This set of Strength of Materials Multiple Choice Questions & Answers (MCQs) focuses on “Core Cross Section”.
1. Beams which are reinforced in both compression and tension sides are called as _______
a) Dual reinforced beam
b) Doubly reinforced beam
c) Composite beam
d) Additional beam
Explanation: The beams which are reinforced in both compression as well as tension sides are known as doubly reinforced beams. These beams are generally provided when the dimensions of the beam are restricted.
2. Doubly reinforced beams are provided when, Mu _____ M.
Explanation: The reinforced beams are generally provided when it is required to resist moment higher than the limiting moment of resistance of a singly reinforced section. The additional moment of resistance required can be obtained by providing compression reinforcement and additional tension reinforcement.
3. The doubly reinforced beams improve the ______ of the beam in earthquake regions.
Explanation: Generally when the depth of beam is restricted due to architectural or any construction problems, the doubly reinforced beams are used. It reduces long term deflection and it also improves the ductility of beam.
4. What is the stress in compression, if d’/d value is 0.1 for Fe415 steel?
a) 355 N/mm2
b) 353 N/mm2
c) 342 N/mm2
d) 329 N/mm2
Explanation: The stress in compression, if d’/d value is 0.1 for Fe415 steel is 353 N/mm2.
|Grade of steel||d’/d value||fsc ( in N/mm2)|
5. The cracks seen on walls are due to _____ failure.
Explanation: The diagonal tension stress caused by shear and combination of shear and bending is likely to cause failure of the section by producing cracks in the walls.
6. Bending is accompanied by _______
Explanation: Usually, the bending is accompanied by shear. The combination of shear and bending stresses produces the principal stress which causes diagonal tension in the beam section.
7. The variation of shear stress is ____________
Explanation: In homogeneous beams, the variation of shear stress is parabolic.
• It is zero at top and bottom.
• It is maximum at neutral axis.
8. What is the maximum shear stress for M20 grade concrete?
a) 2.5 N/mm2
b) 2.8 N/mm2
c) 3 N/mm2
d) 3.5 N/mm2
Explanation: The maximum shear stress for M20 grade concrete is 2.8 N/mm2.
|Max. Shear stress (N/mm2)||2.5||2.8||3.1||3.5||3.7|
9. ________ has to be provided against diagonal tensile stresses.
a) Longitudinal reinforcement
b) Shear reinforcement
c) Torsional reinforcement
d) Transverse reinforcement
Explanation: Shear reinforcement has to be provided against diagonal tension stress caused by shear force. The inclined shear crack starts at the bottom and extends towards compression zone.
10. Vertical stirrups are a form of _______ reinforcement.
Explanation: Generally the vertical stirrups are provided as two legged or four legged stirrups around the tension reinforcement. Hanger bars are provided to keep vertical steps in position otherwise they may get displaced while concreting.
11. The shear to be resisted by shear reinforcement is given by ___________
a) Vus = Vuc + Vu
b) Vus = Vu + Vuc
c) Vu = Vus – Vuc
d) Vus = Vu – Vuc
Explanation: The shear to be resisted by shear reinforcement is given by
Vus = Vu – Vuc
Where: Vuc = shear resistance of concrete
Vu = ultimate shear force
▪ The number of stirrups cut by 45° crack line is n = d/Sv.
12. The shear resistance of bent up bars shall not exceed _____ the total shear to be resisted.
a) 30 %
Explanation: If the bent up bars or inclined stirrups are provided at spacing, the shear resistance of bent up Bar should not exceed 50% of the total shear to be resisted by the shear reinforcement. Because bent up bars alone (without stirrups) are not effective in preventing shear failure.
13. What is the horsepower of the engine if the power is 219324 W.
Explanation: The rate of doing work is known as power.
Horse power = P/746
= 294 W.
14. A lift carry 10 persons each weighing 60 kg to the top storey of the building 100 m height. Calculate the potential energy acquired by the person.
a) 5.88 × 105 J
b) 4.32 × 105 J
c) 2.34 × 105J
d) 1.16 × 105J
Explanation: Height of building = 100 m
Mass of each person = 60 kg
Mass of 10 persons = 600
Potential energy = mgh
= 600 × 9.8 × 105
= 5.88 × 10 /5 J.
15. Calculate the maximum shear stress of a circular beam of 100 mm diameter, if the average shear stress is 0.63 N/mm2.
a) 0.85 N/mm2
b) 1.2 N/mm2
c) 1.5 N/mm2
d) 2.1 N/mm2
Explanation: For the circular section,
The maximum shear stress = 4/3 × average shear stress
= 4/3 × 0.6366
= 0.85 N/mm2.
Sanfoundry Global Education & Learning Series – Strength of Materials.
To practice all areas of Strength of Materials, here is complete set of 1000+ Multiple Choice Questions and Answers.