A frame to be analyzed by moment distribution as shown in the figure below :

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The distribution factors for members EB, ED and EF will be respectively

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BPSC AE Paper 5 (Civil) 11 Nov 2022 Official Paper
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  1. \(\frac{8}{13}, \frac{3}{13}, \frac{4}{13}\)
  2. \(\frac{3}{10}, \frac{4}{10}, \frac{3}{10}\)
  3. \(\frac{4}{11}, \frac{4}{11}, \frac{3}{11}\)
  4. \(\frac{2}{9}, \frac{4}{9}, \frac{1}{3}\)

Answer (Detailed Solution Below)

Option 3 : \(\frac{4}{11}, \frac{4}{11}, \frac{3}{11}\)
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Detailed Solution

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Concept:

Stiffness of a member is a measure of its resistance to deformation under load. The stiffness of a member in terms of flexural rigidity (EI ) and the length ( ) is calculated using standard formulas for different boundary conditions and configurations.

The distribution factor (DF) is a coefficient used in the moment distribution method to distribute moments to various members connected at a joint based on their relative stiffness. It is calculated using the stiffness of each member.

Stiffness for Different Members:

\( \text{Stiffness of member EB} = \frac{4EI}{\ell} \)

\( = \frac{4E(2I)}{6} = \frac{4}{3}EI \)

\( \text{Stiffness of member ED} = \frac{4EI}{3} \)

\( \text{Stiffness of member EF} = \frac{4EI}{4} = EI \)

Distribution Factors for Different Members:

\( (DF)_{EB} = \frac{\frac{4}{3}EI}{\frac{4}{3}EI + \frac{4}{3}EI + EI} \) \( = \frac{\frac{4}{3}EI}{\frac{4}{3}EI + \frac{4}{3}EI + EI} = \frac{4}{11} \)

\( (DF)_{ED} = \frac{\frac{4}{3}EI}{\frac{4}{3}EI + \frac{4}{3}EI + EI} \) \( = \frac{4}{11} \)

\( (DF)_{EF} = \frac{EI}{\frac{4}{3}EI + \frac{4}{3}EI + EI} \) \( = \frac{3}{11} \)

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