# Force Analysis of Circular Diaphragm Wall Based on Circular Cylindrical Shell Theory
## 4 Circular Cylindrical Shell Theory Calculation Method (CCSTCM)
### Internal Force Expressions of Circular Cylindrical Shell under Normal Loads  
When a circular cylindrical shell is only subjected to normal load \(q_{3}=\)\(q_{3}(\alpha)\), which is symmetric about the central axis of the circular cylindrical shell, and the boundaries are also symmetric about the central axis, the displacements and internal forces will also be symmetric about the central axis and their expressions will only be the functions of \(\alpha\). In this case, the displacement function \(H\) can be set as a function including independent variable \(\alpha\) only, \(H=H(\alpha)\), and the partial differential Equation (18) can be simplified to the ordinary differential Equation (20).  
\[\
abla^{8}H+\frac{E\delta}{R^{2}D}\frac{\partial^{4}H}{\partial\alpha^{4}}= \frac{q_{3}}{RD} \tag{20}\]  
Accordingly, the third equation in Equation (17) can be simplified as Equation (21).  
\[w=R\frac{\mathrm{d}^{4}H}{\mathrm{d}\alpha^{4}} \tag{21}\]  
Equation (22) can be derived through substituting Equation (21) into Equation (20). Equation (22) is a fourth-order ordinary differential equation for \(w\), which can be solved according to two boundary conditions on \(w\) at each boundary.  
\[\frac{\mathrm{d}^{4}w}{\mathrm{d}\alpha^{4}}+\frac{E\delta}{R^{2}D}w=\frac{q_ {3}}{D} \tag{22}\]  
By using Equation (21) and \(H=H(\alpha)\), the Equation (23) of internal forces can be obtained from Equation (19), so that all internal forces can be calculated from the solution of \(w\).  
\[\left.\begin{array}{l}F_{\mathrm{T1}}=0\\ F_{\mathrm{T2}}=\frac{E\delta}{R}w\\ F_{\mathrm{T12}}=F_{\mathrm{T21}}=0\\ M_{1}=-D\frac{\mathrm{d}^{2}w}{\mathrm{d}\alpha^{2}}\\ M_{2}=-\mu D\frac{\mathrm{d}^{2}w}{\mathrm{d}\alpha^{2}}=\mu M_{1}\\ M_{12}=M_{21}=0\\ F_{\mathrm{S1}}=-D\frac{\mathrm{d}^{3}w}{\mathrm{d}\alpha^{3}}\\ F_{\mathrm{S2}}=0\end{array}\right\} \tag{23}\]  
In order to simplify the solution, a constant \(\lambda\) is cited that is expressed as Equation (24) with a dimension of L\({}^{-1}\).  
\[\lambda=\left(\frac{E\delta}{4R^{2}D}\right)^{\frac{1}{4}}=\left[\frac{3(1- \mu^{2})}{R^{2}\delta^{2}}\right]^{\frac{1}{4}} \tag{24}\]  
Then, one coordinate \(\xi\), with a dimension of 1 is cited instead of \(\alpha\).  
\[\xi=\lambda\alpha \tag{25}\]  
Thus, the differential Equation (22) is transformed into Equation (26), as follows.  
\[\frac{\mathrm{d}^{4}w}{\mathrm{d}\xi^{4}}+4w=\frac{4R^{2}}{E\delta}q_{3} \tag{26}\]  
Consequently, the solution to Equation (26) can be obtained as Equation (27).  
\[w=C_{1}\sin\xi\mathrm{sinh}\xi+C_{2}\sin\xi\mathrm{cosh}\xi+C_{3}\cos\xi \mathrm{sinh}\xi+C_{4}\cos\xi\cosh\xi+w_{0} \tag{27}\]  
In Equation (27), \(w_{0}\) is any one particular solution, which can be selected by the function form of the normal load according to the requirements of differential Equation (26).  
The constants of \(C_{1}\), \(C_{2}\), \(C_{3}\) and \(C_{4}\) are determined by the boundary conditions, and Equation (23) of internal forces could be further transformed into Equation (28).  
\[\left.\begin{array}{l}F_{\mathrm{T1}}=0\\ F_{\mathrm{T2}}=\frac{E\delta}{K}w\\ F_{\mathrm{T12}}=F_{\mathrm{T21}}=0\\ M_{1}=-\lambda^{2}D\frac{\mathrm{d}^{2}w}{\mathrm{d}\xi^{2}}\\ M_{2}=-\mu\lambda^{2}D\frac{\mathrm{d}^{2}w}{\mathrm{d}\xi^{2}}=\mu M_{1}\\ M_{12}=M_{21}=0\\ F_{\mathrm{S1}}=-\lambda^{3}D\frac{\mathrm{d}^{3}w}{\mathrm{d}\xi^{3}}\\ F_{\mathrm{S2}}=0\end{array}\right\} \tag{28}\]