Find the stress developed inside a tooth cavi | vedclass

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Question

(A) Given:Temperature of hot tea,$t_2 = 57^{\circ} C$Normal temperature of tooth,$t_1 = 37^{\circ} C$Coefficient of linear expansion,$\alpha = 1.7 	i

Vedclass · Accepted Answer

$1.4 	imes 10^8 \ Nm^{-2}$

Vedclass · Answer

(A) Given:Temperature of hot tea,$t_2 = 57^{\circ} C$Normal temperature of tooth,$t_1 = 37^{\circ} C$Coefficient of linear expansion,$\alpha = 1.7 	imes 10^{-5} {}^{\circ} C^{-1}$Bulk modulus,$B = 140 	imes 10^9 \ Nm^{-2}$Change in temperature,$\Delta t = t_2 - t_1 = 57 - 37 = 20^{\circ} C$Thermal stress is given by the product of Bulk modulus and volumetric strain:$	ext{Stress} = B 	imes \frac{\Delta V}{V}$Since $\frac{\Delta V}{V} = \gamma \Delta t$ and $\gamma = 3\alpha$:$	ext{Stress} = B 	imes (3\alpha) 	imes \Delta t$Substituting the values:$	ext{Stress} = 3 	imes (140 	imes 10^9) 	imes (1.7 	imes 10^{-5}) 	imes 20$$	ext{Stress} = 3 	imes 140 	imes 1.7 	imes 20 	imes 10^4$$	ext{Stress} = 14280 	imes 10^4 = 1.428 	imes 10^8 \ Nm^{-2} \approx 1.4 	imes 10^8 \ Nm^{-2}$.

Find the stress developed inside a tooth cavity filled with copper when hot tea at a temperature of $57^{\circ} C$ is drunk. You can take the body (tooth) temperature to be $37^{\circ} C$ and $\alpha = 1.7 \times 10^{-5} {}^{\circ} C^{-1}$,bulk modulus for copper $B = 140 \times 10^9 \ Nm^{-2}$.

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