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# Strength of Materials MCQs Compressive Stress

For keeping the stress wholly compressive the load may be applied on a circular column anywhere within a concentric circle of diameter _____________

a) D/2

b) D/3

c) D/4

d) D/8

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Answer: c

Explanation: The load application on a circular column affects stress. If it is under D/4 the stress will be wholly compressive.

2. Consider two bars A and B of same material tightly secured between two unyielding walls. Coefficient of thermal expansion of bar A is more than that of B. What are the stresses induced on increasing the temperature?

a) Tension in both the materials

b) Tension in material A and compression in material B

c) Compression in material A and tension in material B

d) Compression in both the materials

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Answer:d

Explanation: Since both the supports are fixed and both bars will try to expand, so rise in temperature will cause compressive stresses in the bars.

What will be the unit of compressive stress?

a) N

b) N/mm

c) N/mm2

d) Nmm

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Answer: c

Explanation: As the stress is the ratio of force to the area, so it will be N/mm2. Here mm is normally used in its calculation most of the time.

A cast iron T section beam is subjected to pure bending. For maximum compressive stress to be 3 times the maximum tensile stress, centre of gravity of the section from flange side is ____________

a) h/2

b) H/3

c) H/4

d) 2/3h

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Answer: c

Explanation: H/4 when the applied moment is sagging. Otherwise, I.e. if the applied moment is hogging it is H/4. as in the options both are not given means we have to take hogging.

A solid circular shaft of diameter d is subjected to a torque T. the maximum normal stress induced in the shaft is ____________

a) Zero

b) 16T/πd3

c) 32T/πd3

d) None of the mentioned

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Answer: b

Explanation: The maximum torque transmitted by a circular solid shaft is obtained from the maximum shear stress induced at the outer surface of the solid shaft and given by T = πD3/16 x normal stress,

So, normal stress = 16T/πd3.

When a rectangular beam is loaded transversely, the maximum compressive stress develops on ____________

a) Bottom fibre

b) Top fibre

c) Neutral axis

d) Every cross-section

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Answer: b

Explanation: Loaded means loaded downwards. In that case, upper fibres will be compressed while lower will be expanded. Hence maximum compressive stress will be developed in top layer.

An axial residual compressive stress due to a manufacturing process is present on the outer surface of a rotating shaft subjected to bending. Under a given bending load, the fatigue of the shaft in the presence of the residual compressive stress is ____________

a) Decreased

b) Increases or decreased, depending on the external bending load

c) Neither decreased nor increased

d) Increases

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Answer: d

Explanation: From the Gerber’s parabola that is the characteristic curve of the fatigue life of the shaft in the presence of the residual compressive stress. The fatigue life of the material is effectively increased by the introduction of compressive mean stress, whether applied or residual.

A steel bar of 40mm x 40mm square cross-section is subjected to an axial compressive load of 200kN. If the length of the bar is 2m and E=200GPa, the elongation of the bar well be ____________

a) 1.25mm

b) 2.70mm

c) 4.05mm

d) 5.40mm

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Answer: a

Explanation: Elongation of the bar = Pl/AE = -200x103x 2000 / ( 1600 x 200 x 103) = -1.25

The minus sign here shows that the stress here is compressive.

### Strength of Materials MCQs Thermal Stress

The length, Young’s modulus and coefficient of thermal expansion of bar P are twice that of bar Q. what will be the ration of stress developed in bar P to that in bar Q if the temperature of both bars is increased by the same amount?

a) 2

b) 8

c) 4

d) 16

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Answer: c

Explanation: Temperature Stress = EαδT

Stress in bar P / Stress in bar Q = (EP / EQ) x (αP / αQ) = 2×2 = 4.

A steel bar 600mm long and having 30mm diameter, is turned down to 25mm diameter for one fourth of its length. It is heated at 30 C above room temperature, clamped at both ends and then allowed to cool to room temperature. If the distance between the clamps is unchanged, the maximum stress in the bar ( α = 12.5 x 10-6 per C and E = 200 GN/m2) is

a) 25 MN/m2

b) 40 MN/m2

c) 50 MN/m2

d) 75 MN/m2

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Answer: d

Explanation: As temperature stress do not depend upon properties of cross section like length and area. They only depends upon properties of the material.

Therefore, σ=αEδT

= 12.5 x 10-6 x 200 x 103 x 30

= 75 MN/m2.

A cube having each side of length p, is constrained in all directions and is heated unigormly so that the temperature is raised to T.C. What will be the stress developed in the cube?

a) δET / γ

b) δTE / (1 – 2γ)

c) δTE / 2 γ

d) δTE / (1 + 2γ)

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Answer: b

Explanation: δV/V = P / K = a3 (1 + aT)3 – a3) / a3

Or P / (E / 3(1-2γ)) = 3αT.

A steel rod 10mm in diameter and 1m long is heated from 20 to 100 degree celcius, E = 200 GPa and coefficient of thermal expansion is 12 x10-6 per degree celcius. Calculate the thermal stress developed?

a) 192 MPa(tensile)

b) 212 MPa(tensile)

c) 192MPa(compressive)

d) 212 MPa(compressive)

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Answer: c

Explanation: αEδT = (12 x 10-6) ( 200 x 103) (100-20) = 192MPa.

A cube with a side length of 1m is heated uniformly a degree celcius above the room temperature and all the sides are free to expand. What will be the increase in the volume of the cube? Consider the coefficient of thermal expansion as unity.

a) Zero

b) 1 m3

c) 2 m3

d) 3 m3

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Answer: d

Explanation: Coefficient of thermal expansion = 3 x coefficient of volume expansion.

The thermal stress is a function of _____________

P. Coefficient of linear expansion

Q. Modulus of elasticity

R. Temperature rise

a) P and Q

b) Q and R

c) Only P

d) Only R

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Answer: d

Explanation: Stress in the rod is only due to temperature rise.

A steel rod is heated from 25 to 250 degree celcius. Its coefficient of thermal expansion is 10-5 and E = 100 GN/m2. if the rod is free to expand, the thermal stress developed in it is:

a) 100 kN/m2

b) 240 kN/m2

c) Zero

d) Infinity

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Answer: c

Explanation: Thermal stress will only develop if the body is restricted.

Which one of the following pairs is NOT correctly matched?

a) Temperature strain with permitted expansion – ( αTl – δ)

b) Temperature thrust – ( αTE)

c) Temperature stress – (αTEA)

d) Temperature stress with permitted expansion – E(αTl – δ) / l

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Answer: a

Explanation: Dimension analysis gives Temperature strain with permitted expansion – ( αTl – δ)is wrong. In other options the dimensions are correctly matched.

A steel rod of length L and diameter D, fixed at both ends, is uniformly heated to a temperature rise of δT. The Youngs modulus is E and the coefficient of linear expansion is unity. The thermal stress in the rod is ____________

a) Zero

b) T

c) EδT

d) EδTL

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Answer: c

Explanation: As α = δl / l δT

So, δl = l x 1 x δT

And temperature strain = δl / l = δT

As E = stress / strain

Stress = E δT.

A uniform, slender cylindrical rod is made of a homogeneous and isotropic material. The rod rests on a frictionless surface. The rod is heated uniformly. If the radial and longitudinal thermal stress are represented by σx and σz, then ___________

a) σx = 0, σy = 0

b) σx not equal to 0, σy = 0

c) σx = 0, σy not equal to 0

d) σx not equal to 0, σy not equal to 0

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Answer: a

Explanation: We know that due to temperature changes, dimensions of the material change. If these changes in the dimensions are prevented partially or fully, stresses are generated in the material and if the changes in the dimensions are not prevented, there will be no stress set up. (Zero stresses).

Hence cylindrical rod Is allowed to expand or contract freely.

So, σx = 0 and σy = 0.

which one of the following are true for the thermal expansion coefficient?

a) αaluminium > αbrass> αcopper > αsteel

b) αbrass > αaluminium > αcopper > αsteel

c) αcopper > αsteel > αaluminium > αbrass

d) αsteel > αaluminium > αbrass > αcopper

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Answer: a

Explanation: Aluminium has the largest value of thermal expansion coefficient, then brass and then copper. Steel among them has lowest value of thermal expansion coefficient.

The length, coefficient of thermal expansion and Youngs modulus of bar A are twice of bar B. If the temperature of both bars is increased by the same amount while preventing any expansion, then the ratio of stress developed in bar A to that in bar B will be ___________

a) 2

b) 4

c) 8

d) 16

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Answer: b

Explanation: Temperature Stress = EαδT

So σ1 / σ2 = E1α1δT1/E2α2δT2

From question, α and E of bar A are double that of bar B.