## Q. 2.6

A tensile test is performed on a brass specimen 10 mm in diameter using a gauge length of 50 mm. When applying axial tensile load of 25 kN, it was observed that the distance between the gauge marks increase by 0.152 mm, calculate modulus of elasticity of brass.

## Verified Solution

The problem is sketched in Figure 2.19.

Modulus of elasticity is given by, $E=\frac{\sigma}{\varepsilon}=\frac{\text { stress }}{\text { strain }}$

\begin{aligned} & \text { Stress }=\frac{P}{A}=\frac{25 \times 10^3}{\pi\left(10^2 / 4\right)}=318.3 N / mm ^2 \\ & \text { Strain }=\varepsilon=\frac{d l}{l}=\frac{0.152}{50}=3.04 \times 10^{-3} \\ & E=\frac{318.30}{3.04 \times 10^{-3}}=104703.95 N / mm ^2 \end{aligned}