Collapse load in a fixed beam of span L, carrying uniformly distributed load over the entire span and having plastic moment capacity M p is
- A6Mp/L
- B8Mp/L
- C8Mp/L²
- D16Mp/L²Correct
Solution
Concept
Collapse mechanism of a fixed beam with UDL
Formula
wu = 16 Mp / L²
Step-by-step solution
1. Hinges form at both ends and at mid-span; if the mid-span rotates by θ each end hinge rotates by 2θ... (two end hinges θ each and the central hinge 2θ). 2. Internal work = Mp(θ + 2θ + θ) = 4Mpθ. 3. External work = wu × (area of triangle) = wu × (L/2 × Lθ/2) = wu L² θ/4. 4. wu L²/4 = 4Mp ⇒ wu = 16 Mp / L².
Explanation
For a fixed beam the collapse load is twice that of a simply supported beam with the same Mp (8Mp/L²).
Why the other options are wrong
8Mp/L² is for a simply supported beam; the options with L have the wrong dimension.
Shortcut
SS: 8Mp/L²; fixed: 16Mp/L².
Common mistake
Using the simply supported value.
Correct answer: D) 16Mp/L²