The given figure shows a frame to be analysed by moment distribution (EB = 6 m with 2I, ED = 3 m with I, EF = 4 m with I). The distribution factors for members EB, ED and EF will be _________ respectively
- A2/9, 4/9, 1/3
- B8/13, 3/13, 4/13
- C3/10, 4/10, 3/10
- D4/11, 4/11, 3/11Correct
Solution
Concept
Distribution factor = relative stiffness / total stiffness
Formula
k = 4EI/L (far end fixed); DF = k / Σk
Step-by-step solution
1. kEB = 4(2I)/6 = 1.333 I; kED = 4I/3 = 1.333 I; kEF = 4I/4 = 1.0 I. 2. Σk = 3.667 I. 3. DFs = 1.333 : 1.333 : 1 = 4/11 : 4/11 : 3/11.
Explanation
Distribution factors at joint E are the stiffness of each member divided by the total stiffness meeting at E.
Why the other options are wrong
Other options do not sum correctly to the ratio 4 : 4 : 3.
Shortcut
Use k = I/L and take the ratio.
Common mistake
Using 3EI/L for far-end fixed members.
Correct answer: D) 4/11, 4/11, 3/11