# Thermal-Structural Coupling Analysis of Subsea Connector Sealing Contact
## 4 Thermal-Structural Coupling Mathematical Model of the Subsea Connector
### Three-Dimensional Stress Caused by the Steady-State Temperature Field
\begin{array}{l}C_{2}=-\frac{kK_{1}^{\prime}r_{o}^{2}\tau_{i}^{2}}{2 (r_{o}^{2}-r_{i}^{2})}\ln\frac{r_{o}}{r_{i}}\\ C_{3}=-\frac{kK_{1}^{\prime}r_{i}^{2}}{8(r_{o}^{2}-r_{i}^{2})}\left[\left(1+2 \ln\frac{r_{o}}{r_{i}}\right)r_{o}{}^{2}-r_{i}^{2}\right]\\ C_{4}=-\frac{k_{1}^{\prime}r_{i}^{2}}{4}\end{array} \right. \tag{45}\]  
whereupon:  
\[\left\{ \begin{array}{l}\sigma_{rr}^{0}=\frac{kK_{1}^{\prime}}{20}\left[\left( \frac{r_{o}}{r_{o}^{2}-r_{i}^{2}}\right)\left(\frac{r_{o}^{2}-r_{i}^{2}}{r_{o} ^{2}-r_{i}^{2}}\right)\ln\frac{r_{o}}{r_{i}}-\ln\frac{r}{r_{i}}\right]\\ \sigma_{\theta\theta}^{0}=\frac{kK_{1}^{\prime}}{2}\left[\left(\frac{r_{o}}{r }\right)\left(\frac{r_{o}^{2}+r_{i}^{2}}{r_{o}^{2}-r_{i}^{2}}\right)-\ln\frac {r_{o}}{r_{i}}-\ln\frac{r}{r_{i}}-1\right]\end{array} \right. \tag{46}\]  
If it is set at \(r=r_{i}\), \(\tau=\tau_{i}\) and \(\tau=0\) at \(r=r_{o}\), we can obtain from Equation (43):  
\[K_{0}^{\prime}=\frac{\tau_{i}}{\ln\left(\frac{r_{o}}{r_{i}}\right)} \tag{47}\]Substituting Equation (47) into Equation (46), we can obtain:  
\[\left\{ \begin{array}{l}\sigma_{rr}=\frac{{\alpha^{*}E}}{1-\
u}\frac{{\tau_{i}-{ \tau_{0}}}}{2\ln({\tau_{o}}/{\tau_{i}})}\Big{[}-\ln\frac{{\tau_{0}}}{r}+\frac{{ \tau_{i}^{2}}}{{\tau_{o}}^{2}-{\tau_{i}}^{2}}\Big{(}\frac{{\tau_{0}}^{2}}{{\tau _{i}^{2}}}-1\Big{)}\ln\frac{{\tau_{0}}}{{\tau_{i}}}\Big{]}\\ \sigma_{\theta\theta}=\frac{{\alpha^{*}E}}{1-\
u}\frac{{\tau_{i}-{\tau_{0}}}}{2 \ln({\tau_{o}}/{\tau_{i}})}\Big{[}1-\ln\frac{{\tau_{0}}}{r}-\frac{{\tau_{i}^{2} }}{{\tau_{o}}^{2}-{\tau_{i}}^{2}}\Big{(}\frac{{\tau_{0}}^{2}}{{\tau _{i}^{2}}}+1\Big{)}\ln\frac{{\tau_{0}}}{{\tau_{i}}}\Big{]}\\ \end{array} \right. \tag{48}\]  
According to the axial force formula in cylindrical coordinate system [23], we can obtain:  
\[\sigma_{zz}=\frac{E{\alpha^{*}}}{1-\
u}\bigg{(}\frac{2}{{r_{o}}^{2}-{\tau_{i}} ^{2}}\bigg{.}\int_{{\tau_{i}}}^{{r_{0}}}\tau\tau\mathrm{d}r-\tau\bigg{)} \tag{49}\]  
The subsea connector belongs to steady-state axisymmetric heat transfer [23], and its equation is:  
\[\frac{d^{2}\tau}{dr^{2}}+\frac{1}{r}\frac{d\tau}{dr}=0 \tag{50}\]  
When \(r=r_{i}\), \(\tau=\tau_{i}\); when \(r=r_{o}\), \(\tau=\tau_{o}\). Therefore, the solution of this equation is:  
\[\tau=\tau_{i}+(\tau_{o}-\tau_{i})\frac{\ln(r/r_{i})}{\ln(r_{o}/r_{i})} \tag{51}\]  
Substituting Equation (51) into Equation (49), we can obtain:  
\[\int_{{\tau_{i}}}^{{r_{o}}}\tau\mathrm{d}r=\frac{{\tau_{o}}{{r_{o}}^{2}}-{\tau _{i}}{{r_{i}}^{2}}}{2}-\frac{(\tau_{o}-\tau_{i})\big{(}{r_{o}}^{2}-{r_{i}}^{2 }\big{)}}{4\ln(r_{o}/r_{i})} \tag{52}\]  
Substituting Equation (52) into Equation (49), we can obtain:  
\[\sigma_{zz}=\frac{{\alpha^{*}E}}{1-\
u}\frac{{\tau_{i}-{\tau_{o}}}}{2\ln({r_ {o}}/{r_{i}})}\bigg{[}1-2\ln\frac{{\tau_{o}}}{r}-\frac{{2{r_{i}}^{2}}}{{{r_{ o}}^{2}-{r_{i}}^{2}}}\ln\frac{{\tau_{o}}}{{r_{i}}}\bigg{]} \tag{53}\]  
The three-dimensional thermal stress \(\sigma_{r}^{\tau}\), \(\sigma_{\theta}^{\tau}\) and \(\sigma_{z}^{\tau}\) obtained above are arranged as follows:  
\[\left\{ \begin{array}{l}\sigma_{r}^{\tau}=\sigma_{rr}=\frac{{\alpha^{*}E}}{ 1-\
u}\frac{{\tau_{i}-{\tau_{o}}}}{2\ln({\tau_{o}}/{r_{i}})}\Big{[}-\ln\frac{{ \tau_{o}}}{r}+\frac{{\tau_{i}^{2}}}{{\tau_{i}^{2}}-{r_{i}^{2}}}\Big{(}\frac{{ \tau_{o}^{2}}}{{\tau_{i}^{2}}}-1\Big{)}\ln\frac{{\tau_{o}}}{{\tau_{i}}}\Big{]} \\ \sigma_{\theta}^{\tau}=\sigma_{\theta\theta}=\frac{{\alpha^{*}E}}{1-\
u}\frac{ {\tau_{i}-{\tau_{o}}}}{2\ln({\tau_{o}}/{r_{i}})}\Big{[}1-\ln\frac{{\tau_{o}}} {r}-\frac{{\tau_{i}^{2}}}{{\tau_{o}}^{2}-{r_{i}^{2}}}\Big{(}\frac{{\tau_{o}^{2 }}}{{\tau_{i}^{2}}}+1\Big{)}\ln\frac{{\tau_{o}}}{{r_{i}}}\Big{]}\\ \sigma_{z}^{\tau}=\sigma_{zz}=\frac{{\alpha^{*}E}}{1-\
u}\frac{{\tau_{i}-{ \tau_{o}}}}{2\ln({\tau_{o}}/{r_{i}})}\Big{[}1-2\ln\frac{{\tau_{o}}}{r}-\frac{ {2{r_{i}}^{2}}}{{\tau_{o}}^{2}-{r_{i}^{2}}}\ln\frac{{\tau_{o}}}{{r_{i}}}\Big{]} \end{array} \right. \tag{54}\]