## 4 Prediction on Ultimate Bearing Capacity
### Bearing Capacity of Flexural-Torsional Buckling Mode  
For axially compressed members with a single axisymmetric cross-section, their deformation exhibits a state of both bending and torsion. For any section on the member, there is the following equilibrium equation:  
\[EI_{y}u^{(4)}+Pu^{\prime\prime}+Py_{0}\varphi^{\prime\prime}=0 \tag{12}\]  
\[EI_{\alpha}\varphi^{(4)}+(Pi_{0}{}^{2}-GI_{t}+\overline{R})\varphi^{\prime \prime}+Py_{0}u^{\prime\prime}=0 \tag{13}\]  
where \(y_{0}\) is the distance from the centroid to the shear center; \(u\) denotes the deflection function of the rod; \(\varphi\) is the rotation angle function of the member; and \(i_{0}\) and \(\overline{R}\) can be calculated by Equations (14) and (15), respectively. \[i_{0}{}^{2}=(I_{x}+I_{y})/A+y_{0}{}^{2} \tag{14}\]  
\[\overline{R}=\int\limits_{A}\sigma_{r}(x^{2}+y^{2})\mathrm{d}A \tag{15}\]  
where \(\sigma_{r}\) is the residual stress distribution function on the cross-section. The boundary conditions for both ends are hinged, and the following relationship can be obtained:  
\[u(0)=u(l)=u^{\prime\prime}(0)=u^{\prime\prime}(l) \tag{16}\] \begin{table}
\begin{tabular}{c c c c} \hline \hline
**Specimen** & **Tested Result/kN** & **Predicted Result/kN** & **Tested/Predicted** \\ \hline LR2-1-1 & 547.03 & & 1.086 \\ LR2-1-2 & 539.22 & & 1.070 \\ LR2-2-1 & 459.59 & & 0.973 \\ LR2-2-2 & 469.27 & & 0.993 \\ LR2-2-3 & 478.95 & & 1.014 \\ LR2-2-4 & 440.41 & & 0.932 \\ LR2-3-1 & 425.01 & & 0.988 \\ LR2-3-2 & 403.51 & & 0.938 \\ LR2-3-3 & 410.86 & & 0.955 \\ LR2-3-4 & 384.55 & & 0.894 \\ LR2-3-5 & 394.31 & & 0.916 \\ LR2-3-6 & 418.19 & & 0.972 \\ LR3-1-1 & 746.68 & & 0.984 \\ LR3-1-2 & 757.13 & & 0.997 \\ LR3-2-1 & 709.69 & & 1.032 \\ LR3-2-2 & 702.65 & & 1.022 \\ LR3-2-3 & 642.44 & & 0.934 \\ LR3-2-4 & 687.04 & & 0.999 \\ LR4-1-1 & 1126.65 & & 1.025 \\ LR4-1-2 & 1071.64 & & 0.975 \\ LR4-2-1 & 1032.59 & & 1.075 \\ LR4-2-2 & 1002.37 & 960.25 & 1.044 \\ LR4-2-3 & 977.92 & & 1.018 \\ LR5-1-1 & 1540.69 & & 1.076 \\ LR5-1-2 & 1513.52 & & 1.057 \\ LR5-2-1 & 1295.01 & & 1.056 \\ LR5-2-2 & 1242.33 & 1225.84 & 1.013 \\ LR5-2-3 & 1361.17 & & 1.110 \\ \hline \hline \end{tabular}
\end{table} 
Table 4: Prediction result of simplified model. \[\varphi(0)=\varphi(l)=\varphi^{\prime\prime}(0)=\varphi^{\prime\prime}(l) \tag{17}\]  
It is assumed that \(u\) and \(\varphi\) should satisfy the shape of a sine curve:  
\[u=C_{1}\sin(n\pi z/l) \tag{18}\]  
\[\varphi=C_{2}\sin(n\pi z/l) \tag{19}\]  
where \(C_{1}\) and \(C_{2}\) are parameters to be determined. The elastic solution for flexural-torsional instability can be obtained by combining Equations (12)-(19):  
\[P_{y\omega}=\frac{\left(P_{y}+P_{\omega}\right)-\sqrt{\left(P_{y}+P_{\omega} \right)^{2}-4P_{y}P_{\omega}[1-\left(y_{0}/i_{0}\right)^{2}]}}{2[1-\left(y_{0} /i_{0}\right)^{2}]} \tag{20}\]  
where \(P_{y\omega}\) is bending-torsional instability critical load. \(P_{y}\) and \(P_{\omega}\) are bending and torsional instability critical load, which can be determined by Equations (21) and (22), respectively:  
\[P_{y}=\pi^{2}E_{Iy}/l^{2} \tag{21}\]  
\[P_{\omega}=\frac{1}{i_{0}{}^{2}}(\frac{\pi^{2}E_{I\omega}}{l^{2}}+GI_{I}- \overline{R}) \tag{22}\]  
where \(G\) is the shear modulus, and \(I_{t}\) and \(I_{\omega}\) are the torsional and warping inertia moments, respectively. Especially for sections where the plates intersect at one point, \(I_{\omega}\approx 0\), if the influence of residual stress can be ignored, it can be obtained by  
\[P_{\omega}=\frac{GI_{I}}{i_{0}{}^{2}}=\frac{G(I_{ta}+I_{tb})}{i_{0}{}^{2}} \tag{23}\]  
where \(I_{ta}\) and \(I_{tb}\) are the torsional inertia moments of the steel angle and reinforced plate, respectively; and \(I_{t}\) is the sum of both and denotes the reinforced steel angle's torsional inertia moment. It can be seen that the critical load of flexural and torsional instability can be represented by the critical load \(P_{y}\) of flexural instability and the critical load \(P_{\omega}\) of torsional instability.