# Incorporating Transverse Normal Strain in the Homogenization of Corrugated Cardboards
## 3 Extended Homogenization Method
### Correction on Out-of-Plane Stiffness  
Corrugated cardboard is more complex than a laminated plate due to the cavities, and the coordinate transformation is a kind of mathematic derivation, which can be improved by mechanical analysis. Therefore, a correction to the out-of-plane stiffness is developed by treating the flute as a three-hinged arch here.  
As shown in Figure 5, the corrugated shape is approximately sinusoidal, with a concentrated load at the center hinge. The three-hinged arch is a statically determinate structure, so its reaction forces and internal forces can be directly obtained by solving the equilibrium equations.  
\[F_{ly}=F_{ry}=\frac{F}{2};\quad F_{tx}=F_{rx}=\frac{pF}{4h_{f}} \tag{27}\]  
\[M=\left\{\begin{array}{cc}\frac{F}{2}x-\frac{Fp}{4h_{f}}g(x),&0\leq x\leq \frac{p}{2}\\ \frac{F}{2}x-\frac{Fp}{4h_{f}}g(x)-F(x-\frac{p}{2}),&\frac{p}{2}<x\leq p\\ \end{array}\right. \tag{28}\]  
\[N=\left\{\begin{array}{cc}-\frac{F}{2}\sin\theta-\frac{Fp}{4h_{f}}\cos \theta,&0\leq x\leq\frac{p}{2}\\ \frac{F}{2}\sin\theta-\frac{Fp}{4h_{f}}\cos\theta,&\frac{p}{2}<x\leq p\\ \end{array}\right. \tag{29}\]  
\[T=\left\{\begin{array}{cc}\frac{F}{2}\cos\theta-\frac{Fp}{4h_{f}}\sin \theta,&0\leq x\leq\frac{p}{2}\\ \frac{F}{2}\cos\theta+\frac{Fp}{4h_{f}}\sin\theta,&\frac{p}{2}<x\leq p\\ \end{array}\right. \tag{30}\]  
where \(M\), \(N\), and \(T\) represent the internal moment, axial, and shear force, respectively. The term \(g(x)\) refers to Equation (11).  
According to [PERSON]'s theorem, the displacement \(\Delta\) corresponding to \(F\) can be obtained by:  
\[\Delta=\int\frac{M}{EI}\cdot\frac{\partial M}{\partial F}ds+\int\frac{N}{EA} \cdot\frac{\partial N}{\partial F}ds+\int\frac{kT}{GA}\cdot\frac{\partial T}{ \partial F}ds \tag{31}\]  
where the right-hand side terms correspond to the contribution of flexural, axial, and sheer deformation, respectively. The term \(ds\) is the differential element of arc length; in Cartesian coordinates: \(ds=\sqrt{1+g^{\prime 2}}dx\). \(E\) is Young's modulus, \(G\) is the shear modulus, \(I\) is moment of inertial \(I=t^{3}/12\), \(A=t\) is the sectional area, and \(k\) is the shear correction coefficient.  
Since bending dominates in this load case, the last two terms can be discarded for simplicity. Additionally, symmetry can be exploited to avoid piecewise functions. Thus, we obtain:  
\[\Delta=2\int_{0}^{p/2}\frac{F(\frac{x}{2}-\frac{p}{4h_{f}}g(x))^{2}}{EI}\sqrt {1+g(x)^{\prime 2}}dx \tag{32}\]  
Figure 5: Simplify the flute to a 3-hinged arch.  
where \(F\) is arbitrary and can be set to 1. The integral can be evaluated numerically.  
Then a nominal equivalent modulus can be calculated as:  
\[E_{z}=\frac{h_{f}F}{p\Delta} \tag{33}\]  
This homogenized modulus is directly related to \(A_{33}\), \(B_{33}\), and \(C_{33}\) in Equation (2). Therefore we can use it to correct these three terms. While substituting \(E_{z}\) into \(Q_{33}\) is feasible, the explicit form of the 3D elastic stiffness matrix \(\mathbf{Q}\) is complicated due to [PERSON]'s ratio. However, a simplification is given here, because on one hand the derivation above is inherently uniaxial, and on the other hand work by [PERSON] [31] indicates that these effects are small. Consequently, it is reasonable to replace \(Q_{33}\) with \(E_{z}\). Replace \(t_{k}\) in Equation (25) with \(h\), since we have homogenized the flute. We thus obtain:  
\[\begin{split} A_{33}^{f}&=E_{z}h\\ B_{33}^{f}&=E_{z}z_{k}h\\ C_{33}^{f}&=E_{z}(z_{k}^{2}h+h^{3}/12)\end{split} \tag{34}\]  
For single-wall corrugated cardboard, \(z_{k}=0\) due to symmetry. The rest of the matrix terms remain unchanged as calculated via the procedures in Section 3.1. Validation in Section 4.3 demonstrates that this scheme performs effectively.