This set of Strength of Materials Multiple Choice Questions & Answers (MCQs) focuses on “Thermal Stress”.
1. 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 increases by the same amount ?
Explanation: Temperature Stress = EαδT
Stress in bar P / Stress in bar Q = (EP / EQ ) x (αP / αQ ) = 2×2 = 4.
2. 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
Explanation: As temperature stress do not depend upon properties of cross section like length and area. They only depends upon properties of the material.
= 12.5 x 10-6 x 200 x 103 x 30
= 75 MN/m2.
3. 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)
Explanation: δV/V = P / K = a3 (1 + aT)3 – a3 ) / a3
Or P / ( E / 3(1-2γ ) ) = 3αT.
4. 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)
d) 212 MPa(compressive)
Explanation : αEδT = (12 x 10-6) ( 200 x 103) (100-20) = 192MPa.
5. 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 volume of the cube? Consider the coefficient of thermal expansion as unity.
b) 1 m3
c) 2 m3
d) 3 m3
Explanation: Coefficient of thermal expansion = 3 x coefficient of volume expansion.
6. The thermal stress is a function of
P. Coefficient of linear expansion
Q. Modulus of elasticity
R. Temperature rise
The correct answer is:
a) P and Q
b) Q and R
c) Only P
d) Only R
Explanation: Stress in the rod is only due to temperature rise.
7. 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
Explanation: Thermal stress will only develop if the body is restricted.
8. 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
Explanation: Dimension analysis gives option a is wrong. In other options the dimensions are correctly matched.
9. 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 cofficeint of linear expansion is unity. The thermal stress in the rod is
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.
10. 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
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.
11. 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
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.
12. 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
Explanation: Temperature Stress = EαδT
So σ1 / σ2 = E1α1δT1/E2α2δT2
From question, α and E of bar A are double that of bar B.
Sanfoundry Global Education & Learning Series – Strength of Materials.
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