The Young’s modulus of the material of a rod is 20×1010Pascal. When the longitudinal strain is 0.04% .The energy stored per unit volume is (in kJ/m3)

Given Y=20×1010 pascal Energy stored per unit volume,=12 stress × strain =12Y( strain )2=12×20×1010×0.041002=16×103 Joule /m3

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The Young’s modulus of the material of a rod is $20 \times 10^{10} Pascal$ . When the longitudinal strain is 0.04% .The energy stored per unit volume is (in $kJ / m^{3}$ )

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answer is 16.

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Given $Y = 20 \times 10^{10}$ pascal

Energy stored per unit volume,

$= \frac{1}{2} stress \times strain = \frac{1}{2} Y (strain)^{2} = \frac{1}{2} \times 20 \times 10^{10} \times {[\frac{0 .04}{100}]}^{2}$

$= 16 \times 10^{3} Joule / m^{3}$

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