A simply supported R.C.C. beam, span 6 m, effective depth d = 50 cm, carries a uniformly distributed load of 2400 kg/m (including self-weight). With lever arm factor j = 0.85 and permissible steel stress σst = 1400 kg/cm², compute the required steel area Ast.

Introduction / Context:Sizing tension steel for a simply supported beam under UDL uses the bending moment at mid-span and the internal lever arm between compression and tension resultants. Working-stress design relates moment capacity to steel stress and lever arm.

Given Data / Assumptions:

• Span L = 6 m.
• Effective depth d = 50 cm.
• UDL w = 2400 kg/m (kgf units).
• Lever arm factor j = 0.85 → z = jd.
• Permissible steel stress σst = 1400 kg/cm².

Concept / Approach:For simply supported beam under UDL, maximum bending moment M = wL²/8. Moment resistance in working-stress: M = Tz where T = Astσst and z = jd.

Step-by-Step Solution:Compute M: M = 2400 * 6² / 8 = 2400 * 36 / 8 = 2400 * 4.5 = 10800 kg·m.Convert to kg·cm: 10800 * 100 = 1,080,000 kg·cm.Lever arm z = jd = 0.85 * 50 = 42.5 cm.Ast = M / (σst * z) = 1,080,000 / (1400 * 42.5) = 1,080,000 / 59,500 ≈ 18.15 cm².Choose the nearest provided value: 18 cm².

Verification / Alternative check:Back-compute capacity with 18 cm²: T = 18*1400 = 25,200 kg; Mres = 25,200 * 42.5 = 1,071,000 kg·cm, close to demand (difference due to rounding). Slight conservatism can be handled with bar selection.

Why Other Options Are Wrong:

• 14–17 cm²: Under-reinforced for the given stress limit and lever arm; moment capacity would be insufficient.

Common Pitfalls:Mixing N and kgf; forgetting to convert m to cm; using overall depth instead of effective depth; misusing j.

Final Answer:18 cm²