Physics - Elasticity Question with Solution | TestHub

When the temperature is raised by $\Delta T$, the free thermal expansion of the steel wire would be $L{\alpha }_S\Delta T$ and for copper $L{\alpha }_C\Delta T$. Since the composite wire is fixed between rigid supports, a compressive force $F$ is developed. The total elastic compression must be equal to the total free thermal expansion: $\frac{FL}{A{Y}_S}+\frac{FL}{A{Y}_C}=L{\alpha }_S\Delta T+L{\alpha }_C\Delta T$. Solving for $F$: $F=\frac{A({\alpha }_S+{\alpha }_C)\Delta T}{\frac{1}{Y_S}+\frac{1}{Y_C}}=A({\alpha }_S+{\alpha }_C)\Delta T\frac{Y_S{Y}_C}{Y_S+Y_C}$. The total elastic potential energy stored is $U=\frac{1}{2}F\times (\text{total elastic deformation})$.$$U=\frac{1}{2}F\left(\frac{FL}{A{Y}_S}+\frac{FL}{A{Y}_C}\right)=\frac{1}{2}{F}^2\frac{L}{A}\left(\frac{1}{Y_S}+\frac{1}{Y_C}\right)$$ Substitute $F$:$$U=\frac{1}{2}{\left[A({\alpha }_S+{\alpha }_C)\Delta T\frac{Y_S{Y}_C}{Y_S+Y_C}\right]}^2\frac{L}{A}\left(\frac{Y_S+Y_C}{Y_S{Y}_C}\right)=\frac{1}{2}AL({\alpha }_S+{\alpha }_C{)}^2(\Delta T{)}^2\frac{Y_S{Y}_C}{Y_S+Y_C}$$