A metallic bar of Young's modulus 0.5 times 1 | vedclass

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Question

(A) The thermal strain produced in the bar is given by $\epsilon = \alpha \Delta T$.Given $\alpha = 10^{-5} \,^{\circ}C^{-1}$ and $\Delta T = 100^{\ci

Vedclass · Accepted Answer

$50 	imes 10^3 \,N$

Vedclass · Answer

(A) The thermal strain produced in the bar is given by $\epsilon = \alpha \Delta T$.Given $\alpha = 10^{-5} \,^{\circ}C^{-1}$ and $\Delta T = 100^{\circ}C - 0^{\circ}C = 100^{\circ}C$.So, $\epsilon = 10^{-5} 	imes 100 = 10^{-3}$.Since the bar is prevented from expanding, the compressive stress $\sigma$ is given by $\sigma = Y 	imes \epsilon$, where $Y$ is Young's modulus.$\sigma = (0.5 	imes 10^{11} \,N \,m^{-2}) 	imes 10^{-3} = 0.5 	imes 10^8 \,N \,m^{-2}$.The compressive force $F$ is given by $F = \sigma 	imes A$, where $A$ is the area of cross-section.$F = (0.5 	imes 10^8 \,N \,m^{-2}) 	imes (10^{-3} \,m^2) = 0.5 	imes 10^5 \,N = 50 	imes 10^3 \,N$.

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