Question

Consider the rigid bar ABC supported by the pin-jointed links BD and CE and subjected to a load \(P\) at the end \(A\), as shown in the figure. The axial rigidities of BD and CE are 22500 kN and 15000 kN, respectively. If CE elongates by 5 mm due to the load \(P\), the magnitude of the downward deflection (in mm) of the end \(A\) would be ........ (rounded off to the nearest integer).
\begin{center} \includegraphics[width=8cm]{51.png} \end{center}

Hint:

When dealing with deflections in systems involving axial rigidity, use the principle of similar triangles to relate the deflections in various members.

Answer 1

Correct Answer: 14

Let \( P_D \) and \( P_C \) be the axial forces in the members BD and CE, respectively. From equilibrium, \[ P_D = 2P, \quad P_C = P \] Given, \[ \Delta_C = 5 \, {mm} \] The elongation of CE is given by: \[ \Delta_C = \frac{P \times 300}{15000} = 5 \, {mm} \] Thus, \[ P = 250 \, {kN}, \quad P_D = 2 \times 250 = 500 \, {kN} \] The downward deflection \( \Delta_D \) for member BD is given by: \[ \Delta_D = \frac{P_D \times L_{DB}}{A \times E_{DB}} = \frac{500 \times 200}{22500} = 4.44 \, {mm} \] Using similar triangles, \[ \frac{\Delta_D}{\Delta_C} = \frac{x}{200 - x} \] From this, \[ 4.44 = \frac{x}{200 - x} \times 5 \] Solving for \(x\), \[ x = 94.117 \, {mm} \] Now, the total deflection at \(A\) is \[ \Delta_A = \frac{200 + x}{\Delta_D} \times \Delta_A \] Substituting the values, \[ \Delta_A = 13.87 \, {mm} \quad \approx 14 \, {mm} \] Thus, the downward deflection of end \(A\) is 14 mm.