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. $\frac{F l}{A E}$
• B. $\frac{2 F l}{A E}$
• C. $\frac{3 F l}{A E}$
• D. $\frac{4 F l}{A E}$

Answer 1

Answer: (D)

Elongation in $I^{st}$ part
$\Delta x_{1} =\frac{\text { Net force } \times L}{A Y} $
$=\frac{(3+2) F L}{A E}$
$=\frac{5 F L}{A E}$
Elongation in II $^{\text {nd }}$ part
$\Delta x_{2}=\frac{(F-2 F) L }{A E}=-\frac{F L}{A E}$
So net elongation
$\Delta x =\Delta x_{1}+\Delta x_{2} $
$=\frac{5 F L}{A E}-\frac{F L}{A E}=\frac{4 F L}{A E}$