Form-finding of Repetitive
Tensegrity Structures
R. Pandia Raj
&
Dr. Simon Guest
Cambridge University Engineering Department
Outline
• Introduction
•
•
•
•
Structural Mechanics of Tensegrity Towers
Symmetry Analysis
Repetitive Tensegrity Towers Form-finding
Conclusions
2
Introduction
A Key Challenge?
3
Repetitive Tensegrity Structures
Needle Tower bottom view
Kenneth Snelson’s Needle Tower
Tensegrity Arch
4
Structural Mechanics of Unit Cell
Simplex Tensegrity
Unit Cell
5
Repetitive Tensegrity Tower
• Equilibrium at node 1
Tension Coefficient = Tension/Length
^ (x ¡ x ) + T
^ (x ¡ x ) + T
^ (x ¡ x ) + T
^ (x ¡ x ) + T
^ (x ¡ x ) = 0
T
h1
1
4
h2
1
6
v1
1
12
v2
1
7
s
1
10
^ (y ¡ y ) + T
^ (y ¡ y ) + T
^ (y ¡ y ) + T
^ (y ¡ y ) + T
^ (y ¡ y ) = 0
T
h1
1
4
h2
1
6
v1
1
12
v2
1
7
s
1
10
^ (z ¡ z ) + T
^ (z ¡ z ) + T
^ (z ¡ z ) + T
^ (z ¡ z ) + T
^ (z ¡ z ) = 0
T
h1 1
4
h2 1
6
v1 1
12
v2 1
7
s
1
10
• Equilibrium for nodes 1 – 12 can be written in a matrix form
Sx = 0
Sy = 0
Sz = 0
6
8
¡T
^
< P
ij
^
Sij =
Tik
:
0
if
if
if
i
i
i
6= j,
= j : summation over all nodes k connected to node i
and j are not connected
The stress matrix, S plays a key role in form-finding, and also can be
considered as the geometric stiffness matrix.
4
3
t^c
ts
t^c + t^c + ^ts
- t^c
- ^ts
- t^c
1
- t^c
t^c + t^c + t^s
- t^c
- t^s
1
^
- ts
^
- tc
tc + t^c + t^s
^
- tc
1
- t^c
- t^s
- t^c
t^c + t^c + t^s
1
^
Struts
2
1
Sij =
Cables
^
=0
7
• Nullity of S (Rank Deficient)
1
1
  must lie in the nullspace + 3 independent vectors in the nullspace
 
for a 3 dimensional structure to exist
Thus nullity of S must be d+1. where d is the dimension of the structure.
• Stability
S – positive semi-definite
S – zeros & positive eigenvalues
How can we find a set of tension coefficients that achieve
stable equilibrium configurations ?
8
Symmetry Analysis
• Symmetry - a characteristic of geometrical shapes
1
3
c
b
120°
2
3
2
1
a
Symmetry operation – C3
9
• Tensegrity with D3 symmetry
• Symmetry operations
Identity (E), rotation by 120°(C3) or rotation by 240°(C3)2,
twofold rotation about the 3 horizontal axes.
Irreducible representations of symmetry group
D3
E
C3
C2
Ca
Cb
A1
A2
1
1
1
1 p
· ¡
¸
p1=2 ¡ 3=2
¡1=2
3=2
1
1 p
· ¡
¸
3=2
p1=2
¡ 3=2 ¡1=2
1
-1
1
-1
E
·
1
0
0
1
¸
3
2
·
1
0
0
¡1
Cc
2
¸
·
¡p1=2
¡ 3=2
D3
2
p
¡ 3=2 ¸
1=2
1
-1
p
· ¡
¸
3=2
p1=2
3=2
1=2
10
z
x, y
• Using a symmetry analysis, we can write an orthogonal
transformation matrix, V in a symmetry adapted
coordinate system that is defined by the irreducible
representations of the symmetry group to which the
structure belongs, so that
~
x = VT x
• The orthogonal transformation matrix helps to find a
symmetry
~ adapted stress matrix: the resultant stress
S
matrix is similar to S, but has a block-diagonal form.
11
¯ ¯matrix can be written as
• The block diagonalised stress
¯ ¯
~ = VT SV
S
~ A1 ¯ = 0
¯S
= 1)
¯ (nullity
¯
¯~A ¯
¯S 2 ¯ = 0
(nullity = 2)
~ A1
S
2£2
~ (12 £ 12) =
S
~ A2
S
2£2
~ E (1)
S
4£4
~ E (2)
S
¯ ¯
¯ ¯
¯~E ¯
¯~E ¯
¯S 1 ¯ = 0 ) ¯S
2¯ = 0
4£4
(nullity = 4)
12
jS
jS
~ E1 j = 0
~ A2 j = 0
Solutions of
and
. The solutions lie on line marked by
^
^
^
^
~ A2
~ E1
Tv =Th
Ts =T
S
S
h
circles, where
is positive,
is negative, and
and
are positive semi-definite.
13
Using methods of Pellegrino and Calladine,
• 7 Internal mechanisms
• 1 State of self-stress
The found form.
Totally symmetric inextensional twisting
mechanism. The structure is shown in the
displaced state: Open circles mark the original
position of the nodes, with the displacement
shown by arrows.
14
• Two and Three Stage Tensegrity Towers
15
Repetitive Tensegrity Tower
Form-finding
Assuming,
[xi ¤ , yi ¤ , zi ¤ ] = [xi , yi , (zi § dz )]
We use +, if i * is above the unit cell, and - if it is
below.
16
• Equilibrium at node 1
^ (x ¡ x )
T
1
4
h1
^ (y ¡ y )
T
+
^ (z ¡ z )
T
4
h1 1
h1
1
4
+
+
^ (x ¡ x )
T
1
6
h2
^ (y ¡ y )
T
+
^ (z ¡ z )
T
6
h2 1
h2
1
6
+
+
^ (x ¡ x )
T
v1
1
12
^ (y ¡ y )
T
+
^ (z ¡ z )
T
v1 1
12
v1
1
12
+
+
^ (x ¡ x )
T
v2
1
7
^ (y ¡ y )
T
+
^ (z ¡ z )
T
v2 1
7
v2
1
7
+
+
^ (x ¡ x )
T
s
1
10
^ (y ¡ y )
T
+
^ (z ¡ z )
T
s
1
10
s
1
10
=
0
+
^ (x ¡ x )
T
v2
1
9
^ (y ¡ y )
T
=
0
+
^ (z ¡ (z ¡ d ))
T
z
v2 1
9
=
0
v2
1
9
connection to adjacent unit cell
• Equilibrium for nodes 1 – 12 can be written in a matrix form
Sx = 0
Sy = 0
Sz + dz T = 0
• Symmetry adapted coordinate system
~x = 0
S~
~y = 0
S~
~ z = ¡d T
~
S~
z
Where,
~ = VT SV, ~
~ = VT T.
S
x = VT x, y~ = VT y, ~z = VT z, and T
17
• Block diagonalised
equilibrium equations in the z direction
¯ ¯
~ A1
S
2£2
¯~A ¯
¯S 1 ¯ = 0
(nullity = 1)
~
Z
A
~
T
A
1£2
~
Z
A
~
T
A
1£2
~
Z
E (1)
~
T
E (1)
1£4
~
T
E (2)
1£4
1
~ A2
S
2£2
1
2
2
~ E (1)
S
4£4
~ E (2)
S
¯ ¯
¯ ¯
¯~E ¯
¯~E ¯
¯S 1 ¯ = 0 ) ¯S
2¯ = 0
~
Z
E (2)
= - dz
4£4
(nullity = 3)
~
~
T
T
A2
• It is found that all the · components become
zero,
except
p
¸
¡
^
6Tv 2
p
~ =d
T
¡ 6(T
A2
z
^ +T
^ +T
^ )
18
v1
v2
s
jS
~ E1 j = 0
Solutions of
. The solutions lie on line marked by circles, where
^
^
^
^
~ E1
Tv =Th
Ts =Th
S
positive,
negative, as well as
positive semi-definite.
19
Conclusions

Symmetry analysis vastly simplifies the
form-finding process for repetitive tensegrity
structures.
20
Thank you
21