## 3 Analytical Solution for Anchor Load Transfer
### Plastic State Solution  
When the pullout load surpasses the elastic limit load (\(P_{tk}\)), the anchorage body transitions into a state that is partially elastic and partially plastic. In this state, the expressions for displacement, axial force, and shear stress of the anchorage body should be segmented. The elastic part can follow Equations (7)-(9) seamlessly, while the subsequent section deals with the plastic part of the solution expressions. Reference [25] indicates that the shapes of the distribution curves of axial force and shear stress in the anchorage body are influenced by the magnitude of the applied load, whereas the shapes of the distribution curves of shear displacement remain stable. Therefore, based on the measured data of the anchorage body shear displacement in the case study by [PERSON] et al. [26], the expression for the shear displacement distribution curve is obtained through fitting. The fitting results, shown in Figure 4, demonstrate an ideal fit. The expression for anchorage body shear displacement is as follows:  
\[s_{\text{s}}(x)=e^{a+bx+cx^{2}} \tag{16}\]  
where \(a\), \(b\), and \(c\) are coefficients to be determined.  
Bringing Equation (16) into Equations (2) and (3) yields expressions for the distribution of axial and shear stresses in the plastic part of the anchorage body:  
\[p_{\text{s}}(x)=-EA\frac{ds}{dx}=-EA(2 cx+b)e^{a+bx+cx^{2}} \tag{17}\]  
\[\tau_{\text{s}}(x)=-\frac{1}{U}\frac{dp}{dx}=\frac{EA}{U}(2c+(2 cx+b)^{2})e^{a+bx +cx^{2}} \tag{18}\]  
From Figure 3b, the boundary conditions for the plastic stage are \(s_{\text{s}}(x)\big{|}_{x=L_{1}}=S_{1}\), \(\tau_{\text{s}}(x)\big{|}_{x=L_{1}}=\tau_{1}\), \(p_{\text{t}}(x)\big{|}_{x=L}=0\), \(s_{\text{t}}(x)\big{|}_{x=L_{1}}=S_{1}\), and \(p_{\text{t}}(x)\big{|}_{x=L_{1}}=p_{\text{s}}(x)\big{|}_{x=L_{1}}\), where \(L_{1}\) is the length of the plastic part. The coefficients to be determined for the elastic part \(C_{1}\), \(C_{2}\) and the coefficients to be determined for the plastic part \(a\), \(b\), \(c\) can be solved as follows:  
\[\left\{\begin{array}{l}C_{1}=\frac{S_{1}}{e^{AL_{1}}+e^{AL_{2}}-L_{1}}\\ C_{2}=\frac{S_{1}e^{2 AL}}{e^{AL_{1}}+e^{2 AL_{2}}-L_{1}}\\ a=\ln S_{1}-\frac{aL_{1}(e^{2 AL_{1}}-e^{4 AL_{2}}-2 aL_{1})-2a^{2}L_{1}^{2}e^{2 AL }}{\left(e^{AL_{1}}+e^{2 AL_{1}}-e^{2 AL_{1}}\right)^{2}}\\ b=-\frac{\alpha\left(e^{AL_{1}}-2 aL_{1}-e^{2 AL_{1}}\right)+4a^{2}L_{1}e^{2 AL }}{\left(e^{AL_{1}}+e^{2 AL_{2}}-4L_{1}\right)^{2}}\\ c=\frac{2a^{2}e^{2 AL}}{\left(e^{AL_{1}}+e^{2 AL_{1}}-e^{2 AL_{1}}\right)^{2}} \end{array}\right. \tag{19}\]  
As the pullout load increases, the anchor will reach the plastic limit state when the displacement at the top of the anchored section reaches \(S_{2}\), which is \(s_{\text{s}}(x)|_{x=0}=S_{2}\); simultaneously, the value of the coefficient to be determined \(a\) can be solved, and at this time the pull-out load is called the plastic limit load \(P_{\text{sk}}\). Bring the coefficient of determination \(a\) into Equation (19) to solve for the value of \(L_{1}\), which is the limiting length of the plastic part in the plastic limit state \(L_{\text{s}}\), and bring \(L_{\text{s}}\) into \(p_{\text{s}}(x)|_{x=0}=P\) to solve the plastic limit load \(P_{\text{sk}}\).  
Figure 4: Displacement curves of anchorage body.