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

Value of Young's modulus is obtained as
`2.07* 10^(11)N/m^(2)`

Step-by-step explanation:

We know from the basic stress strain relationship

`\sigma =E* \epsilon`

Stress is obtained as

`\sigma =(P)/(A)=(5* 10^(3))/((\pi* D^(2))/(4))=(5* 10^(3))/(0.25* \pi * (7.98* 10^(-3))^(2))=99.97MPa`

now the strain is obtained as

`\epsilon =(\Delta L)/(L)=(1.207* 10^(-2))/(25)=4.828* 10^(-4)`

using these values in the above equation we obtain E as

`E=(\sigma )/(\epsilon )=(99.97* 10^(6))/(4.828* 10^(-4))=2.07* 10^(11)N/m^(2)`