CLASS TEST S.No.

: 01_ JP_CE_08052023

Delhi | Bhopal | Hyderabad | Jaipur | Pune | Bhubaneswar | Kolkata

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CIVIL ENGINEERING
STRENGTH OF MATERIALS
Duration : 1:00 hr. Maximum Marks : 50

Read the following instructions carefully

1. This question paper contains 30 objective questions. Q.1-10 carry one mark each and
Q.11-30 carry two marks each.

2. Answer all the questions.

3. Questions must be answered on Objective Response Sheet (ORS) by darkening the appropriate
bubble (marked A, B, C, D) using HB pencil against the question number. Each question has
only one correct answer. In case you wish to change an answer, erase the old answer completely
using a good soft eraser.

4. There will be NEGATIVE marking. For each wrong answer 1/3rd of the full marks of the question
will be deducted. More than one answer marked against a question will be deemed as an
incorrect response and will be negatively marked.

5. Write your name & Roll No. at the specified locations on the right half of the ORS.

6. No charts or tables will be provided in the examination hall.

7. Choose the Closest numerical answer among the choices given.

8. If a candidate gives more than one answer, it will be treated as a wrong answer even if one
of the given answers happens to be correct and there will be same penalty as above to that
questions.

9. If a question is left blank, i.e., no answer is given by the candidate, there will be no penalty for
that question.

DO NOT OPEN THIS TEST BOOKLET UNTIL YOU ARE ASKED TO DO SO

 2 Civil Engineering

Q.No. 1 to Q.No. 10 carry 1 mark each What will be the maximum shear stress
induced on the section?
Q.1 Which of the following statement is correct
(a) 50.93 MPa
regarding assumption in Euler’s column
(b) 30.56 MPa
theory?
(c) 40.74 MPa
1. Initially the column is perfectly straight
(d) 60.36 MPa
and load applied is truly axial.
2. The failure of column occurs due to Q.6 The ratio of reactions RA and RB of the simply
buckling alone. supported beam, as shown in figure below,
3. The length of column is small as will be
compared to its cross-section dimensions. 5t 3t
(a) 1 and 2 (b) 2 and 3 2t/m
A B
(c) 1 and 3 (d) 1, 2 and 3
C E D
RA RB
Q.2 A closely-coiled helical spring of round steel
2m 2m 2m 2m
wire 5 mm in diameter having 12 complete
(a) 1
coils of 50 mm mean diameter is subject to
an axial load of 100 N. What will be deflection 3
(b)
of the spring. [Take modulus of rigidity (G) 2
= 80 GPa.] 1
(c)
(a) 192 mm (b) 24 mm 2
(c) 42 mm (d) 252 mm 2
(d)
Q.3 A copper wire of 2 mm diameter is required 3
to be wounded around a drum. What will
Q.7 For a closed coiled helical spring, choose the
be the minimum radius of drum, if the stress
incorrect statements.
in the wire is not to exceed 80 MPa?
1. Resultant shear stress at outer surface of
[Take modulus of elasticity for copper = 100
the spring is more compared to inner
GPa]
surface.
(a) 1.5 m (b) 3 m
2. If spring of stiffness k is divided into m
(c) 1.25 m (d) 2.5 m
equal parts then stiffness of each resultant
Q.4 Which of the following statement is correct spring is mk.
regarding conjugate beam? (a) 1 and 2
1. Slope at any section in given beam will (b) 1 only
be SF in conjugate beam at the (c) 2 only
corresponding section. (d) None of above
2. Deflection at any section in given beam
Q.8 A rectangular beam 300 mm deep is simply
will be BM in conjugate beam at the
supported over a span of 4 metres. What
corresponding section.
uniformly distributed load the beam may
3. Internal hinge in given beam will be
carry, if the bending stress is not to exceed
internal hinge in conjugate beam also.
120 MPa?
(a) 1 and 2 (b) 2 and 3
[Take I = 225 × 106 mm4]
(c) 1 and 3 (d) 1, 2 and 3
(a) 180 kN/m
Q.5 A section of shaft of 100 mm diameter is (b) 45 kN/m
subjected to combined bending and twisting (c) 90 kN/m
moments of 8 kNm and 6 kNm respectively. (d) 360 kN/m

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• Strength of Materials 3

Q.9 In a flitched beam, one section is reinforced Q.13 A simply supported beam AB of length l with
with another section. The purpose of such a a gradually varying load zero at B and w per
beam is to improve unit length at A is as shown. What will be
(a) shearing resistance of the section the maximum value of bending moment in
(b) moment of resistance of the section beam AB?
(c) appearance of the section
w
(d) all of these

Q.10 The state of stress on an element is shown in

B
the given figure. The value of stresses are σx A

= 32 MPa; σy = 10 MPa and major principle

l
stress σ1 = 40 MPa. The minor principle stress
σ2 is
wl 2 wl 2
σy (a) (b)
τxy 9 3 9
wl 2 wl 2
τxy (c) (d)
2 3 9 2

σx σx Q.14 A beam of triangular cross-section having

each side of 100 mm is subjected to a shear
τxy force of 13 kN. What will be the maximum
value of shear stress induced in the section?
τxy
σy (a) 3 MPa (b) 4 MPa
(c) 4.5 MPa (d) 2 MPa
(a) 2 MPa (b) –12 MPa
(c) 8 MPa (d) –18 MPa Q.15 A metallic cantilever beam with span 2 m
carries a uniformly varying load of 50 N/mm
Q. No. 11 to Q. No. 30 carry 2 marks each at the free end to 150 N/mm at the fixed
Q.11 A 30 mm × 30mm square bar of 100 mm end as shown in the figure. What will be
length is subjected to an axial compressive slope at free end?
load of 90 kN. What will be the change in [Take flexual rigidity of beam as 1013 N-mm2]
length if lateral strains are prevented by the 150 N/mm
application of uniform and equal lateral
external pressure of suitable intensity.
[Take E = 100 GPa and ν = 0.25] 50 N/mm
A B
(a) 0.083 mm (b) 0.042 mm 2m
(c) 0.166 mm (d) 0.053 mm
(a) 0.005 rad (b) 0.01 rad
Q.12 A hollow steel shaft is having external and
(c) 0.02 rad (d) 0.04 rad
internal diameter as 80 mm and 60 mm
respectively. Volume of shaft is 106 mm3. Q.16 A rod of steel is 20 m long at a temp of 20°C.
What will be maximum strain energy stored Temperature of rod is then raised to 65°C.
in the shaft if maximum allowable shear What will be thermal stress produced when
stress is 80 MPa? the rod is permitted to expand only by 5.8
[Take shear modulus (G) = 100 GPa] mm?
(a) 25 N-m (b) 50 N-m [Take α = 12 × 10–6/°C and E = 200 GPa]
(c) 20 N-m (d) 40 N-m (a) 108 MPa (b) 58 MPa
(c) 50 MPa (d) 65 MPa

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 4 Civil Engineering

Q.17 The beam section as shown in figure is Q.20 What will be the deflection at C for the beam
subjected to a maximum bending stress of shown?
90 kg/cm2. What will be the force on the
shaded area? P

15 cm A B
C

12 cm l, I a, I

N A Pa 2 l Pa 3 Pa 2 l Pa 3
(a) + (b) +
2EI 3EI 3EI 2EI
12 cm
Pa 2 l Pa 3 Pa 2 l Pa 3
(c) + (d) +
2EI 2EI 3EI 3EI

(a) 2700 kg (b) 10800 kg Q.21 A flitched timber beam made up of steel and
(c) 5400 kg (d) 1350 kg timber has a section as shown in figure. What
will be the moment of resistance of the beam?
Q.18 A cube of 5 mm side is subjected to load [Take σT = 5 MPa and σS = 100 MPa]
system as shown. What will be FOS according
to maximum shear stress theory if yield
strength of the material is 75 MPa? σs = 100 MPa
y
1.25 kN
Timber
1 kN Steel

1 kN
200 mm

2.5 kN

0.625 kN
60 mm 15 mm 60 mm
z

(a) 1.3 (b) 0.77

(a) 10 kNm (b) 7 kNm
(c) 0.80 (d) 1.25
(c) 14 kNm (d) 17 kNm
Q.19 What is the reaction on the pin C for the beam
Q.22 In a strained material, the principal stresses
shown in the figure below.
in the x and y directions are 120 N/mm2
w per unit length (tensile) and 80 N/mm 2 (compression)
A B respectively. On an inclined plane, the normal
C
to which makes an angle of 30° to the x-axis,
L L
what is the tangential stress (in N/mm2)
3 5
(a) wL (b) wL (a) 25 3 (b) 50
8 8
3 5 (c) 50 3 (d) 100
(c) wL (d) wL
16 16

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• Strength of Materials 5

Q.23 A cantilever beam of length ‘L’ carries a Q.26 Figure shows a rectangular beam section 10
concentrated load W at the mid-span. If the cm wide and 20 cm deep. If the maximum
free end is supported on a rigid prop. Which flexural stress is 80 kg/cm2. What will be total
of the following option(s) is incorrect? force on area shaded and moment of this
force about the neutral axis respectively.
5W
(a) The prop reaction is .
16

5 cm
3L

10 cm
(b) The point of contraflexure is at from
11
5 cm
fixed support.
(c) Magnitude of bending moment at the N A

 3WL 

10 cm
prop is  .
16 
(d) Magnitude of bending moment under

 5WL  10 cm
the concentrated load is  .
32 
(a) 1500 kg, 11250 kg cm
Q.24 A cantilever beam 2.5 m long is loaded with (b) 1000 kg, 11666.67 kg cm
a uniformly distributed load of 10 kN/m (c) 1500 kg, 11666.67 kg cm
over a length of 1.5 m from the fixed end. (d) 1000 kg, 11250 kg cm
What will be the deflection at free end of Q.27 Two similar round bars A and B are each 30
the cantilever? cm long and loaded as shown in the given
[Take EI = 1.9 × 1012 Nmm2] figure. The ratio of the energies stored by
(a) 6.3 mm (b) 3.3 mm
(c) 4.3 mm (d) 5.3 mm UB
the bars A and B, is ________. (Ignore
UA
Q.25 A steel section as shown in figure is subjected
to a shear force of 20 kN. What will be shear self weight of bars).
stress at the centre of section?
P P

2 cm 2 cm
20 mm 10 cm

20 cm

60 mm
20 mm 4 cm 20 cm

4 cm 10 cm

20 mm
P P
80 mm
Bar A Bar B

(a) 20.4 MPa (b) 13.6 MPa (a) 0.5 (b) 1

(c) 16.6 MPa (d) 11.4 MPa (c) 1.5 (d) 2

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 6 Civil Engineering

Q.28 A simply supported beam with over-hanging ends carries transverse loads as shown in the figure.
W W
w/meter
A B
C a 10 m a D

If W = 10w, what will be the overhanging length on each side, such that the bending moment at the
middle of the beam is zero?
(a) 2 m (b) 1.25 m
(c) 2.5 m (d) 1 m

Q.29 A simply supported beam of 5 m span is subjected to a clockwise moment of 15 kN-m at a distance
of 2 m from left end as shown in the figure.

15 kNm

2m 3m

A B

Which of the following options show correct SFD of beam AB?

+ 3 kN

(a) 1 kN – 1 kN (b)
3 kN –

+ 1 kN

(c) 3 kN – 3 kN (d)
1 kN –

Q.30 A beam is loaded as shown in figure below:

w0 w0

A C
L B L
2 2
R1 R2

The moment at B (MB) will be

1
(a) 0 (b) w0 L2
24
1 1
(c) w0 L2 (d) w0 L2
12 18

„„„„

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 CLASS TEST S.No. : 01_ JP_CE_08052023

Delhi | Bhopal | Hyderabad | Jaipur | Pune | Bhubaneswar | Kolkata

Web: www.madeeasy.in | E-mail: info@madeeasy.in | Ph: 011-45124612

STRENGTH OF MATERIALS
CIVIL ENGINEERING
Date of Test : 08/05/2023

ANSWER KEY h

1. (a) 7. (b) 13. (a) 19. (c) 25. (b)

2. (b) 8. (c) 14. (c) 20. (d) 26. (c)

3. (c) 9. (b) 15. (b) 21. (c) 27. (c)

4. (a) 10. (d) 16. (c) 22. (c) 28. (b)

5. (a) 11. (a) 17. (c) 23. (c) 29. (c)

6. (a) 12. (a) 18. (b) 24. (a) 30. (b)

 8 Civil Engineering

D E TA I L E D E X P L A N AT I O N S

2. (b)
We know deflection of spring,

64 WR 3 n
δ=
Gd 4
where, W = 100 N, R = 25 mm, n = 12, G = 80 GPa, d = 5 mm

64 × 100 × (25 )3 × 12
So, δ= = 24 mm
80 × 103 × 5 4

3. (c)
d = 2 mm
σb(max) = 80 N/mm2
E = 100 × 103 N/mm2
Distance between the neutral axis of wire and its extreme fibre

2
y= = 1 mm
2
So, minimum radius of the drum
y  f E
R= σ ·E ∵ = 
b (max)  y R
1
= × 100 × 10 3
80
= 1.25 × 103 mm = 1.25 m

4. (a)
Internal hinge in given beam will become hinged support in conjugate beam.

5. (a)
16
τmax = M2 + T 2
πD3
 16 
=  (8)2 + (6)2  × 10 6
 π(100)
3


16 10 × 10 6
= × = 50.93 MPa
π 106

6. (a)
5t 3t
2t/m
A B
C E D
RA RB
2m 2m 2m 2m

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• Strength of Materials 9

Taking moments about B,

6 3 2
RA =   × 5 + × 4 + 3 ×
8 8 8
= 3.75 + 1.5 + 0.75 = 6t
Taking moment about A,
RB = ( 5 + 3 + 2 × 2 ) − RA
= 12t – 6t = 6t
RA 6t
∴ =1
RB = 6t

7. (b)
P

inner outer
P surface surface
P
Due to combined effect of torque and shear force, inner surface will have more shear stress
compared to outer surface.
On dividing a spring into m parts, the no. of turns on each spring will be m times less. Since
stiffness is inversely proportional to no. of turns, the stiffness will become mk.

8. (c)
I 225 × 106
Z = = = 1.5 × 106 mm 3
y 300
2
M = σmax × Z = 120 × 1.5 × 106
= 180 × 106 N-mm
wl 2
Mmax = M =
8
wl 2 w(4 × 103 )2
So = 180 × 106 =
8 8
180
⇒ w = = 90 N/mm = 90 kN/m
2

9. (b)
Flitched beam has a composite section made of two or more materials joined together in such a
manner that they behave as a unit piece and each material bends to the same radius of curvature.
The total moment of resistance of a flitched beam is equal to the sum of the moments of resistance
of individual sections.

10. (d)
By stress invariant law, σx + σy = σ1 + σ2
⇒ 32 + (–10) = 40 + σ2
⇒ σ 2 = –18 MPa

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 10 Civil Engineering

11. (a)
Direct longitudinal stress,
90 × 103
σx = = 100 MPa (Compressive)
30 × 30
1
∈x =
E
(
 −σx + ν σ y + σz 
 ) ...(i)

1
∈y = ∈z =  −σy + ν (σ x + σz ) = 0 ...(ii)
E
As we know σ y = σz
Also on solving equation (ii)
ν 0.25 σ
σy = σx = σx = x
1− ν 1 − 0.25 3
1 1
So, ∈x =
E
( )
−σx + ν σ y + σz  =  −σx + ν × 2σ y 
E
1 σ  1 0.5σx 
= −σ x + 0.25 × 2 × x  =  −σ x +
E  3 E 3 

1  0.5 × 100  1  250 

=  −100 +  =
100 × 103  − 3 
100 × 103 3 
1  250 
δl = l ∈x =  − 3  × 100
100 × 103
= 0.083 mm (Reduction in length)
12. (a)
τ max
2
 D2 + d 2 
Strain energy stored in hollow shaft, U = V
4G  D2 
80 2  80 2 + 60 2  10000 × 106
= 4 × 100 × 10 3  80 2  × 10 =
6
= 2.5 × 10 4 N-mm
  4 × 10 5

= 25 N-m
13. (a)
Bending moment at A and B is zero. It increases in the form of cubic curve.
Maximum value of bending moment in beam AB occurs where shear force changes sign.
wl
Now we know RA =
3
wl
RB =
6
Let at point x from B, shear force will be zero
1 w
V = RB − × x· x
2 l
wl wx 2
⇒ − =0
6 2l
l
⇒ x=
3

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• Strength of Materials 11

l
So, bending moment at x = from B is
3
1 w x wl l 1 w 1 l3
Mmax = RB· x − × · x· x × = × − × × ×
( 3)
3
2 l 3 6 3 2 l 3

wl 2 wl 2 wl 2 wl 2 2 wl 2
= − = − =
6 3 6×3 3 6 3 18 3 18 3
wl 2
=
9 3
14. (c)
We know that area of triangular beam section

3 2
A= a for equilateral triangle
4
3
= (100)2 = 2500 3 mm 2
4
Average shear stress across the section

F 13 × 10 3
τavg = = = 3 MPa
A 2500 3

So maximum shear stress for triangular section

τmax = 1.5 τavg
= 1.5 × 3 = 4.5 MPa

15. (b)
Let us split up the trapezoidal load into a uniformly distributed load (w1) of 50 N/mm and a
triangular load (w2) of 100 N/mm at A to zero at B.
Now slope at free end
 50 × (2 × 10 3 )3 100 × (2 × 10 3 )3 
w1l 3 w2 l 3  +  rad
θB = + =  6 × 1013 24 × 1013
6EI 24EI 
 
= 0.0067 + 0.0033 = 0.01 rad

16. (c)
Free expansion of rod = δl = αl∆t = 12 × 10–6 × 20 × 103 (65 – 20) = 10.8 mm
When the rod is permitted to expand by 5.8 mm in this case, expansion prevented = 10.8 – 5.8
= 5 mm
Expansion Prevented
∴ Strain prevented = Original length

5 1
= 3
=
20×10 4000
∴ Thermal stress = Strain prevented × E
1
= × 200 × 10 3 = 50 MPa
4000

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 12 Civil Engineering

17. (c)

f f max
From bending equation, y = y
max

f max
∴ f= y ×y
max

f max
∴ Force on shaded area = y × ΣAy
max

f max
= y ( Ay )
max

[where A is shaded area, y = distance of centroid of shaded area from N.A.]

90  15  2
= ×  × 12  × × 12 = 5400 kg
12  2  3
18. (b)
2.5 × 103
σx = = 100 MPa (T)
25
1250
σy = = 50 MPa (T)
25
625
σz = = 25 MPa (T)
25
1000
τxy = = 40 MPa
25
σx + σy 1
(σ )
2
σ 1,2 = ± x − σy + 4 τ2xy
2 2
100 + 50 1
= ± (100 − 50 )2 + 4(40)2
2 2
σ 1 = 122.17 MPa
σ 2 = 27.83 MPa
σ 3 = 25 MPa
Now according to maximum shear stress theory
σy
(τabs )max ≤ 2 FOS
( )
 σ1 − σ 2 σ2 − σ 3 σ 3 − σ1  σy
 , ,  =
 2 2 2 max 2 (FOS )

σ1 − σ 3 σy
= 2 FOS
2 ( )
75
FOS = = 0.77
122.17 − 25

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• Strength of Materials 13

19. (c)
We know that at internal hinge, deflection will be same at just left and right of hinge.
So,

w per unit length RC

A C C B
RC
L L

wL4 Rc L3
(δc)left↓ = − ...(i)
8EI 3EI

Rc L3
(δc)right↓ = ...(ii)
3EI
So from eq. (i) and (ii)
wL4 Rc L3 R L3
− = c
8EI 3EI 3EI

2 Rc L3 wL4
⇒ =
3EI 8EI
3
⇒ Rc = wL
16

20. (d)
The deformation of the beam will be as shown below.

a
A θB B C
θB
∆C1
l

∆C2

Now ∆C1 is produced due to deflection of C as caused due to deformation of AB,

∆C1 = θB (BC) = θBa

MBAl Pal
θB = =
3EI 3EI

Pala Pa 2l
∴ ∆C1 = =
3EI 3EI
∆C2 is produced due to deformation of BC
P

A B C

∆C2

Pa 3
∆C2 =
3EI

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 14 Civil Engineering

So total deflection at C,∆C = ∆C1 + ∆C2

Pa 2 l Pa 3
= +
3EI 3EI

21. (c)
Modulus of section for both timber sections

 60 × (200)2 
 = 8 × 10 mm
5 3
ZT = 2 
 6 

Similarly modulus of section for the steel section

15 × (200)2
ZS = = 10 5 mm 3
6
Now, moment of resistance of timber section,
MT = ZT · σT
= 8 × 105 × 5
= 4 × 106 N-mm = 4 kN-m
Similarly,
Moment of resistance of steel section,
MS = ZS · σS
= 105 × 100
= 10 × 106 N-mm
= 10 kNm
Total moment of resistance of the beam
M = MS + MT = 10 + 4 = 14 kNm

22. (c)
120 − ( −80 )
Radius, R = = 100
2

R
τ
0°
2θ = 6

A O B
(–80, 0) C (120, 0)

Tangential stress, τ = R sin 2θ

= 100 sin 60°
3
= 100 ×
2
= 50 3

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• Strength of Materials 15

23. (c)
W

A B
C
L/2 L/2 RB
11W
16

5W
5WL 16
SFD
32

BMD
3WL
16

Since end B is propped

Net deflection at B is zero.

⇒ (δB↓)W = ( δ B ↑ )R
B

3 2
 L  L
W  W 
 2  2 L R L3
+ × = B
3EI 2EI 2 3EI

WL3 WL3 R L3
+ = B
24EI 16EI 3EI
2 WL3 + 3WL3 R L3
= B
48EI 3EI
5W
RB =
16
5W
∴ RA = W − RB = W −
16
11W
=
16
Bending moment diagram:
5Wx
For BC: Mx = RB x =
16
At x = 0; MB = 0 ⇒ Moment at propped end is zero.
L 5WL
At x = ; MC =
2 32
5Wx  L
For CA: Mx = − W x − 
16  2
L 5WL
At x = ; MC =
2 32

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 16 Civil Engineering

3WL
At x = L; MA = −
16
BMxx = 0
5 Wx WL
= Wx −
16 2
8L
x = (From prop end)
11
 8L  3
∴ The point of contraflexure is at  L −  = from fixed end.
11 11

24. (a)

l = 2.5 m
l1 = 1.5 m
w = 10 kN/m
A B C
∆B
θB
θB
θB × (l – l1)

We know deflection at free end for the case given in question,

wl14 wl13
y = + (l − l1 )
8EI 6EI
where l = 2.5 m, l1 = 1.5 m, w = 10 kN/m = 10 N/mm
4 3
10  1.5 × 10 3  10  1.5 × 10 3 
y = + × [2.5 − 1.5] × 10 3
8 × 1.9 × 10 12 6 × 1.9 × 1012
= 6.29 mm
6.3 mm

25. (b)
Since section is symmetric about x-x and y-y, therefore centre of section will lie on the geometrical
centroid of section.
The semi-circular grooves may be assumed together and consider one circle of diameter 60 mm.

80 × (100)3 π
So, Ixx = − (60)4
12 64
= 6.03 × 106 mm4
Now for shear stress at neutral axis, consider the area above the neutral axis,
π 4 × 30
Ay = [80 × 50 × 25] − (30)2 ×
2 3π
= 100000 – 18000 = 82000 mm3
b = 20 mm

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• Strength of Materials 17

VAy 20 × 10 3 × 82000
So, τ = =
Ib 6.03 × 106 × 20
= 13.60 MPa

26. (c)

f max
(i) Force on shaded area = Ay
y max

where, A is area of shaded portion, y is distance of centroid of shaded area from NA

Ay = 5 × 5 × (5 + 2.5) = 187.5 cm3
80
So, Force = × 187.5 = 1500 kg
10
(ii) Moment of this force about the neutral axis

f max
M = y Io
max

(Io = Moment of inertia of shaded area about neutral axis)

5 × 53 4375
Io = + 5 × 5 × (7.5 )2 = cm 4
12 3

80 4375
So, M = × = 11666.67 kg cm
10 3

27. (c)

P2 L
U = (For axially loaded bar)
2 AE
L1 = 10 cm, L2 = 20 cm, d1 = 2 cm and d2 = 4 cm
P 2 L1 P 2 L2
UA = +
2 A1 E 2 A2 E
P 2 L2 P 2 L1
UB = +
2 A1 E 2 A2 E

L2 L1
+
UB A1 A2
∴ U A = L1 L2
+
A1 A2

L1d12 + L2 d22
=
L1d22 + L2 d12

10 × 2 2 + 20 × 4 2 3
= = = 1.5
10 × 4 2 + 20 × 2 2 2

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 18 Civil Engineering

28. (b)

1
RA = RB = ( 2 W + 10 w ) = W + 5w
2
= 15w (W = 10 w)
Now
W W
w/meter
A B
C a 5m E 5m a D
RA RB
ME = RA × 5 – W (a + 5) – w × 5 × 2.5 = 0
⇒ 15w × 5 – 10w (a + 5) – 12.5w = 0
⇒ 75 – 10a – 50 – 12.5 = 0
⇒ 12.5 = 10a
⇒ a = 1.25 m

29. (c)
15 kNm

2m 3m

A B
RA + RB = 0
∑MA = 0
⇒ RB × 5 = 15
⇒ RB = 3 kN
RA = –3 kN
Now the for SFD will be as shon below.
A B

3 – 3

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• Strength of Materials 19

30. (b)
1 L   w0 L
Total load = 2    × w0  =
  
2 2  2
1
By symmetry, R1 = R2 = × Total load
2
w0 L
⇒ R1 = R2 =
4
Bending moment at B,
L 1 L 2 L
(MB) = R1 × − w0 × ×  
2 2 2 3 2
w0 L L w0 L 2  L 
= × − × ×  
4 2 2 2 3  2
w0L2 w0L2 w0L2
= − =
8 12 24
w0L
4

w0L
SFD 4

2
w0L
24

BMD

„„„„

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