Question

A brass bar, having cross sectional area 10 $c{m}^2$ is subjected to axial forces as shown in figure. Total elongation of the bar is (Take Y = $8\times {10}^{2}t/m^2$).

• A. $0.0775cm$
• B. $7.5cm$
• C. $0.75cm$
• D. $0.075cm$

Answer 1

Answer: (A)

$0.0775cm$

$\text{Since the system is in equilibrium}\sum F_{\text{right}}=\sum F_{\text{left}}$

$\begin{array}{c}\text{Note: at section}B\text{net force}3t\text{is right is present.}\\ \text{Similarly for section}C\text{too.}\end{array}$

$\begin{array}{cc}\Delta l_1& =\frac{P_1{L}_1}{AY}=\frac{\left(5t\right)\left(0.6\right)}{A\times 8\times {10}^2\times t}=\frac{3}{800A}\to \left(1\right)\\ \Delta l_2& =\frac{\left(2t\right)\left(1\right)}{A\times 8\times 100t}=\frac{1}{4004}\to \left(2\right)\\ \Delta l_3& =\frac{\left(1t\right)\left(1.2\right)}{A\times 800t}=\frac{1.2}{800A}\to \left(3\right)\end{array}$

$\begin{array}{c}\text{Since each cases is elongation, we add}\to \\ (1+2+3)=\frac{3+2+1.2}{800A}=\frac{6\cdot 2}{800\times 10\times {10}^{-4}}=0.0775cm\end{array}$