# Axial and Bending Bearing Capacity of Double-Steel-Concrete Composite Shear Walls
## 3 Numerical Analysis
### FE Model
The value of \(m\) is generally between 0.4 and 1.5, and the value of \(n\) is generally between 0.5 and 2.0. [PERSON] and [PERSON] [29] conducted two horizontal push-out shear tests of 19 mm diameter stud. Using the Equation (2) proposed by [PERSON] et al. [28] to fit the experimental value of the shear-slip curve in [29], when \(m=0.425\), \(n=0.5\), the two coincide. Another parameter that needs to be determined is the ultimate shear strength \(V_{u}\). The test results of [29] are in good agreement with the calculation results of Eurocode-4 [30]. The equation proposed by Eurocode-4 was applied in this paper to calculate \(V_{u}\), which is expressed as:  
\[V_{u}=\text{ min}\Big{\{}0.8f_{u}A,\ 0.29 ad^{2}\sqrt{f_{d}E_{cm}}\Big{\}} \tag{3}\]  
where \(f_{u}\) is the ultimate tensile strength of the stud, \(A\) is the section area of the stud, \(d\) is the diameter of the stud, \(f_{ck}\) is the compression strength of concrete cylinder, \(E_{cm}\) is the young's modulus of concrete, \(\alpha=0.2(h/d+1)\), and \(h\) is the height of the stud. The relationship between shear force and displacement can be obtained by Equations (1) and (2) and can be used as the nonlinear spring stiffness. According to [PERSON] et al. 's review [31] of the tensile failure of stud, there are generally three types of failure, namely, stud fracture, concrete cone pull-out failure, and concrete column pull-out failure. According to American Concrete Code 318-08 (ACI 318-08), when the diameter of the stud cap is greater than 1.71 times the diameter of the stud rod, there is a 95% probability that the concrete column will not be pulled out. When the length of the stud is greater than 7.5 times the diameter of the stud, the concrete cone will not be pulled out [31]. Therefore, the brittle failure of the concrete can be completely avoided by construction measures. The damage caused by the stud itself breaking is ductile. The bolt is equivalent to a tension member with one end fixed and one end freely stretched, so the stiffness of the stud can be used to calculate the stiffness of the axial spring, which is expressed as follows:  
\[K=\frac{EA}{h} \tag{4}\]  
where \(E\) is the young's modulus of the stud, \(A\) is the section area of the stud and \(h\) is the height of the stud. The steel plates were embedded in the loading beam and the base beam. The studs and tie bars were embedded in the concrete. The contact feature was used between the elements of the concrete wall and the steel plate that was not inserted in the loading beam and the base beam. The contact properties were frictional contact in the tangential direction and hard contact in the normal direction, respectively. The boundary conditions are shown in Figure 2b. The FE model was fixed on the ground by two springs with the spring stiffness of EA/L, where A and L were the bolt section area and length, respectively. The contact between the base beam and the foundation was simulated by an elastic foundation model. The reaction force coefficient of the concrete foundation is generally 7484.6-14,715 kN/mm\({}^{3}\)[32]. In the simulation, the reaction force coefficient was 10,000 kN/mm\({}^{3}\). At both ends of the model foundation beam, the horizontal displacement of the wall in and out of the wall was constrained, and the vertical displacement was not constrained. The concrete damaged plasticity model was applied to simulate the behavior of concrete. The energy equivalence principle of Sidiroff was applied to calculate the damage factor. The ideal elastoplastic stress-strain curve was applied to model the behavior of steel plates and tie bars.