Overview

The adidas "4DFWD" shoe, which has a special 3D-printed structure embedded in the midsole, has the function of converting stepping force into propulsive force. This article interprets this function as a mechanical metamaterial that converts compression into shear and explains how to design it.

This article explains the first of the following three parts.

(1) Explanation of macro physical properties that convert compression to shear

(2) Design of mechanical metamaterials that convert compression to shear

(3) Designing a midsole that converts stepping force into propulsive force

Explanation of 4DFWD functions

The Adidas 4DFWD shoe converts the force of a step into propulsive force, which has the effect of inhibiting braking during running. This conversion of force is achieved by the complex shape of the 3DP midsole called Lattice. Lattice has a periodic structure in which unit structures called unit cells are repeated in the XYZ direction, and the elastic modulus and other mechanical properties of these unit cells can be regarded as a kind of material property. By devising the geometry of Lattice, it is possible to design material properties that cannot be realized with ordinary materials, for example, in the case of this midsole, the properties that convert the force of stepping into the shoe into propulsive force, or "compression into shear" material properties. Materials with special material properties achieved through geometry are called "mechanical metamaterials.

[citation: adidas Running's Youtube]

Material Properties of Unit Cell

So what are the material properties of 4DFWD's midsole that "convert compression to shear"? We will explain the material properties of 4DFWD's midsole that "convert compression to shear".
By looking at the midsole from the side, we will consider this as a two-dimensional problem.
The degrees of freedom for a 2-dimensional material to deform are as follows.

• \(\varepsilon_x\): elongation strain in the x-direction
• \(\varepsilon_y\): elongation strain in y-direction
• \(\gamma_{xy}=\gamma_{yx}\): shear strain

These are collectively denoted as the strain vector \(\boldsymbol\varepsilon = [\varepsilon_x, \varepsilon_y, \gamma_{xy}]^\mathrm{T}\).

The three stresses corresponding to the above material deformation are as follows.

• \(\sigma_x\): Tensile stress in x-direction
• \(\sigma_y\): tensile stress in the y-direction
• \(\tau_{xy}=\tau_{yx}\): shear stress

These are collectively denoted as the stress vector \(\boldsymbol\sigma = [\sigma_x, \sigma_y, \tau_{xy}]^\mathrm{T}\).

Next, how does the material twist when it is subjected to stress? We consider the following relational equation that says

\(\boldsymbol\varepsilon = \mathbf{S} \boldsymbol\sigma
\tag{1-1}\)

This would look like the following when displayed in component form.

\(\left[
\begin{array}{c}
\varepsilon_x \\\\
\varepsilon_y \\\\
\gamma_{xy}
\end{array}
\right]
=\left[
\begin{array}{ccc}
S_{11} & S_{12} & S_{13} \\\\
S_{21} & S_{22} & S_{23} \\\\
S_{31} & S_{32} & S_{33}
\end{array}
\right]
\left[
\begin{array}{c}
\sigma_x \\\\
\sigma_y \\\\
\tau_{xy}
\end{array}
\right]
\tag{1-2}\)

This relation is called the constitutive law for a two-dimensional linear elastic material, and the matrix \(\mathbf{S}\) is called the compliance matrix.

For isotropic materials with homogeneous material distribution, the compliance matrix is expressed using Young's modulus (longitudinal modulus) \(E\), shear modulus (transverse modulus) \(G\) and Poisson's ratio \(\nu\) as follows.

\(\mathbf{S} = 
\left[
\begin{array}{ccc}
\frac{1}{E} & -\frac{\nu}{E} & 0 \\\\
-\frac{\nu}{E} & -\frac{1}{E} & 0 \\\\
0 & 0 & \frac{1}{G}
\end{array}
\right]
\tag{1-3}\)

For isotropic materials, the shear modulus \(G\) is expressed using Young's modulus \(E\) and Poisson's ratio \(\nu\) as follows.

\(G = \frac{E}{2(1+\nu)}
\tag{1-4}\)

Material mechanics explanation of the conversion of compression into shear

Let me explain what is meant by compression being converted to shear on this relational equation.
Applying a compressive force to the top surface corresponds to applying a stress \(-\sigma_y\) in the y-direction.
Note that the tensile direction is positive and the compressive direction is negative.
In this case, the stress vector is \(\boldsymbol\sigma=[0, -\sigma_y, 0]^\mathrm{T}\) and the resulting shear strain \(\gamma_{xy}\) is expressed by the following equation from \(\text{(1-2)}\).

\(\gamma_{xy} = S_{32} (-\sigma_y)\tag{2-1}\)

Therefore, in order to convert compressive stress into shear strain, material properties must be designed such that \(S_{32}\) is non-zero.
In the case of the isotropic material just described, \(S_{32}=0\), no shear strain is generated.

Now what if we rotate the material?
Rotating the material by an angle \(\theta\) corresponds to multiplying the compliance matrix by the inverse of the stress transformation matrix \(\mathbf{T}_\sigma(\theta)\) from the right and the strain transformation matrix \(\mathbf{T}_\varepsilon(\theta)\) from the left. Correspondingly, the new compliance matrix is \(\mathbf{S}(\theta)\) as follows.

\(\mathbf{S}(\theta)=\mathbf{T}_{\varepsilon}(\theta)^\mathrm{-1}\mathbf{S}\mathbf{T}_{\sigma}(\theta)\tag{2-2}\)

where the transformation matrices \(\mathbf{T}_{\sigma}(\theta)\) and \(\mathbf{T}_{\varepsilon}(\theta)\) are represented as follows.

\(\begin{align}
\mathbf{T}_\sigma(\theta)&=
\left[
\begin{array}{ccc}
\cos^2{\theta} & \sin^2{\theta} & 2\cos{\theta}\sin{\theta} \\\\
\sin^2{\theta} & \cos^2{\theta} & -2\cos{\theta}\sin{\theta} \\\\
-\cos{\theta}\sin{\theta}  & \cos{\theta}\sin{\theta} & \cos^2{\theta} - \sin^2{\theta}
\end{array}
\right]\tag{2-3}\\
\mathbf{T}_\varepsilon(\theta)&=
\left[
\begin{array}{ccc}
\cos^2{\theta} & \sin^2{\theta} & \cos{\theta}\sin{\theta} \\\\
\sin^2{\theta} & \cos^2{\theta} & -\cos{\theta}\sin{\theta} \\\\
-2\cos{\theta}\sin{\theta}  & 2\cos{\theta}\sin{\theta} & \cos^2{\theta} - \sin^2{\theta}
\end{array}
\right]\tag{2-4}
\end{align}\)

Calculating this, \(S(\theta)_{32}\), which converts compressive stress to shear strain, is as follows.

\(S(\theta)_{32} = \sin{4\theta}\cdot\frac{1}{4}\left(\frac{1}{G} - \frac{2(1+\nu)}{E}\right)\tag{2-5}\)

In the case of isotropic materials, the equation \(\text{(1-4)}\) holds, so \(S(\theta)_{32}=0\) after rotation.
Therefore, it is clear that compressive stress cannot be converted to shear strain in the case of isotropic materials.

On the other hand, if we can design special material properties for which the equation \((\text{1-4)}\) does not hold, then the equation \(S(\theta)_{32} \neq 0\) holds and compressive stress can be converted to shear strain. This class of materials with linear elastic properties is called square anisotropic materials.

Below are the directional Young's modulus for isotropic and square anisotropic materials at different angles of rotation.
Isotropic materials have a constant Young's modulus in all directions, while square anisotropic materials show different Young's modulus at different angles. By using materials with different Young's modulus at different angles, it is possible to create mysterious mechanical phenomena, such as "converting compressive stress into shear strain".

Summary

What does "compression is converted to shear" mean? was explained with the help of material mechanics.
We showed that isotropic materials do not have the above property and that square anisotropic materials with special material properties are necessary.

In the next issue, we will introduce unit cell structures with square anisotropic properties and explain how they are specifically deformed.