The deflection of a uniform circular bar of diameter d and length $$l$$, which extends by an amount under a tensile pull W, when it carries the same load at its mid-span, is

Solution (By Examveda Team)

We have a circular bar with diameter d and length l.
It extends by an amount e under a tensile pull W.
We want to find the deflection when the same load W is applied at the mid-span of the bar, making it act like a beam.

1. Relate Elongation (e) to Tensile Stress and Strain:
When the bar is under a tensile pull W, we know:
Stress \( \sigma \) = Force (W) / Area (A), where \( A = \frac{\pi d^2}{4} \).
Strain \( \varepsilon \) = Change in Length (e) / Original Length (l).
Also, Stress \( \sigma \) = Young's Modulus \( E \) * Strain \( \varepsilon \).
Therefore, \( E = \frac{\sigma}{\varepsilon} \).

2. Find Young's Modulus (E) in terms of e, l, d, and W:
\( E = \frac{W}{\left( \frac{\pi d^2}{4} \right)} \div \frac{e}{l} \)
\( E = \frac{4Wl}{\pi d^2 e} \).

3. Deflection Formula for a Simply Supported Beam with a Point Load at Mid-Span:
The standard formula for the deflection (\( \delta \)) of a simply supported beam with a point load (W) at the mid-span is:
\( \delta = \frac{Wl^3}{48EI} \), where \( I \) is the area moment of inertia.

4. Area Moment of Inertia (I) for a Circular Cross-Section:
\( I = \frac{\pi d^4}{64} \).

5. Substitute I and E into the Deflection Formula:
\( \delta = \frac{Wl^3}{48 \times E \times \left( \frac{\pi d^4}{64} \right)} \)
\( \delta = \frac{Wl^3}{48 \times \left( \frac{4Wl}{\pi d^2 e} \right) \times \left( \frac{\pi d^4}{64} \right)} \)
\( \delta = \frac{Wl^3 \times \pi d^2 e \times 64}{48 \times 4Wl \times \pi d^4} \)
\( \delta = \frac{64Wl^3 \pi d^2 e}{192Wl \pi d^4} \)
\( \delta = \frac{e l^2}{3 d^2} \).

Therefore, the deflection is: \( \frac{el^2}{3d^2} \).
Correct answer: Option C \( \frac{el^2}{3d^2} \).