Discussion
For the continuous beam ABC shown consisting of span AB = 4 m and BC = 3 m (Iab: Ibc = 1:2). The total stiffness at joint B is:
Answer: A
The total stiffness at joint B is 3EI.
With A fixed and C a simple end support, member BA has its far end fixed and member BC has its far end pinned. Take \(I_{ab} = I\) and \(I_{bc} = 2I\):
\[ \begin{aligned} K_{BA} &= \frac{4EI}{4} = EI \\ K_{BC} &= \frac{3E(2I)}{3} = 2EI \\ K_B &= EI + 2EI = 3EI \end{aligned} \]
Using 4EI/L for BC as well (treating C as fixed) would give \(EI + 8EI/3 \approx 3.67EI\), not an option. 4EI, 5EI and 6EI come from ignoring the span lengths or the 1:2 ratio of moments of inertia.
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