Question

In the elastic range of a tension test, an extensometer records an extension of 1.207 x 10-2 mm as the load increases by 5 kN. Calculate the value of Young’s modulus in GPa. The diameter of test piece is 7.98 mm, and the extensometer gauge length is 25 mm.

Answer 1

Answer:

Value of Young's modulus is obtained as $2.07\times 10^{11}N/m^{2}$

Explanation:

We know from the basic stress strain relationship

$\sigma =E\times \epsilon$

Stress is obtained as

$\sigma =\frac{P}{A}=\frac{5\times 10^{3}}{\frac{\pi\times D^{2}}{4}}=\frac{5\times 10^{3}}{0.25\times \pi \times (7.98\times 10^{-3})^{2}}=99.97MPa$

now the strain is obtained as

$\epsilon =\frac{\Delta L}{L}=\frac{1.207\times 10^{-2}}{25}=4.828\times 10^{-4}$

using these values in the above equation we obtain E as

$E=\frac{\sigma }{\epsilon }=\frac{99.97\times 10^{6}}{4.828\times 10^{-4}}=2.07\times 10^{11}N/m^{2}$