The edge of an aluminium cube is 10; cm long. | vedclass

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(C) The shear modulus $\eta$ is defined as the ratio of shear stress to shear strain:$\eta = \frac{F/A}{\Delta L/L}$Where $F$ is the force applied,$A$

Vedclass · Accepted Answer

$3.92 	imes 10^{-7}\; m$

Vedclass · Answer

(C) The shear modulus $\eta$ is defined as the ratio of shear stress to shear strain:$\eta = \frac{F/A}{\Delta L/L}$Where $F$ is the force applied,$A$ is the area of the face,$L$ is the side length of the cube,and $\Delta L$ is the vertical deflection.Rearranging for $\Delta L$:$\Delta L = \frac{F L}{A \eta}$Given:$F = mg = 100\; kg 	imes 9.8\; m/s^2 = 980\; N$$L = 10\; cm = 0.1\; m$$A = L^2 = (0.1\; m)^2 = 0.01\; m^2 = 10^{-2}\; m^2$$\eta = 25\; GPa = 25 	imes 10^9\; Pa$Substituting the values:$\Delta L = \frac{980 	imes 0.1}{10^{-2} 	imes 25 	imes 10^9}$$\Delta L = \frac{98}{0.25 	imes 10^9} = 392 	imes 10^{-9}\; m = 3.92 	imes 10^{-7}\; m$The vertical deflection is $3.92 	imes 10^{-7}\; m$.

The edge of an aluminium cube is $10\; cm$ long. One face of the cube is firmly fixed to a vertical wall. $A$ mass of $100\; kg$ is then attached to the opposite face of the cube. The shear modulus of aluminium is $25\; GPa$. What is the vertical deflection of this face?

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