## 2 Theoretical Analysis
### Burst Pressure Solution Based ASSY  
A general stress analysis is presented in the cylinders subjected to inner pressure \(P\) (Figure 6). The solution is applied under closed-end conditions and the three principal stresses are as follows [36]:  
\[\sigma_{1}=\sigma_{\theta}=\frac{P_{i}D}{2t},\sigma_{2}=\sigma_{z}=\frac{P_{i }D}{4t},\sigma_{3}=\sigma_{r}\approx 0 \tag{8}\]  
where \(\theta\), \(r\), and \(z\) mean the hoop, radial, and axial directions, respectively; \(\sigma_{r}\), \(\sigma_{\theta}\), and \(\sigma_{z}\) represent the radial stress, the hoop stress, and the axial stress, respectively; and \(D\) and \(t\) are the pipe mean diameter and mean wall thickness, respectively.  
In general, the axial strain is small and can be ignored in the pipeline, and the three strains are  
\[\varepsilon_{1}=\varepsilon_{\theta}=ln\frac{D}{D_{0}},\varepsilon_{2}= \varepsilon_{2}=0,\varepsilon_{3}=\varepsilon_{r}=ln\frac{t}{t_{0}} \tag{9}\]  
where \(D_{0}\) and \(t_{0}\) represent the initial diameter and wall thickness respectively.  
Figure 5: The power exponential constitutive model.  
When the pipe fails due to burst, large plastic deformation usually occurs; considering the incompressibility of the material volume \(\varepsilon_{1}+\varepsilon_{2}+\varepsilon_{3}=0\), From Equation (9), one has  
\[\varepsilon_{1}-\varepsilon_{3}=2\varepsilon_{1}=ln\frac{D}{t}\frac{t_{0}}{D_{ 0}} \tag{10}\]  
From Equations (6) and (8)-(10), the ASSY equivalent stress and equivalent strain in the pipe are determined as  
\[\sigma_{A}=\frac{1}{2+\sqrt{3}}\Big{(}\sqrt{3}\sigma_{T}+2\sigma_{M}\Big{)}= \frac{2\sqrt{3}}{2+\sqrt{3}}\sigma_{1} \tag{11}\]  
\[\varepsilon_{A}=\frac{1}{2+\sqrt{3}}\Big{(}\sqrt{3}\varepsilon_{T}+2\varepsilon _{M}\Big{)}=\frac{2\sqrt{3}}{2+\sqrt{3}}\varepsilon_{1} \tag{12}\]  
From Equations (11) and (12), it can be known from [PERSON]'s plastic work assumption that \(\sigma_{\Lambda}\varepsilon_{\Lambda}=\sigma_{1}\varepsilon_{1}=\sigma\varepsilon\), and the diameter and wall thickness are derived from Equations (10) and (12) as  
\[\frac{D}{t}=\frac{D_{0}}{t_{0}}e^{(4\sqrt{3}/(2+\sqrt{3}))\varepsilon_{A})} \tag{13}\]  
According to Equations (8)-(13), the internal loading limit of the pipe is a function of the geometric size, the properties of the pipe material, and the equivalent strain, which can be expressed by the following formula:  
\[P=\left(\frac{1}{2}+\frac{1}{\sqrt{3}}\right)\frac{2t_{0}}{D_{0}}e^{-((4\sqrt{ 3}/(2+\sqrt{3})\varepsilon_{A})}(\frac{e}{n})^{n}\sigma_{t}\varepsilon_{A}^{n} \tag{14}\]  
When \(\partial P/\partial\varepsilon_{A}=0\), the maximum value of equivalent strain based on ASSY is \(\varepsilon_{A}^{*}=\Big{(}2+\sqrt{3}\Big{)}n/4\sqrt{3}\). Hence, the burst pressure calculated by ASSY is determined as a function of the size and material of the pipe, including diameter, wall thickness, the \(\sigma_{t}\), and the hardening exponent \(n\):  
\[P_{A}=(\frac{2+\sqrt{3}}{4\sqrt{3}})^{n+1}\frac{4t_{0}}{D_{0}}\sigma_{t} \tag{15}\]  
Figure 6: The force and deformation of the cylindrical unit under internal pressure.  
Similarly, the burst pressure based on the three yield criteria can be determined as follows [36]:  
\[\begin{split} P_{T}&=(\frac{1}{2})^{n+1}\frac{4t_{0}}{D_{ 0}}\sigma_{t}\text{ for Tresca criterion;}\\ P_{M}&=(\frac{\sqrt{3}}{3})^{n+1}\frac{4t_{0}}{D_{ 0}}\sigma_{t}\text{ for Mises criterion;}\\ P_{TS}&=(\frac{\pi}{3})^{n+1}\frac{4t_{0}}{D_{ 0}}\sigma_{t}\text{ for TSSY criterion.}\end{split} \tag{16}\]