A metallic bar of Young's modulus $0.5 imes 10^{11} ext{ N m}^{-2}$, coefficient of linear thermal expansio

Question

A metallic bar of Young's modulus $0.5 	imes 10^{11}	ext{ N m}^{-2}$, coefficient of linear thermal expansion $10^{-5}\ ^\circ	ext{C}^{-1}$, length $1	ext{ m}$ and cross-sectional area $10^{-3}	ext{ m}^2$ is heated from $0^\circ	ext{C}$ to $100^\circ	ext{C}$ without expansion or bending. The compressive force developed in the metallic bar is:

Accepted Answer

When the metallic bar is heated and its expansion is prevented, thermal stress is developed in it.
The thermal strain is given by $\epsilon = \alpha \Delta T$.
According to Hooke's law, Young's modulus $Y = \frac{	ext{Stress}}{	ext{Strain}} = \frac{F / A}{\alpha \Delta T}$.
Therefore, the compressive force developed is $F = Y A \alpha \Delta T$.
Given values:
$Y = 0.5 	imes 10^{11}	ext{ N m}^{-2}$
$A = 10^{-3}	ext{ m}^2$
$\alpha = 10^{-5}\ ^\circ	ext{C}^{-1}$
$\Delta T = 100^\circ	ext{C} - 0^\circ	ext{C} = 100^\circ	ext{C}$
Substituting the values into the formula:
$F = (0.5 	imes 10^{11}) 	imes (10^{-3}) 	imes (10^{-5}) 	imes (100)$
$F = 0.5 	imes 10^{(11 - 3 - 5 + 2)}$
$F = 0.5 	imes 10^5	ext{ N}$
$F = 50 	imes 10^3	ext{ N}$

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