From a tensile test, following information is available within elastic limit. A tensile load of 54 kN produces an elongation of 0.112 mm for a specimen with gauge length 300 mm. The initial diameter of the rod specimen was 30 mm and after the application of the load, it gets reduced by 0.00366 mm.

Question

From a tensile test, following information is available within elastic limit. A tensile load of 54 kN produces an elongation of 0.112 mm for a specimen with gauge length 300 mm. The initial diameter of the rod specimen was 30 mm and after the application of the load, it gets reduced by 0.00366 mm. Find out Poisson’s ratio and values of the three elastic moduli.

Accepted Answer

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

[latex] ext { Stress }=\frac{ ext { Load }}{ ext { Area }}=\frac{54}{(\pi / 4) imes(0.03)^2}=76.4 MN / m ^2 [/latex]

The linear and lateral strain are given by

[latex] ext { Linear strain }=\frac{ ext { Elongation }}{ ext { Gauge length }}=\frac{0.112}{30}=3.73 imes 10^{-4} [/latex]

[latex] ext { Lateral strain }=\frac{ ext { Change in diameter }}{ ext { Original diameter }}=\frac{0.0036}{30}=1.22 imes 10^{-4} [/latex]

Poisson’s ratio is found as

[latex] ν=\frac{1.22 imes 10^{-4}}{3.73 imes 10^{-4}}=0.327 [/latex]

The Young’s modulus is found to be

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

As E = 2G(1+ ν) , G can be found by

[latex]G=\frac{E}{2(1+ν)}[/latex]

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.

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