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InterviewSolution
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A metallic cube whose each side is 10 cm is subjected to a shearing force of 100 kgf. The top face is displaced through 0.25 cm with respect to the bottom ? Calculate the shearing stress, strain and shear modulus. |
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Answer» Correct Answer - `9.8 xx 10^4 Nm^(-2), 0.025, 3.92 xx 10^6 Nm^(-2)` Here, `L = 10 cm = 10 xx 10^(-2)m` `F = 100 kg f= 100 xx 9.8 N` `Delta L = 0.25 cm = 0.25 xx 10^(-2) m,` Shearing stress ` = (F)/(L_2) = (100xx9.8)/((10xx10^(-2))` Sheraing strain ` = (Delta L)/(L) = (0.25xx10^(-2))/(10xx10^(-2))` = 0.025 Shear Modulus of elasticity, `G = ("Shearing stress")/("Shearing strain") = (9.8 xx 10^4)/(0.025)` ` =3.92xx 10^6 Nm^(-2)` |
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