A slab of side 50 text{ cm} and thickness 10 | vedclass

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(B) Given: Side of the slab $L = 50 	ext{ cm} = 0.5 	ext{ m}$, thickness $t = 10 	ext{ cm} = 0.1 	ext{ m}$, shearing force $F = 10^5 	ext{ N}$, d

Vedclass · Accepted Answer

$5$

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(B) Given: Side of the slab $L = 50 	ext{ cm} = 0.5 	ext{ m}$, thickness $t = 10 	ext{ cm} = 0.1 	ext{ m}$, shearing force $F = 10^5 	ext{ N}$, displacement $x = 0.2 	ext{ mm} = 0.2 	imes 10^{-3} 	ext{ m}$.Area of the face subjected to force $A = L 	imes t = 0.5 	ext{ m} 	imes 0.1 	ext{ m} = 0.05 	ext{ m}^2$.Shear stress $= \frac{F}{A} = \frac{10^5}{0.05} = 2 	imes 10^6 	ext{ N/m}^2$.Shear strain $= \frac{x}{L} = \frac{0.2 	imes 10^{-3} 	ext{ m}}{0.5 	ext{ m}} = 0.4 	imes 10^{-3} = 4 	imes 10^{-4}$.Shear modulus $\eta = \frac{	ext{Shear stress}}{	ext{Shear strain}} = \frac{2 	imes 10^6}{4 	imes 10^{-4}} = 0.5 	imes 10^{10} 	ext{ N/m}^2 = 5 	imes 10^9 	ext{ N/m}^2 = 5 	ext{ GPa}$.

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