This set of Strength of Materials Multiple Choice Questions & Answers (MCQs) focuses on “Failure due to Shear”.
1. Calculate the nominal shear stress, if a singly reinforced rectangular beam 230×450 mm effective depth is subjected to a factored load of 60 kN.
a) 0.6 N/mm2
b) 0.55 N/mm2
c) 0.4 N/mm2
d) 0.25 N/mm2
Explanation: B = 230 mm; d = 450 mm
Shear force = Vu = 60 kN
Nominal shear stress = Vu/bd = 60 × 103/ 230 ×450
= 0.6 N/mm2.
2. The minimum shear reinforcement is given by Asv/bSv = _______
a) 0.4 /0.87 fy
b) 0.5 /0.85 fy
c) 0.6 /0.9 fy
d) 0.35/ 0.6 fsc
Explanation: The minimum quantity of shear reinforcement that should be provided for all beams except those of minor importance like lintels is given by IS 456:2000, clause 220.127.116.11 by the equation
Asv/bSv = 0.4/0.87 fy.
3. Bent up bars do not resist diagonal tension.
Explanation: Some of the longitudinal bars can be bent up near supports as a bending moment to be resisted near the supports is very little. Such bent up bars resists diagonal tension effectively.
4. The ultimate shear force at a section of an RCC beam is 300 kN. The shear resisted by concrete is 77.5 kN. What is the shear for which shear reinforcement is required?
a) 213.5 kN
b) 220 kN
c) 222.5 kN
d) 122.5 kN
Explanation: Shear to be resisted by shear reinforcement is Vus = Vu – Vuc
Vus = Vu – Vuc; Where Vu = ultimate shear force and Vuc = shear resistance of concrete
= 300 – 77.5
= 222.5 kN.
5. Bond stress is a stress acting ___________ to the bar on the interface between reinforcement and concrete.
Explanation: Bond stress is the stress acting parallel to the bar on the interface between the reinforcing bar and the surrounding concrete hands it is stress developed between the contact surface of Steel and concrete.
6. ________ is developed due to adhesion between concrete and steel.
Explanation: Bond is developed due to the combined influential effect of adhesion between concrete and steel provided by concrete during setting.
7. Bond is developed due to _________
Explanation: The bond is developed due to the combined effect of friction which is provided by gripping of the bar due to shrinkage of concrete. It resists any force that tries to pull out the rods for the concrete.
8. ___________ depends on grade of concrete and diameter of bar etc.
a) Shear stress
b) Bond stress
Explanation: Bond stress depends on grade of concrete, diameter of the bar, bar profile condition nature of force in the bar, bends and hooks in a bar and grouping of bars.
9. Which of the following bond is also known as a local bond?
a) Anchorage bond
b) Fletched bond
c) Flexural bond
d) Composite bond
Explanation: For transferring the change in bar force along its length due to the variation in bending moment free Irfan comes into account it is also known as a local bond.
10. _____________ bond arises when bar carrying certain force is terminated.
Explanation: An Anchorage Bond arises when a bar carrying certain force is terminated. In such cases, it is obligatory to transfer this force in the bar to the surrounding concrete over a certain length.
11. The development length can be determined easily by _______ test.
a) Push out test
b) Pull out test
c) Grading test
d) Slump cone test
Explanation: The length of the bar required to transfer the force in the bar to the surrounding concrete through bond is known as development length. The development length can be easily determined by pull out test.
12. To improve the anchorage of bars ______ are provided in plain bars.
a) Standard hooks
c) Lateral ties
d) Standard bends
Explanation: In situations where straight anchorage length cannot be provided due to lack of space, to improve the anchorage bars many times standard hooks are provided in plane bars.
13. In case of HYSD bars ___________ are provided to increase anchorage length.
a) Lateral ties
b) Helical reinforcement
c) Standard hooks
d) Standard bends
Explanation: Where straight anchorage length cannot be provided due to lack of space (at supports) In this situation, to improve the anchorage of bars many times the standard bends are provided in deformed bars.
14. Polar moment of inertia is denoted by ___________
Explanation: The polar moment of inertia is the inertia of an area about an axis perpendicular to its plane. It is denoted by “J”.
J = 2 I.
15. Calculate the moment of inertia of a hollow circular section whose external diameter is 60 mm and thickness is 5 mm about centroidal axis.
a) 315 m2
b) 320 m4
c) 330 m4
d) 345 m4
Explanation: External diameter (D) = 60 mm
Internal diameter (d) = 50 mm
Ixx = Iyy = π/64 × (604 -504).
= 330 m4.
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.