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Moment Distribution Method Questions and Answers

1.

Which of the following statements is FALSE about the moment distribution method?

Exam Year(s)

DSSSB JE - 2023,

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.

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2.

The distribution factor for the portion BC from the given beam is:

A beam ABC fixed at both ends A and C and supported at the middle joint B. Span AB has length L and moment of inertia 3I; span BC has length L and moment of inertia I.

Exam Year(s)

DSSSB JE - 2023,

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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