Question

The figure below shows a steel rod of $25m{m}^2$ cross sectional area. It is loaded at four points. K, L, M and N. Assume $E_{steel}=200GPa$. The total change in length of the rod due to loading is

• A. $1\mu m$
  
• B. $-10\mu m$
• C. $1\mu m$
• D. $-20\mu m$

Answer 1

The correct option is B $-10\mu m$

Given Area = $25m{m}^2$; $E_{steel}=200GPa$


$\text{A}xialdeformation=\frac{PL}{AE}$

$\Delta l_{total}=\Delta l_1+\Delta l_2+\Delta l_3$

$\Delta l_1=\frac{100\times 500}{25\times (200\times 10{)}^3}=0.01mm$

$\Delta l_2=\frac{-150\times [1700-\{500+400\}]}{25\times (200\times 10{)}^3}=-0.024mm$

$\Delta l_3=\frac{50\times 400}{25\times (200\times 10{)}^3}=4\times {10}^{3}mm$

$\Delta l_{Total}=\Delta l_1+\Delta l_2+\Delta l_3$

$=0.01-0.024+4\times {10}^{-3}$

$=-0.01mm=0.01\times {10}^{-3}m=-1\times {10}^{-5}m$

$=\frac{-1\times {10}^5\times 10}{10}m=-10\times {10}^{-6}m=-10\mu m$