Question

In the shown figure, length of the rod is $L$, area of cross-section $A$, Young's modulus of the material of the rod is $Y$. Then, $B$ and $A$ is subjected to a tensile force $F_A$ while force applied at end $B,F_B$ is lesser than $F_A$. Total change in length of the rod will be

• A. $F_A\times \frac{L}{2AY}$
• B. $F_B\times \frac{L}{2AY}$
• C. $\frac{(F_A+F_B)L}{2AY}$
• D. $\frac{(F_A-F_B)L}{2AY}$

Answer 1

The correct option is C $\frac{(F_A+F_B)L}{2AY}$

Tension at $x$,
$T=F_A-[F_A-F_B]\frac{x}{L}=F_A(1-\frac{x}{L})+F_B\left(\frac{x}{L}\right)$
Change in length
$△\left(dx\right)=\frac{Tdx}{AY}=[F_A(1-\frac{x}{L})+F_B\left(\frac{x}{L}\right)]=\frac{dx}{AY}$
Total change in length
$=\frac{1}{AY}\left[{\int }_0^1{F}_A\right(1-x/L)dx+{\int }_0^1{F}_B\left(\frac{x}{L}\right)dx]$
$=\frac{1}{AY}[F_A(L-\frac{L}{2})+F_B\times \frac{L}{2}]$
$=\frac{1}{AY}[F_A+F_B]\frac{L}{2}=\frac{L}{2AY}(F_A+F_B)$.