# Mechanical Properties of Disconnectable Coupling Joints for Steel Bracing under Eccentric Load
## 6 Moment-Rotation Model
### Determination of Parameters
#### 6.2.2 Determination of \(\theta_{0}\)  
Because the deformation of the middle plate is slight, the point A is regarded as the rotation center. \(\theta\) represents the angle of the axis rotating around point A; it equals the angle before and after the deformation of the upper surface of the steel wedge, as shown in Figure 26. If \(\Delta_{1}\) is the vertical deformation at point B and \(\Delta_{2}\) is the vertical deformation at point C, then \(\theta\) meets the following requirements:  
\[\theta=\arctan\biggl{(}\frac{\Delta_{1}-\Delta_{2}}{l}\biggr{)} \tag{7}\]  
\(\theta_{0}\) is the rotation angle as the joint begins to produce plastic strain. When the joint reaches \(\theta_{0}\), the corresponding load is called the ultimate bending load. The ultimate bending load \(F_{i}\) meets the following formula:  
\[F_{i}=\frac{M_{i}}{e} \tag{8}\]  
When the eccentricity is less than one-sixth of the steel wedge length, the DC joint is a small eccentric compression member. Otherwise, it is a large eccentric compression member. The different force forms of steel wedges under different eccentricities are shown in Figure 28. Figure 28a shows the force form with small eccentricity, and Figure 28b shows the force form with large eccentricity. The parameters in Figure 28 are calculated using the following formulas:  
\[ \begin{cases}P_{0}=\frac{F}{\varpi l}\bigl{(}1+\frac{6\epsilon}{T}\bigr{)}& \left(e\leq\frac{l}{6}\right)\\ P_{0}=\frac{2F}{3\varpi l}&\left(e>\frac{l}{6}\right)\\ P_{1}=\frac{F}{\varpi l}\bigl{(}1-\frac{6\epsilon}{T}\bigr{)}&\\ \end{cases} \]  
(1)Calculation of \(\Delta_{1}\)  
According to [PERSON]'s equation, the elasticity solution of the displacement at any point in the half-space can be obtained under the action of the vertical force, as follows:  
\[\omega(x,y)=\frac{1-\mu^{2}}{\pi E}\iint\limits_{A}\frac{p(\xi,\eta)}{\sqrt{ \left(x-\xi\right)^{2}+\left(y-\eta\right)^{2}}}d\xi d\eta \tag{9}\]  
If the deformation of the steel wedges at point B is \(\Delta_{1}\), and the coordinates \(x=0\) and \(y=0\) of point B are taken into Equation (9), the calculation of \(\Delta_{1}\) is as follows:  
\[\Delta_{1}=\omega(0,0)=\frac{1-\mu^{2}}{\pi E}\int_{0}^{w}\int_{0}^{b}\frac{p (\xi,\eta)}{\sqrt{\xi^{2}+\eta^{2}}}d\xi d\eta\]  
Figure 28: Force form of steel wedge. The result is as follows:  
\[\Delta_{1}=\frac{1-\mu^{2}}{\pi E}{\int_{0}^{w}{P_{0}\left[\ln\left(\frac{b+\sqrt{ b^{2}+\eta^{2}}}{\eta}\right)\right]}d\eta}+\frac{1-\mu^{2}}{\pi E}{\int_{0}^{w}{z \left(\eta-\sqrt{b^{2}+\eta^{2}}\right)}d\eta}\]  
Order, \(\Delta_{1,1}=\frac{1-\mu^{2}}{\pi E}{\int_{0}^{w}{P_{0}\left[\ln\left(\frac{b+ \sqrt{b^{2}+\eta^{2}}}{\eta}\right)\right]}d\eta}\), \(\Delta_{1,2}=\frac{1-\mu^{2}}{\pi E}{\int_{0}^{w}{z\left(\eta-\sqrt{b^{2}+\eta^ {2}}\right)}d\eta}\). According to the Gauss-Legendre quadrature formula, \(\Delta_{1,1}\) can be obtained as follows:  
\[\Delta_{1,1}=\frac{(1-\mu^{2})P_{0}}{\pi E}\cdot\frac{w}{2}\cdot\left[\frac{5 }{9}f{\left(-\frac{\sqrt{15}}{5}\right)}+\frac{8}{9}f{\left(0\right)}+\frac{5 }{9}f{\left(\frac{\sqrt{15}}{5}\right)}\right]\]  
where \(f(x)=\ln\left[\frac{b+\sqrt{b^{2}+\left(\frac{w}{2}+\frac{w}{2}\right)^{2}}}{ \frac{w}{2}+\frac{w}{2}}\right]\). \(\Delta_{1,2}\) is further solved as follows:  
\[\Delta_{1,2}=\frac{1-\mu^{2}}{\pi E}\cdot z\left[\frac{w^{2}}{2}-\frac{w}{2} \sqrt{b^{2}+w^{2}}-\frac{b^{2}}{2}\ln\left(\frac{w+\sqrt{b^{2}+w^{2}}}{b} \right)\right]\]  
Among them, \(z=\left\{ \begin{array}{ll}\frac{P_{0}}{3a}&e>\frac{l}{6}\\ \frac{P_{0}-P_{1}}{l}&e\leq\frac{l}{6}\end{array} \right.\), \(b= \begin{cases}3a&e>\frac{l}{6}\\ l&e\leq\frac{l}{6}\end{cases} \).