## 3 Thermal Deformation Model for Steel Joint
### Study on Analytical Method of Steel Joint's Thermal Deformation  
The joint is discretized into multiple thin rings, and the thickness of each ring is \(\delta\). Thus, the thermal deformation of the joint becomes a plane stress problem, so \(\sigma_{z}=0\), \(u_{r}\) is the radial displacement, and \(r\) is the radius. As shown in Figure 6, polar coordinates are adopted for the axisymmetric plane stress problem. The balance equation is as follows. Figure 5: Temperature of point B when the assembly is completed. Figure 6: Schematic diagram of the ring. \[\left\{ \begin{array}{l}\frac{\partial\sigma_{r}}{\partial r}+\frac{1}{r}\frac{ \partial\tau_{\theta\theta}}{\partial\theta}+\frac{\sigma_{r}-\sigma_{\theta}}{ r}=0\\ \frac{1}{r}\frac{\partial\sigma_{\theta}}{\partial\theta}+\frac{\partial\tau_{ \theta\theta}}{\partial r}+\frac{2\tau_{\theta}}{r}=0\end{array} \right. \tag{1}\]  
where \(\sigma_{r}\) is the normal stress of the radial; \(\sigma_{\theta}\) is the normal stress of the radian and \(\tau_{\theta\theta}\) is the shear stress. The equilibrium equation is as follows:  
\[\left\{ \begin{array}{l}\frac{\partial\sigma_{r}}{\partial r}+\frac{\sigma_{ r}-\sigma_{\theta}}{r}=0\\ \frac{\partial\sigma_{\theta}}{\partial\theta}=0\end{array} \right. \tag{2}\]  
The geometric equation is:  
\[\left\{ \begin{array}{l}\varepsilon_{r}=\frac{\partial u_{r}}{\partial r}\\ \varepsilon_{\theta}=\frac{\mu_{r}}{r}\end{array} \right. \tag{3}\]  
The physical equation is as follows:  
\[\left\{ \begin{array}{l}\varepsilon_{r}=\frac{1}{r}(\sigma_{r}-\mu\sigma_{ \theta})+\alpha T\\ \varepsilon_{\theta}=\frac{1}{E}(\sigma_{\theta}-\mu\sigma_{r})+\alpha T \end{array} \right. \tag{4}\]  
wherein \(\alpha\) is the linear expansion coefficient; \(\mu\) is Poisson's ratio and \(E\) is the elastic modulus. \[\left\{ \begin{array}{l}\sigma_{r}=\frac{E}{1-\mu^{2}}(\varepsilon_{r}+\mu \varepsilon_{r})-\frac{E\alpha T}{1-\mu}\\ \sigma_{\theta}=\frac{E}{1-\mu^{2}}(\varepsilon_{\theta}+\mu\varepsilon_{r})- \frac{E\alpha T}{1-\mu}\end{array} \right. \tag{5}\]  
Substituting Equation (3) into the formula above we have  
\[\left\{ \begin{array}{l}\sigma_{r}=\frac{E}{1-\mu^{2}}(\frac{du_{r}}{dr}+\mu \frac{u_{r}}{r})-\frac{E\alpha T}{1-\mu}\\ \sigma_{\theta}=\frac{E}{1-\mu^{2}}(\frac{u_{r}}{r}+\mu\frac{du_{r}}{dr})- \frac{E\alpha T}{1-\mu}\end{array} \right. \tag{6}\]  
Making the subtraction for Formula (6) results in  
\[\sigma_{r}-\sigma_{\theta}=\frac{E}{1+\mu}(\varepsilon_{r}-\varepsilon_{ \theta}) \tag{7}\]  
Solving the derivative of \(r\) for the above equation and substituting the geometric Equation (3) into the above equation, we have  
\[\sigma_{r}-\sigma_{\theta}=\frac{E}{1+\mu}(\varepsilon_{r}-\varepsilon_{ \theta})=\frac{E}{1+\mu}(\frac{du_{r}}{dr}-\frac{u_{r}}{r}) \tag{8}\]  
Solving the derivative of \(r\) for Equation (6) we obtain  
\[ \begin{array}{l}\frac{\partial\sigma_{r}}{\partial r}=\frac{E}{1-\mu^{2}}( \varepsilon_{r}^{\prime}+\mu\varepsilon_{\theta}^{\prime})-\frac{E\alpha}{1- \mu}\cdot\frac{dT}{dr}\\ =\frac{E}{1-\mu^{2}}\left[\frac{d}{dr}\left(\frac{du_{r}}{dr}\right)+\mu\frac{ d}{dr}\left(\frac{u_{r}}{r}\right)\right]-\frac{E\alpha}{1-\mu}\cdot\frac{dT}{dr}\\ =\frac{E}{1-\mu^{2}}\left[\frac{d^{2}u_{r}}{dr^{2}}+\mu\left(\frac{u_{r}^{ \prime}}{r}-\frac{u_{r}}{r^{2}}\right)\right]-\frac{E\alpha}{1-\mu}\cdot\frac {dT}{dr}\end{array}  \tag{9}\]  
If we substituted Formulas (8) and (9) into Formula (2), there would be  
\[\frac{E}{1-\mu^{2}}\left[\frac{d^{2}u_{r}}{dr^{2}}+\mu\left(\frac{u_{r}^{ \prime}}{r}-\frac{u_{r}}{r^{2}}\right)\right]-\frac{E\alpha}{1-\mu}\cdot\frac{ dT}{dr}+\frac{E}{1+\mu}(\frac{du_{r}}{r\cdot dr}-\frac{u_{r}}{r^{2}})=0 \tag{10}\]The above equation is themmultiplied by \(\left(1-\mu^{2}\right)/E\). \[ \begin{array}{ll}\alpha(1+\mu)\frac{dT}{dr}&=\frac{d^{2}u_{r}}{dr^{2}}+\frac{1} {r}\frac{d\,u_{r}}{dr}+\frac{1}{r^{2}}u_{r}\\ &=\frac{d}{dr}\left[\frac{1}{r}\frac{d\,(r\cdot u_{r})}{dr}\right]\end{array}