A steel plate of face area 1 , cm^2 and thick | vedclass

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(B) The modulus of rigidity $(G)$ is defined as the ratio of shear stress to shear strain.$G = \frac{	ext{Shear Stress}}{	ext{Shear Strain}} = \frac

Vedclass · Accepted Answer

$5 	imes 10^{-6}$

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(B) The modulus of rigidity $(G)$ is defined as the ratio of shear stress to shear strain.$G = \frac{	ext{Shear Stress}}{	ext{Shear Strain}} = \frac{F/A}{x/L} = \frac{FL}{Ax}$Given:$F = 10 \, kN = 10 	imes 10^3 \, N$$L = 4 \, cm = 0.04 \, m$$A = 1 \, cm^2 = 1 	imes 10^{-4} \, m^2$$G = 8 	imes 10^{11} \, N/m^2$Rearranging the formula to solve for lateral displacement $x$:$x = \frac{FL}{AG}$Substituting the values:$x = \frac{10 	imes 10^3 	imes 0.04}{1 	imes 10^{-4} 	imes 8 	imes 10^{11}}$$x = \frac{400}{8 	imes 10^7} = 50 	imes 10^{-7} = 5 	imes 10^{-6} \, m$Thus,the lateral displacement is $5 	imes 10^{-6} \, m$.

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