A uniform section propped cantilever beam of span L and plastic moment carrying capacity Mp carries a concentrated load W at mid-span. The magnitude of W at the collapse stage will be
- A8Mp /L
- B6M p /LCorrect
- C4Mp /L
- D16M p /L
Solution
Concept
Plastic analysis of indeterminate beams using the mechanism method, where collapse occurs when sufficient plastic hinges form to turn the structure into a mechanism.
Formula
W × (L/2) × θ = Mp × θ + Mp × (2θ) => W = 6Mp / L
Step-by-step solution
1. Identify potential plastic hinge locations for a propped cantilever carrying a mid-span concentrated load W: at the fixed support (1) and directly under the concentrated load (2). 2. Let the rotation at the load point be θ. By geometry, the rotation at the fixed support is 2θ. 3. Write the internal work done at the plastic hinges: Wint = Mp(θ) + Mp(2θ) = 3Mpθ. 4. Write the external work done by the load W: Wext = W × (L/2) × θ. 5. Equate internal and external work: W × (L/2) × θ = 3Mpθ => W = 6Mp / L.
Explanation
For a propped cantilever beam of span L carrying a concentrated load W at mid-span, plastic hinges form at the fixed support and under the load. Using the mechanism method or virtual work, equate the internal work of plastic hinges to the external work done by the load: Mp(θ + 2θ) = W(L/2)θ, which simplifies to W = 6Mp/L.
Why the other options are wrong
A: 8Mp/L is the collapse load for a fixed-fixed beam with a central load. C: 4Mp/L is the collapse load for a simply supported beam with a central load. D: 16Mp/L is incorrect and does not correspond to standard configurations.
Shortcut
Remember the collapse load for a propped cantilever with a central concentrated load is 6Mp/L, whereas for a simply supported beam it is 4Mp/L and for a fixed-fixed beam it is 8Mp/L.
Common mistake
Confusing the propped cantilever formula with a simply supported beam (4Mp/L) or a fully fixed beam (8Mp/L).
Correct answer: B) 6M p /L