# Research on the Transportation of Low-Strength Composites Sheets
## 2 Problem Formulation  
Figure 2 shows the axially moving sheet travelling between two rolls with the constant linear velocity \(v\). The geometric parameters of the sheet and rolls are: \(b\) is the width, \(t\) is the thickness, \(l\) is the sheet length between it and rolls' contact points, \(D_{h}\) is the vertical (\(y\) direction) distance between two rolls' axes, \(2D_{l}\) is the horizontal (\(x\) direction) distance between two rolls' axes. The ratio between thickness \(t\) and length \(l\) is very small: \(t/l\ll O(1)\). The material properties are the mass density \(\rho\), the Young's modulus \(E\). \(\sigma_{x}\) is the stress at position \(x\) in tangent direction and \(\varepsilon_{x}\) is the strain at position \(x\) in tangent direction. This study focuses on the scenario that the tension in the sheet is mainly caused by gravity that is enough to break some low-strength materials. This means that \(\sigma_{x}/(\rho gl)\approx O(1)\). In the international system of units (SI), \(\rho gl\approx O(10^{4})\) and \(E>O(10^{6})\) for most materials. Thus \(\varepsilon_{x}=\sigma_{x}/E<O(10^{-2})\). So the elastic effect can be negligible and it can be considered as a 2D problem. Since that many polymer composites are plastic when processing and \(t/l\ll O(1)\), the bending moment in the sheet is assumed to be 0. The sheet is perpendicular to plane \(xy\) and its projection in plane \(xy\) is \(y=f(x)\). The gravity is in the direction of \(-y\). At equilibrium, the kinetic energy T, the potential energy U are given as:  
\[T=\frac{1}{2}\rho bt\int_{l}\mathbf{v}\times\mathbf{v}\mathrm{d}l \tag{1}\]  
\[U=\rho btg\int_{l}y\mathrm{d}l \tag{2}\]  
When the system is steady, \(f\) satisfies:  
\[f=\arg\min_{f}\int_{l}(\rho btgy+\frac{1}{2}\rho btv^{2})\mathrm{d}l \tag{3}\]  
Figure 2: A schematic drawing showing the geometry of the transporting sheet and rolls. Introduce Lagrange multiplier \(\lambda\), \(L=(\rho btgy+\frac{1}{2}\rho btv^{2}-\lambda)\sqrt{1+\dot{y}^{2}}\). The Euler-Lagrange equation of this problem is:  
\[\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\partial L}{\partial\dot{y}}\right)- \frac{\partial L}{\partial y}=0,\quad\Rightarrow\quad\frac{\mathrm{d}}{ \mathrm{d}x}\left(\frac{\rho btgy\dot{y}+\frac{1}{2}\rho btv^{2}\dot{y}- \lambda\dot{y}}{\sqrt{1+\dot{y}^{2}}}\right)-\rho btg\sqrt{1+\dot{y}^{2}}=0 \tag{4}\]  
Solve the equation for \(y\) :  
\[y(x)=\frac{C_{1}}{\rho btg}\cosh\left(\frac{\rho btg(x-C_{2})}{C_{1}}\right)+ \frac{\lambda}{\rho btg}-\frac{v^{2}}{g} \tag{5}\]  
\(\lambda,C_{1}\) and \(C_{2}\) can be determined by boundary conditions. The free-body diagram of the forces acting on the element is shown in Figure 3. \(T(x)\) is the tension in the sheet at position \(x\) and \(G\) is the gravity. Equilibrium equation of the forces is:  
\[T(x+\mathrm{d}x)\vec{e_{u}}(x+\mathrm{d}x)-T(x)\vec{e_{u}}(x)-G\vec{e_{y}}= \frac{\mathrm{d}\vec{v}}{\mathrm{d}t}\rho bt\sqrt{1+\dot{y}^{2}}\mathrm{d}x \tag{6}\]  
That is:  
\[\frac{\mathrm{d}}{\mathrm{d}x}\left(T(x)\vec{e_{u}}\right)-\rho btg\sqrt{1+ \dot{y}^{2}}\vec{e_{y}}=\frac{\mathrm{d}\vec{v}}{\mathrm{d}t}\rho bt\sqrt{1+ \dot{y}^{2}} \tag{7}\]  
in which, \(\vec{e_{y}}\) is the unit vector in the \(y\) direction. \(\vec{e_{u}}\) and \(\vec{e_{v}}\) is the unit tangent vector and unit normal vector of \(v\), respectively:  
\[\vec{e_{y}}= \begin{bmatrix}0\\ 1\end{bmatrix} ,\quad\vec{e_{u}}= \begin{bmatrix}\frac{1}{\sqrt{1+\dot{y}^{2}}} \\ \frac{\dot{y}}{\sqrt{1+\dot{y}^{2}}}\end{bmatrix} ,\quad\vec{e_{v}}= \begin{bmatrix}- \frac{\dot{y}}{\sqrt{1+\dot{y}^{2}}}\\ \frac{\dot{y}}{\sqrt{1+\dot{y}^{2}}}\end{bmatrix}