Moment Distribution Method Questions and Answers

Answer: D

It is false that the moment distribution method is not suited for plane rigid frames. The method is applicable to continuous beams and to plane frames, including rigid-jointed ones, which is what Hardy Cross devised it for. It is an approximate (iterative) method, since the fixed-end moments are distributed and carried over repeatedly, and the accuracy depends on how many successive cycles are carried out. The first three statements are therefore true.

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:

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.

Answer: A

The distribution factors at a joint always add up to One.

The distribution factor of a member is its stiffness divided by the total stiffness of all members meeting at the joint. The unbalanced moment at the joint is shared among the members in these proportions, and since the whole moment must be distributed, the factors sum to 1.

A sum of zero, less than one or greater than one would leave part of the moment undistributed or distribute more than exists. (At a fixed support the factor is 0, but that is a support, not a joint where members share moment.)

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.