What is the section modulus required to design an angle Purlin which experiences a maximum bending moment of 10 kNm, if the section is made with Fe250 grade steel?
- A4.55 cm³
- B45.57 cm³Correct
- C75.57 cm³
- D7.55 cm³
Solution
Concept
The required section modulus is the bending moment divided by the design bending strength of the steel.
Formula
Zreq = M × γm0 / fy
Step-by-step solution
1. Given: M = 10 kNm = 10 × 106 Nmm and fy = 250 MPa. 2. Take γm0 = 1.1 (limit state method). 3. Zreq = 10 × 106 × 1.1 / 250 = 44000 mm³. 4. Convert units: 44000 mm³ = 44 cm³. 5. The nearest option is 45.57 cm³, so the answer is B.
Explanation
The section must resist the bending moment without yielding. Dividing the moment by the design stress fy/γm0 gives about 44 cm³. Option B (45.57 cm³) is the closest, so B is the expected answer.
Why the other options are wrong
A: 4.55 cm³ is about 10 times too small, a unit error. C: 75.57 cm³ is too large and does not match any standard method. D: 7.55 cm³ is far too small.
Shortcut
Z ≈ M/(fy/1.1): 10 × 106/227 ≈ 44000 mm³ ≈ 45 cm³.
Common mistake
Making a unit error when converting mm³ to cm³ (divide by 1000).
Correct answer: B) 45.57 cm³