剛性異方性構造の簡易的な挙動計算について

\(\boldsymbol{T} = [S] \boldsymbol{W}\\
\begin{bmatrix} d_x \\ d_y \\ \theta_z \end{bmatrix}
=\begin{bmatrix} \frac{l^3}{3EI} & 0 & \frac{l^2}{2EI} \\
                0 & \frac{l}{Ebh} &  0  \\
                 \frac{l^2}{2EI} & 0 &  \frac{l}{EI}\end{bmatrix}
\begin{bmatrix} F_x \\ F_y \\ M_z \end{bmatrix}\)

\(I = \frac{bh^3}{12}\)

\(\boldsymbol{W} = [Q]\boldsymbol{T}\\
\begin{bmatrix} F_x \\ F_y \\ M_z \end{bmatrix}
=\begin{bmatrix}  \frac{12EI}{l^3} & 0 & -\frac{6EI}{l^2} \\
                  0 & \frac{bh}{l} &  0  \\
                  -\frac{6EI}{l^2} & 0 & \frac{4EI}{l} \end{bmatrix}
\begin{bmatrix} d_x \\ d_y \\ \theta_z \end{bmatrix}\)

\(\boldsymbol{W'} = [Q'] \boldsymbol{T'}\)

\(\begin{bmatrix} F_x' \\ F_y' \\ M_z' \end{bmatrix} =
\begin{bmatrix} \cos{\phi} & \sin{\phi} & 0 \\
                -\sin{\phi} & \cos{\phi} & 0 \\
                L_y & -L_x & 1 \end{bmatrix}
\begin{bmatrix} F_x \\ F_y \\ M_z \end{bmatrix}\\\)

\(\begin{bmatrix} F_x \\ F_y \\ M \end{bmatrix} =
\begin{bmatrix} \cos{\phi} & -\sin{\phi} & 0 \\
                \sin{\phi} & \cos{\phi} & 0 \\
                L_x\sin{\phi}-L_y\cos{\phi} & L_x\cos{\phi}+L_y\sin{\phi} & 1 \end{bmatrix}
\begin{bmatrix} F_x' \\ F_y' \\ M_z' \end{bmatrix} \\
\boldsymbol{W }= [N] \boldsymbol{W'}\)

\(N_{31} = (\boldsymbol{L}\times\boldsymbol{n_{x'}})\cdot \boldsymbol{e_3}\\ 
N_{32} = (\boldsymbol{L}\times\boldsymbol{n_{y'}})\cdot \boldsymbol{e_3}\\\)

\(\begin{bmatrix} d_x' \\ d_y' \\ \theta_z' \end{bmatrix} =
\begin{bmatrix} \cos{\phi} & \sin{\phi} & L_x\sin{\phi}-L_y\cos{\phi}\\
                -\sin{\phi} & \cos{\phi} & L_x\cos{\phi}+L_y\sin{\phi} \\
                0 & 0 & 1 \end{bmatrix}
\begin{bmatrix} d_x \\  d_y \\ \theta_z \end{bmatrix} \\
\boldsymbol T' =[N]^\text T \boldsymbol T\)

\(\boldsymbol W = [N][Q'][N]^\text T \boldsymbol T\)

\([Q] =
\begin{bmatrix} {1 \over \sqrt{2}} & -{1 \over \sqrt{2}} & 0 \\
                {1 \over \sqrt{2}} & {1 \over \sqrt{2}} & 0 \\
                {1 \over \sqrt{2}}L_x & {1 \over \sqrt{2}}L_x & 1 \end{bmatrix}
\begin{bmatrix}  \frac{12EI}{l^3} & 0 & -\frac{6EI}{l^2} \\
                  0 & \frac{bh}{l} &  0  \\
                  -\frac{6EI}{l^2} & 0 & \frac{4EI}{l} \end{bmatrix}
\begin{bmatrix} {1 \over \sqrt{2}} & {1 \over \sqrt{2}} & {1 \over \sqrt{2}}L_x\\
                -{1 \over \sqrt{2}} & {1 \over \sqrt{2}} & {1 \over \sqrt{2}}L_x \\
                0 & 0 & 1 \end{bmatrix} \\
= \begin{bmatrix}
{6E \over l^3} + {l\over 2bh} & {6E \over l^3} - {l\over 2bh} & {-3\sqrt{2}E \over l^2}+{6EL_x \over l^3}+{lL_x \over 2bh}\\
& {6E \over l^3} + {l\over 2bh} & {-3\sqrt{2}E \over l^2}+{6EL_x \over l^3}-{lL_x \over 2bh}\\
\text{sym.} & & 
{4E \over l} - {6\sqrt{2}EL_x \over l^2} + {6EL_x^2 \over l^3} + {l L_x^2 \over 2bh}
\end{bmatrix}\\\)

\([Q] =　\begin{bmatrix}
\frac{24EI}{l^3} & 0 & \frac{12EI}{l^3}(L_{x1}+L_{x2}-l)\\
0 & \frac{2bh}{l} &  0  \\
\frac{12EI}{l^3}(L_{x1}+L_{x2}-l) & 0 & 
\frac{12EI}{l^3}(L_{x1}^2+L_{x2}^2-L_{x1}l-L_{x2}l)+ \frac{8EI}{l}\end{bmatrix}\)

\([Q] =　\begin{bmatrix}  
\frac{24EI}{l^3} & 0 & -\frac{12EI}{l^2}\\
0 & \frac{2bh}{l} &  0  \\
-\frac{12EI}{l^2} & 0 & \frac{24EI}{l^3}L^2 + \frac{8EI}{l}\end{bmatrix}\)

Introduction

　In compliant mechanisms and mechanical metamaterials that utilize elastic deformation, deformation in unintended directions may be unfavorable. Therefore, they are required to have sufficient stiffness in directions where deformation is not desired. Compliant mechanisms are designed on the basis with stiffness anisotropy. In this article, we will discuss structures with anisotropic macroscopic stiffness not material anisotropy.

　In order to design a structure with the expected deformation behaviors, it is important to understand the stiffness in each direction, in other words, the relationship between the deformation and the applied load. Although this relationship can be calculated using the finite element method, it is important to use a simplified model to reduce calculation time in order to perform many trials for optimization. This article introduces a simple calculation method for compliant mechanism behavior that utilizes basic material mechanics and rigid body mechanics.

Slider structure

　One of the structures frequently used as a component of compliant mechanisms is a structure called a slider structure (parallel-guiding mechanism). As described in previous article, this structure is rigid against deformations other than unidirectional and induces translational movement. Although this structure can be derived from the concept of constraining the deformation degrees of freedom, I would like to explain the stiffness characteristics using mathematical formulas in this article.

Setting up the problem

　Set the following conditions for considering the structure. In this case, we consider a structure that connects upper and lower rigid parts in a 20 mm cubic design space.

Let us assume that we need a structure that allows the upper part to translate in the x-direction relative to the lower part (and minimize the amount of deformation and rotation in other directions as much as possible). In other words, we consider a structure with low stiffness in the X direction and high stiffness in the other directions.

　We also consider the following conditions

• The beam should be a 2-dimensional extruded shape for ease of fabrication.
• The thickness of the beam should be 1 mm.
• Assuming a one-piece structure, only polypropylene resin (Young's modulus 2.0 GPa) should be used as material.

Deformation of a cantilever beam

　First, we consider a single cantilever beam. For simplicity, we will only consider deformations in the XY plane.

Suppose a uniform beam of width \(b\), thickness \(h\), and length \(L\) is placed in the Y direction as shown in the figure. When a load \(\boldsymbol W=[F_x, F_y, M_z]^T\) is applied to the end of the beam, the displacement and rotation \(\boldsymbol T=[d_x, d_y, \theta_z]^T\) of the end in the infinitesimal deformation state are expressed as follows

\(\boldsymbol{T} = [S] \boldsymbol{W}\\
\begin{bmatrix} d_x \\ d_y \\ \theta_z \end{bmatrix}
=\begin{bmatrix} \frac{l^3}{3EI} & 0 & \frac{l^2}{2EI} \\
                0 & \frac{l}{Ebh} &  0  \\
                 \frac{l^2}{2EI} & 0 &  \frac{l}{EI}\end{bmatrix}
\begin{bmatrix} F_x \\ F_y \\ M_z \end{bmatrix}\)

where \(E\) is the Young's modulus of the material. \(I\) is the sectional second moment, which for a rectangular section is expressed by the following equation

\(I = \frac{bh^3}{12}\)

The \([S]\) is the compliance matrix, which represents the ease of deformation with respect to the load. Also, the stiffness matrix \([Q]\) is the inverse of the compliance matrix and is given by

\(\boldsymbol{W} = [Q]\boldsymbol{T}\\
\begin{bmatrix} F_x \\ F_y \\ M_z \end{bmatrix}
=\begin{bmatrix}  \frac{12EI}{l^3} & 0 & -\frac{6EI}{l^2} \\
                  0 & \frac{bh}{l} &  0  \\
                  -\frac{6EI}{l^2} & 0 & \frac{4EI}{l} \end{bmatrix}
\begin{bmatrix} d_x \\ d_y \\ \theta_z \end{bmatrix}\)

The compliance or stiffness matrix can be used to obtain one of the motion and loads from the other.

　For example, if \(h = 1 \text{ mm}, b = 20 \text{ mm}, l = 20 \text{ mm}\) and 1 N load is applied at different angles, the absolute translation \(|d_x|, |d_y|\) and rotation \(|\theta_z|\) are shown in the graph below.

The displacement in the y-direction is very small compared to the translation in the x-direction. Thus, it can be seen that a uniform beam with a sufficiently large length relative to the thickness has high stiffness in the longitudinal direction and low stiffness in the thickness direction. Also, uniform beams are less stiff against rotation, as we will compare with slider structures below.

  In the above calculation, the rotation is caused by a load in the x-direction. For an unloaded uniform beam, the center of rotation is on \(y= {2 \over 3}l\). To approach translational motion in the x direction, the center of rotation must be farther away in the y direction. Theoretically, it is possible to prevent rotational motion by applying a moment load in the opposite direction, but it is not feasible to strictly control the load in a real product, so translational motion is achieved by utilizing the structural characteristics.

Changing beam positions

　Next, consider the case where the beams are connected at different positions and angles in the XY plane.

In a coordinate system with the origin at the beam tip (let \(\text O'\)), the displacement-load relationship holds as before above.

\(\boldsymbol{W'} = [Q'] \boldsymbol{T'}\)

If the position vector of the \(\text O'\) origin in the original coordinate system (let \(\text O\)) is \(\boldsymbol{L}=(L_x, L_y)\) and the angle is \(\phi\), then when the load \(\text O\) is applied to \(\boldsymbol W\), the load \(\boldsymbol W'\) acting on \(\text O'\) when loaded with \(W\) is expressed as follows.

\(\begin{bmatrix} F_x' \\ F_y' \\ M_z' \end{bmatrix} =
\begin{bmatrix} \cos{\phi} & \sin{\phi} & 0 \\
                -\sin{\phi} & \cos{\phi} & 0 \\
                L_y & -L_x & 1 \end{bmatrix}
\begin{bmatrix} F_x \\ F_y \\ M_z \end{bmatrix}\\\)

Let '$[N]$' be the coordinate transformation matrix from \(\boldsymbol W'\) to \(\boldsymbol W\).

\(\begin{bmatrix} F_x \\ F_y \\ M \end{bmatrix} =
\begin{bmatrix} \cos{\phi} & -\sin{\phi} & 0 \\
                \sin{\phi} & \cos{\phi} & 0 \\
                L_x\sin{\phi}-L_y\cos{\phi} & L_x\cos{\phi}+L_y\sin{\phi} & 1 \end{bmatrix}
\begin{bmatrix} F_x' \\ F_y' \\ M_z' \end{bmatrix} \\
\boldsymbol{W }= [N] \boldsymbol{W'}\)

In this case, if the unit vectors in the X'-direction and \(\boldsymbol{n_{x'}}\) and \(\boldsymbol{n_{y'}}\) are respectively, the first and second components of the third row of the transformation matrix \([N]\) are expressed as follows.

\(N_{31} = (\boldsymbol{L}\times\boldsymbol{n_{x'}})\cdot \boldsymbol{e_3}\\ 
N_{32} = (\boldsymbol{L}\times\boldsymbol{n_{y'}})\cdot \boldsymbol{e_3}\\\)

Also, the displacement of \(\text O\) and the displacement of \(\text O'\) within the same object in motion can be expressed by the following relation.

\(\begin{bmatrix} F_x' \\ F_y' \\ M_z' \end{bmatrix} =
\begin{bmatrix} \cos{\phi} & \sin{\phi} & 0 \\
                -\sin{\phi} & \cos{\phi} & 0 \\
                L_y & -L_x & 1 \end{bmatrix}
\begin{bmatrix} F_x \\ F_y \\ M_z \end{bmatrix}\\\)

The relationship between load, displacement, and rotation in `\(\text{O}\) is then given by the following equation

\(\boldsymbol W = [N][Q'][N]^\text T \boldsymbol T\)

 For example, if the beam is placed at \(\boldsymbol L=(L_x,0)\) with angle \(\phi = {\pi \over 4}\), the stiffness matrix \([Q]=[N][Q'][N]^\text T\) will be as follows.

\([Q] =
\begin{bmatrix} {1 \over \sqrt{2}} & -{1 \over \sqrt{2}} & 0 \\
                {1 \over \sqrt{2}} & {1 \over \sqrt{2}} & 0 \\
                {1 \over \sqrt{2}}L_x & {1 \over \sqrt{2}}L_x & 1 \end{bmatrix}
\begin{bmatrix}  \frac{12EI}{l^3} & 0 & -\frac{6EI}{l^2} \\
                  0 & \frac{bh}{l} &  0  \\
                  -\frac{6EI}{l^2} & 0 & \frac{4EI}{l} \end{bmatrix}
\begin{bmatrix} {1 \over \sqrt{2}} & {1 \over \sqrt{2}} & {1 \over \sqrt{2}}L_x\\
                -{1 \over \sqrt{2}} & {1 \over \sqrt{2}} & {1 \over \sqrt{2}}L_x \\
                0 & 0 & 1 \end{bmatrix} \\
= \begin{bmatrix}
{6E \over l^3} + {l\over 2bh} & {6E \over l^3} - {l\over 2bh} & {-3\sqrt{2}E \over l^2}+{6EL_x \over l^3}+{lL_x \over 2bh}\\
& {6E \over l^3} + {l\over 2bh} & {-3\sqrt{2}E \over l^2}+{6EL_x \over l^3}-{lL_x \over 2bh}\\
\text{sym.} & & 
{4E \over l} - {6\sqrt{2}EL_x \over l^2} + {6EL_x^2 \over l^3} + {l L_x^2 \over 2bh}
\end{bmatrix}\\\)

In this case, \(h = 1 \text{ mm}, b = 20 \text{ mm}, l = 20\sqrt{2} \text{ mm}, L_x=10 \text{ mm}\) and the absolute value of displacement \(|d_x|,| d_y|\) when \(1\text{ N}\) load is applied with different angle and The amount of rotation \(|theta_z|\) is as shown in the graph below.

Compared to a simple cantilever beam, the translation values and the maximum load angle are different. Thus, simply by changing the position and orientation of the beams, the properties of the structure can be significantly altered.

　Then, is it possible to have a structure that achieves translational movement with only one beam? If the beams are placed to connect the upper and lower members within the above design range with \(h = 1 \text{ mm}, b = 20 \text{ mm}\), the possible range of behavior \(d_x, d_y, \theta_z\) under 1 N load in the X direction is as follows.

It can be seen that at any beam position, a rotation angle \(\theta_z\) occurs, and that the structure cannot be realized as a translational movement with only one beam.

Series and parallel beams

　Consider a structure with multiple beams. When beams are in series or parallel, compliance and stiffness are obtained by adding up the compliance and stiffness, just like spring constants.

Using these, the values of translation and rotation \(d_x, d_y, \theta_z\) are calculated for the case of two beams with 1N force in the x-direction, as shown in the figure below. The blue plot shows the series case and the red plot shows the parallel case, each plotting approximately 400,000 points. The black dotted line indicates the area that can be taken in the case of a single beam, which is included in the series case. The parallel case does not include the case where the beams intersect each other.

In the series case, the beams are softer and the amount of rotation increases along with the displacement in the x-direction. It can be seen that the beams need to be connected in parallel in order to reduce the rotation.

Stiffness of slider structure

　Having said all that up front, let's derive the behavior of a slider structure. The slider structure consists of parallel beams, and if the position vectors of the two beam tips are \((L_{x1},0)\) and \((L_{x2},0)\), the stiffness matrix is as follows

\([Q] =　\begin{bmatrix}
\frac{24EI}{l^3} & 0 & \frac{12EI}{l^3}(L_{x1}+L_{x2}-l)\\
0 & \frac{2bh}{l} &  0  \\
\frac{12EI}{l^3}(L_{x1}+L_{x2}-l) & 0 & 
\frac{12EI}{l^3}(L_{x1}^2+L_{x2}^2-L_{x1}l-L_{x2}l)+ \frac{8EI}{l}\end{bmatrix}\)

Let \(L_{x1}=-L_{x2}=L\) , then

\([Q] = \begin{bmatrix}  
\frac{24EI}{l^3} & 0 & -\frac{12EI}{l^2}\\
0 & \frac{2bh}{l} & 0 \\
-\frac{12EI}{l^2} & 0 & \frac{24EI}{l^3}L^2 + \frac{8EI}{l}\end{bmatrix}\)

Components are twice that of the cantilever beam, except for the diagonal components to the rotation \(Q_{33}\) has an additional term proportional to the square of \(L\). This indicates that the cantilever beam is now stiffer against rotation (additional degrees-of-freedom constraint). Also, the center of rotation when no moment load is acting is on \(y= {2 \over 3}l+{2 L^2\over l}\) (you can do the calculation if you are interested), and the larger \(L\) is, the closer to translational motion it is.

　Specifically, let \(h = 1 \text{ mm}, b = 20 \text{ mm}, l = 20 \text{ mm}\) and let us look at the behavior of a beam placed parallel to the beam at \(\pm10 \text{ mm}\). The translation and rotation \(d_x, d_y, \theta _z\) values are located in the figure below.

The behavior of a 1N load at different angles is shown below. Only the scale of the rotation angle differs from the figure shown for the cantilever. The displacements in X and Y direction of the cantilever beam show with dotted lines.

It can be seen that the rotation is much smaller than in the cantilever beam case. Thus, in the slider structure, the beams are arranged to increase the stiffness in all but one direction, thereby inducing translational movement.

  In the slider structure, two beams are connected in parallel, which makes it stiffer than a cantilever beam. The displacement in the y-direction under a load in the x-direction is inversely proportional to the cross-sectional area, which is 1/2 of that of a cantilever beam. On the other hand, the displacement in the X direction is about 1/8 due to the higher rigidity against rotation.

How should the structure be designed if we want to reduce the stiffness in the X direction more? This will be discussed in the next article.

Summary

　Using simple material mechanics and rigid body mechanics, we have shown how to compute the deformation behavior of compliant mechanisms using a slider structure with stiffness anisotropy as an example.
　In the next article, we will show how to make the anisotropy more pronounced and the challenges in doing so.