# 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_  
When \(\sigma_{z}>\sigma_{\theta}>\sigma_{r}\), the yield criterion is formulated as  
\[\sigma_{z(i-1)}+\Delta\sigma_{z(i)}=\alpha_{i}\cdot(\sigma_{r(i-1)}+\Delta \sigma_{r(i)})+Y_{i} \tag{10}\]  
Equation (10) can be rewritten as  
\[\Delta\sigma_{z(i)}=\alpha_{i}\Delta\sigma_{r(i)}+\alpha_{i}\sigma_{r(i-1)}+Y _{i}-\sigma_{z(i-1)} \tag{11}\]  
and the condition of compatibility is taken into account,  
\[\Delta\varepsilon_{r}=d\Delta u/dr \tag{12}\] \[\Delta\varepsilon_{\theta}=\Delta u/r\]  
so we have  
\[\frac{d\Delta\varepsilon_{\theta(i)}}{dr}+\frac{\Delta\varepsilon_{\theta(i) }-\Delta\varepsilon_{r(i)}}{r}=0 \tag{13}\]  
Assuming that the plastic potential uses the same formula as the yield criterion,  
\[G(\sigma_{z},\sigma_{r},\eta)=\sigma_{z}-\beta\cdot\sigma_{r} \tag{14}\]  
the incremental radial and axial plastic strain are written as  
\[\Delta\varepsilon_{r(i)}^{p}+\beta_{i}\Delta\varepsilon_{z(i)}^{p}=0 \tag{15}\]  
The incremental plastic part of tangential strain is  
\[\Delta\varepsilon_{\theta(i)}^{p}=0 \tag{16}\]  
where \(\beta_{i}=\frac{1+\sin\psi_{i}}{1-\sin\psi_{i}}\) and \(\psi_{i}\) is the dilation angle corresponding to the \(i\)th load increment. The strains are usually divided into elastic and plastic parts:  
\[\Delta\varepsilon_{r(i)}=\Delta\varepsilon_{r(i)}^{e}+\Delta\varepsilon_{r(i)} ^{p} \tag{17}\]\[\Delta\varepsilon_{\theta(i)}=\Delta\varepsilon^{e}_{\theta(i)}+\Delta\varepsilon^{p} _{\theta(i)} \tag{18}\]  
\[\Delta\sigma_{z(i)}=\alpha_{i}\Delta\sigma_{r(i)}+\alpha_{i}\sigma_{r(i-1)}+Y _{i}-\sigma_{z(i-1)} \tag{11}\]0  
With \(\Delta\varepsilon^{p}_{\theta(i)}=0\) and \(\Delta\varepsilon_{z(i)}=0\) considered, we have  
\[\Delta\sigma_{z(i)}=\alpha_{i}\Delta\sigma_{r(i)}+\alpha_{i}\sigma_{r(i-1)}+Y _{i}-\sigma_{z(i-1)} \tag{11}\]1  
\[\Delta\sigma_{z(i)}=\alpha_{i}\Delta\sigma_{r(i)}+\alpha_{i}\sigma_{r(i-1)}+Y _{i}-\sigma_{z(i-1)} \tag{11}\]2  
The incremental elastic strains should obey [PERSON]'s law, and they are in the following form  
\[ \begin{array}{l}\Delta\varepsilon^{e}_{r(i)}=\frac{1}{E}\big{[}\Delta \sigma_{r(i)}-\
u(\Delta\sigma_{\theta(i)}+\Delta\sigma_{z(i)})\big{]}\\ \Delta\varepsilon^{e}_{\theta(i)}=\frac{1}{E}\big{[}\Delta\sigma_{\theta(i)}- \
u(\Delta\sigma_{z(i)}+\Delta\sigma_{r(i)})\big{]}\\ \Delta\varepsilon^{e}_{z(i)}=\frac{1}{E}\big{[}\Delta\sigma_{z(i)}-\
u( \Delta\sigma_{r(i)}+\Delta\sigma_{\theta(i)})\big{]}\end{array}  \tag{22}\]  
By substituting Equation (22) into Equations (20) and (21), and then into Equation (23), the following equation is obtained:  
\[r(\frac{d\Delta\sigma_{\theta(i)}}{dr}-\
u\frac{d\Delta\sigma_{r(i)}}{dr}-\
u \frac{d\Delta\sigma_{z(i)}}{dr})=-(1+\
u+\
u\beta_{i})\Delta\sigma_{\theta(i) }+(1+\
u-\
u\beta_{i})\Delta\sigma_{r(i)}+\beta_{i}\Delta\sigma_{z(i)} \tag{23}\]  
Equations (9), (11) and (23) will be used to get \(\Delta\sigma_{z(i)}\), \(\Delta\sigma_{\theta(i)}\) and \(\Delta\sigma_{r(i)}\). Substitution of Equations (9) and (10) into Equation (23) produces a second-order differential equation for incremental radial stress. \[r^{2}\frac{d^{2}\Delta\sigma_{r}}{dr^{2}}+f_{1}r\frac{d\Delta\sigma_{r}}{dr}+ f_{2}\Delta\sigma_{r}+f_{3}=0 \tag{24}\]  
where \(f_{1}=3+\
u(\beta_{i}-\alpha_{i})\), \(f_{2}=\beta_{i}(2\
u-\alpha_{i})\) and \(f_{3}=-\beta_{i}(\alpha_{i}\sigma_{r(i-1)}+Y_{i}-\sigma_{z(i-1)})\). Solving thedifferential equation and the general solution is  
\[\Delta\sigma_{r(i)}=F_{i}r^{n_{1}}+G_{i}r^{n_{2}}+H \tag{25}\]  
with \(n_{1}=\frac{1-f_{1}+\sqrt{(1-f_{1})^{2}-4f_{2}}}{2}\), \(n_{2}=\frac{1-f_{1}-\sqrt{(1-f_{1})^{2}-4f_{2}}}{2}\) and \(H=-\frac{f_{3}}{f_{2}}\). In Equation (25), \(F_{i}\) and \(G_{i}\) are determined by boundary conditions.