Discussion
The distribution factor for the portion BC from the given beam is:
Answer: A
Both far ends A and C are fixed, so the stiffness of each member at joint B is \(4EI/L\). For BA it is \(4E(3I)/L = 12EI/L\), and for BC it is \(4EI/L\). The distribution factor for BC is its stiffness divided by the total at joint B:
\[ \begin{aligned} DF_{BC} = \frac{4EI/L}{12EI/L + 4EI/L} \\ &= \frac{4}{16} = 0.25 \end{aligned} \]The factor for BA is 0.75, which is the third option, and the two factors add to 1. The value 0.5 would apply to equal stiffnesses, and 1.0 to a joint with a single member.
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