Question 2.6: A tensile test is performed on a brass specimen 10 mm in dia......

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}