# Strain-Softening Analyses of a Circular Bore under the Influence of Axial Stress
## 3 Strain-Softening Analyses and Solution Procedure
### \(\sigma_{z}\): _Ist Principle Stress_
To get \(F_{i}\) and \(G_{i}\), Equation (26) is used:  
\[\frac{\Delta u_{(i)}}{r}=\Delta\varepsilon_{\theta(i)}=\Delta\varepsilon^{e}_ {\theta(i)}=\frac{1}{E}[\Delta\sigma_{\theta(i)}-\
u(\Delta\sigma_{z(i)}+ \Delta\sigma_{r(i)})] \tag{26}\]  
From Equations (9) and (11), Equation (26) can be written as  
\[\frac{\Delta u_{(i)}}{r}=\frac{1}{E}(r\frac{d\Delta\sigma_{r(i)}}{dr}+f_{4} \Delta\sigma_{r(i)}+f_{5}) \tag{27}\]  
with \(f_{4}=1-\
u(1+\alpha_{i})\), \(f_{5}=-\
u(\alpha_{i}\sigma_{r(i-1)}+Y_{i}-\sigma_{z(i-1)})\). By substituting Equation (25) into Equation (27), Equation (28) is obtained:  
\[\Delta u_{(i)}=\frac{r}{E}[F_{i}\cdot(n_{1}+f_{4})r^{n_{1}}+G_{i}\cdot(n_{2}+ f_{4})r^{n_{2}}+f_{4}H+f_{5}] \tag{28}\]  
Equations (25) and (28) will be used to obtain incremental stress and displacement in the plastic region. For the \(i\)th load increment, the total number of the plastic annulus is \(i\). In each plastic annulus, \(F_{i},~{}G_{i}\) and the outer radius (\(r_{i}=R\)) of the \(i\)th plastic annulus need to be solved. Therefore, total number of unknown variables is \(2i+1\). The radial stress and displacement are continuous between the adjacent plastic annuli, so there are \(2(I-1)\) equations. In addition, at \(r=r_{0}\), \(\Delta\sigma_{r(i)}=\Delta p_{i}\). At \(r=r_{i}=R\), \(\sigma_{r}=p_{e}\)\(u=u_{c}\). In the first case, \(p_{c}=p_{c1}\), \(u_{c}=u_{c1}=\frac{\kappa}{E}(1+\
u)(\sigma_{0}-p_{c1})\). In total, there are \(2i+1\) equations. So far, we have the same number of equations as that of unknown variables. The abovementioned equations form a set, and the unknown variables are solved by iterative methods. As the pressure \(p_{in}\) decreases further, \(\sigma_{z}\) decreases and \(\sigma_{\theta}\) increases. Once \(\sigma_{\theta}=\sigma_{z}\), the plastic area is divided into two. For the outer plastic zone with \(\sigma_{z}>\sigma_{\theta}>\sigma_{r}\), the solution for radial stress and displacement increments was discussed above and can be obtained from Equations (25) and (28). In the inner plastic zone, \(\sigma_{z}=\sigma_{\theta}>\sigma_{r}\), and the yield criteria governing the inner plastic zone are  
\[\sigma_{z}=\alpha_{r}\cdot\sigma_{r}+Y_{r} \tag{29}\]  
\[\sigma_{\theta}=\alpha_{r}\cdot\sigma_{r}+Y_{r} \tag{30}\]  
For the \(i\)th load increment, Equation (30) is written as  
\[\sigma_{\theta(i-1)}+\Delta\sigma_{\theta(i)}=\alpha_{i}(\sigma_{r(i-1)}+ \Delta\sigma_{r(i)})+Y_{i} \tag{31}\]  
The incremental stress components (\(\Delta\sigma_{\theta(i)}\) and \(\Delta\sigma_{r(i)}\)) can be obtained by the equilibrium Equation (8), yield criterion (31), and \(\Delta\sigma_{z(i)}=\Delta\sigma_{\theta(i)}\). The solution processes for the incremental stress components are the same as those in [PERSON] et al. (2016). Because \(\sigma_{z}=\sigma_{\theta}\) in the inner plastic zone, the stresses lie on the intersection of the two yield surfaces (Equations (29) and (30)). The correct flow rule is obtained by summing the contributions from the two plastic potentials ([PERSON] et al., 2016).