Question 1.2: From a tensile test, following information is available with...

The load applied is 54 kN. The stress is given by

\text { Stress }=\frac{\text { Load }}{\text { Area }}=\frac{54}{(\pi / 4) \times(0.03)^2}=76.4  MN / m ^2

The linear and lateral strain are given by

\text { Linear strain }=\frac{\text { Elongation }}{\text { Gauge length }}=\frac{0.112}{30}=3.73 \times 10^{-4}

\text { Lateral strain }=\frac{\text { Change in diameter }}{\text { Original diameter }}=\frac{0.0036}{30}=1.22 \times 10^{-4}

Poisson’s ratio is found as

ν=\frac{1.22 \times 10^{-4}}{3.73 \times 10^{-4}}=0.327

The Young’s modulus is found to be

\begin{aligned} E &=\frac{\text { Stress }}{\text { Strain }}=\frac{76.4}{3.73 \times 10^{-4}} \\ &=2.05 \times 10^5  MN / m ^2 \\ &=205 \times 10^3  MPa \\ &=205  GPa \end{aligned}

As E = 2G(1+ ν) , G can be found by

G=\frac{E}{2(1+ν)}

Putting values of E and ν we get G = 77 GPa.
Similarly, E = 3K (1- 2ν), which gives K = 197 GPa.
Thus, ν = 0.327, E = 205 GPa, G = 77 GPa and K = 197 GPa.