Thermal Expansion Concept Page - 5

Example

Linear thermal expansion where expansion is constrained

A steel rod of cross section 1 mm2$m{m}^2$ is prevented from expansion by heating through 10∘C${10}^{\circ }C$. The thermal force developed in it is 
(Y=2×1011N/m2;α=10−5/∘C)$(Y=2\times {10}^{11}N/m^2;\alpha ={10}^{-5}{/}^{\circ }C)$ΔLL=αΔT=10−5×(10−0)=10−4$\frac{\mathrm{\Delta }L}{L}=\alpha \mathrm{\Delta }T={10}^{-5}\times (10-0)={10}^{-4}$
F=YAΔLL=2×1011×(1×10−6)×10−4=20N$F={\displaystyle \frac{YA\mathrm{\Delta }L}{L}}=2\times {10}^{11}\times (1\times {10}^{-6})\times {10}^{-4}=20N$

Example

Example of constrained volume thermal expansion

Example:
At a temperature T∘C$T^{\circ }C$, a liquid is completely filled in a spherical shell of copper. If ΔT$\mathrm{\Delta }T$ is the increase in temperature of the liquid and the shell, then find the outward pressure dP$dP$ on the shell due to the increase in temperature. (Given, K =$=$ bulk modulus of the liquids, γ=$\gamma =$ coefficient of volume expansion, α=$\alpha =$  coefficient of linear expansion of the material of the shell ) Solution:
We have,
ΔV1=V×3αΔT$\mathrm{\Delta }{V}_1=V\times 3\alpha \mathrm{\Delta }T$
Also,
ΔV2=V×γΔT$\mathrm{\Delta }{V}_2=V\times \gamma \mathrm{\Delta }T$
The developed stress increases the volume of vessel and decreases the volume of liquid.
Let the change in volume due to stress = ΔV$\mathrm{\Delta }V$
The final volume of both vessel and liquid is the same
∴$\therefore$ The final volume of vessel ⇒V1=V+ΔV1+ΔV$\Rightarrow V_1=V+\mathrm{\Delta }{V}_1+\mathrm{\Delta }V$
And the final volume of liquid ⇒V2=V+ΔV2−ΔV$\Rightarrow V_2=V+\mathrm{\Delta }{V}_2-\mathrm{\Delta }V$
But V1=V2$V_1=V_2$
V+ΔV1+ΔV=V+ΔV2−ΔV$V+\mathrm{\Delta }{V}_1+\mathrm{\Delta }V=V+\mathrm{\Delta }{V}_2-\mathrm{\Delta }V$
∴2ΔV=ΔV2−ΔV1$\therefore 2\mathrm{\Delta }V=\mathrm{\Delta }{V}_2-\mathrm{\Delta }{V}_1$
According to Hooke's law,
K=dpΔVV=VdpΔV$K={\displaystyle \frac{dp}{{\displaystyle \frac{\mathrm{\Delta }V}{V}}}}={\displaystyle \frac{Vdp}{\mathrm{\Delta }V}}$
∴dp=KΔVV=KVΔT2(γ−3α)V$\therefore dp=K{\displaystyle \frac{\mathrm{\Delta }V}{V}}={\displaystyle \frac{K{\displaystyle \frac{V\mathrm{\Delta }T}{2}}(\gamma -3\alpha )}{V}}$
∴dp=K2(γ−3α)ΔT$\therefore dp={\displaystyle \frac{K}{2}}(\gamma -3\alpha )\mathrm{\Delta }T$

Example

Stress due to thermal expansion using modulus of elasticity

Two rods of different materials having coefficients of thermal expansion α1,α2${\alpha }_1,{\alpha }_2$, and Young's modulus Y1,Y2$Y_1,Y_2$ respectively are fixed between two rigid walls. The rods are heated such that they undergo the same increase in temperature. There is no bending of rods. If α1:α2=2:3${\alpha }_1:{\alpha }_2=2:3$, thermal stress developed in the rods are equal for what value of Y1:Y2$Y_1:Y_2$.We know, l2=l1(1+αΔt)$l_2=l_1(1+\alpha \mathrm{\Delta }t)$ where Δt$\mathrm{\Delta }t$ = change in temperature, if dl$dl$ = change in length,
we have Δll=αΔt${\displaystyle \frac{\mathrm{\Delta }l}{l}}=\alpha \mathrm{\Delta }t$ = strain
The stress developed is same for both the rods.
So Y1α1Δt=Y2α2Δt$Y_1{\alpha }_1\mathrm{\Delta }t=Y_2{\alpha }_2\mathrm{\Delta }t$
or
Y1Y2=α2α1=32${\displaystyle \frac{Y_1}{Y_2}}={\displaystyle \frac{{\alpha }_2}{{\alpha }_1}}={\displaystyle \frac{3}{2}}$

Example

Thermal stress in a composite system of rods

Two rods P and Q of different metals having the same area A$A$ and the same length L$L$ are placed between two rigid walls as shown in the figure. The coefficients of linear expansion of P and Q are α1${\alpha }_1$ and α2${\alpha }_2$ respectively and their Young's modulus are Y1$Y_1$ and Y2$Y_2$ . The  temperature of both rods is now raised by T$T$ degrees. The force exerted by one rod on the other isThe stress  and force are same in both rods because area is same.
yet P extends by Δ$\mathrm{\Delta }$ L and Q is compressed by Δ$\mathrm{\Delta }$ L
F=AY(ΔL′)L$F={\displaystyle \frac{AY(\mathrm{\Delta }{L}^{\prime })}{L}}$
FP=FQ=AY1(Lα1T−ΔL)L=AY2L(Lα2T+ΔL)−−(1)$F_P=F_Q={\displaystyle \frac{A{Y}_1(L{\alpha }_{1}T-\mathrm{\Delta }L)}{L}}={\displaystyle \frac{A{Y}_2}{L}}(L{\alpha }_{2}T+\mathrm{\Delta }L)--(1)$
⇒ΔL=LTY1+Y2[Y1α1−Y2α2]$\Rightarrow \mathrm{\Delta }L={\displaystyle \frac{LT}{Y_1+Y_2}}[Y_1{\alpha }_1-Y_2{\alpha }_2]$
now putting ΔL$\mathrm{\Delta }L$ in eqn−−(1)$e{q}^n--(1)$
we get FP=FQ=Y1Y2ATY1+Y2[α1+α2]$F_P=F_Q={\displaystyle \frac{Y_1{Y}_{2}AT}{Y_1+Y_2}}[{\alpha }_1+{\alpha }_2]$

Example

Stress in bimetallic system

A bimetallic strip is heated up and not allowed to bend. What stresses are set up as a result?
Solution:
Let the cross-sectional areas, moduli and coefficients of expansion be as indicated.Suppose the bimetallic strip is subject to a temperature rise T$T$. Then the stresses and strains in each strip are related by Hooke's Law:
ϵ1=1Eσ1+α1T${ϵ}_1={\displaystyle \frac{1}{E}}{\sigma }_1+{\alpha }_{1}T$
ϵ2=1Eσ2+α2T${ϵ}_2={\displaystyle \frac{1}{E}}{\sigma }_2+{\alpha }_{2}T$
where σ1,σ2${\sigma }_1,{\sigma }_2$ are the stresses developed in the strips having areas A1,A2$A_1,A_2$ respectively.
Since the strips are bonded together and there is no bending a constant and equal strain develops in the two strips.
ϵ1=ϵ2${ϵ}_1={ϵ}_2$ 
The total horizontal force on the bimetallic strip is zero:
σ1A1+σ2A2=0${\sigma }_1{A}_1+{\sigma }_2{A}_2=0$
1E1σ1+α1T−1E2σ2−α2T=0${\displaystyle \frac{1}{E_1}}{\sigma }_1+{\alpha }_{1}T-{\displaystyle \frac{1}{E_2}}{\sigma }_2-{\alpha }_{2}T=0$
1E1σ1+α1T−1E2(−σ1A1A2)−α2T=0${\displaystyle \frac{1}{E_1}}{\sigma }_1+{\alpha }_{1}T-{\displaystyle \frac{1}{E_2}}(-{\displaystyle \frac{{\sigma }_1{A}_1}{A_2}})-{\alpha }_{2}T=0$
σ1=−E1E2A2(α1−α2)TE2A2+E1A1${\sigma }_1=-{\displaystyle \frac{E_1{E}_2{A}_2({\alpha }_1-{\alpha }_2)T}{E_2{A}_2+E_1{A}_1}}$  and
σ2=−E1E2A1(α1−α2)TE2A2+E1A1${\sigma }_2=-{\displaystyle \frac{E_1{E}_2{A}_1({\alpha }_1-{\alpha }_2)T}{E_2{A}_2+E_1{A}_1}}$

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