# Multiscale and Failure Analysis of Periodic Lattice Structures
## 2 Upscaling Technique  
Figure 2 shows a lattice structure made of periodic rectangular cells. The unit cell of the lattice structure is also shown in the same figure, and Figure 3 shows the enlarged unit cell with the length \(l\) and the width \(w\). The unit cell lattice is taken from the center of one lattice to the center of the next lattice. Thus, the thickness of the unit cell lattice in Figure 3 is \(t/2\) because it is half of the thickness of the lattice structure in Figure 2, which is equal to \(t\). Figure 3: Unit cell of a rectangular lattice. All the members have the same thickness. Figure 2: Lattice structure consisting of rectangular open cells. To determine the effective elastic modulus along the \(x\)-axis of the unit cell in Figure 3, let us apply a uniform displacement \(d\) along the \(x\)-axis of the rectangular unit cell to its right side lattice while the left side lattice is constrained from any movement. In that case, the two horizontal lattices at the top and bottom sides provide the major stiffness, while the two vertical lattices on the left and right sides provide negligible stiffness as compared to the horizontal lattices. Thus, the stiffness in the \(x\)-axis is  
\[k_{lattice}=\frac{E_{s}t}{lb} \tag{1}\]  
where \(k_{lattice}\) is the stiffness of the unit-cell lattice in the \(x\)-axis, and \(E_{s}\) is the elastic modulus of the lattice material. The homogenized material has the stiffness \(k_{\text{homo}}\) expressed as below:  
\[k_{\text{homo}}=\frac{\left(E_{eff}\right)_{x}w}{lb} \tag{2}\]  
where \(\left(E_{eff}\right)_{x}\) is the effective elastic modulus along the \(x\)-axis. The stiffness of the homogenized material should be equal to that of the unit cell lattice, which leads to the following equation:  
\[\frac{\left(E_{eff}\right)_{x}}{E_{s}}=\frac{t}{w} \tag{3}\]  
Equation (3) states that the ratio of the effective modulus along the \(x\)-axis to the elastic modulus of the lattice material is equal to the ratio of the total lattice thickness of the horizontal lattices and the vertical dimension of the unit cell. The vertical lattices in the unit cell only act to separate the horizontal lattices by a specific distance. Thus, a change in the length \(L\) does not affect the effective modulus along the \(x\)-axis. Likewise, the effective modulus along the \(y\)-axis is expressed as  
\[\frac{\left(E_{eff}\right)_{y}}{E_{s}}=\frac{t}{l} \tag{4}\]  
If diagonal members are added to the rectangular lattice structure, the corresponding unit cell is shown in Figure 4. Let us first assume that the two diagonal members are independent of each other and do not have a direct connection. Then, as before, a uniform displacement is applied to the right side of the unit cell in Figure 4 while the left side is constrained from any movement. Figure 5 shows the displacement of one of the diagonal members with the applied horizontal displacement. The diagonal member has the original location at OA whose initial length is \(l_{d}=\frac{l}{\cos\theta}\) in which \(l\) is the horizontal length of the unit cell, and \(\theta\) is the angle between the horizontal and diagonal directions, as shown in Figure 5.  
If AC in Figure 5 is the horizontal displacement \(\delta\), \(AB=\delta\sin\theta\) and \(BC=\delta\cos\theta\). Among the two components of the displacement of the diagonal bar from OA to OC, AB is caused by bending loading, while AC results from the axial force. Because the diagonal member has a dimension like a beam with a large slenderness ratio, the bending load for AB is much smaller than the axial force for AC. If the slenderness ratio is close to one, the bending should be considered. As a result, the bending load is neglected. The axial force on the diagonal member to induce the deformation AC is expressed as  
\[F_{a}=\frac{t_{d}E_{d}\delta}{l}\cos^{2}\theta \tag{5}\]where \(t_{d}\) is the thickness of the diagonal member and \(E_{d}\) is the elastic modulus of the diagonal member.