# Buckling of Rectangular Composite Pipes under Torsion
## 2 Numerical Calculation
### Energy Method
\right] \tag{10}\]  
and substituting Equation (10) into Equation (7), using Equation (9), and integrating Equation (7), \(U_{p}\) becomes  
\[U_{p}=\frac{1}{2}\!\int\!\!\left\{D_{xx}\!\left(\frac{\partial^{2}w}{\partial x ^{2}}\right)^{2}+2(2D_{ss}+D_{xy})\!\left(\frac{\partial^{2}w}{\partial x^{2 }}\right)\!\left(\frac{\partial^{2}w}{\partial y^{2}}\right)+4D_{xs}\!\left( \frac{\partial^{3}w}{\partial y^{3}}\right)\!\left(\frac{\partial w}{\partial y }\right)+4D_{xs}\!\left(\frac{\partial^{3}w}{\partial y^{3}}\right)\!\left( \frac{\partial w}{\partial x}\right)+D_{yy}\!\left(\frac{\partial^{2}w}{ \partial y^{2}}\right)^{2}\right\}dxdy \tag{11}\]  
For a symmetric layup, \(D_{xx}\), \(D_{xy}\), \(D_{yy}\), and \(D_{ss}\) are  
\[D_{xx}=\frac{E_{x}t^{3}}{12(1-\
u_{x}\
u_{y})},\ D_{yy}=\frac{E_{y}t^{3}}{12(1 -\
u_{x}\
u_{y})},\ D_{xy}=\
u_{x}D_{yy},\ D_{ss}=\frac{G_{xy}t^{3}}{12},\ D_{ xs}=D_{ys}=0 \tag{12}\]  
Generally, \(D_{xs}\) and \(D_{ss}\) are non-zero even in a symmetric layup; when the number of layups is large enough, the following assumption can be used. \[D_{xs}=D_{ys}=0 \tag{13}\]  
For vertical plates, \"\(y\)\" is changed to \"z\" in Equations (7)-(13). The total strain energy \(U\) and the work \(W\) of the four plates is written as  
\[\begin{array}{ll}U&=2\times\frac{1}{2}\!\int_{0}^{b}\int_{0}^{l}\!\left\{D_ {xx}\!\left(\frac{\partial^{2}w}{\partial x^{2}}\right)^{2}+2(2D_{ss}+D_{xy}) \!\left(\frac{\partial^{2}w}{\partial x^{2}}\right)\!\left(\frac{\partial^{2} w}{\partial y^{2}}\right)+4D_{xs}\!\left(\frac{\partial^{3}w}{ \partial x^{3}}\right)\!\left(\frac{\partial w}{\partial y}\right)+4D_{ys}\! \left(\frac{\partial^{3}w}{\partial y^{3}}\right)\!\left(\frac{\partial w}{ \The following equation is obtained from Equation (17). \[ \begin{array}{c}\frac{\pi^{4}}{32\Gamma_{\rm tr}l\hbar}\bigg{[}\bigg{(}\frac{ \hbar}{l}\bigg{)}^{3}\bigg{\{}D_{xx}m^{4}+2(2D_{\rm ss}+D_{\rm xy})\Big{(}\frac {l}{b}\Big{)}^{2}m^{2}n^{2}+D_{yy}\Big{(}\frac{l}{b}\Big{)}^{4}n^{4}\bigg{\}} \\ +\bigg{(}\frac{\hbar}{l}\bigg{)}^{3}\bigg{\{}D_{xx}m^{4}+2(2D_{\rm ss}+D_{\rm xx })\Big{(}\frac{l}{\hbar}\Big{)}^{2}m^{2}n^{2}+D_{zz}\Big{(}\frac{l}{\hbar} \Big{)}^{4}n^{4}\bigg{\}}\bigg{]}\Phi_{mn}\\ -mn^{l}\frac{l}{b}\frac{l}{\hbar}\bigg{\{}\bigg{(}\frac{l}{b}\bigg{)}^{2}+ \bigg{(}\frac{l}{\hbar}\bigg{)}^{2}\bigg{\}}\underset{i}{\overset{\infty}{ \sum}}\underset{j}{\overset{\infty}{\sum}}\underset{(\overset{i}{\vec{p}}_{ij} )}{(\overset{\infty}{\vec{p}}_{i}-m^{2})(\overset{\infty}{\vec{p}}_{-n^{2}})}=0 \end{array}  \tag{18}\]  
where summation indicates only those values of \(i\) and \(j\) for which \(m+i\) and \(n+j\) are odd. Equation (18) can be rewritten as following matrix form for \(m+n\) is even is  
\[ \begin{array}{c}\left[\begin{array}{ccccc}\frac{K_{11}}{\tau}&0&\frac{4}{ \beta}\beta_{1}&0&0&\cdots\\ 0&\frac{K_{13}}{\tau}&-\frac{4}{\beta}_{1}&0&0&\cdots\\ \frac{4}{\beta}\beta_{1}&-\frac{4}{\beta}_{1}&\frac{K_{22}}{\tau}&-\frac{4}{ \beta}_{1}&-\frac{20}{\beta}_{1}&\cdots\\ 0&0&-\frac{4}{\beta}_{1}&\frac{K_{31}}{\tau}&0&\cdots\\ 0&0&-\frac{20}{6\beta}\beta_{1}&0&\frac{K_{15}}{\tau}&\cdots\\ \vdots&\vdots&\vdots&\vdots&\vdots&\ddots\end{array}\right]\left[\begin{array} []{c}\Phi_{11}\\ \Phi_{13}\\ \Phi_{22}\\ \Phi_{31}\\ \Phi_{15}\\ \vdots\end{array}\right]=0,\\ K_{mn}=\frac{\pi^{4}}{32\Gamma\hbar l}\bigg{[}\bigg{(}\frac{\hbar}{l} \bigg{)}^{3}\bigg{\{}D_{xx}m^{4}+2(2D_{\rm ss}+D_{xy})\Big{(}\frac{l}{b}\Big{)} ^{2}m^{2}n^{2}+D_{yy}\Big{(}\frac{l}{b}\Big{)}^{4}n^{4}\bigg{\}}\\ +\bigg{(}\frac{\hbar}{l}\bigg{)}^{3}\bigg{\{}D_{xx}m^{4}+2(2D_{\rm ss}+D_{xz}) \Big{(}\frac{l}{\hbar}\Big{)}^{2}m^{2}n^{2}+D_{zz}\Big{(}\frac{l}{\hbar}\Big{)} ^{4}n^{4}\bigg{\}}\bigg{]}\\ \beta_{1}=\frac{b}{l}\frac{\hbar}{l}\bigg{\{}\bigg{(}\frac{l}{b}\bigg{)}^{2}+ \Big{(}\frac{l}{\hbar}\bigg{)}^{2}\bigg{\}}\end{array}  \tag{19}\]  
and for \(m+n\) is odd is  
\[