## 3 Discussion on the Existing Shear Capacity Formula
### Force Transmission Mechanism  
Shown as in Table 3, for the simulation example in this paper, no matter the lateral prestress level is 1.0 MPa or 8.0 MPa, the calculated shear capacity of the example specimens had no obvious change when the friction coefficient ranged from 0.4 to 1.0. This conclusion is inconsistent with AASHTO provisions.  
\begin{table}
\begin{tabular}{c c c c c c} \hline \hline
**Lateral Preload** & **1** & **2** & **3** & **4** & **6** & **8** \\ \hline AASHTO & 83.37 & 102.58 & 121.78 & 140.99 & 179.40 & 217.81 \\ Turmo & 54.01 & 68.41 & 82.81 & 97.21 & 126.01 & 154.81 \\ ATEP & 69.61 & 92.41 & 115.21 & 138.01 & 181.31 & 229.21 \\ Rombach & 71.10 & 84.10 & 97.10 & 110.10 & 136.10 & 162.10 \\ Bakhoum & 110.56 & 137.76 & 164.96 & 192.16 & 246.26 & 300.96 \\ \(V_{E}\) & 89.70 & 113.87 & - & - & - & - \\ \(V_{\text{F}}\) & 82.3 & 105.3 & 124.4 & 140.2 & 159.5 & 175.7 \\ \hline \hline \end{tabular}
\end{table}
Table 2: Calculation results of various formulas under different lateral prestress.  
Figure 8: Calculation results of various formulas under different lateral prestress.  
The friction force status of the joint interface at the different stage is shown in Figure 9, in which the friction coefficient is 0.6, and the lateral prestress is 0.2 MPa. Apparently, when there was only lateral prestress applied, the friction force on the joint interface is almost zero (Figure 9a). When the direct shear force applied by loading displacement, there was friction appeared on surface 4 (see Figure 1), but not much friction on the other surfaces of the joint interface (Figure 9b). When the joint was reaching its ultimate shear capacity, the maximum friction on surface 4 is up to 16.3 MPa, but zero on the other surfaces (Figure 9c). Accordingly, the force state on the joint interface could be simplified as Figure 9d.  
In Figure 9d, \(F_{\mathbf{s4}}\) and \(F_{\mathbf{n4}}\) are the friction force and normal pressure on surface 4, respectively. \(V_{\mathbf{c}}\) is the vertical force on key root section. Take the key as isolated body, it is easy to know that the vertical resistance on the key root section is equal to the sum of vertical components of \(F_{\mathbf{s4}}\) and \(F_{\mathbf{n4}}\).Integrate the results of finite element simulation analysis under ultimate conditions, there is \(F_{\mathbf{s4y}}=17.1 kN\), \(F_{\mathbf{n4y}}=65.2 kN\), \(V_{\mathbf{n}}=82.3 kN\). Compared with the data listed in Table 1, the ultimate shear capacity of the keyed dry joint \(V_{\mathbf{u}}=V_{\mathbf{c}}=F_{\mathbf{s4y}}+F_{\mathbf{n4y}}\). Besides, according to the value of \(F_{\mathbf{s4y}}\) and \(F_{\mathbf{n4y}}\), the corresponding friction coefficient \(\mathbf{\mu}\) on the interface is 0.35. It means that when friction coefficient is over 0.35, the joint interface is in the static friction equilibrium state. The function of the friction force is only to transfer force between adjacent joints; the ultimate shear capacity of the joint does not increase with the increase of friction coefficient. Namely, the shear capacity of the keyed dry  
Figure 9: Vertical force component of tangential force at the contact surface; (**a**) under lateral prestress; (**b**) 0.004 \(V_{u}\) state; (**c**) under ultimate load \(V_{u}\); (**d**) diagram of forces on the interface.  
joint is only provided by the concrete key, which is related to the average shear stress \(\tau\) and area of the key root section \(A_{k}\).