Design of Beams MCQ with answer and solution

Concept

Required plastic section modulus under limit state

Formula

Z_p = (M_u × γ_m0) / f_y

Step-by-step solution

1. Factored maximum bending moment Mu = w × L² / 8 = 20 × (6)² / 8 = 90 kNm = 90 × 10⁶ Nmm. 2. Under IS 800:2007 limit state design, design bending strength Md = Zp × fy / γ_m0, where γ_m0 = 1.10. 3. Setting Md = Mu gives Zp = (90 × 10⁶ × 1.10) / 250 MPa = 396,000 mm³. 4. Converting to cm³: 396,000 / 1000 = 396 cm³, rounding to standard size 400 cm³. Answer: C

Explanation

For a simply supported beam with L = 6 m and factored load w = 20 kN/m, maximum moment is Mu = w·L² / 8 = 90 kNm. Under IS 800:2007, design moment capacity is Md = Zp × fy / γ_m0. Equating Md = Mu with fy = 250 MPa and γ_m0 = 1.10 gives Zp = (90 × 10⁶ × 1.10) / 250 = 396 cm³, which specifies 400 cm³.

Why the other options are wrong

A) 300 cm³ provides insufficient moment resistance. B) 360 cm³ neglects the partial material safety factor γ_m0 = 1.10 prescribed by IS 800:2007. D) 450 cm³ exceeds the required plastic section modulus.

Shortcut

Z_p = (Mu × 1.1) / fy = (90 × 10⁶ × 1.1) / 250 = 396 cm³ ≈ 400 cm³.

Common mistake

Omitting the partial safety factor γ_m0 = 1.10, obtaining 360 cm³.