Find the bending moment at a distance L/4 from end A of the simply supported beam of length L carrying a uniformly distributed load w over entire span.
- AwL² / 32
- B3wL² / 32Correct
- CZero
- DwL² / 8
Solution
Concept
BM at a section equals the moment of all forces on one side of the section.
Formula
Mx = (wL/2)x - wx2/2
Step-by-step solution
1. Find the reactions: RA = RB = wL/2. 2. Write the BM at distance x from A: Mx = RA·x - w·x·(x/2) = (wL/2)x - wx2/2. 3. Put x = L/4: M = (wL/2)(L/4) - w(L/4)2/2. 4. Simplify: M = wL²/8 - wL²/32. 5. Result: M = 4wL²/32 - wL²/32 = 3wL²/32.
Explanation
Each support carries half the total load, wL/2. Take moments of the reaction and of the UDL on the left of the section at L/4. The net moment is 3wL²/32.
Why the other options are wrong
A: wL²/32 is only the moment of the UDL part; the reaction moment was not added. C: BM is zero only at the supports, not at L/4. D: wL²/8 is the maximum BM, which occurs at mid-span (L/2).
Shortcut
For a full-span UDL, Mx = (w·x/2)(L - x). Put x = L/4: (wL/8)(3L/4) = 3wL²/32.
Common mistake
Writing the maximum BM wL²/8 for any section, or forgetting to subtract the UDL moment.
Correct answer: B) 3wL² / 32