temperature To, while the gas has a temperature Tw. The numbers which are relevant to this question are T_{0}=293 K, \quad T_{w}=600 K, \quad b=0.025 m, \quad \alpha=10^{-5} K^{-1}, E=200 G P a a) The thermoelastic stresses in an isolated sphere are analysed. The linear elastic equations for the solid stresses (orr, 0ee) are related to the strain field (err, E9e)through \left[\begin{array}{c} \sigma_{r r} \\ \sigma_{\theta \theta} \end{array}\right]=\frac{E}{(1+v)(1-2 v)}\left[\begin{array}{cc} 1-v & v \\ v & 1-v \end{array}\right]\left[\begin{array}{c} \epsilon_{r r} \\ \epsilon_{\theta \theta} \end{array}\right]-\frac{E \alpha T}{1-2 v}\left[\begin{array}{l} 1 \\ 1 \end{array}\right] Provide a physical justification for the equilibrium conditions \frac{d \sigma_{r r}}{d r}+\frac{2}{r}\left(\sigma_{r r}-\sigma_{\theta \theta}\right)=0 b) The strain is related to the radial displacement u through \epsilon_{r r}=\frac{d u}{d r}, \epsilon_{\theta \theta}=\frac{u}{r} Show that the radial displacement u satisfies \frac{d}{d r}\left(\frac{1}{r^{2}} \frac{d\left(r^{2} u\right)}{d r}\right)=\alpha\left(\frac{1+v}{1-v}\right) \frac{d T}{d r} and the general solution is u=\frac{(1+v) \alpha}{r^{2}(1-v)} \int_{0}^{r} T r^{2} d r+C_{1} r+\frac{C_{2}}{r^{2}} c) Show that the radial stress due to the temperature field is \sigma_{r r}=\frac{2 \alpha E}{1-v}\left(\frac{1}{b^{3}} \int_{0}^{b} \operatorname{Tr}^{2} \mathrm{~d} r-\frac{1}{r^{3}} \int_{0}^{r} \operatorname{Tr}^{2} d \mathrm{r}\right) and calculate the azimuthal stress. d) The effect of temperature on the stress field is analysed using a simplified temperature distribution of where T varies from To to Tw at a distance R from the centre of the sphere: T = To (r

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