Design of Beams MCQ with answer and solution

Concept

As per IS 456, when the nominal shear stress τ_v is less than the design shear strength of concrete τ_c, only minimum shear reinforcement is provided.

Formula

τ_v = V_u/(b d); A_sv/(b s_v) ≥ 0.4/(0.87 f_y)

Step-by-step solution

1. Find the nominal shear stress τ_v = V_u/(b d). 2. Compare τ_v with τ_c (from the table) and τ_c,max. 3. If τ_v < τ_c, provide minimum shear reinforcement. 4. If τ_c < τ_v < τ_c,max, design the shear reinforcement for V_us = V_u − τ_c b d. 5. If τ_v > τ_c,max, revise the section. 6. So minimum shear reinforcement applies when τ_v < τ_c.

Explanation

Concrete alone can carry the shear when τ_v is less than τ_c. Still, IS 456 asks for minimum stirrups to stop sudden brittle failure and to hold the main bars. This minimum steel is given by A_sv/(b s_v) ≥ 0.4/(0.87 f_y).

Why the other options are wrong

A: In this range, designed shear reinforcement is needed, not just minimum steel. C: If τ_v exceeds τ_c,max, the section must be redesigned. D: If τ_v exceeds τ_c, designed shear reinforcement is needed.

Shortcut

τ_v < τ_c gives minimum stirrups. τ_c < τ_v < τ_c,max gives designed stirrups. τ_v > τ_c,max means revise the section.

Common mistake

Mixing up option A (design shear reinforcement) with the condition for minimum shear reinforcement.