This set of Strength of Materials Multiple Choice Questions & Answers (MCQs) focuses on “Compressive Stress”.
1. 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
View Answer
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
View Answer
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.
3. What will be the unit of compressive stress?
a) N
b) N/mm
c) N/mm2
d) Nmm
View Answer
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.
4. 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
View Answer
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.
5. 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
View Answer
Explanation: In a solid circular shaft subjected to torque (T), the maximum stress experienced is shear stress, not normal stress. Normal stress acts perpendicular to the surface of an object, while shear stress acts parallel to the surface, causing a tangential or shearing force. So, the normal stress is essentially zero at all points within the shaft.
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 shear stress,
So, shear stress = 16T/πd3.
6. 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
View Answer
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.
7. 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
View Answer
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.
8. 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
View Answer
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.
Sanfoundry Global Education & Learning Series – Strength of Materials.
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