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