Question

A bar is subjected to axial forces as shown. If $E$ is the modulus of elasticity of the bar and $A$ is its cross section area. Its elongation will be

• A. $\frac{Fl}{AE}$
• B. $\frac{2Fl}{AE}$
• C. $\frac{3Fl}{AE}$
• D. $\frac{4Fl}{AE}$

Answer 1

Answer: (D)

Elongation in $I^{st}$ part
$\Delta x_1=\frac{\text{ Net force }\times L}{AY}$
$=\frac{(3+2)FL}{AE}$
$=\frac{5FL}{AE}$
Elongation in II ${}^{\text{nd }}$ part
$\Delta x_2=\frac{(F-2F)L}{AE}=-\frac{FL}{AE}$
So net elongation
$\Delta x=\Delta x_1+\Delta x_2$
$=\frac{5FL}{AE}-\frac{FL}{AE}=\frac{4FL}{AE}$