0.2 Tensegrity  (Page 2/2)

Definition 2 A Michell Topology of order $q$ consists of Michell spirals of order less than or equal to $q$ and their conjugate spirals where [link] and [link] hold.

Force equilibrium at a generic node

A force applied to a node can be represented by the vector ${\mathbf{w}}_{ik}$ applied to the node ${\mathbf{n}}_{ik}$ and written in complex form as

where

and ${\theta }_{ik}$ is the angle at which the external force ${\mathbf{w}}_{ik}$ is applied, measured from the radial line to node ${\mathbf{n}}_k$ .

At node ${\mathbf{n}}_{ik}$ , the forces along directions ${\mathbf{m}}_{ik},{\overline{\mathbf{m}}}_{ki},{\mathbf{m}}_{i,k-1},{\overline{\mathbf{m}}}_{k,i-1}$ , respectively, have magnitudes $f_{ik},t_{ki},f_{i,k-1},t_{k,i-1}$ . Therefore, the sum of forces at node ${\mathbf{n}}_{ik}$ yields

Multiplying [link] by the vector $p_{i+k}{e}^{j(k-i)\phi }$ gives

Because both the real and imaginary parts of the above equation must be zero,

Linear propogation of forces

Define a vector ${\mathbf{x}}_{\alpha }\in {\mathbb{R}}^{2(\alpha +1)}$ which contains forces in all members that lie within the radii $r_{\alpha }$ and $r_{\alpha +1}$ . That is,

${\mathbf{x}}_{\alpha }=\left[\begin{array}{c}{t}_{0,\alpha }\\ f_{\alpha ,0}\\ t_{1,\alpha -1}\\ f_{\alpha -1,1}\\ ⋮\\ t_{i,\alpha -i}\\ f_{\alpha -1,i}\\ ⋮\\ t_{\alpha ,0}\\ f_{0,\alpha }\end{array}\right]$

Define a vector ${\mathbf{u}}_{\alpha }$ such that

${\mathbf{u}}_{\alpha }=\left[\begin{array}{c}{w}_{\alpha +1,0}\\ w_{\alpha ,1}\\ w_{\alpha -1,2}\\ w_{\alpha -2,3}\\ w_{\alpha -3,4}\\ ⋮\\ w_{0,\alpha +1}\end{array}\right]$

Theorem 1 Let a truss be arranged according to the Michell topology of order $q$ , satisfying the conditions in the definitions, having external forces ${\mathbf{w}}_{ik}$ applied at the nodes ${\mathbf{n}}_{ik}$ with $i,k\ge 0$ and $i+k\le q$ . Let ${\mathbf{x}}_{\alpha }$ contain the forces(normalized by the member length) in all members within the band of members between radii $r_{\alpha }$ and $r_{\alpha +1}$ . Then the forces propagate from one band to the next according to the linear recursive equation ${\mathbf{x}}_{\alpha +1}={\mathbf{A}}_{\alpha }{\mathbf{x}}_{\alpha }+{\mathbf{B}}_{\alpha }{\mathbf{u}}_{\alpha }$ for some matrices ${\mathbf{A}}_{\alpha }$ and ${\mathbf{B}}_{\alpha }$ .

Michell topologies under a single bending load

If $t_{ik}{f}_{ik}<0$ , then one member of the truss is in compression while the other is in tension, so the truss is said to be bending.

Theorem 2 Let a truss be arranged according to the Michell topology of order $q$ satisfying [link] and with $a$ and $c$ given by [link] . Assume the only force applied on the truss is $\mathbf{w}=w{e}^{j\theta }$ at node ${\mathbf{n}}_{00}$ . If

then the forces at all nodes satisfy

Furthermore, if

then $t_{ik}>0$ for all $i\le q$ , $k\le i$ . Otherwise, if

then $t_{ik}<0$ for all $i\le q$ , $k\le i$ .

First, take the case where $q=1$ . In this case, ${\mathbf{n}}_{00}$ is the only node at which external forces are applied:

In the case when $0\le \beta \le \frac{\pi }{2}$ , there are three cases of interest. The first is when $|{\theta }_{00}|<\beta <\frac{\pi }{2}$ . Then $sin\left(2\beta \right)\ge 0$ and

which implies $t_{00}<0$ and $f_{00}<0$ , or, in other words, that both members of the truss are in tension.

The second case is when $|{\theta }_{00}-\pi |<\beta <\frac{\pi }{2}$ . A similar analysis shows that $t_{00}>0$ and $f_{00}>0$ so both members of the truss are in compression.

Lastly, when $\beta <|{\theta }_{00}|<\pi -\beta$ then $t_{00}{f}_{00}<0$ , indicating that one member of the truss is in tension and the other in compression, so the truss is said to be bending.

Now, consider the case where $q>1$ and assume that $0<\beta <{\theta }_{00}<\pi -\beta$ so that $t_{00}>0$ and $f_{00}<0$ (the opposite case is handled similarly). Consider the node ${\mathbf{n}}_{01}$ . In the force diagram, ${\mathbf{w}}_{01}=0$ by assumption, hence $t_{1,-1}=0$ because ${\mathbf{n}}_{01}$ is at the lower boundary of the truss. Repeat the analysis from before for the fictitious force ${\mathbf{w}}_{01}=\left|f_{00}\right|{\mathbf{m}}_{00}$ . Because the truss has $a<1$ , $\beta <{\theta }_{01}=\phi +\beta <\pi$ , so that $t_{10}>0$ and $f_{01}<0$ . The same idea, applied to all nodes on the upper boundary with $k\ge q$ , proves that $t_{k0}>0$ and $f_{0k}<0$ for all $k\le q$ .

The analysis at node ${\mathbf{n}}_{10}$ is similar. Repeat the above analysis for a fictitious force ${\mathbf{w}}_{10}=-\left|t_{00}\right|{\overline{\mathbf{m}}}_{00}$ . The vector has a negative sign because the angle ${\theta }_{10}$ is measured clockwise, so the vector needs to be flipped to get the same angle $\theta$ as before. Therefore, for all nodes ${\mathbf{n}}_{i0}$ with $i\le q$ , $t_{0k}>0$ and $f_{i0}<0$ for all $k\le q$ .

Continuing on, for the node ${\mathbf{n}}_{11}$ , use a fictitious force ${\mathbf{w}}_{11}=-|t_{01}|{\overline{\mathbf{m}}}_{01}+\left|f_{10}\right|{\mathbf{m}}_{10}$ . Because $a<1$ , one can conclude that $\beta <{\theta }_{11}<\pi -\beta$ to get that $t_{11}>0$ and $f_{11}<0$ , and continuing in this fashion gives $t_{ii}>0$ and $f_{ii}<0$ for all $i\le q$ .

The last type of node to consider is an interior node of the form ${\mathbf{n}}_{ik}$ where $i\ne k\ne 0$ . At the node ${\mathbf{n}}_{ik}$ create the fictitious force ${\mathbf{w}}_{ik}=-|t_{i-1,k}|{\overline{\mathbf{m}}}_{20}+\left|f_{ii}\right|{\mathbf{m}}_{ii}$ . The assumption that $a<1$ leads to the conclusion that $\beta <{\theta }_{ik}<\pi -\beta$ , yielding $t_{ki}>0$ and $f_{ik}<0$ . Therefore, when $\beta <{\theta }_{00}<\pi -\beta$ ,

A similar procedure for $-\beta >{\theta }_{00}>\beta -\pi$ yields

Material volume of optimal structures

Let ${\overline{\sigma }}_j,j=1,...,m$ , denote the yield stress of the material used to construct the $j$ th member, ${\mathbf{m}}_j$ , loaded with force density ${\sigma }_i$ and with constant cross section area $A_j$ . To avoid material failure, the force density at any given member should satisfy

with equality holding in minimal volume structures. Some algebra yields

where $v_j$ is the volume of each member. Therefore, the total volume of a structure, $V$ , is

Let $b_j$ and $s_j$ be the bar and string vectors for the $j$ th compressive or tensile member in a structure with $m_b$ bars and $m_s$ strings. Therefore $V=V_b+V_s$ , where

Assuming that all bars are the same material, ${\overline{\lambda }}_j=\overline{\lambda }$ for all $j=1,...,m_b$ . Likewise, if all strings are made of the same material, then ${\overline{\gamma }}_j=\overline{\gamma }$ for all $j=1,...,m_s$ . Hence

so that

Now, let $J$ be defined by

From [link] and [link] , it is true for a structure in static equilibrium that

Let $B={\sum }_{j=1}^{m_b}{\lambda }_j(\parallel {\mathbf{b}}_j\parallel )\parallel {\mathbf{b}}_j{\parallel }^2$ and $S={\sum }_{j=1}^{m_s}{\gamma }_j(\parallel {\mathbf{s}}_j\parallel )\parallel {\mathbf{s}}_j{\parallel }^2$ Therefore

Thus

Assuming that $\overline{\lambda }=\overline{\gamma }$ , we get that

which is consistent with [link] . This implies that minimization of $J$ is independent of the choice of material properties, and depends only on the variable topology.