If M FAB and M FBA are the fixed moments in a fixed beam of span L, then the rotation at end A is
- A(L/3EI)(MFAB + 2MFBA)
- B(L/3EI)(2MFAB + MFBA)
- C(L/6EI)(MFAB + 2MFBA)
- D(L/6EI)(2MFAB + MFBA)Correct
Solution
Concept
Slope at the end from end moments
Formula
θA = (L/6EI)(2MA + MB)
Step-by-step solution
1. For a span with end moments MA and MB (about its own simply supported conjugate), θA = (L/6EI)(2MA + MB). 2. With fixed-end moments as MA = MFAB and MB = MFBA this gives θA = (L/6EI)(2MFAB + MFBA).
Explanation
The slope at one end of a span depends twice as much on the moment at that end as on the far-end moment.
Why the other options are wrong
Options with L/3EI overstate the slope, and C swaps the factors 2 and 1.
Shortcut
θA = (L/6EI)(2MA + MB).
Common mistake
Using 3EI.
Correct answer: D) (L/6EI)(2M_FAB + M_FBA)