Question

A uniform slender rod of length $L$, cross-sectional area $A$ and Young's modulus $Y$ is acted upon by the forces shown in the figure. The elongation of the rod is:

• A. $\frac{3FL}{5AY}$
• B. $\frac{2FL}{5AY}$
• C. $\frac{3FL}{8AY}$
• D. $\frac{8FL}{3AY}$

Answer 1

The correct option is D $\frac{8FL}{3AY}$

Divide the rod into 2 parts of length $2l/m$ and $l/m$ where the force F is acting.

Now solve the problem for the 2 rods separately.

Force acting on both sides of $2l/3$ is $3F$.

So extension $x_1$ will be $x_1=\frac{[3F\times \frac{2l}{3}]}{AY}$

Force acting on both sides of $l/3$ is $2F$

So extension $x2$ will be $x_2=\frac{[2F\times \frac{l}{3}]}{AY}$

Now total extension x is $x=x_1+x_2=8FL/3AY$