Question

A straight bar of uniform cross section is subject to an axial tension, the cross sectional area of the bar is 50mm^2 and it's length is 2mm under a load of 40kN. Find the modulus of elasticity

Answer 1

$\sigma=E\cdot\epsilon=E\cdot\frac{\Delta l}{l}=\frac{F}{A}\to E=\frac{F}{A}\cdot\frac{l}{\Delta l}$

Assume that $\Delta l=1 mm$

So, we have

$E=\frac{F}{A}\cdot\frac{l}{\Delta l}=\frac{40000}{50\cdot10^{-6}}\cdot\frac{0.002}{0.001}=1600 MPa$