Mechanism creation in tensegrity structures by cellular morphogenesis

Abstract

Tensegrity structures are self-equilibrated statically and kinematically indeterminate structures. Cellular morphogenesis represents a generative process for the composition of complex tensegrity structures based on elementary cells. This article discusses the mechanism creation by the integration of mobility conditions in the generation of tensegrity structures using cellular morphogenesis. The creation of finite and infinitesimal mechanisms in tensegrity structures is described by the adhesion of cells sharing less than d nodes (d being the dimension of the workspace) and by the fusion of cells with the removal of two edges. Parametric descriptions of the infinitesimal displacements in the case of trivial and finite mechanisms are derived by the analysis of rigid assemblies corresponding to the rigid parts of the structure. In addition, an interpretation of the degeneracy of tensegrity structures in the case of self-stressable mechanisms is also presented. Analytical solutions of the degeneracy space for configurations resulting from adhesion and the fusion of two cells with the removal of two edges are described using symbolic calculations on the rigidity matrices of tensegrity structures. Although the study focuses on selected configurations and arrangements, the generalization of the ideas and findings included in this paper can lead to the generation of tensegrity structures with predefined static as well as kinematic properties, thus further enabling the application of tensegrity systems in science and engineering.

Appendices

Appendix A

• Proving that $\mathbf{t }_{{\mathrm{e}}_{\mathrm{i}}}$ is a trivial infinitesimal motion

Let $\mathbf{t }_{{\mathrm{e}}_{\mathrm{i}}}$ be the infinitesimal displacement vector defined by:

$$\begin{aligned} \mathbf{t }_{{\mathbf{e }}_{\mathrm{i}}} = \left( {\begin{array}{c} {\mathbb {O}}_{{n_{v}}}\}x_{1} \\ \vdots \\ {\mathbb {l}}_{{n_{v}}}\}x_{i} \\ \vdots \\ {\mathbb {O}}_{{n_{v}}}\}x_{d} \\ \end{array}} \right) = \mathbf{e }_{\mathbf{i }}\otimes {\mathbb {l}}_{{n_{v}}}, \end{aligned}$$

(A.1)

and $\mathrm {R}$ is the rigidity matrix of the structure defined by:

$$\begin{aligned} \mathrm {R = }\left( {\begin{array}{*{20}c} \mathrm {R}_{\mathrm{1}} &{} \mathrm {R}_{\mathrm{2}} &{} {\cdots } &{} \mathrm {R}_{\mathrm{3}}\\ \end{array} } \right) = \left( {\begin{array}{*{20}c} \mathrm {diag}\left( \mathrm {C}\mathrm {X}_{\mathrm{1}} \right) \mathrm {C} &{} \mathrm {diag}\left( \mathrm {C}\mathrm {X}_{\mathrm{2}} \right) \mathrm {C} &{} {\cdots } &{} \mathrm {diag}\left( \mathrm {C}\mathrm {X}_{\mathrm{d}} \right) \mathrm {C}\\ \end{array} } \right) . \end{aligned}$$

(A.2)

The vector $\mathrm {R}\mathbf{t }_{{\mathrm{e}}_{\mathrm{i}}}$ can be written as:

$$\begin{aligned} \sum \limits _{\mathrm{k=1}}^\mathrm {d} {\mathrm {R}_{\mathrm{k}}\mathbf{t }}_{{\mathbf{e }}_{\mathrm{i}_{\mathrm{k}}}} \end{aligned}$$

(A.3)

where $\mathrm {R}_{\mathrm{k}}\mathrm {=diag}\left( \mathrm {C}\mathrm {X}_{\mathrm{k}} \right) \mathrm {C}$ and $\mathbf{t }_{{\mathrm{e}}_{{\mathrm{i}}_{\mathrm{k}}}}=\delta _{ik}{\mathbb {l}}_{{n_{v}}}$. Equation (A.2) reduces to:

$$\begin{aligned} \mathrm {diag}\left( \mathrm {C}\mathrm {X}_{\mathrm{i}} \right) \mathrm {C} {\mathbb {l}}_{{n_{v}}}={\mathbb {O}}_{{n_{v}}}. \end{aligned}$$

(A.4)

C is the connectivity matrix of the structure also known as the node-branch incidence matrix of the underlying graph $G\left( V,E \right)$. In the graph $G\left( V,E \right)$, directions to the edges $\left( v_{i},v_{j} \right)$ can be assigned arbitrarily such that $\left( v_{i},v_{j} \right) \ne \left( v_{j},v_{i} \right)$. The $n_{e}\times n_{v}$ corresponding connectivity matrix C is indexed by the list of nodes V and the list of oriented edges E, and it is described by:

$$\begin{aligned} c_{\left( v_{i},v_{j} \right) ,v_{k}}=\left\{ {\begin{array}{cc} 1 &{} \quad \mathrm{if}\; v_{i} = v_{k} \\ -1 &{}\quad \mathrm{if}\; v_{j} = v_{k} \\ 0 &{} \quad \mathrm{otherwise} \end{array} }. \right. \end{aligned}$$

(A.5)

Clearly, the sum of the row elements of C is always 0 regardless of the topology of the graph. Consequently, vector ${\mathbb {l}}_{{n_{v}}}$ is always in the right nullspace of C. Thus,$\mathbf{t }_{{\mathrm{e}}_{\mathrm{i}}}$is in the right nullspace of the rigidity matrix $\mathrm {R}$ even when the structure is rigid, proving that it is a trivial infinitesimal translation. In addition, relation (A.6)

$$\begin{aligned} \mathbf{t }_{\mathbf{e }_{\mathrm{i}}}^{\mathbf{T }} \mathbf{t }_{\mathbf{e }_{\mathrm{j}}} = \sum \limits _{k=1}^{n_{v}} {\delta _{ik}\delta _{kj}} {\mathbb {l}}_{{n_{v}}}^{\mathrm{T}}{\mathbb {l}}_{{n_{v}}}=\delta _{ij}n_{v}, \end{aligned}$$

(A.6)

proves that the basis is orthogonal. The number of these vectors is equal to the dimension of the infinitesimal translation space T. Therefore, $\mathcal {B}\left( T \right) =\left( \mathbf{t }_{\mathbf{e }_{\mathbf{i }}} \right) _{\mathbf{e }_{\mathbf{i }}\varvec{\in }\mathcal {B}\left( R^{d} \right) }$ is a basis for T

• Proving that $\mathbf{r }_{x_{i}x_{j}}$is a trivial infinitesimal motion

Let $\mathbf{r }_{x_{i}x_{j}}$ be the infinitesimal displacement defined by:

$$\begin{aligned} \mathbf{r }_{x_{i}x_{j}} = \left( {\begin{array}{c} {\mathbb {O}}_{{n_{v}}}\}x_{1} \\ \vdots \\ - \mathbf{x} _{\mathbf{j }}\}x_{i} \\ \vdots \\ \mathbf{x} _{\mathbf{i }}\}x_{j} \\ \vdots \\ {\mathbb {O}}_{{n_{v}}}\}x_{d} \\ \end{array}} \right) = \left( \mathbf{e }_{\mathbf{j }}\otimes \mathbf{x }_{\mathbf{i }} \right) -\left( \mathbf{e }_{\mathbf{i }}\otimes \mathbf{x }_{\mathbf{j }} \right) . \end{aligned}$$

(A.7)

The vector $\mathrm {R}\mathbf{r }_{x_{i}x_{j}}$ is equal to:

$$\begin{aligned} \mathrm {R}\mathbf{r }_{x_{i}x_{j}} = \sum \limits _{\mathrm{k=1}}^\mathrm {d} {\mathrm {R}_{\mathrm{k}}}\mathbf{r }_{{\mathbf{x _\mathbf{i} }\mathbf{x }}_{{\mathrm{j}}_{\mathrm{k}}}} \end{aligned}$$

(A.8)

where $\mathrm {R}_{\mathrm{k}}\mathrm {=diag}\left( \mathrm {C}\mathbf{x }_{\mathbf{k }} \right) \mathrm {C}$ and $\mathbf{r }_{{\mathbf{x _\mathbf{i} }\mathbf{x }}_{{\mathrm{j}}_{\mathrm{k}}}}={-\delta }_{ik}\mathbf{x }_{\mathbf{j }}+\delta _{jk}\mathbf{x }_{\mathbf{i }}$. Equation (A.7) reduces to:

$$\begin{aligned} diag\left( C\mathbf{x }_{\mathbf{j }} \right) C\mathbf{x }_{\mathbf{i }}-diag\left( C\mathbf{x }_{\mathbf{i }} \right) C\mathbf{x }_{{\varvec{j}}}. \end{aligned}$$

(A.9)

Noticing that $diag\left( \mathbf{u } \right) \mathbf{v }=diag\left( \mathbf{v } \right) \mathbf{u }$, it follows that Eq. (A.7) evaluates to exactly ${\mathbb {O}}_{{n_{v}}}$ proving that $\mathbf{r }_{x_{i}x_{j}}$ is in the right nullspace of the rigidity matrix.

• Proving that $\mathcal {B}\left( M^{t} \right) =\left( \left( \mathbf{t }_{\mathrm{e}_{\mathrm{i}}} \right) _{\mathrm{e}_{i}\in \mathcal {B}\left( R^{d} \right) },\left( \mathbf{r }_{x_{i}x_{j}} \right) _{\left( i,j \right) \in \left( {\begin{array}{l} \llbracket 1,d \rrbracket \\ 2 \\ \end{array}} \right) } \right)$ is a basis of the infinitesimal trivial motions

In the planar case, $\mathcal {B}\left( M^{t} \right) =\left( \mathbf{t }_{\mathbf{e }_{1}},\mathbf{t }_{\mathbf{e }_{2}},\mathbf{r }_{xy} \right)$. Assume now that the structure has at least two distinct nodes. Let’s evaluate:

$$\begin{aligned} {\alpha }_{\mathrm{1}}\mathbf{t }_{\mathbf{e }_{1}}+\alpha _{2}\mathbf{t }_{\mathbf{e }_{2}}+\alpha _{3}\mathbf{r }_{xy} = {\mathbb {O}}_{{n_{v}}}. \end{aligned}$$

(A.10)

Then, consider the components of the summation vector that correspond to two distinct nodes in the structure $v_{1}\left( x_{1},y_{1} \right)$ and $v_{2}\left( x_{2},y_{2} \right)$. The corresponding equations are described by:

$$\begin{aligned} \alpha _{1}\left( {\begin{array}{l} 1 \\ 1 \\ 0 \\ 0 \\ \end{array}} \right) +\alpha _{2}\left( {\begin{array}{l} 0 \\ 0 \\ 1 \\ 1 \\ \end{array}} \right) +\alpha _{3}\left( {\begin{array}{l} -y_{1} \\ -y_{2} \\ x_{1} \\ x_{2} \\ \end{array}} \right) = \left( {\begin{array}{*{20}c} {\begin{array}{l} 1 \\ 1 \\ 0 \\ 0 \\ \end{array}} &{} {\begin{array}{l} 0 \\ 0 \\ 1 \\ 1 \\ \end{array}} &{} {\begin{array}{l} -y_{1} \\ -y_{2} \\ x_{1} \\ x_{2} \\ \end{array}}\\ \end{array} } \right) \left( {\begin{array}{l} \alpha _{1} \\ \alpha _{2} \\ \alpha _{3} \\ \end{array}} \right) = {\mathbb {O}}_{4}. \end{aligned}$$

(A.11)

A row echelon transformation of the matrix above can be described by:

$$\begin{aligned} \left( {\begin{array}{ccc} 1 &{} 0 &{} -y_{1} \\ 0 &{} 1 &{} x_{1} \\ 0 &{} 0 &{} y_{1}-y_{2} \\ 0 &{} 0 &{} x_{2}-x_{1} \\ \end{array}} \right) . \end{aligned}$$

(A.12)

This matrix is full rank when $x_{1}\ne x_{2}$ or $y_{1}\ne y_{2}$, which is always valid when two distinct nodes exist in the structure. Hence, the solution to Eq. (A.11) is $\alpha _{1}=0, \alpha _{2}=$0, and $\alpha _{3}=0$. Thus, the vectors of the set $\mathcal {B}\left( M^{t} \right)$ are linearly independent, and their number is equal to the dimension of $M^{t}$. Consequently, $\mathcal {B}\left( M^{t} \right)$ represents a basis for $M^{t}$ when the non-degeneracy condition is fulfilled.

In three-dimensional space, $\mathcal {B}\left( M^{t} \right) =\left( \mathbf{t }_{\mathbf{e }_{1}},\mathbf{t }_{\mathbf{e }_{2}},\mathbf{t }_{\mathbf{e }_{3}},\mathbf{r }_{xy},\mathbf{r }_{yz},\mathbf{r }_{zx} \right)$. Assume that the structure has at least three distinct nodes $v_{1}$, $v_{2}$, and $v_{3}$ that do not lie on a line. Let’s evaluate:

$$\begin{aligned} {\alpha }_{\mathrm{1}}\mathbf{t }_{\mathbf{e }_{1}}+\alpha _{2}\mathbf{t }_{\mathbf{e }_{2}}+\alpha _{3}\mathbf{t }_{\mathbf{e }_{{\varvec{3}}}}{} \mathbf + \alpha _{4}\mathbf{r }_{xy}{} \mathbf + \alpha _{5}\mathbf{r }_{yz}{} \mathbf + \alpha _{6}\mathbf{r }_{zx} = {\mathbb {O}}_{{n_{v}}}. \end{aligned}$$

(A.13)

Then, consider the components of the summation vector that correspond to $v_{1}\left( x_{1},y_{1},z_{1} \right)$, $v_{2}\left( x_{2},y_{2},z_{2} \right)$ and $v_{3}\left( x_{3},y_{3},z_{3} \right)$. The corresponding equations are described by:

$$\begin{aligned}&\alpha _{1}\left( {\begin{array}{l} 1 \\ 1 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ \end{array}} \right) +\alpha _{2}\left( {\begin{array}{l} 0 \\ 0 \\ 0 \\ 1 \\ 1 \\ 1 \\ 0 \\ 0 \\ 0 \\ \end{array}} \right) +\alpha _{3}\left( {\begin{array}{l} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 1 \\ 1 \\ 1 \\ \end{array}} \right) +\alpha _{4}\left( {\begin{array}{l} -y_{1} \\ -y_{2} \\ -y_{3} \\ x_{1} \\ x_{2} \\ x_{3} \\ 0 \\ 0 \\ 0 \\ \end{array}} \right) +\alpha _{5}\left( {\begin{array}{l} 0 \\ 0 \\ 0 \\ -z_{1} \\ -z_{2} \\ -z_{3} \\ y_{1} \\ y_{2} \\ y_{3} \\ \end{array}} \right) +\alpha _{6}\left( {\begin{array}{l} z_{1} \\ z_{2} \\ z_{3} \\ 0 \\ 0 \\ 0 \\ {-x}_{1} \\ -x_{2} \\ -x_{3} \\ \end{array}} \right) \nonumber \\&\quad = \left( {\begin{array}{*{20}c} {\begin{array}{l} 1 \\ 1 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ \end{array}} &{} {\begin{array}{l} 0 \\ 0 \\ 0 \\ 1 \\ 1 \\ 1 \\ 0 \\ 0 \\ 0 \\ \end{array}} &{} {\begin{array}{l} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 1 \\ 1 \\ 1 \\ \end{array}} &{} {\begin{array}{l} -y_{1} \\ -y_{2} \\ -y_{3} \\ x_{1} \\ x_{2} \\ x_{3} \\ 0 \\ 0 \\ 0 \\ \end{array}} &{} {\begin{array}{l} 0 \\ 0 \\ 0 \\ -z_{1} \\ -z_{2} \\ -z_{3} \\ y_{1} \\ y_{2} \\ y_{3} \\ \end{array}} &{} {\begin{array}{l} z_{1} \\ z_{2} \\ z_{3} \\ 0 \\ 0 \\ 0 \\ {-x}_{1} \\ -x_{2} \\ -x_{3} \\ \end{array}}\\ \end{array} } \right) \left( {\begin{array}{l} \alpha _{1} \\ \alpha _{2} \\ \alpha _{3} \\ \alpha _{4} \\ \alpha _{5} \\ \alpha _{6} \\ \end{array}} \right) = {\mathbb {O}}_{9}. \end{aligned}$$

(A.14)

The row echelon form of the matrix above can be calculated by symbolic calculations in MATLAB which show that the matrix is full rank when the non-degeneracy condition is fulfilled. This proves that the vectors of $B\left( M^{t} \right)$ are linearly independent and their number is equal to the dimension of $M^{t}$, which shows that this is a basis for the trivial infinitesimal motions in the space.

Appendix B

• Proving that $\mathbf{r }_{x_{i}x_{j}}^{{\bot }}$is a trivial infinitesimal motion

Let $\mathbf{r }_{x_{i}x_{j}}^{{\bot }}$ be the infinitesimal displacement defined by:

$$\begin{aligned} \mathbf{r }_{x_{i}x_{j}}^{{\bot }} = \left( {\begin{array}{c} {\mathbb {O}}_{{n_{v}}}\}x_{1} \\ \vdots \\ - \mathbf{(x }_{\mathbf{j }}-{\varvec{\omega }}_{j}\}x_{i} \\ \vdots \\ \mathbf ( {} \mathbf{x} _{\mathbf{i }}-{\varvec{\omega }}_{i}\}x_{j} \\ \vdots \\ {\mathbb {O}}_{{n_{v}}}\}x_{d} \\ \end{array}} \right) = \left( \mathbf{e }_{\mathbf{j }}\otimes {\mathbf{(x }}_{\mathbf{i }}-{\varvec{\omega }}_{\mathbf{i }} \right) -\left( \mathbf{e }_{\mathbf{i }}\otimes {\mathbf{(x }}_{\mathbf{j }}-{\varvec{\omega }}_{{\varvec{j}}} \right) = \mathbf{r }_{x_{i}x_{j}}{} \mathbf + \omega _{j}\mathbf{t }_{\mathbf{e }_{\mathbf{i }}}-\omega _{i}\mathbf{t }_{\mathbf{e }_{\mathbf{j }}}. \end{aligned}$$

(B.1)

It as shown in “Appendix A” that $\mathbf{r }_{x_{i}x_{j}}$ , $\mathbf{t }_{\mathbf{i }}$ and $\mathbf{t }_{\mathbf{j }}$ are in the right nullspace of the rigidity matrix $\mathrm {R}$. Consequently, $\mathbf{r }_{x_{i}x_{j}}^{{\bot }}$ is in the right nullspace of $\mathrm {R}$.

• Proving that $\mathbf{r }_{x_{i}x_{j}}^{{\bot }} {\bot }\mathbf{t }_{\mathbf{e }_{\mathbf{k }}}$ $\varvec{\forall }\left( i\mathbf , j \right) \varvec{\in }\left( {\begin{array}{c} \llbracket 1,d \rrbracket \\ 2 \\ \end{array}} \right)$ and $\forall \mathbf{e }_{\mathbf{k }}\in B\left( R^{d} \right)$

Let’s evaluate the product $\mathbf{t }_{\mathbf{e }_{\mathbf{k }}}^{\mathbf{T }}\mathbf{r }_{x_{i}x_{j}}^{{\bot }}$:

$$\begin{aligned} \begin{array}{l} \mathbf{t }_{\mathbf{e }_{\mathbf{k }}}^{\mathbf{T }}\mathbf{r }_{x_{i}x_{j}}^{{\bot }} = \mathbf{t }_{\mathbf{e }_{\mathbf{k }}}^{\mathbf{T }}\left( \mathbf{r }_{x_{i}x_{j}}{} \mathbf + \omega _{j}\mathbf{t }_{\mathbf{e }_{\mathbf{i }}}-\omega _{i}\mathbf{t }_{\mathbf{e }_{\mathbf{j }}} \right) \\ =\mathbf{t }_{\mathbf{e }_{\mathbf{k }}}^{\mathbf{T }}\mathbf{r }_{x_{i}x_{j}}{} \mathbf + \omega _{j}\mathbf{t }_{\mathbf{e }_{\mathbf{k }}}^{\mathbf{T }}\mathbf{t }_{\mathbf{e }_{\mathbf{i }}}-\omega _{i}\mathbf{t }_{\mathbf{e }_{\mathbf{k }}}^{\mathbf{T }}\mathbf{t }_{\mathbf{e }_{\mathbf{j }}} \\ =\left( \mathbf{e }_{{\varvec{k}}}\otimes {\mathbb {l}}_{{n_{v}}} \right) ^\mathbf{T }\left( \left( \mathbf{e }_{\mathbf{j }}\otimes \mathbf{x }_{\mathbf{i }} \right) -\left( \mathbf{e }_{\mathbf{i }}\otimes \mathbf{x }_{\mathbf{j }} \right) \right) +\omega _{j}\left( \mathbf{e }_{{\varvec{k}}}\otimes {\mathbb {l}}_{{n_{v}}} \right) ^\mathbf{T }\left( \mathbf{e }_{{\varvec{i}}}\otimes {\mathbb {l}}_{{n_{v}}} \right) -\omega _{i}\left( \mathbf{e }_{{\varvec{k}}}\otimes {\mathbb {l}}_{{n_{v}}} \right) ^\mathbf{T }\left( \mathbf{e }_{\mathbf{j }}\otimes {\mathbb {l}}_{{n_{v}}} \right) . \\ \end{array} \end{aligned}$$

(B.2)

Let A, B, C,  and D be 4 matrices such that the products AC and BD are defined. The mixed product property of the Kronecker product states that $(A\otimes B)(C\otimes D)=(AC\otimes BD)$. This property can be used to develop the expression in (B.2) which results in:

$$\begin{aligned} \begin{array}{l} \mathbf{t }_{\mathbf{e }_{\mathbf{k }}}^{\mathbf{T }}\mathbf{r }_{x_{i}x_{j}}^{{\bot }} = \left( \left( \mathbf{e }_{{\varvec{k}}}^{\mathbf{T }}\mathbf{e }_{\mathbf{j }}\otimes {{\mathbb {l}}_{{n_{v}}}^{\mathrm{T}}\mathbf{x }}_{\mathbf{i }} \right) -\left( \mathbf{e }_{{\varvec{k}}}^{\mathbf{T }}\mathbf{e }_{\mathbf{i }}\otimes {{\mathbb {l}}_{{n_{v}}}^{\mathrm{T}}\mathbf{x }}_{\mathbf{j }} \right) \right) +\omega _{j}\left( \mathbf{e }_{{\varvec{k}}}^{\mathbf{T }}\mathbf{e }_{{\varvec{i}}}\otimes {\mathbb {l}}_{{n_{v}}}^{\mathrm{T}}{\mathbb {l}}_{{n_{v}}} \right) -\omega _{i}\left( {\mathbf{e }_{{\varvec{k}}}^{\mathbf{T }}\mathbf{e }}_{\mathbf{j }}\otimes {{\mathbb {l}}_{{n_{v}}}^{\mathrm{T}}{\mathbb {l}}}_{{n_{v}}} \right) \\ =\delta _{jk}\left( \sum \limits _{\lambda =1}^{n_{v}} x_{i\lambda } \right) -\delta _{ik}\left( \sum \limits _{\lambda =1}^{n_{v}} x_{j\lambda } \right) +\omega _{j}\delta _{ik}n_{v}-\omega _{i}\delta _{jk}n_{v}. \\ \end{array} \end{aligned}$$

(B.3)

Obviously, when $k\ne i\ne j$, this product evaluates to 0 since $\delta _{ik}=\delta _{jk}=0$. Notice that $\omega _{i}=\frac{1}{n_{v}}\left( \sum \limits _{\lambda =1}^{n_{v}} x_{i\lambda } \right) , \quad \omega _{j}=\frac{1}{n_{v}}\left( \sum \limits _{\lambda =1}^{n_{v}} x_{j\lambda } \right)$ by the definition of the geometric centroid. Consequently, this product is 0 also when $k=i$or$k=j$. Hence, $\mathbf{r }_{x_{i}x_{j}}^{{\bot }} { \bot }\mathbf{t }_{\mathbf{e }_{\mathbf{k }}}$ $\varvec{\forall }\left( i\mathbf , j \right) \varvec{\in }\left( {\begin{array}{c} \llbracket 1,d \rrbracket \\ 2 \\ \end{array}} \right)$ and $\forall \mathbf{e }_{\mathbf{k }}\in B\left( R^{d} \right)$.

• Proving that $B\left( M^{t} \right) =\left( \left( \mathbf{t }_{\mathrm{e}_{\mathrm{i}}}^{{\bot }} \right) _{\mathrm{e}_{i}\in B\left( R^{d} \right) },\left( \mathbf{r }_{x_{i}x_{j}}^{{\bot }} \right) _{\left( i,j \right) \in \left( {\begin{array}{c} \llbracket 1,d \rrbracket \\ 2 \\ \end{array}} \right) } \right)$ is a basis of the infinitesimal trivial motions

Let’s evaluate the linear combination:

$$\begin{aligned} {\alpha }_{\mathrm{1}}\mathbf{t }_{\mathbf{e }_{1}}^{{\bot }}+\alpha _{2}\mathbf{t }_{\mathbf{e }_{2}}^{{\bot }}+\alpha _{3}\mathbf{t }_{\mathbf{e }_{{\varvec{3}}}}^{{\bot }}{} \mathbf + \alpha _{4}\mathbf{r }_{xy}^{{\bot }}{} \mathbf + \alpha _{5}\mathbf{r }_{yz}^{{\bot }}{} \mathbf + \alpha _{6}\mathbf{r }_{zx}^{{\bot }} = {\mathbb {O}}_{{n_{v}}}. \end{aligned}$$

(B.4)

Equation (B.4) can be rewritten as:

$$\begin{aligned} {\alpha }_{\mathrm{1}}\mathbf{t }_{\mathbf{e }_{{\varvec{1}}}}+\alpha _{2}\mathbf{t }_{\mathbf{e }_{{\varvec{2}}}}+\alpha _{3}\mathbf{t }_{\mathbf{e }_{{\varvec{3}}}}{} \mathbf + \alpha _{4}\left( \mathbf{r }_{xy}{} \mathbf + \omega _{{\varvec{y}}}\mathbf{t }_{\mathbf{e }_{{\varvec{1}}}}-\omega _{{\varvec{x}}}\mathbf{t }_{\mathbf{e }_{{\varvec{2}}}} \right) \mathbf + \alpha _{5}\left( \mathbf{r }_{yz}{} \mathbf + \omega _{{\varvec{z}}}\mathbf{t }_{\mathbf{e }_{{\varvec{2}}}}-\omega _{{\varvec{y}}}\mathbf{t }_{\mathbf{e }_{{\varvec{3}}}} \right) \mathbf + \alpha _{6}\left( \mathbf{r }_{zx}{} \mathbf + \omega _{{\varvec{x}}}\mathbf{t }_{\mathbf{e }_{{\varvec{3}}}}-\omega _{{\varvec{z}}}\mathbf{t }_{\mathbf{e }_{{\varvec{1}}}} \right) = {\mathbb {O}}_{{n_{v}}}\nonumber \\ \end{aligned}$$

(B.5)

When the coefficients of each vector in Eq. (B.5) are grouped together, it becomes:

$$\begin{aligned} \left( {\alpha }_{\mathrm{1}}+\alpha _{4}\omega _{y}-\alpha _{6}\omega _{z} \right) \mathbf{t }_{\mathbf{e }_{1}}+\left( \alpha _{2}+\alpha _{5}\omega _{z}-\alpha _{4}\omega _{x} \right) \mathbf{t }_{\mathbf{e }_{2}}+\left( \alpha _{3}+\alpha _{6}\omega _{x}-\alpha _{5}\omega _{y} \right) \mathbf{t }_{\mathbf{e }_{{\varvec{3}}}}{} \mathbf + \alpha _{4}\mathbf{r }_{xy}{} \mathbf + \alpha _{5}\mathbf{r }_{yz}{} \mathbf + \alpha _{6}\mathbf{r }_{zx} = {\mathbb {O}}_{{n_{v}}}\nonumber \\ \end{aligned}$$

(B.6)

The vectors $\left( \mathbf{t }_{\mathbf{e }_{1}},\mathbf{t }_{\mathbf{e }_{2}},\mathbf{t }_{\mathbf{e }_{3}},\mathbf{r }_{xy},\mathbf{r }_{yz},\mathbf{r }_{zx} \right)$ are already shown to be linearly independent in “Appendix A”. It follows that $\alpha _{4}=\alpha _{5}=\alpha _{6}=0$, and consequently, $\alpha _{1}=\alpha _{2}=\alpha _{3}=0$. Since the number of vectors is equal to the dimension of $M^{T}$, $B\left( M^{t} \right) =\left( \mathbf{t }_{\mathbf{e }_{1}}^{{\bot }},\mathbf{t }_{\mathbf{e }_{2}}^{{\bot }},\mathbf{t }_{\mathbf{e }_{3}}^{{\bot }},\mathbf{r }_{xy}^{{\bot }},\mathbf{r }_{yz}^{{\bot }},\mathbf{r }_{zx}^{{\bot }} \right)$ is a basis for $M^{t}$ which concludes the proof for dimension $d=3$. For the planar case, the proof is similar.

• Geometric interpretation of $\mathbf{r }_{x_{i}x_{j}}^{{\bot }}$

Let ${\Omega }\left( \omega _{1},\cdots ,\omega _{d} \right)$ be the geometric centroid of $n_{v}$ points $\mathbf{p }_{\mathbf{i }}\left( x_{i1},\cdots ,x_{id} \right)$ in a d-dimensional space ($d=2$ or $d=3)$ . $\omega _{k}$ is defined as $\frac{1}{n_{v}}\sum \limits _{\lambda =1}^{n_{v}} x_{\lambda k}$. Consider the finite rotation in the plane around the centroid ${\Omega }$ or in space around an axis $({\Omega , }\mathbf{e }_{\mathbf{1 }})$ with $\mathbf{e }_{\mathbf{1 }}\in B\left( R^{d} \right)$. In both cases, the finite rotation matrix ${\mathrm {Rot}}_{d}$ is described by:

$$\begin{aligned} {\mathrm {Rot}}_{2}=\left( {\begin{array}{cc} \cos \left( \theta \right) &{} -\sin \left( \theta \right) \\ \sin \left( \theta \right) &{} cos\left( \theta \right) \\ \end{array} } \right) , \quad {\mathrm {Rot}}_{\mathrm{3}} = \left( {\begin{array}{ccc} \cos \left( \theta \right) &{} -\sin \left( \theta \right) &{} 0\\ \sin \left( \theta \right) &{} \cos \left( \theta \right) &{} 0\\ 0 &{} 0 &{} 1\\ \end{array} } \right) . \end{aligned}$$

(B.7)

However, the rotational transformation is described by:

$$\begin{aligned} {\mathbf{p' }_{\mathbf{i }}\mathrm {=Rot}}_{d}\left( \mathbf{p} _{\mathbf{i }}-{\varvec{\Omega }} \right) +{\varvec{\Omega }}. \end{aligned}$$

(B.8)

When $\theta$ is infinitesimal, the transformations in Eq. (B.8) can be rewritten as:

$$\begin{aligned}&\mathbf{p' }_{\mathbf{i }} = \left( {\begin{array}{*{20}c} 1 &{} -\delta \theta \\ \delta \theta &{} 1\\ \end{array} } \right) \left( \mathbf{p} _{\mathbf{i }}-{\varvec{\Omega }} \right) +{\varvec{\Omega }}, \nonumber \\&\mathbf{p' }_{\mathbf{i }} = \left( {\begin{array}{*{20}c} 1 &{} -\delta \theta &{} 0\\ \delta \theta &{} 1 &{} 0\\ 0 &{} 0 &{} 1\\ \end{array} } \right) \left( \mathbf{p} _{\mathbf{i }}-{\varvec{\Omega }} \right) +{\varvec{\Omega }} \end{aligned}$$

(B.9)

which gives the infinitesimal displacements:

$$\begin{aligned}&\mathbf{r }_{xy}^{{\bot }}=\delta \theta \left( {\begin{array}{l} {-(y}_{i}-\omega _{y} \\ {(x}_{i}-\omega _{x}) \\ \end{array}} \right) \hbox {in } 2 d, \nonumber \\&\mathbf{r }_{xy}^{{\bot }}=\delta \theta \left( {\begin{array}{c} -{(y}_{i}-\omega _{y}) \\ (x_{i}-\omega _{x}) \\ 0 \\ \end{array}} \right) \hbox {in } 3 d. \end{aligned}$$

(B.10)

This proves that the infinitesimal rotations $\mathbf{r }_{xy}^{{\bot }}$ when integrated describe finite rotations around the centroid of the structure or an axis passing by ${\Omega }$ and in the same direction as $\mathbf{e }_{\mathbf{3 }}$. For the rotations, the interpretations are similar.