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madreJ [45]
3 years ago
15

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. C

alculate 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.
Engineering
1 answer:
CaHeK987 [17]3 years ago
5 0

Answer:

Value of Young's modulus is obtained as 2.07\times 10^{11}N/m^{2}

Explanation:

We know from the basic stress strain relationship

\sigma =E\times \epsilon

Stress is obtained as

\sigma =\frac{P}{A}=\frac{5\times 10^{3}}{\frac{\pi\times D^{2}}{4}}=\frac{5\times 10^{3}}{0.25\times \pi \times (7.98\times 10^{-3})^{2}}=99.97MPa

now the strain is obtained as

\epsilon =\frac{\Delta L}{L}=\frac{1.207\times 10^{-2}}{25}=4.828\times 10^{-4}

using these values in the above equation we obtain E as

E=\frac{\sigma }{\epsilon }=\frac{99.97\times 10^{6}}{4.828\times 10^{-4}}=2.07\times 10^{11}N/m^{2}

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