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GATE 2025 CE (CE1) – Question 37

Structural Analysis · Slope-deflection and moment distribution methods · 2 marks · Multiple choice

In the rigid-jointed frame shown in the figure, the distribution factor of the member AD is closest to

A frame with joint A. Member AB is vertical above A, of length L and rigidity EI, with a hinge support at B. Member AC runs left from A, of length L and rigidity 2EI, with a free end at C. Member AF is vertical below A, of length L and rigidity 2EI, fixed at F. Member AD runs right from A to a fixed end at D: the first L/2 next to A has rigidity 2EI and the last L/2 has rigidity EI.
  1. 0.254
  2. 0.267
  3. 0.398
  4. 0.421

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Correct answer: (C) 0.398

Explanation

The distribution factor of a member is its stiffness divided by the total stiffness at the joint. AB is hinged at B, so $k_{AB} = \frac{3EI}{L}$. AF is fixed at F, so $k_{AF} = \frac{4(2EI)}{L} = \frac{8EI}{L}$. AC is free at C, so $k_{AC} = 0$. AD is fixed at D and has a stepped section. A moment $M$ at A with A prevented from moving gives $\theta = \frac{11}{80}\frac{ML}{EI}$ (found by integrating $\frac{m}{EI}$ over the two parts of the member), so $k_{AD} = \frac{80EI}{11L} = \frac{7.27EI}{L}$. The total is $7.27 + 8 + 3 + 0 = 18.27$, in units of $\frac{EI}{L}$, so the distribution factor of AD is $\frac{7.27}{18.27} = 0.398$.