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GATE 2022 ME (ME1) – Question 37

Mechanics of Materials · Energy methods · 2 marks · Multiple choice

An L-shaped elastic member ABC with slender arms AB and BC of uniform cross-section is clamped at end A and connected to a pin at end C. The pin remains in continuous contact with and is constrained to move in a smooth horizontal slot. The section modulus of the member is same in both the arms. The end C is subjected to a horizontal force $P$ and all the deflections are in the plane of the figure. Given the length AB is $4a$ and length BC is $a$, the magnitude and direction of the normal force on the pin from the slot, respectively, are

a horizontal arm AB of length $4a$ clamped at A on the left, and a vertical arm BC of length $a$ going down from B. The pin at C is in a horizontal slot, and the force $P$ acts on it to the right.
  1. $3P/8$, and downwards
  2. $5P/8$, and upwards
  3. $P/4$, and downwards
  4. $3P/4$, and upwards

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Show answer and explanation

Correct answer: (A) $3P/8$, and downwards

Explanation

The slot is smooth, so it can only give a vertical reaction $R$ at C. The vertical deflection of C must be zero. The vertical bar BC carries $P$ horizontally and only deflects horizontally, so the vertical movement of C is the vertical deflection of the end B of the cantilever AB. At B the cantilever AB gets the moment $Pa$ from the force $P$ acting a distance $a$ below it (it bends B upwards, $\frac{Pa(4a)^2}{2EI} = \frac{8Pa^3}{EI}$) and the vertical force $R$ ($\frac{R(4a)^3}{3EI}$). Setting the total to zero: $\frac{8Pa^3}{EI} + \frac{64Ra^3}{3EI} = 0$ gives $R = -\frac{3P}{8}$, that is a force of $\frac{3P}{8}$ downwards.