# A Minimal GBT Model for Distortional-Twist Elastic Analysis of Box-Girder Bridges
## Appendix A The Vlasov Stresses in Rectangular Box-Girders under Nonuniform Torsion  
The rectangular box-girder in Figure 1 was considered, and stresses due to nonuniform torsion were calculated according to the Vlasov theory. The warping function (Equation (1)) is as follows:  
\[\omega(s)=\begin{cases}\frac{1}{2}\frac{bt_{2}-ht_{1}}{bt_{2}+ht_{1}}hs_{i}&i=1,3\\ -\frac{1}{2}\frac{bt_{2}-ht_{1}}{bt_{2}+ht_{1}}hs_{i}&i=2,4\end{cases} \tag{1}\]  
from which the warping inertia \(I_{\omega}=\oint_{\Gamma}\omega^{2}(s)t(s)ds\) is computed as follows:  
\[I_{\omega}=\frac{b^{2}h^{2}}{24}\frac{\left(ht_{1}-bt_{2}\right)^{2}\left(bt_{ 1}+ht_{2}\right)}{\left(ht_{1}+bt_{2}\right)^{2}} \tag{2}\]  
The normal stresses are evaluated via \(\sigma_{z}(z,s)=-\frac{B(z)}{I_{\omega}}\omega(s)\), with \(B(z)=E\theta^{\prime\prime}(z)\), thus obtaining the following:  
\[\sigma_{z}(z,s)=\begin{cases}E\theta^{\prime\prime}(z)\frac{h}{2}\frac{ht_{2} -ht_{1}}{bt_{2}+ht_{1}}s_{i}&i=1,3\\ -E\theta^{\prime\prime}(z)\frac{b}{2}\frac{bt_{2}-ht_{1}}{bt_{2}+ht_{1}}s_{i} &i=2,4\end{cases} \tag{3}\]The _active_ tangential stresses (indexes omitted) are given by the Bredt formula \(\tau^{a}(z,s)=\frac{M_{t}(z)}{G\Omega_{0}t(s)}\), with \(M_{t}(z)=GJ\theta^{\prime}(z)\), \(\Omega_{0}=bh\) and \(J=\frac{4(bh)^{2}}{2\left(\frac{h}{t_{1}}+\frac{h}{2}\right)}\), leading to the following:  
\[\tau^{a}(z,s)=\begin{cases}G\frac{hht_{2}}{H_{1}+b_{2}}\theta^{\prime}(z)&i=1, 3\\ G\frac{hht_{1}}{H_{1}+b_{2}}\theta^{\prime}(z)&i=2,4\end{cases} \tag{10}\]  
The _reactive_ tangential stresses \(\tau^{r}(z,s)\) must be evaluated by integrating the indefinite equilibrium equation \(\left(\tau^{r}t\right)_{s}+\left(\sigma_{z}t\right)_{z}=0\). By accounting for \(\sigma_{z,z}=-E\theta^{\prime\prime\prime}(z)\omega(s)\) and remembering that \(t(s)\) is step-wise constant, it follows that  
\[\tau^{r}(z,s_{i})=\begin{cases}\left(-\frac{1}{4}\frac{bI_{2}-ht_{1}}{bI_{2}+ ht_{1}}ls_{1}^{2}+c_{1}\right)E\theta^{\prime\prime\prime}(z)&i=1,3\\ \left(\frac{1}{4}\frac{bI_{1}-ht_{2}}{bI_{2}+ht_{1}}bs_{1}^{2}+c_{2}\right)E \theta^{\prime\prime\prime}(z)&i=2,4\end{cases} \tag{11}\]  
where \(c_{1},c_{2}\) are arbitrary constants. By enforcing the continuity of the flow at joints, i.e., \(t_{1}\tau^{r}\left(z,s_{1}=\frac{h}{2}\right)=t_{2}\tau^{r}\left(z,s_{2}=- \frac{h}{2}\right)\), and the equivalence of the reactive stresses to \(M_{t}^{*}(x)=-EI_{\omega}\theta^{\prime\prime\prime}\), i.e.,  
\[h\int\limits_{-b/2}^{b/2}\tau^{r}(z,s_{1})t_{1}ds_{1}+b\int\limits_{-h/2}^{h/2 }\tau^{r}(z,s_{2})t_{2}ds_{2}=-EI_{\omega}\theta^{\prime\prime\prime} \tag{12}\]  
a linear system in the two unknown constants is derived:  
\[\left(\begin{array}{cc}t_{1}&-t_{2}\\ t_{1}&t_{2}\end{array}\right)\left(\begin{array}{c}c_{1}\\ c_{2}\end{array}\right)=\frac{1}{bh}E\left(\begin{array}{cc}\frac{1}{16}b^{ 2}h^{2}\beta(bt_{1}+ht_{2})\\ -I_{\omega}+\frac{1}{48}b^{2}h^{2}\beta(bt_{1}-ht_{2})\end{array}\right) \tag{13}\]  
By solving for \(c_{1},c_{2}\) and coming back to Equation (11), Equation (25) is found. It should be noticed that, if \(\frac{h}{b}=\frac{t_{2}}{t_{1}}\), it is \(M_{t}^{*}=0\), then \(\tau_{z_{3}}^{r}(z,s)\equiv 0\).