A simply supported beam of span 6 m carries an ultimate UDL of 25 kN/m. The plastic moment capacity required for the beam is:
Try it first: pick an option, then read the explanation.
Correct answer: D. 112.5 kN·m
Asked in SSC JE 30/10/2020 (Evening)
Detailed Explanation
A simply supported beam is statically determinate. It becomes a mechanism as soon as one plastic hinge forms, at the section of maximum moment — mid-span for a UDL. So the plastic moment needed equals the maximum moment under the ultimate load.
\[ M_p = \frac{w_u L^2}{8} = \frac{25 \times 6^2}{8} = 112.5\ \text{kN·m} \]
Why “112.5 kN·m” Is Correct
The required plastic moment capacity is 112.5 kN·m — option (d).
Why the Other Options Are Wrong
- A. 102.5 kN·m: 102.5 kN·m is below wL²/8.
- B. 146.85 kN·m: 146.85 kN·m is not obtained from this loading.
- C. 125.45 kN·m: 125.45 kN·m does not follow from wL²/8.
Key Point
SS beam, UDL at collapse: M_p = w_u L²/8 (one hinge at mid-span).
Key Concepts for Students
- Collapse load of SS beam with UDL: w_u = 8M_p/L².
- A fixed beam with UDL needs three hinges: w_u = 16M_p/L².
- Plastic moment M_p = f_y × Z_p (plastic section modulus).
Practice more: this is question 22 in Bending Moment and Shear Force MCQs – Part 2. Try the whole set and check every answer instantly.