## 4 Calculation of Bearing Capacity Based on Yield Hinge Line Theory
### Energy Method  
In the case of large deflection, the tension film effect of the plate occurs after the yield hinge line appears, and the reason for the film effect is the plastic work conducted by the elongation of the yield reinforcement at the section of the yield hinge line [19]. Work conducted by reinforcement elongation when the yield hinge line rotates around axes AB and BC:  
\[\int_{0}^{y_{1}}\int_{0}^{v}T_{xT}dyde(x)=\int_{0}^{x_{1}}\int_{0}^{v}T_{xT} \frac{dx}{\tan\varphi}\Bigg{\{}\sqrt{x^{2}+\left[\frac{x}{x_{1T}}(z+dz)\right] ^{2}}-\sqrt{x^{2}+\left(\frac{x}{x_{1T}}z\right)^{2}}\Bigg{\}}=\frac{T_{xT}}{2 }y_{1}\Bigg{(}\frac{v^{2}}{\sqrt{x_{1T}^{2}+v^{2}}+x_{1  
Work conducted by plastic hinge wire AD rotating around axes AO and AB and steel bar elongation:  
\[\frac{T_{xT}}{2}y_{2}\left(\frac{v^{2}}{\sqrt{x_{1T}^{2}+v^{2}+x_{1T}}}\right) \tag{19}\]  
\[\frac{T_{yT}}{2}x_{2}\left(\frac{v^{2}}{\sqrt{y_{1T}^{2}+v^{2}+y_{1T}}}\right) \tag{20}\]  
Work conducted by yield hinge line CD rotating reinforcement elongation around axes BC and OC:  
\[\frac{T_{xT}}{2}y_{1}\left(\frac{v^{2}}{\sqrt{x_{2T}^{2}+v^{2}+x_{2T}}}\right) \tag{21}\]  
\[\frac{T_{yT}}{2}x_{1}\left(\frac{v^{2}}{\sqrt{y_{2T}^{2}+v^{2}+y_{2T}}}\right) \tag{22}\]  
Work conducted by reinforcement elongation when plastic hinge BD rotates around axes BC and AB:  
\[\frac{T_{xT}}{2}y_{2}\left(\frac{v^{2}}{\sqrt{x_{2T}^{2}+v^{2}+x_{2T}}}\right) \tag{23}\]  
\[\frac{T_{yT}}{2}x_{2}\left(\frac{v^{2}}{\sqrt{y_{2T}^{2}+v^{2}+y_{2T}}}\right) \tag{24}\]  
Work conducted by reinforcement elongation when yield hinge line DE rotates around axes AO and BC:  
\[\int_{x_{1}}^{L-x_{2}}\int_{0}^{\infty}T_{yT}dxde(y)=\int_{x_{1}}^{L-x_{2}}\int _{0}^{\infty}T_{yT}dx\left\{\sqrt{y^{2}+\left[\frac{y}{y_{1T}}(z+dz)\right]^{2 }}-\sqrt{y^{2}+\left(\frac{y}{y_{1T}}z\right)^{2}}\right\}=T_{yT}(L-x_{1}-x_{2} )\left(\frac{v^{2}}{\sqrt{y_{1T}^{2}+v^{2}+y_{1T}}}\right) \tag{25}\]  
\[T_{yT}(L-x_{1}-x_{2})\left(\frac{v^{2}}{\sqrt{y_{1T}^{2}+v^{2}+y_{1T}}}\right) \tag{26}\]  
where \(x_{iT}=x_{i}+\alpha_{1}(T-20)x_{i}\), \(y_{iT}=y_{i}+\alpha_{1}(T-20)y_{i}\), where \(\alpha_{1}\) is coefficient of thermal expansion; \(x_{i}\) and \(y_{i}\) are obtained by the plate balance method.  
The work performed by uniformly distributed external load can be expressed:  
\[W=\frac{1}{6}qlv[3L-(x_{1}+x_{2})] \tag{27}\]  
The work performed by resultant moment:  
\[D_{1}=m_{xT}l\frac{v}{x_{1}}+m_{xT}l\frac{v}{x_{2}}+m_{yT}x_{1}\frac{v}{y_{1} }+m_{yT}x_{2}\frac{v}{y_{1}}+m_{yT}x_{1}\frac{v}{y_{2}}+m_{yT}x_{2}\frac{v}{y_ {2}}+m_{yv}L\frac{v}{y_{1}} \tag{28}\]  
Work performed by reinforcement elongation:  
\[D_{2}=\frac{T_{xT}}{2}l\left(\frac{v^{2}}{\sqrt{x_{1T}^{2}+v^{2}+x_{1T}}} \right)+\frac{T_{xT}}{2}l\left(\frac{v^{2}}{\sqrt{x_{2T}^{2}+v^{2}+x_{2T}}} \right)+\frac{T_{xT}}{2}(2L-x_{1}-x_{2})\left(\frac{v^{2}}{\sqrt{y_{1T}^{2}+v ^{2}+y_{1T}}}\right)+\frac{T_{xT}}{2}(2L-x_{1}-x_{2})\left(\frac{v^{2}}{\sqrt{ y_{2T}^{2}+v^{2}+y_{2T}}}\right) \tag{29}\]  
Work performed by external force is equal to work done by internal force:  
\[W=D_{1}+D_{2} \tag{30}\]The limit carrying capacity of two-way slabs with three simply supported edges and one clamped edge can be obtained from Equations (27)-(29):  
\[q=\frac{6}{l[3L-(x_{1}+x_{2})]}\Bigg{\{}\frac{m_{xT}l}{x_{1}}+\frac{m_{xT}l}{x_{2 }}+\frac{m_{yT}L}{y_{1}}+\frac{m_{yT}L}{y_{2}}+\frac{m_{ye}L}{y_{1}}+\frac{m_{ ye}L}{y_{2}}+\sum_{i=1}^{4}m_{pi}\Bigg{\}} \tag{31}\]  
\[m_{p1}=\frac{T_{yT}}{2}(2L-x_{1}-x_{2})\left(\frac{v-v_{0}}{\sqrt{y_{2T}^{2}+v ^{2}+y_{2T}}}\right) \tag{32}\]  
\[m_{p2}=\frac{T_{xT}}{2}l\left(\frac{v-v_{0}}{\sqrt{x_{1T}^{2}+v^{2}}+x_{1T}}\right) \tag{33}\]  
\[m_{p3}=\frac{T_{yT}}{2}(2L-x_{1}-x_{2})\left(\frac{v-v_{0}}{\sqrt{y_{1T}^{2}+ v^{2}}+y_{1T}}\right) \tag{34}\]  
\[m_{p4}=\frac{T_{xT}}{2}l\left(\frac{v-v_{0}}{\sqrt{x_{2T}^{2}+v^{2}}+x_{2T}}\right) \tag{35}\]