# Study on Load-Slip Curve of a PBL Shear Key at a Steel-Concrete Composite Joint
## 3 Formula for the Load-Slip Curve of the PBL Shear Key
### Analytical Formula of Load-Slip Curve
#### 3.2.1 Elastic Stage  
In the elastic state, the steel bar is equivalent to the elastic foundation beam, and the surrounding concrete is equivalent to the elastic foundation. In order to not significantly increase the workload and difficulty of calculations and analyses, the following assumptions are introduced: (1) The horizontal shear force of the PBL shear key acts directly on the coordinates of the origin rather than the interface position between the joint plate and concrete. (2) The joint plate's orifice can well restrain the concrete in the hole and the penetrating steel bar, and the interface between the penetrating steel bar and the joint plate does not rotate during the entire process of the PBL shear key's stress (rotation angle \(\Phi_{x=0}=0\)). (3) Concrete is a continuous elastic body, and concrete can be regarded as the elastic foundation for penetrating steel bars. (4) Bonding actions between steel and concrete and all vertical actions are not considered. (5) The penetrating steel bar is in accordance with the plane section assumption in the process of deformations. The calculation diagram of the free segment is shown in Figure 5, where \(\overline{p}\) is the unit length reaction force of the steel bar at any point \(x\), and it is a function of deflection \(y\) of the steel bar. \(\overline{q}(x)\) indicates the reverse action of steel bars per unit length. Taking element \(dx\), as shown in Figure 5, and combining the mechanical analysis of the material, the flexural differential equation for the steel bar is obtained:  
\[EI\frac{d^{4}y}{dx^{4}}=-\overline{p}(x,y)+\overline{q}(x) \tag{1}\]  
\[p(x,y)=\frac{\overline{p}(x,y)}{d_{\text{PBL}}} \tag{2}\]  
Among them, \(E\) denotes the elastic modulus (N/mm\({}^{2}\)), and \(I\) denotes the moment of inertia of the cross section of the steel bar (mm\({}^{4}\)). Considering that the characteristic of concrete is different from that of soil, the back load, \(\overline{q}(x)\), in Formula (1) is 0. With regard to the calculation method of concrete reaction \(p(x,y)\), referring to the foundation reaction's method of consideration, the elastic foundation reaction method is adopted. By using the bending theory of the beam, the following formula is obtained:  
\[p(x,y)=K(x)y^{n} \tag{3}\]  
Figure 5: Free section calculation diagram. where \(K(x)\) denotes any function of depth \(x\), and \(n\) denotes the index of deflection, \(0<n\leq 1\). Because of the characteristics of concrete, \(K(x)\) is constant in the formula, and both the steel bar and concrete are in a stage of linear elasticity; thus, the index is \(n=1\) in Formula (3). \[p(x,y)=ky \tag{4}\]  
Among them, we have the concrete reaction coefficient (N/mm\({}^{3}\)). In order to calculate the value of \(k\), concrete is regarded as an elastic body, and the formula of concrete reaction coefficient \(k\) is obtained with reference to [PERSON] [15] and [PERSON]'s [16] solution:  
\[k=\frac{22.25E_{c}(1-v_{c})}{(1+v_{c})(3-4v_{c})\Big{(}4.6\log_{10}\frac{2l}{ d_{\text{PBL}}}-0.44\Big{)}d_{\text{PBL}}} \tag{5}\]  
In the formula, \(v_{c}\) denotes the concrete's Poisson's ratio; \(l\) denotes the calculated length of the free section of the steel bar (mm). By introducing Equation (4) into Equation (1), a fourth-order differential equation with a constant coefficient can be obtained:  
\[EI\frac{d^{4}y}{dx^{4}}+d_{\text{PBL}}ky=0 \tag{6}\]  
The general solution can be easily obtained as follows:  
\[y=e^{gx}(C_{1}\cos\beta x+C_{2}\sin\beta x)+e^{-\beta x}(C_{3}\cos\beta x+C_{4 }\sin\beta x) \tag{7}\]  
\[\beta=\sqrt[4]{\frac{d_{\text{PBL}}k}{4 EI}} \tag{8}\]  
In the formula, \(C_{1}\), \(C_{2}\), \(C_{3}\), and \(C_{4}\) denote integral constants, and \(\beta\) denotes the eigenvalue of penetrating steel bars (mm\({}^{-1}\)). The special point of the force acting on the joint penetrating steel bar, the assumption in this paper, and the boundary conditions can be determined as follows:  
\[\left\{ \begin{array}{l}\varphi_{x=0}=0\\ S_{x=0}=F_{s}\\ M_{x=l}=0\\ S_{x=l}=0\end{array} \right.