## 2 Analysis of Pre-Tightening Effects on U-Shaped Steel Clamping
### Calculation of Joint Slip Resistance  
Considering the inner U-steel section as a free-body diagram for force analysis, the cable-clip connector was connected to the U-steel via two bolts. The total pre-tightening force was \(2F\), and its action line coincided with the center line of the U-steel arch. The angle between the inclined surface of the U-steel and the vertical line is \(\alpha\). The supporting force of the outer U-steel section on the inner U-steel section was divided into two parts: 1 The force acting under the upper U-steel ear, \(n_{1}\), is the force distribution coefficient between the U-steel sections, \(n_{1}F\); 2 The force \(N\) acts on the side of the upper U-steel section, and its direction is perpendicular to the inclined surface of the side. According to the equilibrium condition, we can obtain the following:  
\[2F-2n_{1}F-2N\sin\alpha=0 \tag{1}\]  
Therefore, the following can be expressed:  
\[N=(1-n_{1})\frac{F}{\sin\alpha} \tag{2}\]  
Taking the outer U-steel section as a free-body diagram for force analysis, as shown in Figure 2b, according to the law of action and reaction, the forces acting on both sides of the outer U-steel section from the inner U-steel section are \(N\) and \(n_{1}F\). The supporting force of the cable-clip on the lower U-steel section is divided into two parts; 1 the force acting under the ear of the outer U-steel section, \(n_{2}\), is the supporting force distribution coefficient of the cable-clip between the U-steel sections, \(n_{2}F\), and 2 the force P acts on the side of the outer U-steel section in a direction perpendicular to the inclined surface of the side. According to the equilibrium condition, we can obtain the following:  
\[2n_{1}F+2(1-n_{1})\frac{F}{\sin\alpha}\sin\alpha=2n_{2}F+2P\sin\alpha \tag{3}\]  
Therefore, the following also applies:  
\[P=(1-n_{2})\frac{F}{\sin\alpha} \tag{4}\]  
Figure 2: Mechanical model of pre tensioned cable clamping nodes. The connector connects the two U-steel sections at the joint into one body. Suppose the friction coefficient between the two U-steel sections is \(f\), then the friction force between the connector and the inner U-steel section is \(2F\!f\), and the friction force between the connector and the outer U-steel section is \(2\left(n_{2}+\frac{(1-n_{2})}{\sin\alpha}\right)F\!f\). Therefore, the friction force between the connector and the inner U-steel section is less than that between the connector and the outer U-steel section, and the connector and the inner U-steel section experience relative sliding. At this moment, the joint slip resistance is as follows:  
\[S=2(2 Ff+2n_{1}Ff+2 Nf) \tag{5}\]  
Substituting Equation (2) into Equation (5) and simplifying, we can obtain:  
\[S=4F\!f\!\left(1+n_{1}+\frac{1-n_{1}}{\sin\alpha}\right) \tag{6}\]  
The relationship through the bolt connection between the pre-tightening force \(F\) of the joint bolt and the pre-tightening torque \(M\) was calculated. The torque \(M_{1}\) applied by the torque wrench is balanced by the torque \(M_{2}\) that prevents the nut from rotating on the bolt and the friction torque that prevents the nut from rotating within the connector. Thus, we can obtain the following:  
\[F=\frac{M}{tg(\lambda+\rho^{\prime})\frac{d_{2}}{2}+f\!\left(\frac{D_{1}+d_{0 }}{4}\right)} \tag{7}\]  
where \(\lambda\) is the thread lead angle, \(\lambda=tg^{-1}\frac{l_{0}}{\pi d_{2}}\); \(\rho^{\prime}\) is the equivalent friction angle, \(\rho^{\prime}=tg^{-1}f^{\prime}\); \(f^{\prime}\) is the equivalent friction coefficient, \(f^{\prime}=1.155f_{1}\); \(t_{0}\) is the thread lead; \(d_{2}\) is the pitch diameter of the thread; \(f_{1}\) is the friction coefficient between the bolt and the steel plate, generally taken as 0.15-0.2; \(d_{0}\) is the bolt hole diameter on the connected object; \(D_{1}\) is the diameter of the inscribed circle on the hexagonal nut.