (150 )2 + ( 400 )2

σ1 = 677.2 MPa (maximum normal stress)
σ2 = 177.2 MPa (minimum normal stress)
445. The principal strains at a point in a body under
biaxial state of stress are 1000 × 10–6 and –600
× 10–6. What is the maximum shear strain at
that point?
(a) 200 × 10–6 (b) 800 ×10–6
–6
(c) 1000 × 10 (d) 1600 × 10–6 In both condition Mohr's stress circle takes the shape of
RPSC 2016 a point.
ESE 2009 448. The state of stress at a point under plane stress
GATE 1996 conditions is σxx = 40 MPa, σyy = 100 MPa; τxy
Ans : (d) Given, = 40 MPa (notations have usual meaning).
The radius of the Mohr's circle representing
∈1=1600 × 10 –6
the given state of stress in MPa is
∈2= – 600 × 10–6 (a) 40 (b) 50
Maximum Shear strain is given by (c) 60 (d) 100
γ max ∈1 − ∈2 OPSC AEE 2019 Paper-I
= RPSC ADE 2016
2 2
∴ γ max =∈1 − ∈2 GATE 2012, ESE 2005
Ans. (b) : Radius of Mohr's circle = max shear stress
= 1000 × 10–6 – (–600×10–6) 2
= 1600 × 10–6  σx − σy  2
=   + τ xy
446. Mohr's circle construction is valid for both  2 
stresses as well as the area moment of inertia, 2
 40 − 100  2
because =   + 40
 2 
(a) both are tensors of first-order
= 50 MPa
(b) both are tensors of second- order
449. On principal plane the principal stress is :
(c) both are axial vectors
(a) Maximum/Minimum
(d) both occur under plane stress condition
(b) Zero
BPSC Poly. Lect. 2016 (c) Unity
Ans : (b) Mohr's circle construction is valid for both (d) Depends on applied strain.
stresses as well as the area moment of inertia because HPPSC Lect. (Auto) 23.04.2016
both are tensors of second order.
Ans. (a) : Principle plane is a plane on which principal
447. In which of the following two dimensional state stress is either maximum or minimum but shear stress
of stress, Mohr's stress circle takes the shape of on this is plane is zero. such plane is known as principal
a point. plane or plane of zero shear stress.
450. The Mohr’s circle for the following loading
condition has center coincide with the origin of
the normal stress-shear stress co-ordinate
system
(a) Uni-axial tension
(b) Uni-axial compression
(c) Bi-axial tension
UPPSC AE 12.04.2016 Paper-I (d) Pure torsion
CSE Pre-1996 GPSC Asstt. Prof. (Prod.) 21.08.2016
Strength of Materials 228 YCT
Ans. (d) : (c) θp = 33.7° and σmax = + 132 MPa, σmin = + 22
MPa
(d) θp = 33.7° and σmax = + 132 MPa, σmin = + 28
MPa
(e) None of these
CGPSC AE 16.10.2016
Ans. (d) : As we know that,
Principal stresses are :
σx + σy  σ − σy 
2

In case of pure torsion only shear force will come into σ1,2 = ±  x  + τ xy
2

play.
2  2 
Then centre of circle (Mohr's) will be origin of stress  2τxy 
shear stress co-ordinate system. And tan 2θp =  
451. A cylindrical elastic body subjected to pure  σx − σy 
torsion about its axis develops : 1  2 × 48 
So, θp = tan −1   = 33.7°
(a) Tensile stress in a direction 45° to the axis 2  100 − 60 
(b) No tensile or compressive stress 2
100 + 60 100 − 60 
± 
(c) Max shear stress along the axis of shaft
σ1,2 =  + 482
(d) Max shear stress at 45° to the axis 2  2 
HPPSC Asstt. Prof. 18.11.2016 σmax = 132 MPa σmin = 28 MPa
Ans. (a) : 453. At a point in two-dimensional stress system σx
= 100 N/mm2, σy = τxy = 40 N/mm2. What is the
radius of the Mohr circle for stress drawn with
a scale of : 1 cm = 10 N/mm2?
(a) 3 cm (b) 4 cm
(c) 5 cm (d) 6 cm
1 (e) None of the above
2 
σ1,2 = σ x + σ y ± ( σ x − σ y ) + 4τxy
2
CGPSC Poly. Lect. 22.05.2016
2 
ESE 2005
1
σ1,2 =  0 ± 4τ2xy  Ans. (c) : Given
2  σx = 100 N/mm2
σ1,2 = ± τxy σy = τxy = 40 N/mm2
2τ xy 2
tan(2θp ) = =∞  σx − σ y  2
σx − σy Radius Mohr's circle, (R) =   + τ xy

 2 
π
2θ = 2
 100 − 40 
 + ( 40 )
2 = 
2
θ = 45°  2 
452. For the state of plane surface shown,
= ( 30 ) + ( 40 )
2 2
determine the principal planes and the
principal stress? = 900 + 1600 = 2500
= 50 N/mm2
[∵ given, 1cm = 10N/mm2]
R = 5cm
454. A cantilever beam is loaded as shown here.
Which one of the Mohr's circle shown here
represents the stress element at point A?

(a) θp = 35.7° and σmax = + 132 MPa, σmin = + 28

MPa
(b) θp = 33.7° and σmax = + 122 MPa, σmin = + 28
MPa UPRVUNL AE 21.08.2016
Strength of Materials 229 YCT
Ans. : (d) (a) 3 (b) 2.5
(c) 7.5 (d) 1.25
VIZAG MT 2015
Ans. (a) : Given,
σ x = 100N / mm 2
σ y = 40N / mm 2
τxy = 40N / mm 2
2
σx + σy  σ − σy 
(Mohr's circle diagram) σ1,2 = ±  x  + τ xy
2

455. A body is subjected to a pure tensile stress of 2  2 

200 units. What is the maximum shear 2
produced in the body at some oblique plane 100 + 40  100 − 40 
= ±   + (40)
2
due to the above? 2  2 
(a) 0 units (b) 50 units
(c) 100 units (d) 150 units = 70 ± (30)2 + (40) 2
(e) 200 units σ1,2 = 70 ± 50
CGPSC AE 26.04.2015 Shift-I σ1 = 120N / mm 2
Ans. (c) : Maximum shear stress
2
σ 2 = 20N / mm 2
 σ −σ y 
τ max =  x  + τ xy
2
σ1 − σ2 Syt
2 ≤
  2 2FOS
here σy = 0, τxy = 0, σx = 200 units 300
σ 200 (120 − 20 ) ≤
τ max = x = = 100 units FOS
2 2 300
456. Plane stress at a point in a body is defined by FOS =
principal stresses 3σ and σ. The ratio of the 100
normal stress to the maximum shear stresses FOS = 3
on the plane the maximum shear stress is : 458. A machine element is subjected to the normal
(a) 1 (b) 2 stress of 80 MPa and shear stress of 60 MPa.
(c) 3 (d) 4 The yield strength of material is 320 MPa. The
TSPSC Managers, 2015
factor of safety using maximum shear stress
ESE 2000
theory will be
Ans. (b) : (a) 2.85 (b) 4
(c) 2.29 (d) none of these
VIZAG MT 2015
Ans. (c) : Given σ x = 80MPa
σy = 0
τ xy = 60MPa
2
80 − 0  80 − 0 
 + ( 60 )
2
σ1,2 = ± 
2  2 
3σ − σ σ1 = 112.11N / mm 2
Maximum shear stress (τmax) = =σ
2 σ 2 = −32.11N / mm 2
3σ + σ σ1 − σ2 Syt
Normal stress (σ n ) = = 2σ ≤
2 2 2FOS
(At maximum shear stress) Now,
112.11 − ( −32.11) 320
σn 2σ ≤
= =2 2 2FOS
τmax σ
320
457. The state of stress at a point is given as FOS =
144.22
σ x = 100N/mm 2
FOS = 2.21 ≈ 2.29
σ y = 40N/mm 2 The closest answer is option (c).
and τ xy = 40N/mm 2 459. For any state of stress it is always possible to
If the yield strength Syt of the material is 300 define a new co-ordinate system, which has
MPa, the factor of safety using maximum shear axis perpendicular to the planes on which the
stress .......
Strength of Materials 230 YCT
 (a) maximum compressive stress act Ans : (d)
(b) maximum shear stress act
(c) maximum normal stress act
(d) None of the above
KPSC ADF 2015
Ans. (b) : The orientations of maximum and minimum
shear planes are at right angles to each other.
Maximum shear stress, it can be seen that the planes of
maximum and minimum shear stress occur at an angle
of 450 and 1350 with respect to the plane of maximum
and minimum normal stress respectively.
460. The co-ordinate of any point on Mohr's circle
represent :  σx + σ y   σ x − σ y 
(a) State of stress at a point with reference to any ( σ n )max =  +  cos 2θ + τxy sin 2θ
arbitrary set of orthogonal axes passing  2   2 
through that point ( σn )max = Maximum Normal stress
(b) Principal stresses at a point
(c) One of the two direct stresses and shearing τxy = Shear stress
stress at a point Given σ x = σ y = p and θ = 450
(d) Two direct stresses at a point
(HPPSC AE 2014) σx = p, σ y = −p
Ans : (c) The coordinate of any point on Mohr's circle p−p p+p
represent one of the two direct stresses and shearing ( σn )max =  +
0
 cos 90 − 0
stress at a point.  2   2 
461. Ellipse of stress can be drawn only when a ( σn )max = 0
body is acted upon by :
(a) one normal stress 464. The shear strength of every materials tends to
(b) two normal stresses be about ____ their tensile strength
(c) one shear stress (a) half (b) equal
(d) two normal stresses and one shear stress (c) double (d) one third
(HPPSC AE 2014) MPSC HOD (Govt. Poly. Colleges) 04.10.2014
Ans : (b) Ellipse of stress is used to find resultant stress Ans. (a) : Shear strength
and the angle of obliquity on any plane within a stressed σ yt
body. In 2-D, it is called Ellipse of stress. In 3-D it is τsy = = 0.5σ yt (Guest & Tresca's theory)
called Ellipsoid of stress. 2
The axis of ellipse are the two principle stresses. σ yt
τsy = = 0.577σ yt (Von-Mises theory)
462. State of plane stress in x-y plane is 3
accompanied by strains along: 465. Strain gauge transducers are used to measure :
(a) x, y and z- axes (b) x and y- axes (a) Tensile Torsion and Compressive forces
(c) x and z- axes (d) y and z- axes (b) Tensile forces alone
HPPSC Asstt. Prof. 2014 (c) Compressive forces alone
Ans. (a) : (d) Torsional forces alone
TNPSC AE (Industries) 09.06.2013
Ans. (a) : Strain gauge transducers are used to measure
tensile torsion and compressive forces.
466. The principal stresses at a point across two
perpendicular planes are 75 MN/m2 (Tensile)
and 35 MN/m2 (Tensile). Find the tangential
stresses stress and it's obliquity on a plane at 20º with
In plane stress condition, σx ≠ 0, σy ≠ 0, but σz = 0 the major principal plane :
(a) 40 sin 40° (b) 20 sin 40°
and strains
(c) 40 sin 20° (d) 20 sin 20°
εx ≠ 0, εy ≠ 0 & εz ≠ 0
TNPSC AE (Industries) 09.06.2013
i.e. stain along x, y and z all exist.
Ans. (b) : Given,
463. Normal stresses of equal magnitude “p” but of σ = 75 ΜPa (tensile)
x
opposite signs, act at a point of strained σ = 35 ΜPa (tensile)
y
material in perpendicular direction. What is
the magnitude of the resultant normal stress on τ =  σ x − σ y  sin 2θ
plane inclined at 450 to the applied stresses?  2 
(a) 2p (b) p/2
 75 − 35 
(c) p/4 (d) 0 =  sin ( 2 × 20 ) = 20 sin 40°
UPRVUNL AE 2014, ESE 2005  2 
Strength of Materials 231 YCT
467. The normal stresses on two perpendicular 470. A shaft subjected to torsion experiences a pure
planes through a point in a stressed material shear stress on the surface. The maximum
are 80 MPa (tensile) and 20 MPa (tensile) principle stress on the surface which is at 45°
respectively. A shear stress of 40 MPa is also to the axis will have a value
acting on these planes. What is the value of (a) τcos45° (b) 2τcos45°
maximum shear stress at the point? (c) 2τsin45°cos45° (d) τcos245°
(a) 50 MPa (b) Zero Vizag Steel MT (Re-Exam) 24.11.2013
(c) 40 MPa (d) 60 MPa GATE 2003
RPSC AEN Pre-2013 Ans. (c) : It is a pure shear stress case,
GATE 2015 σx = σy = 0 and τxy = τ and θ = 45°
Ans. (a) : σx = 80 MPa (Tensile) Maximum principal stress,
σy = 20 MPa (Tensile), τxy = 40 MPa (0 + 0)  0 − 0 
Maximum shear stress τmax in plane– σp = +  cos(2 × 45°) + τ× sin(2 × 45°)
2  2 
 σ − σy 
2 σp = τ.sin2 × 45°
τmax =  x  + ( τ xy ) 2
σp = 2τ.sin45°.cos45° (sin2θ = 2.sinθ.cosθ)
 2  ∴ The resulting normal stress (tensile or compressive)
2 on planes inclined at 45° to the direction of the shear
 80 − 20  stresses is equal to τ.
=   + (40)
2

 2  471. If the principal stresses at a point in a strained

= (30) 2 + (40) 2 body are σx and σ y (σ x > σ y ), resultant stress
= 50 MPa on a plane carrying the maximum shear stress
is equal to
468. A solid circular shaft is subjected to a
maximum shear stress of 140MPa. Magnitude (a) σ2x + σ2y (b) σ2x − σ2y
of maximum normal stress developed in the
shaft is:- σ 2x + σ 2y σ 2x − σ 2y
(a) 60 MPa (b) 90 MPa (c) (d)
2 2
(c) 110 MPa (d) 140 MPa APPSC AEE 2012
UKPSC AE-2013, Paper-I
Ans : (c) If the principal stresses at a point in strained
CSE Pre-1995
APPSC AEE 2012 body are σ x and σ y then the resultant stress on a plane.
APPSC AE 04.12.2012 Center of Mohr's Circle
Ans. (d) : Maximum normal stress,
σx + σ y 1
( )
2
σ1,2 = ± σ x − σ y + 4τ2xy
2 2
σx = 0, σ y = 0, τxy = 140 MPa
1
4 × (140 )
2
σ1,2 = 0 ±
2
σ1,2 = ±140 MPa
Then, Maximum normal stress is 140 MPa.  σx + σ y 
469. The state of plane-stress at a point is given by  ,0 
σx = –200 MPa, σy = 100 MPa and τxy = 100  2 
MPa. The maximum shear stress in MPa is Resultant stress on a plane
(a) 111.8 (b) 150.1 σR = σn2 + τmax
2

(c) 180.3 (d) 223.6 2 2

Vizag Steel MT (Re-Exam) 24.11.2013  σx + σ y   σx − σ y 
σR =   + 
GATE 2010
 2   2 
Ans. (c) : Given, σx = –200 MPa, σy = 100 MPa
τxy = 100 MPa  σ2 + σ2y + 2σx σ y + σ2y + σ2y − 2σx σ y 
σR =  x 
2
 4 
 σ − σy 
τmax =  x  + τ xy
2

 2  2  σ2x + σ2y  σ2x + σ2y

2
σR = =
−200 − 100  4 2
τmax =   + (100 )
2
472. Angle between the principal planes is
 2 
(a) 2700 (b) 1800
= 150 2 + 100 2 (c) 900 (d) 450
APPSC AEE 2012
τmax = 180.3 MPa Ans : (c) Angle between the principal plane is 900.
Strength of Materials 232 YCT
473. σx + σy = σx' + σy' = σ1 + σ2 (a) The principal stresses at a point are
The above relation is called  σ − σ  2 
σ1 + σ2 2
(a) independency of normal stresses P1 ,P2 = ±  1 2
 +q 
(b) constancy of normal stresses 2  2  
(c) first invariant of stress (b) The position of principal planes with the
(d) all the above three plane of stress σ 1, are
UKPSC AE 2012 Paper-I 1 −1 2q
θ1 = tan ; θ2 = θ1 + 45°
Ans. (c) : first invariant of stress 2 σ1 − σ2
474. When a body is subjected to direct tensile (c) Maximum shear stress is (σt)max =
stresses (σx and σy) in two mutually
 σ − σ  2 
perpendicular directions, accompanied by a ±  1 2
 +q 
2
simple shear stress τxy, then in Mohr’s circle  2  
method, the circle radius is taken as (d) Planes of maximum shear are inclined at 45°
σx − σ y to the principal planes.
(a) + τxy UKPSC AE 2012 Paper-I
2
Ans. (b) : The position of principal planes with the
σx + σ y
(b) + τxy plane of stress σ1, are
2 1 2q
θ1 = tan −1 ; θ2 = θ1 + 45°
1
( ) σ1 − σ2
2
(c) σx − σ y + 4τ2xy 2
2 477. A tension member with a cross-sectional area
of 30 mm2 resists a load of 60 kN. What is the
1
( )
2 2
(d) σx + σ y + 4τxy normal stress induced on the plane of
2
UKPSC AE 2012 Paper-I maximum shear stress ?
(a) 2 kN/mm2 (b) 1 kN/mm2
2
(d) 3 kN/mm2
1
( ) (c) 4 kN/mm
2
Ans. (c) : R = σx − σ y + 4τ2xy
2 UKPSC AE 2012 Paper-I
475. Choose the correct relationship in the given Ans. (b) : Given, A = 30 mm2
statements of Assertion (A) and Reason (R). P = 60 kN
P 60
Assertion (A) : A plane state of stress does not σx = = = 2 N/mm2
necessarily result into a plane state of strain. A 30
Reason (R) : Normal stresses acting along X and σy = 0
Y directions will also result into strain along the σx − σ y 2−0
Z-direction. τmax = = = 1kN / mm 2
2 2
Code : 478. In an axially loaded bar of magnitude P and
(a) Both (A) & (R) are correct. (R) is the correct cross sectional Area A, the shear stress on an
explanation of (A). inclined plane at an angle θ is
(b) Both (A) & (R) are correct. (R) is not the (a) Zero (b) P/A
correct explanation of (A). (c) P/A cos (2 θ) (d) P/A sin θ cos θ
(c) (A) is true, but (R) is false. Jharkhand Urja Vikas Nigam Ltd. AE 2017
(d) (A) is false, but (R) is true. Ans. (d) : The shear stress
UKPSC AE 2012 Paper-I
Ans. (a) : Both (A) & (R) are correct. (R) is the correct
explanation of (A).
476. A body is subjected to two unequal like direct σx − σy
stresses σ1 and σ2 in two mutually perpendicular τ= sin 2θ + τxy cos 2θ
2
planes along with simple shear stress q
P  P 
τ= sin 2θ σx = , σ y and τxy = 0 
2A  A 

P
( 2sin θ.cos θ )
τ=
2A
P
τ = sin θ.cosθ
A
479. If the Mohr’s circle for a state of stress
becomes a point, the state of stress is
Which among the following is then a wrong (a) Pure shear state of stress
statement ? (b) Uniaxial state of stress
Strength of Materials 233 YCT
 (c) Identical principal stresses Ans. (a) : State of stress
(d) None of the above Mohr's circle diagram
UKPSC AE 2012 Paper-I
Ans. (c) : Identical principal stresses
480. In a stressed field, the change in angle between
two initially perpendicular lines is called
(a) Normal strain (b) Shear strain
(c) Principal strain (d) Poisson’s ratio Normal stress on plane of maximum shear stress
UKPSC AE 2012 Paper-I = Normal stress at the centre of mohr circle
Ans. (b) : Shear strain = Radius of Mohr circle
100 − 50 50
481. A body is subjected to a pure tensile stress of = =
100 units. What is the maximum shear 2 2
produced in the body at some oblique plane = 25 MPa
due to the above? 484. A short cast iron cylindrical piece of cross-
(a) 100 units (b) 75 units sectional area 100 mm2 and length 25 mm is
(c) 50 units (d) 0 unit tested in compression and fails at a load of 50
TRB Asstt. Prof., 2012 kN. The shear strength of cast iron is.
ESE 2006 (a) 500 N/mm2 (b) 400 N/mm2
2
Ans. (c) : (c) 300 N/mm (d) 250 N/mm2
UPSC JWM Advt. No.-50/2010
Ans. (d) : As we know,
Maximum shear stress given by–
σ = 100 units σ − σ2
τmax = 1 ⇒ ∴ σ2 = 0
σ 100 2
Max shear stress, τmax = = σ 50 × 103
2 2 τmax = 1 = = 250 N / mm 2 [σ1 = P/A]
τmax = 50 Units 2 2 × 100
482. At a point on a plane there is normal stress + 485. The two-dimensional state of stress at a point
100 MPa and a shear stress 'q', on another is σx = 100 N/mm2, σy = 20 N/mm2 and τzy. If
the larger principal stress at the point is 110
plane perpendicular to this plane there is N/mm2, the smaller principal stress will be
normal stress + 20 MPa and a shear stress 'q'. (a) 10 N/mm2 (b) 15 N/mm2
2
If the maximum principal stress at the point is (c) 16 N/mm (d) 18 N/mm2
110 MPa, what is maximum shear stress at the UPSC JWM Advt. No.-50/2010
point : Ans. (a) : Given, σx = 100 N/mm2, σy = 20 N/mm2
(a) 50 MPa (b) 30 MPa σ1 = 110 N/mm2, σ2 = ?
(c) 10 MPa (d) None of these σx + σ y 2
 σ − σy 
PSPCL AE, 2012 σ1 = +  x  + τxy
2

Ans. (a) : Max principal stress, 2  2 

σx + σ y 100 + 20 100 − 20 
2

σ1 = + radius of mohr circle 110 = +   + τxy

2

2 2  2 
100 + 20 τxy = 30 N/mm2
110 = + Radius
2 σx + σ y  σ − σy 
2

Radius of mohr's circle = max. shear stress = 50 MPa σ2 = −  x  + τxy

2

2  2 
483. For the state of stress shown in the above
figure, normal stress acting on the plane of = 60 − 402 + 30 2 = 60 − 50
maximum shear stress is σ2 = 10 N / mm 2
486. Which one of the following represents the state
of stress on a surface element of a shaft
subjected to a torque?

(a)
(a) 25 MPa tension
(b) 75 MPa compression
(c) 25 MPa compression
(d) 75 MPa tension
ISRO Scientist/Engineer 2011 (b)
ESE 2015
Strength of Materials 234 YCT
 Ans. (d) : A strain gauge only measures strain in one
direction in order to get principal strains it is necessary
to use a strain rosette. A strain rosette is a cluster of 3
(c) strain gauges oriented at different angles. A strain gauge
rosette is an arrangement of two or more closely
positioned the normal strains along different directions
in the underlying surface of the test part.
490. Principal planes are the planes, on which the
resultant stress is the
(a) shear stress (b) normal stress
(d) (c) tangential stress (d) none of these
TNPSC AE, 2008
Ans. (b) : Principal planes are the planes, on which the
UPSC JWM Advt. No.-50/2010 shear stress is the zero and resultant stress is the normal
Ans. (d) : Pure shear stress only – stress.
491. If at a point in a body σx = 70 MPa, σy = 60
MPa and τxy = –5 MPa then the radius of the
Mohr's circle is equal to
(a) 5 5 MPa (b) 2 5 MPa
As we know, when a shaft subjected to torque should (c) 5 2 MPa (d) 25 MPa j
have only shear stress & not any normal stress. DRDO Scientists 2008
So, the correct option is (d). Ans. (c) : Radius of Mohr circle
487. Which one of the following statements is true? 2
(a) In the principal planes, both normal and  σx − σ y 
=   + τ xy2
shear stresses are always zero 2 
(b) In the principal planes, shear stress may or 
may not be zero and in the maximum shear  70 − 60 
2

planes, normal stresses may or may not be =   + 25

zero  2 
(c) In the principal planes, shear stress is always = 5 2 MPa
zero and in the maximum shear planes,
normal stresses are always zero 492. For a body subjected to direct stresses σx, σy
(d) In the principal planes, shear stresses are and σz the direct strain εx is x direction is
always zero but in the maximum shear (where E and γ are Young's Modulus and
planes, normal stresses may or may not be Poisson's ratio respectively)
zero 1
DRDO Scientists 2009 (a) ε x = {σ x + γ (σ y − σ z )}
Ans. (d) : In the principal planes, shear stresses are E
always zero but in the maximum shear stress planes, 1
(b) ε x = {σ x − γ (σ y + σ z )}
normal stress may or may not be zero. E
488. At a point on a component loaded under plane 1
stress condition, the stresses are σx = 120 MPa, (c) ε x = {σ x − γ (σ y − σ z )}
E
σy = 40 MPa and τxy = 30 MPa. The possible 1
principal stresses are (d) ε x = {σ x + γ (σ y + σ z )}
(a) 130 MPa and –30 MPa E
(b) 150 MPa and –50 MPa ISRO Scientist/Engineer 2007
(c) 130 MPa and 30 MPa Ans. (b) :
(d) 150 MPa and 50 MPa
DRDO Scientists 2009
2
σx + σ y  σx − σy 
Ans. (c) : σ1 , σ 2 = ±   + τxy
2

2  2 
2
120 + 40  120 − 40 
= ±   + 30
2

2  2  E and γ are young's modulus and Poisson ratio.

σ1 = 130 MPa; σ2 = 30 MPa So, strain in 3-direction in a body.
489. A Rosette gauge is employed for the 1
measurement of ε x = [σ x − γ (σ y + σ z )]
(a) absolute pressure E
(b) low pressure variations 1
(c) strain in one direction ε y = [σ y − γ (σ x + σ z )]
E
(d) strain in more than one direction 1
TNPSC AE, 2008 ε z = [σ z − γ (σ x + σ y )]
TNPSC AE 2017 E
Strength of Materials 235 YCT
493. The state of stress at a point in a loaded
member is of pure shear stresses of ± 800 MPa 3. Shear Force and Bending
on two given planes at right angles to each
other. The principal stresses are
Moment Diagram
(a) 800 MPa and – 800 MPa 497. The point of contraflexure is a point, where
(b) 800 MPa and 800 MPa (a) shear force changes sign
(c) – 800 MPa and – 800 MPa (b) bending moment changes sign
(d) 1600 MPa and – 1600 MPa (c) shear force is maximum
OPSC Civil Services Pre 2006 (d) bending moment is maximum
Ans. (a) : APPSC AEE SCREENING 17.02.2019
GPSC Asstt. Prof. 28.08.2016
Rajasthan Nagar Nigam AE 2016, Shift-I
KPCL AE 2016, ESC 2005
GPSC Lect. 23.10.2016
VIZAG MT 2015
ISRO Scientist/Engineer 11.10.2015
Max principal stress OA = + 800 MPa KPSC ADF 2015
Min. principal stress OB = – 800 MPa ISRO Scientist/Engineer 24.05.2014
494. If principal stress of a point in plane stressed Haryana PSC Civil Services Pre, 2014
element are σx = σy = 5000 N/cm , then the 2 TRB Asstt. Prof., 2012
TNPSC ACF 2012
normal stress on the plane inclined at 45° to x- TNPSC AE, 2008
axis.
(a) 0 (b) 5000 N/cm 2 Ans. (b) : Point of contraflexure– In a beam if the
(c) 7070 N/cm 2
(d) 500 N/cm 2 bending moment changes its sign at a point, the point it
self having zero bending moment, the beam changes
WBPSC AE 2003 curvature at this point of zero bending moment and this
Punjab PSC (Lect.) 06.08.2017 point is called point of contraflexure.
Ans. (b) : Given, σx = σy = 5000 N/cm2 498. The bending moment diagram for a cantilever
Normal stress on the plane at 45° with U.D.L. over the complete span will show
(σn)θ = σ x cos 2 θ + σ y sin 2 θ [also σ n = σ cos 2 θ ] the curve of a (an)
= σx cos (45°) + σy sin (45°)
2 2 (a) Ellipse (b) Rectangle
2 2 (c) Triangle (d) Parabola
 1   1  Vadodara Muncipal Corp. DEE, 2018
= 5000 ×   + 5000 ×   RPSC ADE 2016
 2  2
2 GPSC Lect. 23.10.2016
= 5000 N/cm
Rajasthan Nagar Nigam AE 2016, Shift-I
495. The maximum shear stress induced in a APPSC AEE Mains 2016 (Civil Mechanical)
member which is subjected to an axial load is Kerala PSC IOF 19.04.2016
equal to MPPSC State Forest Service Exam, 2014
(a) Maximum normal stress Mizoram PSC AE/SDO 2014, Paper-II
(b) Twice the maximum normal stress APPSC AEE 2012
(c) Thrice the maximum normal stress J&K PSC Civil Services Pre, 2010
(d) Half of maximum normal stress GATE 2001
(e) None of the above Ans. (d)
CGPSC AE 16.10.2016
HPPSC AE (PWD) 10.09.2016
Ans. (d) : We know that shear stress on an oblique
plane at an angle 'θ' to the cross-section of a body,
which is subjected to a direct tensile stress 'σ' is equal to
σ
sin 2θ .
2
Thus maximum value of θ = 45°.
σ σ σ − σ2
τmax = sin 90° = = 1
2 2 2
496. The maximum shear due to an axial
compression of 100 MPa is
(a) 50 MPa (b) 100 MPa
(c) 150 MPa (d) 75 MPa
APPSC AE Subordinate Service Civil/Mech. 2016
Ans. (a) : Given – Axial compression = 100 MPa
σ 100 Cantilever beam with uniformly distributed load
τmax = = = 50 MPa
2 2 (W/unit length).

Strength of Materials 236 YCT

499. A concentrated load of P acts on a simply 501. The point of contraflexure occurs in :
supported beam of span L at a distance L/3 (a) Cantilever beam only
from the left support. The bending moment at (b) Simply supported beam only
the point of application of the load is given by : (c) Overhanging beam only
(a) PL/3 (b) 2 PL/3 (d) Continuous beam only
(c) PL/9 (d) 2PL/9 HPPSC ADF 05.03.2019
ISRO Scientist/Engineer 22.04.2018 OPSC AEE 2019 Paper-I
GPSC ARTO Pre 30.12.2018 RPSC Vice Principal ITI 2018
GPSC Executive Engineer 23.12.2018 HPPSC Asstt. Prof. 20.11.2017
Karnataka PSC Lect. 27.05.2017 WBPSC AE, 2017
TSPSC AEE 2017 UPPSC AE 2016
ISRO Scientist/Engineer (RAC) 07.05.2017 UKPSC AE-2013, Paper-I
GPSC AE 26.07.2015 APPSC AEE 2012
GATE 2003 Ans. (c) : The point of contraflexure occurs mainly in
2L overhanging beam in which there is opposite magnitude
P× moment acting on another side of overhanging part
Ans. (d) : R A = 3 = 2P
which makes the net moment to change its sign, by
L 3 virtue of this the point of contraflexure come into
L picture.
P×
RC = 3 =P But it is not always in overhanging beam it can
L 3 also occur in simply supported and fixed beam and
point of contraflexure always occurs in fixed beam.
2P L 2PL
MB = × = 502. A simply supported beam of length l carries a
3 3 9 uniformly distributed load of w per unit
length. It will have maximum bending moment
at midpoint of beam and the value will be:
wl 2 wl 2
(a) (b)
2 4
wl 2 wl 2
(c) (d)
6 8
wl 2
(e)
16
CGPSC AE 15.01.2021
ISRO Scientist/Engineer (RAC), 10.03.2019
(CGPSC Polytechnic Lecturer 2017)
CGPSC AE 26.04.2015 Shift-I
UKPSC AE-2013, Paper-I
500. When shear force at a point is zero, then J & K PSC Screening, 2006
bending moment is __________ at the point. Ans. (d) : We know that
(a) Zero (b) Minimum wl
(c) Maximum (d) None of the above RA = RB =
2
Assam PSC AE (IWT) 14.03.2021
RPSC IOF, 2020
Nagaland PSC (CTSE) 2018, Paper-I
WBPSC AE, 2017
Sikkim PSC (Under Secretary), 2017
APPSC AE Subordinate Service Civil/Mech. 2016
Haryana PSC Civil Services Pre, 2014 Taking a section x-x from a point A at a distance x.
MPSC HOD (Govt. Poly. Colleges) 04.10.2014 Then taking moment at x-x
APPSC Poly. Lect. 2013 x
Mx-x = R A x − wx ⋅
UKPSC AE-2013, Paper-I 2
APPSC AEE 2012 wl x2
Ans. (c) : Mx-x = x−w
2 2
Relation between shear force (V) & bending moment For max. bending moment
(M) dM x − x
=0
dM dx
V=
dx wl x
− 2w = 0
If shear force is zero (V = 0) means first derivative of 2 2
bending moment is zero means at that point bending w l
= wx
moment will be maximum. 2
Strength of Materials 237 YCT
 l 505. A beam is loaded as cantilever, if the load at
x= the end is increased, then the failure will occur
2
(a) in the middle
l
At x = [at mid] bending moment will be max. so (b) at the tip below the load
2 (c) anywhere
wl l w × l 2 (d) at the support
M l= × −
x=
2 2 2 4 × 2 Vizag Steel (MT) 2017
wl 2
wl 2 Karnataka PSC Lect., 27.05.2017
Ml = − Rajasthan AE (Nagar Nigam) 2016 Shift-3
2
4 8
UKPSC AE-2013, Paper-I
wl 2 RPSC AE GWD, 2011, CSE PRE -2005
Ml / 2 =
8 Ans. (d) : A beam is loaded as cantilever, if the load at
503. At point of contraflexure in beam subjected to the end is increased then the failure will occur at the
loading : support.
(a) Shear force is zero
(b) Bending moment is zero
(c) Bending moment is maximum
(d) Shear force is unity
Punjab PSC (Lect.) 06.08.2017
HPPSC AE (PWD) 10.09.2016 At the fixed end shear force and bending moment is
HPPSC Lect. (Auto) 23.04.2016
TNPSC AE (Industries) 09.06.2013 maximum then failure will occur at the fixed end.
APPSC AE 04.12.2012 506. Maximum bending moment in a cantilever
APPSC AEE 2012 carrying a concentrated load at the free end
TSPSC AEE 28.08.2017 (Civil/Mechanical) occurs
DRDO Scientists 2008 (a) at the fixed end (b) at the free end
Ans. (b) : A point of contraflexure is a point where the (c) at the mid span (d) None of these
curvature of the bending moment changes sign. It is
HPPSC Workshop Suptd. 08.07.2021
sometimes referred to as a point of inflection and will
be shown later to occur at the point or points on the GPSC Lect. 23.10.2016
beam where the bending moment is zero. HPPSC LECT. 2016
504. Consider the following statements : APPSC AEE 2012
If at section distance from one of the ends of WBPSC AE 2008
the beam, M represents the bending moments. Ans : (a)
V the shear force and w the intensity of
loading, then
dM dV
1. =V 2. =w
dx dx
dw
3. = y (deflection of the beam at the
dx
section)
Of these statements
(a) 1 and 3 are correct
(b) 1 and 2 are correct
(c) 2 and 3 are correct
(d) 1, 2 and 3 are correct
(e) None of the above
Haryana PSC AE (PHED) 05.09.2020, Paper-II Maximum bending moment in a cantilever beam
SJVN ET 2019, ISRO Scientist/Engineer 2011 carrying a concentrated load at the free end occurs at the
CGPSC Poly. Lect. 22.05.2016 fixed end.
UKPSC AE 2012 Paper-I 507. Bending moment M and torque T is applied on
APPSC AEE 2012 a solid circular shaft. If the maximum bending
Ans. (b) : Bending moment = M stress equals to maximum shear stress
Shear force = V
developed, then M is equal to :
load intensity = w
(a) T/2 (b) T
dM (c) 2T (d) 4T
shear force (V) =
dx RPSC IOF, 2020
dV Vadodara Muncipal Corp. DEE, 2018
load intensity (w) =
dx PSPCL AE, 2012
Strength of Materials 238 YCT
Ans. (a) :Bending equation
σ M E
= =
y I R

M
σ= y
I
M d
= 4×
πd 2
64
32M 509. A simply supported beam of span (l) carries a
σ max = ….(i) point load (W) at the centre of the beam. The
πd 3
Torsion equation shear force diagram will be :
(a) a rectangle
T Gθ τ
= = (b) a triangle
J ℓ R (c) two equal and opposite rectangles
(d) two equal and opposite triangles
TSPSC Manager (Engg.) HMWSSB 12.11.2020
CGPSC AE 16.10.2016
TR HPPSC W.S. Poly. 2016
τ= Ans : (c) A simply supported beam of span (l) carries a
J point load (W) at the centre of the beam. The shear
T d force diagram will be two equal and opposite
= ×
π 4 2 rectangles.
d
32
16T
τmax = 3 ….(ii)
πd
According to question.
(σ) max = τmax
(∵ given, τmax = σmax)
32M 16T
= 3
πd 3 πd
16T
M=
32
M = T/2
508. If the shear force acting at every section of a
beam is of the same magnitude and of the same
direction then it represents a :
(a) Simply supported beam with a concentrated 510. A mass less beam has a loading pattern as
load at the centre. shown in Fig. The beam is of rectangular cross-
(b) Overhung beam having equal overhang at section with a width of 30 mm and height of 100
both supports and carrying equal mm
concentrated loads acting in the same
direction at the free ends.
(c) Cantilever subjected to concentrated load at
the free end.
(d) Simply supported beam having concentrated
loads of equal magnitude and in the same
direction acting at equal distance from the
supports.
GPSC ARTO Pre 30.12.2018 the maximum bending moment occurs at
HPPSC Asstt. Prof. 2014 (a) Location B
APPSC AEE 2012 (b) 2500 mm to the right of A
APPSC IOF, 2009 (c) 2675 mm to the right of A
ESE 1996 (d) 3225 mm to the right of A
Ans. (c) : In a cantilever subjected to concentrated load UPPSC AE 13.12.2020, Paper-I
at the free end, the shear force at every section of a MPPSC AE 2016
beam is of same magnitude and of same direction. GATE 2010
Strength of Materials 239 YCT
Ans : (b)

Bending moment in the portion BC of the beam is a

non-zero constant.
RA + RC = 6000N 513. A uniform rigid rod of mass 'm' and length 'L'
Taking moment about A is hinged at one end as shown in the figure. A
6000 × 3 – RC × 4 = 0 2L
RC = 4500 N force 'P' is applied at a distance of from
3
RA = 1500 N the hinge so that the rod swings to the right.
Taking any section from A The reaction at the hinge is :
Shear force = 1500 – 3000 (x – 2)
and maximum bending moment occur where shear
force = 0
1500 – 3000 (x – 2) = 0
x = 2.5 m
or 2500 mm from A
511. The moment diagram for a cantilever beam
whose free end is subjected to a bending
moment
(a) Triangle (b) Rectangle
(c) Parabola (d) Cubic parabola (a) – P (b) 0
Vizag Steel (MT) 2017
Nagaland PSC CTSE 2016, Paper-I (c) P/3 (d) 2P/3
WBPSC AE 2003 OPSC Civil Services Pre. 2011
Ans. (b) : GATE-2009
Ans. (b) :

The shear force will be zero all along the span.

The bending moment will be constant all along the span
and equal to M.
512. The given figure shows the shear force diagram
for the beam ABCD. Bending moment in the
portion BC of the beam FBD

(a) is zero
(b) varies linearly from B to C
(c) parabolic variation between B and C
(d) is a non–zero constant
APPSC AEE 2012
ISRO Scientist/Engineer 2008
Ans : (d)

Considering rotational moment about support

T=I×α
2L mL2
P× = ×α
3 3
Strength of Materials 240 YCT
 a (b) a cantilever having udl over its whole length
α= (c) simply supported beam with a concentrated
r
load at its mid-length
 mL2 L mL2  (d) simply supported beam having udl over its
 Iendpoint = + mK 2 , K = , I =  whole length.
 12 2 3 
L 2a TSPSC AEE 28.08.2017 (Civil/Mechanical)
and r = therefore, α = UPSC JWM Adv. No-16/2009, CSE PRE 2005
2 L Ans. (b) : For a cantilever beam having uniform
2
2L mL 2a distributed load (UDL) over its whole length, the shear
P× = ×
3 3 L force (SF) diagram is a right angled triangle having its
P base representing the length of the beam.
a=
m
considering horizontal equilibrium
⇒ P – RH = m × a
P
⇒ −R H + P = m ×
m
RH = 0
514. The maximum bending moment in a simply
supported beam carrying total load W
uniformly distributed on the entire span ℓ, is
equal to
(a) Wℓ/2 (b) Wℓ/3
(c) Wℓ/4 (d) Wℓ/8 Shear Force (Fx) = wx
at x = 0 ; FB = 0
WBPSC AE, 2017 & at x = L ; FA = wL
Mizoram PSC AE/SDO 2014, Paper-II
Ans. (d) : 516. A simply supported beam of span l carries a
uniformly variable load of intensity w0x over
its entire span. Maximum bending moment in
the beam is
w 0l 2 w 0l 2 3
(a) (b)
27 27
2
w 0l 2 w 0l 2
(c) (d)
9 9
UPPSC AE 13.12.2020, Paper-I
RPSC VPITI 14.02.2016
Wx 3 WLx
Ans. (b) : BMat x = –
6L 6

wℓ ℓ wℓ ℓ
(BM)max = × − ×
2 2 2 4
wℓ2  1 1 
=  − 
2 2 4
wℓ 2 dM
= For maximum bending moment, =0
8 dx
(∵ wℓ = W) dBM d  Wx 3 WLx 
So, =  – 
Wℓ dx dx  6L 6 
( BM )max =
8 −3Wx 2 WL
515. Which one of the following statement is Shear force = +
6L 6
correct? If the SF diagram for a beam is a 2
right angled triangle having its base 3Wx WL
− + =0
representing the length of the beam, the beam 6L 6
is : L
(a) a cantilever having a concentrated load at its x=
free end 3
Strength of Materials 241 YCT
 L3 L Ans. (b) : Draw a propped cantilever beam subjected to
−W WL × U.D.L.
3 3+ 3
( BM )maximum =
6L 6

Let,
W = Uniform distributed load.
ℓ = Length of cantilever beam
WL2 w ℓ2 3
( BM )max = + or 0 R = Reaction at the prop
9 3 27 ∴ Deflection at point B due to U.D.L. in cantilever
517. A simply supported beam is loaded as shown beam in downward direction
Wℓ 4
in fig. The maximum shear force in the beam
will be :
δB =
8EI
(↓ )
Deflection at point B due to reaction (R) in upward
direction
Rℓ 3
δ'B =
3EI
(↑)
(a) Zero (b) W ∴ At equilibrium condition - δ'B = δ B
(c) 2W (d) 4W Rℓ3 Wℓ 4 3Wℓ
APPSC AE 04.12.2012 ⇒ = ⇒ R=
3EI 8EI 8
APPSC AEE 2012, ESE 1998
519. The point of contraflexure is also called as :
Ans. (c) : A simply supported beam - (a) the point of inflexion
(b) a virtual hinge
(c) Either of the above
(d) None of the above
APPSC AE 04.12.2012
APPSC AEE 2012
Ans. (c) : The point of contraflexure is also called as-
the point of inflection or a virtual hinge.
Because at the point of contraflexure Bending moment
changes sign and Bending moment is zero.
520. A continuous beam is one which is
(a) fixed at one end and free at the other end
(b) fixed at one end and free at the other end
(c) supported on more than two supports
(d) extending beyond the supports
OPSC AEE 2019 Paper-1
TSPSC AEE 2015
(Shear force Diagram- SFD)
Ans : (c) A continuous beam is one which is supported
From the figure - on more than two supports.
→ Let reaction at A and B in the beam is RA and RB
respectively.
∴ Due to symmetrical loading - RA = RB … (i)
& RA + RB = 4W … (ii)
∴ From equation (i) & (ii) 521. The shear force diagram of a loaded beam is
RA = RB = 2W shown in the following figure. The maximum
Therefore, from SFD; maximum shear force = 2W bending moment in the beam is
518. The reaction at the prop in a propped
cantilever beam subjected to u.d.l. is :
Wl 3Wl
(a) (b)
4 8
5Wl 6Wl
(c) (d) (a) 16-kN-m (b) 11-kN-m
8 7 (c) 18-kN-m (d) 18-kN-m
APPSC AE 04.12.2012 UPPSC AE 12.04.2016 Paper-I
APPSC AEE 2012 CSE Pre-1997
Strength of Materials 242 YCT
Ans : (a) 523. The BM diagram of the beam shown in the
figure will be:

(a) A rectangle (b) A triangle

Maximum Bending Moment:- (c) A trapezium (d) A parabolic
( BM )max = 16kN − m (e) A hyperbolic
Nagaland PSC CTSE 2017, Paper-I
We know that (CGPSC Polytechnic Lecturer 2017)
dM
= shear force (S ) Ans. (b) : RA = RB =
W
dx 2
dM = Sdx Shear force S, At point A
C
∫A
dM = Area under Shear force diagram
1
(MC – MA) = 2 × 2 + × 2 ×12
2
= 4 + 12 = 16 kN-m
MA = 0 [At Reaction]
So M C = 16kN − m
and also
1 
MB – MC = −13 × 1 +  × 1× ( −6 ) 
2 
= – 13 – 3 = – 16 kN-m
MB = 0 [At reaction]
M C = 16 kN − m
So maximum bending moment will be 16 kN-m at W
point C. SA = +
2
522. A cantilever OP is connected to another beam W
PQ with a pin joint as shown in figure. A load SC = +
2
of 10 kN is applied at the mid-point of PQ. The W W
magnitude of bending moment in (kNm) at SC = + − W = −
fixed end O is– 2 2
W
SB = −
2
W W
SB = − + =0
2 2
Bending Moment diagram
MA = MB = 0
(a) 2.5 (b) 5 L W×L
MC = R A × =
(c) 10 (d) 25 2 4
PPSC Asstt. Municipal Engineer 15.06.2021 524. In a beam where shear force is maximum the
RPSC 2016 bending moment will be–
GATE 2020 (a) maximum
(b) zero
Ans : (c) By similarity, (c) minimum
RP = RQ = 5kN (d) there no such relation between the two
Nagaland PSC CTSE 2017, Paper-I
Nagaland PSC CTSE 2016, Paper-I
Ans. (d) : There no such relation between the two
525. Variation of bending moment in a cantilever
carrying a load, the intensity of which varies
uniformly from zero at the free end to w per
unit run at the fixed end, is by :
From FBD (a) cubic law (b) parabolic law
Moment at point O (c) linear law (d) none of these
Mo = RP × 2 = 5 × 2 = 10 kN-m WBPSC AE, 2017
(HPPSC AE 2014)
Strength of Materials 243 YCT
Ans : (a) Variation of bending moment in a cantilever mm and height of 100 mm. The maximum
carrying a load, the intensity of which varies uniformly magnitude of bending stress (in MPa) is given by
from zero at the free end to w per unit run at the fixed
end is by cubic law.
Section x–x talking from

(a) 60.0 (b) 67.5

(c) 200 (d) 225
ISRO Scientist/Engineer 11.10.2015
GATE 2010
Ans. (b) : Given,
b = 30 mm, d = 100 mm, w = 3000N/m
Taking moment at
x–x
1 wx 2 x
M x −x = ×
2 l 3
wx 3
M x−x =
6l
RA + RC = 6000 .....(i)
Bending moment at B Taking moment about point 'C'
x = 0, MB = 0
RA × 4 = 6000
6000
RA = = 1500 N
4
RA + RC = 6000
RC = 6000 – 1500 = 4500 N
Bending Moment at A Taking moment about x – x
x=l w(x − 2) 2
2 Mxx = RA⋅x –
wl 2
MA =
6 dM 2w ( – 2)
x
526. The shear force of a cantilever beam of length = RA –
dx 2
'L' with a point load 'W' at its free end is in the For maximum bending moment,
shape of the following:
dM
= 0 ⇒ R A = w ( x – 2)
dx
1500
+2= x
3000
x = 2.5
Maximum bending moment M at x = 2.5 m from point
A.
UJVNL AE 2016
TSPSC AEE 2015
Ans : (a)

(0.5) 2
M = (1500 × 2.5) – 3000 ×
2
M = 3375 N-m
I
Where Z=
y
I bd 3
Section modulus (Z) = =
y 12 × d
2
527. A massless beam has a loading pattern as 3
shown in the figure. The beam is of 30  100 
×  × 2000
rectangular cross section with a width of 30
= 1000  1000 
12 × 100
Strength of Materials 244 YCT
 2.5 × 10−6
= × 2000 × m3 = 5 × 10–5 m3
100
Bending stress, σmax.
M 3375
σmax = max = = 67.5 × 106 N / m 2
Z 5 × 10−5
= 67.5 MPa
528. For a particular load distribution and support
condition in a beam of length 'L', bending
moment at any section 'x' (0 < x < L) is given dM
by M(x) = Ax - Bx2, where A and B are 531. If for a beam = 0 for its whole length, the
constants. The shear force in the beam will be dx
beam is a cantilever.
zero at 'x' equal to (a) Subjected to udl extending from fixed end to
(a) A/2B (b) A/B its mid-length.
2
(c) 2A/B (d) A /B (b) Subjected to udl over its whole length
ISRO Scientist/Engineer 12.05.2013 (c) Subjected to an end moment
DRDO Scientists 2009 (d) Having a concentrated load at its free end
Ans. (a) : Bending moment TSPSC AEE 28.08.2017 (Civil/Mechanical)
M = Ax − Bx 2 ; [0 < x < L] UPSC JWM Advt. No.-52/2010, CSE Pre 2007
dM Ans. (c) : Given,
Shear force F= dM
dx Change in bending moment, =0
dM dx
= A − 2 Bx (for whole length of a cantilever beam)
dx where, M = Constant Bending Moment
If shear force of beam (F) = 0
dM
F= = A − 2 Bx = 0
dx
A - 2Bx = 0
A
x=
2B
529. Shear force is
(a) Rate of change of loading dM
(b) Sum of bending moments If for a beam = 0 for it whole length, the
(c) Rate of change of bending moment dx
(d) None of the above beam is cantilever which is subjected to an end moment.
Shear force is zero throughout whole length.
MPSC HOD (Govt. Poly. Colleges) 04.10.2014
532. A uniformly distributed load (UDL) (in kN/m)
ISRO Scientist/Engineer 2010, 2007 is acting over the entire length of a 3m long
Ans. (c) : Shear force is rate of change of bending cantilever beam. If the shear force at the
moment. midpoint of cantilever is 6 kN, what is the
dM value of UDL (in kN/m)?
Shear force (V) = (a) 2 (b) 3
dx (c) 4 (d) 5
530. A cantilever of length L carries a point load W OPSC AEE 2019 Paper-I
at the free end. The bending moment diagram GPSC ARTO Pre 30.12.2018, ESE 2009
will be a Ans. (c) :
(a) Parabola with maximum ordinate at centre of
the beam
(b) Parabola with maximum ordinate at the
cantilever end
(c) Triangle with maximum ordinate at the free
end
(d) Triangle with maximum ordinate at the
cantilever end At mid point of cantilever beam, shear force is 6 kN
(e) None of the above L
w× = 6
CGPSC AE 15.01.2021 2
CGPSC AE 16.10.2016 3
Ans. (d) : The maximum bending moment of a w× = 6
2
cantilever beam of length 'L' & carrying a point load at
free end, is lies at fixed end & straight in nature. w = 4 kN / m

Strength of Materials 245 YCT

533. A cantilever beam having 5m length is so Ans. (c) : For the shear force to be uniform throughout
loaded that it develops a shearing force of 20 T the span of simply supported beam, it should carry a
couple anywhere within its span.
and a bending moment of 20 T-m at a section
2m from the free end. Maximum shearing
force and maximum bending moment
developed in the beam under this load are
respectively 50 T and 125 T-m. The load on the
beam is :
(a) 25 T concentrated load at free end
(b) 20 T concentrated load at free and
(c) 5 T concentrated load at free end and 2 T/m
load over entire length 535. A simply supported beam of length 3 m carries
(d) 10 T/m uniformly distributed load over entire a concentrated load of 12 kN at a distance of
length 1m from left support. The maximum bending
moment in the beam is.
GPSC ARTO Pre 30.12.2018 (a) 12 kNm (b) 24 kNm
APPSC AEE 2012 (c) 8 kNm (d) 16 kNm
ESE 1995 ISRO Scientist/Engineer (RAC), 10.03.2019
MPPSC State Forest Service Exam, 2014
Ans. (d) : By observing data, we see shear force is Ans. (c) :
varying with length. Hence option (a) and (b) get
eliminated for concentrated load at the ends of
cantilever, shear force remains constant.
Now, considering option (d).

536. A beam under pure bending is subjected to

(a) Same bending stress at all sections
(b) No shearing stress at any section
(c) Same shearing stress at all sections
(d) Uniformly increasing bending moment from
the support to the mid-section
Karnataka PSC AE, 10.09.2017
RA = 10 × 5 = 50T WBPSC AE 2008
Ans. (b) : For pure bending there should be no
5 presence of shear stress (τ = 0) at any sections.
MA = 10 × 5 × = 125T-m
2 537. Bending moment at any point on a beam
534. For the shear force to be uniform throughout subjected to transverse loading will be equal to
(a) Algebraic sum of moments of forces and
the span of a simply supported beam, it should moments of reactions on one side of the
carry which one of the following loadings? section (point)
(a) A concentrated load at mid-span (b) Algebraic sum of moments of forces and
moments of reaction on both sides of the
(b) UDL over the entire span section (point)
(c) A couple anywhere within its span (c) Algebraic sum of moments of forces on both
(d) Two concentrated loads equal in magnitude sides of the section (point)
(d) Algebraic sum of moments of reaction on
and placed at equal distance from each both sides of the section (point)
support. GPSC DEE, Class-2 (GWSSB) 04.07.2021
GPSC ARTO Pre 30.12.2018 Ans. (a) : Bending moment at any point on a beam
subjected to transverse loading will be equal to
TSPSC AEE 28.08.2017 (Civil/Mechanical) algebraic sum of moments of forces and moments of
MPPSC State Forest Service Exam, 2014 reactions on one side of the section (point).
Strength of Materials 246 YCT
538. Which of the following are the statically Ans. (a) : In case of simply supported beam, if the
determinate beams? position of point of application of bending moment is
(a) Cantilevers (b) SSB
(c) Overhanging beams (d) All of the above changed towards left, over the length of the beam, the
APPSC AEE 2012 value of the reactive force at the left end of the beam
Ans. (d) : Examples of determinate structures are- will remain unchanged.
• Simply supported beams (SSB) 542. A cantilever beam 1.5 long is loaded with a
• Cantilever beams uniformly distributed load of 2 kN/m run over
• Single and double overhanging beams a length of 1.25 m from the free end. It also
• Three hinges arches etc. carries a point load of 3 kN at a distance 0.25
539. A simply supported beam of span 8 meters m from the free end. The shear force acting on
carry concentrated loads of 4 kN, 10 kN and 7
kN at distances of 1.5 meters, 4 meters and 6 the fixed end is ______.
meters from the left support. The shear force (a) 7.5 kN (b) 2 kN
at right support will be (c) 3.5 kN (d) 5.5 kN
(a) 4 kN (b) 10 kN VIZAG Steel MT 24.01.2021, Shift-I
(c) 11 kN (d) 7 kN Ans. (d) :
GPSC DEE, Class-2 (GWSSB) 04.07.2021
Ans. (c) :

RA + RB = 21 kN Shear force at fixed point 'A',

∑ MA = 0 (SF)A = 3 + 2.5 = 5.5 kN
R B × 8 = 4 × 1.5 + 10 × 4 + 7 × 6 543. Maximum bending on a cantilever beam of
R B × 8 = 6 + 40 + 42 length 'L' carrying a uniformly varying load
RB = 11 kN zero at free end and 'w' per metre length
RA = 21 –RB W.L2 W.L
RA = 21 – 11 = 10 kN (a) M max = (b) M max =
The shear force at right support will be 11 kN. 9 6
540. The differential equation W.L 2
W.L2
4 4 (c) M max = (d) M max =
EI (d y/dx ) gives 6 9 3
(a) Bending moment (b) Shear force VIZAG Steel MT 24.01.2021, Shift-I
(c) Deflection (d) Rate of loading
GPSC DEE, Class-2 (GWSSB) 04.07.2021 Ans. (c) :
Ans. (d) : EI (y) – deflection
 dy 
EI   − slope
 dx 
d2 y
EI 2 − Bending moment
dx
d3 y
EI 3 − shear force
dx
d4 y
EI 4 − Rate of loading
dx
541. In case of simply supported beam, if the
position of point of application of bending
moment is changed towards left, over the
length of the beam, the value of the reactive Wx 2
force at the left end of the beam will Shear force (SF) =
2L
(a) remain unchanged.
(b) increase.
(c) decrease.
(d) be doubled of the previous value.
PPSC Asstt. Municipal Engineer 15.06.2021
Strength of Materials 247 YCT
544. A 250 mm diameter pulley is mounted on the Ans. (c) : P. a
shaft midway between two supporting Beam is symmetrically loaded
bearings that are 750 mm apart. The pulley is So, R1 = R2 = P
driven by a belt, both strands pulling vertically Maximum Bending moment
upwards. If the tension in the tight side of the M=P×a
belt is 2.7 kN and the slack side is 900 N, what
is maximum bending moment?
(a) 225 N-m (b) 675 N-m
(c) 300 N-m (d) 1000 N-m
CGPSC AE 15.01.2021
Ans. (b) : Given,
dia of pulley = 250 mm = 0.25 m
L = 750 mm = 0.75 m
Total load on pulley = 2.7 kN + 900 N
(W) = 2700 + 900
(W) = 3600 N
WL 3600 × 0.75
Max. Bending Moment = =
4 4
(BM) max = 675N − m
545. A simply supported beam of length 9 m
carrying a uniformly distributed load of 10
kN/m for a distance of 6 m from the left end.
The bending moment acting on the beam at 6
m from its left end is ________.
(a) 40 kNm (b) 80 kNm
(c) 120 kNm (d) 60 kNm
VIZAG Steel MT 24.01.2021
Ans. (d) :
547. The part of a member is said to be in pure
bending if
(a) no bending moment exists in that part
(b) no shear force and no bending moment extra
in that part
(c) shear force and bending moment are
maximum
(d) No shear force exists in that part
GPSC DEE, Class-2 (GWSSB) 04.07.2021
Ans. (d) : The part of a member is said to be in pure
bending if no shear force exists.
548. Reaction RB at support B for the shown simply
supported beam will be :
R1 + R2 = 60
∴ ∑MA = 0
60 × 3 = R2 × 9
60 × 3
R2 = = 20kN
9 (a) zero (b) 3M
R1 = 60 – 20 = 40 kN (c) Ml (d) 3 M/l
∴ The bending moment on beam at 6m from left end JPSC AE 10.04.2021, Paper-II
(BM) = R1 × 6 – 60 × 3 Ans. (d) :
= 40 × 6 – 60 × 3
= 240 – 180
= 60 kN-m
546. Maximum bending moment for the shown
overhang beam with loads will be :

 l l ΣMA = 0
(a) P  a +  (b) P. M + 2M = RB × l
 2  2
(c) Pa (d) P (2a + l) 3M
RB =
JPSC AE 10.04.2021, Paper-II l
Strength of Materials 248 YCT
549. The beam shown here has variable load. The Ans. (d) :
reactions at A and B in kN are ____ and ____, Shear force, V = 1 + 2 × 2 + 1
respectively. = 6 kN (max)
Max. Bending moment occurs at the fixed end
Mmax = 1 × 5 + 2 × 2 × 3 + 1 × 1
= 5 + 12 + 1
= 18 kN-m (hogging)
552. If a beam is subjected to a constant bending
moment along its length then the shear force
will
(a) be maximum at the center and zero at the
(a) 15.76, 26.84 (b) 26.84, 15.76 ends
(c) 6.84, 5.76 (d) 5.76, 6.84 (b) zero at the center and maximum at the ends
AAI Jr. Executive 26.03.2021 (c) be zero at all sections along the beam
BHEL ET 2019 (d) has a constant value everywhere along its
Ans. (c) : w = wo – kx3 length
At x = 0, w = wo = 2400 RPSC Lect. (Tech. Edu. Dept.) 13.03.2021, Paper-I
At x = 6, w = 1200 ESE 1997
1200 = 2400 – k(6)3 Ans. (c) : The shear force will zero along the beam at
k = 5.55 all section, there is no shear force only it is a pure
d2M bending.
Now, − w xx =
dx 2 dM
dM Shear force V =
Shear force, SF = − ∫ (w xx )dx dx
dx Given, M = Constant
= − ∫ (w o − kx 3 )dx  dM 
So shear force V = 0 ∵ = 0
dM kx 4  dx 
= − w o x + c1 553. Support is non-yielding means
dx 4
(a) Slope of beam at support is zero
kx 5 w o x 2
M= − + c1x + c 2 (b) Support is frictionless
20 2 (c) support holds the member firmly
At x = 0, M = 0 (d) None of the above
c2 = 0 GPSC Engineer, Class-II Pre-19.01.2020
At x = 6, M = 0
c1 = 6840.036 Ans. (a) :
554. A simply supported beam PQ is loaded by a
kx 4
∴ SF = − w o x + 6840.036 moment of 1 kN-m at the mid span of the beam
4 of length 1m. The reaction forces RP and RQ at
Reaction at A, x = 0 supports P and Q respectively are
RA = (SF)A = 6.84 kN (a) 1 kN downward, 1 kN upward
Reaction at B, x = 6, (b) 0.5 kN upward, 0.5 kN downward
RB = (SF)B = 5.76 kN (c) 0.5 kN downward, 1 kN upward
550. In a cantilever beam subjected to a point load (d) 0.5 kN upward, 0.5 kN upward
at its free end has maximum bending moment ISRO Scientist/Engineer (RAC) 12.01.2020
________.
(a) fixed end of the beam Ans. (a) :
(b) at a distance of 0.6 × length of the beam
(c) free end of the beam
(d) mid span of the beam
HPPSC Workshop Suptd. 08.07.02021
∑Fv = 0
Ans. (a) : fixed end of the beam RP – RQ = 0
551. For a cantilever, loaded as shown in figure, ∑MP = 0
maximum shear force and maximum bending
moments are- RQ × 1 – 1 = 0
RQ = 1 kN (↑)
RP = 1 kN (↓)
555. In a beam where shear force changes its sign,
the bending moment will be
(a) minimum (b) zero
(c) maximum (d) same as shear force
(a) 4kN, 14 kN-m sagging ISRO Scientist/Engineer (RAC) 12.01.2020
(b) 4kN, 14 kN-m hogging dM
(c) 6kN, 18 kN-m sagging Ans. (c) : Shear force = changes its sign, the
(d) 6kN, 18 kN-m hogging dx
RPSC Lect. (Tech. Edu. Dept.) 13.03.2021, Paper-I bending moment will be maximum.
Strength of Materials 249 YCT
556. A simply supported beam of length 6 m carries Ans. (d) :
a point load of 8 kN at the middle of its span.
The bending moment at the middle of the span
is ____ .
(a) 6 kNm
(b) 24 kNm
(c) 12 kNm
(d) 8 kNm
VIZAG MT, 14.12.2020
Ans. (c) :

Area of SFD = 50 × 0.2 + 0.4 × P1

16 = 50 × 0.2 + P1 × 0.4
6
P1 =
M at mid point = 4 × 3 0.4
= 12 kN-m P1 = 15
557. A simply supported beam of span 8 m is P1 + P2 = 50
carrying a uniformly distributed load of 4 15 + P2 = 50
kN/m over a length of 4 m from its right end. P2 = 50 – 15
The support reactions are ______. P2 = 35 N
(a) 12 kN right support, 4 kN left support
(b) 6 kN left support, 10 kN right support 559. The beam that would generate the bending
moment diagram shown below is
(c) 4 kN left support, 12 kN right support
(d) 8 kN left support, 8 kN right support
VIZAG MT, 14.12.2020 (a) Simply supported beam with a central point
Ans. (c&a) : load
(b) Cantilever with uniformly distributed load
(c) Simply supported beam with uniformly
decreasing load from the midspan towards
the supports
(d) Simply supported beam with uniformly
distributed load
Haryana PSC AE (PHED) 05.09.2020, Paper-II
Ans. (d) : In case of simply supported beam UDL
produce parabola in BMD and straight line in SFD.
Simple supported beam has zero BM at support.
16 × 2
R1 = = 4 kN
8
16 × 6
R2 = = 12 kN
8
558. For the cantilever AB shown in the following
figure, if the reaction force at the fixed support
A is 50 N and the area of the shear force
diagram is 16 N-m, the value of the
concentrated load P2 is

560. A cantilever of span l is subjected to a point

(a) 16 N load W at the free end. The bending moment
(b) 64 N at the free end will be
(c) 15 N (a) Wl (b) zero
(d) 35 N (c) W/l (d) W
Haryana PSC AE (PHED) 05.09.2020, Paper-II TSPSC Manager (Engg.) HMWSSB 12.11.2020
Strength of Materials 250 YCT
Ans. (b) : Wl 1 Wx
Vx = − x.
6 2 l
Wl Wx 2
Vx = −
6 2l
Max. Bending moment occurs at,
Vx = 0
Wl Wx 2
− =0
6 2l
Bending moment (M) at free end = 0 Wx 2 Wl
=
Bending moment at fixed end = W×l 2l 6
561. A mass less beam has a loading pattern shown l
in figure. Find the bending moment at mid x=
span? 3
563. When there is a sudden increase or decrease in
shear force diagram between any two points, it
indicates that there is
(a) No loadings between the two points
(a) 1 kN-m (b) 3 kN-m (b) Point loading at the two points
(c) 2 kN-m (d) 0.0 kN-m (c) Uniformly varying load between the two
ISRO Scientist/Engineer 12.01.2020 points
Ans. (a) : (d) Uniformly distributed load between the two
points
UPPSC AE 13.12.2020, Paper-I
RPSC IOF, 2020
VIZAG Steel MT 24.01.2021
Ans. (b) : When there is a sudden increase or decrease
in shear force diagram between any two points, it
indicates that there is Point loading at the two points.
4 × 1.5
RA = = 3kN
2
(B.M)B = RA × 1 – w × 0.5 = 3 ×1– 4 × 0.5 = 1 kN-m
562. A simply supported beam of length l, carries a
load w(x) = w0(x) over the entire span.
Maximum bending moment in the beam at x
will be 564. A simply supported beam of span L is
l l L
(a) (b) subjected to a moment M0 at a distance of
3 3 4
l 3 l from the left end. Magnitude of the maximum
(c) (d) bending moment in the beam is
2 2 M0
UPPSC AE 13.12.2020, Paper-I (a) M0 (b)
2
Ans. (b) : A simply supported beam carrying load M0 3M 0
(c) (d)
4 4
UPPSC AE 13.12.2020, Paper-I
Ans. (d) :

Wl l Taking moment about point A

× RB × L – M0 = 0
Wl
Reaction at A, R A = 2 3 = M
l 6 RB = 0
Wl 2l L
× RA + RB = 0
Reaction at B, R B = 2 3 = Wl RA = – RB
l 3 M
Vx = + R A − Area of APQ RA = − 0
L
Strength of Materials 251 YCT
 Ans. (d) : Given data,
For simply supported beam length (L) = 6 m
Square cross - section side (b = d) = 100 mm = 0.1 m
Bending stress (σb) = 8 MN/m2 = 8 × 106 N/m2
Let,
P = load acting at point C from A distance = 1 m
RA & RB is reaction at A and B
RA + RB = P .......... (1)
Taking moment about A, then –
P
RB × 6 = P × 1 ⇒ R B =
6
5P
∴ RA =
6
Bending moment –
For section AC –
Mx = RA.x .......... (1)
For section BC –
5P.x
3M 0 Mx = RAx – P (x–1) = − Px + P
Maximum bending moment at (x = L/4) = 6
4
Px
565. A simply supported beam carrying a Mx = − +P ..........(2)
uniformly distributed load over its whole span, 6
is propped at the centre of the span so that the
beam is held to the level of the end supports.
The reaction of the prop will be
(a) half the distributed load
(b) 3/8th the distributed load
(c) 5/8th the distributed load
(d) 0 distributed load
RPSC IOF, 2020
Ans. (c) : The beam held to the level of end means
deflection equal to zero at mid span

From the S.F.D; shear force is zero at point C

4
i.e., at that point C (x=1) bending moment is maximum.
5wl −P × 1 5P
Deflection from distributed load = .......(i) ∴ Mx = 1 = +P= N−m
384EI 6 6
Pl 3 ∵ For square X-sectional (b=d)
Deflection from prop load = ........(ii)
bd 3 ( 0.1) × ( 0.1)
3
48EI
From equation (i) and (ii) ∵I= =
12 12
5wl 4 Pl 3
= d 0.1
384EI 48EI y= =
2 2
5
P = wl Now, from bending equation –
8 M σb
wl ⇒ distributed load =
th
I y
So, The reaction (P) of the prop =   the distributed
5
bd3
8 σ × 3
12 = σ b × d = 8 × 10 × ( 0.1)
4 6
σ ×I b
load. M= b =
y d/2 d 6
566. A simply supported beam, 6.0 m long, carries a 12 ×
single concentrated load of P at 1.0 m from one 2
end. The cross-section of beam is square with 8 × 106 × 10 −3 8 × 103
M= =
the side of 100 mm. What will be the maximum 6 6
value of P so that stress in material should not
5P 8 ×10 3
exceed 8 MN/m2? ⇒ =
(a) 1 kN (b) 1.2 kN 6 6
(c) 1.4 kN (d) 1.6 kN P = 1.6 × 103 N ⇒ P = 1.6 kN
APPSC Poly Lect. 13.03.2020
Strength of Materials 252 YCT
567. A simply supported beam of length L carrying Max shear force at fixed = W × l
a concentrated load W at a section which is at 1 
a distance of 'x' from one end. What will be the Max value of bending moment =  W × l  × l
value of bending moment at this section? 2 
(a) W (x - x2) (b) W (x2 - xL) Wl 2
=
 x2  2
(c) W  x −  (d) Wx
 L  569. Calculate the shear force and bending moment
 at the mid point of the beam
SJVN ET 2019
Ans. (c) :
W

A C B

RA RB (a) 0 kN, 0 kN-m (b) -20 kN, -20 kN-m

RA + RB = W .............(i) (c) 20 kN, -20 kN-m (d) 20 kN, 0 kN-m
Taking moment about A, APPSC AEE SCREENING 17.02.2019
RB×L = Wx Ans. (a) :
Wx
RB =
L
Wx
RA = W – RB = W −
L
Taking moment about C,
 x2 
MC = RA× x = W  x − 
 L
568. A cantilever beam of length l carry a UDL of
W per length across the whole span. What will
be the value of maximum shear stress and
maximum bending moment on the beam
respectively?
Wl
(a) , Wl 2 (b) Wl, Wl2
2 Due to symmetry, RB (or) RC
Wl Wl 2 Wl 2 Total load 10 × 8
(c) , (d) Wl , = = = 40 kN
2 2 2 2 2
SJVN ET 2019 At mid span
HPPSC AE 2018 (SF)E = 10 × 4 - 40 = 0
Punjab PSC SDE 12.02.2017 (BM)E = 40 (2) - 10 × 4 × 2 = 0
GMB AAE 25.06.2017 SF and BM both will be zero at mid point.
WBPSC AE, 2017
570. Consider the following:
Ans. (d) : In question they asked max shear stress that 1. Bending moment is a moment about the
should be max shear force.
longitudinal axis of a beam.
2. A structural component cannot have axial
force and shear force together.
(a) Only 1 is correct
(b) Only 2 is correct
(c) Both 1 and 2 are correct
(d) Both 1 and 2 are incorrect
APPSC AEE SCREENING 17.02.2019
Ans. (d) : • Bending moment is the moment about
neutral axis, but not about longitudinal axis.
• Structural components can have axial forces, shear
forces together when inclined loads are acting, on a
beam.
571. Out of the options given below, which one is
the correct shear force diagram? B is an
internal hinge
Strength of Materials 253 YCT
 (a) Both 1 and 2 are correct
(b) Both 1 and 2 are wrong
(c) Only 1 is correct
(d) Only 2 is correct
APPSC AEE SCREENING 17.02.2019
Ans. (a) : For analysing statically indeterminate
structures both equilibrium and compatibility equations
are required.
(a) Fixed beam has no degree of freedom.
∴ It is kinematically determinate beam.
574. For the beam shown below, the vertical
reactions at A and B are respectively
(b)

(c)

(d) None of the above (a) 2 kN, 3 kN (b) 3 kN, 2 kN

APPSC AEE SCREENING 17.02.2019 (c) -1 kN, 1 kN (d) 1 kN, 1 kN
Ans. (c) : APPSC AEE SCREENING 17.02.2019
Ans. (a) :
5 kN
5 kN-m

A B
C D
RA RB
2m 1m 2m
∑ Fy = 0
RA + RB = 5 kN
∑MA = 0
RB (5) = 5 × 2 + 5 = 15/5
∴ RB = 3 kN and RA = 2 kN
575. A simply supported beam of length 3.5 m
carries a triangular load as shown in the figure
below. Maximum load intensity is 7.2 N/m. The
location of zero shear stress from point A is :
SFD
572. Find out the Static indeterminacy of the beam
in the figure below

(a) 0 (b) 2 (a) 3 m (b) 1.5 m

(c) 3 (d) 6 (c) 2 m (d) 2.5 m
APPSC AEE SCREENING 17.02.2019 BHEL ET 2019
Ans. (b) : Degree of static indeterminacy of beam Ans. (c) : From section 'A C'-
DS = 3m – 3j + Re – Rr
m = Number of member
j = Number of joints
Re = external reaction of support
Rr = Number of internal hinge
In the given beam,
m = 2, j = 3, Re = 6, Rr = 1
DS = 3 × 2 – 3 × 3 + 6 – 1
=6–9+5 RA + RB = 10.8 + 1.8 = 12.6
Ds = 2 ΣM B = 0
573. Consider the following: 2
RA × 3.5 – 10.8 × 1.5 – 1.8 × × 0.5 = 0
1. In addition to equilibrium equations, 3
compatibility equations are also required RA × 3.5 – 16.2 – 0.6 = 0
for solving indeterminate structure.
2. A fixed beam (two ends are fixed) is a 16.8
RA =
kinematically determinate structure. 3.5
Strength of Materials 254 YCT
 RA = 4.8 kW Ans. (a) :
RB = 12.6 – 4.8
= 7.8 kW
from section (A C) –

w = Load per unit length

We know that At B, SF = 0
L→W At A, SF = w × L = wL
W 578. The maximum bending moment of a simply
1→
L supported beam of span L and carrying a
W point load W at the centre of beam is,
x→ x WL WL
L (a) (b)
Load at section (x - x) 4 2
1 W WL2
W = ×x× x (c) WL (d)
2 L 2
W 2 Nagaland PSC (CTSE) 2018, Paper-I, ESE 1996
= x Ans. (a) :
2L
(SF)A = 4.8
Wx 2
(SF)C = 4.8 –
2L
zero shear force mean 'maximum bending moment'
7.2x 2
4.8 – =0
2×3
W
x = 2m So from figure, R A = = RB ∴ RA + RB = W
2
576. Which of the following statements is/are So for BM, at centre,
correct? W L
1. In uniformly distributed load, the nature of BM max = ×
shear force is linear and bending moment is 2 2
parabolic. WL
M max =
2. In uniformly varying load, the nature of shear 4
force is linear and bending moment is 579. For a simply supported beam on two end
parabolic. supports the Bending Moment is maximum
3. Under no loading condition, the nature of (a) On the supports
shear force is linear and bending moment is (b) At mid span
constant. (c) Where the deflection is maximum
Select the correct answer using the code given (d) Where there is no shear force
below. GPSC Executive Engineer 23.12.2018
(a) 1 and 2 (b) 1 and 3 Ans. (d) : Where there is no shear force.
(c) 2 only (d) 1 only
ESE 2019 580. A simply supported beam AB has the bending
Ans. (d) : (i) In UDL, shear force is linear and bending moment diagram as shown in the following
moment is parabolic. Hence statement first is correct. figure
(ii) In UVL, shear force is parabolic and bending moment is
cubic. Hence statement second is incorrect.
(iii) In no loading, shear force is constant and bending
moment is linear. Hence statement third is also incorrect.
577. The shear force diagram of a cantilever beam
of length ℓ and carrying uniformly distributed
load of w per unit length will be
(a) a right angled triangle
(b) an issoscles triangle The beam is possibly under the action of
(c) an equilateral triangle following loads.
(d) a rectangle (a) Couples of M at C and 2M at D
Nagaland PSC (CTSE) 2018, Paper-I (b) Couples of 2M at C and M at D
Strength of Materials 255 YCT
 (c) Concentrated loads of M/L at C and 2M/L at 583. The reaction at the support of a beam with
D fixed end is referred as
(d) Concentrated loads of M/L at C and couple (a) fixed end moment
of 2M at D (b) fixed end couple
GPSC ARTO Pre 30.12.2018
(c) floating end moment
Ans. (a) : The beam is possibly under the action of
couples of M at C and 2M at D. (d) floating end couple
581. A curved member with a straight vertical leg is TNPSC AE 2018
carrying a vertical load at Z as shown in the Ans. (a) : The reaction at the support of a beam with
figure. The stress resultants in the XY segment fixed end is referred as fixed end moment.
are: 584. The beam ABC is supported at A (hinge
support) and B (roller support). If a force of
100 N is applied at C as given in figure, then
the reaction at the supports will be given by :

(a) R A = 50N ( ↓ ) ; R B = 150N ( ↑ )

(a) bending moment, shear force and axial force (b) R A = 50N ( ↑ ) ; R B = 100N ( ↓ )
(b) bending moment and axial force only
(c) bending moment and shear force only (c) R A = 150N ( ↑ ) ; R B = 50N ( ↓ )
(d) axial force only
GPSC ARTO Pre 30.12.2018 (d) R A = 150N ( ↓ ) ; R B = 50N ( ↑ )
Ans. (d) : The stress resultants in X-Y segment is only (e) R A = 50N ( ↓ ) ; R B = 150N ( ↓ )
axial force as load passes through axis of segment.
For bending moment in the section, there is some CGPSC AE 25.02.2018
eccentricity required with the load. Ans. (a) : RA + RB = 100
582. The shear force diagram for a simply MA = 0 ⇒ 100 × 1.5 = RB × 1
supported beam carrying uniformly distributed
load of w per unit length, consists of
(a) One right angled triangle
(b) One equilateral triangle
(c) Two right angled triangle
(d) Two rectangles R B = 150N ( ↑ )
ISRO Scientist/Engineer (RAC) 22.04.2018 RA = – 50N
Ans. (c) : RA + RB = wL
R A = 50N ( ↓ )
wL
RA = RB =
2 585. Which one of the following is the correct
bending moment diagram for a beam which is
hinged at the ends and is subjected to a
clockwise couple acting at the mid-span?

(a)

(b)

(c)

(d)
wL2
BMmax = at mid point [L/2]
8 ESE 2018
Strength of Materials 256 YCT
Ans. (c) : On balancing vertical force 587. For a simply supported beam on two end
supports the bending moment is maximum
(a) usually on the supports
(b) always a mid span
(c) where there is no shear force
(d) where the deflection is maximum
RA + RB = 0 .....(i) Karnataka PSC Lect., 27.05.2017
By taking moment about point A, Ans (c) : Bending moment will be maximum if
RB × L = M dM
= 0 i.e. V = 0
M dx
RB = .....(ii) Bending moment will be maximum at the point at
L which shear force is zero (SFD changes its nature)
From equation (i) and (ii) If M = Bending moment at the section of beam.
M
RA = − dM
= Slope of bending moment diagram
L dx
Bending moment
dM
MA = MB = 0 = V (V = shear force at section of beam)
L M dx
(Me)Right = R B × = (Sagging) 588. A point of contraflexure occurs in
2 2 (a) A simply supported beam
L M (b) A fixed beam
(Mc)Left = R B × − M = − (Hogging) (c) A cantilever
2 2
From the above computed values, bending moment (d) Fixed beam and cantilever
diagram is drawn in the figure below. Karnataka PSC Lect., 27.05.2017
Ans (b) :In a beam if the bending moment changes its
sign at a point, the point itself having zero bending
moment, the beam changes curvature at this point of
zero bending moment and this points is called point of
contraflexure or point of inflection.
A point of contraflexure always occurs in a over
hanging beam.
589. Calculate the bending moment at the mid-
point of a 6 m long simply supported beam
carrying a 20 N point load at the mid-point.
586. A beam carrying a uniformly distributed load (a) 20 Nm (b) 30 Nm
rests on two supports 'b' apart with equal (c) 45 Nm (d) 60 Nm
overhang 'a' at each end. The ratio b/a for zero BPSC AE Mains 2017 Paper - VI
bending moment at the mid span is Ans : (b) : By symmetry,
(a) 1/2 (b) 1 R1 = R2 = 10 N
(c) 3/2 (d) 2
RPSC Vice Principal ITI 2018
Ans. (d) :

(BM)C = R1 × 3 = 10 × 3
(BM)C = 30 Nm
590. A 3 m long beam, simply supported at both
ends, carries two equal loads of 10 N each at a
(2a + b)w distance of 1 m and 2 m from one end. The
RA = RB = shear force at the mid-point would be
2 (a) 0 N (b) 5 N
B.M. at middle of AB (c) 10 N (d) 20 N
w b  b  b 1 BPSC AE Mains 2017 Paper - VI
BM middle = −(2a + b) × + w  a +  a +  × Ans : (a) :
2 2  2  2 2
For zero B.M. at middle
2
w b w b
0 = −(2a + b) × + a + 
2 2 2 2
2ab − b 2 b 2 2ab By symmetry,
= − + a2 + + R1 = R2 = 10 N
2 4 2
2 2
b b
= − ab − + a 2 + + ab
2 4
b
=2
a Hence shear force at midpoint is zero.
Strength of Materials 257 YCT
591. A simply supported beam of span carries over Ans. (d) :
its full span a load varying linearly from zero
at each end to W N/m at mid span. The
maximum bending moment is
WL2 WL2
(a) (b)
12 8
WL2 WL2
(c) (d)
4 2
JWM 2017
Ans. (a) : Consider equilibrium of beam AB total load
1 L 
on beam is 2 ×  × × W 
 2 2 
WL Two identical concentrated loads equidistant from the
Total load = mid span and close to supports.
2 dM
As the beam symmetric, the total load equally 593. For a cantilever = constant for its whole
distributed on both the support. dx
length. What is the shape of the SF diagram
for the beam?
(a) Rectangle (b) Triangle
(c) A parabola (d) A hyperbola
TSPSC AEE 28.08.2017 (Civil/Mechanical), CSE Pre 2008
dM x
Ans. (a) : = Vx
dx
WL dM x
RA = RB = For = constant
4 dx
Bending moment (Mx) Shear force,
L Vx = constant
In the region 0 < x <
2 ∴ Shape of SFD is rectangular.
WL 1 2Wx   x  594. SF diagram for a simply supported beam is a
Mx = .x −  × x × ×  rectangle with its longer side equal to beam
4  2 L  3 length. What type of load is acting on the
WL Wx 3 beam?
= x− (a) Concentrated load at it mid span
4 3L
For maximum bending moment (b) UDL over its whole span
(c) Concentrated load along with a couple at a
d (M x ) point on beam length
=0
dx (d) Couple at a point on the beam length
TSPSC AEE 28.08.2017 (Civil/Mechanical), CSE Pre 2008
WL Wx 2
− =0 Ans. (d) :
4 L
L
∴x =
2
L WL2
At. x = M max =
2 12
592. The bending moment diagram for a simply 595. A cantilever beam of span/carries a uniformly
supported beam is a rectangle over a larger varying load of zero intensity at the free end
portion of the span except near the supports. and w per metre length at the fixed end. What
What type of load does the beam supports. does the integration of the ordinate of the load
What type of load does the beam carry? diagram between at the limits of free and fixed
(a) A uniformly distributed symmetrical load ends of the beam give?
over a larger portion of the span except near (a) Bending moment at the fixed end
the supports (b) Shear force at the fixed end
(b) A concentrated load at mid span (c) Bending moment at the free end
(c) Two identical concentrated loads equidistant (d) Shear force at the free end
from the supports and close to mid-point of TSPSC AEE 28.08.2017 (Civil/Mechanical)
the beam. Ans. (b) :
(d) Two identical concentrated loads equidistant dVx
from the mid span and close to supports = w (Rate of loading)
dx
TSPSC AEE 28.08.2017 (Civil/Mechanical), CSE Pre 2007
Strength of Materials 258 YCT
 Wℓ 2 4  Wℓ 2 
= 2  x (ℓ − x )
12 ℓ  8 
ℓ2
= x (ℓ − x )
6
ℓ2
x2 – ℓx + =0
Shear force between any two points on the loaded beam 6
is equal to total area of the leading diagram between x = 0.2111ℓ
these two points. Directions : Q. No. 599-600 are based of the following
1
(SF )fixed end = × ℓ × w figures:
2
596. The diagram showing the variation of axial
load along the span is called
(a) shear force diagram
(b) bending moment diagram
(c) thrust diagram 599. A beam is simply supported at the ends and
(d) influence line diagram loaded as shown in figure. The shearing force
TSPSC AEE 28.08.2017 (Civil/Mechanical) at the mid-point C will be:
Ans. (c) : Diagram showing the variation of axial load W
along the span is called thrust diagram. (a) (b) W
2
597. The variation of the bending moment due to (c) 2W (d) Zero
the moving load on a fixed ended beam occurs PTCUL AE 25.06.2017
(a) linear (b) parabolic Ans. (d) :
(c) cubic (d) constant
TSPSC AEE 28.08.2017 (Civil/Mechanical)
Ans. (b) : The variation of the Bending moment due to
a moving load on a fixed ended beam occurs parabolic.

598. A prismatic beam of length L is fixed at both

ends carries a uniformly distributed load. The
distance of points of contra flexure from either
end is
(a) 0.207 L (b) 0.211 L
(c) 0.277 L (d) 0.25 L 600. The bending moment between the point A and
B is:
TSPSC AEE 28.08.2017 (Civil/Mechanical) (a) Increasingly linearly
Ans. (b) : (b) Decreasingly linearly
(c) Varying parabolically
(d) Constant
PTCUL AE 25.06.2017
Ans. (d) :

By symmetric–
Reaction at supports is RM = W, RN = W
For SFD: (From left side)
Wℓ 2 Shear force at C, SFM = RM = W
y= Shear force at A (just left of A), SFA = W
12
4h Shear force at A (just right of A), SFA = W – W = 0
y = 2 x (ℓ − x) Shear force at B (just left of B), SFB = 0
ℓ Shear force at B (just right of B), SFB = –W
Strength of Materials 259 YCT
Shear force at C, SFN = RN = –W Ans. (a) : At point of contraflexure bending moment
changes its sign from positive to negative and vice-
versa and at this point value of B.M. becomes zero.
dM
For BMD: (From left side) Shear force (S) = = rate of change of bending
BM at A, BMA = Wa dx
BM at B, BMB = Wa moment.
dS
Intensity of loading (w) = = rate of change of shear
dx
force.
604. The rate of change of shear force at any
section represents
From SFD and BMD it is clear that there will be no (a) Point of contraflexure
shear force between A and B, but there will be a (b) Shear force at that section
uniform bending moment between A and B. (c) Bending moment at that section
601. A beam of length 2 m is pinned at both ends (d) Rate of loading at that section
and is subjected to a concentrated moment 10 Jharkhand Urja Vikas Nigam Ltd. AE 2017
kNm at its center. The maximum bending Ans. (d) : Consider a simply supported beam subjected
moment in the beam is to UDL for the entire span consider a free body diagram
(a) 10 kNm (b) 5 kNm at small portion elemental length example.
(c) 20 kNm (d) 15 kNm This rate of change of shear force at any section is equal
GPSC EE Pre, 28.01.2017 to the intensity of loading.
Ans. (b) : Given- 605. A beam is fixed at one end and is vertically
supported at the other end. What is the degree
of statical indeterminacy?
(a) 1 (b) 2
Length (l) = 2 m (c) 3 (d) 4
Concentrated moment = 10 kN-m ISRO Scientist/Engineer 07.05.2017
M l Ans. (a) :
Maximum bending moment = ×
l 2
10
(BM)max = = 5 kN − m
2
602. The number of points of contraflexure in a
simple supported beam carrying uniformly
distributed load, is
(a) One (b) Two
(c) Zero (d) None of these
GPSC EE Pre, 28.01.2017
• A beam is to be statically indeterminate, if the
Ans. (c) : Zero number of reaction is greater than static equilibrium
condition equation.
• A beam is said to be statically determinate, if
the number of reaction is equal to static equilibrium
condition equation.
For propped cantilever beam—
Four reaction RH, RA, RB and MA.
And three equation of static i.e. ∑ H = 0 , ∑ V = 0 and
∑M = 0
In a bending moment diagram, the point of contra So, hence beam is statically indeterminate to one
flexure is the point on beam where bending moment degree.
changes sign.
So, total No. of Reaction = 4
603. Match the following :
P. Point of i. Bending moment No. of independent equilibrium equation = 3
contraflexure changes sign So, degree of static indeterminacy will be = 4 - 3 = 1.
Q. Load ii. Shear force changes 606. A simply supported beam of length L is
intensity sign subjected to a varying distributed load
iii. Rate of change of sin(πx/L) N/m, where distance x is measured
bending moment
iv. Rate of change of from the left support. The magnitude of the
shear force vertical reaction in N at the left support is
(a) P-i, Q-iv (b) P-ii, Q-iv (a) 3L/π (b) 2L/π
(c) P-i, Q-iii (d) P-ii, Q-iii (c) L/2π (d) L/π
APGENCO AE, 2017 ISRO Scientist/Engineer 07.05.2017
Strength of Materials 260 YCT
Ans. (d) : So,
δM + δP = 0 (given)
4 P (2 L − x).L 8 PL3
2
− =0
2 EI 3EI
4 P (2 L − x).L2 8PL3
RA = ? =+
2 EI 3EI
 πx  4
Load = sin   N / m 2L − x = L
 L  3
L
 πx  2
0
∫
Total load = sin   dx
 L 
x = L = 0.66 L
3
L
608. A propped cantilever is indeterminate
L   πx   externally to
= −  cos    (a) The second degree (b) The third degree
π   L  0
(c) The fourth degree (d) The fifth degree
L
Total load = − [ −1 − 1] TNPSC AE 2017
π Ans. (a) : A propped cantilever is indeterminate
2L externally to the second degree.
=
π
2L 1 L
RA = RB = × =
π 2 π
607. A force P is applied at a distance 'X' from the
end of the beam as shown in the figure. What
would be value of 'X' so that the displacement
For general loading, the total reaction components (R)
at 'A' is equal to zero? are equal to (3 + 2) = 5, while the total number of
conditions (r) are equal to 3. The beam is statically
indeterminate externally to second degree.
Note- For vertical loading, the beam is statically
determinate to single degree (In figure).
609. The difference between member of a truss and
of a beam is:
(a) 0.5 L (b) 0.25 L (a) The members of a truss take their loads along
their length whereas a beam takes loads at
(c) 0.33 L (d) 0.66 L right angles to its length
ISRO Scientist/Engineer 07.05.2017 (b) The member of the truss takes load lateral to
Ans. (d) : From FBD— its length whereas the beam along the length
(c) The member of the truss can be made of C.I
where as the beam is of structural steel only
(d) The member of the truss can have a circular
cross-section whereas the beam can have any
cross-section
JWM 2017
Ans. (a) : Beams support their loads in shear and
bending where as truss support loads in tension and
compression.
The members of a truss take their loads along their length
where as a beam takes loads at right angle to its length.
where, 610. Shear force at any point of the beam is the
M = P(2L - x) algebraic sum of
(a) All vertical forces
Deflection at point (A) due to moment M.
(b) All horizontal forces
M (2 L) 2 4 ML2 (c) Forces on either side of the point
δM = = (↑)
2 EI 2 EI (d) Moment of forces on either side of the point
Vizag Steel (MT) 2017
4 P (2 L − x).L 2
= Ans. (c) :
2 EI Shear force at any point of the beam is the algebraic
Deflection at point (A) due to point load P sum of forces on either side of the point.
P(2 L) 3
8PL 3 Bending moment at any point of the beam is the
δP = = (↓) algebraic sum of moment of forces on either side of
3EI 3EI the point.
Strength of Materials 261 YCT
611. A cantilever carries a concentrated load (W) at (a) Shear force at the section
its free end. Its shear force diagram will be: (b) Rate of loading
(c) Zero always
(d) Bending moment at section
UPRVUNL AE 07.10.2016
(a) APPSC Poly. Lect. 2013
Ans. (a) :
d2y
EI =M
(b) dx 2
We know that
dM
= shear force
dx
then
(c) d 3 y dM
EI 3 = = shear force at section
dx dx
614. Shear force at a section of a horizontal beam is
the
(d) (a) algebraic sum of the vertical forces to any
CIL MT 26.03.2017 one side of the section
Ans. (d) : RA = w (b) algebraic sum of the vertical forces on both
sides of the section
(c) algebraic sum of the moments of all forces
on any one side of the section
(d) algebraic sum of the moments of all forces
on both sides of the section
APPSC AEE Mains 2016 (Civil Mechanical)
Ans. (a) : Shear force at a section of horizontal beam is
the algebraic sum of the vertical forces to any one side
of the section.
615. Bending moment is maximum at a section of a
beam where
Vx = R A = w (a) Shear force changes from –ve to +ve
VA= w, VB = w (b) Shear force is zero
Hence, S.F.D. is as shown in figure below (c) Shear force changes from +ve to –ve
(d) All the given answers
APPSC AEE Mains 2016 (Civil Mechanical)
Ans. (d) : Bending moment at any section x
dM x
612. A cantilever beam of span 2 m with a point = Vx (shear force)
load of 4 kN at its free end will have a constant dx
shear force of _____ throughout the span. The bending moment is maximum at a section of a
(a) 2 kN (b) 4 kN beam where shear force changes its sign or the value of
(c) 6 kN (d) 8 kN shear force is zero.
(e) 10 kN 616. Which of the following is giving correct
(CGPSC Polytechnic Lecturer 2017) relation between load(w), shear force (F) and
Ans. (b) : It have constant shear force of 4 kN. bending moment (M)
dF dM
(a) M = (b) F =
dx dx
dM dw
(c) w = (d) M =
dx dx
APPSC AEE Mains 2016 (Civil Mechanical)
Ans. (b) : As we know, The Rate of change of Bending
moment with respect to 'x' is equal to the shearing force
or the slope of the moment diagram at a given point is
the shear at that point.
613. Basic equation of deflection (y) of the beam is
dM
2
d y So shear force, F =
represented by El 2 = M, where El flexural dx
dx 617. A beam with equal overhangs is carrying UDL
d3y as shown in the following figure. The bending
rigidity and M Bending moment, then El 3 moment at the centre of the beam will be zero
dx
for the condition
Strength of Materials 262 YCT
 WL 1  2Wx 
= − x 
4 2  L 
WL Wx 2
Vx = − ( Parabolic )
(a) a = L/3 (b) a = L/2 4 L
(c) a = 3L/2 (d) a = L
WL Wx 3
APPSC AEE Mains 2016 (Civil Mechanical) Mx = x− ( Cubic Parabolic )
Ans. (b) : We know, 4 3L
619. A simply supported beam of span L and
constant width b carries a point load W at mid
span. The depth of the beam required at the
mid span for maximum extreme fibre stress P.
3WL 3WL
(a) d = (b) d =
 L  2bp 2bp
RA = Wa + 
 2 3WL 3WL
(c) d 3 = (d) d =
W 2bp 2
Mx = RAx − (a + x ) 2bp
2

2 APPSC AEE Mains 2016 (Civil Mechanical)

 L W WL
= W  a +  x − (a + x )
2
Ans. (b) : Bending moment, M max =
 2 2 4
We know, Maximum extreme stress (Bending stress) = P
L d
At x = , M = 0, M
2 My 2
Bending stress, σb = P = =
a 1 L I  bd 3 
So, We get = ⇒ a=  
L 2 2  12 
618. A simple supported beam is carrying a linearly  WL  d 
varying load from zero at either ends to the   
 4  2  = 12 WL
maximum value at the mid span. Then the P =
shape of shear force diagram (SFD) is bd 3 8 bd 2
12
3 WL
P=
2 bd 2
3WL
(a) Rectangle d=
2bP
(b) Triangular
(c) Second Degree Parabola 620. The number of contra flexure points that
(d) cubic occur in a cantilever beam subjected uniformly
distributed load is
APPSC AEE Mains 2016 (Civil Mechanical) (a) 3 (b) 2
1 WL (c) 1 (d) 0
Ans. (c) : Total load = × L × W =
2 2 APPSC AEE Mains 2016 (Civil Mechanical)
Ans. (d) : Point of contraflexure in a beam is a point at
which bending moment changes it sign from positive to
negative and vice versa. But in cantilever there is no
change in sign of bending moment. So, point of
contraflexure is zero.
621. A simply supported beam of span L carries a
point load P at center, the slope at left end is
(a) PL3 /(16EI) (b) PL2 /(48EI)
2
(c) PL /(16EI) (d) PL3 /(48EI)
APPSC AEE Mains 2016 (Civil Mechanical)
Ans. (c) : For a simply supported beam, L, The general
equation for this is
Wx 2Wx dy Px 2 PL2
wx = = EI = −
L L
2 dx 4 16
Shear force at section x, So, the slope of left end.
WL 1 dy PL2
Vx = − xw x =
4 2 dx 16 × EI
Strength of Materials 263 YCT
 Ans. (a) :

As S.F. is due to load only and here there is no load on

the beam so there will not be any S.F. in beam.

622. When equations of statics are not sufficient to

determine all the reactive forces at the
supports, such beams are called
(a) statically indeterminate
(b) statically determinate
(c) Imperfect 625. A simply supported beam with a gradually
(d) Defective varying load from zero at 'B' and 'w' per unit
APPSC AEE Mains 2016 (Civil Mechanical) length at 'A'. The shear force at 'B' is equal to
Ans. (a) : In statics and structural mechanics a structure wl 3wl
(a) (b)
is statically indeterminate when the static equilibrium 6 2
equations - force & moment equilibrium conditions are 2wl
(c) wl (d)
insufficient for determining the internal forces & 3
reactions on that structure. APPSC AE Subordinate Service Civil/Mech. 2016
623. A simply supported beam of span length 6 m Ans. (a) :
carries a uniformly distributed load of 1.5 w/unit
kN/m. What is the maximum value of bending
moment?
(a) 13.5 kN-m (b) 6.75 kN-m
(c) 27 kN-m (d) 9.25 kN-m
UPRVUNL AE 21.08.2016
Ans. : (b)

Total load = Area of loading diagram

1 wl
= × w ×l =
2 2
This load will act at the C.G. of loading diagram
l
i.e from support A,
3
wl 2l
×
Reaction at A, R A = 2 3
l
wl
RA =
3
wl l
×
wl 2 Reaction at B, R B = 2 3
( BM )max = l
8 wl
1.5× 62 RB =
( BM )max = 6
8 626. In a simply supported beam loaded as shown
( BM )max = 6.75kN−m in figure, the maximum bending moment in
Nm is
624. A cantilever of length 4m is subjected to
clockwise couple of 12kN-m at free end. What
is the shear force at fixed end?
(a) zero (b) 12 kN
(c) 3 kN (d) 4 kN
APPSC AE Subordinate Service Civil/Mech. 2016

Strength of Materials 264 YCT

 (a) 25 (b) 30 Ans. (a)
(c) 35 (d) 60
ISRO Scientist/Engineer 03.07.2016
GATE 2007
Ans. (b) :

RA + RB = 2P
RA = P, RB = P

RA + RB = 100 Hence, the shear force for the beam portion between
Taking moment about point A the supports is zero.
RB × 1 = 100 × 0.1 + 100 × 0.5 629. The reaction at a roller support of beam will be
RB = 60N (a) Tangential to support
RA = 40N (b) Normal to support
(c) Unknown in direction
(d) Inclined to support
APPSC AEE Screening Test 2016
KPSC ADF 2015
Ans. (b)

Moment at point (B)

MB = 0
Moment at point (A) MA = 0 Only normal/perpendicular support = Rv
Moment at point C = 40 × 0.5 = 20 RH = 0
Concentrated moment (10 N-m) also act in the same ΣM = 0
direction so
630. For a beam subjected to point loads, the shear
Max. Bending moment at C = 20 + 10 = 30 N-m. force between point loads
627. At a hinge in a beam (a) is constant
(a) Shear force is zero (b) varies linearly
(b) Bending moment is maximum (c) has parabolic variation
(c) Bending moment is zero (d) always remain zero
(d) Bending moment changes sign APPSC AEE Screening Test 2016
APPSC AEE Screening Test 2016 Ans. (a) : For a beam subjected to point loads, the
Ans. (c) shear force between point loads is constant.
631. Which of the following statements is true for
shear force (SF) and bending moment (BM)
diagram (where, w = weight per unit length)
(a) Change in BM over a small length (dM) =
Area of SF diagram under that length (Vdx)
Bending moment (BM) = 0 (b) Change in BM over a small length (dM) =
628. The bending moment for a beam with equal Rate of change of SF under that length
overhangs and carrying equal point loads each (dV/dx)
of P at the free ends, the shear force for the (c) Rate of change of Change in BM over a
beam portion between the supports is small length (dM/dx) = Rate of change of SF
(a) Zero under that length (dV/dx)
(b) equal to P (d) Change in SF over a small length (dV) is
(c) Linearly varying from–P to +P greater than area of loading diagram over that
(d) Linearly varying from 0 to P length (wdx)
APPSC AEE Screening Test 2016 RPSC LECTURER 16.01.2016
Strength of Materials 265 YCT
Ans. (a) : We know that Ans. (c) : RA + RB = wL
dM
= shear force (SF)
dx
dM = (SF) × dx
So,
change in BM over a small length (dM) = Area of SF
diagram under that length taking moment at point A
632. A beam is subjected to a force system shown in L
figure. This force system can be reduced to: RB × L = wL ×
2
wL
RB = RA =
2
taking moment at point C
L L L
M c = RA × − w × ×
(a) A single force of 50 N (downward) at 2.5 m 4 4 8
from A wL L L L 3
(b) A single force of 50 N (downward) at 2.5 m Mc = × − w × = wL2
from D 2 4 4 8 32
(c) A single force of 50 N (upward) at 2.5 m 634. If the shear force diagram for a beam is
from D triangle with length of the beam as its base, the
(d) A single force of 50 N (upward) at 2.5 m beam is
from A (a) A cantilever with a point load at its free end.
UPRVUNL AE 07.10.2016 (b) A cantilever with uniformly distributed load
over its whole span.
Ans. (a) : (c) A simply supported beam with a point load at
its mid-point.
(d) A simply supported beam with uniformly
distributed load over its whole span.
UPPSC AE 12.04.2016 Paper-I
Ans : (b) If the shear force diagram for a beam is
triangle with length of the beam as its base, the beam is
a cantilever with uniformly distributed load over its
whole span.

For equilibrium beam

For x-distance from B moment will be zero.

For equilibrium
∑ M CW = ∑ M ACW
50 + 25 = 50x
x = 1.5 or 635. A railway sleeper is like a beam having
From A at a distance of 2.5 uniformly distributed load acting upwards (as
reaction) and supports at rails with loads
acting downwards due to the passing rail, such
that:
(a) Greatest bending moment is at the centre of
sleeper
Then this force system can be reduced to single force of (b) Greatest bending moment is at the end of
50 N (downward) at 2.5 m from A. sleeper
(c) Greatest bending moment is at the contact of
633. Bending moment at distance L/4 from one end rails
of a simply supported beam of length (L) with (d) Greatest bending moment is as small as
uniformly distributed load of strength w per possible
unit length is given by GPSC Asstt. Prof. 28.08.2016
7 2 5 2 Ans. (d) :
(a) wL (b) wL
32 32 636. Two similar beams of same length and same
3 1 load acting at the centre. First one is fixed at
(c) wL2 (d) wL2 ends and second one is simply supported. The
32 32 maximum bending moment of first with
UPRVUNL AE 07.10.2016 respect to second is
Strength of Materials 266 YCT
 (a) Double (b) Same
(c) Half (d) One fourth
GPSC Asstt. Prof. 28.08.2016 (c) (d)
Ans. (c) : IInd maximum BM for simply supported beam
PL
at centre =
4
PL (e) None of these
Ist maximum BM for fixed end beam at centre =
8 CGPSC AE 16.10.2016
PL Ans. (b) :
(BM)I 1
Ratio = = 8 =
(BM) II PL 2
4
So, Ist to IInd is
1
Maximum (BM)ss = Maximum (BM) fixed
2
637. …….. has the unit 'N-m'.
(a) Kinetic Energy (b) Potential Energy
(c) Shear Force (d) Bending Moment
HPPSC Lect. (Auto) 23.04.2016 640. A beam is loaded as shown :
Ans. (d) : Bending moment has the unit 'N-m'
• Kinetic energy – Joule (J)
• Potential energy – Joule (J)
• Shear force – Newton (N)
638. For a cantilever beam of length 'L' having a The points of contraflexure within the span AB
point load 'P' at its free end, the bending will be
(a) 3m from A and 3 m from B
moment at the mid-span of the beam is given (b) 3m from A and 1.67 m from B
by : (c) 1.67m from A and 3 m from B
(a) PL (b) (P/2) L (d) 1.67m from A and 1.67 m from B
(c) P/L (d) L/P (e) None of the above
HPPSC Lect. (Auto) 23.04.2016 CGPSC AE 16.10.2016
Ans. (b) : Ans. (b) :

Mxx = P × x
At x = L (Fixed point) RA + RB = 3800 N
∑MA = 0
Bending moment (Mat fixed end = PL)
800 × 3 + RB × 8 = 2000 × 5 + 1000 × 10
L
At, x = (Mid point) R B = 2200 N
2
RA = 3800 – 2200
L R A = 1600 N
M mid span = P.
2
639. Draw the SFD and BMD diagram for the beam
and loading shown.

(a) (b)

Strength of Materials 267 YCT

Moment in section AD (x1-x1) Ans. (b) : Formula for max bending moment for simply
M=0 WL
800 × x – 1600 × (x – 3) = 0 supported be is =
x=6m 4
Distance from A = 6 – 3 = 3 m
Moment in section DB (x2-x2)
M=0
1000x = 2200 × (x – 2)
x = 3.666
Distance from B = 3.666 – 2
= 1.667 m
641. Slope of tangent to shear force diagram gives
(a) Bending moment
(b) Couple moment
(c) Support reactions
(d) Rate of loading
(e) Rate of bending moment
CGPSC AE 26.04.2015 Shift-I
dF
Ans. (d) : According to relation = −W (constant)
dx
So, slope of tangent to shear force diagram gives rate of
loading.
642. For a simply supported beam of length L, the ΣFV = 0
bending moment M is described as M = a (x – RA + RB = 500
x3/L2), 0 ≤ x ≤ L; Where a is constant. The take moment about 'B'
shear force will be zero at: RA × 10 = 500 × 5
L RA = 250 N ...(i)
(a) The supports (b) x = RB = 250 N ...(ii)
3 moment about 'C' (will be max.)
(c) X = L (d) X =
L MA = RA × 5 = 250 × 5
2 = 1250 Nm
TSPSC Managers, 2015 644. A simply supported beam carries a load 'P'
Ans. (*) : Given – Bending moment (M) through a bracket as shown in the figure. The
 x3  maximum bending moment in the beam is
M = a  x − 2 , 0 ≤ x ≤ L
 L 
We know that,
dM
shear force (F) =
dx
2
dM 3x
So, = a − 2 ×a Pl Pl ap
dx L (a) (b) +
According to question says that shear force is zero. 2 2 2
Pl Pl
dM (c) + aP (d) − aP
F=0= 2 2
dx ISRO Scientist/Engineer 11.10.2015
3x 2 UPPSC AE 12.04.2016 Paper-I
a − 2 ×a = 0
L Ans. (*) :
3x 2
or =1
L2
3x = L2
2

L
x=
3
643. A simply supported beam of span 10 m
carrying a load of 500 N at the midspan will
have a maximum bending moment of :
(a) 500 Nm (b) 1250 Nm
(c) 2500 Nm (d) 5000 Nm RA + RC = P .....(i)
Kerala PSC AE 06.08.2015 Take a moment about 'C'

Strength of Materials 268 YCT

 ℓ RB × 10 = (20 × 8.5) + (40 × 7.5) + (20 × 5) +
RA × ℓ + Pa = P ×
2  5
Pℓ  20 × 5 × 
RA × ℓ + Pa =  2
2 10RB = 820
Pℓ RB = 82 kN
− Pa RA = 98 kN
P Pa
RA = 2 = −
ℓ 2 ℓ
From equation (i)– RA + RC = P
RA + RC = P
P Pa
RC = P – RA = P − +
2 ℓ
P Pa
RC = +
2 ℓ
SFD–

Bending moment is maximum or minimum where shear

force changes sign
98 2
=
x 5− x
x = 4.9 m
4.9
BM x = 4.9 = 98 × 4.9 − 20 × 4.9 ×
2
= 480.2 - 240.1
= 240.1 N-m
646. The maximum bending moment of a simply
supported beam of span 2m and carrying a
 P Pa  ℓ point load 80 kN at the centre of the beam is
BM (Mmax) =  −  + Pa (a) 160 kN-m (b) 80 kN-m
 2 ℓ 2
Pℓ Pa (c) 320 kN-m (d) 40 kN-m
= − + Pa TSPSC AEE 2015
4 2
Ans : (d)
Pℓ Pa 1  Pℓ 
= + = + Pa 
4 2 2  2 
Note-ISRO given answer (c).
645. Find the maximum bending moment and the
position from the support A.
wl
Maximum Bending moment =
4
80 × 2
(BM)max =
4
(BM)max = 40 KN-m
(a) 240 Nm and 4.9 m 647. The bending moment for a certain portion of a
(b) 240.1 Nm and 4.9 m beam is constant. For that section, shear force
(c) 240 Nm and 5 m would be :
(d) 185 Nm and 7.5 m (a) zero (b) increasing
(e) 260 Nm and 4.9 m
CGPSC AE 26.04.2015 Shift-I (c) decreasing (d) constant
Ans. (b) : Assam PSC CCE Pre 2015
Ans. (a) : The bending moment for a certain portion of
a beam is constant. For that section shear force would
be zero.
648. The maximum bending moment of a simply
supported beam of span l and carrying a point
load W at the centre of the beam is
RA + RB = (20 × 5) + 20 + 40 + 20 (a) Wl/4 (b) Wl/2
RA + RB = 180 ...(1) (c) Wl (d) Wl2/4
taking moment about point A
Assam PSC CCE Pre 2015
Strength of Materials 269 YCT
 Ans. (a) For given load condition and beam.
Maximum bending moment (Mmax) = Wl/4

651. The area-moment theorem …….. respect to

Simply supported beam carrying a point load W at the
bending of beams states that the area of the
centre of the beam.
M/EI diagram between two sections of a beam
649. A simply supported beam of span ' ℓ ' is
subjected to a transverse point load 'W' acting gives :
at a distance of 'a' from the left support A. The (a) the difference in slopes between those two
reaction on the left support will be equal to sections
(a) W.a / ℓ (b) W. ( ℓ -a)/ ℓ (b) the difference in the maximum bending
(c) W/2 (d) w. ℓ /a strains between those two sections
VIZAG MT 2015 (c) the difference in deflections between those
Ans. (b) : two sections
(d) the difference in strain energies between
those two sections
BPSC Asstt. Prof. 29.11.2015
Ans. (a) : Moment Area method (Mohr's method)–
W × BC
RA = This method is suitable for cantilevers and simply
AB supported beams carrying symmetrical loadings and
W × (ℓ − a ) beams fixed at both ends i.e. those beams for which the
=
ℓ area and CG area of BMD can be found easily. This
W × AC means this method is not suitable for triangular loading
RB =
AB and irregular loading. This can also be used for non-
W×a prismatic bars.
=
ℓ This method is suitable for :
650. When the shear force diagram is a parabolic (i) Centilevers (because slope at the fixed end is zero)
curve between two points, it indicates that (ii) simply supported beams carrying.
there is (iii) Beams fixed at both ends (slope at each end is zero.
(a) a point load at two points
(b) no loading between the two points 652. A simply supported beam of 2 m length is
(c) a uniformly distributed load between the two applied with uniformly distributed load of 5
points kN/m through out. It is also applied with a
(d) a uniformly varying load between the two point load of 1 kN at its centre. The maximum
points bending moment in the beam will be–
Assam PSC CCE Pre 2015 (a) 2 kN-m
Ans. (d) : Simply supported beam with a gradually (b) 3 kN-m
varying load, from zero, at one end to w per metre at (c) 4 kN-m
the other end then the shear force diagram is a (d) 5 kN-m
parabolic curve between two points. TSGENCO AE 14.11.2015
Strength of Materials 270 YCT
Ans. (b) : Given – (a) 1/3 (b) 1/ 3
Length (l) = 2m, W = 5 kN-m (c) 1/6 (d) 1
4
(HPPSC AE 2014)
Ans : (d)

Maximum bending moment (case - I) due to point load

to centre of SSB
WL 1 × 2 fig (d) shows the B.M diagram obtained by super-
(BM)1 = = = 0.5 kN-m
4 4 imposing and µ'diagram. At any point distance x from
Maximum bending moment (case - II) due to uniformly A
load. Wx WL
2
M x = µ x + µ′x = −
WL 5× 2 × 2 2 8
(BM)II = = = = 2.5 kN-m L W L WL WL
8 8 At x = , M c = . − =
So, Total (case -I + case - II) 2 2 2 8 8
BM(max) = 2.5 + 0.5 = 3 kN-m Thus, the central B.M. is half of the B.M. for a freely
653. Choose the wrong statement supported beam. For point of Contraflexure
(a) The shear force at any section of a beam is Wx WL
equal to the total sum of the forces acting on Mx = − =0
2 8
any one side of the section. L
(b) The magnitude of the bending moment at any This gives x =
section of a beam is equal to the vector sum 4
of the moments (about the section) due to the 655. The three-moment for continuous beams was
forces acting on the beam on any one side of forwarded by :
the section. (a) Bernoulli (b) Clapeyron
(c) A diagram which shows the values of shear (c) Castigliano (d) Maxwell
forces at various sections of structured (HPPSC AE 2014)
member is called a shear force diagram. Ans : (b) The three moment theorem for continuous
(d) A simply supported beam is one which is beam was forwarded by Clapeyron.
supported on more than two supports. 656. A cantilever beam 6 meter long as shown in
MPPSC AE 08.11.2015
figure is subjected to a linearly varying loading
Ans. (d) : A simply supported beam is one which is not which has a maximum ordinate of 360 N/m at
supported on more than two supports. fig (a) and fig (b) the fixed end on the right. The moment as a
function of x is

If the beam is supported more than two supports called

continuous beam.
654. In fixed beam of length (L) with a concentrated
central load two points of contraflexure will
occur, each from supports at a distance of :
Strength of Materials 271 YCT
 (a) M = -3x3 (b) M = -6x3 R A x 2 wx 4
(c) M = -10x3 (d) M = -20x3 EIy′ = − + C1
ISRO Scientist/Engineer 24.05.2014 2 24L
Ans. (c) : R x 3 wx 5
EIy = A − + C1 x + C2 .....(1)
6 120L
Using conditions (a) at x = 0, y = 0, thus C2 = 0; and (b)
at x = L, y′ = 0, we get
wL3 R A L2
C1 = − ;C2 = 0
24 2
Putting above value in Eq. (1) for x = L, y = 0 we get
R x 3 wx 5 wL3 R A L2
− wx EIy = A − + −
Wx − x = ...(1) 6 120L 24 2
L
− wx x −R A L wL
3 4

( SF ) xx = × ...(2) ⇒ 0= +
L 2 3 30
wL
− wx 2
⇒ RA =
= 10
2L
658. A simply supported beam is shown in the
− wx 2 x
( BM ) xx = × figure below. The location is the point of
2L 3 maximum moment in the beam from reaction
− wx 3 RL is
= ...(3)
6L
So, here, w = 360 N/m
L=6m
−360 3
M = ×x
6×6 3
M = -10x 3 (a) meter
8
657. Consider the beam shown in figure. The beam
is simply supported at its left end A and fixed 8
(b) meter
at its right end B. It carries a load that varies 3
in intensity from zero at support A to 'w' at 18
support B according to the relation wx = (c) meter
(x/L)w. The reaction component at A is 5
21
(d) meter
5
ISRO Scientist/Engineer 24.05.2014
Ans. (b) :

(a) wL/10 (b) 2wL/5

(c) 5wL/2 (d) wL/5
ISRO Scientist/Engineer 24.05.2014
Ans. (a) : Moment at distance x from A is given as Taking section x-x from RL at a distance x
So at a location of max. bending moment, shear force
will be zero so
Sxx = 8 – wx = 8 – 3x = 0
8
x= m
3
659. A cantilever of length l carries a uniformly
1 wx x wx 3
M(x) = R A x − x × × = RAx − distributed load for a distance a (a<L) from the
2 L 3 6L free end. The shear force diagram will be
By double integration method (a) A rectangle
wx 3 (b) A triangle
EIy″ = R A x −
6L (c) A trapezium
Integrating once and twice we get (d) A parabola
Kerala LBS Centre For Sci. & Tech. Asstt. Prof. 2014
Strength of Materials 272 YCT
 Ans. (c) (a) 2 kN/m (b) 3 kN/m
(c) 4 kN/m (d) 5 kN/m
UKPSC AE-2013, Paper-I
Ans. (c) :

From shear force diagram

3w 6
660. With the usual sign convection for bending =
moment, i.e. a positive bending moment causes 3 1.5
compression in top fibre, what is sign, w = 4 kN/m
magnitude and location of numerically largest 663. A beam 10 m long hinged at both the ends is
bending moment in the beam shown in the subjected to a clockwise moment of 50 kNm, at
figure?
a distance of 3 m from one end. What is shear
force at the centre of the beam :
(a) 0 kN (b) 3 kN
(c) 5 kN (d) None of these
PSPCL AE, 2012
Ans. (c) : Shear force to constant throughout a length
(a) (+) 10 kN.m at C (b) (–) 8 kN.m at A with value of 5 kN.
(c) (–) 16 kN.m at A (d) (+) 4 kN.m at A
RPSC AEN Pre-2013
Ans. (b) :

Bending moment at A
(B.M)A = –3 × 4 + 4 = –8 kN.m
661. A simply supported beam of span 4 m with
hinged support at both the ends. It is carrying
the point loads of 10, 20 & 30 kN at 1, 2 and 3
m from left support. The RA & RB are
(a) 27.5 kN, 32.5 kN (b) 15 kN, 45 kN
(c) 25 kN, 35 kN (d) 32.5 kN, 27.5 kN 664. A beam A carries a point load at mid-span.
TNPSC AE 2013 Another identical beam B carries the same
Ans. (c) : load but as uniformly distributed load over the
entire span. The ratio of maximum bending
moment in beam A to that in beam B will be :
3 1
(a) (b)
2 2
(c) 3 (d) 2
APGENCO AE 2012
RA + RB = 60 ....(1) Ans. (d) : C – 1
Taking moment about point A
RB × 4 = 30 × 3 + 20 × 2 + 10 × 1
RB = 35 kN
RA = 25 kN
662. Uniformly distributed load ‘w’ act over per
unit length of a cantilever beam of 3m length.
If the shear force at the midpoint of beam is
6kN, what is the value of ‘w’:-
Strength of Materials 273 YCT
 Wl Ans : (b) In a double overhanging beam carrying UDL
Maximum Bending moment ( BM ) max = throughout its length, the number of points of
4
C–2 contraflexure are two.
Points of contraflexure:– Where the bending moment
will change sign from negative to positive or vice versa.
Such a point, where the bending moment change sign, is
known as a point of contraflexure.

wl 2
Maximum Bending moment ( BM ) max =
8
(∵ W = wL)
WL
( BM ) max =
8
( BM ) A ( BM )1 WL 8
= = ×
( BM ) B ( BM )2 4 WL
=2
665. In a propped cantilever beam, the number of
points of contra-flexure is
(a) 1 (b) 2
(c) 3 (d) 4
APPSC AEE 2012
Ans : (a)

In a propped cantilever beam, the number of point of

contra-flexure is one.
666. If a fixed beam is subject to a point load at mid 668. A cantilever is subjected to udl throughout the
length. If the maximum shear force is 200kN
span, total number of points of contra-flexure and maximum bending moment is 400kN, the
are span "L" of the beam in meters is
(a) 1 (b) 2 (a) 3 (b) 2
(c) 3 (d) zero (c) 4 (d) 8
APPSC AEE 2012 APPSC AEE 2012
Ans : (b) In a fixed beam is subjected to a point load at Ans : (c)
mid span total number of points of contra-flexure are
two.
W
RA = RB =
2
P
M x = ( 4x − L )
8
 L
 x < 2 
PL
M x =0 = M B = M A = −
8
P L  PL
M x=L / 2 = M C = 4 × − L =
8 2  8 Maximum shear force = 200kN
In this bending moment diagram bending moment wL = 200kN ……..(i)
Maximum Bending moment
change it's sign at two points.
667. In a double overhanging beam carrying udl wL2
=400kN ….....(ii)
throughout its length, the number of points of 2
contra flexure are for equation (i) and (ii)
(a) 1 (b) 2 wL2
(c) zero (d) 3 2 = 2 L = 4m
APPSC AEE 2012 wL
Strength of Materials 274 YCT
669. A cantilever beam AB of length l is subjected to Ans. (c) : An anticlockwise moment ‘M’ at C
an anticlockwise couple of 'M' at a section C,
distance 'a' from support. Then the maximum
shear force is equal to
M
(a) M (b)
2
(c) Zero (d) Ma
APPSC AEE 2012
Ans : (c)

673. Two cantilever steel beams of identical length

and of rectangular section are subjected to
A cantilever beam AB of length l is subjected to an same point load at their free end. In one beam,
anticlockwise couple of M at a section C, distance 'a' the longer side of section is vertical, while in
from support, then the maximum shear force is equal to the other, it is horizontal. Beams defect at free
zero. Because No transverse load acting on the beam. end:
670. At section of a beam sudden change in BM (a) equally irrespective of their disposition.
indicates the action of (b) more in case of longer side vertical.
(a) point load (b) couple (c) less in case of longer side horizontal.
(c) point load or couple (d) udl (d) less in case of longer side vertical.
APPSC AEE 2012 UKPSC AE 2012 Paper-I
Ans : (b) At section of a beam sudden change in BM Ans. (d) : less in case of longer side vertical.
indicates the action of couple. 674. The bending moment diagram for a simply
671. For uniform shear force throughout the span supported beam loaded in its centre is
of a simply supported beam, it should carry (a) a right angled triangle
(a) a concentrated load at the mid-span (b) an isosceles triangle
(b) a couple anywhere in the sections (c) an equilateral triangle
(c) udl over its entire span (d) a rectangle
(d) two concentrated loads equally spaced TNPSC ACF 2012
APPSC AEE 2012 Ans. (b) :
Ans : (b) For uniform shear force throughout the span
of a simply supported beam, it should carry a couple
anywhere in the section.
672. The bending moment diagram for a simply
supported beam AB of length ‘L’ is shown
below :
P×ℓ/2
RA =
ℓ
RA = P , RB = P
2 2

M
CD1 = CD2 =
2
Sagging moment : positive
Hogging moment : negative
What is the load acting on beam AB ?
M 675. In a beam where shear force is maximum, the
(a) An upward concentrated load at C.
2 bending moment will be
M (a) zero
(b) A downward concentrated load at C.
2 (b) minimum
(c) An anticlockwise moment ‘M’ at C (c) maximum
(d) A clockwise moment ‘M’ at C. (d) none of these
UKPSC AE 2012 Paper-I TNPSC ACF 2012
Strength of Materials 275 YCT
Ans. (d) : When shear force is zero, bending moment 680. Consider the following statements :
maximum. Shear force in beams is caused by
dM 1. Lateral forces acting on the beam
F=
dx 2. Variation of bending moment along beam
In a beam, where shear force is maximum the bending length
moment depends upon the type of loading and type of 3. A couple acting on the beam
beam.
676. For the beam shown in the fig, the elastic curve Which of the above statements is/are correct ?
between the supports B and C will be (a) 1 only (b) 1 and 2
(c) 2 and 3 (d) 1 and 3
UPSC JWM Advt. No.-50/2010
Ans. (b) : As we know, shear force in beams is caused
by the lateral forces acting on the beam and variation of
(a) circular (b) parabolic bending moment along beam length.
(c) elliptic (d) a straight line 681. Consider the following statements :
APPSC AEE 2012, ESE 1998
Ans. (a) : Form simple bending theory equation If a beam is to bend without being twisted,
σ M E 1. Plane of a load must contain one of the axes
= = of symmetry of the beam
y I R 2. The beam cross-section must have at least an
1 M
= = constant axis of symmetry
R EI Which of the above statements is/are correct ?
Therefore, the elastic curve is the deflection curve
which is circular between the support B and C. (a) 1 only (b) 2 only
677. Bending moment at any point is equal to the (c) Both 1 and 2 (d) Neither 1 nor 2
algebraic sum of : UPSC JWM Advt. No.-50/2010
(a) Moments of all vertical forces Ans. (a) : It is the concept of shear centre that of a beam
(b) Moments of all horizontal forces is to bend without twisting then plane of the load must
(c) Moment of forces on either side of the point
(d) All the above. contain one of the axis of symmetry of beam.
RPSC AE GWD, 2011
Ans. (c) : Bending moment at any point is equal to the
algebraic sum of moments of forces on either side of the
point.
678. What is the beam fixed at one end and free at 682. An overhang beam ABC of length 4m is
the other end called as? supported at A and B 3m apart. It is loaded
(a) Simply supported beam with UDL of 5kN/m along its entire length.
(b) Fixed beam Find the value of load 'P' at C such that the
(c) Overhanging beam reactions A and B are equal and opposite.
(d) Cantilever beam
VIZAG STEEL MT 2011
Ans. (d) : Cantilever beam

679.
(a) P = 10 kN (b) P = 30 kN
(c) P = 20 kN (d) P = 25 kN
ISRO Scientist/Engineer 2009
Ans. (c) : Overhang Beam ABC
length L = 4m
The magnitude of reaction at support B is :
(a) 2 kN (b) 4 kN
(c) 5 kN (d) 6 kN
UPSC JWM Advt. No.-52/2010
Ans. (b) :
As mentioned reactions at A and B are equal and
opposite. So, upward force P will be equal to total
uniformly distributed load.
Moment about A Moment about C RA + RB = P ⇒ 5 × 4 = 20
RD × 5 = 5 ×6 RB × 6 = 6 × 4 But RA = − RB
R D = 6kN R B = 4kN P = 20 kN

Strength of Materials 276 YCT

683. RB + RD = 3P
3P
RB = RD = (figure is symmetric)
2
∑MC = 0

P × − R B ×  − a  = 0
ℓ ℓ
The bending moment diagram for a beam AB
2 2 
ℓ 3P  ℓ
×  − a 
is shown in the above figure. What is the
P× =
nature of the beam? 2 2 2 
(a) Simply supported with a concentrated load at
ℓ ℓ ℓ
its mid length = −a ⇒ a =
3 2 6
(b) Simply supported and is subjected to a
couple M at its mid point 685. Couple M is applied at C on a simply
(c) Simply supported and carries a uniformly supported beam AB. The maximum shear on
varying load from zero at the support to a AC will be
maximum at its mid length
(d) A cantilever subjected to end moment M
UPSC JWM Adv. No-16/2009
Ans. (b) : The nature of the beam is simply supported
and is subjected to a couple M at its mid point. (a) Zero (b) M
2M M
(c) (d)
3 3
ISRO Scientist/Engineer 2008
Ans. (d) :

R1 + R2 = 0
∑MA = 0
684. A beam of length 'l' overhangs equally from its
R2 × 3 = -M
two supports. It carries three identical
concentrated loads at its ends and at mid M
R2 = −
length. If the bending moment at midpoint of 3
the beam is zero, what is the overhang length? M
R1 =
l l 3
(a) (b)
8 6 686. A beam AB is hinge-supported at its ends and
l L is loaded by couple P.c. as shown in the figure.
(c) (d)
4 3 The magnitude of shearing force at a section x
UPSC JWM Adv. No-16/2009 of the beam is
Ans. (b) : Let, a = Over hanging length
b = length of BD
l = Total length of beam

(a) 0
(b) P
(c) P/2L
(d) P.c/2L
∑Rv = 0 ISRO Scientist/Engineer 2008
Strength of Materials 277 YCT
Ans. (d) : Shear force at section (a) 200 KN (b) 150 KN
RA + RB = 0 (c) 100 KN (d) none of the above
∑MB = 0 WBPSC AE, 2007
RA × 2L = Pc Ans. (d) :
Pc
RA =
2L

For centre of cantilever becomes the point of inflexion,

687. A discontinuity or jump in the shear force (BM)at centre = 0
diagram occurs whenever a beam is loaded by ΣMC = 0
(a) a distributed moment 5
P × 5 = 20 × 5  
(b) a distributed force 2
(c) a concentrated moment
P = 50 kN
(d) a concentrated force
DRDO Scientists 2008 690. A simply supported beam of length 'L' acted
Ans. (d) : A discontinuity or jump in the shear force upon by a couple of magnitude 'M' at a length
diagram occurs whenever a beam is loaded by a 2
of L from one of the ends. The magnitude of
concentrated force. 3
688. For the beam loaded as shown in figure the maximum Bending moment in the beam is
magnitude of bending moment at the roller 1 2
support is (a) M (b) M
3 3
3
(c) M (d) M
2
OPSC Civil Services Pre 2006
Ans. (b) :
(a) 100 Nm (b) 1000 Nm
(c) 0 (d) 2000 Nm
DRDO Scientists 2008
Ans. (a) :

RA + RBV = 0
ΣMB = 0
– 100 + RA × 20 = 0
RA = 5 N
RBV = –5 N

RA + RB = 0
RB × l + M = 0
M
Bending moment at roller support, RB =
l
= 100 N-m
689. A cantilever 10 m long carries a uniformly M
RA = –
distributed load of 20 KN/m run throughout l
its length. It is propped by a force P at its free from fig. magnitude of Maximum bending moment is
end so that the centre of the cantilever
2
becomes the point of inflexion, then the M.
magnitude of P is 3
Strength of Materials 278 YCT
 Ans. : (c) In an I-section of a beam subjected to
4. Bending Stresses and Shear transverse shear force, the maximum shear stress is
developed at the centre of the web.
Stresses in Beams Shear stress distribution for I-section
691. The shear stress distribution over a rectangular
cross section of a beam is
(a) linear (b) cubic
(c) parabolic (d) none of these
OPSC AEE 2019 Paper-I
GPSC Executive Engineer 23.12.2018
TSPSC AEE 28.08.2017 (Civil/Mechanical)
694. Which one of the following is the preferable
APPSC AEE Mains 2016 (Civil Mechanical) cross-section of a beam for bending loads?
GPSC Lect. 23.10.2016, APPSC AEE 2012 (a) Circular (b) Annular circular
UKPSC AE 2012 Paper-I, ESE 2004 (c) Rectangular (d) I-section
Ans. (c) : For rectangular section beam, the shear stress RPSC Lect. (Tech. Edu. Dept.) 13.03.2021, Paper-I
distribution is parabolic and maximum shear stress is at TSPSC AEE 28.08.2017 (Civil/Mechanical)
neutral axis of the section.
TANGEDCO AE 2015, TNPSC ACF 2012
CSE Pre-2006
Ans. (d) : For a given cross-sectional area material and
bending moment applied.

'Z' value decrease → MR decreases →

692. Which one of the following statements in Bending stress (σb)max increases
correct? A beam is said to be uniform strength, I section is preferred for bending applications.
if : 695. The section modulus is expressed as–
(a) The bending moment is the same throughout
the beam (a) I/Y (b) E/I
(b) The shear stress is the same throughout the (c) M/1 (d) EI
beam OPSC AEE 2019 Paper-I
(c) The deflection is the same throughout the Nagaland PSC CTSE 2017, Paper-I
beam CGPSC AE 26.04.2015 Shift-I
(d) The bending stress is the same at every APPSC AEE 2012, TNPSC ACF 2012
section along its longitudinal axis Ans. (a) : Section modulus is the ratio of moment of
Assam PSC AE (IWT) 14.03.2021 Inertia about N.A. upon the for test point of section
RPSC IOF, 2020 from Neutral Axis (N.A.).
OPSC AEE 2019 Paper-I
I
GPSC Asstt. Director of Transport 05.03.2017 Z=
TSPSC Managers, 2015 Y
OPSC AEE 2015 Paper-I 696. Which is the correct relation in a beam?
TNPSC AE 2014, ESE 2006 M I R M y E
(a) = = (b) = =
J & K PSC Screening, 2006 σ y E I σ R
Ans : (d) : A beam is said to be uniform strength if, the M σ E M E σ
bending stress is the same at every section along its (c) = = (d) = =
longitudinal axis. I y R y R I
693. In an I-section of a beam subjected to OPSC AEE 2019 Paper-I
transverse shear force, the maximum shear Nagaland PSC (CTSE) 2018, Paper-I
stress is developed at TNPSC ACF 2012, UKPSC AE 2007 Paper -I
(a) The bottom edges of the top flange. ISRO Scientist/Engineer 2006
(b) The top edges of the top flange. Ans : (c) : Bending Formula –
(c) The centre of the web M σ E
(d) The upper edges of the bottom flange. = =
I y R
TSPSC Manager (Engg.) HMWSSB 12.11.2020 Where M = Bending moment
TSPSC AEE 28.08.2017 (Civil/Mechanical) I = Second moment of inertia of cross section
APPSC AE Subordinate Service Civil/Mech. 2016 about neutral axis.
UPPSC AE 12.04.2016 Paper-I σ = Bending stress on any layer
TNPSC AE 2014 Y = Distance of any layer from neutral layer.
ISRO Scientist/Engineer 2007 E = Young’s Modulus
ESE 2008 R = Radius of curvature of the neutral

Strength of Materials 279 YCT

697. The neutral axis of the cross-section a beam is 32M
that axis at which the bending stress is
σ b for circular cross-section
= πd
3
(a) zero (b) minimum
(c) maximum (d) infinity σ b for square cross-section 6M
UPRVUNL AE 05.07.2021 a3
APPSC AE Subordinate Service Civil/Mech. 2016 32  a 3 
KPCL AE 2016 =  3
π d 
Haryana PSC Civil Services Pre, 2014 3
APPSC AEE 2012 32  π   a 3   π 2 2
=   .  ∵ 4 d = a 
Ans. (a) : π  4   d 3 
∴ σb for circular-section > σb for square-section
Stress in circular beam is more so it will be less strong
as compare to square beam.
699. The maximum transverse shear stress in a
circular beam is (V-shear force, A-cross section
area)
3V 4V
(a) (b)
4A 3A
1V 4V
(c) (d)
3A πA
N–A shows neutral axis- at neutral axis there is no Haryana PSC AE (PHED) 05.09.2020, Paper-II
bending stress. Oil India Senior Officer 23.12.2020
M TSPSC AEE 28.08.2017 (Civil/Mechanical)
σbending = ×y at neutral axis (y = 0) ESE 2006
I
4 4V
σ bending = 0 Ans. (b) : τmax = τavg =
3 3A
Note-Inner fibre always subjected to maximum stress. 700. The ratio of average shear stress to the
698. Two beams of equal cross section area are maximum shear stress in a beam with a square
subjected to equal bending moment. If one cross section is :
beam has square cross section and the other (a) 2 (b) 1.5
has circular section, then: (c) 2/3 (d) 1
(a) Both the beams will be equally strong OPSC AEE 2019 Paper-I
(b) Square cross section beam will be stronger TSPSC AEE 28.08.2017 (Civil/Mechanical)
(c) Circular section beam will be stronger J&K PSC Civil Services Pre, 2010
(d) Depends on the loading condition GATE 1998
RPSC AE (GWD) 06.05.2016 Ans. (c) : For square cross-section –
NPCIL ET, 2015, TSPSC AEE 2015 Maximum shear stress (τmax) =
ISRO Scientist/Engineer 2011
ESE 1999
3
2
(
× average shear stress τavg )
Ans. (b) : Since, cross-section area of square and
circular cross-section are equal. τmax 3
=
Let d = diameter of the cross-section are equal τavg 2
a = side of square
τavg 2
[Given, Equal cross-section] or =
π 2 τmax 3
d = a2
4 701. The rectangular beam A has a length λ and
Now, σb for circular cross-section width b but depth d is double that of A. The
elastic strength of beam B will be ____ as
d compared to beam A.
M.y max M. 2 32M (a) same (b) double
σb = = =
I NA πd 4 πd 3 (c) one-fourth (d) four times
64 ISRO Scientist/Engineer (RAC) 12.01.2020
σb for square cross-section Nagaland PSC (CTSE) 2018, Paper-I
a TSPSC AEE 2015, JPSC AE 2013, Paper-V
M. Ans : (d) : We know that section modulus for beam A,
6M
σ b = 42 = 3 I bd 3
/12
a a ZA = A =
12 yA d /2

Strength of Materials 280 YCT

 bd 2 Ans. (c) Shear stress distribution across the rectangular
= section.
6 3
And section modulus for beam B τmax. = τavg = 1.5 τavg
2
b ( 2d ) /12 2bd 2
3
I
ZB = B = =
yB 2d / 2 3
(Depth is double that of A)
Since elastic strength of beam is directly proportional to
their respective section modulus, therefore,
ZB 2bd 2 6
= × 2 or ZB =4ZA
ZA 3 bd
705. A beam cross-section is used in two different
ZB = 4ZA orientations as shown in figure :
702. If the depth is kept constant for a beam of
uniform strength, then its width will vary in
proportional to (where M = Bending moment)
(a) M (b) M
(c) M2 (d) M3
Nagaland PSC (CTSE) 2018, Paper-I Bending moments applied in both causes are
Nagaland PSC CTSE 2016 Paper-I same. The maximum bending stresses induced
in cases (A) and (B) are related as
TSPSC AEE 2015, ESE 1995
(a) σA = σB (b) σA = 2σB
Ans. (a) : We know that
M σ σB σ
= (c) σA = (d) σA = B
I y 2 4
TSPSC Manager (Engg.) HMWSSB 12.11.2020
bd 3 APPSC AEE 2012, ESE 1997
I=
12 Ans : (b)
So M ∝ b
If depth is constant then M is directly proportional to its
width [b].
For a beam of uniform strength bending stress σ will be
also constant
(A) (B)
703. The ratio of maximum shear stress to average For first case:–
shear stress in a beam of rectangular section is 3
(a) 5/1 (b) 2/3 1 b b4
IA = × b  =
(c) 3/2 (d) 1.0 12  2  96
ISRO Scientist/Engineer 12.01.2020 M σA M ×12 × 8 σ A
= ⇒ =
ESE 2016, APPSC AEE 2012, GATE 1994 IA y b4 b/4
Ans. (c) :
24M
σA = ……….. (i)
Cross section τmax τ NA τmax b3
τavg τavg τ NA For second case
b 1
IB =   × ( b ) ×
Rectangle/square 3/2 3/2 1 3

Circle 4/3 4/3 1 2 12

4
Triangle 3/2 4/3 9/8 b
IB =
Diamond 9/8 1 9/8 24
704. A wooden rectangular beam, subject to M σB
=
uniformly distributed load, has an average IB y
shear stress (τavg) across the section. The
M σ
maximum shear stress (τmax) at natural axis is: 4
= B
(a) τmax = 0.5 τavg (b) τmax = 1.0 τavg b b
(c) τmax = 1.5 τavg (d) τmax = 2.0 τavg 12 × 2 2
(e) τmax = 2.5 τavg 12M
(CGPCS Polytechnic Lecturer 2017) σB = ……… (ii)
b3
APPSC AEE Screening Test 2016
For equation first and second σ A = 2σB
APPSC AEE 2012, ESE 2009
Strength of Materials 281 YCT
706. Which one of the following represents the As we know,
shearing stress distribution over a cross section In case of beams with circular section, the ratio of
of a T-beam ? maximum shear stress to average shear stress
4
(a) (b) = τavg = 1.33 τavg
3
The maximum shear stress developed in the beam is
33% greater than the average shear stress.
709. A rectangular strut is 150 mm wide and 120
mm thick. It carries a load of 180 kN at an
(c) (d) eccentricity of 10mm in a plane bisecting the
thickness as shown in the figure. The maximum
and minimum intensities of stress in the section
UPSC JWM Advt. No.-50/2010 will be respectively.
CSE Pre-2000
ESE 2000
Ans. (d) :

As we know that,
F AY
τs = s
I NA b
1 (a) 10 MPa and 4 MPa (b) 14 MPa and 4 MPa
τs ∝
b (c) 10 MPa and 6 MPa (d) 14 MPa and 6 MPa
707. Circular beam of uniform strength can be CGPSC AE 15.01.2021, 16.10.2016
made by varying the diameter in such way Ans. (d) : Given,
that: P = 180 kN
(a) M/Z is constant (b) σ/Y is constant Bending moment, M = 180 × 103 × 10
(c) M/R is constant (d) E/R is constant
My
VIZAG MT 2015 Bending stress ⇒ σ b =
UKPSC AE 2007 Paper -I I
Ans. (a) : According to bending equation–  150 
180 × 104 ×   ×12
M E σ  2 
= = b =
I R y max 120 × (150)3

Uniform strength ( σ b ) =
M σ b = 4 MPa
Z Direct stress,
M
is constant P 180 ×103
Z σd = = = 10MPa (compressive)
708. In case of beam of circular cross-section A 120 ×150
subject to transverse loading, the maximum Maximum intensity of stress = –σb + σd
shear stress developed in the beam is greater = – 4 – 10
than the average shear stress by = |–14| MPa = 14 MPa
(a) 10% (b) 25% Minimum intensity of stress = + σb + σd
(c) 50% (d) 33% = + 4 – 10
RPSC Lect. (Tech. Edu. Dept.) 13.03.2021, Paper-I
= | – 6 MPa | = 6 MPa
APGCL AM, 2021
710. When a rectangular beam is loaded
Ans. (d) :
transversely, the maximum compressive stress
is developed on the
(a) top layer (b) bottom layer
(c) neutral axis (d) every cross-section
Sikkim PSC (Under Secretary), 2017
Haryana PSC Civil Services Pre, 2014
Strength of Materials 282 YCT
Ans. (a) : For simply supported beam– 32M
σ max =
πD 4
16T
τmax
= πD =
3 T
Right answer – (a) σ max 32 M 2M
Haryana PSC gives (b) πD 3
Note-In case of cantilever beam compressive stress is 713. Which one of the following conditions will
obtained at bottom layer. qualify for a constant strength beam?
711. In the case of an l-section beam maximum (a) Bending moment along the whole length of
shear stress is at beam is constant
(a) The junction of Top flange and web (b) Shear force at every section of the beam is
(b) Middle of the web constant
(c) Both (A) and (B) (c) The beam should be of uniform section
(d) The junction of bottom flange and web (d) Ratio of bending moment to its section
APGCL AM, 2021, ESE 2008 modulus at every section must be constant.
Ans. (b) : UKPSC AE 2012 Paper-I
UPSC JWM Advt. No.-52/2010
Ans. (d) : Ratio of bending moment to its section
modulus at every section must be constant for a
constant strength beam
The maximum shear stress in I section of beam is at M
middle of the web. Strength of beam = =
Z
712. If a circular shaft is subjected to a torque T Since, the strength of a beam depends upon its section
and a bending moment M, the ratio of the modulus (Z).
maximum shear stress to maximum bending 714. A shaft was initially subjected to bending
stress is : moment and then it was subjected to torsion. If
(a) 2 M/T (b) T/2M the magnitude of bending moment is found to be
(c) 2T/M (d) M/2T the same as that of the torque, then the ratio of
ISRO Scientist/Engineer (RAC), 10.03.2019 maximum bending stress to shear stress will be.
ISRO Scientist/Engineer 12.05.2013 (a) 0.25 (b) 0.50
16T (c) 2 (d) 4
Ans. (b) : τ max = APPSC IOF, 2009, WBPSC AE 2003
πd 3
T Gθ τ M σ E 
= = Ans. (c) :  = = 
J l R  I y R
TR T × D / 2 M σ σI
τ= = from = ⇒ M=
J π 4 I y y
D
32 σ πd 4
M= × [y = d / 2]
y 64
σ × 2 × π × d 4 πσd 3
16T M= = ...(i)
τmax = d × 64 32
πD 3
T τ Gθ
32 M ∵ = =
σ max = J R L
πd 3
τ.J π 4
σ M E from T = ∵ R = d/2 & J = d
= = R 32
y I R
τ × 2 × πd 4 πd 3 × τ
T= = ....(ii)
d × 32 16
bending moment = torque
M M=T
σ= ×y
I πd 3 σ πd 3 .τ
=
M 32 16
σ= ×D/2 σ
π 4 =2
D
64 τ

Strength of Materials 283 YCT

715. The variation of bending stress in a curved Ans. (d) : We know that
beam is ............ in nature.
(a) Linear (b) Cubic
(c) Parabolic (d) Hyperbolic
HPPSC AE 2018
I
ISRO Scientist/Engineer 24.05.2014 Z = N−A
Ans. (d) : The variation of bending stress in curved Y max.

beam is given as d
Ymax =
My 2
σb =
A.e.[ R n − y ]
3
bd
IN-A =
Where, 12
M = Bending moment acting at the given section about bd 3 × 2
the centroidal axis. Z=
A = Area of cross-section 12 × d
e = Distance from the centroidal axis to the neutral bd 2
axis R – Rn Z =
6
R = Radius of curvature of the centroidal axis
Rn = Radius of curvature of the neutral axis 718. When a beam is subjected to a transverse
y = Distance between the neutral axis to the considered shearing force, the shear stress in the upper
fibre which bending stress needed to be calculated. fibers will be
This equation shows that the stress distribution is (a) Maximum (b) Minimum
Hyperbolic in nature. (c) Zero (d) Depends on other data
716. An I-section of a beam is shown in the figure Nagaland PSC CTSE 2017, Paper-I
below. If the shear stress at point P which is ISRO Scientist/Engineer 2008
very close to bottom of the flange is 12 MPa, Ans. (c) :
the shear stress at the point Q close to the
flange is:

Shear stress varies parabolically. At top & bottom fibres

it's value is zero & at neutral axis it's value is maximum.
719. In the case of a curved beam subjected to pure
bending, which of the following is true?
(a) 40 MPa (b) 12 MPa (a) Neutral axis coincides with the centroidal axis
(c) Indeterminate (d) 60 MPa
(b) Neutral axis lies between the centroidal axis
BHEL ET 2019, ESE 2001 and the center of curvature.
Ans. (d) : Given - shear stress of flange (c) Location of neutral axis depends upon the
( τ flange ) = 12MPa magnitude of bending moment
(d) There is no neutral axis
shear stress of web (τweb) =?
width of flange (bt) = 100 mm DRDO Scientists 2008
width of web (bw) = 20 mm Ans. (b) : In case of curved beam subjected to pure
bending, neutral axis lies between the centroidal axis
from -
( τ flange ) = ( b web ) and the center of curvature.
( τ web ) ( b flange ) In case of normal beam, subjected to pure bending,
neutral axis coincides with the centroidal axis.
12 20 1
= = ( τ ) = 60MPa 720. Neutral plane of the beam
( τ web ) 100 5 , web (a) Is in the middle
(b) Is one whose length remains unchanged
717. Section of the modulus (Z) for a rectangular
during deformation
section with width (b) and depth (d) is given
by: (c) Passes through centers of gravity
(d) Lies at the top most fiber
bd 3 bd 2
(a) Z = (b) Z = ISRO Scientist/Engineer (RAC) 07.05.2017 GPSC
12 12 Asstt. Director of Transport 05.03.2017
3 2
bd bd Ans. (b) : At neutral plane there are no bending stress
(c) Z = (d) Z = (σb = zero) so whose length unchanged during
6 6
deformation.
bd 2
(e) Z =
8
CGPCS Polytechnic Lecturer 2017
TNPSC AE 2013
Strength of Materials 284 YCT
721. A beam of square section, side of square 'a' is a4
used with its diagonal horizontal the modulus ISH =
of section of beam is given by 12
a
a4 a4 y=
(a) (b) 2
6 2 12 2 a3
ZSH =
a3 a3 6
(c) (d)
6 2 12 2 a3
APPSC AEE SCREENING 17.02.2019 ZSH =
6
MPSC Lect. (Prod.) 2018
Ans. (c) a4
ISD =
12
a
y=
2
a4 × 2 a3
a4 ZSD = =
I= 12 × a 6 2
12 we know that
a M∝Z
Ymax =
2 MSH ZSH a 3 6 2
= = × 3 = 2
a4 M SD ZSD 6 a
I a3 724. For a circular section the ratio of Maximum
Section modulus (Z) = = 12 = shear stress to the average shear stress is :
Ymax a 6 2
2 (a) 1.13 (b) 1.23
(c) 1.33 (d) 1.43
722. If E= elasticity modulus, I = moment of inertia
about the neutral axis and M = bending GMB AAE 25.06.2017
moment in pure bending under the symmetric APPSC AE Subordinate Service Civil/Mech. 2016
loading of a beam, the radius of curvature of Ans. (c) : The ratio of maximum shear stress to average
the beam : shear stress of a circular section.
(i) Increases with E (ii) Increases with M 4
(iii) Decreases with I (iv) Decreases with M τmax = τavg
Which of these are correct? 3
(a) (i) and (ii) (b) (ii) and (iii) 725. The maximum bending stress in a curved beam
(c) (iii) and (iv) (d) (i) and (iv) having symmetrical section always occurs at
OPSC AEE 2019 Paper-I, ESE 2013 which one of the following?
Ans : (d) : In pure bending (a) Inside fibre (b) Outside fibre
σ M E (c) Neutral axis (d) Centroidal axis
= = CGPSC AE 15.01.2021
y I R
E .I UPSC JWM Adv. No-16/2009
R= Ans. (a) :
M
R increase with E
R decrease with M
Statement (i) and (iv) are correct.
723. The ratio of flexural strength of a square
section with its two sides horizontal to its
diagonal horizontal is
(a) 2 (b) 2
2 The bending stress in a straight beam is linear whereas
(c) 2 2 (d)
5 that in a curved beam is hyperbolically varying over the
APPSC AEE 2012, ESE 2012 cross-section.
Ans : (a) • If the section is symmetrical, the maximum bending
stress always occurs at radius ri of the inside fiber.
• If the section is unsymmetrical, the stress may be
maximum either at Ri or Ro.
• The neutral axis where the bending stress is zero is
located between the centroidal axis at radius R and
the neutral axis at radius Ri or Ro.

Strength of Materials 285 YCT

• The distance ho is to be considered positive for (d) have the same strength on the original beam
distances towards the center of curvature and because the cross-sectional area remains the
negative for distances away from the center of same
curvature. TSPSC AEE 28.08.2017 (Civil/Mechanical)
726. Consider the following statements : CSE Pre-1999
Two beams of identical cross-section but of Ans. (b) : Flexural strength of beam depends upon it's
different materials are subjected to equal section modulus.
bending moments at a certain section along
1
their length: × 0.6 ×13
1. Maximum bending stress at the section of two Z1 = 12 = 0.1m3
beams is same 0.6 / 2
2. Maximum shearing stress at the sections of two 1
×1× 0.63
( 0.6 ) = 0.6 × 0.1
2
beam is same.
3. Maximum bending stress at the section of two Zll = 12 =
0.6 6
beams will depend on the elastic modulus of the 2
beam materials = 0.06 m3
4. Curvature of the beam having greater value of 0.06
E will be larger. Z11 < Z1 = = 0.6
Which of the above statements is/are correct? 0.1
(a) 1 and 2 (b) 1 and 3 Beam would be weakened by 0.6 times.
(c) 2 and 3 (d) 4 only 729. Three beams of 100 mm × 1200 mm section are
UPSC JWM Adv. No-16/2009, CSE Pre 2007 made from RCC, aluminium and timber
Ans. (a) : According to bending equation – respectively. These three beams are loaded
identically with similar supports, Maximum
M σb E bending stress will occur in which beam?
= = …(i)
I y R (a) Timber beam
(b) Aluminium beam
M .y M
⇒ σb = or …(ii) (c) RCC beam
I Z (d) The three beams will have same maximum
Bending stress = σb bending stress
If two beams of identical cross-section but of different TSPSC AEE 28.08.2017 (Civil/Mechanical)
materials, when subjected to the equal bending moment CSE Pre-2008
at a certain section along their length then according to
bending equation, the bending stress not depend on the Ans. (d) : Bending stress does not depend upon
elastic properties of the material i.e., Maximum bending material.
stress of the material i.e. Maximum bending stress at the 730. A beam of rectangular cross-section is to be cut
section of two beams is same. from a circular beam of diameter D. What is
And also Maximum shearing stress at the section of two the ratio of the depth of the beam to its width
beams is same. for maximum moment of resistance?
727. A square section with side 'X' of a beam is (a) 3 (b) 2
subjected to a shear force 'S'. The magnitude
of shear stress at the top edge of the square is 3 3
(c) (d)
1.5S S 2 2
(a) (b)
X2 X2 TSPSC AEE 28.08.2017 (Civil/Mechanical)
0.5S APGENCO AE 2012
(c) 2
(d) Zero
X Ans. (b)
APPSC AEE 2012, WBPSC AE, 2007
Ans. (d) : For square section,

Shear stress variation is parabolic shear stress at the top b 2 + d 2 = D2

edge of the square is zero.
d 2 = D2 – b 2
728. A cantilever beam of rectangular cross-section
is 1 m deep and 0.6 m thick. If the beam were for the strongest beam in bending its 'z' should be
to be 0.6 m deep and 1 m thick, then the beam minimum.
would
(a) be weakend 0.5 times
(b) be weakend 0.6 times
(c) be strengthed 0.6 times
Strength of Materials 286 YCT
 dz
=0
db
bd 2
z=
6

z=
(
b D2 − b2 )
Section modulus about neutral axis (NA)
6
D 2 − 3b 2 D bd 3
=0⇒b= bd 2
6 3 Z NA = 12 =
d 6
b 2 + d 2 = D2
2
D2 2 2
d 2 = D2 – ⇒d= D bd 2
3 3 Z N.A =
6
d = b. 2
733. When a shaft with diameter (d) is subjected to
depth pure bending moment (Mb), the bending stress
= 2
width (σb) induced in the shaft is given by:
731. If a shaft is simultaneously subjected to a  32M b   64M b 
torque T and bending moment M, the (a) σ b =  3 
(b) σ b =  3 
maximum shear stress is  π d   πd 
16  64M b   32M b 
(a) (M + T) (c) σ b =  2 
(d) σ b =  2 
πD3  πd   πd 
16 APPSC Poly Lect. 13.03.2020
(b) (M + M 2 + T 2 ) CIL (MT) 2017 IInd Shift
πD 3

16 Ans. (a) : In case of pure bending, the bending stress

(c) (M − M 2 + T 2
πD3  32M b 
induced in the shaft σ b =  3 
.
16
M2 + T2  πd 
(d)
πD3 734. The bending stress in a beam is ______ section
Nagaland PSC CTSE 2017, Paper-I modulus.
APPSC AEE Mains 2016 (Civil Mechanical) (a) inversely proportional to two times
Ans. (d) : We know, (b) directly proportional to
M I (c) inversely proportional to
σb = b & Z = (d) None of the above
Z Y Assam PSC AE (IWT) 14.03.2021
M d 32M Ans. (c) : The bending moment equation
σb = × =
 πd 4  2 πd 3 M σ E M M
  = = σ= =
 64  I y max R I / y max Z
16T From the equation we said that bending stress inversely
τ = 3 ⇒ ∴ Tequivalent = M 2 + T 2 proportional to the section modulus
πd
16 × Teq Section modulus Z = I/ymax
16
τ max = = × M2 + T2 735. The rectangular tube shown is extruded from
πd 3
πD3 an aluminum alloy for which σy = 150 MPa, σU
732. Section modulus (Z) of a beam depends on : = 300 MPa and E = 70 GPa. Neglecting the
(a) The geometry of the cross-section effect of fillets, the bending moment M for
(b) Weight of the beam which the factory of safety will be 3.00 is :
(c) Only on length of the beam
(d) Density of the beam
HPPSC Lect. (Auto) 23.04.2016
HPPSC LECT. 2016
Ans. (a) : Section modulus is a geometrical
phenomenon.
My max
Bending stress (σbending) =
I NA
(a) M = 9.20 kN-m (b) M = 3.0 kN-m
I (c) M = 6.1 kN-m (d) M = 15 kN-m
Section modulus (ZN.A) = NA
y max CGPSC AE 15.01.2021
Strength of Materials 287 YCT
 M σ σ = 20 kN-m
Ans. (a) : = → M = ⋅I = 20 × 106 Nmm
I y y
80 × 1203 (80 − 2 × 8)(120 − 2 × 8)3 20 × 106  200 
I= − = × 
100 × ( 200 )  2 
3
12 12
I = 5.52 × 10–6 m4 12
σ 300 σ b = 30 MPa
σallowable = U = = 100 MPa 738. For a cast iron simply supported beam with
FOS 3
−6 low strength in tension and high strength in
5.52 × 10 compression, due to the transversely applied
M = 100 × 106 × = 9.2 kN-m
60 × 10−3 load, with beam bent concave upwards,
736. At the neutral axis of the beam, the layers ___. following section will be better.
(a) do not undergo any strain
(b) are subjected to tension
(c) are subjected to compression
(d) are subjected to maximum bending stress
VIZAG Steel MT 24.01.2021
Ans. (a) : At the neutral axis of the beam, the layers do
not undergo any strain and any stress. Therefore,
bending stress at neutral axis is zero and no load acting
at neutral axis.

737. A simply supported beam of width 100 mm,

height 200 mm and length 4 m is carrying a
uniformly distributed load of intensity 10
kN/m. The maximum bending stress (in MPa)
in the beam is ______ (correct to one decimal JPSC AE 10.04.2021, Paper-II
place). Ans. (d) : If simply supported beam bend concave
upwards then compressive stress developed at top fibre
and tensile stress developed at bottom fibre.

(a) 30 (b) 120 Cast iron is low strength in tension and high strength in
(c) 90 (d) 60 compression so bottom fibre has to be more area (more
Assam Engg. College AP/Lect. 18.01.2021 material) and top fibre is strong in compression so no
Ans. (a) : need to add more material on top fibre so better section
will be trapezoidal section.

Given,
b = 100 mm
D = 200 mm 739. Variation of bending stresses at any section on
RA + RB = 40 a beam is related to the bending moment by
RA × 4 = 10 × 4 × 2 _______.
RA = 20 kN (a) St Venant's formula (b) flexural formula
RB = 20 kN (c) torsional formula (d) bending formula
HPPSC Workshop Suptd. 08.07.02021
bd 3 100 × ( 200 )
3

I= = Ans. (b) : flexural formula

12 12
740. A wooden beam AB supporting two concentrated
σ M E
= = loads P has a rectangular cross-section of width =
y I R 100 mm and height = 150 mm. The distance from
M  D 200  each end of the beam to the nearest load is 0.5 m.
σ = ×y y = =  If the allowable stress in bending is 11 MPa and
I  2 2  the beam weight is negligible, the maximum
Mmax = 20 × 2– 10 × 2 × 1 permissible load will be nearly
Strength of Materials 288 YCT
 (a) 5.8 kN (b) 6.6 kN 3 3
(c) 7.4 kN (d) 8.2 kN b3 = πd
ESE 2020 16
1/ 3
 3π 
Ans. (d) : Mmax = P × 0.5 = P × 500 Nmm b =   ⋅d
6M max M max  16 
σmax = = 2 2 2/3
bd 2 bd Wsquare As × ℓ b2 4b 4  3π 
6 = = =   =  
Wcircular A circular × ℓ π
d2 π  d  π  16 
4
Wsquare 1
6 × P × 500 =
11 = Wcircular 1.12
100 ×1502 743. The average shear stress in a circular cross
P = 8250 N = 8.25 kN section of a beam is equal to:
741. When a beam is subjected to bending moment (a) 0.85 times the maximum shear stress
alone, the bending stress at any point from the (b) 0.55 times the maximum shear stress
neutral axis (c) 0.75 times the maximum shear stress
(a) equal (d) 0.25 times the maximum shear stress
(b) varies directly proportional with respect to NLCIL GET 17.11.2020, Shift-II
distance from neutral axis Ans. (c) : Maximum shear stress occurs at neutral axis
(c) varies inversely proportional with respect to and y = 0
distance from neutral axis
4
(d) stress is independent of the distance Max shear stress = average shear stress
ISRO Scientist/Engineer (RAC) 12.01.2020 3
For circular section
Ans. (b) : For pure bending condition,
3
Avg shear stress = Maximum shear stress
4
= 0.75 times the maximum shear stress.
744. When a rectangular beam is loaded
transversely, the maximum tensile stress is
developed on the
(a) top layer (b) bottom layer
(c) neutral axis (d) every cross-section
RPSC IOF, 2020
Ans. (b) : When a rectangular beam is loaded
transversely, the maximum tensile stress is developed
on the bottom layer.

My
( σb )max =
I NA
( σb ) ∝ y
Compressive stress developed on the top layer.
• The bending stress at any point from the neutral axis Stress is zero at the neutral axis.
varies directly proportional with respect to distance
745. In a triangular section, the maximum shear
from neutral axis. stress occurs at
742. A square beam and a circular beam of same (a) Apex of the triangle (b) Mid of the height
material have the same length, same allowable (c) 1/3 of the height (d) base of the triangle
stress and the same bending moment. The ratio RPSC IOF, 2020
of weights of the square beam to that of
circular beam is Ans. (b) : In a triangular section, the maximum shear
stress occurs at mid of the height.
(a) 1/2 (b) 1
→ For rectangular section Fmax = h/2
(c) 1/1.12 (d) 1/3
→ For triangular section, the shear stress has a
ISRO Scientist/Engineer (RAC) 12.01.2020 parabolic variation.
M
Ans. (c) : Bending stress, σ =
Z
For same stress and moment,
Zsquare = Zcircular
b3 π 3
= d
6 32
Strength of Materials 289 YCT
746. For a standard crane hook (as shown in figure) Ans. (c) : Bending Equation–Every element of the
subjected to a constant load F, the tensile stress beam is bend with the some radius of curvature
at a given section A-A will be maximum at___ M σ E EI
= = or R =
I y R M
A non-uniform bending is deformation is the presence
of shear force and BM change along the axis of beam.
748. The normal stresses in a cross section of a
beam under lateral loading
(a) varies linearly with distance from the neutral
axis
(b) varies quadratically with distance from the
neutral axis
(c) does not vary with distance from the neutral
axis
(d) varies inversely with distance from the
neutral axis
(a) innermost edge (b) neutral axis Haryana PSC AE (PHED) 05.09.2020, Paper-II
(c) outermost edge (d) centroidal axis Ans. (a) : Lateral load produce bending. Bending stress
APPSC Poly Lect. 13.03.2020 developed in cross section is normal to cross section so
Ans. (a) : For a standard crane hook (as shown in called normal stress. Bending stress linearly varies with
distance from the neutral stress.
figure) subjected to a constant load F, the tensile stress
at a given section A-A will be maximum at innermost 749. The shear in a cross section for a rectangular
edge. beam under lateral loading
(a) does not vary with distance from the neutral
axis
(b) varies linearly with distance from the neutral
axis
(c) varies quadratically with distance from the
neutral axis
(d) varies inversely with distance from the
neutral axis
Haryana PSC AE (PHED) 05.09.2020, Paper-II
Ans. (c) : Shear stress in the cross section for any
section beam under lateral loading varies quadratically
(parabolically) with distance from neutral axis.
750. Product of allowable bending stress and
sectional modulus is
(a) Moment of rigidity
Stress distribution in a given section- (b) Radius of gyration
(c) Moment of resistance
(d) Moment of inertia
CIL MT 27.02.2020
Ans. (c) : The maximum bending moment which can be
carried by given section for a given maximum value of
stress is known as moment of resistance (Mr)
I
F MY Section modulus (Z) =
Where, σi = σ t = + y
A Ae R
I
F MY Mr = σ M r = σZ
σo = σc = − y,
A AeR
Moment of resistance=Bending stress×Section modulus
747. A non uniform bending refers to flexure of a
beam 751. A beam of span 3 m and width 90 mm is loaded
as shown in the figure. If the allowable bending
(a) in presence of uniformly distributed loaded in stress is 12 MPa, the minimum depth required
the beam for the beam will be
(b) in presence of uniform bending moment
(c) in presence of shear forces
(d) in absence of shear forces
Haryana PSC AE (PHED) 05.09.2020, Paper-II
Strength of Materials 290 YCT
 (a) 218 mm (b) 246 mm ex. : Chain links with tensile load, crane hooks
(c) 318 mm (d) 346 mm subjected to downward load, ring with uniform tensile
ESE 2020 load.
Ans. (b) : 754. Section modulus of hollow circle with average
RA + RB = 29 diameter 'd' and with thickness 't' is equal to
RA × 3 = 12 × 2.4 + 5 × 1.5 + 12 × 0.6 4 4
(a) td 2 (b) t 2 d 2
RA = 14.5 5 5
RB = 14.5 4 2 5
Mmax = 14.5 × 1.5 – 12 × 0.9 (c) t d (d) td 2
= 10.95 kNm 5 4
6
= 10.95 × 10 N mm TNPSC AE 2019
MY 6 M max  πd 3 
Bending stress σ b = = Ans. (a) : I x − x = I y − y =  ×t
I bd 2
 8 
 
6 × 10.95 × 10 6
12 =
90 × d 2
d = 246.64 mm
752. A vertical hollow aluminium tube 2.5 m high
fixed at the lower end, must support a lateral
load of 12 kN at its upper end. If the wall
1
thickness is th of the outer diameter and the
8
allowable bending stress is 50 MPa, the inner
diameter will be nearly d d
(a) 186 mm (b) 176 mm Ro = + t ≃
2 2
(c) 166 mm (d) 156 mm
 πd 2 
ESE 2020 Z x−x = Z y−y  × t
Ans. (d) :  4 
1 4
t= d 0 ≈ td 2
8 5
di = d0 – 2t 755. Pure bending means :
1 3 (a) The bending beam shall be accompanied by
= d0 − 2 × d0 = d0
8 4 twisting
di 3 (b) Shear force is zero
= (c) There is no twisting
d0 4 (d) None of these
32M OPSC AEE 2019 Paper-I
σ b = 3 max 4
πd 0 (1 − K ) Ans : (b) : Pure bending- Pure bending is a condition
of stress where a bending moment is applied to a beam
32 × 12 ×103 × 2500 without the simultaneous application of axial shear or
50 =
  
3
4
 torsional forces. Beam that is subjected to pure bending
πd30 1 −    means the shear force in the particular beam is zero and
  4   no torsional or axial loads are presented. Pure bending
d0 = 207.54 is also the flexure (bending) of a beam that under a
3 constant bending moment therefore pure bending only
d i = × 207.54 = 155.66 mm occurs therefore pure bending only occurs when the
4
shear force in equal to zero.
753. Which of the following is NOT an example of
756. The point within the cross-sectional plane of
curved beam?
beam through which the resultant of the
(a) Chain links with tensile load external loading on the beam has to pass
(b) Crane hooks subjected to downward load through to ensure pure bending without twisting
(c) Rings with uniform tensile load of the cross-section of the beam is called :
(d) Cantilever beam with point load (a) Moment centre (b) Centroid
APPSC Poly Lect. Automobile Engg., 2020 (c) Shear center (d) Elastic center
Ans. (d) : Curved beams : Beam whose axis is not OPSC AEE 2019 Paper-I
straight and is curved in the elevation is said to be a Ans : (c) : Shear center is the point within the cross
curved beam is the applied loads are along the y- sectional plane of a beam through which the resultant of
direction and the span of the beam is along the x- the external loading on the beam has to pass through to
direction. The axis of the beam should have a ensure pure bending without twisting of cross section of
curvature in xy plane. the beam.

Strength of Materials 291 YCT

757. A rectangular strut is 150 mm wide and 120 1.887 × 108
mm thick. It carries a load of 180 kN at an =
eccentricity of 10 mm in a plane bisecting the 200
thickness as shown in the figure = 943573.33 mm3
Permissible stress
M
σmax =
Z
Bending moment M = σmax × Z
= 150 × 943573.33
= 1.415 × 108 N-mm
759. A hollow circular bar used as a beam has its
outer diameter thrice the inside diameter. It is
subjected to a maximum bending moment of 60
The maximum intensity of stress in the section MN m. If the permissible bending stress is
will be limited to 120 MPa, the inside diameter of the
(a) 14 MPa (b) 12 MPa beam will be.
(c) 10 MPa (d) 8 MPa (a) 49.2 mm (b) 53.4 mm
ESE 2019 (c) 57.6 mm (d) 61.8 mm
Ans. (a) : Given, ESE 2019
P = 180 kN Ans. (c) : Given,
b = 150 mm do = 3 di, M = 60 MN-mm = 60 × 106 N-mm
d = 120 mm
e = 10 mm d
ymax = o
Resultant normal stress is maximum at the right side 2
fiber of the cross section, because the line of action of σmax ≤ σ
eccentric axial compressive load is nearer to this fiber. M
Maximum intensity of stress × y max ≤ σ
I
= σc + σb
−P M 60 ×106 d
σmax = − ⋅ o ≤ 120
π 4
A Z
64
( d o − d i4 ) 2
−P P × e −P 6P × e
= − = − 60 × 106 × 64 d
A  db2  bd db 2 × o = 120
   d   2 4
 6  πd 4o 1 −  i  
−P  6 × e  180 × 103  6 × 10    d o  
= 1 + = − 1+
bd  b  120 × 150  150  384 × 107
= –14 MPa = 14 MPa (compressive) = d 3o
 1  4
758. The cross-section of the beam is as shown in the 2 × 120 × π 1 −   
figure.  3 
do = 172.79 mm
di = do/3
di = 57.6 mm
Note : Bending moment value should be in MN-mm but
in question it is given in MN-m.
760. In a beam of I-section, which of the following
parts will take the maximum shear stress when
subjected to traverse loading?
1. Flange 2. Web
If the permissible stress is 150 N/mm2, the Select the correct answer using code given below.
bending moment M will be nearly (a) 1 only (b) 2 only
(a) 1.21 × 108 N mm (b) 1.42 × 108 N mm (c) Both 1 and 2 (d) Neither 1 or 2
8
(c) 1.64 × 10 N mm (d) 1.88 × 108 N mm ESE 2019
ESE 2019 Ans. (b) :
Ans. (b) : Moment of inertia of I-section
I = Iweb – 2 (Iflange)
 200 × 4003 96 × 3803 
I= −2×
 12 12 
= 1.887 × 10 mm8 4

I Shear stress is max at the neutral axis of I-section i.e. in

Section modulus (Z) =
y max the web portion.

Strength of Materials 292 YCT

761. A pull of 100 kN acts on a bar as shown in the Ac = At
figure in such a way that it is parallel to the bar 1 1
1 
axis and is 10 mm away from xx: × b × b1 =  × 2h × h 
2 2 2 
h2
b1 =
b
from similar triangles
b → b1
h → 2h
The maximum bending stress produced in the b × 2h = b1h
bar at xx is nearly. b
(a) 20.5 N/mm2 (b) 18.8 N/mm2 b= 1
2
(c) 16.3 N/mm (d) 14.5 N/mm2 2
ESE 2019 h2
b=
Ans. (b) : b× 2
h2
b2 =
2
h
b=
2
763. Calculate the shear force and bending moment
P = 100 kN (Tensile) at point B for the beam AB subjected to
Bending moment (M) = 100 × 103 × 10 linearly varying load as shown in the figure.
= 106 N-mm The value of the linearly varying load at the
80 point is 6 kN/m and 4 kN/m, respectively. Point
ymax = = 40 mm
2 B is an internal hinge.
bd 3 50 × 803
I= =
12 12
Maximum bending stress
M 106
σmax = ⋅ y max = × 40
I  50 × 803  (a) 2.67 kN and 0 kN-m (b) 4 kN and 0 kN-m
  (c) 4 kN and 1.33 kN-m (d) 1.33 kN and 0 kN-m
 12 
σmax = 18.8 N/mm2 APPSC AEE SCREENING 17.02.2019
762. The beam of triangular cross-section as shown Ans. (d) :
in the figure below, is subjected to pure
bending. If a plastic hinge develops at a section,
determine the location of neutral axis (distance
b from top) at that section. The beam material
is elastic-perfectly plastic (i.e., yield stress is
constant)

Consider BC part
1 2
×4×2×
SFB = RB = 2 3 = 4 = 1.33kN
2 3
R B = 1.33kN
h h
(a) (b) BMB = 0 (at hinge moment is zero)
3 2 764. A beam of rectangular section 200mm ×
h h 300mm carries certain loads such that bending
(c) (d)
2 3 moment at a section A is M and at another
APPSC AEE SCREENING 17.02.2019 section B it is (M + ∆M). The distance between
section A and B is 1 m and there are no
Ans. (c) : external loads acting between A and B. If ∆M is
20 kNm, maximum shear stress in the beam
section is
(a) 0.5 MPa (b) 1.0 MPa
(c) 1.5 MPa (d) 2.0 MPa
APPSC AEE SCREENING 17.02.2019
Strength of Materials 293 YCT
Ans. (a) : 768. Two beams have the same length, same
allowable stresses and the same bending
moment. The cross sections of the beams are a
square and a rectangle with depth twice that of
width. The ratio of weight of square beam to
the weight of rectangular beam is:
(a) 0.85 (b) 0.75
dM (c) 0.95 (d) 1.25
=F
dx ISRO Scientist/Engineer (RAC), 10.03.2019
20kN − m Ans. (d) : l1 = l2
=F
1m δ1 = δ 2
F = 20 kN
For rectangular cross-section M1 = M 2
3
τ max = τ avg 
2
3  F  3  20 × 103 
τ max =   =   = 0.5 MPa
2  bd  2  200 × 300 
765. A mild steel flat of width 100 mm and thickness From bending equation
12 mm is bent into an arc of a circle of radius M σ E
10 m by applying a pure moment M. If Young's = =
I y R
modulus E = 200 GPa, then the magnitude of
M is M
σ= ×y
(a) 72 Nm (b) 144 Nm I
(c) 216 Nm (d) 288 Nm σ1 = σ 2
APPSC AEE SCREENING 17.02.2019
M1 M
Ans. (d) : Given, Radius of curvature (R) = 10,000 mm × y1 = 2 × y2 (M1 = M 2 )
Modulus of elasticity (E) = 200 × 103 MPa I1 I2
Thickness (t) = 12 mm  y1  y 
From bending equation   = 2 
M E σ  I1   I 2  sq.
= = rec .

I R Y 2b / 2 a/2
= 4
M 200 × 103 (2b) 3
a
= b×
100 × 123  10, 000 12 12
 12  2b a
  =
M = 288000 N-mm 8b 4 a 4
= 288 N-m 1 1
766. A rectangular beam section with depth 400 mm =
4b 3 a 3
and width 300 mm is subjected to a bending
moment of 60 kN-m. The maximum bending a 3 = 4b3
stress in the section is a = 1.587b
(a) 7.50 MPa (b) 2.50 MPa weight of square Beam m g ρV g V
(c) 1.56 MPa (d) 0.42 MPa = 2 = 2 2 = 2
APPSC AEE SCREENING 17.02.2019 weight of rectangular Beam m1 g ρ1V1 g V1
Ans. (a) : Maximum bending stress in the beam a2 × l
=
M M 60 × 106 2b 2 × l
f max = = = = 7.5 MPa
Z  bd 2   300 × 4002  (1.587b) 2 × l1
 6    = (l1 = l2 )
   6  2b 2 × l2
767. In case of pure bending, the beam will bend = 1.25
into an arc of a/an 769. A beam with rectangular cross section (10mm
(a) parabola (b) hyperbola × 20mm) is subjected to maximum shear force
(c) circle (d) ellipse of 5kN. The maximum shear stress induced is :
APPSC AEE SCREENING 17.02.2019 (a) 54.5 MPa (b) 25 MPa
Ans. (c) : If a beam is subjected to pure bending the (c) 37.5 MPa (d) 50 MPa
elastic curve is a circular arc with constant radius. Oil India Limited Sr. Engineer (Drilling) 30.11.2019
Strength of Materials 294 YCT
Ans. (c) : Given, Note- Bending stress of both beam will be same
Area of rectangular cross section (σ b )s = (σ b ) c
A = 10 mm × 20mm 6M s 32M c
=
Shear force (Fs) = 5 kN D3 πD 3
M s 16
=
M c 3π
771. If the Bending moment is increased three times,
then to keep the stress in the beam same,
sectional modulus shall be
(a) decreased 3 times (b) increased 3 times
3 (c) unchanged (d) increased 6 times
τmax =τavg ISRO Scientist/Engineer (RAC) 22.04.2018
2
3 F Ans. (b) : Bending stress
= × s M
2 A σ=
Z
3 5 × 103
= × M
2 10 × 20 For σ = = constant
Z
τmax = 37.5MPa
If M ↑→ 3M
770. The ratio of moment carrying capacity of a then Z also increase 3 times to keep the stress in the
square cross section beam of dimension D to beam same.
the moment carrying capacity of a circular 772. A beam with rectangular section of 120 mm ×
cross section of diameter D is : 40 mm is placed horizontally by mistake, (with
16 16 width as 120 mm and depth as 40 mm) whereas
(a) (b) it was designed to be placed vertically, (with
3π π width as 40 mm and depth as 120 mm). The
16 8 ratio of section modulus will be
(c) (d) (a) 1/3 (b) 1/2
5π 3π
(c) 1/6 (d) 1/8
BHEL ET 2019
ISRO Scientist/Engineer (RAC) 22.04.2018
Ans. (a) : Ans. (a) : Case-I

D3
Section modulus of square cross - section Z s = I NA
6 Z=
Ymax
πD 3
Section modulus of circular cross - section Z c = 120 × (40)3
32 I NA =
12
40
Y=
2
120 × 40 × 40 × 40 × 2
Z1 =
12 × 40
M
σb = Z1 = 32000 mm3
Z NA
Case-II
For square cross section–
M 6M
( σ b ) s = 3s = 3 s
D D
6
For circular cross section–
M 32M c
( σ b ) c = c3 =
πD πD 3
32

Strength of Materials 295 YCT

 40 × (120)3 9 × a3 a3
IN−A = Zsquare = =
12 12 6
 120  9
ymax =  
 2  2
= 0.167 a3 ………..(iii)
40 × 120 × 120 × 120 × 2
Z2 =  I πR 4
120 × 12 Zcircle =   =
Z2 = 96000 mm3  Y Circle 4
Z1 32000 1 R
= =
Z 2 96000 3 π 3 π d3
Zcircle = R = ×
773. Neutral axis of a beam is; 4 4 8
(a) Layer subjected to tensile stress d
∵R =
(b) Layer subjected to compressive stress 2
(c) Layer subjected to torsion stress From equation (i), we get,
(d) Layer subjected to zero stress 3
π  29 
ISRO Scientist/Engineer (RAC) 22.04.2018 Zcircle = × 
Ans. (d) : Neutral axis of a beam is layer subjected to 32  π 
zero stress. = 0.141 a3 ………..(iv)
M From equation (iii) and (iv),
σb = × y (at neutral axis, y = 0) (Z)circle < (Z)square
I
Hence from equation (i), we get
σb = 0
σcircle > σsquare
774. The beams, one having square cross section
and another circular cross section, are 775. The relationship between the radius of
subjected to same amount of bending moment. curvature R, bending moment M and flexural
If the cross section area and the material of rigidity EI is given by
both the beams are the same then M EI
(a) R = (b) M =
(a) Both beams will experience same EI R
deformation R MI
(b) Maximum bending stress developed in both (c) EI = (d) E =
the beams is the same M R
(c) The circular beam experiences more bending TSPSC AEE 28.08.2017 (Civil/Mechanical)
stress than the square one. M E
=
(d) The square beam experiences more bending Ans. (b) : I R
stress than the circular one EI
GPSC Executive Engineer 23.12.2018, GATE 2003 ⇒ M =
R
Ans. (c) : Area of square and circle is same,
776. Due to some point load anywhere on a fixed
π 2
× d = a2 beam, the maximum free bending moment is
4 M. The sum of fixed end moment is
Let, (a) M (b) 1.5 M
a = Side of square (c) 2.0 M (d) 3.0 M
d = diameter of circle TSPSC AEE 28.08.2017 (Civil/Mechanical)
2a Ans. (a) : Due to some point load any where on a fixed
d= …………..(i)
π beam, the maximum free bending moment is M. The
Bending stress is given as, sum of fixed end moment is M.
M 777. A pipe of external diameter 3 cm and internal
σ= ……...……(ii)
Z diameter 2 cm and length 4m supported at its
Where, M = Maximum bending moment ends. If carries a point load of 65 N at its
Z = Section modulus center-section, modulus of pipe will be
Section modulus 65π 3 65π 3
(a) cm (b) cm
I 64 32
Z=
Y 65π 3 65π 3
(c) cm (d) cm
Greater value of section modulus stronger be the section 96 124
and as per equation (ii) lower will be bending stress. TSPSC AEE 28.08.2017 (Civil/Mechanical), CSE Pre 2002
 I Ans. (c) : External diameter D0 = 3 cm
ZSquare =  
 Y Square Internal diameter Di = 2 cm

Strength of Materials 296 YCT

Section modulus, 780. A timber beam is simply supported at the ends
I and carries a concentrated load at mid span.
Z= The maximum longitudinal stress 'f' is 12
y max N/mm2 and the maximum shear stress 'q' is 1.2
π N/mm2. The ratio of span to depth would be
I=
64
( )
D40 − Di4 (a) 10 (b) 6
(c) 5 (d) 4
ymax = D0/2
TSPSC AEE 28.08.2017 (Civil/Mechanical)
(distance of extreme fiber from neutral axis)
Ans. (c)
π 4 4
3 − 2 
Z= 64
3/ 2
65π 3
Z= cm Maximum shearing stress q = q avg
3
96 2
778. In a loaded beam under bending (for rectangular such)
(a) both the maximum normal and maximum 3 W  1
shear stresses occur at the skin fibers. q =  ×
(b) both the maximum normal and maximum 2  2  bd
shear stresses occur at the neutral axis 3W
q= ––––––(1)
(c) the maximum normal stress occurs at the skin 4bd
fibres and the maximum shear stress occurs at M
the neutral axis Maximum bending stress f =
Z
(d) the maximum normal stress occurs at the
neutral axis while the maximum shear stress W
ℓ
3Wℓ
occurs at the stress fibers f = 42 = ––––––(2)
TSPSC AEE 28.08.2017 (Civil/Mechanical), CSE Pre 2003 bd 2bd 2
6
Ans. (c) : f 2ℓ ℓ 1 12
= = = × =5
q d d 2 1.2
781. The moment of inertia of a given rectangular
area is minimum about
(a) its longer centroidal axis
(b) its polar axis
(c) its axis along the diagonal
Normal stress occurs maximum at extreme fiber while (d) its shorter centroidal axis
maximum shear stress occurs at neutral axis. TSPSC AEE 28.08.2017 (Civil/Mechanical)
779. A beam has rectangular section 100 mm × 200 1
mm. If it is subjected to a maximum BM of 4 × Ans. (a) : Ixx = bd 3
12
107 Nmm, then the maximum bending stress
1 3
developed would be I yy = db
(a) 30 N/mm2 (b) 60 N/mm2 12
(c) 90 N/mm 2
(d) 120 N/mm2 since b < d
TSPSC AEE 28.08.2017 (Civil/Mechanical) ∴ Ixx > Iyy
Hence MOI of a given rectangular section is minimum
Ans. (b) : Rectangular section = 100 mm × 200 mm about longer centroidal axis.
1
I = × 100 × 2003 782. A beam of uniform strength refers which
12 the following?
200 (a) A beam in which extreme fibre stresses are
ymax = = 100 same at all cross-section along the length of
2
the beam
1
I × 100 × 2003 2 (b) A beam in which the moment of inertia about
Z= = 12 = × 106 mm3 the axis of bending is a constant at all cross-
y max 100 3 section of the beam
MR = σmax × Z (c) A beam in which the distribution of bending
stress across the depth of cross-sections of the
M R 4 × 107 beam
σ max = = N–mm
z 2 (d) A beam in which the bending stress is
× 10 6
uniform at the maximum bending moment
3
cross-section
σmax = 60 N/mm2
TSPSC AEE 28.08.2017 (Civil/Mechanical)
Strength of Materials 297 YCT
Ans. (a) : Beams of uniform strength are those beams 0.75P P
whose cross-section varies along length of beam, such τmax = ∴ Pmax =
h2 × 2 2
that maximum bending stress at each cross-section
τmax 0.375
reaches permissible stress and is constant through out = = 0.01
the length of beam. σmax 37.5
783. A mild steel plate is subjected to a moment M 786. The stress produced in a wire of diameter 6
each at its ends such that it bends into an arc of mm, when it is bend around a large cylinder of
a circle of radium 10 m. The plate has width 60 diameter 2 m can be obtained as (Young's
mm and thickness 10 mm. E = 2 × 105 N/mm2. modulus of both wire and cylinder = 200 GPa,
What is the maximum bending stress produced the stress strain curve of the material is given
in the plate? below)
(a) 100 MPa (b) 200 MPa
(c) 300 MPa (d) 400 MPa
TSPSC AEE 28.08.2017 (Civil/Mechanical)
Ans. (a) : Thickness of plate = 10 mm
Radius of curvature = 10 m
E = 2 × 105 N/mm2
E σ (a) 600 MPa (b) 300 MPa
Flexure formula =
R y (c) 100 MPa (d) 200 MPa
E×y ISRO Scientist/Engineer 17.12.2017
σ= Ans. (b) : Given,
R
σ = 300 N/mm2
10
2 × 105 × E = 200 × 103 N/mm2
σ= 2 N / mm 2 D 2
10 × 1000 R = = = 1 m = 1000 mm
2 2
σ = 100 N/mm2 = 100 MPa
6
784. In the design of beams for a given strength, ymax = = 3 mm
consider that the conditions of economy of used 2
From bending equation
of the material would avail as follows:
1. Rectangular cross-section is more economical M σ E
= =
than square section of the same cross-sectional I y R
area of the beam. E
2. Circular section is more economical than square σ = ymax
R
section.
200 × 103
3. I-section is more economical than a rectangular = 3×
section of the same depth. 1000
Which of the above are correct? = 600 MPa
(a) 1, 2 and 3 (b) 1 and 2 only But σcritical = 300 MPa
(c) 2 and 3 only (d) 1 and 3 only ∴ σ = 300 MPa
ESE 2017 787. A cantilever beam of rectangular cross section
Ans. (d) : For same material and same volume (depth 'd', width 'b' & length 'L') is subjected
ZI > Zrect > Zsq > Zcir to a shear force 'F' at its free end as shown in
figure. The beam is made of material which
Square section is more economical than circular section
follown maximum shear stress theory. Which
therefore statement second is wrong.
of the following is TRUE?
785. Consider a simply supported beam of length,
50h, with a rectangular cross-section of depth,
h, and width, 2h. The beam carries a vertical
point load, P, at its mid point. The ratio of the
maximum shear stress to the maximum
bending stress in the beam is (a) The beam section has a uniform shear stress,
(a) 0.02 (b) 0.10 F
(c) 0.05 (d) 0.01 bd
GPSC Asstt. Director of Transport 05.03.2017 (b) Maximum shear stress in section and
maximum bending stress in beam is equal
Ans. (d) :
when L = d/2
M ( PL / 4 ) P × 50h / 4 P (c) For L > d/2, bending stress decides the
σ max = max = = = 37.5 2
z ( h / 3)
3 3
h /3 h failure stress
(d) Bending stress developed in the beam is
3 Pmax 3Pmax P independent of 'b'
τmax = = = 0.75 max
2 A 4h 2 h2 ISRO Scientist/Engineer 17.12.2017
Strength of Materials 298 YCT
Ans. (c) (a) Shear stress varies parabolically in a The maximum shear stress will occur at neutral axis of
rectangular cross-section. the section in inverted T-section when subjected to a
1.5F shear force.
(b) Maximum shear stress = ≠ maximum bending791. A steel wire of 10 mm diameter is bent into a
bd
stress. circular shape of 5 m radius. What will be the
(d) Bending stress depends on b. maximum stress induced in the wire, when E =
200 GPa?
788. A solid rod of 12 mm diameter was tested for (a) 50 MPa (b) 100 MPa
tensile strength with the gauge length of 50 (c) 150 MPa (d) 200 MPa
mm. Final length = 80 mm; Final diameter = 4 (e) 250 MPa
mm; Yield load = 1130 N. What is the nearest (CGPCS Polytechnic Lecturer 2017)
yield stress and % reduction in area
respectively? Ans. (d) : Data given,
(a) 10 MPa and 10% (b) 90 MPa and 90% d = 10 mm
R = 5 m = 5 × 103 mm
(c) 10 MPa and 90% (d) 90 MPa and 10%
E = 200 GPa = 200 × 103 N/mm2
ISRO Scientist/Engineer 07.05.2017
σ=?
Ans. (c) : Given, di = 12 mm, Li = 50 mm, Lf = 80 mm We know that pure bending equation
df = 4 mm, P = 1130 N, σy = M σ E
P = =
σyield stress = I Y R
A E d 10
So σ = × y max. ymax. = = = 5 mm
1130 1130 R 2 2
= = = 10 MPa
π
×12 × 12 π× 3 × 12 σ=
200 × 103
×5
4 5 × 103
Percentage reduction– σ = 200 N/mm2
A − Af σ = 200 MPa
% reduction area = i ×100
Ai 792. A rectangular beam 100 mm wide is subjected
π 2 π 2 to a maximum shear force of 50 kN. If the
di − df maximum shear stress is 3 MPa, the depth of
=4 4 the beam will be ______.
π 2 (a) 100 mm (b) 150 mm
di
4 (c) 200 mm (d) 250 mm
144 − 16 128 (e) 275 mm
= × 100 = × 100 = 88.88 ≈ 90% CGPCS Polytechnic Lecturer 2017
144 144
789. A beam of uniform strength has constant : Ans. (d) : We know that
(a) Shear force (b) Bending moment 3
τmax. = τavg. [For rectangular beam]
(c) Cross-sectional area (d) Deflection 2
TRB Polytechnic Lecturer 2017 Shear Force 50 × 103
Ans. (b) : A beam of uniform strength is subjected to τavg = =
Cross - Sectional Area 100 × d
constant bending stress.
3 50 × 103
M 3= ×
σ = ×Y 2 100 × d
I
For particular beam I & Y remains same so, 3 × 50 × 103
d=
σ = M = Constant 2 × 3 × 100
Hence, bending moment also will be constant. d = 250 mm
790. An inverted T-section is subjected to a shear force
F. The maximum shear stress will occur at:
(a) Top of the section
(b) Junction of web and flange
(c) Neutral axis of the section
(d) Bottom of the section
TRB Polytechnic Lecturer 2017
Ans. (c) : 793. The maximum shearing stress induced in the
beam section at any layer at any position along
the beam length (shown in the figure) is equal to

Strength of Materials 299 YCT

 (a) 30 kgf/cm2 (b) 40 kgf/cm2 (b) Moment of inertia about the neutral
(c) 50 kgf/cm2 (d) 60 kgf/cm2 axis/distance of the most distant point from
ESE 2017 neutral axis
Ans. (a) : For rectangular cross section (c) Bending moment/Moment of inertia
3 (d) None of the above
τmax = τavg ISRO Scientist/Engineer (RAC) 07.05.2017
2
3 2000 Ans. (b) : Section Modulus—It is defined as the ratio
= × kgf / mm 2 of moment of inertia about the neutral axis to distance
2 50 × 200 of the most distant point from neutral axis.
= 30 kgf/cm2 It is denoted by Z.
794. A beam of rectangular section (12 cm wide × 20
cm deep) is simply supported over a span of 12 I
Z = N−A
m. It is acted upon by a concentrated load of ymax
80kN at the mid span. The maximum bending 797. Weldments in fabricated steel beams are
stress induced is. designed for
(a) 400 MPa (b) 300 MPa
(c) 200 MPa (d) 100 MPa (a) Bending stress at the flange
ESE 2017 (b) Shear stresses in transverse plane
(c) Combination of bending and shear
Ans. (b) : Given,
(d) Bending and tensile stress
b = 12 cm = 120 mm
d = 20 cm = 200 mm TSPSC AEE 2017
Ans. (c) : Weldments in fabricated steel beams are
designed for combination of bending and shear for
optimum design.
798. A beam with a rectangular section of 120 mm ×
L = 12 m = 12000 mm 60 mm, designed to be placed vertically is
W = 80 kN = 80 × 103 N placed horizontally by mistake. If the
WL 80 × 103 × 12000 maximum stress is to be limited, the reduction
Mmax = = in load carrying capacity would be
4 4
= 240 × 106 N−mm (a) 1/4 (b) 1/3
(c) 1/2 (d) 1/6
bd 3 120 × 2003 GPSC EE Pre, 28.01.2017
I= =
12 12 Ans. (c) : Given, depth (h) = 120 mm
= 80 × 106 mm4 width (b) = 60 mm
d 200 According to bending equation
y= = = 100 mm
2 2 M E σ
MY = = b
Maximum bending stress (σ) = I R y max
I
M × y max
240 × 10 × 100
6
σb =
= I
80 × 106
= 300 MPa h
795. In theory of simple bending an assumption is
( M max )1 ×
I case σ b1 =
st 2
made that plane sections before bending 1 3
remain plane even after bending. This bh
12
assumption implies that
(a) Strain is uniform across the section 6 ( M max )1
σ b1 =
(b) Stress is uniform across the section bh 2
(c) Stress in any layer is proportional to its b
distance from the neutral axis ( M max )2 ×
(d) Strain in any layer is directly proportional to II case σ b2 =
nd 2
its distance from the neutral axis 1 3
hb
JWM 2017 12
Ans. (d) : The assumption made in theory of simple 6 ( M max ) 2
bending that plane sections before bending remain plane =
hb 2
even after bending means the strain in any layer is
directly proportional to its distance from the neutral At given- σ1 = σ2
axis. 6 ( M max )1 6 ( M max )2
⇒ =
796. Section modulus is defined as bh 2
b2h
(a) Moment of inertia about the neutral ( M max )1 bh 2
axis/square of the distance of neutral axis =
from farthest point ( M max )2 b2h

Strength of Materials 300 YCT

 ( M max )1 h 802. A long of wood is of 3 m diameter circular
= section. Then the width of the stronger
( M max )2 b rectangular section is bending that can be cut
( M max )2 h 60 1 out from this log of wood is
== =
( M max )1 b 120 2 (a) 6m (b) 3m
Note- Answer given by commission (a) but actual (c) 1/ 3m (d) 2 / 3m
answer is (c). APPSC AEE Mains 2016 (Civil Mechanical)
799. For high speed engines, a rocker arm of....... Ans. (b) : As we know,
should be used. for strongest rectangle
(a) rectangular section (b) I-section
1
(c) T-section (d) circular b/h =
GPSC EE Pre, 28.01.2017 2
Ans. (b) : A rocker arm is an oscillating lever that
conveys radial movement from the cam lobe into linear
momentum at the poppet valve to open it. It may be
made of cast iron, cast steel as malleable iron. In a order
to reduce inertia of the rocker arm I-section is used for
the high speed engines and it may be rectangular section
for low speed engines.
800. If the average shear stress in a rectangular
section beam is 5 N/sq. mm, then maximum d 3
b= = = 3m
shear stress for the circular section of the equal 3 3
area is
(a) 7.50 N/sq.mm (b) 6.65 N/sq.mm 803. The bending force required for V-bending. U-
(c) 10 N/sq.mm (d) 2.50 N/sq.mm bending and Edge-bending will be in the ratio
APPSC AEE Mains 2016 (Civil Mechanical) of
Ans. (b) : For Circular, (a) 1 : 2 : 0.5 (b) 1 : 1 : 1
(c) 2 : 1 : 0.5 (d) 1 : 2 : 1
UPRVUNL AE 21.08.2016
Ans. : (a) The bending force required for V-bending. U-
bending and Edge-bending will be in the ratio of 1:2:
0.5
804. A cantilever of length 6 m is carrying a
uniformly distributed load of W per unit run
For Circular, for a distance 4m from the free end. The value
4 V 4 of maximum bending moment is
τmax = × = × τavg
3 A 3 (a) 16W (b) 22W
4 (c) 18.5W (d) 21W
τmax = × 5 = 6.65 N / mm 2
3 APPSC AE Subordinate Service Civil/Mech. 2016
801. A horizontal beam of square section is placed Ans. (a) :
with one diagonal (d) placed horizontally. The
average shear stress occurs at
(a) d/8 from neutral axis
(b) d/4 from neutral axis
(c) neutral axis Total load = 4W and act at distance 2 m from free end
(d) 3d/8 from neutral axis for bending moment calculation.
APPSC AEE Mains 2016 (Civil Mechanical)
M C= O
Ans. (a) :
MB = 4W × 2 = 8 W
MA = 4W × 4 = 16 W = Mmax

D D 
h= − y ⇒ Base ⇒ b = 2  − y 
2  2 
D
The Average shear stress occur at from Neutral Axis
8

Strength of Materials 301 YCT

805. A simply supported beam of length 7m is Ans. (a) : Given,
carrying a load whose intensity varies
σ max = 110N / mm 2
uniformly from zero at each end to W per unit
run at the mid span. The value of maximum d = 260 mm
shear force is E = 2 × 105 MPa
(a) 1.75 W (b) 2.38 W From bending equation
(c) 2.96 W (d) 3.42 W M σ E
= =
APPSC AE Subordinate Service Civil/Mech. 2016 I y R
Ans. (a) :
E.y 2 ×105 × 130
R= = = 236363.63 mm
σ 110
R = 236.36m
808. A hollow cast iron pipe is of external diameter
Total load = Area of loading diagram 60 mm and 12 mm thick. The section modulus
1 is
= × 7 × W = 3.5W
2 (a) 18448.13 mm3 (b) 17017 mm3
3
Reactions at A & B will be equal due to symmetry of (c) 19362.24 mm (d) 16496.75 mm3
load, so this total load (3.5 W) is divided equal on both APPSC AE Subordinate Service Civil/Mech. 2016
the supported. Ans. (a) :
3.5W
RA = RB = = 1.75W
2
for uniformly varying loading, the maximum shear
force (S.F.) will be at supports.
So, Max. S.F. = 1.75W
806. A cantilever of length of 2 m fails, when a load
of 2300 N is applied at free end. What is the External dia of pipe (D) = 60 mm
stress at failure, if the section of cantilever is 50
Thickness of pipe (t) = 12 mm
mm × 70 mm. Internal dia. of pipe (d) = D – 2t = 60 – 2 × 12 = 36 mm
(a) 112.65 MPa (b) 121.38 MPa Section modulus of hollow circular section,
(c) 132.93 MPa (d) 141.64 MPa I
APPSC AE Subordinate Service Civil/Mech. 2016 Z=
y
Ans. (a) :
π
I=
64
(
D4 − d 4 )
D
y=
2
From bending equation,
π  4
M max z= D − d4 
Bending stress at failure = σf = 32D  
Z
π  4
Mmax = 2300 × 2 × 103 Z= 60 − 364  = 18448.13 mm3
2 2 32 × 60  
bd 50 × 70 taking π = 3.14
Z= =
6 6 Z = 18448.13 mm3
3
2300 × 2 ×10 × 6 809. For a solid circular section, the shear stress
σf =
50 × 70 2 distribution is according to a
2 (a) parabolic law (b) constant
= 112.65 N / mm
(c) cubic law (d) linear law
= 112.65 MPa APPSC AE Subordinate Service Civil/Mech. 2016
807. A beam of symmetrical section is subjected to Ans. (a) : Shear stress distribution for solid circular
bending moment which produces a maximum section.
stress of 110 N/mm2. If the beam is 260 mm
deep find radius to which the longitudinal axis
of beam will be bent. Take E= 2 × 105 MPa
(a) 236.36 m (b) 21.46 m
(c) 192.71 m (d) 28.9 m
APPSC AE Subordinate Service Civil/Mech. 2016
Strength of Materials 302 YCT
810. A beam is triangular in section and is of base Ans. (a) :
'b' and height 'h'. If it is placed with its base
horizontal, what is maximum shear stress if the
shear force is S.
8S 3S
(a) (b) Slope at C = slope at B
3bh 4bh 3
4S 8S wa 3
(c) (d) θC = θ B =
5b 2 h 5bh 6EI
APPSC AE Subordinate Service Civil/Mech. 2016 Deflection at C,
Ans. (a) : wa 4
yC =
8EI
Deflection at B,
yB = yC + (ℓ – a)θC
wa 4 wa 3
yB = + × (l − a )
8EI 6EI
bh 3 812. A hinged support can transmit
MOI about N.A. = I =
36 (a) two reactive forces
Let the shear stress intensity be at a depth 'y' from the (b) one reactive forces
top. (c) three reactive forces
b (d) four reactive forces
Width of beam at depth 'y' from the top = b' = y APPSC AE Subordinate Service Civil/Mech. 2016
h
Ans. (a) : A hinged support can transmit two reactive
 y by  2 2 
S  3 h − 3 y  forces.
τ=
Say
=  2 h   A roller support can transmit one reactive force.
Ib  bh 3   by  A fixed support can transmit three, reactive force.
   
 36   h 
 12S 
τ =  3  y ( h − y)
 bh 
dτ  12S  h
For τ to be max, = ( h − 2y ) = 0 ∴y =
dy  bh 3  2 813. The curvature of a beam at a section is given by
 12S  2 h M σ
τ max =  3  h× (a) (b)
 bh  3 3 EI y
8 S ε MI
τmax = (c) (d)
3 bh y E
2 APPSC AE Subordinate Service Civil/Mech. 2016
To find the shear stress at N.A, put y = h
3 Ans. (a) : Bending equation
12S 2 h M σ E
τ N.A. = 3
× h× = =
bh 3 3 I y R
8S 1 M
τ N.A. = = curvature =
3bh R EI
811. A cantilever of length ℓ is carrying a uniformly 814. A rectangular bar of width b and height d is
being used as a cantilever. The loading is in a
distributed load of w per unit run of a plane parallel to the side b. The section
distance's from fixed end. The slope at the free modulus is
end is (where EI is the flexural rigidity)
b2d bd 3
wa 3 wa 4 (a) (b)
(a) (b) 6 12
6EI 8EI
bd 2 bd 2
wa 4 wa 3 (c) (d)
(c) (d) 6 8
6EI 8EI APPSC AE Subordinate Service Civil/Mech. 2016
APPSC AE Subordinate Service Civil/Mech. 2016 APPSC AEE Screening Test 2016
Strength of Materials 303 YCT
Ans. (a) : Given, Ans. (a) : Given-
Width = b A = 100 mm2
Height = d Section modulus (Z) = 5 × 103 mm3
I P = 10 kN
Section modulus = e = 10 mm
y
P M
b3d b σ1,2 = ±
I= , y= A Z
12 2
P M
b3 d 2 σmax = +
Z= × A Z
12 b 10 × 103 10 × 10 × 103
= +
b2d 100 5 × 103
Z=
6 = 100 + 20
815. Assumption made in the theory of bending is = 120 MPa (Tensile)
(a) Radius of curvature is small P M
σmin = –
(b) Radius of curvature is large A Z
(c) Transverse sections of the beam do not = 100 – 20
remain plane after bending = 80 MPa (Tensile)
(d) Doesn't follow Hooke's law in bending 818. Bending equation is derived by considering:
APPSC AEE Screening Test 2016 (a) Maximum bending moment on the beam
Ans. (b) Assumption made in the theory of bending (b) The beam is simply supported
(i) The materials of the beam is homogeneous and (c) The beam is a cantilever
isotropic. (d) Constant pure bending moment on the beam
(ii) The value of Young's Modulus of elasticity is
GPSC Asstt. Prof. 28.08.2016
same in tension and compression.
(iii) The transverse sections which were plane before Ans. (d) : Assumptions of Bending Equation–
bending remain plane after bending also. (1) The beam should be isotropic and homogeneous.
(iv) The beam is initially straight and all longitudinal (2) The constant pure bending moment on the beam.
filaments bonds into circular arcs with a common (3) The section remains plane before and after bending.
centre of curvature. 819. Section modulus of a square section of side 'b'
(v) The radius of curvature is large compared to the is equal to
dimensions of the cross-section. (a) b3/6 (b) b2/6
(vi) Each layer of the beam is free to expend or (c) b/6 (d) b3/3
contract, independently of the layer above or TSPSC AEE 2015
below it.
816. A beam strongest in flexural is one which has b4
Ans : (a) Square section moment of inertia (I) =
(a) maximum bending stress 12
(b) maximum area of cross section Section modulus = I/y
(c) maximum section modulus b4 × 2
(d) maximum moment of inertia Zsq =
APPSC AEE MAINS 2016, PAPER-III 12 × b
Ans. (c) : A beam strongest in flexural is one which has b3
Zsq =
maximum section modulus. Section modulus is also 6
known as strength of beam. 820. The maximum bending moment of a square
I N−A
Z= 2003
y max. beam of section modulus mm3 is 20 × 106
6
817. A component has a cross section area of 100 N-mm. The maximum bending stress induced
mm2 and section modulus of 5 × 103 mm3. It is in the beam is
subjected to an eccentric axial tensile load of 10 (a) 30 N/mm2 (b) 7.5 N/mm2
2
kN with an eccentricity of 10 mm. What will be (c) 45 N/mm (d) 15 N/mm2
the maximum and minimum stresses developed TSPSC AEE 2015
in the member? 6
(a) 120 MPa (tensile) and 80 MPa (tensile) Ans : (d) bending moment (M) = 20×10 N-mm
(b) 120 MPa (tensile) and 80 MPa 2003
section modulus (Z) = mm3
(compressive) 6
(c) 80 MPa (tensile) and 80 MPa (compressive) M σ max E
= = ( bending equation )
(d) None of these I y R
RPSC AE (GWD) 06.05.2016
Strength of Materials 304 YCT
 M I 825. Two loads P act at right angles to one another
=
σ max y at the free end of a cantilever beam having
M square cross-section d × d and length l on the
σ max = vertical and horizontal faces. Maximum
Z
bending stress in the beam will be equal to-
20 × 10 6
σ max = (a) 6Pl/d3 (b) 24Pl/d3
200 3 / 6 3
(c) 12 Pl/d (d) 18Pl/d3
σ max = 15N / mm 2
ISRO Scientist/Engineer (RAC) 29.11.2015
821. Area moment of inertia of the cross-section of Ans : (a) According to this figure bending stress due to
the beam resist only vertical force not due to horizontal force so
(a) the applied normal force
(b) both the torque and bending moment
(c) the applied torque
(d) the applied bending moment of the beam
TANGEDCO AE 2015
M σ
Ans. (d) : =
I y
The beam having cross sectional area and area moment
of inertia resist the applied bending moment of the
beam. M σ
822. The radius of gyration 'k' of plane figure is =
I y max
determined by the relation
(a) k= (A/I)1/2 (b) k= (I/m)1/2 M σ
1/2 =
(c) C.k= (m/I) (d) k= (I/A)1/2 d  d 
4

VIZAG MT 2015    
 12   2 
Ans. (d) : I = Ak 2

6Pℓ
 I
1/ 2
σ= 3
k=  d
A
826. For the component loaded with a force F as
823. The stress distribution in a curved beam, when
subjected to pure bending couple is : shown in the figure, the axial stress at the
(a) linear corner point P is
(b) non-linear and hyperbolic
(c) non-linear and parabolic
(d) None of the above
BPSC Asstt. Prof. 29.11.2015
Ans. (b) : The stress distribution in a curved beam,
when subjected to pure bending couple, is non-linear
and hyperbolic.
824. A beam of rectangular cross-section of breadth
10 cm and depth 20 cm is subjected to a F(3L − b) F(3L + b)
bending moment of 20 kNm. Stress developed (a) 3
(b)
at a distance of 10 cm from the top face of the 4b 4b3
beam is : F(3L + 4b) F(3L − 2b)
(c) (d)
(a) Zero (b) 10 kPa 4b 3
4b3
(c) 20 kPa (d) 30 kPa ISRO Scientist/Engineer 11.10.2015, GATE 2008
TSGENCO AE 14.11.2015
Ans. (d) :
σ M
Ans. (a) : = (Bending equation)
y I
M
σ =  ⋅ y
 I 
At 10 cm from top face,
y=0
σ=0
y = measured from neutral axis

Strength of Materials 305 YCT

FBD (Free Body Diagram)– Ans. (a) : Maximum bending stress—Maximum
bending stress will occur at mid-span of beam on either
top or bottom fiber of beam

σ = σa + σb
σa = Axial stress
F wl 2
σa = M max =
2b × 2b 8
σb = (Bending Stress) M max
σ max = y
M F × (L − b) I
σb = =
Z Z wl 2 d
σ max = ×
1 3 8I 2
I (2b)(2b)
Where Z = = 12 w × (4 ×103 ) 2  300 
120 = × 
Y  2b  8 × 8 ×106  2 
 
 2  w = 3.2 N/mm
8b3 4 3 829. Bending stress in a beam cross section at a
Z= = b
6 3 distance of 15 cm from neutral axis is 50 MPa.
So, σ = σa + σb Determine the magnitude of bending of a
F F(L − b) distance of 10 cm from neutral axis.
= 2+ (a) 50 MPa (b) 30.43 MPa
4b 4
b3 (c) 33.33 MPa (d) 75 MPa
3
(e) 45.53 MPa
F 3F(L − b)
= 2+ CGPSC AE 26.04.2015 Shift-I
4b 4b3
Ans. (c) :
Fb + 3FL − 3Fb 3FL − 2Fb
= 3
= 3
4b 4b
F(3L − 2b)
=
4b3
827. A circular log of timber has diameter D. Find
the dimension of the strongest rectangular
section which can be cut from it.
(a) D/ 3 wide and ( 2 / 3) D deep M
Bending stress (σ ) = y
2
(b) D / 3 wide and ( 2 / 3) D deep I
σ∝y
(c) D/ 2 wide and ( 2 / 3) D deep σ 2 y2
(d) D/ 3 wide and ( 1/ 3) D deep =
σ 1 y1
(e) πD/ 3 wide and ( 2 / 3) D deep σ 2 10
CGPSC AE 26.04.2015 Shift-I =
50 15
Ans. (a) : For strongest beam σ2 = 33.33 MPa
D
width of beam = 830. Which of the following statements is correct, in
3 case of a curved beam?
2 (a) Neutral axis is nearer to centre of curvature
depth of beam =   D compared to centroidal axis.
3
(b) Neutral axis is farther to centre of curvature
828. A rectangular beam 300 mm deep, is simply
supported over a span of 4 m. Determine the compared to centroidal axis.
uniformly load per meter, which the beam can (c) Neutral axis matches with centroidal axis.
carry, if the bending stress does not exceed 120 (d) Neutral axis shifts above or below centroidal
N/mm2. Take moment of inertia of the beam = axis depending on loading conditions.
8 × 106 mm4. ISRO Scientist/Engineer 24.05.2014
(a) 3.2 N/mm (b) 1.2 N/mm
(c) 4.2 N/mm (d) 4.5 N/mm Ans. (a) : In case of a curved beam neutral axis is
(e) 2.2 N/mm nearer to centre of curvature compared to centroidal
CGPSC AE 26.04.2015 Shift-I axis.
Strength of Materials 306 YCT
831. A flat spiral mode of strip of breadth 5 mm, 835. Which one of the following statements is true?
thickness 1 mm and length 1.5 m has been (a) A flexure formula is used for pure bending
subjected to a winding couple which induces a only
maximum stress of 150 N/mm2. The magnitude (b) A flexure formula is used for bending with
of winding couple is nearest to shear only.
(a) 20.8 Nmm (b) 41.6 Nmm (c) A flexure formula is used for bending as well
(c) 62.5 Nmm (d) 83.3 Nmm as for bending with shear
TNPSC AE 2014 (d) None of the above statements are true
Ans. (c) : Data given HPPSC Asstt. Prof. 2014
b = 5 mm t = 1 mm Ans. (c) : Flexure formula or bending formula,
L = 1.5 m
σb = 150 N/mm2 M σ E
= =
M=? I y R
We know that is used for bending as well as for bending with shear.
12M 836. Which of the following assumptions is not
σb =
bt 2 made in curved beam?
12 × M (a) Limit of proportionality is not exceeded
150 = (b) Radial strain is not negligible
5 × (1)
2
(c) Plane to transverse sections remain plane
M = 62.5 N − mm after bending
(d) Material considered isotropic and obeys
832. Moment of Inertia of the rectangle of base 80 Hooke's law
mm and height 10 mm about its centroidal HPPSC Asstt. Prof. 2014
(Ixx) axis
4
(a) 6666.66 mm = Ixx (b) 5827.21 mm = Ixx 4 Ans. (b) : Assumptions in curved beam:
4
(c) 7777.22 mm = Ixx (d) 6826.11 mm = Ixx 4 (1) Material is homogeneous and isotropic.
TNPSC AE 2014 (2) Modulus of elasticity in tension and compression
are equal.
Ans. (a) : We know that
(3) Plane sections remain plane after bending
80 × [10]
3
80 × 1000
[ I x − x ] CG = = (4) Obey Hooke’s law (σαε) means limit of
proportionality is not exceeded.
12 12
(5) Radial strain is negligible.
837. A cantilever beam is subjected to a UDL. The
cross section of the beam is a H-Section placed
as shown in Fig. The bending stress
distribution across the cross section will be

[ I x − x ] CG = 6666.66 mm 4
833. In circular plates with edges clamped and with
a uniformly distributed load, the maximum
radial stress occurs at :
(a) clamp edge (b) the centre
(c) the mean radius (d) none of these
(HPPSC AE 2014) (a)
Ans : (b) In circular plates with edges clamped and with
a uniformly distributed load, the maximum radial stress
occurs at the centre.
834. Equivalent moment of inertia of the cross- (b)
section in terms of timber of a flitched beam
made up of steel and timber is (m = Es/Et) :
(a) (It + m/Is) (b) (It + Is/m)
(c) (It + mIs) (d) (It + 2mIt)
(HPPSC AE 2014)
(c)
Ans : (c) Equivalent moment of inertia of the cross-
section in terms of timber of a flitched beam made up-of
steel and timber
is It + mIs
m = Es/Et
Es = Modulus of elasticity of steel. (d)
Et = Modulus of elasticity of timber. ISRO Scientist/Engineer 12.05.2013
Strength of Materials 307 YCT
Ans. (c) : We know that bending stress,
M
σb = ⋅y
I N −A
Where M & IN–A remain constant throughout the beam.
M
σb ∝ y ⇒ σb = c⋅y c=
I N −A
At neutral axis y = 0 Ess = 1 × 105 N/mm2, Ec = 1 × 105 N/mm2, Ass =
600 mm2, Ac = 200 mm2
σb = 0
(a) σc = 20 (Compressive), σss = 30 (Tensile)
So option (b) and (d) ruled out and varies linearly
(b) σc = 30 (Compressive), σss = 20 (Tensile)
throughout the section so option (a) also ruled out
(because change its shape). (c) σc = 30 (Tensile), σss = 20 (Compressive)
(d) σc = 30 (Tensile), σss = 20 (Tensile)
838. A rod of 20 dia is fixed to the ceiling of a roof ISRO Scientist/Engineer 12.05.2013
on one end. A rotor of 50 kg mass is attached to
Ans. (b) : 1. Forces on both metals are equal i.e.
the free end with bearings. The CG of the rotor
σ1A1 = σ2A2
is 10 mm away from the shaft axis. The rotor is
2. Final expansion of metal 1 is equal to final expansion
rotating at 600 rpm. The max tensile stress (in
of metal 2 i.e.
N/Sq. mm) in the rod is nearly equal to
σ σ
α 2 tℓ 2 − 2 ℓ 2 = α1tℓ1 − 1 ℓ1
E2 E1
We know that,
σc × 2Ac = σss × Ass
σc × 400 = σss × 600
σc : σss = 3 : 2
or σc : σss = 30 : 20
In such question there are two equation, and nature of
forces on both material will be opposite hence option
(d) is ruled out so option (b) is right option.
840. Beams with four unknown reaction is
(a) π/2 (b) 200 π (a) In-Determinate Beams
(c) 300 π (d) 400 π (b) Determinate Beams
ISRO Scientist/Engineer 12.05.2013 (c) Propped Beams
Ans. (d) : (d) In- Propped Beams
TNPSC AE 2013
Given d = 20mm, e = 10mm, m = 50 kg
Ans. (a) : A structure is statically indeterminate when
N = 600 rpm
the static equilibrium equation [Force and moment
2π N 2 × 3.14 × 600 equilibrium equation condition] are insufficient for
ω= =
60 60 determining the internal forces and reaction on that
structure.
= 20π
For in-determinate beam No. of equilibrium equation
F = mω2e < No. of Reactions.
= 50 × (20π ) 2 × 10 × 10 −3 For determinate beam No. of equilibrium equation =
No. of reactions.
= 200π2 N 841. The section modulus of hollow circular section
πd3 is
Let Section modulus (Z) =
32 π π
3
(a) (D4 − d 4 ) (b) (D4 − d 4 )
= 250π mm 16 D 32 D
Maximum bending moment in rod = Mmax. π π
(c) ( D3 − d 3 ) (d) ( D3 − d 3 )
= F.L = 200π × 500 = 100000 π N-mm.
2 2
32 D 16 D
M 100000 2 TNPSC AE 2013
∴ σ max = max = π
Z 250π I π  D − d 
4 4

Ans. (b) : Z = =
= 400 πN/mm 2
Ymax 64  D 
839. The cross section of a compound bar 1m long is  2 
as shown in figure. If the temperature is raised π( D − d )
4 4

by 80ºC determine the stresses (in N/mm2) in Z=

each metal 32D
Strength of Materials 308 YCT
842. A 50 mm diameter shaft is made from carbon d2 ∝ x
steel having ultimate tensile strength of 630
MPa. It is subjected to a torque which d∝ x
fluctuates between 2000 N- m to –800 N-m. The
mean shear stress will be As all other things remain constant, depth of the beam
(a) 24.4 N/mm2 (b) 48.8 N/mm2 varies as square root of distance x.
2
(c) 12.2 N/mm (d) 6.1 N/mm2 845. Two simply supported beams of equal lengths,
APPSC Poly. Lect. 2013 cross sectional areas, and section moduli, are
Ans. (a) : d = 50mm, τu = 630 MPa, Tmax = 2000 N-m, subjected to the same concentrated load at its
mid-length. One beam is made of steel and
Tmin = –800 N-m
other is made of Aluminium. The maximum
The mean or average Torque bending stress induced will be in
Tm = (Tmax + Tmin)/2 (a) Steel beam
2000 + (−800) (b) Aluminium beam
=
2 (c) Both beams of equal magnitude
( 2000 – 800 ) (d) The beams according to their Elastic Moduli
= magnitude.
2 UKPSC AE 2012 Paper-I
= 600 N-m = 600 × 103 N-mm
Ans. (c) : Both beams of equal magnitude
→ Mean or average shear stress
16.Tm
τm =
πd 3

τm =
(16 × 600 × 103 )
= 24.4 N/mm 2
PL
π× ( 50 )
3
M max =
4
843. The bending moment at a section tends to bend MY
or deflect the beam and the internal stresses σb =
resist bending. The resistance offered by the I
internal stresses to the bending is called • Since σb is depend only on geometrical shape of the
(a) compressive stress (b) shear stress beam so both beam has equal magnitude.
(c) bending stress (d) elastic stress 846. The flexural rigidity is the product of
JPSC AE 2013, Paper-V (a) modulus of elasticity and mass moment of
Ans : (c) : The bending moment at a section tends to inertia
bend or deflect the beam and the internal stresses resist (b) modulus of rigidity and area moment of
bending. The resistance offered by the internal stresses inertial
to the bending is called bending stress. (c) modulus of rigidity and mass moment of
844. For a beam of uniform strength having inertia
constant width, the depth of the beam at a (d) modulus of elasticity and area moment of
distance of x from the support varies with inertia
(a) x (b) x BPSC AE 2012 Paper - VI
(c) x1/4 (d) x3/4 Ans : (d) : The flexural rigidity is the product of
APPSC Poly. Lect. 2013 modulus of elasticity and area moment of inertia.
Ans. (a) : Assume the beam as cantilever beam 847. A beam is having a triangular cross section
with its neutral axis XX. The section modulus
about XX is given by

Point load is acting at a distance x

Bending moment(BM)= Px (a) bh2/6 (b) bh2/36
My (c) bh2/24 (d) hb2/36
σb =
I ISRO Scientist/Engineer 2012
d Ans. (c) :
P.x.  
σb = 2
b × d3
12
6Px
σb = 2
bd

Strength of Materials 309 YCT

 bh 3 Ans. (b) :
I xx =
36
I I
Z xx = xx = xx
ymax 2h
3 Shear stress is acting in the parallel to the plane and
bh3 Parabolically varies
At Neutral Axis, y = 0
Z xx = 36
2h 3  P
τ max = τ avg ∵τ avg = A 
3 2  
bh3 3 bh 2 850. A steel plate 50mm wide and 100mm thick is to
= × =
36 2h 24 be bent into a circular arc of radius 10m. If
848. A rectangular column is subjected to an E = 2 × 105 N / mm 2 , then the maximum
eccentric load P at distance 'e' from centroidal bending stress induced will be
Axis. The stress diagram at cross-section will be (a) 200 N/mm2 (b) 100 N/mm2
2
(c) 10,000 N/mm (d) 1000 N/mm2
APPSC AEE 2012
Ans : (d) Given, b=50mm, d= 100mm, R = 10m
(a) (b) E= 2×105 N/mm2
Bending equation :-
M σ E
= =
I y R
y = 50mm
σ E
(c) (d) =
y R
E
σ = ×y
R
ISRO Scientist/Engineer 2012 2 × 105
σ= × 50
Ans. (d) : 10 × 103
σ = 1000N / mm 2
851. Radius of curvature of the beam is equal to
ME M
(a) (b)
I EI
EI MI
(c) (d)
From the figure stress at any section will be M E
−P Pe APPSC AEE 2012
= ± ⋅y
A I Ans : (c) Bending equation
Hence compressive load is taken as negative M σ E
= =
849. Shear stress distribution for a rectangular I y R.
beam subjected to transverse loading is M E
=
I R.
EI
Radius of curvature of the (R) =
(a) (b) M
852. A beam of Z-section is called a
(a) doubly symmetric section beam
(b) singly symmetric section beam
(c) a-symmetric section beam
(d) none of the above
UKPSC AE 2012 Paper-I
(c) (d) Ans. (c) : a-symmetric section beam
853. A uniform metal bar of weight ‘W’, length ‘l’,
cross-sectional area ‘A’ is hung vertically with
its top end rigidly fixed. Which section of the
ISRO Scientist/Engineer 2012 bar will experience maximum shear stress ?
Strength of Materials 310 YCT
 (a) Top-section (b) Mid-section 1 2
(c) Bottom-section (d) l/3 from top (a) Sa2 + Sb2 (b) Sa + Sa + Sb2
2
UKPSC AE 2012 Paper-I
Ans. (a) : Top-section (c) 4S2a + Sb2 (d) None of these
854. In theory of simple bending of beams, which RPSC AE GWD, 2011
one of the following assumptions is incorrect ? 2
(a) Elastic modulus in tension and compression Ans. (d) : Maximum shear stress τ =  σ1 − σ 2  + τ2
are same for the beam materials.   s
 2 
(b) Plane sections remain plane before and after
2 2
bending.  sb − 0   sb 
(c) Beam is initially straight. =   + s 2
a =   + sa
2

 2  2
(d) Beam material should not be brittle.
UKPSC AE 2012 Paper-I 1 2
τ= s b + 4s a2
Ans. (d) : Beam material should not be brittle. 2
855. The shear force of a cantilever beam of length 859. Which one of the following assumptions is not
‘ℓ’ carrying a uniformly distributed load of ‘ ω’ valid for the derivation of flexure formula?
per unit length is ……… at the free end M σ E
= =
(a) zero (b) ωℓ/3 I y R
(a) The beam material is linearly elastic
(c) ωℓ/2 (d) ωℓ
(b) Sections which are plane before bending
TNPSC ACF 2012 remain plane after the bending
Ans. (a) : (c) The beam has large initial curvature
(d) Elastic modulus of the beam material are
same in tension and compression.
UPSC JWM Advt. No.-52/2010
Ans. (c) : Explanation : The beam has large initial
curvature is not a valid assumption for the derivation of
flexure formula.
At end B the shear force is zero M σ E
= =
RA = ωℓ RB = 0 I y R
856. The strength of the beam mainly depends on Assumptions made in the theory of pure Bending-
(a) bending moment (b) C.G. of the section (1) The material of the beam is homogenous and
(c) Section modulus (d) its weight isotropic.
APPSC AEE 2012 (2) The value of Young's modulus of elasticity is same
Ans. (c) : Strength of the beam mainly depends upon in tension and compression.
section modulus. (3) The Transverse sections which were plane before
⇒ Strength ∝ Z bending, remain plane after bending also.
Section modulus Z = I/y Where Z = Section modulus (4) The beam is initially straight and all longitudinal
I = Area M.O.I. filaments bend into circular arcs with a common
y = beam width centre of curvature.
857. In a beam of I section, the maximum shear (5) The radius of curvature is large as compared to the
force is carried by dimensions of the cross-section.
(a) the upper flange (b) the web (6) Each layer of the beam is free to expand or contract
(c) the lower flange (d) Any of these independently of the layer, above or below it.
APPSC AEE 2012 860. A cantilever beam of square cross-section
Ans. (b) : Shear force at top and bottom surface is (100mm × 100mm) and length 2 m carries a
minimum and maximum at the centroid of web. concentrated load of 5 kN at its free end. What
is the maximum normal bending stress at its
mid-length cross-section ?
(a) 10 N/mm2 (b) 20 N/mm2
2
(c) 30 N/mm (d) 40 N/mm2
UPSC JWM Advt. No.-50/2010
Ans. (c)

858. The maximum shear stress induced in a shaft M σb

subjected to a shear Sa and bending stress Sb =
will be: I y

Strength of Materials 311 YCT

 5 × 103 × 1000 × 12 × 50 Ans. (c)
= σb (A) Membrane Stress → Thin cylindrical shell
100 × 1003
subjected to internal fluid pressure.
σb = 30 N/mm2
(B) Torsional and direct shear stress → Closed
861. A horizontal beam with square cross-section is coiled helical spring under axial load.
simply supported with sides of the square (C) Double shear stress → Rivets in butt joining
horizontal and vertical and carries a having two cover plates.
distributed loading that produces maximum
(D) Maximum shear stress → Neutral axis of beam.
bending stress σ in the beam. When the beam is
placed with one of the diagonals horizontal the 864. At a particular cross-section of a straight bar of
maximum bending stress will be : rectangular cross-section, the bending moment
is 10000 N-mm and the axial force acting
(a) (1/ 2 ) σ (b) σ through the centroid of the cross-section is 1000
N. If the area of cross-section is 100 mm2 and the
(c) 2σ (d) 2σ
area moment of inertia about the centroidal axis
APPSC IOF, 2009 is 1000 mm4, the distance between the centroidal
M axis and the neutral axis is
Ans. (c) : = Bending stress (a) 0.1 mm (b) 1 mm
Z
For rectangular beam with sides horizontal & vertical, (c) 2 mm (d) 4 mm
DRDO Scientists 2009
a3
Z1 = P My
6 Ans. (b) : σ = ±
a3 A I
For same section with diagonal, Z2 = At the neutral axis, σ = 0,
6 2
My P
Ratio of stresses = 2 =
I A
862. Which one of the following conditions will PI 1000 × 1000
cause bending of the beam without being or y= =
twisted? MA 10000 × 100
(a) The applied load is vertical = 1 mm
(b) The beam cross-section is doubly symmetric 865. A cantilever beam has the cross-section of an
(c) The plane of applied load contains the axis of isosceles triangle and is loaded as shown in
symmetry of the beam section figure. If the moment of inertia of the cross-
(d) The beam cross-section is singly symmetric 1 4
section I zz = m , then the maximum tensile
UPSC JWM Adv. No-16/2009 36
Ans. (a) : The conditions which causes bending of the bending stress is
beams without being twisted when the applied load is
vertical.
863. Match List-I with List-II and select the correct
answer using the code given below the lists :
List-I List-II (a) 1/16 MPa (b) 72 MPa
(Stress Induced) (Component) (c) 36 MPa (d) 1/36 MPa
(A) Membrane stress 1. Neutral axis of DRDO Scientists 2008
beam Ans. (c) :
(B) Torsional and direct 2. Close coiled helical
shear stress spring
under axial load
(C) Double shear stress 3. Thin cylindrical
shell subjected to My
internal fluid pressure σb =
I
(D) Maximum shear 4. Rivets in butt joint
1
stress having two y= m
cover plates 3
Code : bd 3 1× 13 1 4
A B C D I= = = m
36 36 36
(a) 3 4 2 1 M = 3 × 106 Nm
(b) 1 2 4 3
3 × 106 ×1
(c) 3 2 4 1 σb = = 36MPa
1
(d) 1 4 2 3 3×
UPSC JWM Adv. No-16/2009 36

Strength of Materials 312 YCT

866. A beam of I section of depth 20 cm is subjected Ans. (a) : Flexural rigidity = EI
to a bending moment M. The flange thickness kN
is 1 cm. If the minimum stress developed in the Unit of flexural rigidity = 2 × m 4 = kNm 2
2
1 section be 100 N/mm , the stress developed at m
the inner edge of the flange is 869. A cantilever beam of rectangular cross-section
(a) 95 N/mm2 (b) 90 N/mm2 is subjected to a concentrated load at its free
(c) 47.5 N/mm2 (d) 60 N/mm2 end. The state of stress at a point P on the
WBPSC AE, 2007 centroidal longitudinal axis of beam is given by
Ans. (b)

(a)

(b)

We know that under sagging top fibre is in compression

and bottom fibre is in tension
(c)
Hence
(σb)min = –100 N/mm2 at top fibre
(σb)max = +100 N/mm2 at bottom fibre
Bending stress is given by (d)
m×y
σb =
I
where y = distance from neutral axis UPSC JWM Advt. No.-50/2010
⇒ σb α y Ans. (d) : From question the longitudinal axis is where,
Let (σ1)2 = Bending stress of inner fibre of top flange. y=0
(σ b )Top YT M σb
So, =
⇒ = I y
(σ b ) 2 y2
For y=0
100 10 ×10 σb = 0
=
(σ b ) 2 9 × 10 So only shear stresses are there
900
(σ b ) 2 = = 90
10
(σ b ) 2 = 90 N / mm 2
867. A beam of rectangular section of breadth b and
depth d is tested under bending and Mult is the
ultimate bending moment recorded. The
5. Torsion of Shafts
modulus of rupture is given by
6M ult 6M ult 870. Compare the strength of solid and hollow
(a) (b) shafts both having outside diameter D and
bd 3 bd 2 hollow shaft having inside diameter of D/2 in
12M ult 12M ult torsion. The ratio of strength of solid to hollow
(c) 2
(d) 3 shafts in torsion will be
bd bd
WBPSC AE, 2007 (a) 0.5 (b) 0.75
Ans. (b) : Modulus of Rupture is given by (c) 15/16 (d) 1/16
MR = σper × z VIZAG Steel MT 24.01.2021
RPSC IOF, 2020
bd 2 Vadodara Municipal Corp. DEE, 2018
Mult = σ per ×
6 ISRO Scientist/Engineer (RAC) 22.04.2018, 07.05.2017
GPSC EE Pre, 28.01.2017
6 × M ult
σ per = UPRVUNL AE 21.08.2016
bd 2 Kerala PSC IOF 19.04.2016
868. Flexural rigidity (EI) is expressed in units of ISRO Scientist/Engineer (RAC) 29.11.2015
(a) kNm2 (b) MNm MPPSC State Forest Service Exam, 2014
(c) kNm–2 (d) MNm4 UKPSC AE 2012 Paper-I
APPSC AE Subordinate Service Civil/Mech. 2016 TRB Asstt. Prof., 2012, ESE 2001, GATE 1993
Strength of Materials 313 YCT
Ans. (c) : For solid shaft polar moment of inertia τπd 3 2πN
πD 4 P= ×
J= 16 60
32 P∝d 3

PA d 3A (d)3 d3 1
πD 4 2 π = 3 = = 3=
Ts = τ × = × τD3 PB d B (2d) 8d3
8
32 D 16
For hollow shaft PA : PB = 1: 8
π
J = ( D4 – d4 ) ∴ d =
D
32 2 872. A solid circular shaft is subjected to a bending
moment M and twisting moment T. What is the
π  D4 – d4  equivalent twisting moment Te which will
TH = .τ  
16  D  produce the same maximum shear stress as the
above combination?
According to question d = D/2
(a) M 2 + T 2 (b) M+1
π  D 4 − D 
4

TH = ⋅τ 16 (c) M2 + T2 (d) M–1

16  
 D  GPSC DEE, Class-2 (GWSSB) 04.07.2021
Karnataka PSC AE (WRD) 31.07.2021
π
TH = ⋅ τ  D3 
15
RPSC ADE 2016
16 16  GPSC ARTO 01.05.2016
π Kerala PSC IOF 19.04.2016
τD3
TS 16 RPSC AE GWD, 2011
=
TH π 15 ISRO Scientist/Engineer 2011, ESE 2007
⋅ τ ⋅ D3
16 16 32M 16T
Ans. (c) : σ = ,τ= 3
TS 16
= πd 3 πd
TH 15 2
 σ−0 2
TH 15 Max shear stress =  2  +τ
=  
TS 16
= radius of mohr circle
871. Two shafts A and B are made of the same 2 2 2
material. The diameter of shaft B is twice that =  σ  + τ 2 =  16M  +  16T 
2  3   3
of shaft A. The ratio of power which can be    πd   πd 
transmitted by shaft A to that of shaft B is
16
1 1 τ max = 3 M 2 + T 2
(a) (b) πd
2 4
16Teq 16
1 1 = 3 × M2 + T2
(c) (d) πd 3 πd
8 16
ISRO Scientist/Engineer (RAC), 10.03.2019 Teq (Equivalent torque) = M 2 + T 2
Nagaland PSC (CTSE) 2018, Paper-I 873. Maximum shear stress developed on the
APGENCO AE, 2017 surface of a circular shaft under pure torsion is
Karnataka PSC AE, 10.09.2017 240 MPa. If the shaft diameter is doubled, then
Kerala PSC IOF 19.04.2016 the maximum shear stress developed
CGPSC AE 16.10.2016 corresponding to the same torque will be -
OPSC AEE 2015 Paper-I (a) 120 MPa (b) 60 MPa
ISRO Scientist/Engg. 11.10.2015 (c) 30 MPa (d) 15 MPa
GPSC Lect. (Poly.) 07.12.2014 PPSC Asstt. Municipal Engineer 15.06.2021
TRB Poly. Lect., 2012 TSPSC AEE 2017, 2014
ISRO Scientist/Engg. 2012 HPPSC Asstt. Prof. 18.11.2016
ESE 2001, GATE 1994 NPCIL ET, 2015
VIZAG STEEL MT 18.08.2013
Ans. (c) : As we know that power P=T × ω
VIZAG STEEL MT 10.06.2012
πN
P = T×2 GATE 2003
60
Ans. (c) : τ1 = 240 MPa
16T τπd 3 When shaft diameter is doubled, maximum shear stress
and τ = 3 ⇒ T = (τ = shear stress)
πd 16 = τ2
dA=d, dB=2d Let d be the original diameter

Strength of Materials 314 YCT

  T1  Ans : (b) Torsional Rigidity– The torque required to
  produce a twist of one radian per unit length of the
τ1  ZP1  shaft
=
τ2  T2  G.θ.J
  T=
 ZP 2  ℓ
(Since torque in both the cases is constant) T.ℓ
Torsional Rigidity ( GJ) =
τ1 T  Z  (Z ) θ
= 1 ×  P2  = P 2
τ2 ZP1  T2  ( ZP1 ) 876. A solid shaft can resist a bending moment of
3.0 kNm and a twisting moment of 4.0 kNm
π ( 2d )
3
together. The maximum torque that can be
τ1 π 16 applied is
= 163 = × 8d 3 × 3 (a) 7.0 kNm (b) 3.5 kNm
τ2 πd 16 πd (c) 4.5 kNm (d) 5.0 kNm
16 CIL MT 27.02.2020
τ1 Vadodara Muncipal Corp. DEE, 2018
=8
τ2 HPPSC Asstt. Prof. 18.11.2016
Kerala LBS Centre For Sci. & Tech. Asstt. Prof. 2014
1
τ 2 = 240 × = 30 MPa TRB Poly. Lect. 2012
8 Ans. (d) : M= 3.0 kNm, T= 4.0 kNm, Te=?
τ2 = 30 MPa
874. When a shaft of diameter D is subjected to a Equivalent torque = M 2 + T 2 = (3) 2 + (4) 2
twisting moment T and bending moment M, Te = 25 = 5kNm
then equivalent bending moment Me is given by
877. A shaft can safely transmit 90 kW while
(a) M 2 + T 2 rotating at a given speed. If the diameter of the
(b) M 2 − T 2 shaft is doubled and speed is halved the

( )
1 previous shaft, the power that can be
(c) M + M 2 +T2 transmitted by new shaft is
2 (a) 360 kW (b) 90 kW
(d)
1
2
(
M − M 2 +T 2 ) (c) 720 kW (d) 180 kW
Karnataka PSC AE (WRD) 31.07.2021
Nagaland PSC (CTSE) 2018, Paper-I BPSC Poly. Lect. 2016
PTCUL AE 25.06.2017 CGPSC Poly. Lect. 22.05.2016
MPPSC AE 2016 VIZAG MT 2015
Kerala PSC IOF 19.04.2016 TRB Asstt. Prof., 2012, ESE 2002
JPSC AE 2013, Paper-V Ans : (a) We know that
APPSC AE 04.12.2012
2πNT 2πN πd 3 τ
APPSC AEE 2012 P= = ×
WBPSC AE 2003, ESE 2008, 1996 60 60 16
Ans : (c) : Equivalent bending moment, (Me) P ∝ N,d 3

P = kNd3
(
1
Me = M + M +T
2
2
)2 If diameter of the shaft is double and halved the speed.
N
Equivalent Twisting moment, (Te), d2 = 2d1, N2 = 1
2
Te = T 2 + M 2 P1  N1  d1 
3 3
1 1
∴ =   = 2  =
875. What is torsional rigidity? P2  N 2  d 2  2 4
(a) The torque required to produce a twist of one P2 = 4P1 = 4 × 90 = 360 kW
degree per unit length of the shaft
(b) The torque required to produce a twist of one 878. When a circular shaft is subject to torque only,
radian per unit length of the shaft the torsional shear stress is :
(c) The torque required to produce a twist of one (a) Maximum at the axis of rotation and zero at
radian per unit area of the shaft outer surface.
(d) The torque required to produce a twist of one (b) Maximum at outer surface and zero at axis of
degree per unit area of the shaft rotation.
VIZAG Steel MT 24.01.2021, Shift-I (c) Uniform from axis of rotation to outer surface
TSPSC Manager (Engg.) HMWSSB 12.11.2020 (d) None of the above
APPSC AEE Mains 2016 (Civil Mechanical) JPSC AE 10.04.2021, Paper-II
Rajasthan AE (Nagar Nigam) 2016 Shift-3 OPSC AEE 2019 Paper-I
UPRVUNL AE 2014, WBPSC AE, 2007 RPSC Vice Principal ITI 2018, ESE 2012
Strength of Materials 315 YCT
Ans. (b) : According to torsion equation, T2 = 8T1
T Gθ τ
= = Thus after increasing the diameter of shaft and all other
J ℓ R parameter remains unchanged, the torque will be 8
TR times of previous torque.
τ=
J 880. Bending moment 'M' and torque 'T' is applied
Tr on a solid circular shaft. If maximum bending
τ= stress equals to maximum shear stress
J developed, then M is equal to
At axis of rotation,
(a) T (b) T/2
(c) 2T (d) 4T
GPSC Engg. Class-II Pre 19.01.2020
APPSC AE Subordinate Service Civil/Mech. 2016
r=0 GPSC Lect. (Poly.) 07.12.2014
τ=0 ISRO Scientist/Engg. 2010
At outer surface 32M
Ans. (b) : σb max =
r=R πd 3
TR 16T
τ= τmax = 3
J πd
879. The diameter of shaft is increased from 50 mm σ b max
= τ max
to 100 mm all other conditions remaining 32M 16T
unchanged. How many times the torque = 3
carrying capacity increases? πd 3 πd
(a) 2 times (b) 4 times M = T/2
(c) 8 times (d) 16 times 881. When two shafts of same length, one of which is
GPSC ARTO 01.05.2016, ESE 2016 hollow, transmit equal torques and have equal
KPSC AE. 2015 maximum stress, then they should have equal
Kerala PSC AE 06.08.2015 (a) polar modulus
TANGEDCO AE 2015 (b) diameter
(c) polar moment of inertia
1 (d) angle of twist
Ans. (c) : Given, d1 = 50 mm = mm
20 ESE 2018
1 Rajasthan AE (Nagar Nigam) 2016 Shift-3
d2 = 100 mm = mm
10 APPSC AEE 2012
∵ equation for the shaft APPSC AE 04.12.2012
T τ Gθ CSE Pre-1994
= = ...(i)
J R l Ans. (a) : Given, T1 = T2 & τ1 = τ2
π 4 T τ
moment of inertia I = d Torsional formula =
64 J R
⇒ Polar moment of inertia [J = I1 + I 2 ] J
T = .τ
π 4 R
J= d T = τ ZP (∴ ZP = Polar modulus)
32
For transmit equal torque (T = T1 = T2) and equal
T τ
Now from = maximum stress (τ1 = τ2)
J R T =τZ , T =τ Z
1 1 P1 2 2 P2
τ× 2 × πd 4 τ × πd 3
T= = Z P1 = Z P 2
d × 32 16
T∝d 3
882. A solid shaft of diameter d and length L is fixed
3 at both the ends. A Torque T0 is applied at a
 1  distance L/4 from the left end as shown in the
T1 =  
 20  figure given below:
3
1
T2 =  
 10 
T1 1
=
T2 8

Strength of Materials 316 YCT

 The maximum shear stress in the shaft is Ans. (a) : General torsion equation,
(a) (16T0/πd3) (b) (12T0/πd3) T τ Gθ
3 = =
(c) (8T0/πd ) (d) (4T0/πd3) J r L
NPCIL ET, 2015 T = Torque or twisting moment
OPSC Civil Services Pre. 2011 J = Polar moment of inertia
GATE 2009, ESE 2004 τ = Shear stress at outer fibre
Ans. (b) : r = Radius of shaft
G = Modulus of rigidity
θ = Angle of twist
L = Length of shaft
D = Diameter of shaft
For solid shaft,
πD 4
J=
32
3L 885. Two shafts of same length & materials are
T0 × joined in series. If the ratio of their diameter is
T1 = 4 = 3T0
2, then the ratio of their angle of twist will be :
L 4
(a) 2 (b) 4
L (c) 8 (d) 16
T0 ×
T2 = 4 = T0 ISRO Scientist/Engg. 22.04.2018
L 4 J&K PSC Civil Services Pre, 2010
16T WBPSC AE 2003, CSE Pre-2003, 1995
τmax =
πD 3
d
Ans. (d) : Ratio of diameter = 1 = 2
 3T  d 2
16 ×  0 
 4  ΤL
τmax = Angle of twist θ =
πD3 GJ
Let two shaft have diameter d1 and d2, its length L1 and
12T0 L2.
τmax =
πD 3
d
Given, L1 = L2, 1 = 2
883. For a round shaft subjected to pure torsion, the d2
shear stress at the centre (axis) will be
T1L1
(a) maximum
(b) minimum θ1 GJ
= 11
(c) zero θ2 T2 L 2
(d) The information provided is insufficient G 2J2
BPSC AE Mains 2017 Paper - VI For same material,
TSPSC AEE 28.08.2017 (Civil/Mechanical) G1 = G2
GPSC ARTO 01.05.2016 And for series joined shaft if torque applied T1 = T2 for
UKPSC AE-2013, Paper-I both in shaft
θ J
Ans : (c) : So, 1 = 2
T τ Gθ θ2 J1
= =
J R ℓ π 4
d 4 4
T×R θ1 32 2  d 2   1 
τ= = =  = 
J θ2 π 4  d1   2 
d1
T×0 32
τ0 = =0
J θ1 1
=
Hence, shear stress at centre will be zero for round shaft θ2 16
subjected to pure torsion. θ2
884. The value of J in equation T/J = τ/r = G.θ/L for θ = 16
1
a circular shaft of diameter D will be
886. If the diameter of the shaft is increased two
(a) πD /32
4
(b) πD /64
4
times, the torque transmitted will be
(c) πD3/32 (d) πD3/16 (a) Two times (b) Four times
NSPSC AE 2018, RPSC AE 2018 (c) Eight times (d) Sixteen times
Punjab PSC SDE 12.02.2017 APGCL AM, 2021, JPSC AE Paper-II 10.04.2021
WBPSC AE, 2017 RPSC IOF, 2020
Strength of Materials 317 YCT
Ans. (c) T1 L1 T2 L 2
θ= +
T τ GJ1 GJ 2
=
J R
T  L1 L 2 
T ∝ Zp =  + 
G  J1 J 2 
πd1 3 3
T1 d   d  1  
= 16 =  1  =  1  =
T2 πd 2  d 2   2d1  8 T  L1 L2 
θ=  4 + 4 
16 G  πd1 πd 2 
T2 = 8T1  32 32 
 L L/2  T
887. A machine element XY, fixed at end X, is θ= +
subjected to an axial load P, transverse load F, π(2d) 4
π(d)4  G
 
and a twisting moment T at its free end Y. The  32 32 
most critical point from the strength point of T  32L 32 L / 2 
view is_____ =  +
G  π × 16d 4
π × d 4 
T  2L 16L 
=  4 + 4
G  πd πd 
T 18L 
(a) a point on the circumference at location X θ=  4
G  πd 
(b) a point at the center at location Y
18TL 18TL
(c) a point on the circumference at location Y θ= ⇒ d4 =
(d) a point at the center at location X Gπd 4
πGθ
Assam Engg. College AP/Lect. 18.01.2021 18TL
PPSC Asstt. Municipal Engineer 15.06.2021 d = 1/ 4
πGθ
GATE 2016
889. A solid circular shaft needs to be designed to
Ans. (a) : Most critical loading point will be a point on transmit a torque of 50 Nm. If the allowable
circumference at location X shear stress of the material is 140 MPa,
Torque (τ) and axial load (P) will be same for element assuming a factor of safety 2, the minimum
XY but vertical force (F) will produce maximum stress allowable design diameter in mm is:
at a point on circumference at X. (a) 8 (b) 16
888. A torque T is applied at the free end of a (c) 24 (d) 32
stepped rod of circular cross-sections as shown Assam Engg. College AP/Lect. 18.01.2021
in the figure. The shear modulus of the NPCIL ET, 2015, GATE 2012
material of the rod is G. The expression for d to Ans. (b) : Given,
produce an angular twist θ at the free end is T = 50 N-m
τallowable = 140 MPa
Factor of safety (FOS) = 2
τ 140
τsafe = allowable = = 70 MPa
FOS 2
We know that,
32TL 18TL
(a) 1/ 4 (b) 1/ 4 T
πθG πθG τsafe = (z p = Polar section modulus)
zp
16TL 2TL
(c) 1/ 4 (d) 1/ 4 πd 3
πθG πθG zp =
Vadodara Muncipal Corp. DEE, 2018 16
NPCIL ET, 2015, GATE 2011 16T
τsafe = 3
Ans. (b) : πd
16T
d3 =
π× τsafe
16 × 50 × 103
d= 3
π× 70
Given- T1 = T2 = T = 15.8 mm
θ = θ1 + θ2 d ≈ 16 mm

Strength of Materials 318 YCT

890. A twisting moment 'T' and Bending moment Ans. (d) : Bending moment (M) = Torque (T)
'M' are acting on a circular shaft. What is the 32M
ratio of maximum shear stress to maximum Bending stress ( σb ) =
bending stress? πd3
(a) 2 M/T (b) 2T/M 16T σ b 32M / πd 3 32
(c) T/M (d) T/2M Shear stress ( τ ) = 3 = = =2
πd τ 16T / πd 3 16
RPSC Lect. (Tech. Edu. Dept.) 13.03.2021, Paper-I (∴ M = T)
Gujarat PSC AE 2019
UPPSC AE 12.04.2016 Paper-I σ b = 2τ
σ M 893. A circular shaft can transmit a torque of 5 kN-
Ans. (d) : As we know, b = m. If the torque is reduced to 4kN-m, then the
y I
maximum value of bending moment that can
M be applied to the shaft is :
σb = × y
I (a) 1 kN-m (b) 2 kN-m
M d 32M (c) 3 kN-m (d) 4 kN-m
= 4× = RPSC ACF & FRO, 26.02.2021
πd 2 πd3
KPSC AE 2015, APPSC IOF, 2009
64
16T Ans. (c) : Given,
τmax = 3 , Teq = 5 kN-m
πd
T = 4 kN-m
τ T M=?
So max =
σb 2M
Teq = M 2 + T 2
891. A circular shaft subjected to torsion undergoes
a twist of 1° in a length of 120 cm. If the 5 = M 2 + 42
maximum shear stress induced is limited to M = 3 kN-m
1000 kg/cm2 and if the modulus of rigidity, G =
0.8 × 106 kg/cm2, then the radius of shaft in cm 894. A solid shaft steel of 100 mm diameter and 1.0
should be m long is subjected to a twisting moment T.
This shaft is to be replaced by a hollow shaft
π π having outer and inner diameters as 100 mm
(a) (b)
18 27 and 50 mm respectively. If the maximum shear
18 27 stress induced in both the shafts is same, the
(c) (d) twisting moment T transmitted by hollow shaft
π π
ISRO Scientist/Engineer (RAC) 12.01.2020 must be reduced by
TRB Poly. Lect. 2012 (a) T / 4 (b) T / 8
ISRO Scientist/Engg. 2007 (c) T / 16 (d) T /12
Ans. (d) : From torsion equation, BPSC Poly. Lect. 2016, ESE 2012, ESE 1998
T τ Gθ Ans : (c) Solid Shaft hollow shaft
= = d = 100mm do = 100 mm
J r ℓ ℓ = 1.0 m di = 50 mm.
τ Gθ τ L ℓ = 1.0 m.
= r= ×
r ℓ θ G Ts = Twisting moment for solid shaft Th = twisting
π moment transmitted by hollow shaft
θ = 1° = , L = 120cm, τ = 1000 kg / cm 2 same material then τ = τs = τh
180
for solid shaft:-
G = 0.8 × 106 kg/cm2
πd 3τ
1000 kg / cm 2 120cm Ts =
r= × kg / cm 2 16
π /180 0.6 × 106
for hollow shaft:-
27
r = cm
π
892. A shaft was initially subjected to bending
moment and then was subjected to torsion. If
the magnitude of bending moment is found to
be the same as that of the torque, then the ratio
of maximum bending stress to shear stress
would be
(a) 0.25 (b) 0.50
(c) 1.0 (d) 2.0
Punjab PSC (Lect.) 06.08.2017
APPSC AEE 2012, APPSC IOF 2009
Strength of Materials 319 YCT
 15 898. Angle of twist of a solid shaft of torsional
Th = Ts rigidity GJ, length L and applied torque T will
16
Reduce twisting moment = Ts – Th be given by:
15 T L
= T− T (a) (b)
16 GJL GJ T
T TL GJ
Reduce twisting moment = (c) (d)
16 GJ TL
895. The polar section modulus of a solid circular SJVN ET 2019, UKPSC AE 2007 Paper -I
shaft of diameter 'd' about an axis through its τ T Gθ
centre of gravity is Ans. (c) : = =
r J ℓ
π 3 π 3 Gθ T
(a) d (b) d =
4 12 ℓ J
π 3 π 3 TL
(c) d (d) d θ=
16 32 GJ
Nagaland PSC (CTSE) 2018, Paper-I
899. A shaft of diameter (d) is subjected to torque
GPSC Lect. (Auto) 16.10.2016
(T) and bending moment (M). The value of
WBPSC AE 2008 maximum principle stress and maximum shear
J stress is given respectively by:
Ans. (c) : Polar section modulus ZP =
16  16
M + M2 + T2 ; 3  M2 + T2 
R (a)
π×d 4
πd 3   πd  
ZP = ×2
32 × d (b)
16  16
M + M2 + T2 ; 4  M 2 + T2 
π πd 4   πd  
ZP = d3
16  16
16 (c) M 2 + T 2  ; 3 M + M 2 + T 2 
πd 
3  
 πd  
896. In case of a circular shaft subjected to torque
the value of shear stress 16  16
(a) Is uniform through out (d) M 2 + T 2  ; 4 M + M 2 + T 2 
πd 4  
 πd  
(b) Has maximum value at axis
32  32
M 2 + T 2  ; 4 M + M 2 + T 2 
(c) Has maximum value at the surface
(e)
(d) Is zero at the axis and linearly increases to a πd 4  
 πd  
maximum value at the surface of the shaft
CGPSC AE 25.02.2018
Nagaland PSC CTSE 2017, Paper-I
GPSC Asstt. Director of Transport 05.03.2017 Nagaland PSC CTSE 2016 Paper-I
Nagaland PSC CTSE 2016 Paper-I Ans. (a) : Maximum principle stress σ max .
Ans. (d) : In case of a circular shaft subjected to torsion 16
σ max = 3  M + M 2 + T 2 
πd  
the value of shear stress is zero at the centre and
linearly increase to a maximum value at the surface of
the shaft. maximum shear stress
16
897. For a circular shaft of diameter D subjected to τ max = 3  M 2 + T 2 
torque T, the maximum value of the shear πd  
stress is
900. The diameter of shaft is increased from 30 mm
(a) (64T/πD3) (b) (32T/πD3)
to 60 mm, all other conditions remaining
(c) (16T/πD3) (d) (128T/πD3)
3 unchanged. How many times its torque
(e) (8T/πD )
carrying capacity increases?
RPSC 2016, CGPSC AE 26.04.2015 Shift-I
(a) 2 (b) 4
GATE 2006
(c) 8 (d) 16
Ans. (c) : From torsion equation we have
OPSC AEE 2019 Paper-I
T τ
= MPSC HOD (Govt. Poly. Colleges) 04.10.2014
J r Ans. (c) : From torsion equation.
where T = torque
T Gθ τ
J = polar moment of inertia = =
τ = shear stress J ℓ R
r = radius τ× J
T=
The maximum value of shear stress is given by R
T τ 16T π
= , τ= 3 τ × d4
 πD   D 
4
πD 32
   2  T=
 32  d/2
Strength of Materials 320 YCT
 τd 3 π 32 M M
T= (a) (b) 16 2.
16 πD 3 πD 3
T ∝ d3 16M 16 M
(c) (d) . 3
T2  d 2 
3
πD 3 2 πD
=  UPSC JWM Advt. No.-52/2010
T1  d1 
UKPSC AE 2007 Paper -I
3
T2  60  Ans. (b) : Solution : Given :
=   = 23 = 8
T1  30  Combined effect of Bending moment and Twisting
901. A hollow shaft of length L is fixed at its both moment. Both have equal magnitude M.
ends. It is subjected to a torque T at a distance
of L/3 from its one end. What is the reaction
∴ Shear stress (τ) =
16
(
πD 3
M2 +T2 )
torque at the other end of the shaft? 16
(a) 2T/3 (b) T/2 ∴ τmax = × 2M 2 (∵ M = T )
πD 3
(c) T/3 (d) T/4
M
TSPSC AEE 28.08.2017 (Civil/Mechanical) τmax = 16. 2. 3
UPSC JWM Advt. No.-52/2010 πD
Ans. (c) : Let; T is the torque acting at B from the end 903. Torque in a solid shaft of diameter 'd' and
A at a distance L/3 shear stress 'τ' is given by
π 3 π 3
(a) τd (b) τd
8 16
π 3 π 3
(c) τd (d) τd
12 32
UPPSC AE 12.04.2016 Paper-I
Let; TA and TC reaction torque on both side of fixed WBPSC AE 2008
end of the hollow shaft due to applying torque T. Ans. (b) : From Torsion equation
(1) For section – (1) (1) & section (2)- (2) T τ Gθ
= =
Torque ; T1 = TA ; T2 = TA – T or TC …(i) J r L
(2) Net torque; TA + TC = T …(ii) T τ
(3) θ1 + θ2 = 0 …(iii) =
J r
(for both end fixed) Tr T × d × 32
T1L1 T2 L 2 ∵ GJ = constant τ= =
+ =0  J 2 × π × d4
G1 J1 G 2 J 2 i.e. G1J1 = G 2 J 2 = GJ 16T
⇒ T1L1 + T2L2 = 0 τ= 3
πd
L 2L
⇒ TA × + ( TA − T ) × =0
3 3 πd3
T= τ
⇒ TA + 2TA – 2T = 0 16
⇒ 3TA = 2T Where symbols have their usual meaning.
2T 904. In case of a solid shaft the strain energy in
TA =
3 torsion, per unit volume, is equal to
From eqn (ii) ; TA + TC = T τ2 τ2
(a) (b)
2T 2C 4C
⇒ + TC = T
3 τ2 τ2
(c) (d)
T 6C 8C
⇒ TC =
3 ESE 2013, WBPSC AE 2008
TC = T/3 ; Ans. (b) For prismatic bar under torsion.
is the torque at the other end of the shaft. τ2
U= ⋅V
902. A uniform solid shaft of diameter D is 4C
subjected to equal magnitude of bending
U τ2
moment and twisting moment M. What is the =
maximum shear stress induced in the shaft? V 4C

Strength of Materials 321 YCT

905. Two shafts A and B of solid circular cross- I XX
section are identical expect for their diameter Z XX =
dA and dB. The ratio of power transmitted by Y
the shaft A to that of shaft B is π 1
= ×d4 ×
64 d /2
(a)
dA
(b)
(dA ) 2

π 3 3
dB ( d B )2 ∴ Z XX =
32
d mm

(c)
( d A )3 (d)
( d A )4 908. A hollow shaft of the same cross-section area
( d B )3 ( d B )4 and material as that of a solid shaft, transmits:
Karnataka PSC Lect., 27.05.2017 (a) Same torque (b) Lesser torque
KPSC ADF 2015 (c) More torque (d) None
Ans. (c) : UJVNL AE 2016, RO Scientist/Engineer 2007
ESE 2005
π 3
∵ Power (P) = T.ω ∵ Torque ( T ) = dτ Ans : (c) A hollow shaft of the same cross- sectional
16 area and materials transmits more torque than solid
π 3 shaft.
P= d .τ.ω given As = Ah
16
π 3 π 2 π 2
For shaft A; PA = d A .τ.ω
 τA = τB = τ  d = d o − d i2 
  4 4
16 ωA = ωB = ω
d 2 = d o2 − d i2 .............. (i)
π 3
For shaft B; PB = d B .τ.ω from equation (i) we can say that
16 do > d
PA ( d A )
3
πd 3o × τ 1 − k 4 
∴ =
PB ( d B )3 Th
= 16
Ts  πd3 
906. A solid shaft transmits a torque of T. The  × τ
allowable shear stress is τ . What is the  16 
diameter of the shaft ? Th d o 1 − k 
3 4

16T 32T =
(a) 3 (b) 3 Ts d3
πτ πτ If we assume that [hit and trail method]
16T T di = 3 unit
(c) 3 (d) 3
do = 5 unit
τ π
d = 4 unit
OPSC AEE 2015 Paper-I
RPSC AE GWD, 2011, ESE 2008   3 4 
125 1 −   
Ans : (a) Torsion equation Th   5  
T Gθ τ = = 1.7
= = Ts 64
J ℓ r Th = 1.7 Ts
32T 2τ 16T
= = =τ So we can say that Th > Ts.
πd 4 d πd3 909. Angle of twist of a shaft of diameter ‘d’ is
16T inversely proportional to :
d=3 (a) d (b) d2
πτ (c) d 3
(d) d4
907. The section modulus of a circular section of OPSC AEE 2019 Paper-I
diameter (d) is ISRO Scientist/Engg. (RAC) 22.04.2018
(a) π/32 d2 (b) π/32 d3 Ans : (d) : By using torsion formula,
(c) π/64 d 3
(d) π/64 d4
Tℓ Tℓ 1
Nagaland PSC (CTSE) 2018 Paper-I θ= = ⇒θ ∝ 4
GMB AAE 25.06.2017 GJ G × π × d 4 d
Ans. (b) : Section Modulus – Section modulus is the 32
ratio of moment of inertia about N.A. upon the for test 910. A solid circular shaft is subjected to pure
point of section from Neutral axis (N.A) torsion. The ratio of maximum shear to
maximum normal stress at any point would be-
I
Z= (a) 1 : 1 (b) 1 : 2
Y (c) 2 : 1 (d) 2 : 3
for circular section RPSC AE 2018, APPSC AEE 2012, ESE 1996
Strength of Materials 322 YCT
Ans. (a) : A solid circular shaft is subjected to pure 4PL3
torsion, then Ans. (c) :
πR 4 G
1
τ max = (σ x − σ y ) 2 + 4τ xy2
2
σx = σy = 0
τ max = τ xy
σx +σy 1
σ max = + (σ x − σ y ) 2 + 4τ xy2
Vertical deflection at point 'A'
2 2 (X) = θ × L
σ max = τ xy X = θ × L ..............(i)
then, T G θ τ
= = T = P × 2L
τ max J L R
= 1:1 TL
σ max θ=
GJ
911. Two shafts, one solid and the other hollow, x = θ × L
made of the same material, will have the same
TL2
strength and stiffness, if both are of the same =
(a) length as well as weight GJ
(b) length as well as polar modulus P × 2L × L2
=
π
G × ( D4 )
(c) weight as well as polar modulus
(d) length, weight as well as polar modulus 32
ESE 2017, Nagaland PSC CTSE 2016 Paper-I 2 PL3
=
Ans. (b) : Solid shaft and hollow shaft are same π
G ( 2R )
4
material, so G is same. 32
T τ Gθ 4PL 3
= = x =
J r ℓ GπR 4
T GJ Vertical deflection at point 'A'
kt = =
θ ℓ 913. The polar moment of inertia of a hollow shaft
T of outer diameter (D) and inner diameter (d) is
τmax = × rmax π 3 3 π
J
∴ For strength and stiffness to be same, both must have
(a)
16
D –d( ) (b)
16
D4 – d4( )
π π
same polar moment of inertia (J) and same length (ℓ). (c)
32
4
D –d(4
) (d)
64
(
D4 – d 4 )
912. A rigid horizontal rod of length 2L is fixed to a MECON MT 2019
circular cylinder of radius R as shown in the Sikkim PSC (Under Secretary), 2017
figure. Vertical forces of magnitude P are Ans. (c) : Square section Polar Moment of Inertia
applied at the two ends as shown in the figure.
a4
J= = 0.1667a 4
6
Rectangular Section

J=
(
bd b 2 + d 2 )
12
Circular section
The shear modulus for the cylinder is G and
πD 4
the Young's modulus is E. J= = 0.098D 4
The vertical deflection at point A is 32
Tube section
PL3 PL3
π
( )
(a) (b)
πR 4 G πR 4 E J= D4 – d 4
3 3
32
4PL 2PL Triangular section
(c) (d)
πR 4 G πR 4 E
Assam Engg. College AP/Lect. 18.01.2021 3 4
J= S = 0.036S4
GATE 2016 48
Strength of Materials 323 YCT
914. When a shaft is subjected to twisting moment, Torque carrying capacity,
every cross-section of the shaft will be under πD3 (1 − P 4 )
(a) tensile stress (b) compressive stress TH = τmax .....(1)
(c) shear stress (d) bending stress 16
Maximum shear stress for solid shaft,
Vizag Steel (MT) 2017,
16T
Assam PSC CCE Pre 2015 τmax =
Ans. (c) πD3
Torque carrying capacity,
T Gθ τ πD3
Torsion equation – = = TS = τmax .....(2)
J ℓ r 16
π Equation (2) ÷ equation (1)
T = τd 3
16 TS
=
1
TH (1 − P 4 )
917. The greatest twisting moment which a shaft
section can resist is equal to
(a) polar modulus × τ (b) polar modulus/τ
(c) τ/polar modulus (d) None of the above
RPSC VPITI 14.02.2016
APPSC Poly. Lect. 2013
Ans. (a) : The maximum twisting moment which a shaft
16T
Shear stress in shaft, τ = can resist is product of the permissible shear stress and
πd3 polar modulus.
When a shaft is subjected to twisting moment, every T τ Gθ
cross-section of the shaft will be under shear stress. = =
J R ℓ
915. Torsional Stiffness is _______ . Where, T = Maximum Torque,
(a) quotient of shear modulus and polar moment J = Polar Moment of Inertia
of inertia τ = Shear stress, R = Radius,
(b) product of shear modulus and polar moment G = Modulus of rigidity, θ = Angle of twist
of inertia
(c) torque produced per unit moment of inertia 918. For the same material, length and given torque,
(d) torque produced per unit angle of twist a hollow shaft weighs _____ a solid shaft :
(a) less than (b) more than
VIZAG MT, 14.12.2020 (c) equal to (d) None of the above
APPSC AE Subordinate Service Civil/Mech. 2016 APPSC AE 04.12.2012, APPSC AEE 2012
Ans. (d) : Torsional equation, Ans. (a) : For the same material, length and given
T Gθ τ torque, a hollow shaft weighs less than a solid shaft.
= =
J ℓ R 919. Power transmitted by a shaft (kW) subjected to
GJ an average torque T (kNm), rotating at N rpm
T= θ is equal to
ℓ
T = KTθ 2πΝΤ 2πΝΤ
(a) (b)
3600 4500
GJ
KT = ← Torsional stiffness 2πΝΤ πΝΤ
ℓ (c) (d)
60 3060
T APPSC AEE Mains 2016 (Civil Mechanical)
KT = ← Torsional stiffness HPPSC Lect. 2016
θ
Torsional stiffness is torque produced per unit angle of Ans. (c) : As we know,
twist. Work done per minute = force × distance
= Average torque × Angular displacement
916. If 'P' is the Ratio of inside to outside diameter
of a shaft, the ratio of torque carrying capacity 2πΝ
= T×
of solid shaft to that of hollow shaft is given by: 60
(a) (1 – P4)–1 (b) 1 – P4 2 πΝΤ
P= kW
1 60
(c) P4 (d) 4
P 920. Consider the arrangement shown in the figure
Kerala PSC Poly. Lect., 2017, ESE 2008 below where J is the combined polar mass
Ans. (a) : Maximum shear stress for hollow shaft, moment of inertia of the disc and the shafts. K1,
16TH D K2 and K3 are the torsional stiffness of the
τ max = ∵ P= i respective shafts. The natural frequency of the
πD3 (1 − P 4 ) D0 torsional oscillation of the disc is given by
Strength of Materials 324 YCT
 By torsional formula:-
T Gθ τ
= =
J l r
Where, J = Polar moment of inertia
 K1 + K 2 + K 3  l = Length of shaft
(a)  
 J  r = Radius of shaft
maximum shear stresses will occur when
 K1K 2 + K 2 K 3 + K 3K1  r = rmax = R
(b)  
 J(K1 + K 2 )  T
τ max = × rmax
 K1K 2 K 3  J
(c)  J(K K + K K + K K )  T
 1 2 2 3 3 1  τ max = × R ( at surface of shaft )
J
 K1K 2 + K 2 K 3 + K 3K1  922. Maximum shear stress in a solid shaft of
(d)  
 J(K 2 + K 3 )  diameter D and length L twisted through an
Vizag Steel MT (Re-Exam) 24.11.2013 angle θ is τ. A hollow shaft of the same material
GATE 2003 and length having outside and inside diameters
D
Ans. (b) : K1 and K2 are in series and K3 in parallel with of D and respectively is also twisted through
equivalent resistance between 1 and 2. 2
Equivalent stiffness between 1 and 2 (series) the same angle of twist θ. The value of maximum
shear stress in the hollow shaft will be
1 1 1
= + 16 8
K12 K1 K 2 (a) τ (b) τ
15 7
K ⋅K 4
K12 = 1 2 (c) τ (d) τ
K1 + K 2 3
Equivalent stiffness between K12 and K3 is (parallel) UPPSC AE 13.12.2020, Paper-I, ESE 1997
Keq = K12 + K3 τ max Gθ
K1K 2 Ans. (d) : =
K eq = + K3 rmax ℓ
K1 + K 2 τmax ∝ rmax
K1K 2 + K 2 K 3 + K 3 K1 ( τmax )H ( rmax )H ( D / 2 )
K eq =
(K1 + K 2 ) = =
( τmax )S ( rmax )S ( D / 2 )
The natural frequency for torsional oscillation is given
by– (τmax)H = (τmax)S = τ
923. A line shaft rotating at 200 rpm is to transmit
K eq 20 kW. The allowable shear stress of shaft
ω=
J material is 42 MPa. Neglecting the bending
moment, the diameter of the shaft needs to be :
K1K 2 + K 2 K 3 + K 3 K1
ω= (a) 35 mm (b) 50 mm
J(K1 + K 2 ) (c) 65 mm (d) 75 mm
921. The shear stress at a point in a shaft subjected RPSC ACF & FRO, 26.02.2021
to a torque APPSC Poly. Lect. 2013
(a) directly proportional to the polar moment of Ans. (b) : Given, N = 200 rpm
inertia and to the distance of the point from P = 20 kW
the axis τ = 42 MPa
(b) directly proportional to the applied torque and d=?
inversely proportional to the polar moment of
inertia 2πNT
P=
(c) directly proportional to the applied torque and 60
the polar moment of inertia 2π × 200
(d) inversely proportional to the applied torque 20 × 10 =
3
×T
60
and the polar moment of inertia
GPSC Executive Engineer 23.12.2018 T = 954.92 Nm
JPSC AE 2013, Paper-V, CSE Pre-1995 T = 954.92 × 103 Nmm
Ans. (b) : The shear stress produced by torque vary 16T
from zero at centre of cross section to maximum at τ= 3
πd
surface in circumferential directions. Shear stresses in
16T
direction normal to the cross-section plane are always d3 =
complimentary in nature and have equal magnitude. πτ

Strength of Materials 325 YCT

 16 × 954.92 ×103 Maximum principal stress-
d3 = 2
π × 42 σb σ 
σ max = +  b  + τ2
16 × 954.92 ×103 2  2 
d=3
π× 42 50  50 
2

+   + ( 50.93)
2
d = 48.74 mm ≈ 50 mm =
2  2
924. For a hollow shaft of inner diameter 'd' and
252 + ( 50.93 )
2
outer diameter 'D' the modulus of rupture is : = 25 +
(a) 8TD/π(D4 – d4) (b) 16TD/π(D4 – d4) σ max = 82 MPa
4 4
(c) 32TD/π(D – d ) (d) 64TD/π(D4 – d4)
SJVN ET 2019, RPSC VPITI 14.02.2016 927. A circular shaft subjected to twisting moment
results in maximum shear stress of 90 MPa.
Ans. (b) : The maximum shear stress calculated by
Then the maximum compressive stress in the
using the torsion formula is called modulus of rupture material is
T×D/ 2 (a) 130 MPa (b) 100 MPa
τ max =
J (c) 80 MPa (d) 90 MPa
T×D/ 2 RPSC Lect. (Tech. Edu. Dept.) 13.03.2021, Paper-I
τmax = Ans. (d) : Mohr circle for pure shear, So, the
π
32
( D4 – d4 ) compressive stress would be equal to applied shear
stress.
16TD
τmax =
π ( D4 – d 4 )
925. Two shaft A and B under pure tension are of
identical length and identical weight and are
made of the same material. The shaft A is solid
and the shaft B is hollow. We can say that.
(a) Shaft B is better than shaft A
(b) Shaft A is better than shaft B σ1 = σ (Tensile)
(c) Both the shafts are equally good σ2 = σ (Compressive)
(d) Both the shafts are not good τ = σ = 90 MPa
RPSC ACF & FRO, 26.02.2021 928. A solid circular shaft, under pure torsion,
Ans. (a) : Case- I – External diameter same develops maximum shear stress of 10 MPa on
In this case solid shaft will have more strength. the surface. If the shaft diameter is halved,
what will be the maximum shear stress
Case- II – Weight is same.
developed corresponding to the same torque?
In this case hollow shaft will have more
(a) 05 MPa (b) 40 MPa
strength
(c) 20 MPa (d) 80 MPa
Case- III – Cross-sectional area same.
RPSC Lect. (Tech. Edu. Dept.) 13.03.2021, Paper-I
In this case solid shaft transmit more torque.
Ans. (d) : As we know,
In question it is given that weight is identical or same so
Hollow shaft (B) will have more strength, so shaft B is 16T
Maximum shear stress = 3
better than A. πd
926. A solid circular shaft of diameter 100 mm is 16T
10 = 3 = τ1
subjected to an axial stress to 50 MPa. It is πd
further subjected to a torque of 10 kNm. The Max. shear stress, when diameter is halved.
maximum principal stress experienced on the 16T
shaft is closest to τ2 = 3
(a) 41 MPa (b) 82 MPa d
π 
(c) 164 MPa (d) 204 MPa 2
VIZAG STEEL MT 18.08.2013  16T 
τ2 = 8  3 
Vizag Steel MT Mechanical, 2013 Re-Exam  πd 
VIZAG STEEL MT 10.06.2012 τ2 = 8 × τ1
Ans. (b) : Shear stress (τ) =
16T τ2 = 80 MPa
πd 3 929. While transmitting the same power by a shaft,
16 × 10000 if its speed is reduced by half, what should be
τ= = 50.93 MPa its new diameter if the maximum shear stress
π× ( 0.1)
3
induced in the shaft remains same?
Strength of Materials 326 YCT
 (a) (2)1/2 of the original diameter Ans. (c) : Given-
(b) (1/2)1/2 of the original diameter We know,
(c) twice of the original diameter 16T
(d) (2)1/3 of the original diameter Maximum shear stress = = 320MPa
πd 3
RPSC Lect. (Tech. Edu. Dept.) 13.03.2021, Paper-I So, Max. shear stress developed when diameter is
Ans. (d) : Given- doubled.
Power = Torque (T) × Angular speed (ω) 16T 1
If P is const. τnew = = × 320
π ( 2d ) 8
3
1
T∝ τnew = 320 / 8 = 40MPa
ω
T' ω ω 932. A stepped steel shaft shown below is subjected
If = = =2 to 10 N-m torque. If the modulus of rigidity is
T ω'  ω  80 GPa, the strain energy in the shaft in N-mm
 
2 is :
16T
Shear Stress, τ = 3
πd
T ∝ d3
3
T '  d' 
= 
T d
3
 d'  (a) 4.12 (b) 3.46
2=  (c) 1.73 (d) 0.86
d
CGPSC AE 15.01.2021
d'
= (2)1/ 3 Ans. (c) : Strain Energy in the shaft
d UTotal = UAB + UBC
930. The ratio of Shear stresses of solid shaft
(diameter'd') to hollow shaft (internal diameter
'd'
, external diameter 'd') When subjected to
3
same twisting moment
81 80
(a) (b) T2 L T2 L
80 81 = AB + BC
26 27 2GJ 2GJ
(c) (d)
27 26 (10 × 103 ) 2 × 100 (10 × 103 )2 ×100
RPSC Lect. (Tech. Edu. Dept.) 13.03.2021, Paper-I = +
π  π 
Ans. (b) : d = solid shaft dia do = d = hollow shaft 2 × 80 × 103 ×  (50)4  2 × 80 × 103 ×  (25) 4 
 32   32 
external dia = 1.73 N-mm
d
di = = internal dia. 933. A solid circular shaft is subjected to bending
3 moment of 3000N and a torsion of 10,000N-m.
16T The equivalent bending moment would be
τs = 3 − − − − − − − (i) ________.
πd
(a) 6600 N-m (b) 5920 N-m
16Td
τH = − − − − − − − − (ii) (c) 10440 N-m (d) 6720 N-m
 4  d 4  VIZAG Steel MT 24.01.2021
π d −   
   
3 Ans. (d) : Bending moment (M) = 3000 N
By comparing both eq. Torsional moment (T) = 10,000 N
τs 80 Equivalent bending moment,
= 1
τH 81 (Me) =  M + M 2 + T 2 
2
931. Maximum shear stress developed on the
1
surface of a solid circular shaft under pure = 3000 + (3000)2 + (10,000) 2 
torsion is 320 MPa. If the shaft diameter is 2
doubled, then what is the maximum shear 1
stress developed corresponding to the same = [3000 + 9000000 + 100000000 ]
torque? 2
1
(a) 60 MPa (b) 120 MPa = [3000 + 10440.3065]
(c) 40 MPa (d) 20 MPa 2
RPSC Lect. (Tech. Edu. Dept.) 13.03.2021, Paper-I = 6720.153 N-m

Strength of Materials 327 YCT

934. The torsional rigidity of a shaft is the torque 937. The efficiency in resisting torsional moments
required to _________. by hollow shaft when compared to solid shaft is
(a) produce maximum shear stress (a) less (b) more
(b) produce unit twist at unit length (c) equal (d) cannot be determined
(c) produce maximum twist GPSC DEE, Class-2 (GWSSB) 04.07.2021
(d) produce zero twist Ans. (b) : The efficiency in resisting torsional moments
VIZAG Steel MT 24.01.2021 by hollow shaft when compared to solid shaft is more.
Ans. (b) : According to torsional equation– 938. Two shafts will have equal strength, if
T Gθ τ (a) diameter of both the shafts is same
= = (b) angle of twist of both the shafts is same
J ℓ R
Gθ T (c) material of both the shafts is same
= (d) twisting moment of both the shafts is same
ℓ J GPSC DEE, Class-2 (GWSSB) 04.07.2021
For, θ = 1 radian, length (ℓ) = 1 Ans. (d) : Two shafts will have equal strength, if
T twisting moment of both the shafts is same.
G=
J 939. In case of a bar under torsion if the force is
applied at the outer surface of the bar and
G∝T
longitudinally, the shear stress is maximum at
935. For a hollow circular shaft under torsion T (a) the centre of the bar.
only the shear stress, τ will be expressed by : (b) just before the outer surface.
16T 32T (c) the outer surface.
(a) (b) (d) the mid-point of radius of the bar.
π(d o + d i )
3 3
π(d o + d i )
3 3

PPSC Asstt. Municipal Engineer 15.06.2021

16 M 2 + T 2 16Td 0 Ans. (c) : The shear stress varries linearly zero at the
(c) (d)
πd o3
π(d 4o − d i4 ) centre and maximum at outer surface.
Where do, di are outer and inner diameters. 940. A solid circular shaft is subjected to a bending
JPSC AE 10.04.2021, Paper-II moment of 6000 Nm and a torque of 8000 Nm.
The equivalent bending moment of the shaft is
Ans. (d) : Consider, _______.
Outside dia = do (a) 14000 Nm (b) 10000 Nm
Inside dia = di (c) 6000 Nm (d) 8000 Nm
Torque = T VIZAG Steel MT 24.01.2021, Shift-I
Ans. (d) : Given,
Bending moment, M = 6000 N-m
Torque, T = 8000 N-m
∴ Equivalent bending moment,
then shear stress (τ) on hollow shaft.
1
T Gθ τ
= = Me =  M + M 2 + T 2 
J ℓ r 2 
1
Tr 6000 + ( 6000 ) + ( 8000 ) 
2 2
τ= =
2 
 
J
1
=
T(d 0 / 2) Me = [ 6000 + 10000]
π 4 2
(d 0 − d i4 ) 16000
32 = = 8000 N − m
16Td 0 2
τ= 941. The polar moment of inertia of a solid circular
π(d 04 − d i4 )
shaft of diameter d is ______.
936. A circular shaft when subjected to equal πd 3 πd 4
opposite end couples whose axes coincide with (a) (b)
axis of the shaft is said to be in 32 32
(a) pure torsion πd 4
πd 3
(b) pure bending (c) (d)
64 64
(c) combined bending and torsion VIZAG Steel MT 24.01.2021, Shift-I
(d) none of the above Ans. (b) : For solid circular shaft–
GPSC DEE, Class-2 (GWSSB) 04.07.2021 diameter = d
Ans. (a) : A circular shaft when subjected to equal
opposite end couples whose axes coincide with axis of I = πd
4

xx
the shaft is said to be in pure torsion. 64
Strength of Materials 328 YCT
 πd 4 a4 64
Iyy = = ×
64 12 π × 16r 4
polar moment of Inertia, Izz = Ixx + Iyy (
put a = r π )
πd 4 πd 4
= +
( r π ) × 64
4
64 64
=
πd 4 12π × 16r 4
Izz =
32 π 2 × r 4 × 64
=
942. Units of section modulus is 12 × π × 16r 4
(a) mm (b) mm4 I1 π
3 =
(c) mm (d) mm2 I2 3
GPSC DEE, Class-2 (GWSSB) 04.07.2021 944. Two shafts of equal length and similar material
Ans. (c) : mm3 in which one is hollow and other is solid are
943. The cross-sections of two solid bars made of the transmitting same level of torque. If the inside
same material are shown in the figure. The 2
square cross-section has flexural (bending) diameter is of the outside diameter of the
3
rigidity I1, while the circular cross section has hollow shaft, the ratio of weight of hollow shaft
flexural rigidity I2, Both sections have the same to weight of solid shaft is
I1 (a) 0.642 (b) 0.358
cross-sectional area. The ratio is
I2 (c) 0.732 (d) 1.444
UPPSC AE 13.12.2020, Paper-I
Ans. (a) : TH = TS
(τallowable × ZP)H = (τallowable × ZP)S
π 3 π
1 2 d o (1 − k 4 ) = d 3
(a) (b) 16 16
π π 3
π π  do  1 1
(c) (d)   = =
 d  1− k 4 4
1 −  
3 6 2
Assam Engg. College AP/Lect. 18.01.2021 3
π do
Ans. (c) : = 1.076
3 d
Weight = ρg (Volume)
= ρg (A × ℓ)
π 2
WH A H 4 o (1 − k )
d 2
Area of cross section is same. = =
A1 = A2 WS AS π 2
d
a2 = πr2 4
a=r π d 
2
=  o  (1 − k 2 )
Flexural rigidity of square I1 = E1I m1  d 
Flexural rigidity of circle I 2 = E 2 I m2  2

= (1.076 )2 1 −  2  
Where, E1 = E2 = E = Young's modulus (same material)  3 
I m1 & I m2 are moment of inertia of respective figures. = 0.643
I1 EI m1 945. What is the maximum torque transmitted by a
= hollow shaft of external radius 'R', internal
I 2 EI m2 radius 'r' and maximum allowable shear
a 4 stress τ?
I1 a 4
64 π 3 3 π
= 124 = × 4 (d = 2r) (a) (R − r )τ (b) (R 4 − r4 ) τ
I 2 πd 12 πd 16 2R
π π
64 (c) (R 4 − r4 ) τ (d) (R 4 − r4 ) τ
I1 a 4
64 8R 32
= × UPPSC AE 13.12.2020, Paper-I
I 2 12 π ( 2r )4
ESE 2006
Strength of Materials 329 YCT
 π  ( 2R )4 − ( 2r )4  949. A pump is used to transport the water. The
Ans. (b) : Torque ( T ) = τ × maximum power required through the shaft is
16  2R

 62.8 W, when it rotates at 100 rpm. What will
π  R4 − r4  be the diameter of shaft (in cm) if the
T = τ× × 16 
 2R 
16 maximum permissible shearing stress is 100
MN/m2? [π = 3.14]
π
T=τ [R 4 − r4 ]  96 
1/ 3 1/ 3
 196 
2R (a)  (b) 
 
946. A shaft is subjected to a bending moment M =  314   314 
0.75 kNm and a twisting moment T = 1 kNm. 1/ 3 1/ 3
The magnitude of equivalent bending moment  296   396 
(c)   (d)  
in shaft is  314   314 
(a) 1.25 kNm (b) 1.125 kNm APPSC Poly Lect. 13.03.2020
(c) 1.0 kNm (d) 0.75 kNm Ans. (a) : Given data : Power (P) = 62.8 Watt
UPPSC AE 13.12.2020, Paper-I Rotational speed (N) = 100 rpm
Ans. (c) : Given, M = 0.75 kNm shear stress (τ)max = 100 MN/m2
T = 1 kNm π = 3.14
1 Let,
Equivalent bending moment =  M + M 2 + T 2 
2 d = dia of the shaft
1 2πNT
=  0.75 + ( 0.75 ) 2 + 12  ∵ P = Tω =
2 60
= 1 kNm 60P
947. A solid shaft of diameter 'D' carries a twisting ⇒ T=
2πN
moment that develop maximum shear stress τ.
16T πd 3
If this shaft is replaced by a hollow one of τ max = 3 ⇒ T = × τmax
outside diameter D and inside diameter D/2, πd 16
then the maximum shear stress will be π 60P
(a) 1.067 τ (b) 1.145 τ ∵ d 3τ =
16 2πN
(c) 1.335 τ (d) 2 τ 60P × 16 60 × 62.8 × 16
ISRO Scientist/Engineer (RAC) 12.01.2020 ⇒ d3 = =
πτ × 2πN 3.14 × 100 × 106 × 2 × 3.14 ×100
D 1 96 × 2
Ans. (a) : k = i = ⇒ d3 =
Do 2
2 × 100 × 3.14 × 106
16T 96
For solid shaft, τs = ⇒ d3 = × 10−6 m
πD 3 314
16T 1/ 3
For hollow shaft, τ H =  96 
πD3 [1 − k 4 ] ⇒ d3 =   × 10−2 m
 314 
16T
τH =  96 
1/ 3
  1 4  ⇒d =  cm
πD3 1 −     314 
 2 
16T 16 950. Torsional-sectional-modulus of a shaft is also
τH = × called as:
πD3 15 (a) Sectional modulus (b) Polar modulus
τH = 1.066 τ (c) Torsion modulus (d) Torsional rigidity
948. Two solid shafts 'A' and 'B' are made of the CIL MT 27.02.2020
same material. The shaft 'A' is 50 mm diameter
and shaft 'B' is 100 mm diameter. The ratio of Ans. (b) : Polar modulus is defined as the ratio of the
strength of shaft 'B' compared to shaft 'A' is polar moment of inertia to the radius of shaft. It is also
(a) 1/2 (b) 2 called as torsional section modulus (ZP).
(c) 4 (d) 8 π
IP = D 4
ISRO Scientist/Engineer (RAC) 12.01.2020, ESE 2016 32
Ans. (d) : dA = 50 mm π
dB = 100 mm ZP = D3
16
Strength of shaft is proportional to the polar section 951. A solid circular shaft needs to be designed to
modulus. transmit a torque of 55 N m. If the allowable
Z PB  d B 3 shear stress of the material is 280 N/mm2, find
=  =8 the diameter (in mm) of the shaft required to
Z PA  d A 
transmit torque. (Assume, π = 22/7)
Strength of Materials 330 YCT
 (a) 5.62 (b) 10 Ans. (a) : Given,
(c) 31.62 (d) 25.0 Allowable shear stress for shaft = 60 MPa
ISRO Scientist/Engineer 12.01.2020 Transmitted torque,
Ans. (b) : For circular solid shaft, T = 5000 N-m
16T = 5000 × 103 N-mm
τ=
πD 3 = 5 × 106 N-mm
16T 16 × 55 × 103 16T
D3 = = τshaft = 3
πτ per 22 πd
× 280 16 × 5 × 106
7 60 =
3
D = 1000, D = 10 mm πd3shaft
952. A solid bar of circular cross-section having a 16 × 5 × 106
diameter of 40 mm and length of 1.3 m is d 3shaft =
subjected to torque of 340 Nm. If the shear π× 60
modulus of elasticity is 80 GPa, the angle of dshaft = 75.15 mm
twist between the ends will be For pin, allowable shear stress = 28 MPa
(a) 1.26° (b) 1.32°
T = (2F) ×  
d
(c) 1.38° (d) 1.44°
ESE 2020 2
Ans. (a) : Angle of twist By convention, 2 Pin/key are used and both pin/key are
under double shear.
Tℓ
θ=
T = 2 × (2F) ×  
d
GJ So,
2
340 × 103 × 1300
= 75.15 × 10−3
π 5000 = 2 × (2 × F) ×
80 × 10 × × 40
3 4
2
32
= 0.02198 rad, θ = 1.26º F = 33.33 × 10 3
N
So, same force is applied in both shaft and pin/key
953. Two shafts, one hollow and the other solid, are
of same material, weight and length. If the π
F = τpin × Apin = τpin × d p2
inside diameter of the hollow shaft is half of its 4
outer diameter, the strength of the hollow shaft π
to that of solid shaft will be about 33.33 ×103 = 28 ×106 × d 2p
4
(a) 1.067 (b) 1.33
dp = 38.89 mm
(c) 1.44 (d) 1.67
TSPSC Manager (Engg.) HMWSSB 12.11.2020 dp ≈ 40 mm
Ans. (c) : When the hollow and solid shafts are of equal 955. A solid shaft is used to transmit a power of 120
weights. In this case torsional strength is compared. π kW at 120 rpm. The torque transmitted by
the shaft is
Thollow 1 + n2 (a) 30 kNm (b) 60 kNm
=
Tsolid 1 – n2 (c) 90 kNm (d) 120 kNm
Gujarat PSC AE 2019
D
Given as Di = o Ans : (a) : Given-
2 P = 120 πkW
Di N = 120 rpm
n=
Do T2πN
P = Tω =
1 60
= = 0.5 T2 π × 120
2 120π =
Thollow 1 + 0.52 60
= = 1.44 T = 30 kNm
Tsolid 1 – 0.52 956. Hollow shafts are stronger than solid shafts
954. An universal coupling is used to connect two having same weight because
mild steel shafts transmitting a torque of 5000 (a) the stiffness of hollow shaft is less than that
N-m. Assuming that the shafts are subjected to of solid shaft
torsion only, find the diameter of the shafts and (b) the strength of hollow shaft is more than that
pins. The allowable shear stresses for the shaft of solid shaft
and pin may be taken as 60 MPa and 28 MPa (c) the natural frequency of hollow shaft is less
respectively. than that of solid shaft
(a) 75 mm, 40 mm (b) 95 mm, 50 mm (d) in hollow shafts, material is not spread at
(c) 40 mm, 85 mm (d) 75 mm, 60 mm large radius
RPSC IOF, 2020 ESE 2019
Strength of Materials 331 YCT
Ans. (b) : Strength of hollow shaft is more than that of 2π NT
solid shaft because polar section modulus of hollow Ans. (b) : P =
shaft is greater than that of solid shaft. 60
2π (200)(1800)
957. A propeller shaft is required to transmit 45 kW =
power at 500 r.p.m. It is a hollow shaft having 60
inside diameter 0.6 times the outside diameter. = 12000 (π)
It is made of plain carbon steel and the = (12 π) kW
permissible shear stress is 84 N/mm2. The inner 960. A solid circular shaft of length 4 m is to
and outer diameters of the shaft are nearly. transmit 3.5 MW at 200 rpm. If permissible shear
(a) 21.7 mm and 39.1 mm stress is 50 MPa, the radius of the shaft is:
(b) 23.5 mm and 39.1 mm (a) 1.286 mm (b) 12.86 mm
(c) 21.7 mm and 32.2 mm (c) 0.1286 mm (d) 128.6 mm
(d) 23.5 mm and 32.2 mm BHEL ET 2019
ESE 2019 Ans. (d) : Given -
Ans. (b) : Given, Length = L = 4m
P = 45 kW P = 3.5 MW
N = 500 rpm N = 200 rpm
di = 0.6 do τ = 50 MPa, radius = r = ?
d τ = 50 × 103 kPa
k = i = 0.6 Power P = Tω
do
τmax = 84 N/mm 2 2πN
3.5 × 106 = T ×
2πNT 60
P= 2 π × 200
60 = T×
2π× 500 × T 60
45 × 103 = 3.5 × 106 = 20.943 T
60 T = 167.120 kN-m
T = 859.87 N-m = 859.43 × 103 N-mm Torsion equation-
For hollow shaft
T T
16T =
τmax = J r
 d   4

πd 3o 1 −  i   τJ
r=
  d o   T
16 × 859.43 × 103 πd 4
84 = 50 ×
πd3o [1 − 0.64 ] r= 32
3 167.120
d o = 59896.95
d 50 × πd 4
do = 39.12 mm =
di = 23.47 mm 2 5347.84
958. In case of a torsional problem the assumption- 5347.84
= d3
"Plane sections perpendicular to longitudinal 50 × 2 × π
axis before deformation remain plane and d = 0.25724
perpendicular to the longitudinal axis after 2r = 0.25724
deformation" holds true for a shaft having r = 0.12862 m
(a) circular cross-section
r = 128.6 mm
(b) elliptical cross-section
(c) circular and elliptical cross-section 961. The maximum shear stress developed on the
(d) any cross-section surface of a solid circular shaft under pure
APPSC AEE SCREENING 17.02.2019 torsion is 160 MPa. If the shaft diameter is
Ans. (a) : In case of a torsional problem the doubled, then the maximum shear stress
assumption- developed corresponding to the same torque
"Plane sections perpendicular to longitudinal axis will be:
(a) 10 MPa (b) 30 MPa
before deformation remain plane and perpendicular to
(c) 40 MPa (d) 20 MPa
the longitudinal axis after deformation" holds true for a
shaft having circular cross-section only. It is not valid BHEL ET 2019
for other shape of cross-section. Ans. (d) : Given
959. A shaft turns at 200 rpm under a torque of ( τ max ) = 160MPa
1800 Nm. The power transmitted is d1 = d
(a) 6π kW (b) 12π kW d2 = 2d
(c) 24π kW (d) 36π kW
same torque T1 = T2 = T
APPSC AEE SCREENING 17.02.2019
Strength of Materials 332 YCT
 16T 16T π 1
( τ max ) 1 = ⇒ 160 = 3 ...(1) θ = 0.25 × ×
πd 3 πd 180 1000
when diameter is doubled. θ = 4.363 × 10−6 rad/mm
16T 16T 60P
( τmax )2 = = ...(2) T=
2πN
π(2d)3 8πd 3
from equation (1) / equation (2) 60 × 4 × 103
= = 47.77 N-m
160 16T 8πd 3 2π× 800
= 3× = 47.77 × 10 N-mm
3
( τ max )2 πd 16T
θ T
( τ max )2 = 20 MPa =
L GJ
962. A 50 mm diameter solid shaft is subjected to 4.363 × 10−6 47.77 × 103
both, a bending moment and torque of 300 kN- =
1 84 × 103 × J
mm & 200 kN-mm respectively. The maximum J = 130343.90 mm4
shear stress induced in the shaft is : π 4
(a) 10.22 MPa (b) 14.69 MPa d = 130343.90
(c) 146.9 MPa (d) 102.2 MPa 32
BHEL ET 2019 d = 33.949 mm
d = 34 mm
Ans. (b) : Given diameter of shaft d = 50 mm
Bending moment M = 300 kN-m 965. In a propeller shaft, sometimes apart from
Torque T = 200 kN-m bending and twisting, end thrust will also
develop stresses which would be
Te = M 2 + T 2 be the equivalent torque, which acts (a) Tensile in nature and uniform over the cross-
alone producing the same maximum shearing stress section
T 16 (b) Compressive in nature and uniform over the
τ max = e 3 = 3 Te cross-section
πd πd (c) Tensile in nature and non-uniform over the
16 cross-section
16 ×10 3 (d) Compressive in nature and non-uniform over
( 300 ) + ( 200 )
2 2
τ max = the cross-section
π× ( 50 )
3
ESE 2019
16 × 360.555 × 10 3 Ans. (b) : Due to end thrust, stresses would be
=
π× 125000 compressive in nature and uniform over cross section.
5768880 966. If a shaft is required to transmit twice the
= power at twice the speed for which it is
392699.081
= 14.69 MPa designed, its diameter must
(a) increase two times (b) reduce two times
963. Which of the following is correct for flexible
shaft? (c) remain same (d) increase three times
(a) it has very low rigidity both in bending and ISRO Scientist/Engineer (RAC) 22.04.2018
torsion Ans. (c) : We know that power transmitted by shaft
(b) it has very high rigidity in bending and low P = T.ω
rigidity in torsion
π D3
(c) it has low rigidity in bending and high rigidity T= τ
in torsion 16
(d) It has very high rigidity both in bending and π × D3 ×τ P
torsion =
SJVN ET 2019 16 2π N
Ans. (c) : flexible shaft has low rigidity in bending and If P' = 2P, N' = 2N
high rigidity in torsion. then D' = ?
964. A steel spindle transmits 4 kW at 800 r.p.m. π × D '3 × τ 2P P π × D3 ×τ
= = =
The angular deflection should not exceed 16 2 × 2π × N 2π N 16
0.25°/m length of the spindle. If the modulus of D' = D
rigidity for the material of the spindle is 84
Diameter must be same.
GPa, the diameter of the spindle will be
(a) 46 mm (b) 42 mm 967. The power transmitted by a shaft 60 mm
(c) 38 mm (d) 34 mm diameter at 180 RPM, if the permissible stress
ESE 2019 is 85 N/mm2
Ans. (d) : Given, P = 4 kW (a) 68 kW (b) 650 kW
N = 800 rpm (c) 1200 kW (d) 7 kW
G = 84 GPa ISRO Scientist/Engineer (RAC) 22.04.2018
Strength of Materials 333 YCT
Ans. (a) : d = 60 mm, N = 180 RPM, τ = 85 MPa Ans. (b) : Torsion bars are in parallel if they have equal
We know that angles of twist and applied torque apportioned between
π d3 2 ×π × N them.
P = T ×ω = ×τ × T = T1 + T2
16 60 θ = θ1 = θ2
π × 60 × 85 2 × π × 180
3
971. Shearing stress produced on the surface of a
P= ×
16 60 solid shaft of diameter (d0) is τ. The shear stress
P = 67952226.3 N-mm/sec produced on the surface of hollow shaft of same
P = 67.9 kW material, subjected to same torque, and having
968. Torque, T is applied at the free end of a the outer diameter d0 and internal diameter d0x
stepped rod of circular cross section as shown is given as: [Where x < 1]
in figure. If the shear modulus of the material τ
is G, the angular twist, θ at free end will be (a)
1− x 4
( )
(b) 1 − x 4 τ

τ
(
(c) 1 − x 2 τ ) (d)
1− x 2
τ
12TL 24TL (e)
(a) (b) 1 − 2x 4
Gπ d 4 Gπ d 4
CGPSC AE 25.02.2018
36TL 48TL
(c) (d) Ans. (a) : Shear stress in solid shaft
Gπ d 4 Gπ d 4
d
ISRO Scientist/Engineer 22.04.2018 T× 0
Ans. (c): We know that, When Shafts are connected in τs = 2 =τ
π
series × d 04
θ = θ1 + θ2 32
Shear stress in hollow shaft is
 TL   TL  T × ( d o / 2)
=  + 
 GJ 1  GJ  2 τΗ =
π 4
[ d i = xd o ]
4
T × L × 32 T × 2 L × 32 d 1 − x
32  
0
= +
G × π× d 4
Gπ× 2 × d
4 4
T × (d 0 / 2) 1
T × L × 32  2 τΗ = ×
= + 1 +
Gπ × d 4  2 4 
π
32
d 04( ) ( )
1− x 4
36TL
θ= τ
Gπd 4 τH =
969. The angle of twist of shaft is 1− x 4
(a) directly proportional to (shaft diameter)2 972. A solid shaft is subjected to bending moment of
(b) inversely proportional to (shaft diameter)2 3.46kN-m and a torsional moment of 11.5kN-
(c) directly proportional to (shaft diameter)4 m. For this case, the equivalent bending
(d) inversely proportional to (shaft diameter)4 moment and twisting moment are
Nagaland PSC (CTSE) 2018, Paper-I (a) 7.73kN-m and 12.0kN-m
Ans. (d) : As we know, (b) 14.96kN-m and 12.0kN-m
TL (c) 7.73kN-m and 8.04kN-m
Angle of twist, φ = (d) 14.96kN-m and 8.04kN-m
JG
ESE 2018
πD 4
J= Ans. (a) : Given,
32 M = 3.46 kN-m
1 T = 11.5 kN-m
φ∝
J Equivalent bending moment is given by
1 1
⇒ φ∝ 4 Me =  M + M 2 + T 2 
D 2
So, φ is inversely proportional to (shaft diameter)4.(d4)
= 3.46 + ( 3.46 ) + (11.5) 
1 2 2
970. Torsion bars are in parallel 2
(a) if same torque acts on each = 7.734 kN-m
(b) if they have equal angles of twist and applied Equivalent twisting moment is given by
torque apportioned between them
(c) are not possible Te = M 2 + T 2
(d) if their ends are connected together
= ( 3.46 ) + (11.5 ) = 12 kN-m
2 2
TNPSC AE 2018
Strength of Materials 334 YCT
973. A solid shaft can resist a bending moment of 6 From torsion equation,
kN m and a torque of 8 kN m applied together. 16T
The maximum torque that the shaft can resist =τ
πd3
when applied alone is
16 375 × 103
(a) 7 kN m (b) 48 kN m × = 100
(c) 10 kN m (d) 14 kN m πd 3 6π
ISRO Scientist/Engineer 22.04.2018 16 × 375 × 103
d3 = 2
Ans. (c): T = 8kN-m π × 6 × 100
M = 6kN-m d3 = 1013.21
Maximum torque means equivalent torque Teq. d = 10.04 mm
976. A torque T is applied at the free end of a
Teq = M 2 + T 2
stepped rod of circular cross-sections as shown
= 82 + 6 2 in the figure. The shear modulus of the
material of the rod is G. The expression for
= 100 diameter 'd' to produce an angular twist θ at
Teq = 10 kN-m the free end is
974. The diameter of shaft in any power
transmission system is proportional to:
(a) speed of the shaft
(b) horse power to be transmitted
(c) torque to be transmitted
1/ 4 1/ 4
(d) allowable shear stress of the material  30TL   33TL 
(a)   (b)  
Kerala PSC AE (Irrigation) 05.04.2017  πθG   πθG 
Ans. (b) : The diameter of shaft in any power 1/ 4 1/ 4
 18TL   24TL 
transmission system is proportional to horse power to be (c)   (d)  
2πN  πθG   πθG 
transmitted power P = T × ISRO Scientist/Engineer 07.05.2017
60
Ans. (b) :
16T τπd 3
and τ = 3 or T =
πd 16
τπd 3 2πN
P= ×
16 60
480P T Gθ τ
d3 = 2 = =
π JN J L R
1/ 3
P TL
d∝  θ=
N GJ
So, diameter is directly proportional to power. T = T 1 = T2

975. A 3 m long steel shaft has to transmit 7.5 kW at θ = θ1 + θ2

3600 rpm. Required shaft diameter is given by L
(Take allowable shear stress = 100 N/mm2, π2 = T×
T1L1 T2 L 2 TL 2
= + = +
10)? G1J1 G 2 J 2 πd 4 G × π (16d 4 )
(a) 10 mm (b) 7.5 mm G
32 32
(c) 25 mm (d) 12.5 mm
ISRO Scientist/Engineer 17.12.2017 TL  1 32 
= 32 + × 
Ans. (a) : Given, Gπd 4  2 16 
L = 3 m, P = 7.5 kW = 7500 W TL
θ= [33]
N = 3600 rpm τ = 100 N/mm2 G π d 4

π2 = 10 TL
We know that d4 = [33]
Gπθ
2πNT
P= 1
60  TL ( )  4
d= 33 
2π× 3600 × T  Gπθ 
7500 =
60 977. The torque (T) transmitted by sleeve coupling
375 375 × 103 with D and d as outer and inner diameters,
T= N−m = N − mm
6π 6π respectively, is given by:

Strength of Materials 335 YCT

 π  D4 − d 4  Ans. (a) : d = 60 mm
(a) T = × τ  N = 150 rpm
8  D 
τ = 80 MPa = 80 N/mm2
π  D4 − d 4  We know that
(b) T = × τ 
12  D  πd3 π× 60 × 60 × 60
T= ×τ= × 80 N − mm
π  D4 − d 4  16 16
(c) T = × τ  T = 1.08π kN-m ≃ 1.1π kN-m
16  D 
980. A solid shaft is to transmit 20 kW at 200 r.p.m.
π  D4 − d 4 
(d) T = × τ   The ultimate shear stress for the shaft material
32  D  is 360 MPa and the factor of safety is 8. The
π  D4 − d4  diameter of the solid shaft shall be
(e) T = × τ   (a) 42 mm (b) 45 mm
64  D  (c) 48 mm (d) 51 mm
(CGPCS Polytechnic Lecturer 2017)
ESE 2017
Ans. (c) : Torque transmitted by sleeve coupling
Ans. (c) : Given,
π  D4 − d 4  P = 20 kW = 20 × 103 W
T=  ×τ
16  D  N = 2000 rpm
τu = 360 MPa
60P 60 × 20 3
T= = = kN − m
2πN 2π× 200 π
3
16 ×
16T τu π = 360
τ= 3 = =
πd FS π× d3 8
978. The boring bar of a boring machine is 25 mm d = 47.6 mm
in diameter. During operation, the bar gets
twisted though 0.01 radians and is subjected to 981. Maximum shear stress in the shaft if subjected
a shear stress of 42 N/mm2. The length of the to combine twisting moment (Mt) and bending
bar is (Taking G = 0.84 × 105 N/mm2) moment (Mb)–
(a) 500 mm (b) 250 mm 16
(a) (M b )2 + (M t ) 2
(c) 625 mm (d) 375 mm πd 3
GPSC EE Pre, 28.01.2017 16
(b) (M b )2 − (M t )2
Ans. (b) : Given- πd 3
Diameter (D) = 25 mm 32
25 (c) (M b )2 + (M t ) 2
R= mm πd 3

2 32
Twist (θ) = 0.01 radian (d) (M b )2 − (M t )2
πd 3
τ = 42 N/mm2 Nagaland PSC CTSE 2017, Paper-I
G = 0.84 × 105 N/mm2
Ans. (a) : Maximum, shear stress in the shaft.
From the equation of torsion
T τmax Gθ 16
= = τmax = 3 (M b )2 + (M t ) 2
J R l πd
τmax Gθ 982. A hollow shaft of 20 mm diameter and 16 mm
= inside diameter is subjected to a torque of 40
R l
Nm. The shear stress at outside of the shaft will
Gθ× R
l= be:
τmax (a) 53.12 N/mm2 (b) 43.13 N/mm2
2
25 (c) 62.52 N/mm (d) 34.50 N/mm2
0.84 × 105 × 0.01× ISRO Scientist/Engineer (RAC) 07.05.2017
= 2
42 Ans. (b) : D = 20 mm
l = 250 mm d = 16 mm
979. A circular shaft of 60 mm diameter is running Torque T = 40 N-m
at 150 r.p.m. What will be the torque
transmitted by the shaft if (τ = 80 MPa)?
(a) 1.1 π kN-m (b) 1.6π kN-m
(c) 2.1π kN-m (d) 2.6π kN-m
(e) 3.1π kN-m
(CGPCS Polytechnic Lecturer 2017)
Strength of Materials 336 YCT
 Shear stress τ Also angle of twist at coupling is equal for both shafts
16T Ts ℓ Tb ℓ
= =
π D 3 (1 − K 4 ) Gs j G b J
where (ℓ, are equal for both shafts)
d 16
K= = Ts G s
D 20 = =2 –––– (2)
Tb G b
16 × 40 × 1000
τ= Tb =
500
= 250Nm
  16 4  2
π × (20)3 × 1 −   
  20   985. When subjected to a torque, a circular shaft
τ = 43.13 N/mm2 undergoes a twist of 1° in a length of 1200 mm,
and the maximum shear stress induced is 80
983. Two steel shafts, one solid of diameter D and
N/mm2. The modulus of rigidity of the
other hollow of outside diameter D and inside
diameter D/2 are twisted to the same angle of material of the shaft is 0.8 × 105 N/mm2. What
twist per unit length. Ratio of maximum shear is the radius of the shaft?
stress in solid shaft to that in the hollow shaft is (a) 90/πmm (b) 108/πmm
4 8 (c) 180/πmm (d) 216/πmm
(a) (b) TSPSC AEE 28.08.2017 (Civil/Mechanical)
9 7
16 Ans. (d) : τ = 80 N/mm2
(c) (d) 1 θ 1° 1 π
15 = = × (in radian)
TSPSC AEE 28.08.2017 (Civil/Mechanical) ℓ 1200 1200 180
CSE Pre-1998 G = 0.8 × 105 N/mm2
Ans. (d) : Torsion formula Torsion formula
τ Gθ τ Gθ
= =
rmax ℓ rmax ℓ
for solid shaft rmax = D/2 80 0.8 × 105 × 1 × π
for hollow shaft rmax = D/2 =
rmax 1200 × 180
Since both are made of steel
1200 × 180 216
∴ G remains same rmax = =
Given angle of twist per unit length is ame 105 π π
θ 986. Power is transmitted through a shaft, rotating
∴ for both shafts is same at 2.5 Hz (150 rpm). The mean torque on the
ℓ shaft is 20 × 103 Nm. What magnitude of power
δs ( r )
max s D/2 in kW is transmitted by the shaft?
= = =1
δh ( rmax ) h D / 2 (a) 50π (b) 120π
(c) 100π (d) 150π
984. Steel shaft and brass shaft of same length and
diameter are connected by a flange coupling. TSPSC AEE 28.08.2017 (Civil/Mechanical)
The assembly is rigidly held at its ends and is Ans. (c) : N = 150 rpm
3
twisted by a torque through the coupling. T = 20 × 10 Nm
Modulus of rigidity of steel is twice that of 2πNT
brass. If torque of the steel shaft is 500 Nm, P = kW
then the value of the torque in brass shaft will 60 × 1000
be 2π× 150 × 20 × 103
=
(a) 250 Nm (b) 354 Nm 60 × 1000
(c) 500 Nm (d) 708 Nm 100π× 103
TSPSC AEE 28.08.2017 (Civil/Mechanical) = = 100π kW
CSE Pre-2001 1000
987. A shaft of circular section is said to be in pure
Ans. (a) :
torsion when it is subjected to equal and
opposite end couples whose axes coincides with
the
(a) axis of the shaft
(b) normal line to the axis of the shaft
(c) any line passing through the section of the
GS = 2 GB shaft
TS = 500 NM (d) parallel line to the axis of the shaft
TS + Tb = T ––––– (1) APPSC AE Subordinate Service Civil/Mech. 2016
Strength of Materials 337 YCT
Ans. (a) : A shaft of circular section is said to be in pure 989. A hollow circular shaft of outside and inside
torsion when it is subjected to equal and opposite end diameter 100 mm and 90 mm is subjected to a
couples whose axis coincides with the axis of the shaft. torque T = 3π kNm. Polar moment of inertia of
circular section is π × 10-6 m4. Maximum shear
stress on the shaft is given by:
(a) 175 MPa (b) 100 MPa
(c) 150 MPa (d) 125 MPa
UPRVUNL AE 07.10.2016
988. A tubular shaft, having an inner diameter of 30 Ans. (c) : Data given,
do = 100 mm
mm and an outer diameter of 40 mm, is to be
di = 90 mm
used to transmit 80 kW of power. The speed of
T = 3π k N-m
rotation of the shaft so that the shear stress will
J = π × 10-6 m4
not exceed 50 MPa is
(a) 29.6 rpm (b) 3557.4 rpm T τ Gθ
= =
(c) 1778.7 rpm (d) 59.2 rpm J  o d l
2
ISRO Scientist/Engineer 03.07.2016  
Ans. (c) : Inner diameter di = 30 mm 3π × 103 × 0.05
τ=
Outer diameter do = 40 mm π ×10−6
P = 80 kW τ = 150 MPa
τ = 50 MPa 990. For shaft in torsion
P(kW) × 60 q T Cθ T
T= × 106 (N-mm) (a) = (b) =
2πN(rpm) r J ℓ J
80 × 60 q Cθ
T= × 10 6
(c) = (d) All the above
2π× N r ℓ
2400 × 106 24 × 108 APPSC AEE Screening Test 2016
T= = (N-mm) Ans. (d)
π× N πN
T Cθ q
T Gθ τ = =
= = J ℓ r
J L R Where,
C = Modulus of rigidity
T τ
= ℓ = Length
J R
T = Torque
For solid– q = Shear stress
16T
τmax = 3 991. The ratio of maximum shear stress of a solid
πd shaft of diameter D to that of a hollow shaft
16T having external diameter D and internal
τmax =
πd 3 (1 − k 4 ) diameter 0.5D is
(a) 0.50 (b) 1.00
d (c) 1.50 (d) 2.00
Where k = i
do APPSC AEE Screening Test 2016
16 × 24 × 108 Ans. (*) : Given,
Inner diameter = 0.5D
50 × 106 = πN
Outer diameter = D, solid shaft dia = D
πd 0 (1 − k )
3 4

d i 30 16T
k= =
d o 40 Ratio = τsolid = πD3
τ hollow 16T
k4 = (0.75)4   d 4 
16 × 24 × 10 8 πD3 1 −   
50 × 106 =   D  
π2 × (40)3 × 1 − (0.75) 4  × N   0.5D 4 
πD3 1 −   
16 × 24 × 10 2   D  
50 = =
π × 64000 × 0.68 × N
2
πD3
16 × 24 × 100 = [1–(0.5)4]
N=
50 × 64000 × 0.68 × π2 = (1–0.0625) = 0.9375 ≃ 1
N = 1778.7 rpm Note-Official answer given by the commission is (d).

Strength of Materials 338 YCT

992. Which of the following assumption in the 995. The torque transmitted by a solid shaft of
theory of pure torsion is false diameter 40 mm if the shear stress is not to
(a) All radii get twisted due to torsion exceed 400 N/cm2, would be :
(b) the twist is uniform along the length (a) 1.6 × π N-m (b) 16π N-m
(c) the shaft is uniform circular section through (c) 0.8 × π N-m (d) 0.4 × π N-m
out HPPSC W.S. Poly. 2016
(d) cross section plane before torsion remain
plane after torsion Ans : (b) Shaft diameter = 40mm
APPSC AEE MAINS 2016, PAPER-III Shear stress (τmax) = 400N/cm2
T Gθ τ
Ans. (a) : Assumption in the theory of Torsion– = =
1. The material of shaft is uniform throughout the J ℓ r
length. τ.J
2. The twist along the shaft is uniform. T=
r
3. The shaft is of uniform circular section throughout
the length. πd3 τ
T=
4. Cross-section of the shaft, which are plane before 16
twist remain plane after twist. π × 403 × 400
5. All radii which are straight before twist remain T=
straight after twist. 16 × 1000 × 100
T = 16π N-m
993. In a rectangular shaft is subjected to torsion,
the maximum shear stress will occur 996. A shaft of 20 mm diameter and length 1 m is
(a) Along the diagonal subjected to a twisting moment, due to which
(b) At the corners shear strain on the surface of the shaft is 0.001.
(c) At the centre The angular twist in the shaft is
(d) At the middle of the longer side (a) 0.1 radian (b) 0.01 radian
Nagaland PSC CTSE 2016 Paper-I (c) 0.05 radian (d) 0.5 radian
Ans. (d) : For a rectangular shaft is subjected to torsion UPPSC AE 12.04.2016 Paper-I
the maximum shear stress will occur at the middle of Ans : (a)
the longer side.
994. Three shafts (spring constant k1, k2, k3) are
connected in series such that they carries the
same torque (T), then spring constant (k) for
composite shaft will be
(a) k = k1 + k2 + k3 ℓ = Shaft length = 1000 mm
(b) k = (k1k2 + k2 k3 + k3 k1 )1/ 2 θ = twist angle
1 1 1 1 φ = shear strain = 0.001
(c) = + + r = shaft radius
k k 1 k2 k3
1/ 2
 k1k2 k3  AA'
(d) k =   φ= .............( i )
 k1 + k2 + k3  ℓ
RPSC LECTURER 16.01.2016
Ans. (c) : In series, same torque acting on each three AA'
θ= .............( ii )
shafts r
T1 = T2 = T3 = T for equation (i) and (ii)
θ = θ1 + θ2 + θ3 ...(1) φℓ = θ r
GJ
θ= φℓ
TL θ=
Equation No. (1) divided by T r
θ θ1 θ 2 θ3 0.001 × 1000
= + + θ=
20
T T T T
2
1 1
1 1
= + + θ = 0.1 radian
T   T   T   T 
       
 θ   θ1   θ 2   θ 3  997. A shaft of J polar moment of Inertia and C
modulus of rigidity is fixed at one end and
T GJ subjected to torque T at the free end and the
=K=
θ L same torque at mid length in opposite direction
1 1 1 1 as shown in figure, then the difference in the
= + + twist between the free end and the midpoint is
k k1 k2 k3 equal to
Strength of Materials 339 YCT
 τ σ
corresponds to
r y
Gθ E
corresponds to
l R
T 3T 1000. The unit of torsional stress in SI units is :
(a) (b)
2CJ 2CJ (a) Kg/m2 (b) Kg-m
T T (c) N/m2 (d) N-m
(c) (d) HPPSC Lect. (Auto) 23.04.2016
CJ 4 CJ
APPSC AEE Mains 2016 (Civil Mechanical) Ans. (c) : The unit of torsional stress in SI units is N/m2
torsional shear stress
Ans. (a) : Beam divide this Shaft into two parts AB &
BC 16T
τ= 3 T = torque transmitted
πd
d = diameter of shaft
1001. The torsional stiffness of individual sections of
lengths AB, BC and CD are 40 N-m/rad, 60 N-
m/rad and 120 N-m/rad respectively.

θC − θΒ is asked θΒ = 0 ∴ [ TB = TC ] What is the angular deflection between the

& No torque at AB portion. ends A and D of the shaft?
ΤL (a) 0.5 rad (b) 0.1 rad
∴⇒ θ = (c) 1 rad (d) 5 rad
CJ
(e) 10 rad
L
Τ× CGPSC Asstt. Workshop Supt., 17.07.2016
θC − θΒ = 2 = TL
Ans. (c) : As we know,
CJ 2CJ
998. Polar moment of inertia of a hollow tube of
diameter 'd' and thickness 't' is
πd 3 t πd 3 t
(a) (b) T
32 16 = Torsional stiffness
θ
πd 3 t πd 4 1× 20 1× 20
(c) (d) θAB = , θBC =
4 16 40 60
RPSC ADE 2016
1× 20
Ans. (c) : θCD =
3
120
πd t 1× 20 1× 20 1× 20
IP = So, θAD = θAB + θBC + θCD = + +
4 40 60 120
d = diameter
1 1 1
t = thickness = + + = 1rad
999. The following torsional terms is analogous to 2 3 6
bending terms except 1002. A stepped steel shaft shown in the figure below
(a) T/J corresponds to M/I is subjected to 20 Nm torque.
(b) G/ℓ corresponds to E/R
(c) τ/r corresponds to σ/y
(d) Gθ/ℓ corresponds to E/R
RPSC VPITI 14.02.2016
M σ E
Ans. (b) : Bending moment equation = =
I y R If the modulus of rigidity is 80 GPa, then what
is the strain energy, in N-mm, in the shaft?
T τ Gθ
Torsion equation = = (a) 6.4 (b) 10.8
J r l (c) 12.7 (d) 16.9
T M (e) 23.7
corresponds to
J I CGPSC Asstt. Workshop Supt., 17.07.2016
Strength of Materials 340 YCT
Ans. (d) : Given, 1005. Torsional flexibility is the reciprocal of
Torque (T) = 20 Nm (a) Torsional stiffness (b) Torsional rigidity
Modulus of rigidity (G) = 80 GPa (c) Polar modulus (d) Twisting moment
From figure, APPSC AE Subordinate Service Civil/Mech. 2016
T1 = T2 = T = 20 Nm Ans. (a) : Torsional flexibility → It is defined as the
θ = θ1 + θ2 angle of twist produced by the unit torque applied. It is
TL T L reciprocal of torsional stiffness.
θ= 1 1 + 2 2 L θ
J1G1 J 2 G 2 Torsional flexibility = =
Here, T1 = T2 = 20 × 103 N-mm GJ T
L1 = L2 = 100 mm GJ T
Torsional stiffness = =
G1 = G2 = 80 × 103 N/mm2 L θ
TL  1 1  1006. A shaft is subjected to the combined bending
θ= +
G  J1 J 2  load 'M' and tensional load 'T', the maximum
shear stress at the outer surface of the shaft
20 ×1000 × 100  32 32  will be equal to
θ=  π × 20 4 + π × 404 
80 × 103 (a) 16T / πd 3
20 × 1000 ×100 × 32  1 1  (b) 32M / πd 3
θ=  + 4
π × 80 ×10 3
 20 40 
4
(c) 16(M 2 + T 2 )1/ 2 / πd 3
θ = 1.69102 × 10–3 (d) 32(M 2 + T 2 )1 / 2 / πd 3
1
Strain energy = T × θ VIZAG MT 2015
2
16  2
1 Ans. (c) : τmax = 3 M + T 2
= × 20 × 1000 ×1.69102 ×10−3 πd  
2
Strain energy = 16.91 N-mm 1007. In a shaft with a transverse hole, as the hole to
the shaft diameter ratio increases, the torsional
1003. If the diameter of shaft is doubled its stiffness stress concentration factor-
increases by a factor of: (a) Increases (b) Remains constant
(a) 2 (b) 6 (c) Decreases (d) Is equal to one
(c) 16 (d) 24 NPCIL ET, 2015
HPPSC Asstt. Prof. 29.10.2016
Ans. (c) : • As the hole to shaft diameter increase then
GJ G π 4 the torsional stress concentration factor decrease.
Ans. (c) : Torsional stiffness (kt) = = d
ℓ ℓ 32 1008. A composite shaft, consisting of two stepped
kt ∝ d4 portions having spring constants of K1 and K2,
is held firmly at one end and other end is free
d2 = 2d1 and subjected to a torque T. Its equivalent
( k t )2  d 2  spring constant is:
4

= = 16 (a) (K1 + K2)/2 (b) (K1 K2) / (K1 + K2)

( k t )1  d1  (c) (K1 + K2) (d) (K1 + K2) / (K1 K2)
( k t )2 = 16 ( k t )1 NPCIL ET, 2015
1 1 1
1004. Polar modulus is the ratio of Ans. (b) : = +
K eq K1 K 2
(a) Polar moment of inertia of the shaft to the
maximum radius 1 K + K2
= 1
(b) Moment of inertia of the shaft to the K eq K1K 2
maximum radius
(c) Polar moment of inertia of the shaft to the K 1K 2
K eq =
permissible shear stress K 1 + K2
(d) Moment of inertia of the shaft to the 1009. For two geometrically similar shafts of the
maximum shear stress same material having diameter ratio 2 : 1, if the
APPSC AE Subordinate Service Civil/Mech. 2016 smaller one has a torque capacity of 1 kg-m,
Ans. (a) : Polar modulus is defined as the ratio of polar then the larger one will carry.
moment of inertia of the shaft to the maximum radius of (a) 2 kg-m
the shaft. It is also called as torsional section modulus. (b) 4 kg-m
J (c) 8 kg-m
Zp = (d) 1 kg-m because of similarity
R
BPSC Asstt. Prof. 29.11.2015
Strength of Materials 341 YCT
 π From torsion equation
Ans. (c) : T1 = τ( d1 )3 T Gθ
16 =
π J l
T2 = τ( d 2 )3 T.l 2000 × 10
16 θ= = = 0.157 radian
GJ 100 × 109 × π (0.06)4
π
τ(d1 )3 32
T1 16
= θ = 0.157 ×
180
= 9degree
T2 π
τ( d 2 ) 3 π
16 1013. In case of shaft design, one of the following
T1 (d1 )3 equation is known as stiffness equation:
=
T2 (d 2 )3 T τ M σ
(a) = (b) =
T1 (2) 3 J R J R
= 3 M τ T Gθ
1 (1) (c) = (d) =
I R J L
T1 = 8 kgm
M Gθ
1010. In a hollow circular shaft of outer and inner (e) =
I L
diameters of 20 cm and 10 cm respectively, the
shear stress is not to exceed 40 N/mm . The 2 CGPSC AE 26.04.2015 Shift-I
maximum torque which the shaft can safely  T Gθ 
transmit is Ans. (d) : Equation  =  is known as stiffness
J L 
(a) 58.9 kN-m (b) 57.9 kN-m equation.
(c) 56.9 kN-m (d) 58.7 kN-m 1014. 100 kW is to be transmitted by each of two
MPPSC AE 08.11.2015 separate shafts 'A' and 'B'. 'A' is rotating at
Ans. (a) : Given, for hollow circular shaft 250 rpm and 'B' at 300 rpm. Which shaft must
Outer dia. (do) = 20 cm = 0.2 m have greater diameter.
Inner dia. (di) = 10 cm = 0.1 m (a) A
Shear stress (τ) = 40 N/mm2 = 40 × 106 N/m2 (b) B
Let, T = maximum torque (c) Both will have same diameter
(d) Unpredictable
π
16
( )
∴ T = d o .3 1 – K 4 .τ ,  Where, K = di = 0.1 = 1 
 d o 0.2 2  Ans. (a) : We know that
CGPSC AE 26.04.2015 Shift-I

  1 4  1
π N∝
= × ( 0.2 ) × 1 –    × 40 × 106
3
D
16   2   D N 300
= 58.90 × 103 N-m then, A = B =
D B N A 250
T = 58.90 kN-m
then, DA > DB
1011. Angle of twist allowed in case of camshaft is :
1015. The strength of a hollow shaft for the same
(a) Dependent on its length length, material and weight is .......... a solid
(b) Restricted to ½ degree irrespective of length shaft:
of the shaft (a) Less than (b) More than
(c) Depending on the torque acting on it (c) Equal to (d) None of these
(d) Dependent on the nature of the engine (i.e. 4 OPSC AEE 2015 Paper-I
stroke or 2 stroke)
Ans : (b) The strength of a hollow shaft for the same
OPSC AEE 2015 Paper-I length, material and weight is more than a solid shaft.
Ans : (c) Angle of twist allowed in case of camshaft is When the shaft is subjected to pure torsional moment
Depending on the torque acting on it. (T). the torsional shear stress is given by
1012. A solid circular shaft of 60 mm diameter and 16T
10 m length, transmits a torque of 2000 N-m. for solid shaft : τ = 3
The value of maximum angular deflection, if πd
the modulus of rigidity is 100 GPa is 16T
For hollow shaft : τ = 3
(a) 18 degree
(c) 15 degree
(b) 13 degree
(d) 9 degree (
πd 0 1 − C4)
(e) 5 degree 1016. A shaft of 10 mm diameter, whose maximum
CGPSC AE 26.04.2015 Shift-I shear stress is 48 N/mm2 can produce a
Ans. (d) : Given, d = 60 mm maximum torque equal to
l = 10 m = 10000 mm (a) 2000 π N - mm (b) 4000 π N - mm
G = 100 GPa = 100 × 103 N/mm2 (c) 1000 π N - mm (d) 3000 π N-mm
T = 2000 N-m = 2000 × 103 N-mm TSPSC AEE 2015
Strength of Materials 342 YCT
Ans : (d) Shaft dia (d) = 10 mm Ans. : (a) Equivalent twisting moment (Te)
Max shear stress (τmax) = 48 N/mm2
( Km × M ) + (K t ×T)
2 2
16T Te =
τmax = 3
πd M = Bending moment = 500 N-m
τ × πd3 T = Twisting moment = 1000 N-m
T= Km = fatigue factor for bending = 1.2
16
Kt = fatigue factor for torsion = 2
48 × π× 103
T=
(1.2 × 500 ) + ( 2 ×1000 )
2 2
16 Te =
T = 3000π N-mm. Te = 2088 N-m
1017. The equivalent twisting moment to design a
shaft subjected to the fluctuating loads will be 1020. Strain energy stored in a solid circular shaft is
given by proportional to (GJ is torsional rigidity)
(a) GJ (b) 1/GJ
(a) ( K t M ) 2 + ( K m T )2 (c) 1/(GJ) 2
(d) (GJ)2
Mizoram PSC AE/SDO 2014, Paper-II
(b) ( K m M ) 2 + ( K t T )2 Ans. (b) : Strain energy stored in a solid circular shaft is
KmM + (KmM) + (Kt T)
2 2 proportional to–
(c)
1 T2 L
1 U=
(d) KmM + ( K m M ) + ( K t T ) 
2 2
2 GJ
2  
TSPSC AEE 2015 1
U∝
Ans : (b) (i) Equivalent Bending moment GJ
1 2
M e = K m M + ( K m M ) + ( K t T ) 
2
1021. Shaft AB is 3 meter long and its free at the left
2  end and fixed torsionally at the right end. It is
(ii) Equivalent twisting moment subjected to variable torque as shown in the
Te = ( K m M ) + ( K t T )
2 2 figure. The intensity of the torsional loading is
given by q = 2x. The internal torque T, as a
Km = Shock and fatigue factor for bending
kt = shock and fatigue factor for twisting function of x is
1018. If a body is transmitting torque T kgm at
angular speed of θ radians/sec, then h.p.
transmitted will be-
(a) Tθ/75 (b) T/θ (a) T = x2 (b) T = 2x2
(c) Tθ/50 (d) Tθ
(c) T = 4x2 (d) T = 16x2
ISRO Scientist/Engineer (RAC) 29.11.2015
ISRO Scientist/Engineer 24.05.2014
Ans : (a) We know that
Ans. (a) :
 N−m
Power (P) = T × ω = T × ω × g  
 sec 
1 h.p.= 746W
1
1W = hp
746 The intensity of the torsional loading q = 2x
9.81T × θ The internal torque 'T' as the function of 'x'.
P= ω = θ radian / sec
746 T = ∫ q dx
T×θ
P= h.p. = ∫ 2 x dx
75
1019. A shaft subjected to fluctuating loads for which  2 x2 
the normal torque (T) and bending moment = 
(M) are 1000 N-m and 500 N-m respectively. If  2 
the combined shock and fatigue factor for T = x2
bending is 1.2 and combined shock and fatigue
factor for torsion is 2 then the equivalent 1022. The variation of shear in a circular shaft
twisting moment for the shaft is______ subjected to torsion is :
(a) 2088 N-m (b) 2050 N-m (a) linear (b) parabolic
(c) 2136 N-m (d) 2188 N-m (c) hyperbolic (d) None of these
(HPPSC AE 2014) MPPSC State Forest Service Exam, 2014
Strength of Materials 343 YCT
Ans. (a) : When a shaft is subjected to a torque or 1026. A solid shaft is replaced by a hollow shaft with
twisting a shearing stress is produced in the shaft. The outer diameter same as the diameter of the
shear stress varies from zero in the axis to a maximum solid shaft. The internal diameter of the hollow
at the outside surface of the shaft. The variation due to shaft is kept as 3/4th of its outer diameter. What
concentration loads is linear. is the ratio of torque transmitting capacity of
hollow shaft to that of the solid shaft?
(a) 175/256 (b) 3/4
(c) 9/16 (d) 27/64
RPSC AEN Pre-2013
Ans. (a) : Given –
For Hollow shaft-
1023. If pure torsion is applied to a piece of Inner diameter (ID) = Di
classroom chalk, it may crack along a 45° Outer diameter (OD) = Do
helical surface due to _______. For solid shaft,
(a) maximum shear stress Diameter (D) = Do
(b) maximum principal stress 3 D 3
(c) maximum tensile stress Di = D o ⇒ i =
(d) none of these 4 Do 4
MPSC HOD (Govt. Poly. Colleges) 04.10.2014 16T πD3 τ
We know that, τ = ⇒ T=
Ans. (b) : Chalk is a brittle material hence it will fails πD 3
16
by maximum principle stress theory. ∵ T∝D 3

in case of pure torsion.

For hollow shaft,
σx = 0 & σy = 0
τ xy   D 4 
tan ( 2θP ) = =∞ D 1 −  i  
3

σx − σy  D 
4
 THollow   D o  
THollow ∝ D3 1 −  i   =
2
  Do   TSolid D3
2θP = 90° ,
4
θP = 45° T  3 
Hollow
= 1−  
1024. The torisonal stiffness of a circular shaft is
TSolid 4
(a) Directly proportional to its length 81
= 1−
(b) Independent of its length 256
(c) Inversely proportional to its length
THollow 175
(d) Proportional to square of its length =
Kerala LBS Centre For Sci. & Tech. Asstt. Prof. 2014 TSolid 256
Ans. (c) 1027. For the two shafts connected in parallel, which
T of the following in each shaft is same?
Torsional Stiffness (kt) = (a) Torque (b) Shear stress
θ
(c) Angle of twist (d) Torsional stiffness
T Gθ τ T GJ
= = → = UKPSC AE-2013, Paper-I
J L R θ L Ans. (c) : When the two shafts connected in parallel
GJ then both shafts are subjected to same angle of twist.
kt = θ = θ1 = θ2
L T = T1 + T2
1025. A torque of 50 N-m applied on the wheel 1028. A hollow shaft has external and internal
operating a valve. If the wheel is rotated diameters of 10cm and 5cm respectively.
through two revolutions, work done in Newton- Torsional section modulus of shaft is:-
metres is given by (a) 375 cm3 (b) 275 cm3
3
(a) 100 (b) 25 (c) 184 cm (d) 84 cm3
(c) 314 (d) 628 UKPSC AE-2013, Paper-I
TNPSC AE 2014 Ans. (c) : We know that torsional section modulus for
Ans. (d) : T = 50 N-m hollow shaft
one revolution = 360° π
θ = 4π (720o) Zp = D3 1 − K 4 
16  
Then work done will be -
d
W = Tθ = 50 × 4π where K =
D
W = 628 N − m putting d = 5 cm, D = 10 cm

Strength of Materials 344 YCT

 From Equation (1)
3   5 
4
π
Zp = × (10 ) × 1 −    T = TAl + Tss
16   10   GAl J Al
= Tss + Tss
Zp = 184 cm3 Gss J ss
1029. Which part of the shaft cross section has  G J + GAl J Al 
T = Tss  ss ss 
maximum shear stress ?  Gss .J ss 
(a) 1 mm from the centre of the surface
TAl L T .L
(b) Centre of the surface ∴ θ= = SS
(c) Outermost surface GAl J Al GSS J SS
(d) 2 mm from the centre of the surface TL
θ Al = θ SS = radian
TNPSC AE (Industries) 09.06.2013 (Gss J ss + G Al J Al )
Ans. (c) : Maximum shear stress of the shaft occurs at Distance of circumferential point.
outermost surface.  D
d = 2R mm  R = 2 
1030. A 'SS' tube is inserted into an 'Al' tube. They
are permanently fixed at one end. The other
TDL
end is attached to a rigid plate. A torque 'T' is ∴ d= mm
applied to the rigid plate. The circumference of 2(G AL J Al + Gss J ss )
the 'Al' tube at dia 'D' at the plate and with 1031. Torsional rigidity of a solid cylindrical shaft of
respect to the fixed and rotates by a distance diameter ‘d’ is proportional to
............... mm due torque 'T'. The polar MOI & (a) d (b) d2
Rigidity modulus of Al & SS are JAl, GAl and Jss,
1
Gss respectively. (c) d4 (d) 2
d
UKPSC AE 2012 Paper-I
Ans. (c) : TR = GJ
TR ∝ J
π
TR ∝ d 4
32
TR ∝ d4
DLT DLT 1032. Polar moment of inertia of an equilateral
(a) (b)
2(GAl − Gss J ss ) 2G Al J Al triangle of side ‘x’ is given by
DLT 2 DLT x4 x4
(c) (d) (a) (b)
2(GAl J Al + Gss J ss ) Gss J ss 16 16 3
4
ISRO Scientist/Engineer 12.05.2013 x x4
Ans. (c) : At end, Torque = T (c) (d)
32 64
Circumference of the Al tube Diameter = D UKPSC AE 2012 Paper-I
The polar moment of Inertia (MOI) and Rigidity of Al 4
and SS are JAl, GAl and Jss, Gss. Ans. (b) : x
Here, Al and SS are connected in parallel. 16 3
So, In parallel, 1033. Torsional rigidity of a shaft is given by
T = TAl + Tss ...(1) T T
(a) (b)
Let angle of twist of Al tube is θAl and angle of twist of ℓ J
SS tube is θss due to applied torque acting at cross- T T
section. (c) (d)
θ r
TAl L
θ Al = APPSC AEE 2012
GAl J Al Ans : (c) Torsion Equation
T L T Gθ τ
θ ss = ss = =
Gss J ss J ℓ r
In parallel, T = Torque in N-mm
l = length of the shaft in mm
θ Al = θ ss = θ R = Radius of the circular shaft in mm
TAl L Tss L G = Modulus of rigidity of shaft material in N/mm2
∴ = T Gθ τ
GAl J Al Gss J ss = =
J ℓ r
GAl J Al T ℓ
∴ TAl = Tss GJ =
Gss J ss θ

Strength of Materials 345 YCT

G J is known as torsional rigidity of the shaft. It is Ans : (c) Power develop (P) = T ω
important to note that the relative stiffness of two shafts Case 1st:-
is measured by the inverse ratio of the angles of twist is
equal length of shafts when subjected to equal torques. πd 3τ
T=
1034. Shear stress for a circular shaft due to torque 16
varies πd 3 τ
(a) from surface to centre parabolically P1 = ×ω
16
(b) from surface to centre linearly Case 2nd :-
(c) from centre to surface parabolically
π(2d)3 × 2τ
(d) from centre to surface linearly P2 = ×ω
APPSC AEE 2012 16
assume, both case angular speed same then
Ans : (d) Shear stress for a circular shaft due to torque
varies from centre to surface linearly. P  1
Ratio of power developed  1  =
 P2  16
P1 1
=
P2 16
1037. The shear stress developed at a radial distance
r is q . The shear stress developed at a radial
T Gθ τ r
= = stress is :
J l r 2
The above relation states that the intensity of shear (a) 0.75 q (b) 0.5 q
stress at any point in the cross-section of a shaft
(c) q (d) 0.25 q
subjected to pure torsion is proportional to its distance
from the centre. APGENCO AE 2012
1035. A circular shaft subjected to torsion undergoes Ans. (b) : Shear stress (τ) ∝ radial distance (r)
a twist of 10 in a length of 1.2 m. If the τ ∝ r → In radial distance
maximum shear stress induced is 100 MPa and τ1 r1
the rigidity modulus is 0.8×105 MPa, the =
τ2 r2
radius of the shaft in mm should be
270 π q r q 2
(a) (b) = ⇒ =
π 270 τ2 r τ2 1
180 π 2
(c) (d) q
π 180 τ 2 = = 0.5q
APPSC AEE 2012 2
1038. The maximum shear stress occurs on the outer
Ans : (a) Twist angle (θ) = 10 most fibers of a circular shaft under torsion. In
length of shaft = 1.2 m a close coiled helical spring the maximum shear
Maximum shear stress = 100 MPa stress occurs on the
Modulus of rigidity (G) = 0.8 × 105 MPa. (a) Outermost fibers
T Gθ τ (b) Fibers at mean diameter
Torsion equation = = (c) Innermost fiber
J l r
(d) End coils
100 × 106 0.8 × 105 × π × 106
= RPSC Lect. (Tech. Edu. Dept.) 2011
r 180 × 1.2
Ans. (c) : Helical spring the maximum shear stress
270 occur on the inner side of the spring for both tensile &
r= mm
π compressive load on spring, this is because, inner shear
stress is the sum of shear stress due to torsion and shear
1036. Two shafts are of same length and same stress due to force or applied load.
material. The diameter and maximum shear
stress of the second shaft is twice that of the 1039. A shaft revolving at ' ω' radians/s transmits
first shaft. Then the ratio of power developed torque 'T' N-m. the power developed is :
between the first and second shaft is (a) T.ω. Watts (b) 2πT. ω Watts
16 (c) 2T. ω Watts (d) 2π T/ ω Watts
(a) 16 (b) RPSC AE GWD, 2011
3 3
Ans. (a) :
1 3
(c) (d) 2πNT
16 16 Power P = T × ω =
APPSC AEE 2012 60

Strength of Materials 346 YCT

1040. A solid shaft of 100mm diameter transmits 160 θ = θ1 + θ 2 + θ 3
HP at 200 rpm. The modulus of rigidity G = 8 ×
Tle Tl Tl Tl
105 kg/cm2. Then the maximum angle of twist = 1 + 2 + 3
for a length of 6 meter is GJ GJ1 GJ 2 GJ 3
(a) 5º (b) 2.5º le l l l
(c) 3.2º (d) 2º = 1 + 2 + 3
π 4 π 4 π 4 π 4
ISRO Scientist/Engineer 2011 de d1 d2 d3
32 32 32 32
Ans. (b) : P = 160 HP le l l l
N = 200 rpm 4
= 14 + 24 + 34
d d d d
G = 8 × 105 kg/cm2 e 1 2 3

1   d1  
4 4
l = 6m le  d1 
θ = ? d = 0.100 m =  l1 + l2  + l 3  
d e4 d14   d2   d3  
P = 160 × 746 
P = 119360 W We assume, d1 = de
4 4
= 119.360 kW d  d 
then, we get le = l1 + l2  1  + l3  1 
2π NT  d2   d3 
P=
60 1042.
2 × 3.14 × 200 × T
119.36 =
60
T = 5.70 kN/m
T Gθ τ A solid circular shaft, of polar moment of
= = inertia J and modulus of rigidity of the
J l r material G, is fixed at one end and loaded by
T Gθ two torques as shown in the figure. The twist at
=
J l the free end of the shaft will be zero when
(a) T2 = 0.5 T1 (b) T2 = T1
5.70 8 × 10 × 9.81× 10 × θ
5 4
= (c) T2 = 2T1 (d) T1 = 0
π 4 6 ×1000 UPSC JWM Advt. No.-50/2010
d
32 Ans. (c) :
Or θ = 0.044 rad
0.044 × 180
=
π
θ ≈ 2.5º As we know that,
1041. Two heavy rotating masses are connected by According to torsion equation –
shafts of lengths l1, l2 and l3 and the T Gθ
=
corresponding diameters are d1, d2 and d3. This J L
system is reduced to a torsionally equivalent TL
length of the shafts is θ=
GJ
 d1 
4
 d1 
4
⇒ G & J are constant or same.
(a) l1 + l2   + l3   So, the equilibrium is zero,
 d2   d3  ∴ ∑T = 0
3 3
 d1   d1  ℓ
T1 × ℓ − T2 × = 0
(b) l1 + l2   + l3  
 d2   d3  2
T2 = 2T1
l +l +l
(c) 1 2 3
3 1043. For solid shaft subjected to a torque of 18000
Nm having a permissible shear stress of
(d) l1 + l2 + l3 60N/mm2, the diameter of shaft is
ISRO Scientist/Engineer 2011 (a) 115 mm (b) 121 mm
Ans. (a) : (c) 149 mm (d) 108 mm
ISRO Scientist/Engineer 2010
Ans. (a) : T = 18000 N-m
permissible shear stress τ = 60 N/mm2
16T
τ=
πd3
Strength of Materials 347 YCT
 16 × 18000 × 1000 T τ
60 = =
πd3 J R
πd = 16 × 3000 × 100
3
T = τ×
J d
⇒k= i =
1
16 × 3000 × 100 R do 2
d3 =
π π 4 J π
JH = d o (1 − k 4 ) ⇒ H = d 3o (1 − k 4 )
32 R 16
16 × 3000 × 100
d=3 For solid,
3.141 d = 30 mm
d = 115 mm τs = ?
1044. A compound rod is formed by tightly inserting π
an aluminium rod inside a steel tube. The Js = d4
length of this compound rod is l and its ends 32
are welded together so as to prevent any d
∴ R=
relative motion. This rod is subjected to a 2
torque T, applied to its ends in opposite π 4
directions. If polar moment of inertia of the d
J 32
tube and rod are same and modulus of rigidity =
of steel is three times that of aluminium, what R d/2
is the ratio of torque shared by steel tube to π
that of aluminium rod? = d3
16
(a) 4 : 1 (b) 3 : 1 J π
(c) 2 : 1 (d) 1 : 1 = d3
UPSC JWM Advt. No.-50/2010 R 16
π
Ans. (b) : τh × × d o3 (1 − k 4 )
Th 16 403 (15 /16) ∵ Given, 
= =  τ =τ 
Ts π 303  S
τs × × d 3 H

16
Th 20
=
We know that, Ts 9
T Gθ 1046. Assertion (A) : Stress at a point is completely
=
J L defined by specifying its magnitude, nature,
Given, direction and orientation of the plane containing
the point.
ℓAℓ = ℓSt = L, JSt = JAℓ
Reason (R) : Stress is a tensor of second order.
TL Codes :
θ= ⇒ GSt = 3 GAℓ
GJ (a) Both A and R are individually true and R is
For parallel connection the correct explanation of A
θ = θSt = θAℓ (b) Both A and R are individually true and R is
not the correct explanation of A
Ts ℓ s TAℓ ℓ Aℓ T G (c) A is true but R is false
So, = ⇒ s = s =3
J s G s J A ℓ G Aℓ TAℓ G Aℓ (d) A is false but R is true
UPSC JWM Advt. No.-50/2010
Ts 3
= Ans. (a) : As we know, stress is a tensor of second
TAℓ 1 order therefore, it is completely defined by specifying
1045. A hollow shaft of outside diameter 40 mm and its magnitude, nature, direction & orientation of the
inside diameter 20 mm is to be replaced by a solid plane containing point.
shaft of 30 mm diameter. If the maximum shear 1047. Determine the diameter of solid shaft which
stresses induced in the two shafts are to be equal, will transmit 90 kW at 160 rpm, if the shear
what is the ratio of the maximum resistible torque stress in the shaft is limited to 60 N/mm2
in the hollow to that of solid shaft ? (a) 50 mm (b) 60 mm
10 20 (c) 77 mm (d) 70 mm
(a) (b)
9 9 ISRO Scientist/Engineer 2009
30 40 Ans. (c) : Power P = 90kW
(c) (d) N = 160 rpm
9 9
UPSC JWM Advt. No.-50/2010 Shear stress
Ans. (b) : Hollow cylinder τ = 60 N/mm2
do = 40 mm, di = 20 mm 2πNT
P=
τH = ? 60
Strength of Materials 348 YCT
 2π × 160 × T Ans. (d) :
90 × 103 =
60
Torque, T = 5371.47 N-m
16T
Shear stress, τ = 3
πd
16 × 5371.47 × 103
d3 =
π × 60 1
d = 76.96 mm τα
r
d = 77 mm τmax τmin
=
1048. One-half length of 50 mm diameter steel road is r1 r2
solid while the remaining half is hollow having r
a bore of 5 mm. The rod is subjected to equal τmin = 2 τmax
and opposite torque at its ends. If the r1
maximum shear stress in solid portion is τ, the 40
maximum shear stress in the hollow portion is : = ×120 = 80 MPa
60
15
(a) τ (b) τ 1050. In torsion, which of the following sections will
16 be best?
4 16 (a) triangular (b) rectangular
(c) τ (d) τ
3 15 (c) hollow circular (d) solid cylinder
ISRO Scientist/Engineer 2009 WBPSC AE 2008
π 4
D (1 − k 4 )
ESE 2003, 1997
Ans. (c) : J =
Ans. (b) : 32
• Hollow circular section is best in torsion as the
polar moment of inertia is higher for hollow
circular section (same amount of material).
1051. When a solid circular shaft is in pure tension
16T and deforms elastically, the shearing stress in
τ solid = the shaft
πD 3

(a) is inversely proportional to the shear modulus

Tr
τ hollow = of elasticity
J (b) varies linearly with the radial distance from
D the axis of the shaft
T×
2 (c) varies linearly with length of the shaft
=
π 4 (d) is inversely proportional to the diameter of
[ D − (0.1× D) 4 ] the shaft
32
16TD DRDO Scientists 2008
=
π D 4 (1 − 0.14 ) τ T Gθ
Ans. (b) : x = =
16T 1 rx J ℓ
τ hollow = ×
π D 0.9999
3 r x – radial distance from axis of shaft.
τ 1052. If 'T' is the torque and 'M' is the moment, then
τ hollow = solid the shaft is subjected to:
0.9999
(a) bending stress and direct stress
⇒ τhollow ≈ τsolid
(b) shearing stress and direct stress
1049. A hollow cylindrical shaft is 1 m long and has (c) principal stress and shearing stress
inner diameter and outer diameter as 40 mm (d) principal stress and direct stress
and 60 mm, respectively. When this shaft is NLCIL GET 17.11.2020, Shift-II
subjected to a pure torsion of certain Ans. (b) : If T is the torque and M is the moment, the
magnitude, maximum shear stress induced is 16T
120 MPa. The corresponding minimum value shaft is subjected to shearing stress =
of shear stress induced will be πd3
And direct stress,
(a) 20 MPa (b) 40 MPa
(c) 60 MPa (d) 80 MPa 32M
Bending stress =
DRDO Scientists 2009 πd 3
Strength of Materials 349 YCT
1053. A bending moment of 400 kNm and a twisting
moment of 300 kNm is applied on a circular 6. Thick and Thin Cylinders
solid shaft. If direct stress based on maximum
principal stress theory is σ and shear stress and Spheres
based on maximum shear stress theory is τ,
σ 1055. The ratio of hoop stress to longitudinal stress
then is: is___ for a thin cylinder subjected to internal
τ pressure.
5 5 (a) 2 : 1 (b) 3 : 2
(a) (b)
9 3 (c) 1 : 1 (d) 1 : 2
3 9 RPSC Lect. (Tech. Edu. Dept.) 13.03.2021, Paper-I
(c) (d)
5 5 Assam PSC AE (IWT) 14.03.2021
AAI Jr. Executive 26.03.2021 GPSC Executive Engineer 23.12.2018
ESE 2000 GPSC ARTO Pre 30.12.2018
GPSC Asstt. Director of Transport 05.03.2017
Ans. (d) : Given, Bending moment = 400 kN-m
GATE 2017, ESE 2017, WBPSC AE, 2017
Twisting moment = 300 kN-m
APPSC AE Subordinate Service Civil/Mech. 2016
Direct stress (σ) based on max. principal stress theory RPSC VPITI 14.02.2016, TSPSC AEE 2015
16 MPPSC State Forest Service Exam, 2014
σ = 3  M + M 2 + T 2 
πd GATE 2013, UKPSC AE 2012 Paper-I
16 APPSC AEE 2012, APPSC AE 04.12.2012
σ = 3  400 + (400) 2 + (300)2  J&K PSC Civil Services Pre, 2010
πd CSE Pre-1994
16
= 3 [ 400 + 500] Ans. (a) : We know,
πd pd
 16  hoop stress =
σ =  3  × (900) 2t
 πd  &
 16  pd
Shear stress, τ =  3  ( M 2 + T2 ) Longitudinal stress =
 πd  4t
 
 3  ( (400) 2 + (300) 2 )
⇒ 16 pd
 πd 
So, = 2t = 2 :1
 16  pd
=  3  × ( 500 )
 πd  4t
σ 9 1056. Auto frettage is the method of
∴ = (a) a type of corrosion
τ 5
(b) method to reduce stress concentration
1054. At a certain cross-section, a shaft of 100 mm (c) a type of pre-stressing
diameter is subjected to a bending moment of 4 (d) a type of heat treatment
kNm and a twisting moment of 8 kNm.
HPPSC Asstt. Prof. 18.09.2017
Maximum principal stress induced (in N/mm2)
in the section is. APGENCO AE, 2017
APPSC AEE Screening Test 2016
(a) 72.8 (b) 6.17
GPSC AE 26.07.2015, UPRVUNL AE 2014
(c) 65.9 (d) 68.6 APPSC AEE 2012, APPSC AE 04.12.2012
HPPSC ADF 05.03.2019 ESE 2005, ESE 1996
Ans. (c) : d = 100 Ans. (c) : Auto frettage is the method of a type of pre-
Bending Moment, M = 4 kN-m stressing of thick walled cylinders to increase its strength.
Twisting Moment, T = 8 kN-m 1057. Longitudinal stress in a thin cylinder is
Max principal stress, (a) Equal to the hoop stress
16
σ1 = 3  M + M 2 + T 2  (b) Half of the hoop stress
πd   (c) Twice the hoop stress
4 + ( 4 ) + (8) 
16  2
(d) Three fourth of the hoop stress
= 2

π (100 )  
3
VIZAG Steel MT 24.01.2021, Shift-I
16 TSPSC Manager (Engg.) HMWSSB 12.11.2020
3[
= 4 + 8.94] MECON MT 2019
π (100 ) Rajasthan AE (Nagar Nigam) 2016 Shift-3
16 RPSC LECTURER 16.01.2016
σ1 = × 12.94 × 106
π (100 ) APPSC AEE MAINS 2016, PAPER-III
3

WBPSC AE 2008
Max principal stress, σ1 = 65.9 N/mm2 ISRO Scientist/Engineer 2006
Strength of Materials 350 YCT
Ans. (b) : Longitudinal stress in a thin cylinder is half pd pd
of the hoop stress. (a) (b)
4t 16t
Pd pd pd
Longitudinal stress σ L = (c) (d)
4t 8t 2t
Pd Haryana PSC AE (PHED) 05.09.2020, Paper-II
Hoop stress σH = VIZAG Steel MT, 14.12.2020
2t
APPSC AE Subordinate Service Civil/Mech. 2016
σ L Pd 2t Vizag Steel MT (Re-Exam) 24.11.2013
= ×
σ H 4t Pd DRDO Scientists 2009, CSE Pre-1996
σL 1 Ans. (c) : σc =
pd
σℓ =
pd
=
σH 2 2t 4t
σc − σℓ pd
σH Maximum shear stress = =
σL = 2 8t
2 σc = Circumferential stress
Where, σL → longitudinal stress
σℓ = Longitudinal stress
σH → Hoop stress
p = internal pressure
1058. A thin cylinder of inner radius 500mm and
thickness 10mm is subjected to an internal d = internal diameter
pressure of 5 MPa. The hoop stress is equal to t = thickness
(a) 1000 (b) 500 1061. Circumferential and longitudinal strains in the
(c) 250 (d) 125 cylindrical boiler under internal steam
Vadodara Muncipal Corp. DEE, 2018 pressure are ε1 and ε2 respectively. Change in
GPSC ARTO Pre 30.12.2018 the volume of the boiler cylinder per unit
GPSC Executive Engineer 23.12.2018 volume will be
UPRVUNL AE 21.08.2016 (a) ε 12ε 2 (b) ε1ε 22
HPPSC Asstt. Prof. 18.11.2016 (c) 2ε 1 + ε 2 (d) ε 1 + 2ε 2
VIZAG STEEL MT 18.08.2013 ESE 2018, UPRVUNL AE 07.10.2016
VIZAG STEEL MT 10.06.2012 APPSC AE 04.12.2012
GATE 2011 ISRO Scientist/Engineer 2009
Ans. (c) : Given ISRO Scientist/Engineer 2007
ri = r = 500mm Ans. (c) : Circumferential strain = ε1
t = 10mm longitudinal strain = ε2
P = 5 MPa change in the volume of the boiler cylinder per unit
Pd volume
Hoop stress (σh) = volume of cylinder
2t
5 × (2 × 500) π
= V = D2 × l
2 × 10 4
dV dD dl
σ h = 250 MPa =2 +
V D l
1059. Thick cylinders are designed by: ev = 2ε1 + ε 2
(a) Lame's equation
(b) calculating radial stress which is uniform 1062. Hoop stress in a thin cylinder of diameter 'd' and
thickness 't' subjected to pressure 'p' will be
(c) thick cylinder theory
(d) thin cylinder theory pd pd
(a) (b)
HPPSC Asstt. Prof. 18.09.2017 4t t
RPSC AE 2016 pd 2pd
APPSC AEE Screening Test 2016 (c) (d)
2t t
TSPSC AEE 2015 MPPSC AE 2016, SJVN ET 2013
UPRVUNL AE 2014 CSE Pre-2009, TNPSC AE, 2008
UKPSC AE 2012 Paper-I J & K PSC Screening, 2006
RPSC ACF-2011 Ans. (c) : For a thin cylinder –
Ans. (a) : When the ratio of inner diameter of the Pd
cylinder to the thickness is less than 20, the cylinder is Hoop stress/circumferential stress ( σc ) =
called thick walled cylinder or simply thick cylinder. 2t
Thick cylinders are designed by Lame's equation. Where; P = Pressure
1060. Maximum shear stress in any point in a thin d = dia of the cylinder
cylinder is given by_____. t = thickness of the cylinder

Strength of Materials 351 YCT

1063. In a thick cylinder pressurized from inside, the 1066. A thin cylinder of diameter D, wall thickness t is
hoop stress is maximum at subjected to an internal fluid pressure p. If E is
(a) The centre of the wall thickness the Young's modulus and υ is the Poisson's ratio
(b) The outer radius for the material of the cylinder, the expression
(c) The inner radius for volumetric strain of the cylinder is
pD pD
(d) Same at both inner and outer radii (a) ( 5 − 4µ ) (b) ( 4 − 5µ )
GPSC AE 26.07.2015, APPSC AEE 2012 4tE 4tE
APPSC AE 04.12.2012, ESE 1998 pD pD
Ans. (c) : Maximum hoop stress in thick cylinder
(c) ( 5 − 4µ ) (d) ( 4 − 5µ )
2tE 2tE
pressurized from inside at the inner surface. Max hoop RPSC LECTURER 16.01.2016
stress in thick cylinder under external pressure occurs of WBPSC AE, 2007, CSE Pre-2006
the outer surface. ESE 2003, CSE Pre-1997
1064. The longitudinal stress within cylindrical shell Ans. (a)
of diameter (D), length (L) and thickness (t), δv δℓ 2δd
when subjected to an internal pressure (P) is εv = = +
(a) PD/(4t) (b) PD/(2t) v ℓ d
(c) 2PD/t (d) 4PD/t δℓ σ µσ pd µpd
εℓ = = L− h = −
TNPSC AE 2019 ℓ E E 4tE 2tE
APPSC AE Subordinate Service Civil/Mech. 2016 pd
ISRO Scientist/Engineer (RAC) 29.11.2015 ε1 = (1 − 2µ)
4tE
Kerala LBS Centre For Sci. & Tech. Asstt. Prof. 2014
δd σh σ L Pd Pd
Ans. (a) εd = = − = −
Safe condition d E E 2tE 4tE
σL ≤ Sut Pd
εd = (2 − µ)
Limiting condition 4tE
σL = Sut Pd 2Pd
or Ff ≤ FR (OR) Ff = FR εv = (1 − 2µ) + (2 − µ)
4tE 4tE
π 2
P × D = σL × πDt Pd
4 =
4tE
[1 − 2µ + 4 − 2µ]
PD
σL = Pd
4t εv = (5 − 4µ)
4tE
1065. A thin gas cylinder with an internal radius of 1067. A thin walled spherical shell is subjected to an
100 mm is subject to an internal pressure of 10 internal pressure. If the radius of the shell is
MPa. The maximum permissible working increased by 1% and the thickness is reduced
stress is restricted to 100 MPa. The minimum by 1%, with the internal pressure remaining
cylinder wall thickness (in mm) for safe design the same, the percentage change in the
must be ----------- circumferential (hoop) stress is
(a) 8 mm (b) 10 mm (a) 0 (b) 1
(c) 12 mm (d) 14 mm (c) 1.08 (d) 2.02
(e) None of the above UPPSC AE 13.12.2020, Paper-I
OPSC AEE 2019 Paper-I TSPSC AEE 2017, GPSC ARTO 01.05.2016
CGPSC Poly. Lect. 22.05.2016 GATE 2012
Nagaland PSC CTSE 2016 Paper-I Ans. (d) : Hoop stress for a thin spherical shell (σh)
GATE 2014 Pr
=
Ans. (b) 2t
radius r = 100 mm By applying logarithm on both sides,
d = 200 mm P
pressure (P) = 10 MPa log ( σ h ) = log   + log ( r ) – log ( t )
permissible stress = 100 MPa  2
For safe design– dσ h 1 1
= 0 + dr − dt
Pd σ h r t
≤ σpermissible
2t dr dt
Percentage change = × 100 + × 100
10 × 200 r t
= 100 = 1% + 1% = 2%
2t
Percentage increase will be 2%.
t = 10mm Note-Option (d) is given by commission.
Strength of Materials 352 YCT
1068. What will happen in winter to a large Ans. (a) : In case of thin walled cylinders shell, the
cylindrical vessel which was sealed in summer? ratio of hoop strain to longitudinal strain is,
(a) It buckles and collapses P.d  1  ____ (i)
(b) It becomes lighter εc = 1 − 
2tE  2m 
(c) It explodes
(d) Nothing happens to it P.d  1 1  ____ (ii)
εℓ =  − 
GPSC Asstt. Director of Transport 05.03.2017 2tE 2 m
ISRO Scientist/Engineer (RAC) 07.05.2017 from eqn (i) & (ii)
Rajasthan Nagar Nigam AE 2016, Shift-I ε 2m − 1
Ans. (a) : Large cylindrical vessel which was sealed in ε = m − 2
c

ℓ
summer, is buckles and collapses in winter due pressure
variation between inside and outside. 1072. A boiler shell of 200 cm diameter and plate
thickness 1.5 cm is subjected to internal
1069. A metal pipe of 1 m diameter contains a fluid pressure of 1.5 MN/m2 then hoop stress is:
2
having a pressure of 10 kgf/cm . If the (a) 30 MN/m2 (b) 50 MN/m2
permissible tensile stress in the metal is 200 (c) 100 MN/m 2
(d) 200 MN/m2
kgf/cm2, then the thickness of the metal
required for making the pipe would be KPSC AE 2015, SJVN ET 2013
(a) 5 mm (b) 10 mm Ans. (c) : Given,
(c) 25 mm (d) 20 mm d = 200 cm = 200 × 10-2 m
APPSC AEE 2012 , ISRO Scientist/Engineer 2007
t = 1.5 cm = 1.5 × 10-2 m
WBPSC AE 2003
P = 1.5 MN/m2
Pd
Ans. (c) : d = 1 m σ=
2 2t
P = 10 kgf/cm
circumferential stress or hoop stress 1.5 × 200 ×10 −2
σ= −2
= 100 MN / m 2
Pd 2 × 1.5 × 10
σh = = 200
2t σ = 100 MN / m 2
10 ×1000
= 200 1073. A thin cylinder of radius r and thickness t
2×t when subjected to an internal hydrostatic
t = 25 mm pressure 'p' causes a radial displacement 'u'.
Note: Longitudinal stress or Axial stress Then the tangential strain caused is
Pd du 1 du
σl = (a) (b) .
4t dr r dr
1070. A thin cylinder with both ends closed is u 2u
subjected to internal pressure p. The (c) (d)
r r
longitudinal stress at the surface has been APPSC AEE 2012, ESE 2002
calculated as σ0 . Maximum shear stress at the Ans. (c) :
surface will be equal to :
(a) 2σ0 (b) 1.5σ0
(c) σ0 (d) 0.5σ0
APPSC AEE Mains 2016 (Civil Mechanical)
OPSC AEE 2015 Paper-I, ESE 1999
Pd
Ans : (d) Longitudinal stress ( σ2 ) = = σ0
4t
Pd
Hoop stress ( σ1 ) = = 2σ 0
2t
Principal stresses = 2σ0 , σ0 Data given-
2σ 0 − σ 0 σ 0 Radius = r
Shear stress = = = 0.5σ0 Thickness = t
2 2
Internal hydrostatic pressure = P, σr = radial stress
1071. In case of thin walled cylinders the ratio of
hoop strain to longitudinal strain is Radial displacement = u, σt = tangential stress
2m − 1 2m − 1 du
(a) (b) (i) Radial strain (εr) =
m−2 m −1 dr
m−2 m−2 (ii) Circumferential/Tangential strain (εt) = u/r
(c) (d)
2m − 1 2(m − 1) σr σ σ 
Nagaland PSC (CTSE) 2018, Paper-I (iii) Axial strain (εA) = – u r + t 
Nagaland PSC CTSE 2016 Paper-I E E E
Strength of Materials 353 YCT
1074. Pressure vessels are made of 1078. A steel hub of 100 mm internal diameter and
(a) non-ferrous materials uniform thickness of 10 mm was heated to a
(b) Sheet metal (steel) temperature of 300°C to shrink-fit it on a shaft.
(c) cast iron On cooling, a crack developed parallel to the
(d) Any of the above direction of the length of the hub. Consider the
APPSC AEE 2012, APPSC AE 04.12.2012 following factors in this regard :
Ans. (b) : Many pressure vessels are made of steel. To 1. Tensile hoop stress
manufacture a cylindrical or spherical pressure vessel, 2. Tensile radial stress
railed and possibly forged parts would have to be 3. Compressive hoop stress
welded together. Some mechanical properties of steel, 4. Compressive radial stress
achieved by rolling or forging, could be adversely The cause of failure is attributable to :
affected by welding unless special precautions are (a) 1 alone (b) 1 and 3
taken. (c) 1, 2 and 4 (d) 2, 3 and 4
1075. The stresses in a thick cylinder subjected to GPSC ARTO Pre 30.12.2018
uniform pressure vary proportional to : ESE 2016
(a) r (b) 1/r
Ans. (a) : In a shrink fit, the hub is subjected to tensile
(c) r2 (d) 1/r2
or hoop stress (which is greater as compared to radial
APPSC AE 04.12.2012 stress) and the shaft is subjected to compressive stress.
APPSC AEE 2012 Therefore, the failure (or crack on surface) is due to
Ans. (d) : From the Lame's equation for thick cylinder - tensile hoop stress.
1079. A thick cylinder subjected to an internal
b Where; pressure of 60MPa. If the hoop stress on the
σr = − a σ = radial stress outer surface is 150 MPa, then the hoop stress
r2 r
on the internal surface is :
b σC = Circumfrential stress
and σ C = 2 + a (a) 105 MPa (b) 180 MPa
r a & b is constant (c) 210 MPa (d) 135 MPa
1076. As per Lame's equation, the hoop stress for MPPSC AE 08.11.2015, APGENCO AE 2012
thick cylinder at any points is: Ans. (c) : If internal pressure = Pi
(a) (A + B/r3) (b) (A/r – B) External pressure = 0
(c) (A + B/r2) (d) (A + Br)2
Pr 2  r 2 
HPPSC ADF 05.03.2019 Circumferential or hoop stress (σc ) = 2 i i 2  02 + 1
TNPSC AE, 2008 r0 − ri  ri 
Ans. (c) : According to Lame's equation, the hoop stress At, Pi = 60MPa, σc = 150 MPa
for thick cylinder and r = r0
B r2  r2 
Hoop stress (σ h )hoop = A + 2 ∴ 150 = 60 2 i 2  o2 + 1
r r0 − ri  ro 
B
p=A–A+ 2  r 2

r 150 = 2 × 60  2 i 2 
B  ro − ri 
Radial stress (σ r ) = –p = A – 2 2
r ri 2
150 5  r0  9
A, B Lame's constant or = = ⇒   =
ro2 − ri 2 120 4  ri  5
1077. A thin walled pipe contains water at a pressure ∵ at r = r
i
of 2N/mm2 and discharges water into a tank. If
the pipe diameter is 25 mm and wall thickness ri 2  r02 
2.5 mm, the longitudinal stress induced in the σ c = 60 ×  + 1
r0 − ri 2  ri 2 
2

pipe is
(a) 0 (b) 2 N/mm2 5 9 
2 2 = 60 × ×  + 1
(c) 5 N/mm (d) 10 N/mm 4 5 
UPSC JWM Advt. No.-52/2010 σc = 210 MPa
CSE Pre-2007 1080. Hoop stress and longitudinal stress in a boiler
Ans. (c) : Solution : Given : Pressure (P) = 2N/mm2 shell under internal pressure are 100 MN/m2
Diameter of pipe (d) = 25 mm and 50 MN/m2 respectively. Young's modulus
Thickness of pipe (t) = 2.5mm of elasticity and Poisson's ratio of the shell
material are 200 GN/m2 and 0.3 respectively.
Pd
∴ Longitudinal stress (σL) = The hoop strain in the boiler shell is :
4t (a) 0.425 × 103 (b) 0.5 × 10–3
2 × 25 (c) 0.5750–3 (d) 0.7 × 103
σL = = 5 N / mm 2
4 × 2.5 APPSC IOF, 2009, ESE 1995
Strength of Materials 354 YCT
Ans. (b) : σh = 100 MN/m2, σℓ = 50 MN/m2 1083. The typical ratio of diameter to wall thickness
for a thin vessel is
E = 200 GN/m2, µ = 0.3 (a) Less than 20 (b) More than 10
1 1 (c) Less than 2 (d) More than 20
ε h = [ σh − µσℓ ] = [100 − 50 × 0.3] × 106
E 200 × 109 Haryana PSC AE (PHED) 05.09.2020, Paper-II
= 0.425 × 10–3 APPSC AE Subordinate Service Civil/Mech. 2016
Nearest option is (b). D
Ans. (d) : For thin vessel D > 20t or > 20
1081. A thin cylindrical shell of diameter (d) and t
thickness (t) is subjected to an internal Where, D = Diameter
pressure (p). The ratio of longitudinal strain to t = Thickness of shell
volumetric strain is 1084. In a thick cylinder, subjected to internal and
m –1 2m –1 external pressures, let r1 and r2 be the internal
(a) (b) and external radii respectively. Let u be the
2m –1 m –1
radial displacement of a material element at
m–2 m–2 radius r, r2 ≥ r ≥ r1. Identifying the cylinder
(c) (d)
3m – 4 5m – 4 axis as z axis, the radial strain component is :
Sikkim PSC (Under Secretary), 2017 (a) u/r (b) u/θ
UPPSC AE 12.04.2016 Paper-I (c) du/dr (d) du/δθ
Pd APPSC IOF, 2009, ESE 1996
Ans. (d) : Longitudinal strain ( εℓ ) = (1 – 2µ ) .....(i) Ans. (c) : The strains εr and εθ may be given by
4tE
∂u 1
Volumetric strain ( ε v ) =
Pd ε r = r = [ σ r − vσθ ] since σz = 0
( 5 – 4µ ) .............(ii) ∂r E
4tE
eqn (i) ÷ eqn (ii) ( r + u r ) ∆θ − r∆θ = u r 1
εθ = = [ σθ − vσr ]
r∆θ r E
Pd
ε ℓ 4tE (
1 – 2µ ) du
= ε rr =
εv Pd dr
( 5 – 4µ ) 1085. A cylindrical tank with closed ends is filled with
4tE compressed air at a pressure of 500 kPa. The
ε ℓ 1 – 2µ inner radius of the tank is 2m, and it has wall
=
ε v 5 – 4µ thickness of 10 mm. The magnitude of
maximum in-plane shear stress (in MPa) is ___
1
∵µ = (a) 100 (b) 50
m (c) 250 (d) 25
2 Assam Engg. College AP/Lect. 18.01.2021
1– GATE 2015
∴= m
4 Ans. (d) : The maximum in plane shear stress
5– Pd
m τ max =
εℓ m–2 8t
=
ε v 5m – 4 500 × 103 × 2 × 2
=
1082. A thin spherical shell of diameter (d) and 8 ×10 × 10 –3
thickness (t) is subjected to an internal = 25 × 106 N/m2
pressure (p). The stress in the shell material is τmax = 25 MPa
(a) pd/t (b) pd/2t 1086. The hoop or circumferential stress in a riveted
(c) pd/4t (d) pd/8t cylindrical shell, when subjected to an internal
Sikkim PSC (Under Secretary), 2017 pressure p is equal to :
Karnataka PSC AE, 10.09.2017 pD pD
(a) (b)
Ans. (c) : Thin spherical shell 4tηc 2tηc
pD pD
(c) (d)
2tηl 4tηl
Where D = internal diameter, ηl = efficiency of
longitudinal joint and ηc = efficiency of circumferential
joint.
CGPSC AE 15.01.2021, CGPSC AE 16.10.2016
π
σ1 × πdt = p × d 2 pD
4 Ans. (c) : Hoop or circumferential stress (σc ) =
pd 2tηℓ
σ= where, ηℓ = Efficiency of longitudinal joint
4t
Strength of Materials 355 YCT
1087. A thin cylinder of 'D' internal diameter, is Ans. (b) :
subjected to an internal pressure of 'P'. If the
permissible tensile stress is σt' the cylinder
wall thickness should be
2σ t PD
(a) (b)
PD σt
PD PD
(c) (d) 1090. The maximum shear stress in a cylinder shell of
4σ t 2σ t
4 m diameter and plate thickness 3 cm,
RPSC Lect. (Tech. Edu. Dept.) 13.03.2021, Paper-I subjected to an internal pressure of 3 N/mm2 is
RPSC AE GWD, 2011 __________.
Ans. (d) : As we know, (a) 200 N/mm2 (b) 50 N/mm2
2
Permissible tensile stress σt = Hoop stress in thin (c) 100 N/mm (d) 400 N/mm2
cylinder. VIZAG Steel MT 24.01.2021
PD Ans. (c) : Given,
Hoop stress =
2t Diameter (d) = 4 m = 4000 mm
PD Thickness (t) = 3 cm
σt = = 30 mm
2t
Internal pressure (P) = 3 N/mm2
PD
t= Pd
2σ t Maximum shear tress (τmax) =
4t
1088. A thick cylinder with internal diameter d and 3 × 4000
outside diameter 2d is subjected to internal =
4 × 30
pressure P. Then the maximum hoop stress
developed in the cylinder is τ max = 100 N / mm 2
(a) P (b) (2/3)P 1091. The ratio of volumetric strain of sphere to
(c) (5/3)P (d) 2 P strain in its diameter will be
RPSC Lect. (Tech. Edu. Dept.) 13.03.2021, Paper-I (a) 2 (b) 1/3
ESE 2003 (c) 3 (d) 1/2
Ans. (c) : di = d [internal diameter] GPSC DEE, Class-2 (GWSSB) 04.07.2021
do = 2d [outer diameter] APPSC AE Subordinate Service Civil/Mech. 2016
P = Pressure (internal) Ans. (c) :
In thick cylinder, maximum hoop stress, Volumetric strain of sphere = 3 × diametric strain
r22 + r12 Volumetric strain of cylinder = longitudinal strain +
σ hoop = P × 2 2 [r1 = d, r2 = 2d]
r2 – r1 (2 × lateral strain)
 (2d) 2 + (d)2  1092. A 800 mm diameter pipe contains a fluid at a
= P 2
pressure of 3 Nmm-2. If the safe stress in
 (2d) 2
− (d)  tension is 150 Nmm-2, the minimum thickness
 5d 
2 of the pipe is
= P 2  (a) 16 mm (b) 2 mm
 3d  (c) 4 mm (d) 8 mm
5P GPSC DEE, Class-2 (GWSSB) 04.07.2021
σ hoop =
3 Ans. (d) :
1089. Pick the correct statement about the hoop 1093. In a thick cylindrical shell, the maximum radial
stress in a thick cylinder subjected to internal stress at the outer surfaces of the shell is
pressure. (a) Zero (b) P
(a) It varies linearly form zero at the outer (c) –P (d) 2P
surface to maximum at the inner surface. RPSC IOF, 2020
(b) It varies parabolically from a minimum stress Ans. (a) : The radial stress across the thickness of a
at the outer surface (with non-zero thick cylinder is maximum at the inner surface and zero
magnitude) to a maximum value at the inner at the outer surface.
surface.
(c) It varies parabolically from zero at the outer
surface to a maximum at the inner surface.
(d) It remains constant across the thickness.
HPPSC Asstt. Prof. 2014
RPSC AEN Pre-2013
Strength of Materials 356 YCT
1094. A thin spherical shell of diameter 200 mm is Pd
subjected to an internal pressure of 2 MPa, If ∈hoop = ( 2 − µ ) _______ ( ii )
admissible tensile stress is 40 MPa, then 4tE
Dividing equation (i) by equation (ii),
thickness of the shell is:
∈ℓ (1 − 2µ ) 60 × 10−6
(a) 2 mm (b) 5 mm = =
(c) 10 mm (d) 2.5 mm ∈hoop 2−µ 255 × 10−6
CIL MT 27.02.2020 µ = 0.3
Ans. (d) : Diameter d = 200 mm E = 2G(1+µ)
= 2×77×(1+0.3)
Internal pressure (P) = 2 MPa = 200.2 GPa
Tensile stress σ = 40 MPa Putting the respective values in equation (i)
Pd P × 600
σ= 60×10-6 = × (1 − 2 × 0.3)
4t 4 × 18 × 200.2
Pd 2 × 200 P = 3.603 MPa
t= = = 2.5 mm ≈ 3.6 MPa
4σ 4 × 40
t = 2.5 mm 1098. A compressed air spherical tank having an inner
1095. A cylindrical pressure vessel has diameter 200 diameter of 450 mm and a wall thickness of 7
mm and thickness 2 mm. Find the hoop and mm is formed by welding. If the allowable shear
axial stress (N/mm2) in the cylindrical vessel, stress is 40 MPa, the maximum permissible air
when it is subjected to an internal pressure of 5 pressure in the tank will be nearly
(a) 3 MPa (b) 5 MPa
MPa.
(c) 7 MPa (d) 9 MPa
(a) 125, 125 (b) 125, 250
ESE 2020
(c) 250, 250 (d) 250, 125
ISRO Scientist/Engineer 12.01.2020 Pd
Ans. (b) : σ1 = σ 2 =
pd 5 × 200 4t
Ans. (d) : σ hoop = = = 250 MPa Pd
2t 2× 2 ∴ τ max =
1 8t
σlong or σaxial = σ hoop = 125MPa P × 450
2 40 =
1096. Which of the following statements regarding 8× 7
thin and thick cylinders, subjected to internal P = 4.978 MPa
pressure only, is/are correct? P ≈ 5 MPa
1. A cylinder is considered thin when the 1099. A thin spherical shell is subjected to an
ratio of its inner diameter to the wall external pressure po. The volumetric strain of
thickness is less than 15. the spherical shell is
2. In thick cylinders, tangential stress has (where, d is the diameter of shell t is the
highest magnitude at the inner surface of thickness of the shell E is Young's modulus of
the cylinder and gradually decreases elasticity of shell material
towards the outer surface µ is Poisson's ratio of shell material)
(a) 1 only (b) 2 only
pd 3po d
(c) Both 1 and 2 (d) Neither 1 nor (a) o ( 5 − 4µ ) (b) (1 − µ )
ESE 2020 4tE 4tE
Ans. (b) : If the thickness of the wall of shell is greater 3po d −3po d
than 1/10 to 1/15 of its diameter. (c) (1 − 2µ ) (d) (1 − µ )
4tE 4tE
1097. A cylindrical storage tank has an inner UPPSC AE 13.12.2020, Paper-I
diameter of 600 mm and a wall thickness of 18 Ans. (d) :
mm. The transverse and longitudinal strains
induced are 255 × 10-6 mm/mm and 60 × 10-6
mm/mm, and if G is 77 GPa, the gauge
pressure inside the tank will be
(a) 2.4 MPa (b) 2.8 MPa
(c) 3.2 MPa (d) 3.6 MPa
ESE 2020
Ans. (d) : Given, Po d
d = 600 mm Circumferential or hoop stress ( σc ) = −
t = 18 mm 4t
∈ℓ = 60 × 10−6 3σ
Volumetric strain E v = (1 − µ )
∈ hoop = ∈ transverse E
= 255×10-6 3P d
Pd = − o (1 − µ )
∈ℓ = (1 − 2µ ) _______ ( i ) 4tE
(–ve sign is due to external pressure)
4tE
Strength of Materials 357 YCT
1100. Internal and external radii of a thick cylinder Ans : (b) :
are a and b. It is subjected to an internal A
pressure of pi. The radial stress at a radius r in Hoop stress, σ h = +B
the cylinder is r2
Hence hoop stress will be maximum at inner radius and
a 2 pi  a 2  a 2 pi  b 2  minimum at outer radius.
(a)  −  (b) 1−
( b 2 − a 2 )  r 2 
1
( b2 − a 2 )  r 2  1104. Thin cylindrical pressure vessel of 500 mm
diameter is subjected to an internal pressure of
b 2 pi  a 2  b 2 pi  b 2  2 N/mm2. If the thickness of the vessel is 20
(c)  −  (d) 1−
( b 2 − a 2 )  r 2 
1
( b2 − a 2 )  r  2
mm, the hoop stress is
UPPSC AE 13.12.2020, Paper-I (a) 10 (b) 12.5
Ans. (b) : Given, Internal radius = a (c) 25 (d) 50
External radius = b Gujarat PSC AE 2019
Internal pressure = Pi Ans : (c) :
External pressure = 0 Pd
Hoop stress ( σ h ) =
Circumferential or hoop stress is given by with internal 2t
radius (ri) & external radius (r0) P = 2 N/mm2
P r2  r2  d = 500 mm
= 2 i i 2 1 − o2  t = 20 mm
ro − ri  r 
2 × 500 1000
Pa 2
 b2  σh = =
Similarly radial stress (σ r ) = i
1− 2 × 20 40
2
b − a  r 2 
2 
σ h = 25 N / mm 2
1101. A thin cylinder of internal diameter D = 1 m
and thickness t = 12 mm is subjected to internal 1105. A spherical shell with internal diameter 320
pressure of 2 N/mm2. Determine the hoop stress mm and external diameter 640 mm is subjected
(N/mm2) developed. to an internal fluid pressure of 75 N/mm2. The
(a) 83 (b) 102 hoop stress developed at the outer surface will
(c) 56 (d) 79 be:
Oil India Senior Officer 23.12.2020 (a) 15.132 N/mm2 (b) 16.071 N/mm2
2
(c) 14.067 N/mm (d) 17.173 N/mm2
Pd 2N 1000mm
Ans. (a) : σ h = = × HPPSC ADF 05.03.2019
2t mm 2 2 × 12mm Ans. (*) : d = 320 mm
= 83.33 N/mm2 i
d0 = 640 mm
1102. A thin cylindrical shell of internal diameter D t = d0 – di = 320 mm
and thickness t is subjected to internal pressure p = 75 N/mm2
p, E and µ are respectively the Elastic modulus
and Poisson's ratio. The change in diameter is pd 75 × 640
σh = 0 =
pD 2
pD 2 2 × t 2 × 320
(a) (1 − 2µ) (b) (2 − µ) σ h = 75N / mm 2
4tE 4tE
pt 2 pt 2 1106. A spherical steel pressure vessel 400 mm in
(c) (2 − µ) (d) (1 − 2µ) diameter with a wall thickness of 20 mm, is
4 DE 4 DE coated with brittle layer that cracks when
APPSC AEE SCREENING 17.02.2019 ,CSE Pre-1998 strain exceeds 100 × 10-7. What internal
Ans. (b) : Circumferential strain in thin cylinder pressure will cause the layer to develop cracks?
σh  µ  ( E = 200GPa, µ = 0.3)
∈h =  1 − 
E 2 (a) 0.057 MPa (b) 5.7 MPa
 PD  (c) 0.57 MPa (d) 57 MPa
 where σ h =  BHEL ET 2019
 2t 
Ans. (c) : Diameter (d) = 400 mm
δ D PD  2 − µ 
∴ =   Thickness (t) = 20 mm
D 2tE  2  Strain ( ∈ ) = 100 × 10-7
PD 2 E = 200 GPa, µ = 0.3
∴ δD = (2 − µ ) Pd
4tE ∈= (1 − µ )
1103. Where does the maximum hoop stress in a 4tE
P × 400
thick cylinder under external pressure occur? 100 × 10 −7 = [1 − 0.3]
(a) At the outer surface 4 × 20 × 200 × 10 3
(b) At the inner surface 10 −5 × 16 ×10 6 = p × 400 × 0.7
(c) At the mid-thickness
(d) At the 2/3rd outer radius 160
= P .. P = 0.57 MPa
OPSC AEE 2019, 2015 Paper-I 400 × 0.7 ,
Strength of Materials 358 YCT
1107. A welded steel cylindrical drum made of a 10 (a) 60 mm (b) 50 mm
mm thick plate has an internal diameter of 1.20 (c) 40 mm (d) 30 mm
m. Find the change in diameter that would be ESE 2019
caused by an internal pressure of 1.5 MPa. Ans. (d) : Maximum stress = Hoop stress in cylinder
Assume that Poisson's ratio is 0.30 and E = 200 σmax = σh
GPa (longitudinal stress, σy = PD/4t
circumferential stress , σx = PD/2t).
σu pd
=
(a) 4.590 mm (b) 0.459 mm FOS 2t
(c) 45.90 mm (d) 0.0459 mm 340 15 × 250
=
BHEL ET 2019 5 2× t
Ans. (b) : t = 27.57 mm ≃ 30 mm
1111. A spherical shell of 1.2 m internal diameter and
6 mm thickness is filled with water under
pressure until volume is increased by 400 × 103
mm3. If E = 204 GPa, Poisson’s ratio v = 0.3,
neglecting radial stresses, the hoop stress
developed in the shell will be nearly
t = 10 mm (a) 43 MPa (b) 38 Mpa
P = 1.5 MPa (c) 33 Mpa (d) 28 Mpa
d = 1.20 m = 1200 mm
ESE 2019
µ = 0.30
E = 200 GPa Ans. (a) : Given,
Pd 1.5 × 1200 d = 1.2 m = 1200 mm
σy = = r = 600 mm
4t 4 × 10 t = 6 mm
σ y = 45 MPa ∆v = 400 × 103 mm3
Pd 1.5 × 1200 E = 204 GPa = 204 × 103 MPa
σx = = = 90 MPa µ = 0.3
2t 2 × 10
σ σ ∆v 400 × 103 400 × 103
εx = x − µ y ∈v = = =
E E v 4 3 4
πr π(600)3
90  45  3 3
= − 0.3   ∈v = 4.42 × 10−4
20000  20000  Volumetric strain (∈v) = 3 × hoop strain
εx = 3.825×10–4 4.42 × 10−4 = 3 × ∈h
∆D ∈h = 1.47 × 10−4
εx =
D σ
∆D = ε x × D ∈h = n (1 − µ)
E
= 3.825 × 10–4 × 1200 = 0.459 mm
σh
1108. Pressure vessel is said to be thin cylindrical 1.47 × 10−4 = (1 − 0.3)
shell, if the ratio of the wall thickness of the 204 × 103
shell to its diameter is σh = 42.94 ≃ 43 MPa
(a) equal to 1/10 (b) less than 1/10 1112. Consider the following statements:
(c) more than 1/10 (d) none of these 1. In case of a thin spherical shell of diameter d
Gujarat PSC AE 2019 and thickness t, subjected to internal pressure
p, the principal stresses at any point equal
Ans : (b) A thin cylindrical shell is the ratio of the wall
thickness of the shell to its diameter is less than 1/10. pd
1109. The thickness of thin cylinder is determined on 4t
the basis of 2. In case of thin cylinders the hoop stress is
(a) Radial stress determined assuming it to be uniform across
(b) Longitudinal stress the thickness of the cylinder
(c) Circumferential stress 3. In thick cylinders, the hoop stress is not
(d) Principal shear stress uniform across the thickness but it varies
Gujarat PSC AE 2019 from a maximum value at the inner
circumference to a minimum value at the
Ans : (c) : The thickness of thin cylinder is determined outer circumference.
on the basis of circumferential stress. Which of the above statements are correct?
1110. The inner diameter of a cylindrical tank for (a) 1 and 2 only (b) 1 and 3 only
liquefied gas is 250 mm. The gas pressure is (c) 2 and 3 only (d) 1, 2 and 3
limited to 15 MPa. The tank is made of plain
ESE 2018
carbon steel with ultimate tensile strength of
2
340 N/mm , Poisson’s ratio of 0.27 and the Ans. (d) : (i) In case of a thin spherical shell of
factor of safety of 5. The thickness of the diameter d and thickness t, subjected to internal
cylinder wall will be. pressure p, the principal stresses at any point is given by

Strength of Materials 359 YCT

 pd 1115. A boiler shell 180cm diameter and plate
σ= thickness of 12mm is subjected to an internal
4t pressure of 1.2MN/m2. The hoop stress will be
(ii) In case of thin cylinders, the hoop stress is (a) 45 N/mm2 (b) 90 N/mm2
determined assuming it to be uniformly distributed (c) 135 N/mm 2
(d) 155 N/mm2
over the thickness of the wall, provided that the
thickness is small compared to radius. WBPSC AE, 2017
(iii) For a thick cylinder hoop stress is given by pd
Ans. (b) : σ h =
pR 2  r 2 + x 2  2t
σh = 2   (here r ≤ x ≤ R)
r − R2  x2  1.2 ×106 ×180 × 10−2
σh =
where r and R are inner and outer radius, respectively. 2 × 12 × 10−3
σh is maximum when x is minimum. = 90N / mm 2
σh is maximum at x = r
i.e. at inner surface. 1116. In case of thin walled cylinders the ratio of
Hence all the statement are correct. hoop stress to radial stress is
1113. A thin walled cylinder (diameter = D, length = 1 1
(a) (b)
L, thickness of cylinder material = t, modulus 2 4
of elasticity = E, Poission's ratio v) is subjected (c) 2 (d) None of the above
to fluid pressure (P) inside it. The total volume WBPSC AE, 2017
of fluid that an be stored in the cylinder will be: σ h hoop stress
π 2  PD Ans. (d) : =
(a) D L 1 + ( 5 − 4ν ) σR radial stress
4  4tE  σR = 0
π 2  PD
(b) D L 1 + ( 5 + 4ν ) • Radial stress in thin cylindrical shell and sphere
4  4tE  radial stress is neglected and assumed zero.
π 2 However its value at inner phase equal to outer
(c) D L pressure and outer phase it is equal to atmospheric
4 pressure is zero but for all practical purpose its
π 2  PD
(d) D L ( 5 − 4ν )  value taken zero from inner to outer surface.
4  4tE  1117. A cylinder is said to be thin if the thickness to
π 2  PD diameter ratio is less than
(e) D L 1 + (1 − ν )  (a)
1
(b)
1
4  4tE 
5 10
CGPSC AE 25.02.2018 1 1
Ans. (a) : We know that volumetric strain in thin walled (c) (d)
cylinder is given as: 15 20
Karnataka PSC Lect., 27.05.2017
dV Pd
= [ 5 − 4ν ] Ans (d) :
t
≤
1
––– Thin cylinder
V 4tE D 20
Pd
dV = [ 5 − 4ν ] × V t
>
1
––– Thick cylinder
4tE D 20
Then total volume of thin walled cylinder 1118. A thin cylinder of 4 m diameter has a wall
VTotal = V + dV thickness of 50 mm. It is subjected to an
π Pd π internal pressure of 2 MPa, What is the
= D2 ×L + [ 5 − 4ν ] × D 2 × L maximum shear stress induced in an element
4 4tE 4 considered on the inner surface?
π 2 
( 5 − 4ν )
Pd (a) 80 MPa (b) 41 MPa
VTotal = D × L 1 + (c) 40 MPa (d) 160 MPa
4  4tE 
APGENCO AE, 2017
1114. The Hoop stresses are acting across the
(a) Circumferential section Ans. (c) : In thin cylinder inner surface is subject to
(b) Longitudinal section three stresses
(c) Radial section pd
σc =
(d) None of the above 2t
RPSC Vice Principal ITI 2018 pd
HPPSC Asstt. Prof. 20.11.2017 σ( l ) = and radial pressure (Negligible)
4t
Ans. (b) : σ −0
The Hoop stress are acting across the longitudinal Max shear stress at inner surface = c
section. 2
The Longitudinal stress acting when two cross
pd 2×106 ×4
sectional areas of the cylinder are subjected to equal τ= =
and opposite forces the stress experienced by the 4t 4×50×10−3
cylinder. , τ = 40 MPa

Strength of Materials 360 YCT

1119. A 300 mm diameter basketball of 2mm wall (a) 10.5 mm (b) 12.5 mm
thickness has been inflated to a pressure of 100 (c) 14.5 mm (d) 16.5 mm
kPa. The value of hoop stress developed is (e) 18.5 mm
equal to (CGPCS Polytechnic Lecturer 2017)
(a) 10 MPa (b) 5 MPa
Ans. (b) : Pressure vessel diameter = 1 m = 1000 mm
(c) 3.75 MPa (d) 2.75 MPa
Steam pressure = 1.4 N/mm2
Punjab PSC (Lect.) 06.08.2017
Longitudinal stress (σℓ) = 28 MPa = 28 N/mm2
Ans. (c)
Thickness (t) = ?
Pd
Hoop stress = We know that
4t
Pd
100 × 1000 × 300 σℓ =
= 4t
4× 2
P × d 1.4 × 1000
107 × 3 t= =
= 4 × σℓ 4 × 28
4× 2
= 3.75 MPa t = 12.5 mm
1120. The purpose of longitudinal butt-joint in boiler 1124. A thin cylindrical pressure vessel of 500 mm
shell is : internal diameter is subjected to an internal
(a) to make cylindrical ring from steel plate pressure of 2 N/mm2. What will be the hoop
(b) to increase the length of boiler shell by stress if the thickness of the vessel is 20 mm?
connecting one ring to another. (a) 25 N/mm2 (b) 23 N/mm2
2
(c) to make diameter and length of boiler shell (c) 27 N/mm (d) 29 N/mm2
(d) to connect openings to shell CIL (MT) 2017 IInd Shift
HPPSC Asstt. Prof. 20.11.2017 Ans. (a) : Hoop stress in the cylindrical pressure vessel
Ans. (a) : The purpose of longitudinal butt joint in pd 2 × 500
boiler shell is to make cylindrical ring from steel plate. = = = 25 N / mm 2 .
1121. A seamless pipe of 600 mm diameter contains a 2t 2 × 20
fluid under a pressure of 3 N/mm2. If the 1125. A 2.0 m diameter penstock pipe carries water
permissible tensile stress be 100 N/mm2, the under a pressure head of 100 m. If the wall
minimum thickness of the pipe is thickness is 7.5 mm, the tensile stress in the
(a) 8 mm (b) 9 mm pipe wall, in MPa, is
(c) 11 mm (d) 10 mm (a) 65.3 (b) 130.5
Jharkhand Urja Vikas Nigam Ltd. AE 2017 (c) 231.0 (d) 1305.0
Ans. (b) : Given, d = 600 mm TSPSC AEE 28.08.2017 (Civil/Mechanical)
P = 3 N/mm2 Ans. (b) : D = 2 m
Permissible tensile stress = 100 N/mm2 H = 100 m
Pd t = 7.5 mm = 0.0075 m
Hoop stress =
2t P = ρgh (N/m2)
Hoop stress = permissible tensile stress
PD (1000 × 9.81× 100 ) × 2
Pd σt = =
= 100 2t 2 × 0.0075
2t = 130.8 × 106 N/m2
3 × 600
t= = 9 mm = 130.8 MPa
2 × 100 1126. In design of thin cylinder, the thickness of
1122. The reason why the pressure vessels are not cylinder plate is calculated from the relations:
made of rectangular shape is
PD PD
(a) These are difficult to fabricate (a) T = (b) T =
(b) It has been a practice to use cylindrical ft 2ft
vessels 2PD
(c) They are not economical (c) T = (d) None of these
ft
(d) None of the above
GPSC Asstt. Director of Transport 05.03.2017 Where:
D = Internal diameter of cylinder
Ans. (c) : Because gas cylinders contain high internal
pressure, so we cannot use shapes like rectangle and T = Thickness of cylinder plate
cube because there will be many stresses associated ft = Tangential or hoop stress
with such shapes. They are not economical. P = Internal pressure
1123. A thin cylinder pressure vessel of 1m diameter PTCUL AE 25.06.2017
generates steam at a pressure of 1.4 N/mm2. PD
What will be the wall thickness when Ans. (b) : T =
longitudinal stress does not exceed 28 MPa? 2ft

Strength of Materials 361 YCT

1127. A cylinder of internal diameter 200 mm is T 2000 × 103
subjected to internal pressure of 15 MPa. τ xy = = = 25.46 MPa
Determine the thickness of the cylinder if the 2πr t 2π(50) 2 (5)
2

material has a yield strength of 300 MPa. Take Pr 10 × 50

σx = = = 100 MPa
Factor of safety =2. t 5
(a) 10 mm (b) 5 mm Pr 100
(c) 7.5 mm (d) None of these σy = = = 50 MPa
2t 2
RPSC AE (GWD) 06.05.2016 Principal stress,
Ans. (a) : Diameter D = 200 mm
P = 15 MPa σx + σy  σ − σy 
2

σ1,2 = ±  x  + τ xy
2
Thickness t = ? 2  2 
Yield strength (Sys) = 300 MPa 2
100 + 50   100 − 50  + (25.46)2
Factor of safety (N) = 2 =  ±  
PD  2   2 
σhoop =
2t σ1,2 = 75 ± 35.68
Or σ hoop ≤ ( σ yt ) Max. principal stress ⇒ σ1 = 75 + 35.68 = 110.68 MPa
Min. principal stress ⇒ σ2 = 75 – 35.68 = 39.32 MPa
PD Sys 1130. The volumetric strain in case of thin spherical
=
2t N shell when subjected to internal pressure P is
15 × 200 300 Pd  1 1  Pd  1 
= (a)  −  (b) 1 − 
2× t 2 2tE  2 m  2tE  2m 
3000 Pd  1 3Pd  1 
t= (c) 1 −  (d) 1 − 
300 4tE  m  4tE  2m 
t = 10 mm (e) None of these
1128. A cylinder is called a thick wall cylinder when CGPSC AE 16.10.2016
the ratio of inner diameter of the cylinder to Ans. (e) : As we know,
wall thickness is For thin cylinder,
(a) between 20 to 30 (b) less than 15 Pd
(c) between 25 to 30 (d) greater than 25 ∈v = ∈ℓ + 2 ∈h ∴∈h = (1 − v)
Rajasthan Nagar Nigam AE 2016, Shift-II 4Et
Pd
Ans. (b) : When the ratio of the inner diameter of the = (5 − 4v)
1 1 4tE
cylinder to the wall thickness is less than to , the For spherical thin shell,
10 15
cylinder is called thick walled cylinder. ∈v = 3 ∈h ∴ ∈ℓ = ∈h
Example – Hydraulic cylinder 3Pd
High pressure pipe = (1 − v)
4tE
Gun barrels
1131. In thick cylinder, if hoop stress is plotted w.r.t.
1129. A thin cylinder of 100 mm internal diameter and
5 mm thickness is subjected to an internal 1
 2  , then the curve will be
pressure of 10 MPa and a torque of 2000 Nm. r 
What is the magnitude of the principal stresses? (a) Parabolic (b) Hyperbolic
(a) 75 MPa, 25 MPa (b) 75 MPa, 50 MPa (c) Linear (d) Elliptical
(c) 100 MPa, 50 MPa (d) 125 MPa, 50 MPa RPSC LECTURER 16.01.2016
(e) 125 MPa, 70 MPa Ans. (c) : In thick cylinder, if hoop stress is plotted
CGPSC Asstt. Workshop Supt., 17.07.2016 1
w.r.t.  2  , then the curve will be linear.
Ans. (*) : Given, d = 100 mm; r = 50 mm, t = 5 mm r 
P = 10 MPa 1132. A thick cylinder, having ro and ri as outer and
Torque, T = 2000 N-m inner radii, is subjected to an internal pressure
P. The maximum tangential stress at the inner
surface of the cylinder is
P ( ro2 + ri2 ) P ( ro2 − ri2 )
(a) (b)
ro2 − ri2 ro2 + ri2
2Pri2 P ( ro2 − ri2 )
(c) (d)
(r2
o − ri2 ) ri2
UPPSC AE 12.04.2016 Paper-I
Strength of Materials 362 YCT
Ans. : (a) For a thick cylinder, the hoop stress is 1137. A thin cylindrical shell of diameter (d), length
maximum at the inner surface and is given by. (ℓ) and thickness (t) is subjected to an internal

σ=
( d + d ) .p
2
o
2
i pressure (p). The ratio of the longitudinal
(d − d )
2
o
2
i
strain to hoop strain is
m−2 2m − 1
(a) (b)
σ = P.
(r + r )
2
o i
2
2m − 1 m−2
m−2 m+2
(r − r )
2
o i
2
(c) (d)
2m + 1 2m + 2
1133. Compound tubes are used in internal pressure ISRO Scientist/Engineer 11.10.2015
cases, for following reasons ESE 2004
(a) For increasing the thickness.
1
(b) For increasing the outer diameter o the tube. Ans. (a) : Longitudinal strain (∈ℓ) = [ σℓ − µσ h ]
(c) The strength is more. E
(d) It evens out stresses. 1 σ 
=  h − µσ h 
UPPSC AE 12.04.2016 Paper-I E 2 
Ans. : (d) Compound tubes are used in internal pressure σ
∈ℓ = h [1 − 2µ ]
cases, for it evens out stresses 2E
1134. In a thick cylinder subjected to internal 1
pressure, the hoop stress is ______ at the outer Hoop strain (∈h) = [ σ h − µσℓ ]
E
surface and is ______ at the inner surface
1 µσ 
(a) min, min (b) max, max = σ h − h 
(c) max, min (d) min, max E 2 
σh
APPSC AEE Mains 2016 (Civil Mechanical) ∈h = [2 − µ]
Ans. (d) : The hoop stress is minimum at outer surface 2E
& its maximum at inner surface (σh) Required ratio (∈ℓ/∈h) =
Longitudinal strain
Hoop strain
σh
[1 − 2µ]
= 2E
σh
[ 2 − µ]
2E
(1 − 2µ )
=
(2 − µ)
1135. If a steel tyre is heated and struck on a rigid  1
wheel, after cooling the tyre will be subjected to 1 − 2 × m   1
(a) Bending (b) Torsion =  ∵µ= 
 1  m
(c) Hoop stress (d) Compression  2 − m 
APPSC AEE Mains 2016 (Civil Mechanical)
m−2
Ans. (c) : If a steel tyre is heated and struck on a rigid =
2m − 1
wheel, after cooling the tyre will be subjected hoop
stress, due to shrinkage from outside & expansion on 1138. A spherical vessel with an inside diameter of 2
inside, due to temperature variation property varies. m is made of material having an allowable
stress in tension of 500 kg/cm2. The thickness of
1136. If the hoop stress in a thin cylinder is 24
a shell to withstand a pressure of 50 kg/cm2
N/sq.mm, then its longitudinal stress is equal to
should be-
(a) 36 N/sq.mm (b) 24 N/sq.mm
(a) 5 cm (b) 2.5 cm
(c) 12 N/sq.mm (d) 6 N/sq.mm
(c) 10 cm (d) 1.25 cm
APPSC AEE Mains 2016 (Civil Mechanical)
ISRO Scientist/Engineer (RAC) 29.11.2015
Ans. (c) : As we know, for cylinder
Ans : (a) For spherical vessel
Pd Pd
σΗ = & σL = PD
2t 4t σ h = σℓ =
4t
So, σΗ = 2σ L
50 × 200
24 = 2 × σ L 500 =
4× t
σ L = 12 N / mm 2 t = 5cm

Strength of Materials 363 YCT

1139. The cross section area of a hollow cylinder has Ans : (a) Lame's equation is used to determine
an internal diameter of 50 mm and a thickness cylindrical shell with closed end
of 5 mm. Moment of inertia of the cross-section b b
about its centroidal axis is σx = 2 + a px = 2 − a
(a) 2.848 × 105 mm4 (b) 4.294 × 105 mm4 r x
5
(c) 1.424 × 10 mm 4 5
(d) 1.647 × 10 mm 4 Where p x = Compressive Nature
(e) 3.294 × 105 mm4 σx= Tensile Nature
CGPSC AE 26.04.2015 Shift-I 1142. A thin-walled cylinder with closed ends for
which inner radius is 0.50 meter and outer
Ans. (e) :
radius is 0.52 meter, is subjected to internal
pressure 2 MPa. The absolute maximum
shearing stress on the inner surface of the
cylinder shall be
(a) 37.5 MPa (b) 75 MPa
(c) 26 MPa (d) 24 MPa
ISRO Scientist/Engineer 24.05.2014
Ans. (c) : Inner Radius r = 0.50 m
Outer Radius R = 0.52 m
Internal pressure P = 2 MPa
Given di = 50 mm Pd
do = 60 mm τ max =
4t
d
k= i So, outer diameter D = 2 × 0.52
d0 = 1.04 m
Thickness t=R–r
π 4   di  
4

Moment of inertia (I) = d 0 1 −    = 0.52 - 0.50

64   d o   t = 0.02 m
 
  50   4 2 × 1.04
π τ max =
= 604 1 −    4 × 0.02
64   60   = 26 MPa.
= 329209.37 = 3.29 × 105 mm4 1143. The initial hoop stress in a thick cylinder when
1140. A thin cylinder of internal diameter D = 1 m it is wound with a wire under tension will be :
and thickness t = 12 mm is subjected to internal (a) zero (b) tensile
pressure of 4 N/mm2. Determine hoop stresses (c) compressive (d) bending
developed. (HPPSC AE 2014)
(a) 83.33 N/mm2 (b) 83.33×10-3 N/mm2
Ans : (c) The initial hoop stress in a thick cylinder
(c) 166.67×10-3 N/mm2 (d) 166.67 N/mm2 when it is wound with a wire under tension will be
2
(e) 166.67 N/m compressive.
CGPSC AE 26.04.2015 Shift-I Analysis of Thick Cylinders/Lame's Theorem:-
Ans. (d) : Given * Lame's assumption
D = 1 m = 1000 mm (i) Material of shell is homogeneous
t = 12 mm (ii) Plane section of cylinder, perpendicular to
P = 4 N/mm2 longitudinal axis remains plane under pressure.
In a thin cylinder pressure vessel, hoop * Subjected to internal pressure
(circumferential) stress P  R 02 + R i2 
Pd (i) x = R i , σ h =
σ= R 02 − R i2
2t
2PR 2
4 × 1000 (ii) x = R 0 , σ h = 2 i 2
= R0 − Ri
2 ×12
= 166.67 N/mm2 1144. Shrinking a thick cylinder over another helps:
1141. Which of the following equation is used to (a) reduce the magnitude of tensile hoop stress
determine the thickness of thick cylindrical (b) reduce the difference between the higher and
shell with closed ends and made of brittle lower magnitude of tensile hoop stress
material? (c) remove the longitudinal stress
(a) Lame’s equation (d) reduce the cost
(b) Clavarino’s equation (HPPSC AE 2014)
(c) Birnie’s equation Ans : (b) Shrinking a thick cylinder over another helps
(d) Barlow’s equation reduce the difference between the higher and lower
UPRVUNL AE 2014 magnitude of tensile hoop stress.
Strength of Materials 364 YCT
1145. A thick cylinder of inner dia 'D', wall thickness 1147. A thin spherical shell of 1.5 m diameter is 8
t2 and length 'L' is sealed at its both ends with mm thick is filled with a liquid at the pressure
caps. The thickness of the cap is t1. Allowable of 3.2 N/mm2. The stress induced in the shell is
tensile yield stress = σy and allowable shear (a) 75 N/mm2 (b) 150 N/mm2
2
stress = τy. A gas is pumped this cylinder at (c) 200 N/mm (d) 180 N/mm2
pressure 'p'. The cap will yield in shear at TNPSC AE 2013
circumference of diameter 'D' when the gas Ans. (b) : Data given-
pressure applied is more than d = 1.5 m = 1500 mm
t = 8 mm
P = 3.2 N/mm2
Pd 3.2 × 1500
σl = σH = =
4t 4×8
σ H = 150 N / mm 2
1148. In thick cylinder, the radial stresses in the wall
thickness:-
(a) is zero
(b) negligible small
4t1τ y 8t1τ y
(a) (b) (c) varies from the inner face to outer face
D D (d) None of the above
4t2σ y 2t1τ y UKPSC AE-2013, Paper-I
(c) (d) Ans. (c) : In thick cylinder, the radial stresses in the
D D wall thickness varies from inner face [maximum] to
ISRO Scientist/Engineer 12.05.2013 outer face [zero].
Ans. (a) : Busting of pressure variation (P.V.) occurs 1149. A cylindrical shell of diameter 200 mm and
"circumferential/Axial" wall thickness 5 mm is subjected to internal
fluid pressure of 10 N/mm2. Maximum
shearing stress induced in the shell in N/mm2, is
(a) 50 (b) 75
(c) 100 (d) 200
UKPSC AE 2012 Paper-I
Ans. (d) : Data Given
Diameter of shell (d) = 200 mm
A = π Dt1 thickness (t) = 5 mm
pressure (P) = 10 N/mm2
Safe Condition then, maximum shear stress
F ≤ FR Pd
F = FR τ=
2t
π 10 × 200
P × D 2 = τ y × π × D × t1 =
4 2×5
4t1τ y τ = 200 N/mm2
or, P= 1150. The bursting pressure for a cold drawn
D
seamless steel tubing of 60 mm inside diameter
1146. A pipe of diameter 800 mm contains fluid with 2 mm wall thickness is (The ultimate
2
under a pressure of 2 N/mm . If the tensile strength of steel is 380 MN/m2)
2
strength is 100 N/mm , the thickness of the pipe (a) 25.33 MN/m2 (b) 24.33 MN/m2
is (c) 26.33 MN/m2 (d) 50.66 MN/m2
(a) 16 mm (b) 4 mm APPSC AEE 2012
(c) 8 mm (d) 10 mm Ans. (a) : Given, d = 60 mm, t = 2m, σh = 380 MN/m2
TNPSC AE 2013 pd σ × 2t 380 × 2 × 2
Ans. (c) : Data given - Solution- σ h = ⇒P= h =
2 2t d 60
d = 800 mm, P = 2 N/mm P = 25.33 MN/m 2
σH = 100 N/mm 2
1151. A solid thick cylinder is subjected to an
we know that external hydrostatic pressure 'P'. The state of
Pd stress in the material of the cylinder is
σH = represented as :
2t
2 × 800
t=
2 × 100 (a) (b)
t = 8 mm

Strength of Materials 365 YCT

 1156. A thin cylindrical shell is subjected to internal
pressure, such that hoop strain developed in
shell is 850 microstrain. If Young's modulus is
(c) (d) 100 GPa and Poisson's ratio is 0.3, what is hoop
stress developed in shell :
(a) 200 MPa (b) 100 MPa
APPSC AE 04.12.2012
(c) 50 MPa (d) None of these
Ans. (c) : A solid thick cylinder is subjected to an
PSPCL AE, 2012
external hydrostatic pressure 'P'. The state of stress
1
represented by – Ans. (b) : ε c = ( σc − µσl )
E
1  σ 
850 × 10−6 = × σ − 0.3 × c 
9  c
100 ×10  2 
pd
σc = 2σl =
In case of hydrostatic state of stress only pure 2t
compressive stress acts on all the faces. σc = 100 MPa
1152. Thin cylinders are frequently required to 1157. A thick cylindrical shell is subjected to internal
operate under pressures upto : pressure 'p', such that maximum hoop stress
(a) 5 MN/m2 (b) 15 MN/m2 developed in shell is 100 MPa. If external
2
(c) 30 MN/m (d) 250 MN/m2 radius of shell is two times the internal radius,
APPSC AE 04.12.2012 what is the magnitude of 'p'.
Ans. (c) : Thin cylinders are frequently required to (a) 100 MPa (b) 60 MPa
operate under pressure upto 30 MN/m2. (c) 40 MPa (d) None of these
1153. The bursting pressure for a cold drawn PSPCL AE, 2012
seamless steel tubing of 60 mm inside diameter B
with 2 mm wall thickness is (The ultimate Ans. (b) : σh = A + 2
strength of steel is 380 MN/m2) x
(a) 25.33 MN/m2 (b) 24.33 MN/m2 ri = Inner radius
(c) 26.33 MN/m 2
(d) 50.66 MN/m2 ro = Outer radius
APPSC AE 04.12.2012 Using boundary condition –
Ans. (a) : For thin cylinder - Pr 2
A= 2 i 2
Pd ro − ri
σc =
2t
P× r 2 × r 2
P = Bursting pressure = ? B = 2 i 2o
Given, ro − ri
d = Inside diameter of steel tube = 60mm Given, ri = r
t = Wall thickness of seamless steel tube = 2mm ro = 2r
σc = Circum perential stress (or) Bursting stress (or)
= Ultimate strength of steel = 380 MN/m2 P × r2 P
A= 2 2 =
4r − r 3
σ × 2t 380 MN / m 2 × 2 × 2mm
∴P= c = P × 4r × r
2 2
4 P× r 2
d 60mm B = =
P = 25.33 MN/m2 4r − r
2 2
3
1154. For a thick-walled shell, the diameter-thickness P 4P × r 2
5P
ratio is σh = + = (σh = 100 MPa)
3 3× r2 3
(a) less than 20 (b) greater than 20
5P
(c) equal to 20 (d) equal to 20 100 =
BPSC AE 2012 Paper - VI 3
Ans : (a) : For a thick walled shell the diameter- P = 60 MPa
thickness ratio is greater than 20 1158. A compound cylinder with inner radius 50 mm
1155. In a thick cylinder, radial stress at inner and outer radius 70 mm is made by shrinking
surface is one cylinder to the other cylinder. The junction
(a) independent of fluid pressure radius is 60 mm and the junction pressure is 11
(b) more than fluid pressure N/mm2. The maximum hoop stress developed
(c) less than fluid pressure in the inner cylinder would be ?
(d) equal to fluid pressure (a) 3.6 N/mm2 compression
BPSC AE 2012 Paper - VI (b) 3.6 N/mm2 tension
Ans : (d) : The radial stress for a thick walled cylinder (c) 7.2 N/mm2 compression
is equal and opposite to the gauge pressure on the inside (d) 7.2 N/mm2 tension
surface and zero on the outside. VIZAG STEEL MT 2011
Strength of Materials 366 YCT
Ans. (c) : P = 1.1 N/mm2 1161.
Junction radius, R2 = 60 mm
Inner radius, R1 = 50 mm
−2PR 22
Hoop stress, σh =
R 22 − R12 Principal stresses on the outside surface
element of a thin cylindrical shell subjected to
−2 × 1.1 × (60)2 internal fluid pressure as shown in the figure,
= are represented by
(60)2 − (50) 2
σh = 7.2N / mm 2 (Compressive)
1159. A thin cylinder contains fluid at a pressure of (a) (b)
30 kg/cm2. The inside diameter of the shell is 60
cm and the tensile stress in the material is to be
limited to 900 kg/cm2. The shell must have
minimum wall thickness of
(c) (d)
(a) 1 mm (b) 2.7 mm
(c) 10 mm (d) 9 mm
ISRO Scientist/Engineer 2011 UPSC JWM Advt. No.-50/2010
Ans. (c) : P = 30 kg/cm 2 Ans. (d) : As we know that,
d = 60 cm PD
Hoop stress, σ H =
σ = 900 kg/cm 2 2t
Pd PD
σ= Longitudinal stress, σ L = ⇒ σH > σ L
2t 4t
So,
Pd
Hoop stress (Circumferential) σ =
2t
30 × 60
900 =
2×t
30 × 60
t=
900 × 2
t = 1cm = 10mm 1162. A compound cylinder with inner radius 5 cm
1160. A long thin cylindrical shell is fabricated by and outer radius 7 cm is made by shrinking one
making longitudinal and circumferential weld cylinder on to the other. The junction radius is
6 cm and the junction pressure is 11 kg/cm2.
seams. The efficiencies of longitudinal and The maximum hoop stress developed in the
circumferential weld seams are 80% and 75% inner cylinder is :
respectively. Now this shell is subjected to (a) 36 kgf/cm2 compression
internal fluid pressure. The ratio of the larger (b) 36 kgf/cm2 tension
to smaller principal stresses induced in the (c) 72 kgf/cm2 compression
shell material is (d) 72 kgf/cm2 tension
15 APPSC IOF, 2009
(a) 2 (b)
8 −2P R 22
Ans. (c) : Hoop stress = 2
(c) 1.75 (d) 1.50 R 2 − R12
UPSC JWM Advt. No.-50/2010 R1 = 50 mm, R2 = 60 mm
Ans. (b) : We know that, for riveted & welded, P = 11 N/mm2
Hoop/circumferential stress, = 72 kgf/cm2 (Compression)
pd pd pd 1163. A thin cylindrical shell of mean diameter D and
σh = = = thickness t is subjected to internal fluid
2tηℓ 2 × 0.8 × t 1.6t pressure pi. The ends of this shell are rigidly
pd pd pd closed by flat circular plate. If the longitudinal
Given, longitudinal stress, σ L = = = stress in the shell is to be zero what external
4tηh 4 × 0.75 × t 3t pressure pe must be applied at the ends?
σh pd 3t 15 (a) 2pi (b) 1.5pi
So, = × = (c) p (d) 0.5pi
σL 1.6t pd 8 i
UPSC JWM Adv. No-16/2009
Strength of Materials 367 YCT
Ans. (c) : If the longitudinal stress in the shell is zero 1167. Which of the following is usually considered as
i.e., In equilibrium condition, the internal fluid pressure thin cylinders?
(Pi) is equal to external (Pe). (a) Boiler (b) Tanks
Pe = Pi (c) Steam pipes (d) All of the above
1164. A 10 m radius thin spherical tank is to be used WBPSC AE 2008
to store gas. If the wall thickness of the tank is Ans. (d) : Boilers, Tanks and steam pipes are usually
10 mm and the allowable tensile stress for the considered as thin cylinder.
material of the tank is 125 MPa, the maximum
possible gas pressure (neglecting radial stress) is 1168. Which of the following assumptions is made
(a) 0.25 MPa (b) 0.125 MPa while solving problems of thick cylinders using
(c) 0.5 MPa (d) 1 MPa Lame's theory
DRDO Scientists 2008 (a) the material is homogeneous and isotropic
Ans. (a) : For spherical tank hoop stresses : (b) the material is stressed within the elastic limit
pd (c) plane sections perpendicular to the
σ1 = ≤ 125 longitudinal axis of the cylinder remain plain
4t after the application of internal pressure
4t (d) all of the above
p ≤ 125 ×
d WBPSC AE 2008
4 ×10 Ans. (d) : Assumptions in Lame's theory for thick
or p ≤ 125 × cylinder–
20000
or p ≤ 0.25 MPa 1. Material is homogeneous and isotropic in which
hooke's law is valid.
pmax = 0.25 MPa
2. The plane section perpendicular to longitudinal
1165. A thin cylindrical pressure vessel with mean axis remain plane after fluid pressure is applied. It
diameter 10 m and wall thickness 20 mm is means there will be no bending i.e. longitudinal
subjected to an internal fluid pressure of 0.4 strain is constant.
MPa. If the yield strength of the material of the
cylinder is 200 MPa, the factor of safety 3. The material is stressed within the elastic limit and
according to maximum shear stress theory each fibre of material is free to contract and
(neglecting radial stress) is expand individually.
(a) 0.5 (b) 1 1169. When a thin cylinder shell is subjected to
(c) 1.5 (d) 2 internal fluid pressure. Which of the following
DRDO Scientists 2008 stress is developed in. its wall?
Ans. (d) : Maximum shear stress in thin cylindrical (a) circumferential stress (b) longitudinal stress
pressure vessel, (c) both of (A) and (B) (d) none of the above
PD 0.4 × 10 WBPSC AE 2008
τmax = =
4t 4 × 20 × 10−3 Ans. (c)
τmax = 50 MPa Pd
According to shear stress theory, σhoop or σcircumferential =
2t
σy σy
τmax = ⇒ FOS = σlongitudinal =
Pd
2 FOS 2 τmax 4t
200 where
FOS = =2
2 × 50 P = internal pressure
1166. In thick cylinders the radial stress in the wall d = diameter of thin cylinder
thickness t = thickness of cylinder
(a) zero 1170. If water in closed pipe freezes, the pipe will
(b) negligibly small (a) Rupture along a plane perpendicular to the
(c) not negligible and varies from the inner axis of pipe
surface to outer surface
(b) Rupture of zig-zag along the length
(d) any of the above
WBPSC AE 2008 (c) Rupture along weakest circumferential
section
Ans. (c)
(d) Rupture along a line running longitudinally
along the pipe
ISRO Scientist/Engineer 2007
Ans. (d) : The failure will take place along the
In thick cylinder the radial stress in the wall thickness longitudinal joint because of the circumferential/hoop
varies from the inner surface to outer surface. stress.
Strength of Materials 368 YCT
1171. In a thick cylindrical shell subjected to internal APPSC AEE 2012, APPSC AE 04.12.2012
fluid pressure, the state of stress at the outer RPSC AE GWD, 2011, TNPSC AE, 2008
surface is WBPSC AE 2007
(a) three-dimensional (b) two-dimensional Ans. (b) :
(c) isotropic (d) none of the above Conditions Effective length
UKPSC AE 2007 Paper -I
(i) Both end fixed ℓ/2
Ans. (b) : Two-dimensional
1172. The design of this cylindrical shells is based on: (ii) Both end hinge ℓ
(a) hoop stress (iii) One fixed, one hinge ℓ/ 2
(b) longitudinal stress (iv) One fixed, other free 2ℓ
(c) volumetric stress
(d) average of hoop and longitudinal 1175. For a long slender column of uniform cross
section, the ratio of critical buckling load for
J & K PSC Screening, 2006 the case with both ends clamped to the case
Ans. (a) : In thin cylindrical shell loop stress is more as with both ends hinged is :
compared to longitudinal stress so hoop stress is used (a) 1 (b) 2
for designing of thin cylinder. (c) 4 (d) 16
Pd Gujarat PSC AE 2019, BHEL ET 2019
Hoop stress σ h =
2t GPSC ARTO Pre 30.12.2018
Pd UP Jal Nigam AE 2016, RPSC 2016
Longitudinal stress σ L = TSPSC AEE 2015
4t BPSC Asstt. Prof. 29.11.2015
1173. A thin cylindrical shell has closed VIZAG STEEL MT 18.08.2013
hemispherical ends. The material properties VIZAG STEEL MT 10.06.2012
are Young's modulus (E) and Poisson's Ratio GATE 2012
(v). For no distortion to take place at the
Ans. (c) :
junction of cylindrical and hemispherical Ratio of critical buckling
portions, the ratio of shell thickness of
spherical portion to that to cylindrical portion Critical load with both ends clamped
=
is Critical load with both ends hinged
1– v 1+ v  π2 EI 
(a) (b)
2+ v 2–v  2 
Pcr  ℓ e clamped column
1– v 1+ v 1 =
(c) (d) Pcr  π2 EI 
2–v 2+ v 2  2 
OPSC Civil Services Pre 2006  ℓ e  hinged column
Ans. (c) : t1 = thickness in cylindrical portion
t2 = thickness in spherical portion  
circumferential strain in cylindrical portion  2 
 π EI 
Pd 1  1  ℓ 2
∈c1 = × 2 – 
4t1 E  m    
Pcr   2   clamped column
circumferential strain in hemispherical portion 1 =
1 Pd  1 
Pcr  π2 EI 
∈c2 = 1– 2  2 
E 4t 2  m   ℓ hinged column
For no distortion at junction ∈c1 =∈c2 4
=4 =
t2 m –1 1– v 1
= = 1176. Slenderness ratio of a column is defined as the
t1 2m – 1 2 – v ratio of its length to its
(a) least radius of gyration
7. Theory of Columns (b) least lateral dimension
(c) maximum lateral dimension
1174. A column of length ‘l’ is fixed at both the ends. (d) maximum radius of gyration
The equivalent length of the column is:- Nagaland PSC (CTSE) 2018, Paper-I
(a) 2 l (b) 0.5 l HPPSC Asstt. Prof. 18.09.2017
(c) 4 l (d) l Vizag Steel (MT) 2017, CILMT 26.03.2017
RPSC Lect. (Tech. Edu. Dept.) 13.03.2021, Paper-I OPSC AEE 2015 Paper-I
TNPSC AE 2018 TRB Asstt. Prof., 2012, APPSC AEE 2012
UPPSC AE 12.04.2016 UPSC JWM Advt. No.-52/2010
UKPSC AE-2013, Paper-I ESE 2003
Strength of Materials 369 YCT
Ans. (a) : Slenderness Ratio (S) = Stress due to axil load (σa)
σb = σa
Effective Length of column
S= M 4W
Least Radius of Gyration =
Z π ( D2 − d 2 )
Le 64 × W × e × D 4W
S= =
K min (
2π D − d
4 4)
π D2 − d 2 )
(
I min D2 + d 2
Where k= e=
A 8D
1177. If the diameter of a long column is reduced by 1179. In Rankine's formula, the material constant for
20 percent, the reduction in Euler buckling mild steel is
load in percentage is nearly:-
(a) 4 (b) 36 1 1
(a) (b)
(c) 49 (d) 59 9000 5000
HPPSC W.S. Poly. 2016 1 1
TSPSC AEE 2015 (c) (d)
UKPSC AE-2013, Paper-I 1600 7500
APPSC AEE 2012 RPSC Lect. (Tech. Edu. Dept.) 13.03.2021, Paper-I
APPSC AE 04.12.2012 Karnataka PSC AE (WRD) 31.07.2021
ESE 2000 Karnataka PSC AE, 10.09.2017
Ans. (d) : We know that GPSC Lect. Poly. 07.12.2014, APPSC AEE 2012
Ans : (d)
π EI min.
2
Pe = Material Value of 'a'
L2e wrought iron 1/9000
Pe ∝ I min cast iron 1/1600
Mild Steel 1/7500
π 4
Pe ∝ d Timber 1/750
64 1180. For a column of length L is fixed at both the
Pe ∝ d 4 ends, corresponding Euler's critical load is
Pe1 = kd 4 (a) π 2 EI / L2 (b) 2π 2 EI / L2
(c) 3π EI / L
2 2
(d) 4π 2 EI / L2
Pe2 = k ( 0.8 ) .d
4 4
ISRO Scientist/Engineer 22.04.2018
Pe2 = 0.4096 Pe1 APPSC AEE Mains 2016 (Civil Mech.)
ISRO Scientist/Engineer 2011
Then percentage reduction in Euler buckling ISRO Scientist/Engineer 2007
load OPSC Civil Services Pre 2006
Pe1 − Pe2 Ans. (d) : Euler's critical load
= × 100
Pe1 π 2 E I min
Pcr =
= kd 4
[1 − 0.4096] ×100 L2e
Le = effective length
kd 4
= 59% Both end fixed
1
1178. A short column of external diameter D and Le = l
internal diameter d carries an external load W. 2
The greatest ecentricity which the load can Critical load
have without producing tension on the cross
section of the column is-
π 2 E I min
P cr =
(a) (D+d)/8 (b) (D2+d2)/8 1 
2

2 2
(c) (D +d )/8D 2 2
(d) (D +d )/8d   l
2 
GPSC Engg. Class-II Pre-19.01.2020
4π 2 EI min
TSPSC AEE 28.08.2017 (Civil/Mechanical) Pcr =
ISRO Scientist/Engineer 12.05.2013 l2
ISRO Scientist/Engineer 2008 1181. Euler’s formula holds good only for
ISRO Scientist/Engineer 2006 (a) short columns
ESE 1999 (b) long columns
Ans. (c) : External diameter = D (c) both short and long columns
Internal diameter = d (d) weak columns
External load = W, ecentricity e = ? Sikkim PSC (Under Secretary), 2017
Stress due to bending moment (σb) Haryana PSC Civil Services Pre, 2014

Strength of Materials 370 YCT

 UKPSC AE-2013, Paper-I, BPSC AE 2012 Paper-VI
π2 EI π2 EI
APPSC AE 04.12.2012, APPSC AEE 2012 (a) (b)
Ans. (b) : L2 2L2
(1) Euler's Formula applicable only for long column πEI 2π2 EI
which fails due to buckling. (c) (d)
L2 L2
π2 EImin GATE 2018
Pb = APPSC AE Subordinate Service Civil/Mech. 2016
le2
APPSC AEE Screening Test 2016
(2) Rankine formula is applicable for both long and CGPSC 26th April Shift-I
short column means they consider both failure buckling APPSC AEE 2012, GATE 2006
as well as crushing.
Ans. (a) :
1 1 1 Eff.
= +
PR Pc Pb Crippling Eff. length
Description
load length as per
PR → Rankine load IS code
Pc → Crushing load
Pb → Buckling load.
1182. Match List I with List II and select the correct 2π2 EI leff =
l
0.8 l
2 2
answer using the codes given below the lists : L
List I List II
(End conditions (Lowest critical
of columns) load)
(A) Column with (I) (π2EI)/L2 4π2 EI l
leff = 0.65 l
both ends hinged 2 2
L
(B) Column with (II) (2π2EI)/L2
both ends fixed
(C) Column with one (III) (4π2EI)/L2
end fixed and π2 EI leff = 2l 2l
other end hinged 4L2
(D) Column with one (IV) (π2EI)/4L2
end fixed and
other end free
(a) A-I, B-II, C-III, D-IV π2 EI leff = l l
(b) A-III, B-II, C-I, D-IV L2
(c) A-I, B-III, C-II, D-IV
(d) A-II, B-IV, C-III, D-I
1184. A column has a rectangular cross-section of 10
Punjab PSC (Lect.) 06.08.2017 mm × 20 mm and a length of 1 m. The
TRB Asstt. Prof. 2012 slenderness ratio of the column is close to :
APPSC IOF, 2009, ESE 1995 (a) 200 (b) 346
Ans. (c) : Column with both end hinged ⇒ L = Le ⇒ (c) 625 (d) 1000
π2 EI Vadodara Municipal Corp. DEE 2018
2 GPSC Asstt. Director of Transport 05.03.2017
L
UP Jal Nigam AE 2016
L 4π2 EI GATE 2011
Column with both ends fixed, L e = ⇒
2 L2 Ans. (b) : Cross sectional area (A) = 10 × 20 = 200
Column with one end fixed and other end hinged, mm2
L 2π2 EI
Le = ⇒
2 L2
Column with one end fixed and other is free, Le = 2L ⇒
π2 EI
4L2
1183. When both ends of the column are hinged the
crippling load at which the column just 10 × 203
buckled is I xx = = 6666.67 mm 4
12
(Where E is Young's modulus of column, I is
20 × 103
the Moment of inertia of column and L is I yy = = 1666.67 mm 4
effective length of column) 12
Strength of Materials 371 YCT
 Least radius of gyration, Ans. (a) : We know that,
I min 1666.67 π2 EImin
K min = = Pe =
A 10 × 20 L2e
= 2.88 mm Where, Le = effective length
Slenderness ratio (λ) is given by :
Imin = Moment of Inertia about centroidal axis.
Effective length of column (ℓ e )
= 1
Least radius of gyration ( K min ) ∵ Pe ∝ 2
Le
1000
λ= = 346.409 (P ) (L ) 2 (L) 2
2.88 Then ratio, e 2 = e 12 ⇒
1185. According to Euler's column theory, the (Pe )1 (Le ) 2 (L / 2) 2
crippling load of a column of length (l), with (Pe)2 = 4(Pe)1
one end is fixed and the other end is hinged is = 4 × 10 kN= 40 kN
(a) π 2 EI/l 2 (b) π 2 EI/4l 2 1188. The section of a column is circular with
(c) 2π 2 EI/l 2 (d) 4π 2 EI/l 2 diameter d. What is the diameter of its core :
Karnataka PSC AE (WRD) 31.07.2021 (a) d/2 (b) d/3
Nagaland PSC CTSE 2018 Paper-I (c) d/4 (d) None of these
APPSC AEE MAINS 2016, PAPER-III
PSPCL AE, 2012
WBPSC AE, 2007
APPSC AE 04.12.2012
Ans. (c) :
End Condition Euler's crippling load OPSC Civil Services Pre-2006
Both ends hinged Ans. (c) :
π2 EI
( le = l ) 2×d d
ℓ2 Diameter of core or kernel= = for circular
8 4
Both ends fixed 4π2 EI  l section
 le = 
ℓ2  2
one end hinged and other 2π2 EI  l 
end fixed
2  le = 
ℓ  2
one end fixed and other π2 EI
end free ( le = 2l )
4ℓ 2
1186. What is the expression for crippling load for a
column of length L, with one end fixed and
other end free?
Core or kernel in column is region of cross section,
(a) P = π 2 EI / 4 L2 (b) P = π 2 EI / L2
when eccentric force is applied with in the region, there
(c) P = 2π 2 EI / L (d) 4π 2 EI / L2 is no tensile force any of the fibre of column. Column
UPRVUNL AE 21.08.2016 are designed to bear compressive load only.
ESE 2012 1189. Four vertical columns of the same material,
ISRO Scientist/Engineer 2009, 2008 height and weight have the same end
ESE 2006, GATE 1994 conditions. The buckling load will be largest
Ans. (a) : Crippling load = Pe for a column having.
Column length = L (a) Square cross section
π 2 EI (b) Hollow circular cross section
P= (c) Circular cross section
( Le ) 2
Le = 2L (one end fixed and other end free) (d) I-section
RPSC Lect. (Tech. Edu. Dept.) 13.03.2021, Paper-I
π 2 EI π 2 EI
P= = MPPSC State Forest Service Exam, 2014
(2 L)2 4 L2 ESE 2009
1187. The buckling load for a column pinned at both Ans. (b) : As we know,
ends is 10 kN. If the ends are fixed, the π2 EImin
buckling load changes to P=
(a) 40 kN (b) 2.5 kN L2
(c) 5 kN (d) 20 kN So, P ∝ Imin
Karnataka PSC AE, 10.09.2017 In this inertia of the hollow circular cross section is
TRB Asstt. Prof., 2012 minimum, can carry least weight, have maximum
WBPSC AE 2003, GATE 1998 buckling.

Strength of Materials 372 YCT

1190. The Euler load for a column is 1000 kN and π2 EI min
crushing load is 1500 kN. The Rankine load is Ans. (b) : Pc =
equal to: L2e
(a) 600 kN (b) 1000 kN 1
(c) 1500 kN (d) 2500 kN Pc ∝
L2e
MPSC Poly. Lect. 2014
Pc2 L2e1 2

=  
MPPSC State Forest Service Exam, 2014 1
=
Ans. (a) : As we know, Pc1 L2e2  2 
1 1 1
= + Pc1
PR PE PC Pc2 =
4
PE × PC 1000 × 1500
PR = = 1194. A long column of length l is hinged at its ends.
PE + PC 1000 + 1500
The equivalent length in buckling is :
PR = 600 kN
(a) 2l (b) l
1191. Rankine Gordon formula for buckling is valid (c) l/2 (d) l/4
for :
(a) long column UKPSC AE 2012 Paper-I
(b) short column UPSC JWM Advt. No.-52/2010
(c) short and long column Ans. (b) : The equivalent Length (Le) of a column
(d) very long column which is hinged at both ends, if length of column is 'l'
APPSC AEE 2019, APPSC IOF, 2009 ∴ Le = l
σc A 1195. A column that fails due to direct stress is called
Ans. (c) : P = σ A = 2 (a) short column (b) long column
1 + a  
L
(c) Medium column (d) Slender column
k
σc = Crushing stress TSPSC Manager (Engg.) HMWSSB 12.11.2020
A = Cross sectional area APPSC AE Subordinate Service Civil/Mech. 2016
a = Rankine's constant Ans. (a)
L = Effective length • Short column fails due to direct stress and failure is
k = Slenderness ratio crushing failure.
It is valid for both short and long column buckling. • Long column fails due to bending stresses and
1192. A column has an effective length twice the failure is buckling failure.
actual length, then the ends of column are: • Intermediate medium column fails due to
(a) Fixed at both the ends combination of both direct stresses and bending
(b) Fixed at one end and free at the other end stress and mode of failure is crushing and buckling.
(c) Hinged at both the ends
(d) Fixed at one end and hinged at the other end 1196. The buckling load will be maximum for a
CIL MT 27.02.2020 column, if
Assam PSC AE (PHED) 18.10.2020 (a) One end of the column is fixed and other end
Ans. (b) : A column has an effective length twice the is hinged
actual length, then the ends of column are fixed at one (b) Both ends are fixed
end and free at the other end. (c) One end is fixed and other end is free
(d) Both ends are hinged
Kerala PSC IOF 19.04.2016
Kerala LBS Centre For Sci. & Tech. Asstt. Prof. 2014
APPSC AEE 2012
Ans. (b)
π2 EImin
Both end Hinged =
L2
1193. If the length of a column is doubled, the
critical load becomes : 4π2 EImin
Both end fixed =
(a) 1/2 of the original value L2
(b) 1/4 of the original value
(c) 1/8 of the original value 2π2 EImin
One end hinged and one end fixed =
(d) 1/16 of the original value L2
HPPSC Asstt. Prof. 18.11.2016 π2 EImin
APPSC AE Subordinate Service Civil/Mech. 2016 One end fixed and other is free =
GATE 1998 4L2
Strength of Materials 373 YCT
1197. In the theory of columns, Rankine's constant is Ans : (d) Euler's formula for a column one end fixed
given by: and other end free
σ σ
(a) 2 c (b) 2c π2 EI π2 EI
π E π Pe1 = = ( le = 2l )
( 2ℓ )
2
4ℓ 2
σc σ2
c
(c) (d) 2 Euler 'sformula for a column both end fixed
E πE
Oil India Ltd. Sr. Engg. (Production) 30.11.2019 4π2 EI  l
Pe1 =  le = 
APPSC AEE 2012 l2  2
Ans. (a) : Design of intermediate column- Intermediate π2 EI
column fails by both crushing and buckling. Rankine's 4 × 10 = ___________ ( i )
l2
formula is used in the design of intermediate column. It
4π2 EI
is an imperical equation (Relation): Pe2 =
If P = Rankine's crippling load l
1 1 1 Pc + Pe ( Pe )2 = 4 × 40
= + =
P Pc Pe Pc Pe ( Pe )2 = 160kN
Pc Pe P π2 EI 1200. Match List I with List II and select the correct
⇒ P= = c but Pc = σc A & Pc = 2 answer using the odes given below the Lists:
Pc + Pe 1 + Pc Le List-I List II
Pe A Both ends 1 L
hinged
σc A   Le  
2

∴ P= ∵ I = Ak & λ =   
2 B One end fixed 2 L/ 2
σA   k   and other end
1 + 2c
π EI free
L2e C One end fixed 3 L/2
and other pin-
σc A joined
P=
σA D Both ends fixed 4 2L
1+ 2 c 2
π EAk Code:
L2e A B C D
(a) 1 3 4 2
σc A (b) 1 4 2 3
P=
σ (c) 3 1 2 4
1+ 2 c 2 (d) 3 1 4 2
π Eλ
APPSC AEE 2012, ESE 2000
σc A σ
P= ⇒ α = 2c Ans. (b) : Joined Equivalent length (le)
1 + αλ 2 πE A. Both end hinged - L
Where α is Rankine constant which depends on type of B. One end fixed and - 2L
material of column and λ is slenderness ratio. other end free
1198. A column is said to be a short column, when C. One end fixed and the - L/ 2
(a) its length is very small other pin-joined
(b) its cross-sectional area is small D. Both end fixed - L/2
(c) the ratio of its length to the least radius of 1201. The safe compressive load on a hollow cast
gyration is less than 80 iron column (one end fixed other hinged) of
(d) the ratio of its length to the least radius of 150 mm external diameter, 100 mm internal
gyration is more than 80 diameter and 10 m length is (Use Euler's
Vizag Steel (MT) 2017 formula with a factor of safety of 5, and E = 95
JPSC AE 2013, Paper-V GN/m2)
(a) 74.8 kN (b) 149.6 kN
Ans : (c) : A column is said to be short column, when
(c) 37.4 kN (d) 299.2 kN
the ratio of its length to the least radius of gyration is
APPSC AE 04.12.2012, APPSC AEE 2012
less than 80.
Ans. (a) : Given- Do = 150 mm, Di = 100 mm, L = 10 m
1199. The bucking load for a column one end fixed F.O.S. = 5, E = 95 GN/m2
and other end free is 10kN. If both ends of this M.O.I.
column is fixed, then what would be the π π
buckling load capacity of this column ? I = ( Do 4 – Di 4 ) = ( 0.154 – 0.14 ) = 1.99418 × 10 –5 m 4
(a) 10 kN (b) 20 kN 64 64
(c) 80 kN (d) 160 kN L 10
Equivalent length le = = = 7.071m
MPPSC AE 2016, PSPCL AE, 2012 2 2
Strength of Materials 374 YCT
 π2 EI π2 × 95 × 109 × 1.99418 ×10 –5
Buckling load Pe = =
le 2 ( 7.071)
2

4
Euler buckling load Pe = 37.4×10 = 374 kN
Pe 374
For safe compressive load Pc = =
F.O.S. 5
Pc = 74.8 kN
1202. What is the cause of failure of a short MS strut
under an axial load?
(a) Fracture stress (b) Shear stress
(c) Buckling (d) Yielding
TSPSC AEE 28.08.2017 (Civil/Mechanical)
CSE Pre-2007
Ans. (d) : The cause of failure of a short MS strut under
an axial load is yielding.
1203. If one end of a hinged column is made fixed
and other end free. how much is the critical
load to the original value?
(a) Four times (b) One-fourth
(c) Half (d) Twice
UPRVUNL AE 07.10.2016
BPSC POLY. TEACH 2016
π2 EI JPSC AE 10.04.2021, Paper-II
Ans : (b) Euler load =
le2 Ans. (c) : For rectangular section
Case– I
Initially condition [Both ends hinged]
π2 EI
(P1)=
le2
Case - II
One end fixed and other and free
π 2 EI π 2 EI P1
P2 = = =
( 2l )
2
4l 2 4
P1
P2 =
4
1204. The effective length of the column of length 2.5
m, when one end is fixed and other end is free
is
h
(a) 5 m (b) 2.5 m e xx =
(c) 1.25 m (d) 1.77 m 6
GPSC DEE, Class-2 (GWSSB) 04.07.2021 b
e yy =
Ans. (a) : Effective length for one end fixed and other 6
free, Le = 2 × L For Circular section –
= 2 × 2.5
=5m
1205. The core of a section in a strut within which
the load P will induce compression over the
entire section for the rectangular section will
be :

1206. Euler's theory assumes that failure within a

column is solely due to _______
(a) torsion (b) buckling
(c) sagging (d) bending
UPRVUNL AE 05.07.2021
Strength of Materials 375 YCT
Ans. (b) : Euler's theory assumes that failure within a Ans. (a) : For both ends hinged,
column is solely due to buckling. π2 EI
1207. ______ is elastic instability which results in ( Pcr )1 = 2 .....(i)
L
sudden large lateral deflection. For both ends clamped,
(a) Sagging (b) Fatigue
4π2 EI
(c) Buckling (d) Failure ( Pcr )2 = 2 .....(ii)
UPRVUNL AE 05.07.2021 L
Ans. (c) : Buckling is elastic instability which results in ( Pcr )1 1
So, = = 0.25
sudden large lateral deflection. ( Pcr )2 4
1208. For long column, which of the following
1212. If the length of a column subjected to
relations holds true?
compressive load is increase by three times its
(a) Crushing load is negligible as compared to
original length, the critical buckling load
buckling load becomes
(b) Crushing load is equal to buckling load
(a) 1/3 of the original value
(c) Crushing load is more than buckling load
(b) 3 times the original value
(d) Crushing load is less than buckling load
(c) 1/9 of the original value
UPRVUNL AE 05.07.2021
(d) 1/27 of the original value
Ans. (c) : Crushing load is more than buckling load
ISRO Scientist/Engineer 12.01.2020
1209. The assumptions made in Euler's column
1
theory is that Ans. (c) : Critical buckling load Pe ∝ 2
(a) the failure of column occurs due to buckling L
alone Le2 = 3Le1
2
(b) the length of column is very large as Pe2  L1   1 
2

compared to its cross-sectional dimensions =     =

Pe1  L 2   3 
(c) the column material obeys Hooke's law
(d) All of the above P
Pe2 = e1
Assam PSC AE (IWT) 14.03.2021 9
Ans. (d) : Assumptions Made in Euler's Theory of 1213. While applying Euler's theory of columns to
Column– calculate the crippling load, equivalent length
(1) The failure of column occurs due to buckling alone. calculations are made by referencing ______.
(2) The length of column is very large as compared to (a) column with both ends are fixed
its cross-sectional dimensions. (b) column with one end fixed and other end
(3) The column material obeys Hooke's law. hinged
(c) column with one end fixed and other free
1210. The relation between equivalent length (L) and
(d) column with both ends hinged
actual length (l) of a column whose one end is
VIZAG MT, 14.12.2020
fixed and other end is hinged is given by :
Ans. (d) : While applying Euler's theory of columns to
(a) L = l (b) L = l/ 2 calculate the crippling load, equivalent length
(c) L = l/2 (d) L = 2l calculations are made by referencing column with both
RPSC ACF & FRO, 26.02.2021 ends hinged.
Ans. (b) 1214. A structural member subjected to an axial
Conditions of columns Equivalent length compressive force is called
Both hinged Le = l (a) beam (b) column
(c) frame (d) strut
One hinged and other fixed l
Le = Assam PSC AE (PHED), 18.10.2020
2 ESE 2008
Both fixed l Ans. (d) : A structural member subjected to an axial
Le = compressive force is called strut.
2
As per definition strut may be horizontal, inclined or
One fixed and other free Le = 2l
even vertical.
1211. For a long slender column of uniform cross The vertical strut is called a column.
section, the ratio of critical buckling load for Beam- Beam is a structural element that primarily
the case with both ends hinged to the case with resists loads applied laterally to the beam axis.
both ends clamped is Frame- The frame structure is the combination of
(a) 0.25 (b) 4.0 beams column and slab. Rigid structural frame refers to
(c) 0.125 (d) 0.5 the frames which are capable in resisting the
ISRO Scientist/Engineer 12.01.2020 deformation.
Strength of Materials 376 YCT
1215. The buckled shape (node shape) of a pinned- Ans : (a) : Using Euler's formula critical buckling load
pinned end column in the first node of is given by,
buckling
π 2 EI min
(a) is a half period sine wave Pcr =
(b) is a full period sine wave ℓ e2
(c) is a quarter period sine wave Given, ℓ = 2m
(d) none of the above E = 13 GPa = 13 × 109 Pa
Haryana PSC AE (PHED) 05.09.2020, Paper-II Pcr = 250 kN = 250 × 103 N
Ans. (a) : Deflect position is a half period sine wave. Imin = ?
π 2 ×13 ×10 9 × I min
250 ×10 3 =
( 2) 2
Imin = 7.79 × 10 m4 -6

= 7.8 × 10-6 m4
1219. The slenderness ratio of a 4 m column with
fixed ends having a square cross-sectional area
of side 40 mm is :
(a) 173 (b) 17.3
(c) 1.73 (d) 100
BHEL ET 2019
1216. The slenderness ratio in the case of column Ans. (a) : Given A = 40 mm × 40 mm
with length L, area moment of inertia l and
cross-sectional area A, is given by
I L
(a) L (b) L
A A
A LA
(c) L (d) L
I I
l=4m
Haryana PSC AE (PHED) 05.09.2020, Paper-II
Le
Length slenderness ratio =
Ans. (c) : Slenderness ratio = k
Radius of Gyration effective length of column = Le = αL
and it is unitless. where α = length fixity coefficient
value 'α' for fixed end column-
I
Radius of gyration (k) = 1
A α=
2
A 1
Slenderness ratio = L × so, Le = × 4 = 2m
I 2
1217. The critical load in buckling of columns L
Slenderness ratio (s) = e
(a) can be increased by increasing the strength of k
the column material
(b) can be increased by decreasing the strength I min a4
k= =
of column material A 12
(c) is independent of the strength of the column a2
material
a2
(d) none of the above =
Haryana PSC AE (PHED) 05.09.2020, Paper-II 12
Ans. (a) : Critical load in buckling of column 40 × 40 × 10 −6
=
π2 EI 12
Pc =
le 2 1600
=
Pc ∝ E 12 × 10 6
k = 0.01154 m
1218. A 2 m long pin ended column having Young's
L 2
modulus (E) equal to 13 GPa can sustain 250 (s ) = e = = 173.310
kN Euler's critical load for buckling. The k 0.01154
permissible cross sectional size (I) of the 1220. Which of the following statements is correct?
column will be (a) Euler's buckling load increases with increase
(a) 7.8 × 10–06 m4 (b) 3.9 × 10–06 m4 in effective length.
(c) 1.95 × 10–06 m4 (d) 0.975 × 10–06 m4 (b) Buckling load of a column does not depend
Gujarat PSC AE 2019 on its cross-section
Strength of Materials 377 YCT
 (c) If free end of a cantilever column is propped Ans. (c) : Euler's formula holds good for long columns
then the buckling load increases only
(d) Two geometrically identical columns made
of different material have same buckling for a short column
load. ℓ
APPSC AEE SCREENING 17.02.2019 < 100
R min
Ans. (c) : If the free end is supported by a propped, the
effective length of column decreases. The load carrying equivalent length of column with one end fixed and
capacity increase. other end hinged is
1 L
PB ∝ ℓe =
Le 2
1221. Euler's formula for a mild steel column is not 1225. Which one of the following columns has
valid if the slenderness ratio is
effective length twice the value of actual
(a) 60 (b) 90
length?
(c) 100 (d) 120
Nagaland PSC (CTSE) 2018, Paper-I (a) Hinged-Hinged column
Ans. (a) : As we know, when slenderness ratio is less (b) Fixed-Fixed column
than 80 it is a short column from Euler's formula for (c) Fixed-Hinged column
mild steel column is not valid if the slenderness ratio is (d) Fixed-Free column
less than 80.
TSPSC AEE 28.08.2017 (Civil/Mechanical)
1222. What is the mode of failure of a short mild
steel column (having slenderness ratio less Ans. (d) : fixed-free column
than 10) under axial compressive load?
(a) Fracture (b) Buckling
(c) Yielding (d) Both (b) and (c)
RPSC AE 2018
Ans. (c) : Yielding is the mode of failure of short mild
steel column under axial compressive load because
short column (SR < 30) always fail in crushing.

1226. Critical Euler bucking load for a long column

of diameter D was evaluated as P. If the
diameter of the section is reduced to D/2, what
In this question, given material of short column is mild is the load carrying capacity of the modified
steel so fracture or bucking failure not occur because column ?
fracture failure related to brittle material and buckling (a) P/2 (b) P/4
failure occurs in long columns (SR > 120)
(c) P/8 (d) P/16
1223. A Column of rectangular section (Ixx > Iyy) is
subjected to an axial load. What is the axis TSPSC AEE 28.08.2017 (Civil/Mechanical)
about which the column will have a tending to π2 EI
buckle? Ans. (d) : Pcr =
(a) X–X ℓ 2e
(b) Y–Y Pcr ∝ I
(c) The diagonal of the section
(d) X–X or Y–Y axis without any preference Pcr ∝ D4
RPSC Vice Principal ITI 2018 Pcr1 D4
Ans. (b) : Column will buckle around the axis with the =
Pcr2  D 4
lowest moment of Inertia.  
1224. Which one of the following statements is 2
correct? P
(a) Euler's formula holds good only for short Pcr2 =
16
columns
(b) A short column is one which has the ratio of 1227. Column C1, has both the ends hinged while the
its length to least radius of gyration greater column C2 has one end hinged and other end
than 100 fixed. What is the ratio of the critical load for
(c) A column with both ends fixed has minimum C1 to that of C2 according to the Euler's
equivalent length or effective length formula?
(d) The equivalent length of column with one
end fixed and other end hinged is half of its (a) 2 (b) 1/2
actual length (c) 4 (d) 1/4
TSPSC AEE 28.08.2017 (Civil/Mechanical) TSPSC AEE 28.08.2017 (Civil/Mechanical)
Strength of Materials 378 YCT
 1 Ans. (d) : The critical buckling load (Pcr) for a column
Ans. (b) : Pcr ∝ having length L, modulus of elasticity E and second
ℓ 2e moment of cross sectional area I is loaded centrally both
2
 L  ends hinged.
( Pcr )hinged −hinged   nπ2 EI π2 EI
=
2 n = 1, Pcr = = 2
= 1: 2
( Pcr )hinge−fixed ( L)
2 L2 L
1231. For thin cylindrical shell structures loaded in
1228. In an axially loaded compressive member with compression, the design is based on
a circular cross-section of radius, r, what is the (a) Yield strength of material
radius of the core section which is proof (b) Ultimate strength of material
against tensile stress? (c) Buckling strength of the structure
(a) r/2 (b) r/3 (d) Shear strength of the structure
(c) r/4 (d) r/6 ISRO Scientist/Engineer 17.12.2017
TSPSC AEE 28.08.2017 (Civil/Mechanical) Ans. (c)
Ans. (c) : Core section which is proof against tensile (1) Yield strength is the point where the material tends
D to deform. For the structures that should not deform
stresses in case of circular reaction is of diameter on usage, the design is based on yield strength of
4
material.
D
∴ Core section radius will be (2) Brittle materials generally fail at the ultimate
8 strength of material.
if radius of circular section is r' then radius of core (3) Buckling under axial compression is normally the
section will be controlling design consideration for thin cylindrical
2r r structures.
r' = = Thin and long cylindrical structures fail before the
8 4
yield strength of the member is reached.
1229. A column of length 4 m, an area of cross- (4) Shear strength of a component is important for
section 2000 mm2. moments of inertia, Ixx = 720 designing the dimensions and materials to be used
cm4, Iyy = 80 cm4, is subjected to a buckling for the manufacture or construction of the
load. Both the ends of the column are fixed. component e.g. beams, plates or bolts.
What is the slenderness ratio of the column?
(a) 200 (b) 120 1232. In designing a connecting rod, it is considered
like.........for buckling about X-axis.
(c) 100 (d) 80
(a) both ends fixed
TSPSC AEE 28.08.2017 (Civil/Mechanical) (b) both ends hinged
Ans. (c) : Length of column L = 4 m area of cross – (c) one end fixed and the other end hinged
section = A = 2000 mm2 (d) one end fixed and the other end free
Ixx = 720 cm4 GPSC EE Pre, 28.01.2017
Iyy = 80 cm4
Ans. (b) : In designing a connecting rod, it is
L
ℓe fixed-fixed = = 2m considered like both end hinged for buckling about x-
2 axis.
ℓ
Slenderness ratio λ = e
k min
Imin
K min =
A
80 cm 4
=
2000 × 10−2 cm 2
80cm 4 1233. Which of the following are correct?
= = 2cm
20cm2 P. Slenderness ratio is defined as the ratio of
the length of the column to the radius of
ℓ 200cm
∴λ = e = = 100 gyration.
k min 2cm Q. Euler's formula is applicable for short
1230. A pin ended column of length L modulus of columns
elasticity E and second moment of the cross R. Rankine's constant for a mild steel column
sectional area I is loaded centrally by a is 1/7500
compressive load P. The critical buckling load (a) Only P and Q are correct
(Pcr) is given by (b) Only Q and R are correct
(a) E/π2L2 (b) π2EI/3L2 (c) Only P and R are correct
(c) πEI/L2 (d) π2EI/L2 (d) P, Q and R are correct
TSPSC AEE 2017 APGENCO AE, 2017
Strength of Materials 379 YCT
Ans. (c) : Euler formula is used for long column which Ans. (c) : For short column
fails due to buckling only. ℓ
< 8, S.R. < 32
Rankine formula is used for long as well as short d
column. Medium column
1234. If both the length and the diameter of a steel ℓ
8 ≥ ≤ 32, 32 < S.R. < 120
column are doubled, the elastic buckling load d
of the column will increase by a factor of Long column
(a) 2 (b) 4 ℓ
> 32, S.R. > 120
(c) 8 (d) 16 d
GPSC Asstt. Director of Transport 05.03.2017 1237. The equivalent length of a column supported
π2 EI firmly at one-end and free at other end is
Ans. (b) : Pcr = 2
Lnew = 2 load, Dnew = 2D (or)
(L) A compression member effectively held in
I π position and restrained in direction at one end
Pcr ∝ I = D4 but not held in position or restrained in
(L) 2 64
4
direction at the other end. If it's actual length
4
D (2D) is L, then its effective length is equal to
Pcr ∝ ⇒ Pcr ∝
(L) 2 (2L) 2 (a) 2L (b) 0.7 L
Pcr new = 4 Pcr old (c) L (d) 0.5L
Nagaland PSC CTSE 2017, Paper-I
1235. Buckling loads (Pi) of 4 columns of equal length
APPSC AEE Mains 2016 (Civil Mechanical)
and cross section, but with different and
conditions are shown below. Which of the Ans. (a) : Equivalent length of a column supported
following is TRUE? firmly at one-end so other end is free,
L=2l
1238. Consider the following assumptions made in
developing Euler's column theory:
1. The column material obeys Hooke's law
2. The failure of the column occurs due to
buckling
3. The column is 'long' compared to its cross-
sectional dimensions
Which of the above statements are correct?
(a) 1 and 2 only (b) 1 and 3 only
(c) 2 and 3 only (d) 1, 2 and 3
(a) P1 < P2 < P3 < P4 (b) P3 < P1 < P4 < P2
JWM 2017
(c) P1 < P4 < P2 < P3 (d) P3 < P4 < P1 < P2 Ans. (d) : Euler's column theory assumption-
ISRO Scientist/Engineer 17.12.2017 (i) The column is perfectly straight and of uniform
Ans. (b) : cross-section.
Diagram End Effective Buckling Load (ii) The material is homogenous and isotropic
(iii) The material behaves elasticaly.
Condition Length (iv) The load is perfectly axial and passes through the
p Both ends ℓe = ℓ π2 EI centroid of column section.
hinged P1 = 2
ℓ (v) The weight of the column is neglected.
q Both ends ℓ 4π2 EI 1239. As the slenderness ratio of column increase, its
fixed ℓ e = P 2 = compressive strength
2 ℓ 2
(a) Increases
r One end fixed ℓe = 2ℓ π2 EI (b) Decreases
and other end P3 = 2 (c) Remains unchanged
free 4ℓ
(d) May increase or decrease depending on
s One end fixed ℓ 2π2 EI length
and other end ℓe = 2 P4 = 2 Vizag Steel (MT) 2017
hinged ℓ
Ans. (b) :
Hence P3 < P1 < P4 < P2 Effective Length
1236. A column is known as short column if Slenderness ratio =
Least Radius of gyration
(a) The length is more than 30 times the So, the slenderness ratio of column increases as its
diameter compressive strength decrease.
(b) Slenderness ratio is more than 120 1240. For Euler formula to be valid for mild steel
(c) The length is less than 8 times the diameter struts which has yield stress of 3200 kg(f)/
(d) The slenderness ratio is more than 32 sq.cm, the slenderness ratio should not be less
Nagaland CTSE 2016, 2017 Ist Paper than
Strength of Materials 380 YCT
 (a) 80.48 (b) 40.28 Ans. (c) :
(c) 160.96 (d) 53.65
APPSC AEE Mains 2016 (Civil Mechanical)
Ans. (a) : As we know, Euler's formula is applicable for
long columns, (Long columns have slenderness ratio is
greater than 80). Middle third rule states that no tension is developed in a
π2 E I min wall or foundation if the resultant compressive force lies
Pc = within the middle third of the structure.
L2e Tensile stress is zero, if the eccentric, compressive load
I = AK2 acts in the middle third of the section.
L 1245. If the length of a column is halved, the critical
= slenderness ratio load ______.
K
1241. Condition for no tension in a solid column of (a) remains the same
dia D the eccentricity e must be less than or (b) becomes 2 times of the original value
equal to (c) becomes 4 times of the original value
(a) D/8 (b) D/6 (d) becomes 8 times of the original value
(c) D/4 (d) D/2 (e) becomes 16 times of the original value
CGPSC Asstt. Workshop Supt., 17.07.2016
APPSC AEE Mains 2016 (Civil Mechanical)
Ans. (c) : As we know,
Ans. (a) : As we know, for no tension condition
σ b ≤ σa π2 EI
Pe = 2
Le
Pe 4p
≤ L
 πd 3  πd 2 L e1 = e
  2
 32  4π2 EI
Pe1 = 2 , Pe1 = 4Pe
D Le1
e≤
8 1246. The critical load for an ideal elastic column is
1242. A column that fails essentially by direct often called the......... load.
crushing at ultimate load is called (a) Gordan (b) Rankine
(a) Euler's column (c) Shanley (d) Euler
(b) long column TSPSC AEE 28.08.2017 (Civil/Mechanical)
(c) short column HPPSC AE (PWD) 10.09.2016
(d) None of the given answers Ans. (d) : Euler's critical load is the compressive load at
APPSC AEE Mains 2016 (Civil Mechanical) which a slender column will suddenly bend of buckle
Ans. (c) : Medium column is a column which fails due π2 EI min
to direct stress or buckling stress. (S.R. = 32 to 120) Pcr =
( Le )
2
Shorter column (S.R is less than 32) is column which
fails due to direct crushing at ultimate load. Where,
1243. The Rankine's constant for a cast iron column Pcr = Euler critical load
with both ends hinged E = Young's modulus
(a) 1/7500 (b) 1/1600 I = Minimum moment of inertia
(c) 1/9000 (d) 0.00016 Le = Effective length of column
APPSC AEE Mains 2016 (Civil Mechanical) 1247. The load at which the member just buckles is
Ans. (b) : As we know, called
The Rankine's constant for cast Iron column with both (a) Critical load (b) crushing load
ends hinged is given by. (c) equilibrium load (d) ultimate load
1 APPSC AE Subordinate Service Civil/Mech. 2016
a= , σC = 550 N / mm 2 Ans. (a) : The critical load is the greatest load that will
1600
not cause lateral deflection (buckling). For loads greater
Because τeffective = Length of column than the critical load, the column will deflect laterally.
1244. A short rectangular column has zero tensile The critical load puts the column in a state of unstable
stress if the eccentric, compressive load acts: equilibrium. A load beyond the critical load causes the
(a) In the middle three quarters of the section. column to fail by buckling.
(b) In the middle half section Crushing load – It is the intensity of load required to
(c) In the middle third of the section crush the material.
(d) It is not possible Ultimate local – Absolute maximum load that a
GPSC Asstt. Prof. 28.08.2016 structure can bear without failing.
Strength of Materials 381 YCT
1248. Eulers critical load is proportional to 1251. The factor of safety considered for Euler's
(Where L = length of the column k = constant, formula for crippling load is
k = constant depends on end conditions and EI (a) 1 (b) 3
= flexural rigidity) (c) 5 (d) 6
(a) EI (b) L2 APPSC AEE MAINS 2016, PAPER-III
(c) 1/EI (d) k2/EI Ans. (a) : The factor of safety considered for Euler's
APPSC AE Subordinate Service Civil/Mech. 2016 formula for crippling load is 1.
1252. Which of the following is true for ideal column
π2 EI compressed by an axial load (P)?
Ans. (a) : Crippling load =
L2 (a) Column will be in stable equilibrium if P >
1 Pcritical
P ∝ 2 ∝ EI (b) Column will be in stable equilibrium if P <
L Pcritical
1249. The slenderness ratio of free standing column (c) Column will be in unstable equilibrium if P <
of length 4 m and 40 mm × 40 mm section is Pcritical
(a) 115.2 (b) 230.4 (d) Column will buckle if P < Pcritical
(c) 692.8 (d) 346.8 UPRVUNL AE 07.10.2016
APPSC AEE Screening Test 2016 Ans. (b) : If axial load P = P critical (Neutral)
If axial load P < Pcritical (Stable equilibrium)
Ans. (c) If axial load P > P (Unstable equilibrium)
critical
1253. The buckling load for a column hinged at both
ends is 15kN. If the ends are fixed, the
buckling load changes to :
(a) 60 kN (b) 30 kN
(c) 45 kN (d) 3.75 kN
ℓe Kerala PSC IOF 19.04.2016
Slenderness (λ) =
k min π2 EI
Ans. (a) Pcr =
effective length (ℓe) = 2ℓ L2e
[∵ free end column] For both end hinged
ℓe = 2 × 4 π2 × E × I
15 =
=8m L2
For both end fixed
ℓe = 8000 mm
L
where Le =
2
ℓe = effective length
4 × π2 × E × I
Pcr =
bd3 a4 L2
I min 12 = 12 Pcr = 4 × 15
kmin = =
A bd a2 Pcr = 60 kN

a4 1 1254. The Euler's load for a column is 1 MN and

= × crushing load is 1.5 MN. The Rankine load is :
12 a 2 (a) 1 MN (b) 0.6 MN
a2 40 × 40 (c) 1.5 MN (d) 2.5 MN
= = Kerala PSC IOF 19.04.2016
12 12
P ⋅P
k min = 11.547mm Ans. (b) PR = c e
Pc + Pe
8000
Slenderness ratio (λ) = = 692.82 1.5 × 1 1.5 3
11.547 = = = = 0.6
1.5 + 1 2.5 5
1250. In case of eccentrically loaded struts 1255. Compression members tend to buckle in the
(a) Hollow section is preferred direction of
(b) Solid section is preferred
(a) least radius of gyration
(c) Composite section is preferred
(d) Any of the above section may be used (b) axis of load
Nagaland PSC CTSE 2016 Paper-I (c) perpendicular to axis of load
Ans. (c) : A structural member subjected to an axial (d) minimum cross-section
compressive stress is called strut. It is case of ISRO Scientist/Engineer (RAC) 29.11.2015
eccentrically loaded struts, a composite section is Ans : (a) Compression members tend to buckle in the
preferred. Strut may be horizontal, vertical or inclined. direction of least radius of gyration.
Strength of Materials 382 YCT
1256. A certain high tensile strength steel has a 1259. Two columns of different lengths having
modulus of elasticity of 2×106 kg/cm2 and a circular cross-sections have the same L/d ratio
yield point stress of 6,000 kg/cm2. Find the as 100. Both have hinged ends and the material
minimum limiting value of the slenderness is the same. The compressive stress P/A at the
ratio for which Euler's equation is valid critical buckling load for the larger column
(a) 99 (b) 80 will be:
(c) 57 (d) 75 (a) less
ISRO Scientist/Engineer (RAC) 29.11.2015 (b) equal
π2 EI (c) greater than that for the smaller column
Ans : (c) Pcr. = 2 I = AK 2
ℓe (d) None of the above
BPSC Asstt. Prof. 29.11.2015
π EA 
2
ℓe 
Pcr. = Slenderness Ratio ( SR ) = P
(SR )  K 
2
Ans. (b) : Compressive stress   at critical buckling
So critical stress –  A
load for both columns are equal.
π2 E
σcr. = π2 EI
( SR )
2
Pcr = 2
le
π2 × 2 ×106
( )
2
SR = π E πD14
2
6000 Pcr =
SR=57.357 64ℓ 2e
1257. A vertical column has two moments of inertia Pcr π2 ED12
Ixx and Iyy. The column will tend to buckle in =
πD1 2
16ℓ 2e
the direction of the
(a) axis of load 4
(b) perpendicular to the axis of load  
(c) maximum moment of inertia  P  π2 E  1 
2
l
σC1 =  cr 2  =   ∵ = 100
 πD1
(d) minimum moment of inertia 16  100  D

ISRO Scientist/Engineer 11.10.2015  4 column1
Ans. (d) : Vertical Column– Similarly,
Moment of Inertia = IXX and IYY
We know that,  
 Pcr  π2 E  1 
2
Euler's formula (Pe) σC 2 =  =
2   
π2 EI min nπ2 EImin  πD2  16  100 
(Pe) = or  4 column2
L2e L2
Pe = Euler's buckling load 1260. A 1.5 m long column has a circular cross-
Imin = min of [IX-X & IY-Y] section of 5 cm diameter. One end of the
L = Actual length of the column column is fixed and other end is free. Taking
factor of safety as 3 and E = 120 GN/m2, safe
Le = Effective length of the column
load according to Euler's theory is
Le = αL
(a) 13.00 kN (b) 13.27 kN
α = Length finite coefficient
(c) 13.47 kN (d) 13.87 kN
Pe ∝ E
MPPSC AE 08.11.2015
Pe ∝ Imin
Ans. (c) : Given, E = 120 GN/m2, Factor of safety = 3
1
Pe ∝ 2 Length of column (L) = 1.5 m
L Diameter of column (d) = 5 cm = 5 × 10–2 m
Pe ∝ I min For one end fixed and other end is free for a column
1258. Rankine's formula is valid upto the effective length of column = Le = 2L = 2 × 1.5 = 3 m
slenderness ratio of Pmax
(a) 180 (b) 240 FOS =
Pe
(c) 300 (d) 60
(e) 120 P
Pe = max
CGPSC AE 26.04.2015 Shift-I 3
Ans. (e) : The Rankine's formula is applicable for both ∴ Euler's load formula-
long and short column.
π2 E.I min π 4
For short column slenderness ratio is less than 32 and Pe = ∵ I min = d
for long column it is more than 120. 3L2e 64

Strength of Materials 383 YCT

 1264. Which of the following strength forms the basis
( )
4
π2 × 120 × 109 × π × 5 × 10 –2 in designing of a push rod?
= (a) Tensile strength
3 × 64 × ( 3)
2
(b) Compression strength
= 1345.75 × 109 × 10–8 (c) Bending strength
= 13457.5 Newton (d) Buckling strength
⇒ Pe = 13.457 kN UPRVUNL AE 2014
1261. The buckling load for a given column depends Ans : (d) : The push rod is designed as a column based
on on buckling criteria, push rods are often subjected to a
(a) least radius of gyration (b) length of column compressive force which results in buckling.
(c) modulus of elasticity (d) all of these 1265. Short column is usually defined as one whose
Mizoram PSC AE/SDO 2014, Paper-II slenderness ratio is less than about ____
Ans. (d) : (a) 2 (b) 5
(c) 8 (d) 10
π3 E I min MPSC HOD (Govt. Poly. Colleges) 04.10.2014
Pe =
L2e Ans. (d) : For short column
Pe = Buckling load Slenderness ratio (λ) ≤ 32
Imin minimum value of Ixx and Iyy For medium column
L = actual length 32 ≤ λ ≤ 100
Pe ∝ E For long column
Pe ∝ Imin λ > 120
1 1266. Secant formula is applicable for :
Pe ∝
Le (a) short columns under axial loading
1262. Rankine's formula for determining the load (b) long columns under axial loading
carrying capacity of a column is valid for (c) short columns under eccentric loading
(a) long column only (b) short column only (d) long columns under eccentric loading
(c) columns with hinged (d) all columns (HPPSC AE 2014)
Mizoram PSC AE/SDO 2014, Paper-II Ans : (d) Secant formula is applicable for long column
Ans. (d) : Rankine's formula for determining the load under eccentric loading.
carrying capacity of column is valid for– Secant formula for ν max and σ max
(i) Long column In the Euler's buckling formula we assume that the load
(ii) Short column P acts through the centroid of the cross - section.
However in reality this might not always be the case :
1263. For a circular column having its ends hinged,
the load P might applied at an offset or the slender
the slenderness ratio is 160. The L/d ratio of
member might not be completely straight.
the column is :
To account for this, we assume that the load P is applied
(a) 80 (b) 57 at a certain distance away from the centroid. This would
(c) 40 (d) 20 obviously change the way we calculate our buckling
MPPSC State Forest Service Exam, 2014 load, which is what the Secant formula is for.
π 1267. If the slenderness ratio for a column is 100,
Ans. (c) : Ak 2 = d 4
64 then it is said to be a _______ column.
π (a) long (b) medium
A = d2 (c) short (d) expand
4
π 2 2 π 4 Haryana PSC Civil Services Pre, 2014
d ×k = d Ans. (b) :
4 64
Effective length of column
d Slenderness ratio (λ) =
k= least radius of gyration
4
(1) Short column – λ < 32
ℓe 32 < λ < 120
Slenderness ratio ( λ ) = (2) Medium Column –
k (3) Long Column – λ > 120
ℓe = ℓ (because both ends are hinged) 1268. A loaded column fails due to
ℓ (a) stress due to direct load
λ= = 160
d/4 (b) stress due to bending
ℓ (c) both (1) and (2)
= 40 (d) None of the above
d
APPSC Poly. Lect. 2013
Strength of Materials 384 YCT
Ans. (c) : A loaded column fails due to both direct & Ans : (c) The formula given by I.S. code in calculating
bending stress. allowable stress for the design of eccentrically loaded
→ The long columns, commonly fail because of columns is based on Perry's formula.
buckling stress, it has direct stress very small 1272. If the flexural rigidity of the column is doubled,
compare to bending stress. then the strength of the column is increased by
→ The short column primarily fails due to direct stress (a) 16 (b) 8
1269. The ratio of the compressive critical load for a (c) 2 (d) 4
long column fixed at both the ends and a APPSC AEE 2012
column with one end fixed and the other end Ans : (c) Case 1st :-
being free is:-
(a) 2 : 1 (b) 4 : 1 π2 EI
Strength of column (P1) = 2
(c) 8 : 1 (d) 16 : 1 L
UKPSC AE-2013, Paper-I (Flexural rigidity of column = E ×I)
ESE 1997 Case 2st :-
Ans. (d) : Case I- When both ends of column are fixed Flexural rigidity of column is doubled.
2
4π EI π2 × 2EI
[ Pe ]1 = 2 strength of column (P2) =
l L2
Case II- When one end fixed and other end is P2 = 2 × P1
free In the flexural rigidity of the column is doubled, then
π2 EI the strength of the column is increased by two times.
[ Pe ]2 = 2 1273. The least radius of gyration for solid circular
4l
column is
Then
[ Pe ]1 16 (a) d (b)
d
(c)
d
(d)
d
= = 16 :1 2 4 3
[ Pe ]2 1 APPSC AEE 2012
1270. The crippling load of a column with one end Ans : (c) The least radius of gyration for solid circular
fixed and other end hinged is column is d/4.
(a) 2 times that of a both ends hinged colum Radius of gyration-Radius of gyration is defined as the
square root of moment of inertia divided by the area of
(b) Two times that of a both ends hinged column
itself.
(c) Four times that of a both ends hinged column
(d) Eight times that of a both ends hinged I I yy
rx = xx ; ry =
column A A
APPSC AEE 2012
π / 64 × d 4
d2 d
Ans : (b) = = =
End Conditions Effective length π/4×d 2
16 4
Both end hinged L 1274. In a mild steel tube 4 m long, the flexural
One fixed other free 2L rigidity of the tube is 1.2 × 1010 N − mm 2 . The
Both end fixed L/2 tube is used as a strut with both ends hinged.
One end fixed and other L/ 2 The crippling load in kN is given by
hinged (a) 14.80 (b) 7.40
(i) Crippling load of a column with one end fixed (c) 29.60 (d) 1.85
and other end hinged. APPSC AEE 2012
π2 EI 2π2 EI Ans : (b) tube length = 4mm
P1 = = …………. (i) flexural rigidity (E×I) = 1.2 × 1010 N – mm2
(L / 2 ) 2 L2
End condition = Both ends hinged
(ii) Crippling load of a column when both end hinged Effective length = l
π2 EI π2 EI
P2 = 2 …………..(ii) Crippling load =
L ℓ2
P1 = 2 × P2 π ×1.2 ×1010
2
Pe =
1271. The formula given by I.S. code in calculating 42
allowable stress for the design of eccentrically
Pe = 7.40 kN
loaded columns is based on
(a) Johnson's parabolic formula 1275. A member under tension is called :
(b) Straight line formula (a) strut (b) tie
(c) Perry's formula (c) strut-tie (d) column
(d) Secant formula APPSC AE 04.12.2012
APPSC AEE 2012 Ans. (b) : A member under tension is called tie.
Strength of Materials 385 YCT
1276. Two long columns A and B are of the same π 2 EI  2L 
material, the same length, and the same cross- Pcr = 2 ∵ Le = 3 
section. Column A is hinged at both ends, ( Le)  
while column B is fixed at one end and hinged π EI
2
9π 2 EI
at the other. The ratio of the Euler-crippling = 2
=
loads for A and B is :  2L  4 L2
(a) 2 (b) 1  
 3 
1 L
(c) 2 (d) II for lower part length =
2 3
VIZAG Steel MT 24.01.2021, Shift-I Conditions- Up one end hinged and other end fixed
APGENCO AE 2012 ∴ Buckling load,
Ans. (d) : Column-A– both ends hinged. π 2 EI  L 
Pcr = ∵ Le = 3 2 
π E I min π EI min
2 2
( Le ) 2
 
( Pcr ) A = =
le2 l2
π EI
2
18π 2 EI
Both ends hinged (le = l) = 2
=
Column-B– One end hinged and other fixed  L  L2
 
 l  3 2 
 le =  Hence, we take the min. of the two buckling load, that is
 2
9π 2 EI
π2 E I min π2 E I min × 2 4 L2
( Pcr ) B = 2
=
 l  l2 1278. Slenderness ratio has dimension of
  (a) cm (b) cm–1
 2 (c) cm2 (d) None
(P ) π2 E I min l2 UKPSC AE 2012 Paper-I
Ratio = cr A = × 2 Ans. (d) : It is a dimensionless quantity.
( Pcr ) B l 2
π E I min × 2
L
=
1 S= e
2 k
1277. A column of length 'L' is fixed at bottom end I
k = min
and hinged at the other end. This column is A
restrained from lateral displacement at 1/3rd 1279. Given that
height. The buckling load is given by PE = the crippling load given by Euler
PC = the load at failure due to direct
compression
PR = the load in accordance with the Rankine's
criterion of failure
Then PR is given by
( PE + PC )
(a) (b) PE × PC
2
PE .PC
(a) 9π 2 EI /(4 L2 ) (b) 4π 2 EI / L2 (c) (d) None of the above
PC + PE
(c) 4π 2 EI /(9 L2 ) (d) π 2 EI / L2 APPSC AEE Mains 2016 (Civil Mechanical)
ISRO Scientist/Engineer 2012 ISRO Scientist/Engineer 2010
Ans. (a) : The column will behaves as 2 columns and Ans. (c) : PR = The load with Rankine's criterion of
restrained means it's hinged. failure.
PC = Load due to direct compression
PE = Crippling load given by Euler.
1 1 1
= +
PR PC PE
PC PE
PR =
PC + PE
2L 1280. Assertion (A) : A long column fails (becomes
(i) for upper part length = elastically unstable) due to excessive bending.
3 Reason (R) : Euler’s buckling load is inversely
Conditions- Both ends hinged Buckling load. proportional to the equivalent length of the
Buckling load, column.
Strength of Materials 386 YCT
 Codes : π2 EI
(a) Both A and R are individually true and R is Ans. (c) : Pcr = 2 (Hinged-hinged end condition)
the correct explanation of A L
Pcr = EI (Taking I as least moment of inertia)
(b) Both A and R are individually true and R is = 100 × 109 × 100 × 10–8
not the correct explanation of A = 100000 N = 100 kN
(c) A is true but R is false
1283. If the slenderness ratio of a column is more
(d) A is false but R is true than 120 it is termed as
UPSC JWM Advt. No.-50/2010 (a) short column (b) medium column
Ans. (c) : As we know, (c) long column (d) none of the above
π2 EI WBPSC AE 2008
Buckling load ( Pe ) = 2 Ans. (c) : For long column slenderness ratio is greater
Le
than 120.
1
Pe ∝ 2
Le 8. Strain Energy Method
Euler’s buckling load is inversely proportional to the
square of equivalent length of the column. 1284. A cantilever beam of length L and flexural
1281. An aluminium column of square cross-section modulus EI is subjected to a point load P at the
(10 mm × 10 mm) and length 300 mm has both free end. The elastic strain energy stored in the
ends pinned. This is replaced by a circular beam due to bending (neglecting transverse
cross-section (of diameter 10 mm) column of shear)
the same length and made of the same material P 2 L3 P 2 L3
(a) (b)
with the same end conditions. The ratio of 4EI 6EI
critical stresses for these two columns PL2 3
P 2 L3
corresponding to Euler's critical load, (c) (d)
(σ ) : (σ ) , is 2EI 8EI
critical square critical circular
Assam Engg. College AP/Lect. 18.01.2021
(a) 1 : 4 (b) 3 : 4 Gujarat PSC AE 2019
(c) 4 : 3 (d) 4 : 1 GATE 2017, MPPSC AE 2016
DRDO Scientists 2009 CGPSC Poly. Lect. 22.05.2016
Ans. (c) : Euler load Pe is given by APPSC AEE Screening Test 2016
APPSC AEE Mains 2016 (Civil Mechanical)
π2 EI (HPPSC LECT. 2016)
P=
le2 ISRO Scientist/Engineer 11.10.2015, 2006
OPSC Civil Services Pre. 2011, ESE 2009
P π2 EI Ans. (b) :
= σc =
A Ale2
I
σc ∝
A 1
Strain energy U = × Pδ
( σc )circular =
( πd 4 ) ( 4 ) =
d 2 (10 )
=
2
2
64 ( πd 2 ) 16 16 1 PL3
= ×P×
2 3EI
a 2 (10 )
2
a4
( σc )square = = =  PL3 
12 ( a 2 ) 12 12 δ = 
 3EI 
( σc )square / ( σc )circular = 4 : 3 P 2 L3
U=
1282. A long slender column is pin-supported at the 6EI
ends and compressed by an axial load P as Or
shown in figure. If I11 = 100 cm4, I22 = 200 cm4 M 2 dx
L

and Young's modulus of elasticity = 100 GPa, Strain energy U = ∫ M = px

0 2EI
then the critical load for buckling is
L ( px ) dx
2

=∫
0 2EI
L
P2  x3 
= × 
2EI  3 0
(a) 500 kN (b) 200 kN
P 2 L3
(c) 100 kN (d) 1000 GN U=
DRDO Scientists 2008 6EI

Strength of Materials 387 YCT

1285. The property of a material to absorb energy Ans. (a) : Strain energy stored in a beam
within elastic limits is known as 1. when beam is subjected to bending moment (M)
(a) Elasticity (b) Toughness M 2 .dx
(c) Tensile strength (d) Stiffness U B.M = ∫
(e) Resilience 2EI
2. when beam is subjected to Twisting moment (T)
SJVN ET 2019
JPSC AE PRE 2019 T 2 .dx
U T.M = ∫
Nagaland PSC (CTSE) 2018, Paper-I 2GI P
(CGPSC Polytechnic Lecturer 2017) 3. When beam is subjected to axial load P
RPSC LECTURER 16.01.2016 P 2 .dx
RPSC 2016 U A.L = ∫
2AE
CGPSC AE 26.04.2015 Shift-I
TNPSC AE (Industries) 09.06.2013 1288. Proof resilience may be defined as
(a) Work done in straining the material
Ans. (e) : (b) Max. strain energy that can be stored in a
material per unit volume
(c) Max. strain energy that can be stored in a
material under elastic condition
(d) Max. load which can be applied to a member
Nagaland CTSE 2017 Paper-I
Resilience—The capacity of material to absorb and Nagaland CTSE 2016 Paper-I
release strain energy within elastic limit. ISRO Scientist/Engineer 2009
Toughness—Strain energy store upto fracture point. Ans. (c) : Proof Resilience
Modulus of toughness—Capacity of material to absorb Maximum strain energy that can be stored in a material
maximum strain energy upto fracture per unit volume is under elastic limit.
called MOT. 1289. The maximum energy that can be absorbed
1286. The strain energy stored in a body, when the per unit volume without creating a permanent
load is gradually applied, is distortion is called _____
(a) σE/V (b) σV/E (a) Resilience
(c) σ2E/2V (d) σ2V/2E (b) Distortion limit
where σ = stress in the material of the body, V = (c) Modulus of resilience
volume of the body and E = modulus of elasticity (d) None of these
of the material. RPSC ACF & FRO, 26.02.2021
RPSC IOF, 2020 VIZAG Steel MT 24.01.2021
Nagaland PSC (CTSE) 2018, Paper-I KPCL AE 2016
Assam PSC CCE Pre 2015 Ans. (c) : Modulus of Resilience- The maximum
TSPSC AEE 2015, TNPSC AE 2014 energy that can be absorbed per unit volume without
TRB Asstt. Prof., 2012 creating a permanent distortion-
ESE 2006, 2002 Strain energy
Modulus of Resilience(MOR) =
1 Volume
Ans. (d) : Strain energy = stress × Strain × Vol.
2 σ2
The strain energy stored in a body, when the load is × Volume
(MOR) = 2E
σ2 V Volume
gradually applied is .
2E σ 2
1287. Strain energy stored in a beam due to bending (MOR) =
2E
is given by (Where M is bending moment, E is
1
modulus of elasticity, I is moment of inertia, G MOR = σEL ε EL (∵ EL = Elastic limit )
is modulus of rigidity, L is the length of the 2
beam, and σ is the tensile strength.) 1290. The strain energy stored in a spring, when
M 2 dx σ2 dx subjected to maximum load, without suffering
(a) ∫ (b) ∫ permanent distortion, is known as
2EI 2EI (a) impact energy
2 2
M dx M dx
(c) ∫ (d) ∫ (b) proof resilience
2GI 2EL (c) proof stress
HPPSC AE 2018 (d) modulus of resilience
UPPSC AE 12.04.2016 Paper-I Sikkim PSC (Under Secretary), 2017
APPSC AE 04.12.2012 Nagaland PSC CTSE 2017, Paper-I
APPSC AEE 2012 Nagaland PSC CTSE 2016 Paper-I
Strength of Materials 388 YCT
Ans. (b) : The strain energy stored in a spring, when Ans. (a) : Strain energy per unit volume = modulus of
subjected to maximum load, without suffering resilience
permanent distortion is known as proof resilience. 2
 200 
1291. Resilience of a material is considered when it is
σa2 ( P / A )
2  
10 × 10 
subjected to = = = 
(a) Fatigue (b) Creep 2E 2E 2 × 200 × 103
(c) Frequent heat treatment (d) Shock Loading 1
RPSC IOF, 2020 = 5 N − mm / mm3 = 10N − m / m3
10
Rajasthan Nagar Nigam AE 2016, Shift-I = 10 J/m3
Ans. (d) : Resilience– It is the strain energy that can 1296. For the state of stress of pure shear τ, the
be absorbed by the material without plastic strain energy stored per unit volume in the
deformation upto elastic limit. Toughness is the elastic, homogeneous, isotropic material
maximum strain energy that can be absorbed by the having elastic constants – Young's modulus, E
material upto fracture.
and Poisson's ratio µ will be
Creep– The permanent deformation of a material over
τ2 τ2
a period of time at constant load and at elevated
temperature is called creep.
(a) (1 + µ ) (b) (1 + µ )
E 2E
Fatigue–The behaviour of the material under 2τ2 τ2
fluctuating loads is called fatigue. (c) (1 + µ ) (d) (2 + µ)
E 2E
1292. "Total Strain Energy" theory of failure is also UPPSC AE 13.12.2020, Paper-I
known as ________.
ESE 1998
(a) Couloumb Guest Theory
(b) Lame Rankine Theory Ans. (a) : Strain energy per unit volume
(c) Von - Mises Theory τ2
U=
(d) Beltrami-Haigh Theory 2G
HPPSC Workshop Suptd. 08.07.02021 E = 2G(1 + µ)
Ans. (d) : Total Strain Energy theory of failure is also E
known as Beltrami-Haigh Theory. 2G =
1+ µ
1293. The property of material whereby it absorbs
energy due to straining actions by undergoing τ2
U=
plastic deformation is called E / (1 + µ )
(a) Elasticity (b) Fatigue
(c) Resilience (d) Toughness τ 2 (1 + µ )
U=
Oil India Senior Officer 23.12.2020 E
Ans. (d) : It is the property of material which absorb 1297. A cantilever beam, 2 m in length is subjected
energy while straining by undergoing plastic to a uniformly distributed load of 10 kN/m. If
deformation without fracture. E = 200 GPa and I = 1000 cm4, the strain
1294. If a gradual tensile load (P) is applied to an energy stored in the beam will be
elastic bar axially, the change in the length of (a) 7 Nm (b) 12 Nm
bar is x. What will be the strain energy stored (c) 8 Nm (d) 40 Nm
in the material of the bar? UPPSC AE 13.12.2020, Paper-I
1 2
(a) Px (b) Px  Wx 2 
ℓ

∫
L

(c) 2 Px
2
(d) Zero Ans. (d) : U =
∫ 0
M 2x dx 
=  2 
0
 dx
APPSC Poly Lect. 13.03.2020 2EI 2EI
Ans. (b) : W2 2

1
Strain energy (U) = × load × change in the length
=
8EI ∫
0
x 4 dx
2 (10 ×103 ) 2  25 
= −2 4
× 
1 8 × 200 × 10 ×1000 × (10 )  5 
9
U = P.x
2 200
=
1295. A square bar of size 10 mm × 10 mm and 5
length 1000 mm is subjected to 200 N axial = 40 N-m
tensile force. The bar is made of mild steel 1298. Two elastic bars of equal length and same
having modulus of elasticity of 200 GPa. Find material, one is of circular cross-section of 80
the strain energy density stored in the bar mm diameter and the other of square cross-
under this state of loading? section of 80 mm side. Both absorbs same
(a) 10 J/m3 (b) 20 J/m3 amount of strain energy under axial force.
3
(c) 2 J/m (d) 5 J/m3 What will be the ratio of stress in circular
ISRO Scientist/Engineer 12.01.2020 cross-section to that of square cross-section?
Strength of Materials 389 YCT
 (a) 0.972 (b) 0.886 1302. Modulus of resilience for the below material is
(c) 1.013 (d) 1.128
SJVN ET 2019
Ans. (d) : Uc = Us
σc2 σ2
× A c × L = s × As × L
2E 2E
σc A s
2
802
= =
σs A c π × 802
2

4
σc 4
2
(a) 200 kN/m2 (b) 100 kN/m2
=
σs2 π (c) 606.25 kN/m 2
(d) 12.29 MN/m2
σc 2 2 ISRO Scientist/Engineer 17.12.2017
= = Ans. (b) : Modulus of resilience
σs π 3.14
= Area under stress–strain diagram up to elastic limit
σc (EL)
= 1.128
σs 1
= σEL × ∈EL
1299. The resilience of steel can be found by 2
integrating stress-strain curve up to the 1
(a) ultimate fracture point (b) upper yield point = × 200 × 0.001 N/mm2
2
(c) lower yield point (d) elastic point = 0.1 × 1000 kN/m2
ESE 2018 = 100 kN/m2
Ans. (d) : Resilience is the ability of a material to 1303. Which one of the following statements is
absorb energy per unit volume without permanent correct?
deformation and is equal to the area under the stress- (a) The strain produced per unit volume is called
strain curve upto the elastic limit. resilience.
1300. A circular shaft of length 'L' a uniform cross- (b) The maximum strain produced per unit
sectional area 'A' and modulus of rigidity 'G; volume is called proof resilience.
is subjected to a twisting moment that (c) The least strain energy stored in a unit
produces maximum shear stress 'τ' in the volume is called proof resilience.
shaft. Strain energy in the shaft is given by the (d) The greatest strain energy stored in a unit
expression AL/kG, where k is equal to volume of a material without permanent
(a) 2 (b) 4 deformation is called proof resilience.
(c) 8 (d) 16 ESE 2017
TSPSC AEE 28.08.2017 (Civil/Mechanical) Ans. (d) : The greatest strain energy stored in a unit
Ans. (b) : Strain energy stored in a shaft due to torque volume of a material with permanent deformation is
called proof resilience.
 τ2  1304. The strain energy stored in a solid circular
U=  vol.
 4G  shaft, under pure torque, per unit its volume
(take max. shear stress = q and modulus of
τ2 rigidity = (C) is expressed as
U= ×A×L
4G (a) q2/C (b) q/(4C)
AL (c) q3/(4C) (d) q2/(4C)
U=
kG APPSC AEE Mains 2016 (Civil Mechanical)
Then k = 4 Ans. (d) : We know, strain energy stored =
1301. The strain energy corresponding to the stress 1 V
= × shear stress 2 ×
at the elastic limit is known as ______ 2C 2
(a) Modulus of resilience (b) Toughness For per unit volume
(c) Proof resilience (d) Resilience under pure torque
APGENCO AE, 2017 1
Ans. (c) : Total strain energy stored within elastic limit = × shear stress 2
4C
is known as resilience.
q2
Total strain energy stored corresponding to elastic limit =
is called proof resilience. 4C
Strain energy stored per unit volume corresponding to 1305. A prismatic bar 1 m long and 4 sq.cm in cross
elastic limit is called modulus of resilience and its value sectional area is compressed by a force of 80
remains constant and not depend upon dimension of kN. If E = 200 kN/sq.mm, the total strain
work piece. energy stored in the bar is equal to
Strength of Materials 390 YCT
 (a) 40 kN-mm (b) 0.05 kN-mm 1308. As compared to uniaxial tension or
(c) 400 kN-mm (d) 80 kN-mm compression, the strain energy stored in
APPSC AEE Mains 2016 (Civil Mechanical) bending is only
(a) 1/3 (b) 1/8
Ans. (a) : For cantilever
(c) 1/4 (d) 1/2
1
Strain energy (u) = × stress × strain × volume APPSC AE Subordinate Service Civil/Mech. 2016
2 Ans. (a) : Strain energy due to tension or compression
Given :
1
F = 80 kN U1 = × stress × strain × volume
L = 1m 2
A = 4 cm 2
σ2
U 1= × Volume .... (i)
F2 L (80 × 103 )2 × 1000 2E
U= = Strain energy due to bending
2EA 2 × 200 × 103 × 400
U = 40kN − mm
L
M 2 dx
L
( Wx )2 W2
L

1306. A uniform rod of length 'L', cross-section area

U2 = ∫
2E
0
= ∫
2E
0
dx = ∫
2E
0
x 2 dx

'A' and modulus of elasticity 'E' is held rigidly 2 3

w L
at the fixed ends and an axial load 'P' is U2 = ........ (ii)
applied at mid length of the rod. The strain 6E
energy stored is : From equation (i) and (ii)
U1 1
P2 L P2 L =
(a) (b) U2 3
EA 4EA
1309. What will be the strain energy stored in the
P2 L
(c) (d) None of these metallic bar of cross sectional area of 2 cm2
8EA and gauge length of 10 cm if it stretches 0.002
RPSC ADE 2016 cm under the load of 12 kN?
Ans. (c) : (a) 10 N-cm (b) 12 N-cm
L = length of rod (c) 14 N-cm (d) 16 N-cm
RPSC LECTURER 16.01.2016
A = cross-section area
Ans. (b) : Data given,
E = modulus of elasticity
A = 2 cm2, l = 10 cm
P = Applied load δl = 0.002 cm
1
U = Wδ P = 12 × 103 N
2 Strain energy
1 WL σ2
W× U= × volume
2 AE 2E
P P
W= σ=
2 A
2 δ l σ
Strain energy =
P L = =e
8EA l E
2 then we get
1307. A specimen 160mm in cross section stretches
1
by 0.06 mm over a 50 mm gauge length under U = × P ×δ
an axial load of 35 kN. The strain energy is 2
(a) 1050 N-mm (b) 780 N-mm 1
= × 12 ×103 × 2 × 10−3
(c) 1435 N-mm (d) 1298 N-mm 2
APPSC AE Subordinate Service Civil/Mech. 2016 = 12 N-cm
Ans. (a) : Given, 1310. Strain energy stored in a prismatic bar
2 suspended from one end due to its own weight
Area of specimen A = 160 mm
(elastic behaviour) [x = specific weight of
Gauge length L = 50mm material, A = cross-sectional area, L = length
Change in gauge length ∆L = 0.06 mm of bar]:
Axial load applied P = 35 kN = 35 × 103 N x A L3 x 2 A2 L2
1 (a) U = (b) U =
Now, strain energy (U) = × P × ∆L 6E 6E
2 x 2 A L3 x 2 A L3
1 (c) U = (d) U =
U = × 35 ×103 × 0.06 = 1050N − mm 3E 6E
2 UPRVUNL AE 07.10.2016
Strength of Materials 391 YCT
Ans. (d) : P 2 1003 ×12 
U2 =  
E  6 × 44 
7812.5P 2
U2 =
E
Ratio of strain energies
U 2 7812.5P 2 × E
=
U1 E × 3.125P 2
U2
Self weight at section - = 2500
U1
W .l
Pl = 1312. The strain energy stored in the beam with
L
We know that strain energy for prismatic bar suspended flexural rigidity EI and loaded as shown in the
from one end due to its self weight figure is:
2
L ( P ) dl
U =∫ l
o 2( AE )
l
AE = constant
2
 Wl  − P 2 L3 2 P 2 L3
  dl
L L 
(a) (b)
=∫ 3EI 3EI
O 2 AE 4 P 2 L3 8P 2 L3
L (c) (d)
W 2
l  3
3EI 3EI
= 2  
2 AE × L  3  O BPSC Asstt. Prof. 29.11.2015
Ans. (c) : According to question,
( xAL )2 L3
= ×
2 AEL2 3
x 2 AL3
U=
6E
1311. A square bar of side 4 cm and length 100 cm is
subjected to axial load P. The same bar is then
used as a cantilever beam and subjected to an Mx = Px – | P (x – L)|
end load P. The ratio of the strain energies, For portion AB and CD,
stored in the bar in the second case to that (M ) = Px = (M )
x AB x CD
stored in first case, is For portion BC,
(a) 16 (b) 400 (Mx) = Px – P(x – L) = Px – Px + PL
(c) 1000 (d) 2500 (Mx)BC = PL
MPPSC AE 2016 Strain energy in the beam due to bending is
Ans : (d) Square bar Area = 16cm2 U = U AB + U BC + U CD
length = 100cm. L 3L
moment of Inertia of square bar (M x )2AB dx (M 2x ) BC dx
bh 3 44
U = 2 ∫0 2EI + ∫L 2 EI
= = = 21.33cm 4
12 12 4 P 2 L3
U=
case1 : −
st
3EI
σ2 1313. Proof resilience is the property of the materials
Strain Energy ( U1 ) = × volume which indicated their capacity to bear :
2E
2 (a) shocks
 P  16 P
U1 =   × × 100 σ=  (b) static compressive loads
 16  2E  16  (c) static tensile loads
3.125P 2 (d) none of these
U1 = VIZAG MT 2015
E
case − II : − Ans. (a) : Proof resilience– The maximum energy
P 2 L3 which a body stores 'upto' elastic limit is called the
U2 = 'proof resilience'. It indicates shock bearing capacity of
6EI material.
Strength of Materials 392 YCT
1314. Proof resilience per unit volume of a material is 1317. Strain energy stored in a body due to a
known as- suddenly applied load compared to when
(a) modulus of resilience(b) resilience applied slowly is:
(c) Proof resilience (d) toughness (a) twice (b) four times
ISRO Scientist/Engineer (RAC) 29.11.2015 (c) eight times (d) half
Ans : (a) Proof resilience per unit volume of a material (HPPSC AE 2014)
is known as modulus of resilience. Ans : (b) Strain energy stored in a body due to
1315. Strain energy stored in beam with flexural suddenly applied load compared to when applied slowly
rigidity EI and loaded as shown figure is is four times.
Stress in a body due to
gradually applied load gradually = σ
stress in a body due to suddenly applied load = 2σ
strain energy due to gradually applied load
(a) (P2L3/3EI) (b) (2P2L3/3EI) σ2
(c) (4P2L3/3EI) (d) (6P2L3/3EI) U GAL = × Volume ...........(i)
2E
(e) (8P2L3/3EI) strain energy due to suddenly applied load
CGPSC AE 26.04.2015 Shift-I
( 2σ )
2
Ans. (c) : USAL = × Volume
2E
 σ2 
USAL = 4 ×  × volume 
 2E 
USAL = 4 × U GAL
1318. Strain energy U, in the case of axially loaded
bar is expressed as (where P is axial load, L is
RA = RB = P bar length, A is cross-section area of the bar, E
Total strain energy, is Young's Modulus of the bar, σ is axial stress,
( Px )
dx ( PL ) ( 2L )
2 2
L δL is elongation due to axial load)
U = 2∫ +
0 2EI 2EI P 2 L2
(a)
P 2 L3 P 2 L3 2 AE (δ L)
U= +
3EI EI (b) Area under stress versus strain diagram of
4 P 2 L3 axially loaded bar
U= σ (δ L)
3 EI (c)
2 L
1316. Strain energy stored in a solid circular shaft is
proportional to P2L
(d)
(a) GJ (torsional rigidity)(b) 1/(GJ)2 2 AE
(c) GJ2 (d) 1/(GJ) ISRO Scientist/Engineer 24.05.2014
(e) 1/ GJ Ans. (d) : Strain energy = U
CGPSC AE 26.04.2015 Shift-I Axial Load = P
Ans. (d) : For solid shaft—Let us suppose that a solid Length = L
shaft is subjected to a torque which increases gradually Cross-Section = A
from zero to a value T. Let θ represent the resultant Young's modulus = E
angle of twist. Then, the energy stores in the shaft is
Axial stress = σ
equal to work done in twisting i.e.
PL
1
U = .T .θ Elongation δL =
2 AE
1  T .l  1
= .T .  Strain Energy U = × Force × deflection
 2
2  GJ 
 T Gθ  1 PL P 2 L
= ×P× =
 =  2 AE 2 AE
J l 
1319. If the area, length and the stress to which a bar
1  T 2l  is subjected be all doubled, then the elastic
=  
2  GJ  strain energy of the bar will be :
(a) Doubled (b) Three times
1
U∝ (c) Four times (d) Sixteen times
GJ J&K PSC Civil Services Pre, 2013
Strength of Materials 393 YCT
Ans. (d) : Energy stored within elastic limit is called (a) 2q2/C (b) 2C/q2
resilience. (c) q2/2C (d) C/2q2
Strain energy stored upto elastic limit is called proof APPSC AEE 2012
resistance. Ans. (c) : Shear stress intensity τ = q
Energy stored upto elastic per unit volume is called modules of rigidity G = C
modulus of resilience. Hence
σ2 σ2 τ2
Strain energy U1 = × volume = × A×l Strain energy stored per unit volume =
2E 2E 2G
U2 =
( 2σ )2 × 2A × 2l = 16 σ2 × A × l S.E q2
2E 2E =
Volume 2c
1320. The unit of strain energy in S.I. unit is
1323.
(a) watt (b) joule
(c) joule/sec (d) joule/m.
TNPSC ACF 2012
Ans. (b) : Strain Energy- It is defined as the energy
stored in a body due to deformation.
1 I = 375 × 10–6 m4
U = Pδ
2 l = 0.5 m
1 PL 1 P 2 L E = 200 GPa
U = ×P× = Determine the stiffness of the beam shown in
2 AE 2 AE the above fig.
Unit of strain energy N-m or Joule. (a) 12×1010 N/m (b) 10×1010 N/m
10
1321. Principal stresses induced in a material are 60 (c) 4×10 N/m (d) 8×1010 N/m
MPa, 30 MPa and –20 MPa, E = 80 GPa, µ = APPSC AE 04.12.2012
0.20. The value of total strain energy per unit Ans. (*) :
volume is:
(a) 24.5 kN-m/m3 (b) 49 kN-m/m3
3
(c) 30.63 kN-m/m (d) 61.25 kN-m/m3
HPPSC ADF 05.03.2019
Ans. (c) : Given,
σ1 = 60 MPa, σ2 = 30 MPa, E = 80 GPa, µ = 0.20 Given,
σ3 = –20 MPa, I = 375×10–6 m4
Totalstrain energy l = 0.5 m
= E = 200 GPa = 200×109 Pa
Volume Let, K = stiffness of beam
1 ∴ Bending moment (Mx) = P.x
= σ 1 + σ 2 + σ 3 – 2µ (σ 1σ 2 + σ 2σ 3 + σ 3σ 1 ) 
2 2 2

2E  l
( Px )
2 2l
( Px )
1 U = ∫ .dx + ∫
l 2E ( 2I )
.dx
3 (
=  60 ) + ( 30 ) + ( –20 )
2 2 2
0 2EI
2 × 80 × 10 l 2l
–2 × 0.20 ( 60 × 30 – 30 × 20 – 60 × 20 )   P2 x3   P2 x3 
=   + 
1  3 × 2EI  0  3 × 2E × 2I  l
= 3600 + 900 + 400 – 0.40 (1800 – 600 –1200 )
2 × 80 × 103  P 2 l 3 P2 ( 8l − l ) P 2 l 3 7ρ2 l 3
3 3

= + = +
1 6EI 12EI 6 EI 12 EI
=  4900 – 0.40 ( 0 ) 
160 × 103  9P 2 l 3 3P 2 l 3
U= = ___ (i)
1 4900 kN − m 12 EI 4EI
= × 4900 =
160 × 103 160 m3 1 1 P2  p
∵ U = P×δ = _____(ii) ∵ δ =
kN − m 2 2 k  k
= 30.625 n
m 3 From eq (1) and (2)
kN – m 1 P 2 3P 2 l 3 2EI
≈ 30.63 3
= ⇒k= 3
m 2 k 4EI 3l
1322. If the cross-section of a member is subjected to 2 × 200 × 10 × 375 × 10−6
9

a uniform shear stress of intensity 'q', then the ⇒ k = = 4×108 N/m

3 × ( 0.5 )
3
strain energy stored per unit volume will be
given by (C = modulus of rigidity) No any option matching with the solution.

Strength of Materials 394 YCT

1324. The strain energy per unit volume of a body TNCSC AE 2020
subjected to a direct stress σ is given by GPSC Executive Engineer 23.12.2018
σ2 σ APGENCO AE, 2017, RPSC 2016
(a) (b) APPSC AEE Screening Test 2016
2E 2E
TNPSC AE (Industries) 09.06.2013
σ2 σ ESE 2008, GATE 2014
(c) (d)
3E 3E Ans. (d) :
WBPSC AE, 2007, ESE 2016 Max.
UKPSC AE 2007 Paper-I Case Max. slope
deflection
Ans. (a) : Strain energy per unit volume (Svol) • Cantilever with point WL2 WL3
1 load (W) at free end
Svol = × σ × ε 2EI 3EI
2
• Cantilever UDL wL3 wL4
1 σ
Svol = × σ × having load intensity
2 E (w) on whole span 6EI 8EI
σ2 • Simply supported WL2 WL3
Svol = point load (W) at
2E middle of beam 16EI 48EI
1325. What will be the total strain energy stored in a • Simple supported
simply supported beam of span 'L' and 1 wL3 5 wL4
UDL having load
flexural rigidity 'EI' subjected to a intensity (w) on 24 EI 384 EI
concentrated load 'W' at the centre? whole span
W 2 L3 W 2 L3 1327. Deflection due to point load at the free end of
(a) (b)
40 EI 60 EI the cantilever due to load P (with usual
notation)
W 2 L3 W 2 L3 (a) Pl3 /(3EI) (b) Pl2/(3EI)
(c) (d) 2
192 EI 96 EI (c) Pl /(2EI) (d) Pl4/(3EI)
WBPSC AE 2003 NLC GET 17.11.2020, Shift-II
Nagaland PSC (CTSE) 2018 Paper-I
WL3
Ans. (d) : We know that deflection δ = for TSPSC AEE 2015
48EI APSC AEE 2012
simply supported beam in concentrated load (W) at mid APSC AE 04.12.2012
span then, TNPSC AE, 2008
WL3 ∂U Ans. (a) : Let,
δ= = (Strain energy = U )
48EI ∂W l = length of cantilever
WL3 ∂U I = moment of inertia.
Deflection, δ = =
48EI ∂W
WL3
∫ ∂U = ∫ 48EI ∂W
Bending moment at the section = Px
W 2 L3 d2y
U= or EI 2 = − P.x
96EI dx

9. Deflection of Beams
1326. What is the maximum deflection in a
cantilever beam carrying a uniformly
distributed load of w per unit length? (L-
length of the beam, EI-flexural rigidity of the
section of the beam)
wL4 wL3 dy Px 2
(a) (b) Integrating EI =− + C1
3EI 192EI dx 2
5wL4 wL4 dy Pl 2
(c) (d) at x = l, = 0 C1 =
384EI 8EI dx 2
JPSC AE 10.04.2021, Paper-II dy Px 2 Pl 2
EI =− +
NLCIL GET 17.11.2020, Shift-II dx 2 2
Strength of Materials 395 YCT
 dy P 2 then deflection at free end will be max . (x = L)
Slope = (l − x 2 )
dx 2EI ML2
Integrating this equation y =
2 EI
Px 3 Pl 2
EI y = − + x + C2 and slope at free end (x = L)
6 2
 dy  ML
at x = l , y = 0 C 2 = −
Pl 3  dx  = EI
3   x=L
3 2 3
Px Pl Pl 1329. A simply supported beam of span l carries a
EI y = − + x−
6 2 3 point load W at mid span. The downward
Thus deflection given by deflection under the load will be :
P Wl 3 Wl 3
y=− (2l 3 − 3l 2 x + x 3 ) (a) (b)
6EI 3EI 8EI
At the free end (x = 0), the slope and deflection are 3
maximum W l 5 Wl 3
(c) (d)
48EI 384 EI
Pl 3
Deflection = − SJVN ET 2019
3EI ISRO Scientist/Engineer (RAC) 07.05.2017
–ve sign represent deflection is downward. MPSC HOD (Govt. Poly. Colleges) 04.10.2014
1328. Maximum deflection in cantilever due to pure UKPSC AE-2013, Paper-I
bending moment M at its ends is- APPSC AE 04.12.2012
ML2 ML2 VIZAG STEEL MT 2015
(a) (b)
2 EI 3EI Wl 3
2 2 Ans. (c) : Deflection, δ at mid span =
ML ML 48EI
(c) (d)
4 EI 6 EI
GPSC ARTO Pre 30.12.2018
TSPSC AEE 2017
UPPSC AE 12.04.2016 Paper-I
TSPSC Managers, 2015
ISRO Scientist/Engineer 2010
ISRO Scientist/Engineer 2006 1330. A simply supported beam of length L is
GATE 2012 subjected to uniformly distributed load w/unit
Ans. (a) : using double integration method, at any length for its entire span. Taking E as Young's
distance x from A, we have modulus and I as moment of inertia, the
d y2 deflection at the mid span is given by:
EI = −M x
dx 2 −5wL3 − wL3
(a) y = (b) y =
384 EI 384 EI
− wL4 −5wL4
(c) y = (d) y =
384 EI 384 EI
MECON MT 2019
Oil India Limited Sr. Engineer (Drilling) 30.11.2019
dy
EI = − Mx + C1 …(i) TRB Polytechnic Lecturer 2017
dx CGPSC AE 26.04.2015 Shift-I
x2 TANGEDCO AE 2015
EI ( y ) = M + C1 x + C2 …(ii)
2 Ans. (d) : We know that
If
A wL3
dy Maximum slope (θ) = =−
At x = 0 , = 0 ; C1 = 0 EI 24 EI
dx
 wL3   5 L 
At x = 0 , y = 0 ; C2 = 0  × . 
AX  24   8 2 
then putting the values of C1 & C2 in Maximum deflection (δ) = =
EI EI
eqn. (ii)
Mx 2 −5 wL4
EI ( y ) = δ=
2 384 EI

Strength of Materials 396 YCT

Note- Area of BMD (shaded) d  dy  d 2 y
=  =
2  L   wL  wL
2 3
dx  dx  dx 2
A = × × =
3  2   8  24 1 d2 y
∴ =
R dx 2
Now we know that equation of theory of bending
M σ E
= =
I y R
M I 1
∴ = , put the value of
EI R R
M d2 y d2 y
= 2 ⇒ M = EI 2
EI dx dx
1332. The simple supported beam 'A' of length 'l'
carries a central point load 'W'. Another beam
'E' is loaded with a uniformly distributed load
such that the total load on the beam is 'W'.
The ratio of maximum deflections between
beams A and B is
1331. The curvature equation of beam is given as: (a) 5/8 (b) 8/5
d3 y dy (c) 5/4 (d) 4/5
(a) M = EI (b) M = EI TSPSC AEE 28.08.2017 (Civil/Mechanical)
dx 3 dx
TSPSC AEE 2015
d2 y d4 y VIZAG Steel MT 2011
(c) M = EI (d) M = EI
dx 2 dx 4 J & K PSC Screening, 2006
Assam PSC AE (PHED), 18.10.2020 WL3
Oil India Ltd. Sr. Engg. (Production) 30.11.2019 Ans. (b) : δ A = (simply supported beam)
48EI
APPSC AEE Screening Test 2016
5wL4 (simply supported with
APPSC AEE 2012 δB =
384 EI uniform distributed load)
Ans. (c)
∵ Given, (W = wL)
5 WL3
Then δ B =
384 EI
δA 8
=
δB 5
1333. A cantilever beam carries a load W uniformly
distributed over its entire length. if the same
load is placed at the free end of the same
cantilever, then the ratio of maximum
deflection in the first case to that in the second
case will be is
(a) 3/8 (b) 8/3
Curvature equation of beam- (c) 5/8 (d) 8/5
Let ds ≅ dx ISRO Scientist/Engg. RAC 12.01.2020
JWM 2017
dx = Rdθ……….(i) APPSC AEE 2012
dy WBPSC AE 2003
tan θ =
dx ESE 1996
θ is very small ⇒ tanθ = θ Ans. (a) : In first case –
dy
θ= .............. ( ii )
dx
From equation (i)
1 dθ wℓ 4
= Deflection δ = {∵ W = wℓ}
R dx 8EI
Strength of Materials 397 YCT
 Wℓ 3 bd 3
δ= ...... (i) where, I =
8EI 12
[∴ Given – If width is doubled (i.e. b1 = 2b)]
In Second Case –
WL3 WL3 WL3
δ1 = = ⇒
48EI  2bd3  48E(2 × I)
48E ×  
 12 
δ
3
δ1 ⇒
Wℓ 2
Deflection δ = ...... (ii)
3EI 1336. The slope at the support of a simply supported
beam having uniformly distributed load
Wℓ3 throughout its span is __________.
δudℓ
= 8EI3 w.L3 w.L4
δconcentrated Wℓ (a) θmax = (b) θmax =
24EI 48EI
3EI
δudℓ 3 w.L3 w.L4
= (c) θmax = (d) θmax =
δconcentrated 8 48EI 24EI
VIZAG Steel MT 24.01.2021
1334. .....................theorem used based on strain
energy concept and can be used for SJVN ET 2019
determining deflection in a beam. WBPSC AE, 2007
(a) Euler's (b) Rankine's Ans. (a) :
(c) Gourdan's (d) Castigliano's
ISRO Scientist/Engineer 17.12.2017
HPPSC Asstt. Prof. 29.10.2016
HPPSC Lect. (Auto) 23.04.2016
4
APPSC AE 04.12.2012 Deflection (δ) = 5wℓ
Ans. (d) : Let, U = strain energy due to bending 384EI
moment. wℓ 3
b
M 2x − x dx Slope (θ) =
U=∫ 24EI
a
2 ( EI )x − x 1337. In a cantilever beam, If the length is doubled
As per this theorem, while keeping the cross-section and the
∂U ∂U concentrated load acting at the free end the
Deflection (y A ) = Slope( θA ) = same, the deflection at the free and will
∂WA ∂M A increase by:
Where, WA = concentrated load at A. (a) 2.66 times (b) 3 times
MA = Concentrated moment at A. (c) 6 times (d) 8 times
Castigliano's theorem is used based on strain energy OPSC AEE 2019 Paper-I
concept and can be used for determining deflection in ISRO Scientist/Engineer (RAC) 22.04.2018
a beam. OPSC AEE 2015 Paper-I
CSE Pre-1996
1335. If the width of a simply supported beam
carrying an isolated load at its centre is Ans : (d) : In a cantilever beam, If the length is doubled
doubled, the deflection of the beam at the while keeping the cross-section and the concentrated
centre is changed by: load acting at the free end the same, the deflection at the
(a) 1/2 (b) 1/8 free and will increase by eight times.
(c) 4 (d) 2
Nagaland PSC (CTSE) 2018, Paper-I
APPSC AE Subordinate Service Civil/Mech. 2016
TSPSC Managers, 2015
Cantilever beam when length = l
Ans. (a) :
wℓ3
Deflection ( δ1 ) =
3EI
Cantilever beam when length = 2l
w ( 2ℓ )
3

Deflection of simply supported beam (SSB) Deflection ( δ 2 ) =

3EI
WL3
(δ) = δ 2 = 8δ1
48EI
Strength of Materials 398 YCT
1338. Two simply supported beams 'A' and 'B' of 1
same breadth and depth and same material δ∝
carries constant central load W at the centre. a4
The span of beam B is double that of beam A. δ1 a 2 4
a 2 = a + 0.19a ⇒ =
The deflection of beam B compared to beam A δ 2 a14
will be
(a) one-half (b) one-eighth (a1 + 0.19a1 )4 a14
= 4 (1 + 0.19 ) = 2.005
4
=
(c) four times (d) eight times a14 a1
ISRO Scientist/Engineer (RAC) 12.01.2020
δ2
CSE Pre-1996 δ1 = 2.0053δ 2 ⇒ Decrease = 1 −
Ans. (d) : δ1
2.0053δ 2 − δ 2
Decrease =
2.0053δ 2
= 50.03%
wL3
δ= ≈ 50%
48EI
δ ∝ L3 1341. A cantilever beam of length L is subjected to
3 an on concentrated load P at a distance of L/3
δB  L B  from free end, what is the deflection at free
= =8
δ A  L A  end of the beam?
1339. In a cantilever, maximum deflection occurs 2PL3 3PL3
(a) (b)
where– 81EI 81EI
(a) bending moment is zero 3
14PL 15PL3
(b) bending moment is maximum (c) (d)
(c) shear force is zero 81EI 81EI
(d) slope is zero RPSC Vice Principal ITI 2018
RPSC 2016 ISRO Scientist/Engineer 03.07.2016
J & K PSC Screening, 2006 ESE 2004
Ans : (a) Ans. (c)

Deflection at free end A is

 L
δA = δBC + θBC ×  
In the cantilever beam loaded as shown, the max  3
3 2
bending moment is at (x = 0) but the max deflection  2L   2L 
occur at (x = l) P  P 
3 δ =  3  +  3  L
×
Wl A
Maximum deflection at x = l is given by = 3EI 2EI 3
3EI 8PL3 4PL3
1340. A cantilever beam having square cross section = +
of side is subjected to an end load. If a 81EI 54EI
increased by 19% the tip deflection decreases 14PL3
approximately δ A =
81EI
(a) 19% (b) 29%
(c) 41% (d) 50% 1342. The method in which a single equation is
GPSC Asstt. Director of Transport 05.03.2017 formed for all loads on the beam and the
RPSC 2016 equation constructed in such a way that the
GATE 2016 integral constants apply to all the sections of
Ans : (d) the beam is
(a) Moment area method
Wl 3 Wl 3
δ= = (b) Conjugate beam method
3EI 3E 1 a 4 (c) Macaulay's method
12 (d) method of super position
 12Wl 3  1
δ =  4 TSPSC Manager (Engg.) HMWSSB 12.11.2020
 3E  a APPSC AEE Screening Test 2016

Strength of Materials 399 YCT

 Ans. (c) : The method in which a single equations is Ans. (d) : For cantilever beam
formed for all loads on the beam and the equation
constructed in such a way that the integral constants
apply to all the sections of the beam is Macaulay's
method.
e.g.
⇓

PL3
Deflection at the free end, δ1 =
3EI
+
1343. A cantilever beam of rectangular cross section
is subjected to load at free end. If the depth of
the beam is double and load is halved,
deflection of the free end as compared to
original will be
(a) 1/2 (b) 1/8
ML2
(c) 2 (d) 1/16 Deflection, δ 2 =
Nagaland PSC CTSE 2017, Paper-I 2 EI
ISRO Scientist/Engineer 2009 So, Total Deflection (I + II)
WL3 = δ1 + δ2
Ans. (d) : For cantilever beam deflection y = PL3 ML2
3EI = +
bd 3 3EI 2 EI
I=
12 1345. A point load 'W' is acting at mid span of
b(2d ) 3 cantilever of length 'l'. If the free end is
I1 = supported on a rigid prop, the reaction of the
12 prop is
8bd 3 5W 5W
= (a) (b)
12 13 11
I1 = 8I
5W 7W
W 3 (c) (d)
L 16 9
y= 2 APPSC AE Subordinate Service Civil/Mech. 2016
3EI × 8
APGENCO AE 2012
WL3
= Ans. (c) :
6 EI × 8
WL3 WL3 1 y
= = × =
48 EI 3EI 16 16
1344. The given figure shows a cantilever of span 'L'
subjected to a concentrated load 'P' and a To find reaction due to prop,
moment 'M' at the free end. Deflection at the Deflection at free end due to load 'W' at mid span
free end is given by = Deflection at free end due to prop reaction
W (l / 2) W (l / 2)
3 2
l Pl 3
+ × =
3EI 2EI 2 3EI
3 3
Wl Wl Pl 3
PL2 ML2 ML2 PL2 + =
(a) + (b) + 24EI 16EI 3EI
2 EI 3EI 2 EI 48EI
2W + 3W P
ML2 PL2 ML2 PL3 =
(c) + (d) + 48 3
3EI 2 EI 2 EI 3EI
5W
UJVNL AE 2016, ISRO Scientist/Engineer 2011 P=
ESE 1996 16

Strength of Materials 400 YCT

1346. A simply supported laterally loaded beam was Ans : (c)
found to deflect more than a specified value.
Which of the following measures will reduce
the deflection?
(a) Increase the area moment of inertia Wℓ 2
(b) Increase the span of the beam When load acts free end then slope =
2EI
(c) Select a different material having lesser
W ( ℓ / 2)
2
modulus of elasticity
When load acts mid span the slope =
(d) Magnitude of the load to be increased 2EI
RPSC Lect. (Tech. Edu. Dept.) 13.03.2021, Paper-I Wℓ 2
GPSC ARTO Pre 30.12.2018 Slope =
8EI
Ans. (a) : For deflection, deflection is inversely
proportional to moment of inertia, 1350. The resultant deflection of a beam under
symmetrical bending is
Wℓ3
Deflection (δ) = (a) perpendicular to the neutral axis
48EI (b) parallel to the axis of symmetry
1 (c) perpendicular to the axis of symmetry
δ∝
I (d) parallel to the neutral axis
Therefore, increase in moment of inertia, reduces Assam PSC AE (IWT) 14.03.2021
deflection. UKPSC AE-2013, Paper-I
1347. For a fixed end of a long column, _________. Ans. (a) : The resultant deflection of a beam under
(a) The deflection and the slope is zero symmetrical bending is perpendicular to the neutral
(b) The deflection is zero, but the slope is axis.
maximum 1351. A simply supported beam is of rectangular
(c) The deflection is maximum, but the slope is section. It carries the uniformly distributed
zero load over the span. The deflection at the centre
(d) The deflection is zero, but the slope is is y. If the depth of the beam is doubled, the
minimum deflection at the centre is ______.
VIZAG Steel MT 24.01.2021 (a) 2y (b) y/8
APPSC AEE 2012 (c) y/2 (d) 4y
Ans. (a) : For a fixed end of a long column, the VIZAG Steel MT 24.01.2021, Shift-I
deflection and slope are zero. The fixed beam is WBPSC AE, 2017
stronger, stiffen and more stable. The slope at the Ans. (b) : Given,
supports is zero. Maximum bending moment at the Figure
centre is reduced because of fixing moments developed
5wl 4
at supports. The deflection is zero for hinged end. Then, deflection (y) =
1348. The flexural rigidity for the deflection of beam 384 EI
is expressed as ∴ If depth of beam is doubled d1 = 2d, b1 = b
I E 5wl 4
(a) (b) Now, y1 =
E I bd 3
384 × E ×
1 12
(c) EI (d)
EI 5wl 4
APPSC AEE Mains 2016 (Civil Mechanical) y1 =
b × ( 2d )
3

APPSC Poly. Lect. 2013 384 × E ×

APPSC AEE 2012 12
Ans. (c) : The flexural rigidity for the deflection of 5wl 4 5wl 4
= =
beam is expressed as EI. bd 3 384 EI8
384 × E × ×8
1349. A cantilever beam of length L, moment of 12
inertia of inertia I and Young modulus E
1 5wl 4 1
carries a concentrated load W at the middle of y1 = × = ×y
its length. The slope of cantilever at the free 8 384 EI 8
end is: y
(a) (Wℓ2)/2EI (b) (Wℓ2)/4EI y1 =
8
(c) (Wℓ2)/8EI (d) (Wℓ2)/16EI 1352. A fixed beam of length l carries a point load W
OPSC AEE 2015 Paper-I at the centre. The deflection at the centre is
ESE 2001 ______.
Strength of Materials 401 YCT
 (a) one-fourth of a deflection of a simply d2 y 12x 2
supported beam EI = 2 × 3 × 2x − −0
(b) half of the deflection of a simply supported dx 2 6
beam d3 y 24x
EI 3 = 12 −
(c) double the deflection of a simply supported dx 6
beam
d4y 24
(d) same as for a simply supported beam EI = 0−
VIZAG Steel MT 24.01.2021, Shift-I dx 4 6
Ans. (a) : d4 y
EI = −4
dx 4
d4 y
q = − EI
deflection at centre, dx 4
Wℓ3 q = − (−4) = 4kN / m
( y )1 =
192EI 1354. A cantilever beam of 3 m is carrying a point
load of unknown mass and produce a slope at
the free end as 0.01 rad. The deflection at the
free end is _________.
Wℓ3
( y2 ) = (a) 20 mm (b) 5 mm
48EI (c) 15 mm (d) 10 mm
∴ So, there deflection at centre of fixed beam, VIZAG Steel MT 24.01.2021
1
( y1 ) = ( y2 ) Ans. (a) :
4
1 Wℓ3
= ×
4 48EI
1
( y1 ) = × deflection of simply supported beam At point 'B'
4
1353. The equation for the deflected shape of a beam Pℓ 2
Slope (θ) =
carrying a uniformly distributed load and 2 EI
simply supported at the ends is given by :
Pℓ 2
1  3 x4  =θ
y=  2x − − 36x  2EI
EI  6 
Given, θ = 0.01 radian, ℓ = 3m
where E is the elastic modulus of beam
material and I is the second moment of beam P × (3) 2
cross-section area about neutral axis. What is 0.01 =
2EI
the intensity of uniformly distributed load over
the beam span? 0.01× 2EI 0.02 EI
P= =
(a) 12 kN/m (b) 8 kN/m 9 9
(c) 6 kN/m (d) 4 kN/m Deflection at free end 'B',
RPSC ACF & FRO, 26.02.2021 Pℓ 3
Ans. (d) : Equation of deflected shape of beam (δ) =
3EI
1  3 x4 
y=  2x − − 36x  0.02EI
EI  6  × 27
δ= 9
 d4 y  3EI
Intensity of loading (q) = −  EI 4 
 dx  0.02 EI × 3
=
1  3 x 4
 3EI
y=  2x − − 36x 
EI  6  δ = 0.02 m
δ = 20 mm
x4
EIy = 2x 3 − − 36x 1355. A force P is applied at a distance x from the
6
end of the beam as shown in the figure. What
dy 4x 3 would be the value of x so that the
EI = 2 × 3x 2 − − 36
dx 6 displacement at A is equal to zero?

Strength of Materials 402 YCT

 6
δc = 0.002 ×
3
δc = 0.004 m
δc = 4 mm
1357. In deflection curve of beams, the point of
contraflexure or inflection point is the point
(a) of maximum curvature
(a) 0.33L (b) 0.25L (b) of zero-curvature
(c) 0.5L (d) 0.66L (c) of zero slope
Assam Engg. College AP/Lect. 18.01.2021 (d) none of the above
GATE 2014 Haryana PSC AE (PHED) 05.09.2020, Paper-II
Ans. (a) : 0.33L Ans. (b) : Point of contraflexure or inflection point
occur in bending moment diagram. At this point
bending moment change its sign from positive to
negative and vice-versa. At this point (the point of
contraflexure) bending moment becomes zero.
σ M E
= =
y I R
R = Radius of curvature
At point of contraflexure, BM = 0
Deflection (displacement) at point 'A' will be zero
M E
So, δA = δ1 + δ2 = 0 =
I R
PL P(L − x)L
3 2
− + =0 1 M
3EI 2EI = =0
R EI
PL P ( L – x ) L
3 2
= 1358. A cantilever beam of length 100 mm is
3EI 2EI subjected an end load of 150 N. If Young's
3 3
PL PL – PL x 2 Modulus is 200 GPa and Moment of Inertia is
= 5000 mm4, the maximum deflection of this
3 2 beam is:
2PL3 = 3PL3 – 3PL2x (a) 5 mm (b) 0.01 mm
3PL2x = PL3 (c) 0.5 mm (d) 0.05 mm
x = L/3 CIL MT 27.02.2020
x = 0.33L
Ans. (d) : Given, ℓ = 100 mm
1356. A simply supported beam of length 6 m is
carrying a point load W at the centre. When Load W = 150 N
the slope at the ends of the beam is 0.002 E = 200 GPa
radians, the deflection at the centre would I = 5000 mm4
be_______.
Wℓ3
(a) 2 mm (b) 1 mm Maximum deflection =
(c) 8 mm (d) 4 mm 3EI
VIZAG MT, 14.12.2020 150 × (100)3
ymax =
Ans. (d) : 3 × 200 × 103 × 5000
ymax = 0.05 mm
1359. Which of the following states, "The deflection
between any two points is equal to the
moments of its area of bending moment
diagram between the two points about the last
Slope at ends of simply supported beam = 0.002 radians point divided by EI"?
Wℓ 2 (a) Macaulay's Method
= 0.002 (b) Mohr's theorem
16EI
(c) Castigliano's first theorem
Deflection at centre of beam δc (d) Double integration method
Wℓ3 VIZAG MT, 14.12.2020
δc =
48EI Ans. (b) : Mohr's Theorem–It states that deflection
Wℓ 2 ℓ between any two points is equal to the moments of its
δc = × area of bending moment diagram between the two
16 EI 3 points about the last point divided by EI.
Strength of Materials 403 YCT
1360. Simply supported beam of span (l) carries 1363. A girder of uniform section and constant depth
UDL (w kgf/m) throughout. Deflection at is freely supported over a span of 3 metres.
center is (y). If length is doubled, deflection at The point load at the midpoint is 30 kN and
center would be: Moment of inertia = 15 × 10–6 m4 and Young's
(a) 2y (b) 4y Modulus = 200 GN/m2. The deflection at centre
(c) 8y (d) 16y will be :
GPSC Engineer, Class-II Pre-19.01.2020 (a) 6.6 cm (b) 8.6 cm
Ans. (d) : For simply supported beam carrying UDL, (c) 5.6 cm (d) 6.6 cm
5 wℓ4 ISRO Scientist/Engineer (RAC), 10.03.2019
Deflection, δ =
384 EI Ans. (c) :
δ ∝ ℓ4
If length is doubled, deflection at centre,
5 w(2ℓ)4
δf =
384 EI
5  wℓ 4 
δf = (16)  
384  EI  Deflection of mid
δfinal = 16 δinitial Wl 3
δB =
1361. Assertion (A) : In a simply supported beam 48 EI
subjected to a concentrated load P at mid- 30 ×103 × (3000)3
span, the elastic curve slope becomes zero =
under the load. 48 × 200 × 103 ×15 ×10−6 × (103 ) 4
Reason (R) : The deflection of the beam is δ B = 5.625mm
maximum at mid-span.
(a) Both (A) and (R) are individually true and 1364. A cantilever beam with rectangular cross-
(R) is the correct explanation of (A) section is subjected to uniformly distributed
(b) Both (A) and (R) are individually true but load. The deflection at the tip is δ1. If the width
(R) is NOT the correct explanation of (A) and depth of the beam are doubled then
(c) (A) is true but (R) is false δ
(d) (A) is false but (R) is true deflection at tip is δ2. Then 2 is
OPSC AEE 2019 Paper-I δ1
Ans : (a) : Elastic curve slope becomes zero at the point (a) 0.0625 (b) 16
of maximum deflection in this case. Hence, of both (c) 0.5 (d) 2
assertion and reason are correct and reason is correct APPSC AEE SCREENING 17.02.2019
explanation of assertion. 1
Ans. (a) : Deflection ∝
I
 bd 3 
 
δ 2 I1 12 
= = 
1
∴ = = 0.0625
δ1 I 2 (2b)(2d )3 16
1362. The value of maximum deflection for a 12
cantilever beam of length l and carrying 1365. In the case of a beam simply supported at both
uniformly varying load from zero at free end ends, if the same load instead of being
to w per unit length at fixed end is : concentrated at centre is distributed uniformly
(a) wl3/8 EI (b) wl3/16 EI throughout the length, then deflection at
4 3
(c) wl /30 EI (d) wl /48 EI centre will get reduced by
ISRO Scientist/Engineer (RAC), 10.03.2019 (a) 1/2 times (b) 1/4 times
GATE 2019 (c) 5/8 times (d) 3/8 times
Ans. (c) : ISRO Scientist/Engineer (RAC) 22.04.2018
Ans. (d) : Case-I
Center point load on SSB, we have deflection.
WL3
δP =
48EI
wℓ 3 Case-II
Slope ( θ) = Deflection for UDL
24EI
5wL4
wℓ 4 δUDL =
Deflection ( δ) = 384 EI
30EI If, W=w×L
Strength of Materials 404 YCT
then 1368. The maximum deflection of a simply supported
3 beam, 4 m long due to mid span concentrated
5 WL 48
δUDL = × × load of 5 kN is
384 EI 48 (Take E = 200 GN/m2, I = 16 × 10–6 m4)
δUDL 5 5 (a) –6.8 mm (b) –3.8 mm
= × 48 = (c) –5.2 mm (d) –2.083 mm
δP 384 8
Karnataka PSC AE, 10.09.2017
So the reduction is given by
Ans. (d) :
5 3
δ P − δUDL = δ P − δ P = δ P
8 8
1366. The ratio of area under the bending moment
diagram to the flexural rigidity between two
PL3
points along a beam gives the change in Deflection ( δc ) =
(a) deflection (b) slope 48EI
(c) shear force (d) bending moment 5 × 103 × (4000)3
TSPSC AEE 28.08.2017 (Civil/Mechanical) =
48 × 200 ×109 ×10−6 × 16 × 10−6 × 1012
CSE Pre-1998 5 × 64 × 1012
2 =
d y M 48 × 200 ×16 × 109
Ans. (b) : =
dx 2 EI = 2.083 mm
Integrating it between two points A and B on elastic Note-(–ve) sign shows the downward deflection of
curve beam.
B 2 B 1369. A simply supported beam AB carries a point
d y M
∫A
dx 2 ∫=
A
EI
load W at a point C at a distance less than half
from the left end A. The maximum deflection
B B will be
 dy  Mdx
∫
=  = (a) at C (b) between A and C
 dx  A A EI (c) between C and B (d) None of the above
B
Karnataka PSC AE, 10.09.2017

∫A
Mdx = (area of bending moment diagram between A Ans. (c) :

and B)
1367. A simply supported beam AB of span L is
subjected to a concentrated load W at the • Deflection is maximum slope is zero in the region
centre C of the span. According to Mohr's between point of application of load and mid span.
moment area method, which of the following • Slope is maximum at a support which is nearer to
gives the deflection under the load? the load.
(a) Moment of the area of M/EI diagram 1370. Governing differential equation for small
between A and C taken about C deflections of elastic beams is given by
(b) Moment of the area of M/EI diagram
d2 y M d3y
between A and B taken about B (a) 2
= (b) EI 3 = V ( x )
(c) Moment of the area of M/EI diagram dx EI dx
4
between A and B taken about A d y
(c) EI 4 = q ( x ) (d) All the above
(d) Moment of the area of M/EI diagram dx
between A and C taken about A ISRO Scientist/Engineer 17.12.2017
TSPSC AEE 28.08.2017 (Civil/Mechanical) Ans. (d) : Differential equation of the elastic line is
Ans. (d) : d2 y
EI 2 = M(x)
dx
d 2 y M(x)
=
dx 2 EI
For shear force
d3 y
EI 3 = V(x)
dx
For shear force density (distributed load)
Moment of the area of M/EI diagram between A and C d4 y
EI 4 = q(x)
taken about A. dx

Strength of Materials 405 YCT

1371. We can find the deflection of beam carrying: Theorem I = The change in slope between tangents
(a) Uniformly distributed load draw to plastic curve at any point K & L is equal to the
(b) Central point load 1
(c) Gradually variable load product of , Multiplied by area of moment.
(d) All of these loads EI
TRB Polytechnic Lecturer 2017 1 L M
θkL = (Area KL) = ∫ dx
Ans. (d) : A subjected to UDL, central point load and EI K EI
also gradually variable load, deflection can find out.
1375. A simply supported beam AB of L span and EI
1372. A uniform bar, simply supported at the ends, flexural rigidity is subjected to MB moment at
carries a concentrated load P at mid-span. If B. Then the rotation at A is given by
the same load be, alternatively, uniformly
ΜL ΜL
distributed over the full length of the bar the (a) θΑ = (b) θΑ =
maximum deflection of the bar will decrease by. 6EI 4EI
(a) 25.5% (b) 31.5% ΜL ΜL
(c) 37.5% (d) 50.0% (c) θΑ = (d) θΑ =
3EI 3EI
ESE 2017 APPSC AEE Mains 2016 (Civil Mechanical)
Ans. (c) :
Ans. (a) : As we know,

PL3 5 PL3
δ1 = δ2 =
48EI 384 EI
δ1 − δ2 1
% decrease in max deflection = ×100 θΑ = (Area AB)
δ1 ΕΙ
ΒM
1 5 θΑ = ∫ dx
− Α EI

= 48 384 We get,
1
ΜL
48 θΑ =
= 37.5% 6EI
1373. A cantilever beam of length L is subjected to a 1376. The width b and depth d of a cantilever beam
point load W at free end. Then the slope at the carrying a point load at its free end are
free end is equal to changed into 0.5b and 2d respectively. Then its
WL3 WL2 maximum deflection for the same load
(a) (b) condition reduces by
3EI 60EI
3 2 (a) one-fourth (b) half
WL WL
(c) (d) (c) three-fourth (d) zero
60EI 2EI APPSC AEE Mains 2016 (Civil Mechanical)
APPSC AEE Mains 2016 (Civil Mechanical)
Ans. (c) : As, we know,
Ans. (d) : For cantilever beam, Max. deflection formula for cantilever
WL2 WL3
Slope of free end (θ) = δ max 1 =
2EI 3EI1
1374. The mathematical expression for moment area
theorem I for angle θKL between tangents bd 3
∴ I1 = …(i)
drawn at K and L to elastic curve is 12
L L
MI M 0.5b × (2d)3 8d 3b 4bd 3
(a) θΚL = ∫ dx (b) θΚL = ∫ dx I2 = = =
K
E K
EI 12 2 × 12 12
M E ∴ I 2 = 4I1
(c) θΚL = (d) θΚL =
EI MI WL3 δmax 1
APPSC AEE Mains 2016 (Civil Mechanical) δ max 2 = =
3EI 2 4
Ans. (b) : As we know,
δ max 1
Ratio ⇒ Reduction = δ max 1 −
4
3δ max 1
=
4

Strength of Materials 406 YCT

1377. The elastic curve of a beam means Wa 2 b 2 Wabl
(a) BMD (b) SFD (a) (b)
3EIl 3EI
(c) Deflection curve (d) Stress-stain curve
APPSC AEE Mains 2016 (Civil Mechanical) Wab Wa 2 bl
(c) (d)
Ans. (c) : Deflection curve, is the elastic curve of a 6EIl 3EI
beam, APPSC AE Subordinate Service Civil/Mech. 2016
1378. A steel beam of breadth 100 mm and height Ans. (a) :
1000 mm is loaded as shown in the figure
below. Assume modulus of elasticity as 200
GPa.

Wb
Reaction at A =
l
What is the maximum deflection in the beam? Wa
(a) 751.25 mm (b) 781.25 mm Reaction at B =
(c) 805.25 mm (d) 821.75 mm l
The strain energy stored by the Beam AB (U)
(e) 851.75 mm
U AB = U AC + U BC
CGPSC Asstt. Workshop Supt., 17.07.2016
a 2 b 2
Ans. (*) : Given, L = 10 m = 10000 mm
 Wb  dx  Wa  dx
E = 200 GPa = 200 × 109 N/m2 =  ∫ x +  x ∫
 l  2EI 0  l  2EI
= 200 × 103 MPa or N/mm2 0
Breath (b) = 100 mm W 2 b 2 a 3 W 2 a 2 b3
Height (h) = 1000 mm = +
6EI.l 2 6EI.l 2
w = 100 kN/m
W 2 a 3 b 2 W 2 a 2 b3
W = 100 × 10 kN = 106 N = +
6EIl 2 6EIl 2
bh 3 0.1 × (1)3
I= = W 2a 2 b2
12 12 = (a + b)
We know, 6EIl 2
∵ a+b =l
5WL4
δ= W 2 a 2 b2
384EI UAB =
6EIl
5 × 106 × (10000)4 × 12
= Deflection under load 'W' is given by –
384 × 200 × 103 × (1000)3 × 100
∂u ( 2W ) a b
2 2
= 78125.0 mm δ= =
Note-Official answer given by the commission (b). ∂W 6EIl
2 2
1379. Which of the following describes a single load Wa b
δ=
or force that has a small contact area as to be 3EIl
negligible compared with the entire surface
1381. A simply supported beam of 10 m span is
area of the supporting member? carrying a load of 4.8 kN at mid span. If
(a) Contact load (b) Concentrated load Young's modulus of elasticity (E) is 2×108
(c) Common load (d) None of these kN/m2 and moment of inertia (I) is 20 cm4, then
KPCL AE 2016 the maximum deflection will be
Ans. (b) : Concentrated load describes a single load or (a) 5.00 mm (b) 2.50 mm
force that has a small contact area to be negligible (c) 0.50 mm (d) 0.25 mm
compared with the entire surface of area of the APPSC AEE Screening Test 2016
supporting member. Ans. (b) : given
I = 20 cm4
E = 2×108 kN/m2

1380. A beam of length l simply supported at the

ends carries a point load W at a distance 'a'
from the left end. Also l -a= b. The deflection
under load is given by :
Strength of Materials 407 YCT
 1384. Maximum slope in case of a cantilever of
Wℓ3 length l carrying a load P at its end is
Maximum deflection (yc) =
48EI (a) Pl2/2EI (b) Pl2/EI
2
(c) Pl /4EI (d) Pl2/6EI
4.8 × 103 × (10)3
= (e) Pl2/8EI
48 × 2 × 108 × 103 × 20 × 10−8 CGPSC AE 26.04.2015 Shift-I
4.8 ×103 × 103 4.8 × 103 Ans. (a) :
= = Type of load Slope at end Maximum
48 × 2 ×10 × 20 48 × 2 × 20
3
deflection
4800 100 Pl 2 Pl 3
= = = 2.5 mm θ= δ=
48 × 2 × 20 40 2 EI 3EI
1382. A cantilever beam rectangular in cross section
is subjected to a load W at its free end, causing wl 3 wl 4
θ= δ=
deflection δ1. If the load is increased to 2W, 6 EI 8EI
δ
causing deflection δ2, the value of 1 would be 1385. The ratio of central deflection due to a central
δ2 load in the case of a beam freely supported at
(a) 1 (b) 2 both ends to the beam fixed at both ends will
(c) 4 (d) 1/2 be-
Nagaland PSC CTSE 2016 Paper-I (a) 1/2 (b) 4
Ans. (d) : A cantilever beam rectangular in cross- (c) 1/4 (d) 2
section subjected, load (W), then deflection (δ1). ISRO Scientist/Engineer (RAC) 29.11.2015
WL3 ____ ( ) Ans : (b) Central deflection for simply supported beam
δ1 = i for central load W.
3EI
WL3
When, Load is increased to (2W), then deflection (δ2) δ1 =
will be, 48EI
Central deflection for fixed beam (at both ends) for
2WL3 ____ ( ) central load W.
δ2 = ii
3EI WL3
from (i) 2 (ii) δ2 =
192EI
δ1 WL3 1
= = Then, δ1 / δ 2 = 4
δ2 3EI 2
2WL3 1386. A frame of two arms of equal length L is shown
in the figure. The flexural rigidity of each arm
3EI of the frame is EI. The vertical deflection at the
1383. If the depth of a rectangular beam is halved, point of application of load P is
the deflection for a beam carrying a mid point
load shall be–
(a) halved (b) doubled
(c) four times (d) eight times
RPSC 2016
d
Ans : (d) Given, d1 = , d2 = d
2 PL3 2PL3
3 3 (a) (b)
δ = WL = WL 3EI 3EI
48EI 3
48E bd (c)
4PL3
(d) None of the above
12
3EI
1
δ∝ ISRO Scientist/Engineer 11.10.2015
d3 GATE 2009
 d Ans. (c) :
If depth is halved  d 2 = 
 2
3
 d1 
δ  2  1
1= =
δ d13 8
2
δ = 8δ
2 1

Strength of Materials 408 YCT

 P 2 L3
UBC =
2EI
∂U BC 2PL3 PL3
= =
∂P 2EI EI
∂U AB ∂U BC
Vertical deflection = +
∂P ∂P
3 3 3
PL PL 4 PL
= + =
3EI EI 3 EI
1387. A concentrated load P is applied at the end of a
cantilever as shown in Fig. The cross section of
For section 'AB' the beam is a square of side 'a' with a hole of
M 2 dx dia 'a/2'. The deflection at the tip of the
UAB = ∫2EI cantilever is given by
L M 2 dx
=
0 ∫ 2EI
L (P × x) 2 dx L P2 × x 2dx
=
0 ∫ 2EI
= ∫0 2EI
L
P2 L P2  x3 
∫ 3P L3 L3
2
= x dx =   1024 P
2EI 0 2EI  3  0 (a) (b)
E a4 (256 − 3π ) E a 4
P 2  L3 
=   1024 P L3 256 P L3
2EI  3  (c) (d)
π 4
(1024 − 3π ) E a 4
P 2 L3 (256 − ) E a
UAB = 64
6EI ISRO Scientist/Engineer 12.05.2013
For Section 'B' Ans. (b) : Firstly find the moment of inertia of a Hollow
cross-section.
Let moment of Inertia I1 and I2

M 2 dx
L
UBC =
0∫ 2EI
I = I1 - I2
L (P × L) 2 dx
=
0∫ 2EI =
a 4 π (a / 2) 4
−
12 64
L P 2 × L2 dx
=
0∫ 2EI =
a4
−
π a4
2 2 12 16 × 64
PL L P 2 L2 L
=
2EI 0 ∫
dx =
2EI
[x]0
=
a4 1 π 
 3 − 4 × 64 
4  
P 2 L3
UBC = a 4  256 − 3π 
2EI I=
So, vertical deflection, 4  4 × 3 × 64 
∂U AB ∂U BC So, Deflection of Cantilever beam
= +
∂P ∂P PL3
δ=
P 2 L3 3EI
UAB =
6EI PL3
=
∂U AB 2PL3 PL3 a  256 − 3π 
4
= = 3E 
∂P 6EI 3EI 4  4 × 3 × 64 

Strength of Materials 409 YCT

 4 × 4 × 64 P × L3  ii 
= From   –
(256 − 3π ) E × a 4 i
1024 P  L3  δ 2 Pl 3 / 2EI 3
∴ Deflection, δ = = = = 1.5
  δ1 Pl 3 / 3EI 2
(256 − 3π ) E  a 4 
1388. Determine the equivalent stiffness of the δ 2 = 1.5δ1
system shown below 1390. The slope and deflection at the centre of a
simple beam carrying a central point load are
(a) zero and zero
(b) zero and maximum
(c) maximum and zero
(d) minimum and maximum
BPSC AE 2012 Paper - VI
Ans : (b) : zero and maximum
(a) 6.35 × 107 N/m (b) 4 × 107 N/m 1391. A beam of length, l, fixed at both ends carries a
(c) 3 × 10 N/m
7
(d) 5 × 105 N/m uniformly distributed load of w per unit length.
Vizag Steel MT (Re-Exam) 24.11.2013 If EI is the flexural rigidity, then the maximum
Ans. (a) : Given, E = 180 × 10 N/m , I = 4.6 × 10 m
9 2 6 4 deflection in the beam is
ks = 1.5 × 10 N/m
7 wl4 wl 4
(a) (b)
L1 = 80 cm = 0.8 m 192EI 24EI
L2 = 10 cm = 0.1 m wl 4
wl4
(c) (d)
PL3 384EI 12EI
Deflection at A, δA =
3EI APPSC AEE 2012
P 3EI Ans : (c) (i)
Beam stiffness, k 0 = = 3
δA L
3EI
Equivalent stiffness, keq = + ks
L31
3 × 180 × 109 × 4.6 × 106
= + 1.5 × 107
(0.8)3 Wℓ 4
= 4851.5625 × 1015 + 1.5 × 107 Deflection ( δc ) =
384EI
≈ 4852 × 1015 N/m (ii)
No option is matching.
But official answer is (a).
1389. If a cantilever beam of length l, the
concentrated load P at the end is replaced by a
bending moment P × l, the deflection will
Wℓ3
increase by : Deflection ( δc ) =
(a) 1 time (b) 1.5 times 192EI
(c) 1.25 times (d) 2 times 1392. A simply supported beam span 3m, is subjected
J&K PSC Civil Services Pre, 2013 to a central point load of 5kN, then the slope at
the mid span is equal to
Ans. (b) : Case-I
25 256
(a) (b)
24EI EI
40
(c) (d) Zero
Deflection at free end, 48EI
Pl 3 APPSC AEE 2012
δ1 = .....(i) Ans : (d) At mid span of the beam, deflection is
3EI
maximum and slope is zero.
Case-II
1393. A list of some theories of failure is given below:
(1) Maximum normal stress theory
(2) Maximum shear stress theory
(3) Maximum distortion theory
Ml 2 (Pl )l 2
δ2 = = .....(ii) Which of these will be used for designing the
2EI 2EI machine components made of Cast Iron?
Strength of Materials 410 YCT
 (a) (2) only (b) (2) and (3) Given,
(c) (1) an (3) (d) (1) only Beam length = 4 m
RPSC AE GWD, 2011 P = 120 kN
EI = 2000 kN-m2
Ans. (d) : Cast iron is a brittle material so for designing Deflection of centre of beam
the machine components made of cast iron maximum
PL3 120 × 43
normal stress theory is the best suited. δ max = = = 0.02m
1394. The maximum deflection of a fixed beam of 192EI 192 × 2000
δmax = 2 cm = 20 mm
length ‘ℓ’ carrying a central point load w is
1397. A simply supported beam with width 'b' and
wℓ3 wℓ3 depth 'd' carries a central load W and
(a) (b) undergoes deflection δ at the centre. If the
48 EI 96EI
width and depth are interchanged, the
wℓ3 wℓ3 deflection at the centre of the beam would
(c) (d)
192EI 384EI attain the value :
2
TNPSC ACF 2012 d d 
(a) δ (b)   δ
Ans. (c) : b b
3 3/ 2
d  d 
(c)   δ (d)   δ
b b
ISRO Scientist/Engineer 2011
wℓ3
Ymax = Ans. (b) : In reactangle
192 EI width = b
1395. The reaction at the prop in a propped depth = d
cantilever beam subjected to u.d.l. is Deflection for S.S.B. when load acting at center of
Wl 3Wl beam.
(a) (b)
4 8 WL3
δ=
5Wl 6Wl 48 EI
(c) (d)
8 7 bd 3
I1 =
APPSC AEE 2012 12
Ans. (b) : RB = Prop reaction. Then Interchanged depth and width
db3
I2 =
12
1 3
δ1 I 2 12 db
Deflection of cantilever beam for u.d.l = =
δ 2 I1 1 3
Wl 4 bd
δB = ...........(i) 12
8EI
δ1  b 
2

ΣδB = 0 = 
δ2  d 
Wl 4 R B .l 3
– =0 2
8EI 3EI d 
δ 2 =   δ1
3 b
R B = Wl
8 δ 1 = δ
2
1396. A beam length 4 m, fixed at both ends carries a d
point load of 120 kN at the centre. If EI for the δ 2 =   δ
beam is 2000 kN m2, deflaction at the centre of b
beam is 1398. If the diameter of a circular section is doubled
(a) 1.0 mm (b) 2.0 mm its deflection is reduced by :
(c) 5.0 mm (d) 10.0 mm (a) 16 times (b) 8 times
APPSC AEE 2012 (c) 4 times (d) 2 times
Ans. (*) : J&K PSC Civil Services Pre, 2010
Ans. (a) : Given, Deflection =δ
Diameter- d
1
δ∝ 4
d

Strength of Materials 411 YCT

 4 4 Flexural rigidity = EI
δ1  d 2   2d1 
=  = 
δ2  d1   d1 
δ1
= 16
δ2 ⇓
1
δ2 = δ1
16 ...(1)
1399. Maximum slope of a cantilever of length 6 m +
carrying load at a distance of 4 m from fixed
end is _______. ...(2)
(a) Fixed end (b) Under the load YA = Y1 + Y2
(c) Free end (d) Can’t predict
PL3
J&K PSC Civil Services Pre, 2010 Y1 = −
3EI
Ans. (*) : Both (c) & (b) are correct.
Slope at a point 'C' = ML2
Y2 =
WL2 2 EI
θC = YA = 0 [deflection at point A is zero)
2EI 3 2
PL ML
W ( 4)
2 − + =0
16W 3EI 2 EI
θC = =
2EI 2EI M=P×a
16W P × a × L2 PL3
Maximum slope, θmax = θC = θB = =
2EI 2 EI 3EI
∴ So, hence maximum slope at free end and under the
a L
load is same. =
2 3
a 2
=
L 3
1401. A cantilever beam of length 'L' carries a
1400. A cantilever beam AB of length L and uniform concentrated load 'P' at its midpoint. What is
flexural rigidity EI has a bracket AC attached the deflection of the free end of the beam?
to its free end shown in Fig. Vertical load is PL3 PL3
applied to the free end C of the bracket. In (a) (b)
24 EI 48EI
order that the deflection of point A to be zero
the ratio a/L should be PL3 5 PL3
(c) (d)
16 EI 48EI
UKPSC AE 2007 Paper -I
Ans. (d) :

1 1 5 PL3
(a) (b) YAB =
2 3 48EI
1402. A cantilever of length 4 m carries a uniformly
1 2
(c) (d) distributed load w throughout its length. If the
4 3 maximum bending moment in the cantilever is
ISRO Scientist/Engineer 2010 80 KNm, and El is the flexural rigidity, the
Ans. (d) : slope at the end of the cantilever is
640 620
(a) (b)
3EI 3EI
160 80
(c) (d)
3EI 3EI
WBPSC AE, 2007
Strength of Materials 412 YCT
Ans. (*) wL3
∴ deflection δ ' = ...(i)
48EI'
wL3 wL3 × 12 × d 2  db3 
Deflection, δ ' = =  I′ = 
db3 48E(db3 ) × d 2  12 
wℓ3 48E
Slope θ = 12
6EI
wL3 ×12 × d 2
wℓ 2 =
∵ Maximum bending moment M = at fixed end 48E × (db3 ) × b 2
2
in cantilever beam with UDL wL3 × d 2
=
48E × (bd 3 ) × b 2
w × (4)2
80 =  wL3  d 2
2 = × 2
160  48EI  b
w= = 10kN / m
10 d2
= δ. 2 ...(ii)
10 × (4)3 b
Slope (θ) = d = 4, b = 6 substitute in equation (ii)
6EI
64 × 10 320 16 4
= = δ' = δ× = δ × = 0.44δ
6EI 3EI 36 9
1403. A beam is simply supported at its ends over a δ ' = 0.44δ
span l . If the load applied at the middle of the
beam is W, the minimum slope in the beam is
10. Theory of Failure
Wl 2 Wl 2
(a) (b)
16 EI 3 EI 1405. Which of the following is applied to brittle
Wl 2 materials?
(c) (d) none of these (a) maximum principal stress theory
2 EI
(b) maximum principal strain theory
WBPSC AE, 2007
(c) maximum strain energy theory
Ans. (a) (d) maximum shear stress theory
JPSC AE 10.04.2021, Paper-II
APPSC Poly Lect. 13.03.2020
OPSC AEE 2019 Paper-I
Gujarat PSC AE 2019
(CGPSC Polytechnic Lecturer 2017)
WL2 RPSC AE 2016
Slope( θ) = UPRVUNL AE 07.10.2016
16EI
RPSC VPITI 14.02.2016
WL3 TSPSC AEE 2015
Deflection ( δ) = ISRO Scientist/Engineer 11.10.2015
48EI
TANGEDCO AE 2015
1404. A simply supported beam of rectangular TRB Poly. Lect., 2012
section 4 cm by 6 cm carries a mid-span RPSC ACF-2011
concentrated load such that the 6 cm side lies
WBPSC AE, 2007
parallel to line of action of loading and the
deflection under the load is δ. If the beam is Ans. (a) : Maximum Principal Stress Theory
now supported with the 4 cm side parallel to (MPST) (Rankine Theory)–
line of action of loading, then the deflection Assumption–
under the load will be- Neglect the shear stress
(a) 0.44 δ (b) 2.25 δ Neglect the other principal stress (σ2)
(c) 1.5 δ (d) 1.75 δ It is suitable for brittle materials
(because brittle material are weak in tension) under all
WBPSC AE 2003
loading condition.
wl3 wl3 It's not suitable for ductile material
Ans. (a) : Deflection at center (δ) = =
48EI  bd 
3
(∵ Shear stress is neglected)
48E  
 12  MPST suitable for ductile material under uni-axial
∵ Question says that the 6 cm side lies parallel to the state of stress and biaxial state of stress when
line of action of loading. principal stress are in same nature.

Strength of Materials 413 YCT

1406. In case of crystalline and brittle materials 1408. Which theory of failure will you use for
criterion of failure, that takes place under aluminum components under steady loading?
maximum principal stress, is put forwarded by (a) principal stress theory
(a) Theory of Rankine (b) principal strain theory
(b) Theory due to Guest (c) strain energy theory
(c) Theory of St. Venant (d) maximum shear stress theory
(d) Theory due to Beltrami and Haigh Karnataka PSC AE, 10.09.2017
GPSC DEE Class-2 GWSSB 04.07.2021 HPPSC Asstt. Prof. 18.11.2016
RPSC IOF, 2020 GPSC ARTO 01.05.2016
Kerala PSC Poly. Lect., 2017 VIZAG Steel MT 18.08.2013
CGPSC AE 16.10.2016 VIZAG Steel MT 10.06.2012
APPSC AE Subordinate Service Civil/Mech. 2016 GATE 1998
UPRVUNL AE 21.08.2016 Ans. (d) : Aluminium is ductile material and loading is
KPSC ADF 2015 static so we will use, maximum shear stress theory
Mizoram PSC AE/SDO 2014, Paper-II only.
APPSC Poly. Lect. 2013 Maximum shear stress theory
TRB Asstt. Prof., 2012
RPSC AE GWD, 2011  σ − σ2 σ2 − σ3 σ1 − σ3 
τmax = max  1 , , 
Ans. (a) :  2 2 2 
(i) Max. Principle Stress Theory/Rankine's Theory/ τmax > Ssy → failed
Lames Theory/Maximum Stress Theory τmax ≤ Ssy → Safe
• Applicable for brittle materials (Best for brittle 1409. A transmission shaft subjected to bending and
materials) torsional moments should be designed on the
• Not suitable for ductile materials basis of
• Not suitable for pure shear case. (a) Maximum principal stress theory
(ii) Maximum Principal Strain Theory/Saint (b) Maximum shear stress theory
Venant's Theory (c) Permissible bearing pressure
• Satisfactory for brittle materials (d) None of the mentioned above
• Not suitable for pure shear case and hydrostatic GPSC EE Pre, 28.01.2017
loading. TNPSC AE 2017
(iii) Maximum Shear Stress Theory/Tresca's Theory MPPSC AE 08.11.2015
/ Guest Theory TNPSC ACF 2012
• Applicable for ductile materials Ans. (b) : Torsional moment is due to shear force, so
• Most conservative theory that the design of shaft based on maximum shear stress
theory, where as bending moment is due to normal
• Not suitable for hydrostatic loading. force, so that the design of shaft based on maximum
(iv) Maximum Strain Energy Theory / Haigh's principal normal stress.
Theory
1410. As per maximum shear stress theory of failure.
• Applicable for ductile material The relation between yield strength in shear
• Not suitable for brittle material and pure shear case. (τy) and yield strength in tension (σt) is:
(v) Maximum Shear Strain Energy Theory / (a) τt = 1.2 σt (b) τt = 0.7 σt
Distortion Energy Theory / Hencky Theory / (c) τt = 0.3 σt (d) τt = 0.5 σt
Von-Mises Theory TNPSC AE 2019, Vizag Steel (MT) 2017
• Most appropriate theory for ductile materials CIL MT 26.03.2017
• Perfect for pure shear case. Ans. (d) : Maximum shear stress theory (or) Guest &
1407. Shear stress theory is applicable to Tresca's Theory-According to this theory, failure of
(a) ductile materials (b) brittle materials specimen subjected to any combination of load when
the maximum shearing stress at any point reaches the
(c) elastic materials (d) plastic materials
failure value equal to that developed at the yielding in
RPSC AE 2016
an axial tensile or compressive test of the same material.
GWSSB DEE 07.07.2016
σ
APPSC AEE Mains 2016 (Civil Mechanical) τy ≤ t
TNPSC AE 2014, RPSC ACF-2011 2
J&K PSC Civil Services Pre, 2010 Where σt is the yield strength in tension and τy is the
WBPSC AE 2008 yield strength in shear.
Ans : (a) Shear stress theory is applicable to ductile 1411. The maximum distortion energy theory of
material. failure is suitable to predict the failure of one
(i) Maximum principal stress theory – Brittle material of the materials is :
(ii) Maximum principal strain theory – Brittle material (a) Brittle material (b) Ductile material
(iii) Maximum shear stress theory – Ductile material (c) Plastics (d) Composite materials
(iv) Maximum total strain energy – Ductile material. UJVNL AE 2016
theory OPSC AEE 2015 Paper-I, ESE 2004
Strength of Materials 414 YCT
Ans : (b)
 σ1 − σ3   σe 
Maximum Principal Stress Theory : Brittle material  2 = 2 
Maximum Shear Stress Theory : Ductile material    
Maximum Strain Energy Theory : Ductile material maximum shear stress is also known as Guest's 'or'
Maximum Shear Strain Energy Theory : Ductile Tresca's theory.
material
Maximum Principal Strain Theory : The theory can 1414. Match the following criterion of material
be applied for ductile and Brittle material both but result failure under biaxial stresses σ1 and σ2 and
are not accurate for either case. yield stress σy, with their corresponding
1412. A shaft subjected to maximum bending stress graphic representations.
of 80N/mm2 and maximum shearing stress A Maximum L.
equal to 30N/mm2 at a particular section. If the . shear stress
yield point in tension of the material is 280 criterion
N/mm2 and maximum shear stress theory of
failure is used, then the factor of safety
obtained will be :
(a) 2.5 (b) 2.8
(c) 3.0 (d) 3.5 B. Maximum M
MPSC Poly. Lect. 2014 distortion .
UPRVUNL AE 2014 energy
APPSC IOF, 2009 criterion
Ans. (b) : σmax = 80 N/mm 2

τxy = 30 N/mm2
σyt = 280 N/mm2
FOS = ? C. Maximum N.
normal stress
2
 σ − σy  criterion
Maximum shear stress =  x  + τxy
2

 2 
2
 80 − 0 
τmax =   + 30
2

 2  (a) A – L, B – N, C – M
τmax = 50MPa (b) A – N, B – L, C – M
From maximum shear stress theory (c) A – M, B – N, C – L
σyt (d) A – N, B – M, C – L
τmax = ISRO Scientist/Engineer 07.05.2017
2FOS GATE 2011, ESE 1997
280
50 = Ans. (b) : Maximum Shear Stress Theory–
2 × FOS
280
FOS =
50 × 2
FOS = 2.8
1413. The failure at a point in a member, when the
maximum principal stress in a bi-axial stress
system reaches the elastic limit of the material,
in a sample tension test. This is according to Maximum Distortion Energy Theory–
(a) Rankine's theory (b) Guest's theory
(c) Lame's theory (d) Goodman's theory
Kerala LBS Centre For Sci. & Tech. Asstt. Prof. 2014
WBPSC AE 2003
Ans. (a) : Rankine's theory– Rankine's theory assumes
that failure will occur when maximum principal stress at
any point reaches a value equal to the tensile stress in a
simple tension specimen at failure. Rankine's theory Maximum Normal Stress Theory–
applied for Brittle material, and not applicable for
ductile material.
Guest's theory- According to this theory, the elastic
failure occurs when the greatest shear stress reaches a
value equal to the shear stress at elastic limit in a simple
tension test. This theory is considered for Ductile
material.
Strength of Materials 415 YCT
1415. Guest's theory of failure is applicable for Ans. (b) : Maximum shear stress theory [Guest and
following type of materials Tresca theory]– Failure of the component occurs when
(a) brittle (b) ductile the maximum shear stress in the complex system
(c) elastic (d) plastic reaches the value of maximum shear stress in simple
(e) tough tension at the elastic limit. This theory is used for
GPSC DEE, Class-2 (GWSSB) 04.07.2021 ductile material.
CGPSC AE 26.04.2015 Shift-I 1419. According to the distortion-energy theory, the
Ans. (b) : Maximum shear stress theory or Guest or yield strength in shear is
Tresca's theory is well justified for ductile material. (a) 0.277 times the yield stress
(b) 0.377 times the maximum shear stress
1416. Which of the following theories of failure is not
(c) 0.477 times the yield strength in tension
suitable for ductile material (d) 0.577 times the yield strength in tension
(a) Maximum shear stress theory ESE 2020
(b) Maximum principal strain theory BPSC AE Mains 2017 Paper - VI
(c) Maximum total strain energy theory
Ans. (d) : According to distortion energy theory,
(d) Maximum principal stress theory
RPSC LECTURER 16.01.2016 Syt
Ssy = = 0.577Syt
MPSC HOD (Govt. Poly. Colleges) 04.10.2014 3
Ans. (d) : Maximum principal stress theory (Rankine's 1420. Von-Mises and Tresca criteria give different
theory) is suitable for brittle material not for ductile yield stress for :
material whereas Maximum shear stress theory (Guest's (a) Uni-axial stress
and Tresca theory), maximum principal strain theory (b) Balanced bi-axial stress
(Saint-Venant theory) and maximum total strain energy (c) Pure shear stress
theory (Haigh's theory) are used for ductile material. (d) All
1417. Region of safety for maximum principal stress Vadodara Muncipal Corp. DEE, 2018
theory under bi-axial stress is shown by: GPSC EE Pre, 28.01.2017
(a) Ellipse (b) Square Ans. (c) : Von-Mises and Tresca criteria gives different
(c) Pentagon (d) Hexagon yield stress for pure shear stress.
APPSC AEE SCREENING 17.02.2019
UPRVUNL AE 07.10.2016 1421. For designing ductile material, which of the
following theories is/are used?
Ans. (b) : Maximum principal stress theory—
(a) Maximum shear stress theory
Maximum principal stress theory or normal stress
(b) Shear strain energy theory
theory says that, yielding occurs at a point in a body,
when principle stress (maximum normal stress) in a (c) Both (A) and (B)
biaxial system reaches limiting yield value of that (d) None of the above
material under simple tension test. This theory used for Assam PSC AE (PHED), 18.10.2020
brittle material. This theory also known as Rankine APPSC AEE SCREENING 17.02.2019
theory. Region of safety for maximum principal stress Ans. (c) : For designing ductile material- Most
theory under bi-axial stress is square important appropriate failure theory is maximum shear
stress theory or shear strain energy theory.
For brittle material- Maximum normal stress or
principal stress theory is more suitable for brittle
material.
1422. Maximum distortion energy theory of failure is
also called :
(a) Rankine's theory
(b) Guest's or Tresca's theory
(c) Von Mises theory
1418. Failure of the component occurs when the (d) St. Venant's theory
maximum shear stress in the complex system CGPSC AE 15.01.2021
reaches the value of maximum shear stress in UPRVUNL AE 05.07.2021
simple tension at the elastic limit. This is
Ans. (c) : (i) Maximum Principal Stress Theory
known as:
(a) Rankine theory ⇒ Rankine's Theory
(b) Guest and Tresca theory (ii) Maximum Shear Stress Theory
(c) Haigh theory ⇒ Guest's or Tresca's Theory
(d) St. Venant theory (iii) Maximum distortion energy Theory
(e) Carnot theory ⇒ Von Mises Theory
CGPSC AE 25.02.2018 (iv) Maximum Principal Strain Theory
ISRO Scientist/Engineer 2008 ⇒ St. Venant's Theory
Strength of Materials 416 YCT
1423. Shown below are four elements a, b, c and d 1426. A cold rolled steel shaft is designed on the
with different states of plane stress. Using basis of maximum shear stress theory. The
maximum shear stress theory yield, which of Principal stresses induced at its critical section
the following will yield first? Assume the are 500 MPa and -50 MPa respectively. If the
material is homogenous and isotropic. shear yield stress for the shaft material is 400
MPa, the factor of safety of the design is
(a) 3 (b) 4
(c) 6 (d) 2
RPSC Lect. (Tech. Edu. Dept.) 13.03.2021, Paper-I
Ans. (d) : As we know,
(a) A (b) B σ t – σc τ
(c) C (d) D = sy
2 FOS
RPSC ACF & FRO, 26.02.2021
500 – ( –50 ) 400
Ans. (d) : According to max shear stress theory, =
2 FOS
 σ σ σ − σ2 
τmax = max  1 , 2 , 1  550
=
400
 2 2 2  2 FOS
σ1 σ FOS = 1.45 ≃ 2
(τmax )A = =
2 2
1427. If a machine member is loaded in such a way
σ σ that the three principal stresses are 600 kPa
(τmax ) B = 2 =
2 2 tension, 800 kPa tension and zero, the
σ1 σ2 σ maximum shear stress at the point is :
(τmax )C = (or) = (a) 1150 N/m2 (b) 1000 N/m2
2 2 2 2
(c) 700 kN/m (d) 400 kN/m2
σ1 − σ2 σ − (−σ)
(τmax ) D = = CGPSC AE 15.01.2021
2 2 Ans. (d) : Given, rincipal stress,
= σ (max)
σ1 = 600 kPa, σ2 = 800 kPa, σ3 = 0
(So, element d will yield first).
 σ − σ 2   σ 2 − σ3   σ1 − σ3  
1424. A tension member of diameter d is designed Max.shear stress =  1 , , 
with F.O.S. of 3. If the load and diameter are  2   2   2  
doubled F.O.S. will be  600 − 800 800 − 0 600 − 0 
(a) Reduced to half (b) Unchanged = , , 
(c) Tripled (d) Doubled  2 2 2 
RPSC Lect. (Tech. Edu. Dept.) 13.03.2021, Paper-I = {100, 400, 300 }
σ τmax = 400kN / m 2
Ans. (d) : FOS = allowable
σ max
1428. A machine element is subjected to bi-axial
σallowable state of stress : σx = 80 MPa; σy = 20 MPa; τxy
=3
(P / A) = 40 MPa. If the shear strength of the material
σallowable is 100 MPa, the factor of safety as per Tresca's
=N Maximum Shear Strength theory is________.
( 2P / 4A )
(a) 1.0 (b) 2.5
1 3 (c) 2.0 (d) 3.3
Hence, =
2 N Assam Engg. College AP/Lect. 18.01.2021
N=6 GATE 2015
Ans. (c) : 2.0
1425. A rod having cross-sectional area 100×10-6 m2
subjected to a tensile load. Based on the Teresa
failure criterion, if the uniaxial yield stress of
the material is 200 MPa, the failure load is :
(a) 10 kN (b) 20 kN
(c) 100 kN (d) 200 kN
RPSC Lect. (Tech. Edu. Dept.) 13.03.2021, Paper-I
Ans. (b) : We know, Tresca's failure,
P σ σx = 80 MPa
τmax = = yt or P = σ yt × A σy = 20 MPa
2A 2
P = 200 × 106 × 100×10–6m2 τxy = 40 MPa
P = 20 × 103N = 20 kN τxy = 100 MPa
Strength of Materials 417 YCT
 2 Select the correct answer using the code given
 σ – σy  below.
Maximum shear stress =  x  + τxy
2
(a) 1, 2 and 3 (b) 1 only
 2  (c) 2 only (d) 3 only
 80 – 20 
2 ESE 2019, 2020
=   + 40 2
Ans. (c) : Since ductile materials, under static
 2  conditions, mostly fail due to shear or distortion,
= 50 distortion energy theory or Von-mises theory produces
Tresca's maximum shear stress theory most accurate results.
τsy 1433. Tresca's theory of failure is also known as,
τmax = (a) Maximum principle stress theory
FOS (b) Maximum principle strain theory
100 (c) Maximum shear stress theory
50 =
FOS (d) Maximum distortion energy theory
FOS = 2 Oil India Senior Officer 23.12.2020
1429. For the condition of principal stresses σ1, σ2 Ans. (c) : Tresca's theory of failure is also known as
positive and σ3 = 0, and also σ1 > σ2 > σ3, the maximum shear stress theory.
equation of boundary line of failure envelope 1434. Which one of the following figures represents
as per maximum shear stress theory will be : the maximum principal stress theory?
(a) σ2 – σ2 = Syt (b) σ1 = Ssy
(c) σ1 = Syt (d) σ2 = Syt
JPSC AE 10.04.2021, Paper-II (a) (b)
Ans. (c) : Given, σ1 > σ2 > σ3 = 0
σ − σ3
maximum shear stress τmax = 1
2
σ1 − σ3 (c) (d)
= τmax
2
σ1 Syt
= OPSC AEE 2019 Paper-I
2 2
Ans : (a)
σ1 = Syt

1430. Failure of material is called fatigue when it

fails
(a) at the elastic limit
(b) at the yield point
(c) below the elastic limit (i) Maximum shear stress theory
(d) below the yield point
GPSC DEE, Class-2 (GWSSB) 04.07.2021
Ans. (d) : Failure of material is called fatigue when it
fails below the yield point.
1431. In theory of failure, maximum strain energy
theory is also called ______. (ii) Maximum shear strain energy theory
(a) Tressca theory (b) Rankine theory
(c) Haigh's theory (d) Von Mises theory
VIZAG Steel MT 24.01.2021, Shift-I
Ans. (c) : In theory of failure, maximum strain energy
theory is ab called Haigh's theory.
2
S 
σ12 + σ22 + σ32 + 2µ ( σ1σ2 + σ2 σ3 + σ3σ1 ) ≤  yt 
N (iii) Maximum strain energy theory
1432. The theory of failure used in designing the
ductile materials in a most accurate way is by
(1) maximum principal stress theory
(2) distortion energy theory
(3) maximum strain theory
Strength of Materials 418 YCT
(iv) Maximum principal theory 1438. The uniaxial tensile stress that would create
the same distortion energy as is created by the
actual combination of applied stresses is called
as
(a) Effective distortion stress
(b) Impact stress
(c) Joint effective stress
1435. Steel machine element at the critical section is (d) Von mises effective stress
in biaxial stress state with two principal stress Haryana PSC AE (PHED) 05.09.2020, Paper-II
being 300 N/mm2 and 300 N/mm2 (equal
magnitude). Find the von Mises stress (in Ans. (d) : Distortion energy is associated with Von-
N/mm2) in the member mises.
(a) 212.1 (b) 600 1439. If S is the applied stress, c is the width of the
(c) 424.2 (d) 300 crack and r the radius of curvature at the tip
ISRO Scientist/Engineer 12.01.2020 of the crack, Griffith's crack theory gives the
concentrated stress Sc as
Ans. (d) : (a) 2S (c/r)1/2 (b) 2S (c/r)1/3
1/2
(c) S (c/r) (d) 2S (2c/r)1/2
ISRO Scientist/Engineer 12.01.2020
Ans. (a) : Griffith crack theory- The Griffith theory
states that a crack will propagates when the reduction in
potential energy that occurs due to crack growth is
greater than or equal to the increase in surface energy
due to the creation of new free surface. This theory is
σ1 = σ2 = 300 applicable to elastic materials that fracture in brittle
By using maximum distortion energy theory, Von- fashion.
Mises Theory- Griffith crack theory gives,
For plane stress condition, c
1/ 2

Sc = 2S  
σ per = σ + σ – σ1σ2
2
1
2
2 r
σper = σ = 300 MPa 1440. At a material point the principal stresses are
1436. The uniaxial yield stress of a material is 300 σ1 = 100 MPa and σ2 = 20 MPa. If the elastic
MPa. According to von Mises criterion, the limit is 200 MPa, what is the factor of safety
shear yield stress (in MPa) of the material is --- based on maximum shear stress theory?
(a) 173.28 (b) 150.55 (a) 1.5 (b) 2
(c) 124 (d) 99.87 (c) 2.5 (d) 3
APPSC AEE SCREENING 17.02.2019
(e) None of the above
CGPSC Poly. Lect. 22.05.2016 Ans. (b) : σ1 = 100 MPa
σ2 = 20 MPa
Ans. (a) : Given,
σ3 = 0 (Minimum principal stress)
300 MPa = Syt
σy = 200 MPa
According to Von - Mises criterion–
According to Maximum Shear Stress theory
Sys 1
= σy
Syt 3 τ max =
2 FOS
Sys 1 σ1 − σ 3 σy
= ⇒ =
300 3 2 2 FOS
300 σy
Sys = ⇒ σ1 − σ 3 =
3 FOS
Sys = 172.20 MPa 200
⇒ 100 − 0 =
FOS
1437. Failure occurs when the maximum shear stress
200
in the part exceeds shear stress in the tensile FOS = =2
specimen at yield is a statement of 100
(a) Distortion energy theory 1441. A point in a structural member with allowable
(b) Maximum shear stress theory yield strength of 300 MPa is subjected to
(c) Maximum strain energy theory Principal stresses σ1 = 200 MPa, σ2 = 50 MPa
(d) Maximum normal stress theory and σ3 = –100MPa. What is the calculated
Haryana PSC AE (PHED) 05.09.2020, Paper-II value of yield stress (σ0) and does yielding
Ans. (b) : Maximum shear stress theory occurs?
Strength of Materials 419 YCT
 (a) σ0 = 150 MPa and yielding occurs (a) 1.75 (b) 1.5
(b) σ0 = 300 MPa and yielding occurs (c) 2.0 (d) 1.2
(c) σ0 = 300 MPa and yielding does not occurs AAI Jr. Executive 29.11.2018
(d) σ0 = 150 MPa and yielding does not occurs Ans. (c) : Shear stress = 30 N/mm2
2
Oil India Limited Sr. Engineer (Drilling) 30.11.2019 Maximum bending moment2= 80 N/mm
Ans. (a) : Using Tresca's criteria σyield = 200 N/mm
2
σ1 − σ3 Syt  80 − 0 
τ  + 30 = 50 N/mm
2 2
≤ max = 
2 2  2 
200 + 100 300 According to maximum shear stress theory –
≤
2 2 σ yield
150 = 150 τmax =
2N
Hence yielding occurs
200
1442. Let σ1, σ2 and σ3 are the principal stresses at a 50 =
material point. If the yield stress of the 2N
material is σy, then according to Von Mises Factor of safety (N ) = 200/100 = 2
theory yielding will not occur if 1445. For a ductile material, the limiting value of
(a) (σ 1 − σ 2 ) + (σ 2 − σ 3 ) + (σ 3 − σ 1 ) < 2(σ y )
2 2 2 2 octahedral shear stress (τo) is related to the
yield stress (Sy) as
(b) max[(σ 1 − σ 2 ), (σ 2 − σ 3 ), (σ 3 − σ 1 )] < σ y
2
(c) (σ 1 ) 2 + (σ 2 )2 + (σ 3 ) 2 < (σ y ) 2 (a) τ o = S y (b) τ o = S y 3 2
3
(d) (σ1 ) 2 + (σ 2 )2 + (σ 3 ) 2 − 2ν (σ1σ 2 + σ 2σ 3 + σ 3σ1 ) < (σ y )2 3
(c) τ o = S y (d) None of the above
APPSC AEE SCREENING 17.02.2019 2
Ans. (a) : Von Mises Yielding Failure theory BPSC AE Mains 2017 Paper - VI
(σ 1 − σ 2 ) + (σ 2 − σ 3 ) + (σ 3 − σ 1 )
2 2 2 Ans : (a) : According to Von-mises (theory of failure)
σy ≥ (σ1 – σ2)2 + (σ2 – σ3)2 + (σ3 – σ1)2 = 2σ2yield ..........(i)
2 The octahedral shear stress can be given by the
(σ 1 − σ 2 ) 2 + (σ 2 − σ 3 ) 2 + (σ 3 − σ 1 ) 2 ≤ 2σ y2 expression,
1
1443. If an element is loaded as shown in figure τ ys = (σ1 − σ 2 ) 2 + (σ2 − σ3 )2 + (σ3 − σ1 ) 2 ...........(ii)
below, the Von Mises stress will be close to 3
from equation (i) and (ii)
1
τ ys = 2σ 2yield
3
2
τ ys = σ yield
3
1446. As per Von Mises theory the shear yield stress
(a) 180 MPa (b) 350 MPa (K) can be related with yield stress (σy) under
(c) 255 MPa (d) 495 MPa state of pure shear by ________
ISRO Scientist/Engineer 22.04.2018 (a) K = σy / (3)0.5 (b) K = σy
Ans. (b) : Data given (c) K = ( σ y ) 0.5
/3 (d) None of the above
σ 1 = −100MPa, σ 2 = 300MPa, σ 3 = 150MPa Punjab PSC SDE 12.02.2017
Ans. (a) : Von Mises theory is also known as maximum
We know that (according to Von Mises theory) shear strain energy theory or distortion energy theory
(σ 1 − σ 2 ) 2 + (σ 2 − σ 3 ) 2 + (σ 3 − σ 1 ) 2 = 2σ y2 this assumes that failure occur when shear strain energy
(distortion energy) in the complex system is equal to
(−100 − 300) + (300 − 150) + (150 − (−100)) = 2σ y
2 2 2 2
that the yield point in tensile compression test.
160000 + 22500 + 62500 = 2 × σ y 2 1447. Distortion energy theory of failure is
applicable to:
1 (a) components made of plain carbon steel
σ y2 = × 245000 = 122500
2 (b) components made of composites
σ y = 350 MPa (c) components made of cast iron
(d) components made of non-metals
1444. A particular section of a shaft is subjected to a HPPSC Asstt. Prof. 20.11.2017
maximum shear stress equal to 30 N/mm2 and Ans. (a) : Maximum distortion energy theory [Von-
2
a maximum bending moment of 80 N/mm . If mises theory]
the yield point in tension for this material is – best theory of failure for ductile material because it
200 N/mm2 and maximum shear stress theory
of failure for static loading is used, then the given safe and economic design.
factor of safety will be: Plain carbon steels are ductile materials.
Strength of Materials 420 YCT
1448. One of the theories of failure generally used is 1451. Pick the wrong statement about various
Maximum principal strain theory. This is also theories of failure
known as (a) Rankine's theory is suitable for brittle
(a) Coulomb's Tresca and Guest’s theory materials.
(b) St. Venant’s theory (b) Maximum shear stress theory is more
(c) Haigh’s theory suitable for ductile materials.
(d) Rankine’s Theory (c) Distortion theory is suitable for ductile
UPRVUNL AE 05.07.2021 materials.
Jharkhand Urja Vikas Nigam Ltd. AE 2017
(d) Distortion energy theory is easy to apply as
Ans. (b) : The maximum principle strain in the complex compared to maximum shear stress theory.
stress system must be less than the elastic in simple
tension if there is to be no failure. RPSC AE (GWD) 06.05.2016
σ − νσ2 − νσ3 Ans. (d)
In the limit ε1 = 1 • Rankine's theory (maximum, principle stress theory)
E
is suitable for brittle materials
σ
ε1 = • Maximum shear stress theory in more suitable for
E
ductile materials.
σ1 − νσ 2 − vσ3 = σ • Distortion theory is suitable for ductile materials.
1449. The most applicable failure theory for metals 1452. Three principal stresses of 10 MPa, 20 MPa
like aluminium is and 30 MPa (tensile in nature) are acting on a
(a) Maximum principal stress theory homogeneous material. Using Von Mises yield
(b) Maximum principal strain theory criterion, what is the value, in MPa, of the
(c) Total strain energy theory estimated shear yield stress?
(d) Maximum distortion energy theory
(a) 16.07 (b) 17.3
ISRO Scientist/Engineer 17.12.2017
(c) 18.8 (d) 19.3
Ans. (d) The most applicable theory of failure for metals
(e) 20.6
(1) Ductile material
(a) Maximum shear stress theory CGPSC Asstt. Workshop Supt., 17.07.2016
(b) Maximum distortion energy theory Ans. (b) : σx = 10 MPa, σy = 20 MPa, σz = 30 MPa
(2) Brittle material Von-Mises, yield criterion,
(a) Maximum principal stress theory
Syt
(b) Maximum principal strain theory = σ12 + σ 22 + σ32 − ( σ1σ 2 + σ 2 σ3 + σ3σ1 )
(c) Total strain energy theory FOS
1450. A cold rolled steel shaft is designed on the basis = 102 + 202 + 302 − (10 × 20 + 30 × 20 + 30 × 10)
of maximum shear stress theory. The principal
stresses induced at its critical section are 60 = 900 + 400 + 100 − (1100)
MPa and -60 MPa respectively. If the yield
stress for the shaft material is 360 MPa, the Syt = 300 MPa = 17.3MPa
factor of safety of the design is_____ Fs
(a) 6 (b) 4 Estimated shear stress,
(c) 2 (d) 3
UPRVUNL AE 21.08.2016, HPPSC W.S. Poly. 2016 τxy Syt
= 0.577 = 0.577 300
OPSC AEE 2015 Paper-I, ESE 2002 Fs Fs
Ans. : (d) Maximum stress (σmax) = 60 Mpa
Minimum stress (σmin) = –60 Mpa 1453. What is the relationship between tensile (σy)
and shear (K) yield stresses as per Tresca and
σ − σmin
Maximum shear stress ( τ max) = max Von mises yield criteria
2 σ σ
60 + 60 (a) K = y and K = y
τ max = = 60 Mpa 2 3
2
According to maximum shear stress theory σ
τy = FOS × τ max (b) K = y and K = 3σ y
2
yield stress
yield point shear stress = σ σ
2 (c) K = y and K = y
τy = 180 MPa 2 3
τ y 180 σy 2
FOS = = =3 (d) K = and K = σy
τ max 60 2 3
FOS = 3 GPSC Asstt. Prof. (Prod.) 21.08.2016
Strength of Materials 421 YCT
Ans. (c) : Tresca's Theory– σx = 100 N/mm2, σy = 40 N/mm2, τxy = 40N/mm2
2
 σx − σ y 
( )
2
τmax =   + τ xy
 2 
τmax = 900 + 1600
Hexagon τmax = 50 N / mm 2
σy According to maximum shear stress theory
K= τy= FOS × τmax
2
K = Shear yield stress Sy= 300 MPa
σy = tensile yield stress Sy
then τy= 150 MPa τy= = 0.5 Sy
Tresca's theory is well used for ductile materials 2
Von-Mises Theory– 150 = FOS × 50
FOS = 3
1456. Which theory of failure is applicable for copper
components under steady load?
Ellipse (a) Principal stress theory
σy (b) Strain energy theory
K= (c) Maximum shear stress theory
3
(d) Principal strain theory
This is the most economical theory of failure used for MPPSC AE 2016
ductile materials.
Ans : (c) Maximum shear stress theory of failure is
1454. In the case of analysis of fracture of brittle applicable for copper components under steady load.
specimens and structures when all three Maximum shear stress theory or Guest and Trecas's
principal stresses are compressive, theory is well justified for ductile materials.
................theory seems to predict lower 1457. A transmission shaft subjected to bending
strengths than are actually obtained. loads must be designed on the basis of
(a) Mcaulay's (b) Von-Tresca's (a) Maximum shear stress theory
(c) Griffith's (d) Albert's (b) Fatigue strength
HPPSC Asstt. Prof. 29.10.2016 (c) Maximum normal stress and maximum shear
Ans. (c) : Griffith Theory of brittle fracture– stress theories
Griffith proposed that a brittle material contains a (d) Maximum normal stress theory
population of fine small cracks and flows that have a MPPSC AE 2016
variety of sizes, geometries and orientation which ESE 1996
produces a stress concentration of sufficient magnitude Ans : (b) A transmission shaft subjected to bending
so that the theoretical cohesive strength is reached in load must be designed on the basis of Fatigue strength
localized regions at a normal stress which is well Note-As per MPPSC, official answer is (d).
below the theoretical value. 1458. The factor of safety in bending for rectangular
1455. The State of stress at a point is given as σx= 100 beams in terms of yield point stress (σyp) and
N / mm2, σy= 40 N / mm2 strength τxy = 40 N / working stress (σw) is
mm2. If the yield strength Sy of the material is σ σ
(a) 1.5 YP (b) 1.3 YP
300 MPa, the factor of safety using maximum σW σW
shear stress theory will be σ σ
(a) 3 (b) 2.5 (c) 1.33 YP (d) YP
σW σW
(c) 7.5 (d) 1.25
APPSC AEE Mains 2016 (Civil Mechanical)
BPSC Poly. Lect. 2016
Ans. (d) : As we know,
Ans : (a)
yield strength
Working stress =
FOS
For ductile material
σ yp
So, FOS =
σw
1459. Consider the following
1. hard materials 2. brittle materials
3. malleable materials 4. ductile materials
5. elastic materials
Strength of Materials 422 YCT
 Of the above, Shear stress theory is applicable Ans. (c) : The slip line theory is developed to analyse
for which material non homogeneous plane strain deformation.
(a) 1 and 2 (b) 2 and 3 1463. The yield locus for von Mises yield criterion on
(c) 3 (d) 4 π- plane is:
(e) 4 and 5 (a) Circle (b) Ellipse
CGPSC AE 26.04.2015 Shift-I (c) Hexagon (d) Circular cylinder
Ans. (d) : Shear stress theory is applicable for ductile HPPSC Asstt. Prof. 2014
material.
Shear stress theory is also know as Guest & Tresca Ans. (b) :
theory.
1460. Tresca theory of failure is applicable for which
of the following type of materials?
(a) Elastomers (b) Plastic
(c) Ductile (d) Brittle
MPPSC AE 08.11.2015 Ellipse
Ans. (c) : Tresca's theory of failure is applicable for 1464. SEQA effective Von Mises stress at phase
ductile materials. difference of 90º is minimum when the ratio of
1461. A solid circular shaft is subjected to a bending torsion stress amplitude and bending stress
moment of 3000 N-m and a torque of 10000 N- amplitude is near to :
m. The shafts is made of 45 C8 steel having (a) 0.2 (b) 0.6
ultimate tensile stress of 700 MPa and an (c) 1 (d) 1.6
ultimate shear stress of 500 MPa. Assume a HPPSC Asstt. Prof. 2014
factor of safety as 6. The diameter of the shaft Ans. (b) : According to Von-Mises
according to the maximum shear stress theory σ yt
is Shear yield stress τxy =
(a) 84 mm (b) 85 mm 3
(c) 86 mm (d) 87 mm τxy = 0.577 σyt
MPPSC AE 08.11.2015 τ xy
Ans. (c) : Given, Bending moment (M) = 3000 N-m = 0.577 ≈ 0.6
σ yt
Torque(T) = 10000 N-m
1465. From a uniaxial tension test, the yield strength
σut = 700 MPa & τut = 500 MPa of steel was found to be 200 N/mm2. A steel
= 500×106 N/m2 shaft is subjected to a torque 'T', and a
Factor of safety (FOS) = 6 bending moment 'M'. The theory of failure
Let d = diameter of the shaft. which gives safest dimensions for the shaft and
2 2 the relationship for design is
∴ Equivalent torque(Te) = M + T (a) Maximum Principal Strain Theory σ1 = σ y
( 3000 ) + (10000 ) = 10440.306 N − m
2 2
(b) Maximum Principal Strain Theory
According to the maximum shear stress theory- σ1 µσ σ y
− 2=
τ 6 E E E
500 × 10 (c) Maximum Shear Stress Theory
τ = ut =
FOS 6 σ1 −σ2 σ y
∴ Diameter of shaft is- =
2 2
3 16Te 16 × 10440.306 σ12 σ22 σ2y
d = = (d) Total Strain Energy Theory + =
π× τ 10 6
2E 2E 2 E
π× 500 ×
6 TNPSC AE 2014
6 × 16 ×10440.306 × 10 –6 Ans. (c) : Maximum shear stress theory [Guest and
= Tresca] of failure gives safest dimensions for the shaft
π× 500 and the relationship for design.
⇒ d3 = 638.06 × 10–6 1466. If a shaft made of ductile material is subjected
d = 8.609 × 10–2 m to combined bending and twisting moments,
d ≈ 0.086 m calculation based on which one of the following
⇒ d = 86 mm theories would give the most conservative
values ?
1462. The slip line theory is developed to analyse: (a) Maximum principle stress theory
(a) non-homogeneous plane stress deformation (b) Maximum shear stress theory
(b) homogeneous plane stress deformation (c) Maximum strain energy theory
(c) non-homogeneous plane strain deformation (d) Maximum distortion energy theory
(d) homogeneous plane strain deformation J&K PSC Civil Services Pre, 2013
HPPSC Asstt. Prof. 2014 ESE 1996, GATE 1988
Strength of Materials 423 YCT
Ans. (b) : Ans. (c) : According to maximum shear stress theory
Material Theory of failure criterion σ1 − σ3 σy
Brittle max principle stress theory ≤
2 N×2
ductile max shear stress theory
σy σy σy
1467. A rod of diameter 20 mm is subjected to a N= = =
tensile load. Based on Tresca's failure σ1 − σ3 P1 − 0 P1
criterion, if the uniaxial yield stress of the Squaring both sides
material is 300 MPa, the failure load is
(a) 20.75 kN (b) 54.00 kN σy2
N2 = − − − − − − − (1)
(c) 94.25 kN (d) 105.50 kN P12
Vizag Steel MT (Re-Exam) 24.11.2013
According to distortion energy theory
Ans. (c) : Maximum shear stress theory (Guest and
Tresca's theory) 1  2 σy2
( σ − σ ) 2
+ ( σ − σ )
2
+ ( σ − σ ) ≤
σy 2   N 2
1 2 2 3 3 1
τmax ≤ (For no failure)
2 2σ y 2
For design max of shear stress, N2 =
 (σ1 − σ 2 ) (σ 2 − σ3 ) (σ3 − σ1 )  ≤  σ y 
( P1 − P2 )2 + P22 + P12
 , ,   2FOS 
2 2 2 σy2
For uni-axial loading, (σ2 = σ3 = 0) N2 = − − − − − − − (2)
P12 + P2 2 − P1P2
σ1 ≤ σyt [∵ FOS = 1]
Equating (1) & (2)
d = 20 mm, σyt = σ1 = 300 MPa
σy2 σy2
Load = 2
σ1 =
Area P12 P1 + P2 2 − P1P2
π P12 = P12 + P2 2 − P1P2
⇒ Load = σ1 × Area = σ1 × d 2
4
π 2 P12 = P1P2
Load = 300 × 20 = 94.25kN
4 P2 = P1
1468. A small element at the critical section of a P1
component is in a bi-axial state of stress with =1
P2
the two principal stresses being 360 MPa and
140 MPa. The maximum working stress
according to Distortion energy theory is : 11. Springs
(a) 220 MPa (b) 110 MPa
(c) 314 MPa (d) 330 MPa
1470. A coil is cut into two halves, the stiffness of cut
APGENCO AE 2012 coil will be:
Ans. (c) : Given, σ1 = 360 MPa (a) Double (b) Half
σ2 = 140 MPa (c) Same (d) None of above
According to maximum Distortion theory - UPPSC AE 13.12.2020 Paper-I
Gujarat PSC AE 2019
σ working = σ12 + σ 22 − σ1σ 2 BPSC AE Mains 2017 Paper - VI
= (360)2 + (140) 2 − 360 ×140 GWSSB DEE 07.07.2016
GPSC Asstt. Prof. 28.08.2016
= 314.32 MPa
OPSC AEE 2015 Paper-I
1469. At a point in a strained material the principal
TNPSC AE 2014, SJVN ET 2013
stresses are P1, P2, and zero. In order that the
factor of safety obtained by using the ESE 2002, CSE Pre-2002
maximum shear stress theory of failure is the Gd 4
Ans. (a) : k =
same as that obtained by using the distortion 64R 3 n
energy theory, the ratio of P1, to P2, should be 1
1 kα
(a) –1 (b) n
2 k1 n 2
(c) 1 (d) –2 =
k 2 n1
WBPSC AE, 2007
Strength of Materials 424 YCT
Where, k1 and k2 are stiffness and n1 and n2 are number The Wahl's stress factor (Kw) = Ks × Kc
of coils in the spring. Ks = stress factor due to shear
When a spring is cut into two equal halves. Kc = Curvature effect factor
n 1473. The spring constant of a helical compression
⇒ n2 = 1 spring DOES NOT depend on :
2
k1 n 2 n (a) Coil diameter
⇒ = = 2 (b) Material strength
k 2 n1 2n 2
(c) Number of active turns
k2 = 2k1
(d) Wire diameter
1471. Two close coiled helical springs with stiffness Assam Engg. College AP/Lect. 18.01.2021
K1 and K2 respectively are connected in series.
Gujarat PSC AE 2019
The stiffness in equivalent spring is given by
GPSC ARTO Pre 30.12.2018
Κ1 K 2 Κ1 − K 2 GPSC Executive Engg. 23.12.2018, GATE 2016
(a) (b)
Κ1 + K 2 Κ1 + K 2 Ans. (b) : Spring constant,
Κ1 + K 2 Κ1 − K 2 Gd 4
(c) (d) k=
Κ1 K 2 Κ1 K 2 8D3 n
GPSC DEE Class-2 (GWSSB) 04.07.2021 Stiffness depends on following –
APPSC AEE SCREENING 17.02.2019 (i) Modulus of Rigidity of spring
PTCUL AE 25.06.2017 (ii) Wire diameter
VIZAG MT 2015, TNPSC AE 2013 (iii) Mean diameter
APPSC AEE 04.12.2012 (iv) Number of active turns.
Ans. (a) : When connected in series – 1474. Due to space constraint 5 coils are to be cut
Equivalent stiffness – and removed from a spring with 25 active
coils. Which one among the following is correct
with respect to stiffness?
(a) 1.25 times that of original spring
(b) 80 percent of that of original spring
(c) The same for both springs
(d) Five percent less than that of original spring
RPSC Lect. (Tech. Edu. Dept.) 13.03.2021, Paper-I
APGENCO AE, 2017, ESE 2004
Ans. (a) : Deflection of spring
64WR 3n
δ=
Gd 4
1 1 1
= + W 1
K e K1 K 2 Stiffness ⇒ ∝
δ n
1 K1 + K 2 1
= stiffness ∝
ke K 1K 2 number of coil (n)
K 1K 2 1
Ke = K1 ∝ [n1 = 25]
K1 + K 2 25
1472. Which one of the following expresses the stress 1
K2 ∝ [n2 = 20]
factor K used for design of closed coiled helical 20
spring? Where C = spring index
25
4C − 1 4C − 1 0.615 K2 ∝ × K1
(a) (b) + 20
4C − 4 4C − 4 C
K 2 = 1.25 K1
4C − 4 0.615 4C − 4
(c) + (d) 1475. Two coiled springs each having stiffness k are
4C − 1 C 4C − 1
placed in parallel. The stiffness of the
VIZAG MT, 14.12.2020, Gujarat PSC AE 2019 combination will be :
CIL MT 26.03.2017, UPRVUNL AE 21.08.2016 (a) 4k (b) 2k
APPSC AE 04.12.2012, ESE 2008
k k
Ans. : (b) Wahl's stress factor for helical spring (c) (d)
2 4
4C − 1 0.615
Kw = + JPSC AE 10.04.2021, Paper-II
4C − 4 C J & K PSC Screening, 2006
C = Spring index (D/d) ESE 2000, CSE Pre-1998
Strength of Materials 425 YCT
Ans. (b) : If springs are connected 1478. An open coiled helical spring of mean diameter
in parallel then d is subjected to an axial force P. The wire of
the spring is subjected to :
(a) Direct shear stress only
(b) Combined shear and bending only
(c) Combined shear, bending and twisting
(d) Combined shear and twisting only
GPSC DEE Class-2 (GWSSB) 04.07.2021
HPPSC Asstt. Prof. 18.09.2017
keq = k + k OPSC AEE 2015 Paper-I
keq = 2k Ans : (d) An open coiled helical spring of mean diameter
1476. A helical spring has N turns of coil diameter D d is subjected to an axial force P. The wire of the spring is
and a second spring, made of same wire diameter subjected to combined shear and twisting only.
and same material, has N/2 turns of coil of 1479. A closed coil, helical spring is subjected to a
diameter 2D. If the stiffness of the first spring is torque about its axis. The spring wire would
k, then stiffness of the second springs will be experience a
(a) k/4 (b) k/2 (a) Direct shear stress
(c) 2k (d) 4k (b) Torsional shear stress
ISRO Scientist/Engineer (RAC) 10.03.2019 (c) Bending stress
TSPSC AEE 2017, ESE 1999 (d) Direct tensile stress
Ans. (a) : (e) Bending stress and shear stress both
CGPSC AE 26.04.2015 Shift-I
W W Gd 4 UPSC JWM Advt. No.-52/2010
Spring stiffness (k1) = = =
δ 8WD 3 N 8D 3 N ESE 1998
Gd 4 Ans. (c) : When a closed coil spring fixed at one end is
Gd 4 subjected to twisting couple about the central axis of the
k1 = spring then the bending moment will be product then
8D3 N spring will be experience bending stress.
[Given, if coil diameter is doubled and no. of turns is
half] 1480. A spring is made of a wire of 2 mm diameter
having a shear modulus of 80 GPa. The mean
Gd 4 coil diameter is 20 mm and the number of
k2 =
8 × (2 D)3 × N / 2 active coils is 10. If the mean coil diameter is
reduced to 10 mm, the stiffness of the spring is:
Gd 4 (a) increased by 16 times
=
N (b) decreased by 8 times
8 × 8D3 × (c) increased by 8 times
2
(d) decreased by 16 times
Gd 4
= RPSC Lect. (Tech. Edu. Dept.) 13.03.2021, Paper-I
8 × 4 D3 × N BHEL ET 2019
1  Gd 4  VIZAG Steel MT (Re-Exam) 24.11.2013
=   GATE 2008
4  8D3 N 
Ans. (c) : D1 = 20 mm
k2 = k1 / 4 D2 = 10 mm
Stiffness of spring
1477. While calculating the stress induced in a closed
helical spring. Wahl's factor is considered to Gd 4
k=
account for 8D 3 n
(a) the curvature and stress concentrated effect 1
(b) shock loading k∝ 3
(c) fatigue loading D
3
(d) poor service conditions k1  D 2 
GPSC ARTO 01.05.2016, TNPSC AE 2014 =  
ESE 2002, ESE 1997 k 2  D1 
3
Ans. (a) : While calculating the stress induced in a k1  10 
closed helical spring. Wahl's factor is considered to =  
k 2  20 
account for the curvature and stress concentrated effect.
4C − 1 0.615 k2 = 8k1
K= + 1481. If both the mean coil diameter and wire
4C − 4 C
diameter of a helical compression or tension
D
where C = [Spring index ] spring be doubled, then the deflection of the
d spring close coiled under same applied load will:
Strength of Materials 426 YCT
 (a) be doubled Coil dia. D
(b) be halved Ans. (a) : Spring Index = =
Wire dia. d
(c) increase four times
(d) get reduced to one-fourth 1484. Stiffness of a close coiled helical spring in
GPSC ARTO Pre 30.12.2018 terms of wire diameter d, modulus of rigidity
OPSC AEE 2019 Paper-I G, number of turns n and mean coil radius R
ESE 2012 is given by
Gd 4 Gd 4
8WD13 n (a) (b)
Ans. (b) : Deflection, δ1 = 16nR 3 32nR 3
Gd14
Gd 4 Gd 4
8WD32 n (c) (d)
Given, D2 = 2D1 δ2 = 64nR 3 96nR 3
Gd 42
APPSC IOF, 2009
8W(2D1 )3 n 8  8WD13 n  WBPSC AE, 2007, ESE 1996
d2 = 2d1 δ2 = =  
G(2d1 ) 4 16  Gd14  Ans. (c)
δ Gd 4
δ2 = 1 K=
2 8D3 n
1482. In helical compression spring design formula : Gd 4
K=
τ = 8FCK /( πd 2 ) 8 × 8R 3 n
Where Gd 4
F = load Stiffness, K =
64R 3n
τ = shear stress
C = spring index 1485. Two closely coiled helical springs 'A' and 'B'
d = wire dia. are equal in all respects but the number of
K = Wahl factor. turns of spring 'A' is half that of spring 'B' The
The Wahl factor takes into account. ratio of deflections in spring 'A' to spring 'B' is
(a) the bending stress and the curvature of the (a) 1/8 (b) 1/4
spring (c) 1/2 (d) 2
(b) the direct shear and the curvature of the Sikkim PSC (Under Secretary), 2017
spring APPSC AE Subordinate Service Civil/Mech. 2016
(c) the elastic constants (modulus of rigidity and
64wR 3 n
Poisson's ratio) of the spring. Ans. (c) : δ A =
(d) the effect of the inactive number of coils at Gd 4
the ends of the spring. 1
Given, if n =
NLC GET 17.11.2020, Shift-II 2
MPPSC AE 2016
64wR 31
BPSC Asstt. Prof. 29.11.2015 δA = .................(i)
4C − 1 0.615 Gd 4 2
Ans. (b) : K w = K sh ⋅ K c = +
4C − 4 C 64wR 3 n
δB =
Where, Kw = Wahl factor Gd 4
Ksh = Direct shear stress factor If, n = 1
Kc = Curvature effect factor
64wR 3 × 1
D δB = ...............(ii)
C = Spring index = Gd 4
d
The Wahl factor takes into account the direct shear and δA
Ratio of deflection
the curvature of the spring. δB
1483. The spring index in helical spring is
64wR 3 × 1
(a) a ratio of coil diameter to wire diameter 4
(b) a ratio of wire diameter to coil diameter = Gd ×3 2
(c) slope of its force-deflection curve 64wR × 1
(d) slope of its deflection force curve Gd 4
Haryana PSC AE (PHED) 05.09.2020, Paper-II δA 1
Kerala PSC Poly. Lect. 09.01.2014 =
δB 2
WBPSC AE 2008
Strength of Materials 427 YCT
1486. A helical spring has spring constant k. If the 1489. A helical coil spring with wire diameter 'd' and
wire diameter, spring diameter and the mean coil diameter 'D' is subjected to axial
number of coils are all doubled then the spring load. A constant ratio of 'D' and 'd' has to be
constant of the new spring is ______ maintained, such that the extension of the
(a) k (b) 0.5 k spring is independent of 'D' and 'd' what is the
(c) 8 k (d) 16 k ratio?
Assam Engg. College AP/Lect. 18.01.2021 (a) D3/d4 (b) d3/D4
MPPSC AE 2016 (c) D4/3/d3 (d) d4/3/D3
Ans. (a) : BPSC Poly. Lect. 2016
Spring Constant VIZAG MT 2015, ESE 2008
W W Gd 4 Gd 4 8WD3n
k= = = = Ans. (a) : Deflection of spring (δ) =
δ 64WR n 64R n
3 3
D
3
Gd 4
64   n
Gd 4 2 If extension of spring is independent of "D" and "d"
4 than the ratio of "D" & "d" is –
Gd
k= 3 D3
8D n
[Given, if coil diameter, wire diameter and no. of coil is d4
doubled]
1490. Spring driven watches and clocks utilize
G ( 2d )
4
16  Gd 4  (a) involute gears (b) cyloid gears
Then, k ' = =  3 
8 ( 2D ) × 2n 16  8D n 
3
(c) epricycloid gears (d) straight rack gears
k' = k Karnataka PSC AE (WRD) 31.07.2021
RPSC ACF-2011
1487. In a helical coil spring, if C is the spring index,
then the shear correction factor is given by : Ans. (b) : Spring driven watches and clock utilize
1 2C cycloid gears whereas spiral spring are used in watches
(a) K s = (b) K s = and toys. Torsion springs are used in places where a
2C 2C + 1 static position is required to be maintained against
2C + 1 C +1 impact and load fluctuation.
(c) K s = (d) K s =
2C 2C 1491. A helical compression spring of stiffness K is
APPSC Poly Lect. 13.03.2020 cut into two pieces, each having equal number
CIL MT 27.02.2020 of turns and kept side-by-side under
Ans. (c) : compression. The equivalent spring stiffness of
2C + 1 this new arrangement is equal to :
Shear stress concentration factor (Ks) =
2C (a) 4K (b) 2K
D (c) K (d) 0.5K
Where; C = = Spring index for helical coil spring. RPSC ACF & FRO, 26.02.2021
d
1488. The deflection of a spring with 20 active turns CGPSC AE 26.04.2015 Shift-I
under a load of 1000 N is 10 mm. The spring is Ans. (a) :
made into two pieces each of 10 active coils
and placed in parallel under the same load.
The deflection of this system is
(a) 20 mm (b) 10 mm
(c) 5 mm (d) 2.5 mm
UPPSC AE 13.12.2020 Paper-I Keep side by side under compression.
TRB Poly. Lect., 2012
GATE 1995
Ans. (d) : ∆x= 10 mm, N= 20 turn , F= 1000N
F 1000
K= = = 100N / mm
∆x 10 K eq = 2K + 2K = 4K
On cutting spring in two equal parts, new 'K' will be
equal 2K 1492. A load applied at centre of the carriage spring
to straighten its leaves is known as
Under parallel arrangement, K of the system will
(a) Yield load (b) Ultimate load
become 2K+2K = 4K. i.e 400 N/mm
(c) Proof load (d) Safe load
1000
Under same force (F), we get ∆x ' = = 2.5mm. TSGENCO AE 14.11.2015
400 APPSC AEE 2012
Strength of Materials 428 YCT
Ans. (c) : Proof Load– Proof load is the maximum load 1
carrying capacity of the spring, without getting mg (h + x) = kx 2
permanently distorted. 2
1
8WD 1000(10 + x) = × 200 × x 2
fs = 2
πd 3
1
1493. If a compression coil spring of stiffness 10 N/m 1000 × 10 = × 200 × x 2 − 1000 x
is cut into two equal parts and the used in 2
parallel the equivalent spring stiffness will be : x 2 − 10x − 100 = 0
(a) 10 N/m (b) 20 N/m
(c) 40 N/m (d) 80 N/m 10 + (10) 2 + 400
x=
Kerala PSC AE 06.08.2015 2
KPSC AE 2015 x = 16.18 cm
Ans. (c) : Given spring – 1495. A torsion bar with a spring constant 'k' is cut
into 'n' equal lengths. The spring constant for
each portion would be
(a) nk (b) kn
(c) k/n (d) k1/n
Kerala PSC Poly. Lect. 09.01.2014
ISRO Scientist/Engineer 12.05.2013
Ans. (a) : We know that
1
K∝
L
∴ previously stiffness of spring was K and after Hence, if we decrease the length into n parts the value
cutting into parts k1 & k2
of K increase by n times.
k1 = 2k k2 = 2k
Connected then in parallel 1496. A 2 kg pan is placed on a spring. In this
condition, the length of the spring is 200 mm.
When a mass of the 20 kg is placed on the pan,
the length of the spring becomes 100 mm. For
the spring, the un-deformed length L and the
spring constant k (stiffness) are
(a) L = 220 mm, k = 1862 N/m
(b) L = 210 mm, k = 1862 N/m
(c) L = 210 mm, k = 1960 N/m
keq = k1 + k2 (d) L = 200 mm, k = 1960 N/m
= 2k + 2k (e) L = 200 mm, k = 2156 N/m
=4k ....(i) CGPSC AE 26.04.2015 Shift-I
Putting, k = 10 N/m ISRO Scientist/Engg. 12.05.2013
[keq = 4 × 10 = 40 N/m] GATE 2005
1494. A body having weight of 1000 N is dropped Ans. (c) :
from a height of 10 cm over a closed coiled
helical spring of stiffness 200 N/cm. The
resulting deflection of spring is nearly :
(a) 5 cm (b) 16 cm
(c) 35 cm (d) 100 cm
GPSC ARTO Pre 30.12.2018
APPSC AE 04.12.2012 Spring force F = - kx
Ans. (b) : By conservation of energy, F = kx (compressive force)
Potential energy lost by body = strain energy stored in Case (i)—When 2 kg pan is placed on spring
spring. Force (F) = mg = 2g N
x = (l - 200) mm
Case (ii)—When 20 kg mass is placed on the pan
Force (F) = mg = 22g N
x = (l - 100) mm
From case (i) and case (ii) spring force
F = kx
2g = k (l - 200) ...(1)
22g = k (l - 100) ...(2)

Strength of Materials 429 YCT

Divide equation (2) by (1) Since wire weight is the same,
22 l − 100 π π
= therefore d12 n1 = d 22 .n 2
2 l − 200 4 4
2 2
11l - 2200 = l - 100 n1  d 2   1 
= =  
n 2  d1   2 
10 l = 2100
l = 210 mm
From equation (1) n
or, 1 = 1/ 4
2g = k (l - 200) n2
2g = k (210 - 200)
K1 d4 n
2 g 2 × 9.81 = × 2
k= = K 2 ( d / 2 ) 4 n1
10 10
k = 1.962 N/mm K1
= 16 × 4
k = 1962 N/m K2
1497. If the wire diameter of a closed coil helical
K1
spring subjected to compressive load is = 64
increased from 1 cm to 2 cm, other parameters K 2
remaining same, then deflection will decrease 1500. A spring used to absorb shocks and vibrations
by a factor of : is
(a) 16 (b) 8 (a) Torsion spring (b) Conical spring
(c) 4 (d) 2 (c) Leaf spring (d) Disc spring
GPSC ARTO Pre 30.12.2018 Assam PSC AE (IWT) 14.03.2021
GPSC AE 26.07.2015 UPPSC AE 13.12.2020, Paper-I
8WD3 n Ans. (c) : Leaf spring is used to absorb shocks and
Ans. (a) : Deflection δ1 = vibrations. Leaf springs are an integral part of vehicle's
Gd14
suspension system. They are installed to help and
Given, d1 = 1 cm support the entire weight of car or truck.
8WD3 n 8WD3 n 1501. In design of helical springs the spring index is
d2 = 2 cm δ2 = =
G(2d1 ) 4 16Gd14 usually taken as
(a) 3 (b) 5
d2 = 2d1 (c) 8 (d) 12
TSPSC AEE 2015
δ1 Kerala PSC Poly. Lect. 09.01.2014
δ2 =
16 Ans. (b) : In design of helical springs the spring index
is usually taken as 5.
1498. When a helical compression spring is subjected
to an axial compressive load, then which one of 1502. A truck of mass 20 tonnes traveling a 1.6 m/s
the following stress will be induced? impacts with a buffer spring which compress 2
mm per kN. The maximum compression of the
(a) Shear (b) Bending
spring is ______.
(c) Compressive (d) Tensile
(a) 160 mm (b) 240 mm
TNPSC AE (Industries) 09.06.2013 (c) 80 mm (d) 320 mm
UPSC JWM Adv. No-16/2009 VIZAG Steel MT 24.01.2021, Shift-I
Ans. (a) : When a helical compression spring is Ans. (d) : Given,
subjected to an axial compressive/tensile load, then
Mass of truck (m) = 20 tonne = 20 × 103 kg
there will be shear stress (direct + torsional) induced in
Velocity of truck (v) = 1.6 ms
the spring.
Buffer spring constant (k) = 2 mm/kN
1499. Two helical tensile springs of the same material x = maximum compression of the spring
and also having identical mean coil diameter
and weight, have wire diameter d and d/2. The 1
We know that kinetic energy of truck (kE) = mv 2
ratio of their stiffness is 2
(a) 1 (b) 4 1
= × 20 × 103 × (1.6 )
2
(c) 64 (d) 128
2
VIZAG STEEL MT 18.08.2013
= 25600 kN-m ……… (i)
VIZAG STEEL MT 10.06.2012
x x
64PR 3n P Gd 4  Now, compression load, W = =
Ans. (c) : δ = or, =
 δ 64R 3n  k 2
Gd 4   = 0.5x kN
d4 ∴ Work done in compressing the spring = Average
Stiffness K ∝
n compressive load × displacement

Strength of Materials 430 YCT

 0.5x 1507. For the shown figure, if k1 = 10 N/m, k2 = 20
= × x = 0.25 x 2 kN − m …..(ii) N/m and θ = 30°, then equivalent stiffness of
2
Since, the entire kinetic energy of the truck is used to the system will be _______ N/m.
compare the spring.
∴ from equation (i) and (ii)–
0.25x2 = 25600
25600
x2 =
0.25 (a) 30 (b) 12
x = 320 mm (c) 6.66 (d) 15
1503. A closed coiled helical spring is subjected to an AAI Jr. Executive 26.03.2021
axial load. the stress which is not produced in Ans. (a) : Stiffness of the system are in parallel,
the wire of the spring is___. keq = k1 + k2
(a) Torsional shear stress
= 10 + 20
(b) Compressive stress
= 30 N/m
(c) Direct shear stress
(d) Bending stress 1508. A helical spring is made of 12 mm diameter
RPSC Lect. (Tech. Edu. Dept.) 13.03.2021, Paper-I wire by winding it on a 120 mm diameter
mandrel. If there are 10 active turns and
Ans. (d) : The stress which are produced in a closed
modulus of rigidity of material is 8.2 × 104
coil helical springs are only torsional shear stress, there
is not bending stress, compressive and direct shear N/mm2 the spring constant will be
stress are produced in wire coiled spring (helical). (a) 2.94 Nmm-1 (b) 3.64 Nmm-1
-1
(tension & compressions) (c) 9.24 Nmm (d) 11.58 Nmm-1
1504. The longest leaf in Semi-elliptic leaf spring is GPSC DEE, Class-2 (GWSSB) 04.07.2021
known as Ans. (*) : Given,
(a) chief leaf (b) master leaf Wire dia, d = 12 mm
(c) major leaf (d) higher leaf Coil dia, D = 120 mm
RPSC Lect. (Tech. Edu. Dept.) 13.03.2021, Paper-I No. of turn, n = 10
Ans. (b) : The longest leaf in semi-elliptic leaf spring is Modulus of rigidity, G = 8.2 × 104 N/mm2
known as master leaf through which the spring from Spring constant, k = ?
one end of bushing to other end.
Gd 4
1505. If a spring deflect 50 mm under a 2 kN load, k=
the energy absorbed by the spring in one 8D3n
gradual application of this load is : 8.2 × 104 × (12) 4
(a) 100 J (b) 50 J =
8 × (120)3 × 10
(c) 75 J (d) 25 J
CGPSC AE 15.01.2021 k = 12.3N / mm
Ans. (b) : Given, Note-Official answer given by commission is (c).
deflection, δ = 50 mm = 0.05 m 1509. Free length for helical compression springs
Load, P = 2kN having
Energy absorbed by the spring, (a) Pn = 2d (b) Pn + 3d
E = 1/ 2 × P × δ (c) 2(p+d) (d) Pn + 4d
1 TNCSC AE 2020
= × 2 × 103 × 0.05
2 Ans. (b) : Free length for helical compression springs
E = 50 J having square ends is given as
1506. Two springs with stiffness 2 N/m and 3 N/m free length (l0) = Pn + 3d
are connected in parallel, then the equivalent Where, Pn = pitch
stiffness is ________. d = wire diameter
(a) 3 N/m (b) 1 N/m
1510. A simple spring mass vibrating system has a
(c) 5 N/m (d) 2 N/m
natural frequency of N. If the spring stiffness
VIZAG Steel MT 24.01.2021 is halved and mass is doubled, then natural
Ans. (c) : Given, k1 = 2 N/m, k2 = 3 N/m frequency will become
Parallel connected, (a) N/2 (b) 2 N
keq = k1 + k2 (c) 4 N (d) 8 N
= 2 + 3 = 5 N/m RPSC IOF, 2020
Strength of Materials 431 YCT
 1 k 1514. With compression of close coiled helical
Ans. (a) : Natural frequency f n = spring, the wire gets subjected to
2π m (a) tension
1 k (b) compression
N= .............(i)
2π m (c) shear and bending
[Given, if spring stiffness is halved and mass doubled] (d) a combination of shear and tension
1 k/2 TSPSC Manager (Engg.) HMWSSB 12.11.2020
fn =
2π 2m Ans. (d) : With compression of close coiled helical
1 k spring the wire gets subjected to a combination of shear
fn = .............(ii) and tension.
2π 4m
From equation (ii) & (ii) 1515. A compression coil spring made of an alloy
steel has mean diameter of coil as 50 mm and
1 k wire diameter as 5 mm? What will be the
N
= 2 π m shear stress factor?
fn 1 k (a) 1.04 (b) 1.05
2π 4m (c) 1.02 (d) 1.03
N NLCIL GET 17.11.2020, Shift-II
= fn
2 Ans. (b) : Given, Coil diameter (D) = 50 mm
N Wire diameter (d) = 5 mm
fn =
2 We know that spring index
So, natural frequency will be come halved D 50
C= = = 10
1511. The Wahl's factor is involved in the design of d 5
(a) Gear (b) Clutch Shear stress factor
(c) Hooke's Joint (d) Spring
1 1
Haryana PSC AE (PHED) 05.09.2020, Paper-II ks = 1 + = 1+
Ans. (d) : Wahl's correction factor is used in shear 2C 2 × 10
stress spring formula to compensate the effect of 21
curvature and direct shear stress. ks = = 1.05
20
1512. For squared - ground ends in helical coil
compression spring, the number of active coils 1516. When two springs are connected in parallel
is theoretically with stiffness ‘k’ and ‘s’, the effective stiffness
(a) Two coils less than the total number of coils is:
(b) One coil less than the total number coils (a) 1/k + 1/s (b) k+s
1 (c) k/s (d) s/k
(c) coil less than the total number of coils
2 NLCIL GET 17.11.2020, Shift-II
(d) Same as the total number of coils
Haryana PSC AE (PHED) 05.09.2020, Paper-II Ans. (b) : Spring stiffness in parallel
Ans. (a) : keq = k1 + k2
Spring ends Number of effective coil Where, k 1=k

Plain end Total coil k 2=s

Plain and grounded Total coil –1 So, kequivalent = k + s

flat/squared total coil –2 1517. If two identical helical coil springs are placed
squared and grounded Total coil –2 one over the other, total stiffness is
1513. The angle of twist for the equivalent bar to (a) doubled (b) halved
spring is given by: (c) increased four times (d) remains same
(a) 16PD2N/Gd4 (b) 8PD2N/Gd4 TSPSC Manager (Engg.) HMWSSB 12.11.2020
3
(c) 6PDN/Gd (d) 4PD2N/Gd4 Ans. (b) : If two helical springs are identical then
CIL MT 27.02.2020 stiffness of both springs are equal.
Ans. (a) : Angle of twist k1 = k2
TL
θ= Hence connection is in series –
GJ 1 1 1
PD / 2 × πDN PD πd 4 = +
θ= [∵ M = , ℓ = πDN, J = ] k eq k1 k 2
G × πd 4 / 32 2 32
16PD 2 N k
θ= Putting k1 = k2 ⇒ k eq = 1
Gd 4 2

Strength of Materials 432 YCT

1518. In which applications is leaf springs mostly When two spring having stiffness k and are connected
used? in series, then the combined stiffness of the spring (keq.)
(a) Two-wheeler front suspensions is given by
(b) In bicycles 1 1 1
(c) Airplanes = +
k eq k k
(d) In four-wheelers
APPSC Poly Lect. Automobile Engg., 2020 1 1+1
=
Ans. (d) : The laminated or leaf spring (also known as k eq k
flat spring or carriage spring) consists of a number of
k
flat plates (known as leaves) of varying length held k eq =
together by means of clamps and bolts. They are 2
mostly used in automobiles. 1522. A helical spring is subjected to an axial load W
1519. A leaf spring is to be made with seven steel and corresponding deflection in the spring is δ.
plates 65 mm wide 6.5 mm thick. Calculate the Now if the mean diameter of the spring is
length of spring to carry a central load of 2.75 made half of its initial diameter keeping the
kN and the bending stress is limited to 160 material, number of turns and wire cross-
MPa section same, the deflection will be
(a) 644.2 mm (b) 64.42 mm
(c) 74.42 mm (d) 744.2 mm δ δ
(a) (b)
TNPSC AE 2019 2 8
Ans. (d) : Data given- δ
σ b = 160 MPa b = 65 mm (c) (d) 2δ
4
t = 6.5 mm P = 2.75 kN APPSC AEE SCREENING 17.02.2019
L=? n=7
We know that 8WD 3n
Ans. (b) : Deflection in spring (δ ) = ⇒ δ α D3
3PL Gd 4
σb = 3
2nbt 2 δ 2  D2 
∴ = 
2nbt 2σ b 2 × 7 × 65 × 6.5 2 × 160 δ1  D1 
L= =
3P 3 × 2.75 × 10 3 D
3

L = 745.64mm ≃ 744.2mm δ2  2 
= 
1520. Which is true statement about Belleville δ1  D 
springs?  
 
(a) These are used for dynamic loads
δ
(b) These are composed of coned discs which ⇒ δ2 =
may be stacked upto obtain variety of load- 8
deflection characteristics 1523. For the spring system given in figure, the
(c) These are commonly used in clocks and equivalent stiffness is
watches
(d) These take up torsional load
TNPSC AE 2019
Ans. (b) : These are composed of coned discs which
may be stacked upto obtain variety of load-deflection
characteristics is true statement about Belleville springs.
1521. When two springs are in series (having
stiffness k), the equivalent stiffness will be
(a) k (b) k/2 (a) 5K/2 (b) 2 K/5
(c) 2k (d) k2 (c) K (d) 3K
Gujarat PSC AE 2019 ISRO Scientist/Engineer 22.04.2018
Ans : (b) : Ans. (c): In this system upper two springs are in parallel
then equivalent spring stiffness is
K1 = K +K = 2K
Then K1 and 2K are in series then equivalent spring
stiffness is K2
1 1 1 1 1
= + = +
K 2 K1 2 K 2 K 2 K
K2 = K

Strength of Materials 433 YCT

1524. Calculate equivalent stiffness of the spring for 1526. Angle of helix in a closed coiled spring is :
the system shown below, which has spring (a) < 10° (b) >10°
stiffness of 1000 N/m (c) = 10° (d) None
Kerala PSC Poly. Lect., 2017
Ans. (a)
• Helix angle of closed coil helical spring is below 10°.
• Helix angle of open coil helical spring is more than
10°.
1527. A Leaf spring in automobiles is used-
(a) To apply forces
(a) 7500 (b) 2000 (b) To measure forces
(c) 750 (d) 4000 (c) To absorb shocks
Vadodara Muncipal Corp. DEE, 2018 (d) To store strain energy
Ans. (c) : Given, GMB AAE 25.06.2017
k = 1000 N/m Ans. (c) : The overall purpose a leaf spring is to provide
keq = ? support for a vehicle. It is also provides for a smoother
1 & 2 in parallel ride absorbing any bumps in the road. Leaf spring are
also used to locate the axle and control the height at
which the vehicle rides and helps keep the tires aligned
on the road.
1528. A 0.5 kg weight is attached to a light spring
elongates it by 0.981 mm. The natural
frequency of the system would be
approximately
(a) 6 cps (b) 11 cps
(c) 16 cps (d) 21 cps
k12 = 2k WBPSC AE, 2017
Now, k12 will be parallel with 3 Ans. (c) : Given, m = 0.5 kg
k13 = 2k + k = 3k Elongation, δ = 0.981 mm
k13 in series with 4 W mg
k= =
3k × k 3k 2
3 δ δ
so, k eq = = = ×k
3k + k 4k 4 0.5 × 9.81
=
3 0.981 × 10−3
= × 1000 k = 5000 N/m
4
= 750N / m 1 k
Natural frequency, fn = ×
1525. A closed coil helical spring has 15 coils. If five 2π m
coils of this spring are removed by cutting the
1 5000 50000
stiffness of the modified spring will : × =
(a) Increase to 2.5 times 2π 0.5 5
(b) Increase to 1.5 times 1 50
fn = × 100 =
(c) Reduced to 0.66 times 2π π
(d) Remain unaffected fn = 15.916
GPSC ARTO Pre 30.12.2018 fn ≈ 16 cps or 16 cycle/sec
Ans. (b) : Given, 1529. Which of the following springs are subjected to
Gd 4 Gd 4 both the torsion and bending?
n = 15; stiffness (k) = 3
= 3 P. Closed coil helical spring
8D n 8D (15)
Q. Closed coil helical spring subjected to
∵ After cutting 5 coils, axial couple
Gd 4
15 R. Open coil helical spring subjected to axial
n = 10 k ' = × force
8D3 (10) 15 S. Open coil helical spring subjected to axial
15  Gd 4  torque
k'=  3  (a) Only P and Q (b) Only Q and R
10  8D (15) 
(c) Only R and S (d) P, Q and R
k ' = 1.5k APGENCO AE, 2017
Strength of Materials 434 YCT
Ans. (c) : Closed coil spring subjected to axial force Ans. (d) : Spring force w = kδ
subjected to torque only. So stiffness (k) = w/δ
Closed coil spring subjected to axial couple subjected to
bending only. 1535. The device that permits variation in the
Open coil helical spring always experience bending and distance between the spring eyes of leaf spring
torsion both irrespective of loading either axial force or as the spring flexes is called :
axial couple. (a) Spring shackle (b) Spring U bolt
1530. The energy stored per unit volume in coil (c) Spring hanger (d) Spring leaf
spring as compared to leaf spring is Kerala PSC Poly. Lect. (Auto) 05.05.2016
(a) Equal amount (b) Double the amount
(c) Four times higher (d) Six times higher Ans. (a) The device that permits variation in the
TNPSC AE 2017 distance between the spring eyes of leaf spring as the
Ans. (d) : The energy stored per unit volume in coil spring flexes is called spring shackle.
spring as compared to leaf spring is six times higher. 1536. In the figure shown, the spring deflects by δ to
1531. Due to addition of extra full length leaves the position A (the equilibrium position) when a
deflection of a semi-elliptic spring mass m is at a position B at some instant. The
(a) increases (b) decreases change in potential energy of the spring mass
(c) does not change (d) is doubled
system from position A to position B is
TNPSC AE 2017
Ans. (b) : Due to addition of extra full length leaves the
deflection of a semi-elliptic spring decreases.
1532. The ratio of total load on the spring to the
maximum deflection is called is
(a) Spring Tension (b) Spring life
(c) Spring efficiency (d) Spring rate
TNPSC AE 2017
Ans. (d) : The ratio of total load on the spring to the
maximum deflection is called is Spring rate is denoted by k.
1 2 1 2
Spring rate also known as stiffness of spring. (a) kx (b) kx − mgx
W 2 2
K= 1 1
δmax . (c) k ( x + δ)
2
(d) kx 2 + mgx
1533. For a spring mass system, the frequency of 2 2
vibration is 'N' what will be the frequency when RPSC ADE 2016
one more similar spring is added in series Ans. (b) :
(a) N/2 (b) N / 2
(c) 2/N (d) 2N
TNPSC AE 2017
Ans. (b) : We know that,
1 s
N=
2π m 1 2
kx − mgx
N∝ s 2
When two similar spring is added in series then 1537. The reason why a laminated spring is made up
s of a series of leaves is to:
sequivalent =
2 (a) reduce the inter leaf friction
then, (b) soften the spring action and increase the
N2 s/2 maximum deflection
= (N1 = N)
N1 s (c) allow the leave to slide during the bump
movement
N
N2 = (d) overcome the weakness at the centre of a
2 single leaf spring
1534. The spring rate or stiffness (k) of the spring is RPSC VPITI 14.02.2016
given by: Ans. (d) : As the laminated spring deflects the plates or
(Where w load and δ deflection of spring)
leaves slide over each other causing inter-plate friction.
(a) k = 2wδ (b) k = δ/w
(c) k = wδ (d) k = w/δ To overcome the weakness at the centre of a single leaf
CIL (MT) 2017 IInd Shift spring.
Strength of Materials 435 YCT
1538. Shear stress in a closed coil helical spring (coil Ans. (b) : F = 200 N
diameter 'D') of wire (diameter 'd') subjected Deflection = 10 cm
to an axial load 'W' is (δ) = 0.1 m
(a) WD/πd3 (b) (H2 – H1)/(H3 – H1) F
(c) (H3 + H1)/(H3 + H2) (d) (H1 + H2)/(H1 + H3) Stiffness k =
δ
RPSC VPITI 14.02.2016 200
Ans. (*) : Note- No answer given by commission. = = 2000 N/m
0.1
1539. The bending stress in full length leaves of a
1 k 1 2000
laminated spring is f= =
(a) same as in graduated leaves 2π m 2 π m
(b) 50% more than that in graduated leaves 2000
(c) 50% less than that in graduated leaves 4π =
m
(d) 25% more than that in graduated leaves 2000
RPSC AE (GWD) 06.05.2016 = 16 π2
m
Ans. (b) : The bending stress in full length leaves of a 2000
laminated spring is 50% more than in graduated m=
leaves. 16π2
m = 12.66 kg
1540. Which of the following types of energy is
m ≈ 12.5 kg
possessed by a tightly wound spring?
(a) Kinetic energy 1543. For a helical spring of mean diameter D. wire
diameter d. number of coils n. modules of
(b) Magnetic energy
transverse elasticity C, when subjected to an
(c) Strain energy axial load W the deflection would be
(d) Electro-magnetic energy
Wn D3 2Wnd 3
KPCL AE 2016 (a) (b)
Cd 4 CD 4
Ans. (c) : When a tightly wound spring then energy is
possessed in form strain energy. But a compressed 8Wn 8WnD3
(c) (d)
spring has the potential energy in it due its compressed CD 4d 3 Cd 4
state when the compressed spring is released the Nagaland CTSE 2016-17 Ist Paper
potential energy changes to kinetic energy. Ans. (d) : In helical spring, when it is subjected to an
1541. When an open coiled helical spring (one end is axial load (W) the deflection (δ) would be,
fixed) in the vertical position is subjected to an 8WD3 n
δ=
axial tensile load and a couple, the strain Cd 4
energy stored will be due to ......................... Where,
(a) bending moment only D = mean diameter of spring
(b) torsional moment only d = wire diameter
(c) both bending moment and torsional moment C = modules of transverse elasticity
(d) shear force only W = axial load
n = no. of turn or coils
HPPSC Asstt. Prof. 29.10.2016
1544. A spring of stiffness 50 N/mm is mounted on
Ans. (c) : In this case both bending moment and top of second spring of stiffness 100 N/mm,
torsional moment will apply. what will be the deflection under the action of
■ Twisting in spring- 500 N?
16T (a) 15 mm (b) 3.33 mm
( τs )max = (c) 5 mm (d) 10 mm
πd 3
UPRVUNL AE 07.10.2016
■ Bending in spring-
Ans. (a) : Spring arrangement in series then,
32M
(σ b ) max = 1
=
1
+
1
=
3
πd 3 keq 100 50 100
1542. The reading of a spring balance is from 0 to
100
200 N and is 10 cm long. A body suspended keq = N/mm
from the spring balance is observed to oscillate 3
vertically at 2 Hz. The mass of the body is Then deflection (δ)
nearly W 500
(a) 22.5 kg (b) 12.5 kg δ= = ×3
keq 100
(c) 37 kg (d) 45 kg
ISRO Scientist/Engineer 03.07.2016 δ = 15 mm
Strength of Materials 436 YCT
1545. When a closely coiled helical spring of mean Ans : (c)
diameter {D} is subjected to an axial load (W),
the stiffness of the spring is given by
(a) Cd4/D3n (b) Cd4/2D3n
(c) Cd4/4D3n (d) Cd4/8D3n
APPSC AEE MAINS 2016, PAPER-III
Ans. (d) : We know that deflection due to axial load
(W) in a closely coiled helical spring is given as,
3 Wℓ
64WR 3 n Maximum bending stress (σmax) =
δ= 2 Nbt 2
Cd 4 l = spring length = 1m
8WD3 n W = central point load = 2000N
δ=
Cd 4 b = Width = 5 cm = 5 × 10-2m
We know that, spring stiffness of spring. t = thickness = 1cm = 1 × 10-2m
W σmax = maximum bending stress = 100N/mm2
K= 3 Wℓ
δ σ=
2 Nbt 2
Cd 4
K= 3 Wℓ
8D3 n N=
2 σbt 2
1546. A composite shaft consisting of two stepped 3 2000 × 1
portions having spring constant k1 and k2 is N= ×
held between two rigid supports at the ends. 2 100 × 10 × 5 × 10−2 × 1×10−4
6

Its equivalent spring constant is N=6

(a) (k1 + k2)/2 (b) (k1 + k2)/k1k2 1549. In a laminated spring the strips are provided
(c) k1k2/(k1+k2) (d) (k1 + k2) in different lengths for
APPSC AEE MAINS 2016, PAPER-III (a) Equal distribution of stress
Ans. (d) : (b) Equal distribution of strain energy
(c) Reduction in weight
(d) All are correct
MPPSC AE 2016
Ans : (a) In a laminated spring the strips are provided in
distribution of bending stress.
k equivalent = k1 + k 2
Different length for equal distribution of bending stress
1547. A closed coiled helical spring is subjected to therefore spring behave just like a uniform strength
axial load (W) and absorbs 100 Nm energy at beam.
4cm compression. The value of axial load will 1550. In a leaf spring, the deflection at its centre is
be: (a) δ = Wl3 /8 Enbt3 (b) δ = Wl3/4 Enbt3
(a) 12.5 kN (b) 5.0 kN
(c) 10.0 kN (d) 2.5 kN (c) δ = 3Wl / 8Enbt
3 3
(d) δ = Wl3/2Enbt3
UPRVUNL AE 07.10.2016 TSPSC AEE 2015
Ans. (b) : Energy absorbs in spring = 100 N-m (Where W = Max. load on the pring
(W + 0) W l - length of the spring
Average load on spring = = n = No. of plates
2 2
Deflection in spring = 4 cm = 0.04 m b - Width of the plates
Then t = thickness of the plates)
W Ans : (c)
× 0.04 = 100
2
200
W= = 5000 N
0.04
W = 5kN
1548. A laminated spring 1 m long carries a central
point load of 2000 N. The spring is made of
plates each 5 cm wide and 1 cm thick. The
bending stress in the plates is limited to 100 M = Maximum bending moment
N/mm2. The number of plates required will be: I = Moment of Inerita.
(a) 3 (b) 5 t
Y = = half thickness of spring.
(c) 6 (d) 8 2
HPPSC W.S. Poly. 2016 n = Number of leaf spring.
Strength of Materials 437 YCT
 Wℓ Equivalent stiffness
Max bending moment = 1 1 1 1
4n = + +
3 Wℓ ke k1 k2 k3
Maximum bending stress =
2 nbt 2 1 1 1 1
3
= + +
3 wℓ ke 30 20 12
maximum deflection =
8 nEbt 3 1 2 + 3+ 5
=
1551. Type of spring used in door hinges is : ke 60
(a) Spiral spring ke = 6 N/m
(b) Multi leaf spring 1554. A helical spring is made from a wire of 6 mm
(c) Helical torsion spring diameter and has outside diameter of 75 mm.
(d) Helical extension spring If the permissible shear stress is 350 N/mm2
GWSSB AAE, 27.12.2015 and modulus of rigidity is 84 kN/mm2, the
Ans. (c) : Helical torsion spring are particularly useful axial load which the spring can carry without
as door hinges from residential doors, garage doors, and considering the effect of curvature.
automobile doors to large, industrial heavy duty (a) 422.5 N (b) 382.5 N
overhead door used at loading docks and ware house. (c) 392.5 N (d) 412.5 N
1552. In a close-coiled helical spring subjected to an (e) 402.5 N
axial load and other quantities remaining the CGPSC AE 26.04.2015 Shift-I
same, if the wire diameter is doubled, then the Ans. (d) : Given,
stiffness of the spring when compared to the d = 6 mm, D0 = 75 mm, τ = 350 MPa, G = 84 kN/mm2
original one will become We know that
(a) two times (b) four times D = D0–d = 75 – 6 = 69 mm
(c) eight times (d) sixteen times D 69
∴ spring index, C = = = 11.5
MPPSC AE 08.11.2015 d 6
Ans. (d) : Let, d = diameter of spring wire. Neglecting the effect of curvature–
P = Axial load and K = stiffness of spring. We know that shear stress factor
For the close - coiled helical spring, if all the quantities 1 1
remaining same except diameter of spring wire. Ks = 1 + = 1+ = 1.043 and maximum shear
2C 2 × 11.5
∴ Stiffness of spring (K) ∝ d . 4
stress induced in the wire (τ)
d 
4 8WD 8W × 69
K 350 = K s × = 1.043 ×
⇒ 1 = 1  ∵ d2 = 2d1 πd 3 π× 63
K 2  d2 
350
4 4 W= = 412.7 N
K1  d1  1 1 0.848
⇒ =  =  = 1555. Which is not a type of ends for helical
K 2  2d1  2 16
compression spring
⇒ K2 = 16 K1 (a) Plain ends
i.e., Sixteen times. (b) square ends
1553. When three springs are in series having (c) half hook ends
stiffness 30 N/m, 20 N/m and 12 N/m, the (d) plain and ground ends
equivalent stiffness will be (e) square and ground ends
(a) 14 (b) 8 CGPSC AE 26.04.2015 Shift-I
(c) 12 (d) 6 Ans. (c) : half hook ends
(e) 10
CGPSC AE 26.04.2015 Shift-I 1556. Spring stiffness is :
Ans. (d) : Let stiffness of the composite spring be ke (a) Load per unit deflection
Given (b) Load carrying capacity of the spring
k1 = 30 N/m (c) Ratio of mean coil diameter to wire diameter
k2 = 20 N/m (d) None
k3 = 12 N/m Kerala PSC Poly. Lect. 09.01.2014
Ans. (a) : Spring stiffness –
W
k=
δ
Where, k → Spring stiffness
W → Load
δ → Deflection

Strength of Materials 438 YCT

1557. A helical spring, having N number of coils, has Ans. (b) : Levy-von Mises flow equation
spring-constant 2000 N/m. If there is another
helical spring of the same material and same σ  σ y σz 
εx = x – V  + 
wire diameter and having N/2 number of coils, E E E
its spring constant will be 
(a) 4000 N/m (b) 2000 N/m So, Levy von Mises equation is the relation between the
(c) 1000 N/m (d) 500 N/m ratios of the component of strain increment and
Haryana PSC Civil Services Pre, 2014 deviatoric stress and shear stresses.
Ans. (a) Given, 1561. A helical spring of wire diameter 6 mm and
Number of Coils = N spring index 6 is acted by an initial load of 750
Spring constant (k) = 2000 N/m N. After compressing it further by 12 mm the
If no. of coils = N/2 stress in the wire is 500 MPa. Find the number
then k = ? of active coils. (given : G = 84000 MPa)
4 (a) 10 (b) 18
Gd
k= (c) 24 (d) 30
8D3 N UPRVUNL AE 2014
1 Ans : (b) Given, C = 6, d = 6 mm, τmax = 500 MPa
k∝
N G = 84000 MPa
Initial load = 750 N
k×N
⇒ 2000 × N = Compression, δ = 12 mm
2 D = C × d = 36
k = 4000 N / m 4 × 6 − 1 0.615
kw = + = 1.2525
1558. The work done in stretching a spring of spring 4×6 − 4 6
constant k by a length ∆ is k ×8× F× D
(a) k ∆ (b) k ∆2 ∵ τmax = w
πd 3
(c) k ∆/2 (d) k ∆2/2 1.2525 × 8 × F × 36
MPPSC State Forest Service Exam, 2014 500 =
π × 63
Ans. (d) : The work done in stretching a spring is equal F = 940.6 N
to the potential energy of stretched spring.
F 940.6 − 750
As we know, k= =
1 δ 12
P.E. in a stretched spring U = k∆ 2 k = 15.83
2 No. of active coil –
Where, k = Stiffness, ∆ = Deflection/elongation
Gd 4
1559. A body weighing 500 kg falls 8 cm and strikes N=
a 500 kg/cm spring. The deformation of the 8D3k
spring will be ________cm. 84000 × 64
(a) 8 (b) 2 =
15.83 × 8 × (36)3
(c) 20 (d) 4
MPSC Poly. Lect. 2014 N = 18.42
Ans. (d) : m = 500 kg N ≃ 18
h = 8 cm 1562. Load pc and p0 respectively acting axially upon
k = 500 kg/cm close coiled and open coiled helical springs of
kinetic energy of body = Energy stored in spring. same wire dia, coil dia, no of coils and material
1 1 2 N to cause same deflection
mv = kx
2
[k = 500 kg/cm = 500 × 9.81 ]
2 2 cm (a) pc/p0 is 1, < 1 or > 1 depending upon α
mv2 = kx2 (b) pc/p0 = 1
500 × ( 2 × 9.81× 8)2 = 500 × x 2 × 9.81 (c) pc/p0 > 1
(d) pc/p0 < 1
x = 4cm
(HPPSC AE 2014)
1560. The Levy-von Mises equation is relation Ans : (d) Load Pc and Po respectively acting axially
between the ratios of: upon close coiled and open coiled helical spring of
(a) The components of strain and stress same wire dia, coil dia, no of coils and material to case
increments same deflection
(b) The components of strain increment and Pc/Po < 1
deviatoric stress and shear stresses
(c) Plastic strain increments and deviatoric stress 1563. The load required to produce and deflection in
and shear stresses a spring is called
(d) The stress increments and plastic strain (a) stiffness constant (b) strain constant
increments (c) rigidity constant (d) torsional constant
HPPSC Asstt. Prof. 2014 Mizoram PSC AE/SDO 2014, Paper-II
Strength of Materials 439 YCT
Ans. (a) : Stiffness : It is defined as the load required 1566. Pick the correct statement about the maximum
for unit deflection in spring. bending stress in the various leaf or laminated
W spring, assuming it has been designed ideally.
k= (a) It increases uniformly from the shortest leaf
δ to the longest leaf.
k → Stiffness constant
(b) It is largest in the longest leaf.
W → Load
(c) It is largest in the shortest leaf.
δ → Deflection
(d) It is same in all leafs.
1564. A closed coil helical spring having mean RPSC AEN Pre-2013
diameter of coil 'D' and wire diameter 'd'
subjected to axial force 'P'. The strain energy Ans. (d) : Leaf springs (sometimes also known as
stored in the spring is laminated springs) are made up of beams of uniform
strength and are normally used in automobiles to
8PC3n 8PD3n absorbs the shock and vibration produced by road
(a) (b)
Gd Gd 4 bumps there by providing comfort to passengers.
4P 2 D3n 8PD 1567. A closely coiled helical spring is cut into two
(c) (d) equal parts. What will be the ratio of the
Gd 4 πd 2 deflection of any of the resulting spring to the
GPSC Lect. (Poly.) 07.12.2014 deflection of the original spring for the same
1 load?
Ans. (c) : Strain energy = T.θ (a) 2 (b) 1/2
2
2 (c) 1 (d) 3/4
1 TL 1 T L
= T. = RPSC AEN Pre-2013
2 GJ 2 GJ 3
8WD n
 D
2 Ans. (b) : Deflection (δ) =
 P  × πDn Gd 4
U= 
1 2 If coil cut into two equal part –
2 G. π .d 4 F
32 Deflection (δ) =
k
4P 2 D3n
U=
Gd 4
deflection of springs.
By castigliano's theorem
∂U
δ= 1
∂P δ∝
k
8PD3n F
δ=
Gd 4 δ1 2k δ 1
= ⇒ 1=
1565. In a spring mass system if one spring of same δ F/ k δ 2
stiffness is added in series, new frequency of
vibration will be:- 1568. A close-coiled helical spring absorbs 80 N mm
of energy while extending by 4 mm. The
ωn 2 stiffness of the spring is :
(a) (b)
2 ωn (a) 5 N/mm (b) 10 N/mm
(c) 16 N/mm (d) 20 N/mm
ω 2 APPSC AE 04.12.2012
(c) n (d)
2 ωn Ans. (b) : For a closed - Coiled helical spring -
UKPSC AE-2013, Paper-I 1
Ans. (a) : Case I - Absorbe energy (U) = W.δ … (i)
2
K Given, δ = deflection of coil under the load W
ωn =
m = 4 mm
Case II - when spring is added in series. U = Absorbs energy
= 80 N-mm
K
K eq = 2U
2 ∴ W= [From eq. (i)]
δ
K 2 × 80N − mm
[ ωn ]II = W= = 40N
2.m 4mm
ω
[ ωn ]II = n ∴ Stiffness of spring (K) =
W 40N
= = 10N / mm
2 δ 4mm
Strength of Materials 440 YCT
1569. Wahl suggested the correction in the stress 1573. In a close-coiled helical spring
factor to account for (a) plane of the coil and axis of the spring are
(a) the additional transverse shear stress closely attached
(b) stress concentration (b) angle of helix is large
(c) plane of the coil is normal to the axis of the
(c) fatigue stress spring
(d) none of these (d) deflection is small
TNPSC ACF 2012 BPSC AE 2012 Paper - VI
Ans. (b) : Wahl suggested the correction in the stress Ans : (c) : plane of the coil is normal to the axis of the
factor to account for stress concentrations. spring.
1570. A close coiled helical spring absorbs 500 Nmm 1574. In a squared and ground helical spring, the
of strain energy while extending by 10 mm. effective number of turns is increased by
What is the stiffness of spring : (a) one (b) two
(a) 50 N/mm (b) 10 N/mm (c) 15 (d) nil
(c) 5 N/mm (d) None of these TNPSC ACF 2012
PSPCL AE, 2012 Ans. (b) : In a squared and ground helical spring the
effective number of turns is increased by-two.
1
Ans. (b) : U = Kx 2 1575. Stiffness of a spring can be increased by
2 (a) increasing the number of turns
1 2 (b) increasing the free length
500 = × K × 10
2 (c) decreasing the number of turns
K = 10 N/mm (d) decreasing the spring wire diameter
1571. Match List – I with List – II and select the TNPSC ACF 2012
correct answer using the code given below the W Gd 4
lists. Ans. (c) : Stiffness of Spring = =
List – I List – II δ 64R 3n
(Characteristic) (Member) 1
A. Kernel of section 1. Helical spring K∝
n
B. Tie and Strut 2. Bending of beams 1576. Due to addition of extra full length leaves the
C. Section modulus 3. Eccentric loading of deflection of a semi-elliptic spring
short column (a) increases (b) decreases
D. Stiffness 4. Roof truss (c) does not change (d) is doubled
Code : TNPSC ACF 2012
A B C D Ans. (b) : Addition of extra full length leaves the
(a) 1 2 3 4 deflection of semi-elliptic spring decreases.
(b) 3 4 2 1 1577. A length of 10 mm diameter steel wire is coiled
(c) 4 1 2 3 to a close coiled helical spring having 8 coils of
(d) 2 3 1 4 75 mm mean diameter and the spring has a
UKPSC AE 2012 Paper-I stiffness K. If the same length of steel wire is
Ans. (b) : coiled to 10 coils of 60 mm mean diameter,
Kernel of section - Eccentric loading of short column then the spring stiffness will be
Tie and Strut -Roof truss (a) K (b) 1.95 K
Section modulus - Bending of beams (c) 1.56 K (d) 1.25 K
Stiffness - Helical spring APPSC AEE 2012
1572. A spring scale reads 20 N as it pulls a 5.0 kg Ans. (c) : wire diameter d = 10mm.
mass across a table. what is the magnitude of Mean diameter of first coil D1 = 75mm
the force exerted by the mass on the spring No. of Coils of first N1 = 8
scale ? Mean diameter of second coil D2 = 60 mm
(a) 4.0 N (b) 5.0 N No. of coils of second N2 = 10
(c) 20.0 N (d) 49.0 N
Gd 4
UKPSC AE 2012 Paper-I Then spring stiffness k =
Ans. (c) : 8D3 N
Gd 4
3
k1 8D13 N1 D32 N 2  60  10
= = =   ×
k2 Gd 4 D13 × N1  75  8
8D32 N 2
k1
• 5 kg mass is balance when 20 N applied then scale = 0.64 ⇒ k 2 = 1.56k1
read 20 N. k2

Strength of Materials 441 YCT

1578. Cracks in helical springs used in railway 1581. A mass is suspended at the bottom of two
carriages usually start on the inner side of the springs in series having stiffness 10 N/mm and
coil because of the fact that 5N/mm. The equivalent spring stiffness of the
(a) It has a lower curvature than the outer side two springs is nearly
(b) It is subjected to higher stress than outer side (a) 0.3 N/mm (b) 3.3 N/mm
(c) It is subjected to higher cyclic loading than
(c) 5 N/mm (d) 15 N/mm
outer side
(d) It is more stretched than outer side during RPSC Lect. (Tech. Edu. Dept.) 2011
manufacturing Ans. (b) : Given,
APPSC AEE 2012 k1 = stiffness of spring 1st = 10 N/mm
Ans. (b) : Cracks in helical springs used in railway k2 = stiffness of spring 2nd = 5 N/mm
carriages usually start on the inner side of the coil For series,
because it is subjected to higher stress than outer side.
1 1 1
1579. If two springs with stiffness K1 and K2 are keq (equivalent stiffness) ⇒ = +
connected in series, then stiffness of the k eq k1 k 2
composite spring is given by k ×k 10 × 5
1 1 So, k eq = 1 2 = = 3.3 N / mm
(a) K1 + K2 (b) + k1 + k 2 15
K1 K 2
1582. A semi-spherical laminated leaf spring
1 1
(c) − (d) K1– K2 consisting of number of full length leaves say
K1 K 2 'nf' and number of graduated leaves say 'ns' is
APPSC AEE 2012 subjected to a central load 'P'. The ratio of
Ans. (b) : If spring in series. stress in full length leaves to the stress in
graduated leaves will be:
(a) 1/1 (b) 2/1
(c) 3/2 (d) ns/nf
RPSC AE GWD, 2011
Ans. (d) : Maximum bending stress in full length leaves
6F L
(σ b ) f = f 2
n f bt
1 1 1 Maximum bending stress in Graduated leaves–
= + 6F L
K eq K1 K 2 ( σ b )s = s 2
n s bt
K1 .K 2
or K eq = Given, Ff = Fs = F
K1 + K 2
⇒ If spring in parallel ( σ b )f n s
⇒ =
Keq = K1 + K2 ( σ b )s n f
1580. Two close-coil springs are made from a small
diameter wire, one wound on 2.5 cm diameter 1583. Assertion (A) : In open-coiled helical springs,
core and the other on 1.25 cm diameter core. if axial load causes normal stress and shearing
each spring had 'n' coils, then the ratio of their stress in the spring wire.
spring constant would be Reason (R) : Helix angle is large in open-coiled
(a) 1/16 (b) 1/8 helical springs.
(c) 1/4 (d) 1/2 Codes :
APPSC AEE 04.12.2012
(a) Both A and R are individually true and R is
Ans. (b) : D1 = 2.5 cm, D2 = 1.25 cm
the correct explanation of A
No. of coils = n
(b) Both A and R are individually true and R is
K1
=? not the correct explanation of A
K2 (c) A is true but R is false
Gd 4 (d) A is false but R is true
K= 3
8D n UPSC JWM Advt. No.-50/2010
1 Ans. (b) : Pitch of helical springs is the axial distance
K∝ 3 between two similar points on the adjacent coils.
D Springs with helix angle larger than 10° is called open
3 3
K1  D 2   1.25  1 coil helical springs and helix angle less then 10° is
=  =  = called closed coil helical spring. Such spring have
K 2  D1   2.5  8 longer pitch.
Strength of Materials 442 YCT
1584. The maximum shear stress induced in the wire Ans. (a)
of a circular section of a helical spring depends
L3
on δspring α
(a) material of the wire t3
(b) size of cross-section t3
(c) the ratio of d/D ⇒K α
L3
(d) all of these
TNPSC AE, 2008 C × t3
or K =
Ans. (b) : For closed coil helical spring under axial pull L3
8K w FC where C = constant
τmax = for same stiffness
πd 2
where; C × t13 C × t 32
= 3
Kw = Wahl's factor L31 L2
C = Spring index (D/d)
t13 t 32
d = Spring wire diameter =
F = Applied force L31 ( 2L1 )3
The maximum shear stress induced in the wire of a
t 32
circular section of a helical spring depends on size of t13 =
cross-section. 8
1585. Two spring of stiffness 'S1' and 'S2' are t 3L = 8t13
connected in parallel, what will be the stiffness
t1 = 2t1
of the composite spring?
(a) S2 + S2 (b) S2 × S2 1588. In an open coiled helical spring an axial load
S (S1 S2 ) on the spring produces which of the following
(c) 1 (d) stresses in the spring wire?
S1 (S1 + S2 )
(a) normal (b) torsional shear
WBPSC AE 2008 (c) direct shear (d) all of the above
Ans. (a) : For parallel – UKPSC AE 2007 Paper -I
equivalent stiffness (Seq) = S1 + S2 Ans. (d) : Under axial load spring will be under both
For series – bending and torsion. Thus is under complex stress
1 1 1 therefore maximum principal stress and max. shear
equivalent stiffness = +
Seq S1 S2 stress are found.
1589. When a helical coiled spring is compressed
1586. Two concentric helical springs with spring
constants 200 N/mm and 160 N/mm when axially, it possesses
subjected to a load of 1800 N will deflect by (a) potential energy (b) kinetic energy
(a) 5 mm (b) 10 mm (c) mechanical energy (d) none of the above
UKPSC AE 2007 Paper -I
(c) 20.25 mm (d) 40.5 mm
DRDO Scientists 2008 Ans. (a) : Potential energy
1590. A closed coil helical spring of stiffness 'K' is
P1 k1 200 5
Ans. (a) : = = = 2
P2 k 2 160 4 cut to of its original length. The stiffness of
3
P1 + P2 = 1800
this spring of reduced length is
 5
1 +  P2 = 1800 (a) K
3
(b) K
 4 2
P2 = 800 N
2 1
P2 = k2x (c) K (d) K
800 = 160 x 3 3
x = 5mm OPSC Civil Services Pre 2006
1587. If l is the length of a leaf spring, t the W Gd 4 1
thickness, of each leaf, then for the same Ans. (a) : Stiffness (K) = = or K ∝
δ 64R 3n n
stiffness, if l is doubled than t should be
where n = no of coils
(a) doubled (b) halved
(c) made four times (d) remain same 3
the new stiffness of spring will be K
WBPSC AE, 2007 2

Strength of Materials 443 YCT