A simply supported beam of effective span 1·5 m is carrying a factored concentrated load of 360 kN at mid span. What is the section modulus of the beam ? (Take fy = 250 N/mm² and γm0 = 1·1)
- A594 × 10³ mm³Correct
- B651 × 10³ mm³
- C768 × 10³ mm³
- D256 × 10³ mm³
Solution
Concept
Moment capacity of a plastic/compact section
Formula
Md = Zp fy / γm0 (βb = 1 for plastic sections)
Step-by-step solution
1. M = W L / 4 = 360 × 1.5 / 4 = 135 kN·m = 135 × 10⁶ N·mm. 2. Zrequired = M γm0 / fy = 135 × 10⁶ × 1.1 / 250 = 594 × 10³ mm³.
Explanation
For a plastic or compact section the design moment capacity is Zp fy/γm0. Setting it equal to the factored moment gives the required section modulus.
Why the other options are wrong
651×10³, 768×10³ and 256×10³ mm³ do not follow from M γm0/fy with M = WL/4.
Shortcut
Z = (WL/4) × 1.1 / 250.
Common mistake
Using W L/8 (UDL moment).
Correct answer: A) 594 × 10³ mm³