Shear Reinforcement (IS 456) – Minimum area of vertical stirrup legs Given beam width b, stirrup spacing x, and characteristic strength of stirrup steel fc (yield strength), the minimum total cross-sectional area of stirrup legs Asv should satisfy:

Introduction / Context:
Minimum shear reinforcement in reinforced concrete beams prevents sudden diagonal tension failure, controls crack widths, and provides ductility even when computed shear demand is low. IS 456 prescribes a minimum stirrup area based on beam width, stirrup spacing, and steel strength. This question asks for the correct expression using the given symbols.

Given Data / Assumptions:

• b = beam width (mm).
• x = stirrup spacing along the beam (mm).
• fc = characteristic/yield strength of stirrup steel (N/mm2).
• Vertical two-legged (or multi-legged) stirrups crossing the shear plane.

Concept / Approach:

IS 456 gives a minimum shear reinforcement requirement in the form Asv/s ≥ 0.4 b / 0.87 fy for vertical stirrups, where fy is the yield strength of stirrup steel. Rearranging for total stirrup area within spacing x (i.e., s = x) gives Asv ≥ (0.4 b x) / (0.87 fy). Mapping fy to the symbol fc yields the requested expression.

Step-by-Step Solution:

Verification / Alternative check:

Try b = 230 mm, x = 200 mm, fc = 415 N/mm2 → Asv ≥ (0.4230200)/(0.87*415) ≈ 51 mm2 per spacing; a 2-legged 8 mm bar (area ≈ 100 mm2) exceeds this, satisfying the minimum.

Why Other Options Are Wrong:

(a) misses the 0.87 factor; (c) and (d) are dimensionally and conceptually incorrect for minimum shear steel; only (b) matches the codal rearrangement.

Common Pitfalls:

Using bar area of a single leg instead of total legs, ignoring hooks/bends for detailing, or confusing design shear with minimum requirement.

Final Answer:

Asv ≥ (0.4 * b * x) / (0.87 * fc)