Limit State Method MCQ with answer and solution

Concept

Limiting flexural capacity of singly reinforced beam

Formula

M_u,lim = 0.133 × f_ck × b × d² (for Fe 500)

Step-by-step solution

1. For Fe 500 grade steel, limiting depth factor xu,max / d = 0.46 (or 0.456). 2. Limiting moment of resistance formula is Mu,lim = 0.36 × (xu,max/d) × [1 - 0.42·(xu,max/d)] × fck × b × d² = 0.133 × fck × b × d². 3. Given b = 250 mm, d = 450 mm, and fck = 25 N/mm². 4. Mu,lim = 0.133 × 25 × 250 × (450)² = 0.133 × 25 × 250 × 202,500 = 168,328,125 Nmm = 168.3 kNm approx 170 kNm. Answer: C

Explanation

As per IS 456:2000, for Fe500 steel, the limiting depth of neutral axis is xu,max = 0.46 d, and the limiting moment factor is Mu,lim = 0.133 × fck × b × d². For M25 concrete, width 250 mm, and effective depth 450 mm, Mu,lim = 0.133 × 25 × 250 × (450)² = 168.3 kNm, which rounds to 170 kNm.

Why the other options are wrong

A) 130 kNm underestimates the concrete grade capacity. B) 150 kNm corresponds to a lower effective depth. D) 190 kNm overestimates capacity by using Fe 250 moment factors.

Shortcut

Mu,lim = 0.133 × fck × b × d² = 0.133 × 25 × 250 × 0.2025 = 168.3 kNm ≈ 170 kNm.

Common mistake

Using 0.138 (for Fe 415) or 0.148 (for Fe 250) instead of the factor 0.133 for Fe 500.