9111-SRM MCET Mechanical Engineering

Unit- I
STRESS, STRAIN AND DEFORMATION OF SOLIDS
Part – A (Q & A)
1. What is meant by Poisson’s ratio? (Nov/Dec’23, Apr/May’24)
Poisson’s ratio is the ratio of lateral strain to the longitudinal strain in a material under
axial loading.

2. Give the relation between three moduli. (Nov/Dec’23)

The relation between three moduli:
E=2G(1+ ν )=3 K (1−2 ν)where E is Young's modulus, G is shear modulus, K is bulk
modulus, and ν is Poisson's ratio.

3. State Hooke’s Law (Apr/May’24, Nov/Dec’21 & Nov/Dec’18)

Hooke’s law states that, “Within the elastic limit, stress is directly proportional to strain”.
σ =E ⋅ε ;Where, σ = stress; ε = strain; E=¿ Young’s modulus.

4. What is meant by temperature stress? (Nov/Dec’20)

Temperature stress is the stress induced in a material due to restriction of thermal
expansion or contraction when the temperature changes.

5. Write a relation for change in length of a hanging freely under its own weight
(Apr/May’19)
2
ρgL
ΔL=
2E
where ρ is density, g is acceleration due to gravity, L is the original length, and E is
Young’s modulus.

6. What does the radius of Mohr’s circle refer to? (Apr/May’19, Nov/Dec’20)
The radius of Mohr’s circle represents the maximum shear stress in the material.

7. Difference Elasticity and Elastic Limit. (Nov/Dec’20)

Elasticity: The property of a material to return to its original shape after removal of load.
Elastic Limit: The maximum stress a material can withstand without permanent
deformation.

8. What is principle of super position? (Nov/Dec’20)

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The total deformation in a structure is equal to the algebraic sum of deformations due to
individual loads acting independently.

9. What is meant by principal stress? (Apr/May’18, Apr/May’17)

Principal stress is the maximum or minimum normal stress acting on a plane where shear
stress is zero.

10. Give the relation between modulus of elasticity and bulk modulus. (Apr/May’18)
E=3 K (1−2 ν )

where 𝐸 is Young’s modulus; 𝐾 is bulk modulus, and 𝜈 is Poisson's ratio.

11. Write an expression of volumetric strain for a rectangular bar subjected to an axial
load P. (Nov/Dec’18)
Volumetric strain for a rectangular bar subjected to an axial load P:
ε v =εx+ εy+ εz=EP(1−2 ν)

12. What is point of contra flexure (Apr/May’19)

It is a point in a beam where the bending moment changes sign, resulting in zero bending
moment.

13. List the assumptions made in theory of simple bending. (Apr/May’19)

 The material is homogeneous and isotropic.
 The cross-section remains plane and perpendicular to the neutral axis after bending.
 The beam is initially straight with uniform cross-section.

14. Define ‘longitudinal strain’ and ‘lateral strain’. (Nov/Dec’20)

Longitudinal strain: The change in length per unit original length in the direction of load.
Lateral strain: The change in dimension perpendicular to the direction of load per unit
original dimension.

15. Define strain energy and write its unit. (Nov/Dec’20)

Strain energy is the energy stored in a body due to deformation under applied load. Unit:
Joule (J).

16. Define the term modulus of resilience (Nov/Dec’21)

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It is the maximum strain energy stored per unit volume within the elastic limit.

17. Define shear strain (Apr/May’18)

Shear strain is the angular distortion caused by shear stress, measured as the change in
angle between two originally perpendicular lines.

18. Write the relationship between the elastic constants. (Apr/May’18)

Relationship between the elastic constants are:
E=2G(1+ ν )E=3 K (1−2 ν )E

19. Mention the relationship between the three elastic constants. (Apr/May’23)
The relationship between the three elastic constants:
1/ K=1 /E+ 2 ν / E ,G=E /2(1+ν )

20. Outline the application of Mohr’s circle of stress. (Apr/May’23)

Mohr’s circle is used to determine principal stresses, principal planes, and maximum
shear stresses graphically.

Part – B (Q&A)
1. Figure 1.1 shows a steel bar is 900 mm long; its two ends are 40 mm and 30 mm in
diameter and the length of each rod is 200 mm. The middle portion of the bar is 15
mm in diameter and 500 mm long. If the bar is subjected to an axial tensile of 15
kN, find its total extension, Take E= 200 GN/m2. (Nov/Dec’23)

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Figure 1.1: Steel bar

Given data

Total length of the bar (L) = 900 mm

Diameter at one end (D1) = 40 mm

Diameter at the other end (D2) = 30 mm

Diameter of the middle portion (D3) = 15 mm

Length of each end portion (L1 and L2) = 200 mm

Length of the middle portion (L3) = 500 mm

Axial tensile load (P) = 15000 N

Young’s Modulus for Steel (E) = 200 x 109 N m2

Step 1: calculate the cross sectional Areas

1. Cross sectional area at the ends:

For diameter D1 = 40 mm.

π 2 π 2
A1 = D1 = ( 40) = 1256.6
4 4

For diameter D2 = 30 mm.

π 2 π 2
A2 = D 2 = (30) = 706.86
4 4

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For diameter D3 = 30 mm.

π 2 π 2
A3 = D3 = (15) = 176.71
4 4

Step 2: Calculate Extensions of Each Section

ΔL = PL /AE

1. Extension in the first section (L1):

Length L1 = 200 mm

Area A1 = 1256.64 mm2

Young’s Modulus E = 200 x 103 N/mm2

Calculate extension:

ΔL1 = (15000 x 200) / (1256.64 x 200000) = 0.0119

2. Extension in the second section (L2):

Length L2 = 200 mm

Area A2 = 706.86 mm2

Young’s Modulus E = 200 x 103 N/mm2

Calculate extension:

ΔL2 = (15000 x 200) / (706.86 x 200000) = 0.0212

3. Extension in the second section (L2):

Length L3 = 500 mm

Area A3 = 176.71 mm2

Young’s Modulus E = 200 x 103 N/mm2

Calculate extension:

ΔL2 = (15000 x 500) / (176.71 x 200000) = 0.212

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Step 3: Total Extension

ΔLtotal = ΔL1 + ΔL2 + ΔL3 = 0.2451 mm

Conclusion:

Total Extension: ΔLtotal = 0.2451 mm

2. Figure 1.2 shows a steel rod of 3 cm diameter is enclosed centrally in a hollow copper
tube of eternal diameter 5 cm and internal diameter of 4 cm. The composite bar is then
subjected to an axial pull of 45000 N. If the length of each bar is equal to 15 cm,
Determine:
(i) The stresses in the rod and the tube and
(ii) Load carried by each bar,
Take E for steel = 2.1 x 105 N/mm2 and for copper 1.1 x 105 N/mm2. (Nov/Dec’23)

Figure 1.2 Steel rod

Given Data

Steel rod Diameter (D1) = 30 mm

Copper Tube External Diameter (D2) = 50 mm

Copper Tube Internal Diameter (d2) = 40 mm

Load (P) = 45000 N

Lenth of each bar (L) = 150 mm

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Young’s Modulus for steel (E1) = 2.1 x 105 N mm2

Young’s Modulus for Copper (E2) = 1.1 x 105 N mm2

Step 1: Calculate Cross Sectional Areas

1. Cross Sectional area of the steel rod (A1):

π 2 π 2
A1 = ( D1 ) =¿ (30) = 706.86 mm
4 4
2. Cross sectional area of the copper tube (A2):

π 2 π 2
Aouter = ( D ) =¿ (50) = 1963.5
4 2 4

π 2 π 2
Ainner = ( d2 ) =¿ ( 40) = 1256.64
4 4

A2 = Aouter - Ainner = 706.853

Step 2: Calculate Stresses in Each Bar

P1 P2
σ1 = ; σ2 = ;
A1 A2

σ1 σ2
=
E1 E2

σ1 E2 = E1 σ2

σ1 = k σ2

Where K = E1 / E2

Step 3: Relate Stresses and Loads

P1 = A1 σ1,

P2 = A2 σ2,

P = A1 σ1 + A2 σ2

Using the relationship: σ1 E2 = E1 σ2

This leads to σ1 = k σ2

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Step 4: Solve for Leads

P = A1 (kσ2) + A2 σ2

P = (A1k + A2) σ2

Thus,

σ2 = P / A1k + A2

k = Es /Ec

K = 210000 / 110000 = 1.909

Substitute the equation

Ac1 k = 1349.12 mm2

A1 k + A2 = 2055.98 mm2

Now substituting into the stress equation:

σ2 = P / A1k + A2

= 21.85 N /mm2

σ1 = k σ2

= 41.76 N /mm2

Step 5: Calculate Load Carried by Each Bar

Now we can find P1 and P2.

P1 = A1 σ1 = 29456 N

P2 = A2 σ2 = 15484 N

Conclusion:

(i) Stresses:

Stress in Steel rod (σ1) = 41.76 N /mm2

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Stress in Copper Tube (σ2) = 21.85 N /mm2

(ii) Load Carried by Each Bar:

Load Carried by Steel Rod (P1) = 29456 N

Carried by Copper Tube (P2) = 15484 N

3. A copper rod of 40 mm diameter is surrounded tightly by a cast iron tube of 80 mm

external diameter, the ends being firmly fastened together. When put to a
compressive load of 30 kN, what load will be shared by each? Also determine the
amount by which the compound bar shortens if it is 2 m long. Take: E cl = 15 N/m2
and Ec = 75 GN/m2. (Apr/May’23)
Given,
Diameter of Copper Rod: dc=40 mm
Diameter of Cast Iron Tube: dt=80 mm (external diameter)
Length of the Bars: L = 2 m = 2000 mm
Total Compressive Load: P = 30 kN = 30000 N
Young's Modulus for Copper: Ec =75 GN m2 = 75×109 N m2
Young's Modulus for Cast Iron: Eci =15 N m2

Calculating Cross-sectional Areas:

π 2 π 2 2
Area of Copper Rod, Ac ¿ d c = (40) =1256.64 mm
4 4
Area of Cast Iron Tube: The area is calculated as the difference between the outer and
inner areas.
Inner diameter of the tube (equal to the diameter of the copper rod): dc = 40 mm
External diameter: dt = 80 mm

π 2 2 π π
Aci =
4
( d t −d c ) = ( 80 −40 ) = ( 6400−1600 )=3769.91 mm
4
2 2
4
2

Calculating Load Distribution

Using the formula for load distribution in composite bars:

A c Ec
Pc =P ×
A c Ec + A ci Eci

Pci =P−Pc

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Calculating Effective Loads
Load on copper rod:
9
1256.64 ×75 ×10
Pc =3000 × 9 12
=29994 N
1256.64 × 75× 10 +3769.91 ×10

Load on cast iron tube:

Pci =P−Pc =30000−29994=6 N

Calculating Shortening of Each Bar

The shortening (δL) for each bar can be calculated using:

PL
δL=
AE

Shortening for copper rod:

Pc L
δL=
Ac Ec

substituting the values,

29994 x 2000
=( 4.76 x 10 ) m
−5
δ Lc =
1256.64 x ( 75 x 10 ) 9

Shortening for cast iron tube:

P ci L
δ Lci =
A ci Eci

Substituting the values,

6 x 2000
=( 3.18 x 10 ) m
−5
δ Lci =
3769.91 x 15

4. Two circular bar A and B of the same material are subjected to the same pull force
(P) and are deformed by the same amount. What is the ratio of their length. If one
of them has a constant diameter of 60 mm and the other uniformly tapers from 80
mm at the one end to 40 mm at the other? (Apr/May’24)
Given Data

1. Bar A:

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Diameter DA = 60 mm
Cross – sectional Area AA
2. Bar B:
Tapering from D1 = 80 mm to
D2 = 40 mm
Average Diameter Davg
Cross – Sectional Area AB
Step 1: Calculate Cross-sectional Areas
For Bar A:
The cross-sectional area AA is calculated using the formula for the area of a circle:

π 2
AA= D
4 A

π 2
Substituting the diameter: A A = (60)
4

π
= (3600)
4

= 2827.43 mm2

For Bar B:
First, calculate the average diameter:

Davg = (D1+ D2)/2 => (80+ 40)/2

= 60 mm

Now, calculate the average cross-sectional area : A B

π 2
AB = D avg
4

π
= ( 60 )2
4

= 2827.43 mm2

Step 2: Relate Lengths Using Elongation Formula

Since both vars are subjected to the same pull force P and experience the same elongation δL, we
can set up the equation:

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L A LB
=
A A AB

Calculate length Ratio

Rearranging gives us:

AA = 2827.43 mm2

AB = 2827.43 mm2

Thus,
LA 2827.43
AA
= 2827.43
=1
Conclusion:
The ratio of the lengths of bars A and B is 1 :1

5. i) A hollow steel tube of external diameter 100 mm, is used to carry a compressive
load of 456 N. Find the internal diameter if the permissible stress is 150 Nmm2
To find the internal diameter of a hollow steel tube iven the eternal diameter,
compressive load, and permissible stress, we can use the following steps:
Given,
External Diameter (D): 100 mm
Compressive Load (P): 456 N
Permissible Stress (σ): 150 N/mm2
Step 1: Calculate the cross-sectional Area
P
The permissible stress is defined as: σ =
A
Where A is the cross-sectional area.
Rearranging gives:
P
A=
σ
Substituting the values:
A = (456 N )/¿ mm2
= 3.04 mm2
Step 2: Determine the cross-sectional area of a hollow cylinder
The cross-sectional area A of a hollow cylinder can e expressed as:
π 2
A = ( D ¿ ¿ 2−d )¿
4

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Where D is the eternal diameter and d is the internal diameter.
Step 3: Substitute and solve for internal Diameter
Substituting for A:
π 2
3.04 = (100 ¿ ¿2−d )¿
4
Step 4: Rearranging the Equation
Multiply both sides y 4:
12.6 = π (10000 - d 2)
Now divide both sides by π:
12.16
10000 - d 2 =
π
12.16
Calculating :
π
10000 - d 2= 3.87
Rearranging gives:
2
d =10000−3.87=9996.13
Step 5: Calculate Internal Diameter
Taking the square root to find d:
d = √ 9996.13 ≈ 99.98 mm.
Conclusion
The internal tube of the hollow steel tube is approximately: d ≈ 99.98
mm.

ii) A 25 mm diameter rod is subjected to an aial load P, and the strain in the rod is
0.00075. Find the tensile force, if E = 2 x 105 N/mm2 . (Nov/Dec’20)
Step 1 : Calculate the cross sectional area
σ = Stress (N/mm2)
Ꜫ = Strain
E = Young’s modulus (N/mm2)
σ = P/A
Diameter of the rod is 25 mm
π
A= (25)2
4
= 490.87 mm2

Step 2: Calculate stress

σ = E. Ꜫ
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= (2 x 105) . (0.00075)

σ = 150 N mm2

Step 3: Calculate Tensile force

σ = P/A

P = σ.A => 73630.5 N.

Conclusion:

Tensile Force = 73630.5 N.

6. A steel cylinder is enclosed in a copper tube as shown in Figure 1.3. The cylinder
and the tube are compressed between rigid parallel plates. Find the stresses in the
steel and copper and also the compressive strain. P = 450 kN, d = 100 mm and D =
200 mm. For steel E = 210 kN/mm2 and for copper E = 110 N/mm2. (Nov/Dec’20)

Figure 1.3: Steel cylinder is enclosed in a copper tube

Given Data

Axial Load (P): 450 kN = 450000 N

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Diameter of steel cylinder (d): 100 mm

Diameter of Copper Tube (D) = 200 mm

Young’s Modulus for Steel (E1): 210 kN/mm2 = 210,000 N/mm2

Young’s Modulus for Copper (E2): 110 kN/mm2 = 110000 N/mm2

Step 1: Calculate Cross- Sectional Areas

a. Cross- Sectional Area of the steel cylinder:

π
Asteel = (d)2 = 7853.98 mm2
4

b. Cross- Sectional Area of the copper tube:

Outer Diameter (D) = 200 mm

Inner diameter d = 100 mm

π π
Asteel = (D)2 - (d)2
4 4

= 2352.52 mm2

Step 2: Calculate Stresses

a. Stress in the steel cylinder (σ 1):

σ 1 = P / Asteel

= 45000 / 7853.98

= 57.3 N mm2

b. Stress in the copper tube (σ 2):

σ 2 = P / Acopper
= 45000 / 23562.52
= 19.1 N mm2
Step 3: Calculate Compressive Strain
a. Strain in the steel (Ꜫ1)

Ꜫ1 = σ 1 /E 1

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= 5.3/210000

= 0.000273

b. Strain in the copper (Ꜫ2)

Ꜫ2 = σ 2 /E 2

= 19.1/110000

= 0.000173

Conclusion:

Stress in the steel = 57.3 N mm2

Stress in the copper = 30.3 N/mm2

Compressive Strain = 0.000273

7. Three bars made of copper, zinc and aluminium are of equal length and have cross
section 50, 5 and 100 square mm respectively. They are rigidly connected at either
end. If this compound member is subjected to a longitudinal pull of 250 kN.
Estimate the proportion of load carried on each rod, and the induced stresses.
Assume Ecu = 10 kN/mm2. Ezinc = 100 N/mm2 and Eal = 80 N/mm2. (Apr/May’19)

Given Data

Total Load (P) = 250 kN = 250000N

Cross Sectional area of copper (A1) = 50 mm2

Cross sectional area of zinc (A2) = 5 mm2

Cross sectional area of aluminium (A3) = 100 mm2

Young’s Modulus for Copper (E1) = 10 kN/mm2

Young’s Modulus for Zinc (E2) = 100 N/mm2

Young’s Modulus for Aluminium (E3) = 80 N/mm2

Step 1: Calculate the Equivalent Stiffness

1/ Eeq = (A1/E1) + (A2/E2) + (A3/E3)

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= 1.305

Thus,

Eeq = 1/ 1.305 = 0.766 N mm2

Step 2: Calculate Load Distribution

Pi = P .[( Ai + Ei )/ Σ( A j+ E j )]

P1 = P .[( A1 + E1 )/ [( A1 + E1 )+ ( A2 + E2 )+ ( A3 + E 3)] ]

= 250000 . [(50 x 10000)/(50 x 10000 + 5 x 100 + 100 x 80)]

Calculating Denominator

Copper: 50 x 10000 = 500000

Zinc: 5 x 100 = 500

Aluminium: 100 x 80 = 8000

Total:

500000 + 500 + 8000 = 508500

P1 = 250000 . (500000/ 508500)

= 250000 x 0.983 = 2457

Load on Zinc (P2)

P2 = P .[( A2 + E2 )/ [( A1 + E1 )+ ( A2 + E2 )+ ( A3 + E 3)] ]

Load on Aluminium (P3)

P3 = P - P1 - P2 = 250000 - P1 - P2

Step 3: Calculate Induced stresses

σ i = Pi + A i

Stress in Copper (σ 1 ¿:

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P1 = 245750 N

σ 1 = 245750/50

= 4915 N/mm2

Stress in Zinc (σ 2 ¿:

P2 = 245 N

2 = 245/5 = 49 N/mm2

Stress in Aluminium (σ 3 ¿ :

P3 = 4 N

σ 3 = 4/100

= 40 N/mm2

Conclusion

Pcopper = 245750 N

σ copper= 4915 N/mm2

PZinc = 245 N

σ zinc = 49 N/mm2

Paluminium = 4 N

σ aluminium= 40 N/mm2

8. A copper flat 500 mm long and 40 mm (width) x 60 mm (thickness) uniform section

is acted upon by the following forces: 50 kN tensile in the direction of the length 300
kN compressive in the direction of the width and 250 kN tensile in the direction of
the thickness. Determine change in dimension and hence change in volume of the
flat. Assume the modules of the elasticity and Posson’s ratio of copper as 120
KN/mm2 and 0.25 respectively. (Apr/May’19)

Given Data
Length (L) = 500 mm

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Width (W) = 40 mm
Thickness (T) = 60 mm
Tensile Force in Length Direction (F1) = 50000 N
Compressive Force in Length Direction (F2) = 3,00,000 N
Tensile Force in Thickness Direction (F3) = 250,000 N
Young’s Modulus for Copper : 120,000 N/mm2
Poisson’s Ratio for Copper (υ) : 0.25
Step 1: Calculate Stresses
1. Stress in Length Direction (σ 1 ¿:
σ 1=F1/ A
= 50000/ (W.T)
= 50000/ (40 x 60) = 20.83

2. Stress in Width Direction (σ 2 ¿:

σ 2=F2/ (L.T)
= 300000/(500 x 60) = 10

3. Stress in Thickness Direction (σ 3 ¿:

σ 3=F3 / (W.L)
= 250000/20000
= 12.5 N
Step 2: Calculate Strains
Ꜫi = σ i/E
1. Strain in Length Direction ( Ꜫ1)

Ꜫ1= σ 1/E
= 20.83 / 120000
= 0.000136
2. Strain in Width Direction ( Ꜫ2)
Ꜫ2= 10/120000
= - 0.00008333
3. Strain in Thickness Direction ( Ꜫ3)
Ꜫ3= 12.5/120000
= - 0.00010417
Step 3: Calculate Change in Dimensions

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1. Change in Length (ΔL):
ΔL = L. Ꜫi
= 500 mm x 0.0001736
= 0.0868 mm
2. Change in Width (ΔW):
ΔW = W (−υ Ꜫ 1)
= 40 mm (-0.25) (0.0001736)
3. Change in Thickness (ΔT):
ΔT = T (−υ Ꜫ 1) + T Ꜫ3
= 60
Step 4: Calculate Change in Volume
Vo = LWT = 500 x 40 x 60 = 1200 mm
Ꜫ v = Ꜫ1 + Ꜫ2 + Ꜫ3
ΔV = Vo x Ꜫ v = 1,200,000 x 0.0001944
= 233.28
Conclusion
Change in Length (ΔL): 0.0868 mm
Change in Width (ΔW): -0.001736 mm
Change in Thickness (ΔT): 0.003646 mm
Change in Volume (ΔV): 233.28 mm3

9. i) A bar 30 mm in diameter is subjected to a tensile load of 54 N and the measured

extension on 300 mm auge length is 0.112 mm and change in diameter is 0.00366
mm. Calculate Poission’s ratio and the values of three moduli. (Nov/Dec’20)
Given Data
Diameter of the bar (d)= 30 mm
Tensile load (F) = 54 N

Extension (ΔL)= 0.112 mm

Original Gauge lenth (Lo) = 300 mm

Change in diameter (Δd)= 0.00366 mm

Ꜫl = ΔL / Lo = 0.112 mm / 300 mm = 0.0003733

Step 1: Calculate Longitudinal Strain

Ꜫt = Δd / d = 0.00366 mm/ 30 mm = 0.000122

Step 2: Calculate Transverse Strain

Step 3: Calculate Poisson’s Ratio

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υ = −(Ꜫt / Ꜫl ) ) = - 0.327
υ = 0.327 (Conventionally Positive)
Step 4: Calculate the Moduli
1. Young’s Modulus E= σ /Ꜫ l
π
A= (30)2 = 706.86 mm2
4

σ =F / A = 54/06.86 = 0.0763 N/mm2

E = 0.0763 / 0.000333 = 204.5 N/mm2

2. Shear Modulus (G):
G = E / 2 (1 + υ) = 204.5 / 2 (1 + 0.327) = 77 .04 N/mm2
3. Bulk Modulus (K):
K = E/ [3(1-2 υ)] = 204.5 / 3(1- 0.654) = 197.6 N/mm2
Conclusion
Poisson’s Ratio υ = 0.327
Young’s Modulus E = 204.5 N/mm2
Shear Modulus G = 77 .04 N/mm2
Bulk Modulus K = 197.6 N/mm2
ii) A tapered circular bar tapers uniformly frm a diameter d at its small end to D at
its bi end. The length of the bar is L. Derive an expression for the elongation of the
bar due to an axial tensile force P. (Nov/Dec’20)

Step 1 : Define the Geomentry

dx = d + [(D-d)/L] x
Step 2 : cross sectional area
π π
A (x) = (d2x) = (d + [(D-d)/L] x)2
4 4
Step 3: Stress and Strain
σ x = P/A(x)
= 4P / π (d + [(D-d)/L] x)2
Ꜫ (x) = σ (x) / E
= [4P / π E (d + [(D-d)/L] x)2] dx
Step 4: Elongation of an Element
dL = Ꜫ (x) dx
= [4P / π E (d + [(D-d)/L] x)2] dx
= [4P / π E (d + [(D-d)/L] x)2] dx
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Step 5: Total Elongation
L

Ltotal = ∫ dL
0
L

= ∫¿¿
0

K = D-d
L
4P
Ltotal =
πE ∫¿¿
0

Step 6: Solve the Integral

Let u = d + k/Lx
du = / Ldx
dx = (L/k)du
When x= 0, u = d, x = L, u =D.
Now, Changing the Integral Limits
D
4 PL 4 PL
Ltotal =
kE ∫u 2
du = -
kE
¿u -1¿ Dd
d

Final expression for elongation

4 PL
Ltotal = - (log (D/d))
E( D−d)

10. A compound tube consists of a steel 10 mm outer diameter and 10 mm thickness and
an outer brass tube 190 mm outer diameter and 10 mm thickness. The two tubes are
the same length. The compound tube carries an axial load 1000 kN. Find the
stresses, the load carried by each tube and forces acting on it. Also find the
deformation of the tube in terms f its length. Length of each tube is 200 mm.
Young’s modulus for steel and brass are 200 N/mm2 and 100 N/mm2 respectively.
(Nov/Dec’20)

Given Data

Steel Tube:

Outer diameter Ds = 10 mm

Thickness ts = 10 mm

Inner diameter ds = Ds – 2ts = 10 – 2 (10) = - 10 mm

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Brass Tube:

Outer diameter Db = 190 mm

Thickness tb = 10 mm

Inner diameter db = Db – 2tb = 190 – 2 (10) = 170 mm

Axial Load (P) – 1000 kN = 1000,000 N

Young’s Modulus:

For steel Es = 200 N mm2 = 200,000 N mm2

For Brass Eb = 100 N mm2 = 100,000 N mm2

Length of each tube (L): L = 200 mm

Step 1: Calculate Cross Sectional Areas

Cross sectional area of the steel tube

π π π
As = Aouter = (D2s) = (102) = (100) = 78
4 4 4

Cross sectional area of the brass tube

π π
Ab = Aouter – Ainner = (D2b) - (d2b)
4 4

π π
= (190)2 - (170)2 = 565.49 mm2
4 4

Step 2: Calculate Load Distribution

Ks = AsEs

Kb = AbEb

Stiffness of steel Tube:

Ks = AsEs = 8.54 x 200 = 15708 N/mm

Stiffness of Brass Tube:

Kb = AbEb = 565.49 x 100 = 56549 N/mm

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Step 3: Total Stiffness and Load Distribution

Ktotal = Ks + kb = 72257 N/mm

Load on Steel Tube

Ps = P (Ks / Ktotal) = 1000000

Load on Steel Tube

Pb = P (Kb / Ktotal) = 1000000

Step 4: Calculate Stresses in Each Tube

Stress in steel tube

σ s= Ps / As = 2763.5 N/mm2

Stress in brass tube

σ b= Pb/ Ab = 1385.5 N/mm2

Step 5: Calculate Deformation of Each Tube

ΔL = PL /AE

Deformation in Steel Tube

ΔLs = Ps L/As Es

= 21000 (200) / 78.54 (200000) = 43400000 / 15708000 = 0.266 mm

Deformation in Steel Tube

ΔLb = Pb L /Ab Eb

= 783000 (200) / 565.49 (100000) = 0.277 mm

Results

Ps = 217000 N

Pb = 783000 N

σ s= 2763.5 N/ mm2

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σ b= 1385.5 N/ mm2

ΔLs = 0.276 mm

ΔLb = 0.277 mm

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