A steel rod of radius 20 mm and length 2 m | vedclass

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Question

(B) Given: Radius $r = 20 \ mm = 20 	imes 10^{-3} \ m$,Length $L = 2 \ m$,Force $F = 400 \ kN = 400 	imes 10^3 \ N$,Young's modulus $Y = 2 	imes 10

Vedclass · Accepted Answer

$3.18 	imes 10^8 \ N \ m^{-2}, 0.16 \%$

Vedclass · Answer

(B) Given: Radius $r = 20 \ mm = 20 	imes 10^{-3} \ m$,Length $L = 2 \ m$,Force $F = 400 \ kN = 400 	imes 10^3 \ N$,Young's modulus $Y = 2 	imes 10^{11} \ N \ m^{-2}$.Stress is defined as force per unit area: $	ext{Stress} = \frac{F}{A} = \frac{F}{\pi r^2}$.Substituting the values: $	ext{Stress} = \frac{400 	imes 10^3}{3.14 	imes (20 	imes 10^{-3})^2} = \frac{400 	imes 10^3}{3.14 	imes 400 	imes 10^{-6}} = \frac{10^9}{3.14} \approx 3.18 	imes 10^8 \ N \ m^{-2}$.Strain is calculated using Young's modulus: $Y = \frac{	ext{Stress}}{	ext{Strain}} \implies 	ext{Strain} = \frac{	ext{Stress}}{Y}$.$	ext{Strain} = \frac{3.18 	imes 10^8}{2 	imes 10^{11}} = 1.59 	imes 10^{-3} = 0.159 \% \approx 0.16 \%$.Thus,the values are $3.18 	imes 10^8 \ N \ m^{-2}$ and $0.16 \%$.

$A$ steel rod of radius $20 \ mm$ and length $2 \ m$ is acted upon by a force of $400 \ kN$ along its length. The values of stress and strain are respectively $(Y_{\text{steel}} = 2 \times 10^{11} \ N \ m^{-2})$.

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