## 2 Theoretical Analysis
\begin{array}{c}\mathbf{N}_{\varphi}\\ \mathbf{N}_{y}\\ \mathbf{N}_{\varphi y}\end{array} \right\}=\int_{-t/2}^{t/2}\left\{ \begin{array} []{c}\sigma_{\varphi}\\ \sigma_{y}\\ \sigma_{\varphi y}\end{array} \right\}dz=\sum_{k=1}^{N}\int_{-z_{k-1}}^{z_{k}} \left\{ \begin{array}{c}\sigma_{\varphi}\\ \sigma_{\varphi y}\\ \sigma_{\varphi y}\end{array} \right\}k \tag{8}\]  
\[\left\{ \begin{array}{c}M_{\varphi}\\ M_{y}\\ M_{\varphi y}\end{array} \right\}=\int_{-t/2}^{t/2}\left\{ \begin{array}{c} \sigma_{\varphi}\\ \sigma_{y}\\ \sigma_{\varphi y}\end{array} \right\}zdz=\sum_{k=1}^{N}\int_{-z_{k-1}}^{z_{k}} \left\{ \begin{array}{c}\sigma_{\varphi}\\ \sigma_{y}\\ \sigma_{\varphi y}\end{array} \right\}zdz \tag{9}\]  
For composite tape springs, commonly used layer configurations are symmetric layups, so \(B_{ij}=0\)  
\[N_{\varphi}=A_{11}\bigg{(}\frac{dv}{dy}+\frac{1}{2}\frac{d^{2}w}{dy^{2}}\bigg{)} +A_{12}\Big{(}-\frac{w}{r}\Big{)} \tag{10}\]  
\[N_{y}=A_{12}\bigg{(}\frac{dv}{dy}+\frac{1}{2}\frac{d^{2}w}{dy^{2}}\bigg{)}+A_{ 22}\Big{(}-\frac{w}{r}\Big{)} \tag{11}\]  
\[M_{\varphi}=D_{11}\frac{1}{r}+D_{12}\bigg{(}\frac{1}{R}-\frac{d^{2}w}{dy^{2}} \bigg{)} \tag{12}\]  
\[M_{y}=D_{12}\frac{1}{r}+D_{22}\bigg{(}\frac{1}{R}-\frac{d^{2}w}{dy^{2}}\bigg{)} \tag{13}\]  
Substituting into the equilibrium equation:  
\[D_{22}\bigg{(}\frac{d^{4}w}{dy^{4}}\bigg{)}+\bigg{(}A_{11}-\frac{A_{12}}{A_{22} }^{2}\bigg{)}\frac{w}{r^{2}}=0 \tag{14}\]  
Simplify:  
\[\bigg{(}\frac{d^{4}w}{dy^{4}}\bigg{)}+4\frac{\Big{(}A_{11}-\frac{A_{12}}{A_{22 }}^{2}\Big{)}r^{2}}{4D_{22}}\frac{w}{r^{4}}=0 \tag{15}\]Let \(n=\sqrt[4]{\left(\frac{A_{11}-\frac{A_{12}{22}^{2}}{A_{22}}\right)^{2}}{4D_{22}}}\) ; therefore, Equation (15) can be simplified as  
\[\left(\frac{d^{4}w}{dy^{4}}\right)+4n^{4}\frac{w}{r^{4}}=0 \tag{16}\]  
Obtaining:  
\[w=e^{\frac{m}{r}}\left(c_{1}\cos\frac{ny}{r}+c_{2}\sin\frac{ny}{r}\right)+e^{- \frac{m}{r}}\left(c_{3}\cos\frac{ny}{r}+c_{4}\sin\frac{ny}{r}\right) \tag{17}\]  
Considering \(w(y)=w(-y)\). \[w=c_{1}\cosh\frac{ny}{r}\cos\frac{ny}{r}+c_{2}\sinh\frac{ny}{r}\sin\frac{ny}{r} \tag{18}\]  
On the boundary \(y=\pm\frac{s}{2}M_{y}=0\), \(\frac{dM_{y}}{dy}=0\)  
Obtaining:  
\[ \begin{array}{l}c_{1}=\frac{1}{2\;a^{2}}\left(\frac{1}{R}+\frac{D_{12}}{D_{2 2}}\frac{1}{r}\right)\frac{\cos\left(\frac{4s}{2}\right)\;\sinh\left(\frac{ \kappa s}{2}\right)-\cosh\left(\frac{4s}{2}\right)\;\sin\left(\frac{\kappa s}{ 2}\right)}{\sin\left(\frac{4s}{2}\right)\cos\left(\frac{4s}{2}\right)\;+\sinh \left(\frac{\kappa s}{2}\right)\cosh\left(\frac{4s}{2}\right)}\\ c_{2}=\frac{1}{2\;a^{2}}\left(\frac{1}{R}+\frac{D_{12}}{D_{22}}\frac{1}{r} \right)\frac{\cosh\left(\frac{4s}{2}\right)\;\sin\left(\frac{\kappa s}{2} \right)+\cos\left(\frac{4s}{2}\right)\;\sinh\left(\frac{\kappa s}{2}\right)}{ \sin\left(\frac{\kappa s}{2}\right)\cos\left(\frac{\kappa s}{2}\right)}+\sinh \left(\frac{\kappa s}{2}\right)\cosh\left(\frac{\kappa s}{2}\right)}\end{array}  \tag{19}\]  
The bending moment required for driving in the bending process of the tape spring is \(M\), and \(M\) can be calculated by integration:  
\[ \begin{array}{l}M=\int_{-\frac{s}{2}}^{\frac{s}{2}}\left(M_{\varphi}-N_{ \varphi}\omega\right)dy\\ =\int_{-\frac{s}{2}}^{\frac{s}{2}}\left[D_{22}\frac{1}{r}+D_{11}\left(\frac{1 }{R}-\frac{d^{2}w}{dy}\right)-\frac{w^{2}}{r}\left(\frac{A_{12}{}^{2}}{A_{22} }-A_{11}\right)\right]dy\\ =\left(D_{22}\frac{1}{r}+D_{11}\frac{1}{R}\right)s-\;D_{11}\frac{dw}{dy}\left| _{-\frac{s}{2}}^{\frac{s}{2}}-\int_{-\frac{s}{2}}^{\frac{s}{2}}\left[\frac{w^ {2}}{r}\left(\frac{A_{12}{}^{2}}{A_{22}}-A_{11}\right)\right]dy\end{array}  \tag{20}\]  
Among:  
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