Topic: Strength of Materials
The extension of a mild steel bar 4m long, 2000 $mm^{2}$ cross section under the action of an axial load of 20kN, if $E=2\times10^{5}N/mm^{2},$ is
- A. 2mm
- B. 0.2mm
- C. 0.5mm
- D. 0.05mm
Answer: B. 0.2mm
Explanation: Given data: $l=4m=4000~mm$, $A=2000~mm^{2}$, $P=20~kN = 20\times10^{3}N$, $E=2\times10^{5}N/mm^{2}$. Extension is given by $\delta = \frac{PL}{AE} = \frac{20\times10^{3}\times4000}{2000\times2\times10^{5}} = 0.2~mm$.