A bar is subjected to axial forces as shown. If $E$ is the modulus of elasticity of the bar and $A$ is its crosssection area. Its elongation will be
Diffcult
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(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 $II^{nd}$ part
$\Delta x_2=\frac{(F-2 F) L}{A E}=-\frac{F L}{A E}$
So net elongation
$\Delta x=\Delta x_1+\Delta x_2$
$=\frac{5 F L}{A E}-\frac{F L}{A E}=\frac{4 F L}{A E}$
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