DEPARTMENT OF MECHANICAL AND AEROSPACE ENGINEERING UNIVERSITY OF FLORIDA
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Transcript of DEPARTMENT OF MECHANICAL AND AEROSPACE ENGINEERING UNIVERSITY OF FLORIDA
DEPARTMENT OF MECHANICAL AND AEROSPACE ENGINEERING
UNIVERSITY OF FLORIDA
Ph.D. DISSERTATION
Presented by: Jahan B Bayat
Summer, 2006
2
Agenda
• Objective• Literature review• Analysis of planar tensegrity mechanisms
– 2-spring planar tensegrity– 3-spring planar tensegrity– 4-spring planar tensegrity (added to proposal)
• Future work, Summary and Conclusion
3
Dissertation Objective
• determine all equilibrium poses for 2-spring, 3-spring, and 4-spring planar tensegrity mechanisms
spatial tensegrity
struts
ties
planar tensegrity
4
Contribution
• analysis of these mechanisms provides a first insight into this class of mechanisms
• the knowledge gained here may assist in the analysis of more complex structures
5
Definitions
• Tensegrity is an abbreviation of tension and integrity.
• Tensegrity structures are formed by a combination of rigid elements in compression called struts and connecting elements that are in tension called ties.
• In three dimensional tensegrity structures no pair of struts touches and the end of each strut is connected to non-coplanar ties, which are in tension.
• In two dimensional tensegrity structures, struts still do not touch.
• A tensegrity structure stands by itself in its equilibrium position and maintains its form solely because of its arrangement of its struts and ties.
• The potential energy of the system stored in the springs is a minimum in the equilibrium position.
6
Tools for Analysis
• Linear Algebra with Matrix manipulation• Polynomials• Sylvester method• Maple• AutoCAD
7
Basic 3D and 2D Tensegrity Structures
triangle
quadrilateral
pentagon
hexagon
Figure 1: Family of Tensegrity Structures
struts
ties
Figure 2: Planar Tensegrity Structure
1
4
2 3a23
a41
a12
a34
8
Literature
• Overview of some basic definitions, geometries and applications.
• An example of practical applications is deployable structures such as Self deployed Space Antenna.
• Tensegrity is a new science (about 25 years).
9
A few of Literatures:
• Roth, B., Whiteley, W., “Tensegrity framework,” Transactions American Mathematics Society, Vol.265, 1981.
• Skelton, R. E., Williamson, D., and Han, J. H., Equilibrium Conditions of Tensegrity Structure, Proceedings of the Third World Conference on Structural Control (3WCSC) Como, Italy, April 7-12, 2002.
• Duffy, J., and Crane, C., Knight, B., Zhang, "On the Line Geometry of a Class of Tensegrity Structures“.
• Ian P. Stern, “Development of Design Equations for Self Deployable N-Struts Tensegrity Systems”.
10
Continue
• Ian P. Stern, “Development of Design Equations for Self Deployable N-Struts Tensegrity Systems”.
• Yin, J. P., Marsh, D., Duffy, J.“Catastrophe Analysis of Planar Three-Spring Systems”.
• Carl D. Crane, Joseph Duffy, Kinematics Analysis of Robot Manipulators.
• Jahan B. Bayat, Carl D. Crane, “Closed-Form Equilibrium Analysis of a Planar Tensegrity Structure”.
• Etc.
11
Agenda
• Objective• Literature review• Analysis of planar tensegrity mechanisms
– 2-spring planar tensegrity– 3-spring planar tensegrity– 4-spring planar tensegrity (added to proposal)
• Future work, Summary and Conclusion
12
Two-Spring Tensegrity System
• given:– a12, a34 strut lengths
– a23, a41 non-compliant tie lengths
– k1, L01 k2, L02 spring parameters
• solution 1, find:– L1, L2 at equilibrium
• solution 2, find:– c4, c1 at equilibrium
1
L2L1
a41
a12
a23
a34
2
3
4
13
Two-Spring Tensegrity, Solution 1
• obtain geometric equation f1(L1, L2) = 0
• write potential energy equation
U = ½ k1 (L1-L01)2 + ½ k2 (L2-L02)2
• evaluate
this may be written as f2(L1, L2) = 0
• solve the two equations for all sets of L1, L2
• resulted in 28th degree polynomial in L1
0dL
dU
1
14
Approach 1
• given:– a12, a34 strut lengths
– a23, a41 non-compliant tie lengths
– k1, L01 k2, L02 spring parameters
• find:– L1, L2 at equilibrium
1
L2L1
a41
a12
a23
a34
2
3
4
15
Figure 3: Planar Tensegrity Structure Figure 4: Triangle 4-3-2
a34
L1
a23
'4
2
3
41
L2L1
a41
a12
a23
a34
2
3
4'4
"44
Geometric Constraint
341
234
21
223
4 aL2
aLa'cos
4 + 4' = + 4" (2-3)(2-1)
16
L1
a12
a41
"4
1
4
2
Figure 5: Triangle 4-1-2 Figure 6: Triangle 4-1-3
4
3
1
4
L2
a41
a34
Geometric Constraint – Cont.
411
241
21
212
4 aL2
aLa"cos
4134
241
234
22
4 aa2
aaLcos
(2-7)(2-5)
17
Geometric Constraint – Cont.
cos (4 + 4') = cos ( + 4")
cos4 cos4' – sin4 sin4' = - cos4"
cos4 cos4' + cos4" = sin4 sin4' .
(cos4)2 (cos4')
2 + 2 cos4 cos4' cos4" + (cos4")2 = (sin4)
2 (sin4')2 .
4 + 4' = + 4"
(cos4)2 (cos’4)2 + 2 cos4 cos’4 cos”4 + (cos”4)2 = (1-cos24) (1-cos2’4)
(2-12)
18
Geometric Constraint – Cont.
• substituting (2-3), (2-5), and (2-7) into (2-12) and gives geometry equation.
where
and
B2, B0, C2, and C0 are expressed in terms ofknown quantities
A L24 + B L2
2 + C = 0 (2-13)
A = L12 , B = L1
4 + B2 L12 + B0, C = C2 L1
2 + C0
19
Potential Energy Constraint
• at equilibrium, the potential energy in the springs will be a minimum
U = ½ k1 (L1-L01)2 + ½ k2 (L2-L02)2
• the mechanism being considered is a one degree of freedom device
• one parameter can be selected as the generalized coordinate for the problem; L1
20
Potential E. Constraint – Cont.
• at a minimum potential energy state,
• dL2/dL1 can be obtained via implicit differentiation of the geometry constraint as
0dL
dL)LL(k)LL(k
dL
dU
1
202220111
1
))]aa)(aa()aaaaL2L(L[L
)]aa)(aa()aaaaL2L(L[L
dL
dL2
342
232
412
122
122
342
412
232
22
12
12
234
241
223
212
212
234
241
223
21
22
221
1
2
(2-19)
(2-20)
21
Potential E. Constraint – Cont.
• substituting (2-20) into (2-19) gives
D L25 + E L2
4 + F L23 + G L2
2 + H L2 + J = 0 (2-21)
where the coefficients D through J are polynomials in L1
• equations (2-13) and (2-21) represent two equations in the two unknowns L1 and L2
• Sylvester’s elimination method is used to obtain values for these parameters that simultaneously satisfy both equations
22
0
0
0
0
0
0
0
0
0
1
L
L
L
L
L
L
L
L
0000C0B0A
000JHGFED
000C0B0A0
00JHGFED0
00C0B0A00
0JHGFED00
0C0B0A000
C0B0A0000
JHGFED000
2
22
32
42
52
62
72
82
Sylvester’s elimination
• determinant of coefficient matrix must equal zero which yields a 28th degree polynomial in L1
23
Symbolic Expansion
• determinant of coefficient matrix must equal zero which yields a 28th degree polynomial in L1
• Maple program used to obtain all coefficients symbolically
• corresponding values of L2 for each value of L1 can be readily obtained from solving (2-13) and then (2-21)
24
2.3.4 Numerical Example
• given:– a12 = 3 in. a34 = 3.5 in.
– a41 = 4 in. a23 = 2 in.
– L01 = 0.5 in. k1 = 4 lbf/in.
– L02 = 1 in. k2 = 2.5 lbf/in.
• find L1 and L2 at equilibrium
25
Numerical Example
• results– coefficients of 28th
degree polynomial in L1
obtained
– 8 real roots for L1 with corresponding values for L2
– 4 cases correspond to minimum potential energy
Case L1, in. L2, in.
1 -5.4854 2.3333
2 -5.3222 -2.9009
3 -1.7406 -1.4952
4 -1.5760 1.8699
5 1.6280 1.7089
6 1.8628 -1.3544
7 5.1289 -3.2880
8 5.4759 2.3938
26
spring in compression with a negative spring length
spring in tension
4
44
41
1 1
1
2
2
2
2
3
33
3
Case 3 Case 4
Case 5 Case 6
Numerical Example Cont.
27
Force Balance Verification
CaseForce in Spring 1,
labForce in Spring 2,
labForce in Strut a12,
lab
Force in Strut a34,
lab
Force in Tie a41,
lab
Force in Tie a23,
lab
3 + 8.9624 + 6.2379 -13.4186 -16.8097 9.1623 18.7570
4 + 8.3040 2.1749 -5.4965 -11.7069 4.1031 11.9485
5 4.5120 1.7722 -4.2448 -7.1095 3.0883 7.4454
6 5.4514 + 5.8860 -11.4149 -11.6983 7.2064 14.1181
28
Conclusion for Approach 1
• closed-form solution to 2 strut, 2 spring tensegrity system presented
• geometric and potential energy constraints gave two equations in the spring lengths L1 and L2
• elimination of L2 resulted in a 28th degree polynomial in the single variable L1
• numerical example presented showing 4 real solutions
• other examples have been investigated, but all gave 4 real solutions
29
Approach 2
• Determine (c4 and c1) to Minimize Potential Energy.
• The objective of this approach is to again investigate, in closed-form, the planar 2-spring tensegrity system.
30
Geometry
4
32
1
4
a41
a12a34L1
L2
a23
1
31
Approach 2
The problem statement is written as:
given: a41, a12, a23, a34
k1, k2, L01, L02
find: cos4 (and corresponding value of cos1) when the system is in equilibrium
32
Two-Spring Tensegrity, Solution 2
• obtain geometric equationsf1(c4, c1) = 0,f2(c4, L2) = 0,f3(c1, L1) = 0,
• write potential energy equation
U = ½ k1 (L1-L01)2 + ½ k2 (L2-L02)2
• evaluate
this may be written as f4(c4, c1, L1, L2) = 0
• use f2 and f3 to eliminate L1 and L2 from f4
• solve this equation and f1 for all sets of c4, c1
• resulted in 32nd degree polynomial in c4
0dc
dU
4
33
Solution Approach 2
• from cosine law for planar quadrilateral
which can be factored as2
aZ
223
41
A c12 + B c1 + D = 0 (2.37)
A = A1 c4 + A2
B = B1 c42 + B2 c4 + B3
where
34
Solution Approach 2
• from derivative of potential energy
C10 c110 + C9 c1
9 + C8 c18 + C7 c1
7 + C6 c16 + C5 c1
5 + C4 c1
4 + C3 c13 + C2 c1
2 + C1 c1 + C0 =0
where the coefficients Ci are functions of c4
(2-55)
35
Solution Approach 2
0
0
0
0
0
0
0
0
0
0
0
0
1
c
c
c
c
c
c
c
c
c
c
c
0CCCCCCCCCCC
CCCCCCCCCCC0
000000000DBA
00000000DBA0
0000000DBA00
000000DBA000
00000DBA0000
0000DBA00000
000DBA000000
00DBA0000000
0DBA00000000
DBA000000000
1
21
31
41
51
61
71
81
91
101
111
012345678910
012345678910
36
Solution Approach 2
• Expansion of the 12×12 determinant yields a 32nd degree polynomial in the parameter c4.
•Using earlier numerical values, there are 8 real values and 20 complex values for c4.
•Four of real values are identical to real values in approach 1.
•Table on next page presents real values of second approach.
37
Solution Approach 2
Case c4 (radian) c1 (radian) L1 (inches) L2 (inches)
1 -0.8144902900 0.2120649698 -5.4853950884 2.3332963547
2 -0.7083900124 0.1385596475 -5.3221641782 -2.9008756697
3 -0.9290901467 -0.9154296793 -1.7405998094 -1.4951507917
4 -0.8840460933 -0.9381755562 -1.5760033787 1.8699490332
5 -0.9046343262 -0.9312332602 1.6280054527 1.7088706403
6 -0.9434161072 -0.8970754537 1.8628443599 -1.3543814070
7 -0.6228159543 0.0544134519 5.1289299905 -3.2880318246
8 -0.8042767224 0.2077177765 5.4758767915 2.3937944296
38
Agenda
• Objective• Literature review• Analysis of planar tensegrity mechanisms
– 2-spring planar tensegrity– 3-spring planar tensegrity– 4-spring planar tensegrity (added to proposal)
• Future work, Summary and Conclusion
39
3-spring planar tensegrity
• closed-form analysis of a three spring, two strut tensegrity system with one non-compliant element
4
32
1
4
a41
a12a34L24
L31
L23
1
40
3-S: 1st approach
The problem statement can be explicitly written as:
Given: a12, a34 lengths of struts,
a41 lengths of non-compliant tie
k1, L01 spring constant and free length, point 4 to 2
k2, L02 spring constant and free length, point 3 to 1
k3, L03 spring constant and free length, point 2 to 3
Find: L1 L2 L3
length of springs at equilibrium position,
41
3-S: 1st approach cont.
42
3-S: 1st approach cont.
Cosine law for triangles 2-4-3 and 4-1-2 :
2
L'cosaL
2
a
2
L 23
4341
234
21
2
a"cosaL
2
a
2
L 212
4411
241
21
4 + 4' = + 4"
(3-1)
(3-3)
(3-7)
43
3-S: 1st approach cont.
3.2.1 Development of Geometric Equation,
G1 L34 + (G2 L2
2 + G3) L32 + (G4 L2
4 + G5 L22 + G6) = 0
3.2.2 Development of Potential Energy Equations
20333
20222
20111 )LL(k
2
1)LL(k
2
1)LL(k
2
1U
(3-13)
(3-16)
44
3-S: 1st approach cont.
• At equilibrium, the potential energy will be a minimum.
0L
L)LL(k)LL(k
L
U
1
303330111
1
0L
L)LL(k)LL(k
L
U
2
303330222
2
]GLGLGLLGLG2[L
]GLGLLGLLGLL2[L
L
L
b32
1a32
2b22
22
1a22
313
a62
2b54
22
3a32
32
2a22
22
11
1
3
]GLGLGLLGLG2[L
]GLGLLGLLGLL2[L
L
L
b32
1a32
2b22
22
1a22
313
c52
1b54
12
3b22
32
1a22
22
12
2
3
(3-17)
(3-18)
45
3-S: 1st approach cont.
Substituting, results (3-21) and (3-24) :
(D1 L22+D2) L3
3 + (D3 L22+D4) L3
2 + (D5 L2
4+D6 L22+D7) L3 + (D8 L2
4+D9 L22+D10) = 0
(E1 L2 + E2) L33 + (E3 L2) L3
2 + (E4 L2
3 + E5 L22 + E6 L2 + E7) L3 + (E8 L2
3 + E9 L2) = 0
(3-21)
(3-24)
46
3-S: 1st approach cont.
Creating solution matrix M52x52 :
M λ = 0
M =
2221
1211
MM
MM
λ = [L27L3
3, L25L3
5, L23L3
7, L27L3
2, L26L3
3, L25L3
4, L24L3
5, L2
3L36, L2
2L37, L2
7L3, L26L3
2, L25L3
3, L24L3
4, L23L3
5, L22L3
6, L2L3
7, L27, L2
6L3, L25L3
2, L24L3
3, L23L3
4, L22L3
5, L2L36, L3
7, L2
6, L25L3, L2
4L32, L2
3L33, L2
2L34, L2L3
5, L36, L2
5, L24L3,
L23L3
2, L22L3
3, L2L34, L3
5, L24, L2
3L3, L22L3
2, L2L33, L3
4, L2
3, L22L3, L2L3
2, L33, L2
2, L2L3, L32, L2, L3, 1]T .
47
3-S: 1st approach cont.
000D0D000000D0D000D0D00000
0D0000D0D00000000000000000
0000D0D0000D0D000000000000
00000000000G0G0G0000000000
0000G0G0G00000000000000000
000G0G0G000000000000000000
0G000000000000000000000000
00000000000000000000000000
0000EEEE0000E0E00000000000
E0000E0E000000000000000000
0000E0E0000000000000000000
E0000000000000000000000000
00000000000000000000000000
00000D0D0000D0D00000000000
0000D0D0000000000000000000
D0000000000000000000000000
G0000000000000000000000000
00000000000000000000000000
00000000000000000000000000
00000000000000000000000000
00000000000000000000000000
00000000000000000000000000
00000000000000000000000000
00000000000000000000000000
00000000000000000000000000
00000000000000000000000000
263815
815
3815
124
124
124
4
235814
814
14
4
3815
15
5
4
11M
48
3-S: 1st approach cont.
0000EE000000E0000EEEE00EE0
0000E00000EEEE000E0E000000
000000EE000000E0000EEEE0EE
E000000E00000EEEE000E0E000
00EEEE0000E0E0000000000000
E00000EEEE0000E0E000000000
000D0D0000D0D00000D0D00DD0
00000D0D0000D0D00000D0D0DD
D0000D0D000000D0D000D0D000
00D0D000000D0D000D0D000000
D000000D0D0000D0D000000000
000000000000G0G00000000GGG
00000G0G0000000000G0G0G000
G00000000000G0G0G000000000
0000G0G0000000000G0G0G0000
0000000000G0G0G00000000000
00000G0G0G0000000000000000
00G0G0G0000000000000000000
000000E00000EEEE000E0E0000
00000EEEE0000E0E0000000000
00000E00000EEEE000E0E00000
000EEEE0000E0E000000000000
EE0000E0E00000000000000000
000E0E00000000000000000000
0000D0D000000D0D000D0D0000
000000D0D0000D0D0000000000
796235814
6235814
796335814
96235814
235814
6235814
49263815
49263815
9263815
263815
63815
35124
35124
5124
35124
124
124
124
6235814
235814
6235814
235814
5814
14
263815
3815
21M
49
3-S: 1st approach cont.
00000000000D0000D00000D0D0
00000D00D000D0D00D0D00000D
000000D000D0000D0D000D0D00
000000000000G0000000000G0G
00000000G00000000G0G000000
0000000G00000000G0G0000000
00000G000000G0G00000000G0G
000G000000G0G00000000G0G0G
00000000000000000EE00000E0
000000000000EE0000E0000EEE
00000000000EE0000E0000EEEE
00000000EE000E000EEEE000E0
000000EE000E000EEEE000E0E0
0000000D000D0000D0D000D0D0
000D00D000D0D00D0D00000D0D
00D0D00D0D0D0D0000D0D000D0
0000G000000G0G00000000G0G0
0G0000G0G000000G0G0G000000
00G0000G0G000000G0G0G00000
0000000EE000E000EEEE000E0E
000EE00E00EEEE00E0E0000000
0000EE00E00EEEE00E0E000000
0D0D00D0D0D0D0000D0D000D0D
0EE0E0EEEE0E0E000000000000
DD0D0DD0D000D0D00D0D000000
G00G0G0000G0G0G00000000000
10749
10749263
74926
635
635
635
63512
635124
796
796235
7962358
79623581
96235814
1074926
107492638
1074926381
63512
635124
635124
796235814
796235814
796235814
10749263815
796235814
10749263815
635124
12
10
7
M
50
3-S: 1st approach cont.
00000000000000000000000000
0000000000000000000000EE00
00000000000000000000000000
0000000000000000000000000E
000000000000000EE00000E000
0000000000000000000EE00000
0000000000000000D00000D000
000000000000000000D00000D0
0000000000000D0000D00000D0
0000000000D0000D00000D0D00
000000000D000D0000D0D000D0
000000000000000000000000G0
000000000000000000G0000000
0000000000000G0000000000G0
00000000000000000G00000000
00000000000G0000000000G0G0
000000000G00000000G0G00000
000000G00000000G0G00000000
000000000000000000000000EE
000000000000000000EE00000E
00000000000000000000000EE0
0000000000000000EE00000E00
0000000000000EE0000E0000EE
0000000000EE0000E0000EEEE0
000000000000D0000D00000D0D
00000000D000D0000D0D000D0D
79
7
796
79
107
107
1074
10749
107492
6
6
63
6
635
635
635
79
796
79
796
79623
7962358
10749
1074926
22M
51
3-S: 1st approach cont.
• In order for a solution for the vector λ to exist, it is necessary for the 52 resulting equations to be linearly dependent. This will occur if the determinant of M equals zero.
• It was not possible to expand the determinant symbolically.
• A numerical case was analyzed and a polynomial of degree 158 in the variable L1 was obtained.
• A numerical example is presented next .
52
3-S: 1st approach cont.
Numerical Example :
• strut lengths:
a12 = 14 in. a34 = 12 in.
• non-compliant tie lengths:
a41 = 10 in.
• spring 1,2 and 3 free length & spring constant:
L01 = 8 in. k1 = 1 lbf/in.
L02 = 2 in. k2 = 2.687 lbf/in.
L03 = 2.5 in. k3 = 3.465 lbf/in.
53
3-S: 1st approach cont.
Solution : (3-13), (3-21), and (3-24)
• 100 L34 + [(-L1
2 + 96) L22 + 44 L1
2 – 52224] L32 + L1
2 L24 + (L1
4 – 440 L1
2 -13824) L22 – 8624 L1
2 + 6773760 = 0 (3-33)
• (2.5 L1 L22 + 90 L1 – 1600) L3
3 + (-8.663 L1 L22 + 381.165 L1) L3
2 +
[-2.5 L1 L24 + (-6 L1
3 + 8 L12 + 1196 L1 – 768) L2
2 + 44 L13 – 352 L1
2 – 30664 L1 + 417792] L3 + 8.663 L1 L2
4 + (17.326 L13 – 3811.651 L1) L2
2 – 74708.354 L1 = 0 (3-34)
• [(2.5 L12 + 160) L2 – 1074.637] L3
3 + (831.633 – 8.663 L12) L2 L3
2
+ [(-7 L12 + 192) L2
3 + (-515.826 + 5.373 L12) L2
2 + (-2.5 L14 + 1188 L1
2 – 69888) L2 – 236.420 L1
2 + 280609.161] L3 + 17.326 L12 L2
3+ (-119755.135 + 8.663 L1
4 – 3811.651 L12) L2 = 0 . (3-35)
54
3-S: 1st approach cont.
• Continuation method is applied and solutions were obtained.
• Only one of the solutions was acceptable (Force Balance).
• Result: L1 = 13.0
L2 = 8.0
L3 = 7.017
• Next, 2nd approach applied to verify the results.
55
3-S: 2nd approach
Problem Statement :
• The problem statement is presented bellow:
• given: a41, a12, a34
k1, k2, k3, L01, L02, L03
• find: cos4, cos1 when the system is in equilibrium
56
3-S: 2nd approach, cont.
Solution :• It is a two d.o.f. system
• obtain expressions for L1, L2, and L3 in terms of cos4 and cos1
• write the potential energy equation
• determine values of cos4 and cos1 such that dU/dcos4 = dU/d cos1= 0
57
3-S: 2nd approach, cont.
• Geometry equation :
quadrilateral 1-2-3-4 can be written as:
Where,
2)(
212
414141241
aZcYsXaZ
X4 = a34 s4
Y4 = -(a41 + a34 c4)
(3.36)
(3.37)
(3.38)
(3.39)
2
LZ
23
41
58
3-S: 2nd approach, cont.
44134
241
234
4 caa2
a
2
aZ
Rewrite the quadrilateral cosine law as:
Square both sides and substitute cosine for sines
(3.40)
(3.41)2
L
2
aZcYasXa
23
212
414121412
59
3-S: 2nd approach, cont.
Obtain an equation w.r.t L23.
L34+ A L3
2 + B = 0 (3.42)
A = A1 c4 + A2 c1 + A3 c1 c4 + A4 (3.43)
Where,
B = B1 c12 + B2 c12 c4 + B3 c1 + B4 c1 c4 + B5 c1 c42 + B6 c42 + B7 c4 + B8
(3.44)
60
3-S: 2nd approach, cont.
A cosine law for triangle 4-1-2 may be written as:
A cosine law for the triangle 3-4-1 may be written as:
(3.46)
(3.47)
2
Lcaa
2
a
2
a 21
14112
241
212
2
Lcaa
2
a
2
a 22
44134
241
234
61
3-S: 2nd approach, cont.
The total potential energy stored in all three springs is given by:
U = ½ k24 (L24-L024)2 + ½ k23 (L23-L023)2 +
½ k31 (L31-L031)2 (3.48)
62
3-S: 2nd approach, cont.
The differentiation of U with respect to c1 and c4 gives:
dU/dc4 = k24 (L24 – L024) dL24/dc4 + k23 (L23 – L023) dL23/dc4 +
k31 (L31 – L031) (3.49)
dU/dc1 = k24 (L24 – L024) dL24/dc1 + k23 (L23 – L023)dL23/dc1 + k31 (L31 – L031) (3.50)
63
3-S: 2nd approach, cont.
dU/dc4 = 0, This equation does not contain L24.
dU/dc1 = 0, This equation does not contain L31.
From (3.46) and (3.47), it is apparent that dL1/dc4 and dL2/dc1 are zero. So,
64
3-S: 2nd approach, cont.
(p1 c12 + p2 c1 + p3) c4
3 + (p4 c1
3 + p5 c12 + p6 c1 + p7) c4
2 + (p8 c1
4 + p9 c13 + p10 c1
2 + p11 c1 + p12) c4 +(p13 c1
4 + p14 c13 + p15 c1
2 + p16 c1 + p17) = 0 (3.65)
(q1 c1 + q2) c44 +
(q3 c12 + q4 c1 + q5) c4
3 + (q6 c1
3 + q7 c12 + q8 c1 + q9) c4
2 + (q10 c1
3 + q11 c12 + q12 c1 + q13) c4 +
(q14 c13 + q15 c1
2 + q16 c1 + q17) = 0 (3.66)
65
3-S: 2nd approach, cont.
Next, factor (3.7) into the following form,
(r1 c1 + r2) c42 +
(r3 c12 + r4 c1 + r5) c4 +
(r6 c12 + r7 c1 + r8) = 0 (3.69)
Note: Equations (3.65), (3.66), and (3.69) represent three equations in the three unknowns L3, c1, and c4.
66
3-S: 2nd approach, cont.
•To apply the Sylvester method, equations 3.65, 3.66, and 3.69 are multiplied by powers of c1 and c4.
•This step is to create a sufficient equation set in the variable c4 and c1 where the parameter L3 is embedded in the coefficients.
67
3-S: 2nd approach, cont.
• Equation (3.65) is multiplied by c1, c12, c4, c4
2, c43, c1
4, c1
2c4, c1c42, and c1
2c42 to obtain 10 equations including it.
• Equation (3.69) is multiplied by c1, c12, c1
3, c14, c4, c4
2, c43,
c44, c1c4, c1
2c4, c13c4, c1
4c4, c1c42, c1
2c42, c1
3c42, c1
4c42, c1c4
3, c1
2c43, c1
3c14, c1c4
4 and c12c4
4 to obtain 22 equations including it.
• Equation (3.66) is multiplied by c1, c12, c1
3, c4, c42, c1c4,
c12c4, c1
3c4, c1c42 and c1
2c42 to obtain 11 equations including
it.
68
3-S: 2nd approach, cont.
To show this 43 by 43 coefficient matrix, M, it is divided as a combination of sub-matrices Mij.
:= Mij
M11 M12 M13 M14
M21 M22 M23 M24
M31 M32 M33 M34
M41 M42 M43 M44
where,
69
3-S: 2nd approach, cont.
M11, M12, M13, M21, M22, M23, M31, M32, M33 are 11 by 11 matrices.
M14, M24, M34 are 11 by 10 matrices.
M41, M42, M43 are 10 by 11 matrices.
M44 is a 10 by 10 matrix.
70
3-S: 2nd approach, cont.
And all sub-matrices are shown bellow:
:= M11
r8 r7 r6 0 0 0 0 r5 r4 r3 0
0 r8 r7 r6 0 0 0 0 r5 r4 r3
0 0 r8 r7 r6 0 0 0 0 r5 r4
0 0 0 r8 r7 r6 0 0 0 0 r5
0 0 0 0 r8 r7 r6 0 0 0 0
0 0 0 0 0 0 0 r8 r7 r6 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 r8 r7 r6
0 0 0 0 0 0 0 0 0 r8 r7
71
3-S: 2nd approach, cont.
:= M12
0 0 0 r2 r1 0 0 0 0 0 0
0 0 0 0 r2 r1 0 0 0 0 0
r3 0 0 0 0 r2 r1 0 0 0 0
r4 r3 0 0 0 0 r2 r1 0 0 0
r5 r4 r3 0 0 0 0 r2 r1 0 0
0 0 0 r5 r4 r3 0 0 0 0 r2
0 0 0 r8 r7 r6 0 0 0 0 r5
0 0 0 0 0 0 0 0 0 0 r8
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 r5 r4 r3 0 0 0 0
r6 0 0 0 0 r5 r4 r3 0 0 0
72
3-S: 2nd approach, cont.
:= M13
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
r1 0 0 0 0 0 0 0 0 0 0
r4 r3 0 0 0 0 r2 r1 0 0 0
r7 r6 0 0 0 0 r5 r4 r3 0 0
0 0 0 0 0 0 r8 r7 r6 0 0
r2 r1 0 0 0 0 0 0 0 0 0
0 r2 r1 0 0 0 0 0 0 0 0
73
3-S: 2nd approach, cont.
:= M14
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 r2 r1 0 0 0 0 0 0
0 r5 r4 r3 0 0 r2 r1 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
74
3-S: 2nd approach, cont.
:= M21
0 0 0 0 0 0 0 0 0 0 r8
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 r8 r7
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
75
3-S: 2nd approach, cont.
:= M21
0 0 0 0 0 0 0 0 0 0 r8
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 r8 r7
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
76
3-S: 2nd approach, cont.
:= M23
0 0 r2 r1 0 0 0 0 0 0 0
0 0 0 r2 r1 0 0 0 0 0 0
r5 r4 r3 0 0 0 0 r2 r1 0 0
0 r2 r1 0 0 0 0 0 0 0 0
0 0 r5 r4 r3 0 0 0 0 r2 r1
0 0 0 r5 r4 r3 0 0 0 0 r2
r8 r7 r6 0 0 0 0 r5 r4 r3 0
0 r8 r7 r6 0 0 0 r5 r4 r3 0
0 0 r8 r7 r6 0 0 0 0 r5 r4
0 0 0 0 0 0 0 r8 r7 r6 0
0 0 0 0 0 0 0 0 r8 r7 r6
77
3-S: 2nd approach, cont.
:= M24
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
r1 0 0 0 0 0 0 0 0 0
0 0 r2 r1 0 0 0 0 0 0
0 0 0 r2 r1 0 0 0 0 0
r3 0 0 0 r2 r1 0 0 0 0
0 0 r5 r4 r3 0 0 r2 r1 0
0 0 0 r5 r4 r3 0 0 r2 r1
78
3-S: 2nd approach, cont.
:= M31
q17 q16 q15 q14 0 0 0 q13 q12 q11 q10
0 q17 q16 q15 q14 0 0 0 q13 q12 q11
0 0 q17 q16 q15 q14 0 0 0 q13 q12
0 0 0 q17 q16 q15 q14 0 0 0 q13
0 0 0 0 0 0 0 q17 q16 q15 q14
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 q17 q16 q15
0 0 0 0 0 0 0 0 0 q17 q16
0 0 0 0 0 0 0 0 0 0 q17
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
79
3-S: 2nd approach, cont.
:= M32
0 0 0 q9 q8 q7 q6 0 0 0 q5
q10 0 0 0 q9 q8 q7 q6 0 0 0
q11 q10 0 0 0 q9 q8 q7 q6 0 0
q12 q11 q10 0 0 0 q9 q8 q7 q6 0
0 0 0 q13 q12 q11 q10 0 0 0 q9
0 0 0 q17 q16 q15 q14 0 0 0 q13
q14 0 0 0 q13 q12 q11 q10 0 0 0
q15 q14 0 0 0 q13 q12 q11 q10 0 0
q16 q15 q14 0 0 0 q13 q12 q11 q10 0
0 0 0 0 q17 q16 q15 q14 0 0 0
0 0 0 0 0 q17 q16 q15 q14 0 0
80
3-S: 2nd approach, cont.
:= M33
q4 q3 0 0 0 0 q2 q1 0 0 0
q5 q4 q3 0 0 0 0 q2 q1 0 0
0 q5 q4 q3 0 0 0 q2 q1 0 0
0 0 q5 q4 q3 0 0 0 q2 q1 0
q8 q7 q6 0 0 0 q5 q4 q3 0 0
q12 q11 q10 0 0 0 q9 q8 q7 q6 0
q9 q8 q7 q6 0 0 0 q5 q4 q3 0
0 q9 q8 q7 q6 0 0 0 q5 q4 q3
0 0 q9 q8 q7 q6 0 0 0 q5 q4
q13 q12 q11 q10 0 0 0 q9 q8 q7 q6
0 q13 q12 q11 q10 0 0 0 q9 q8 q7
81
3-S: 2nd approach, cont.
:= M34
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 q2 q1 0 0 0 0 0 0 0
0 q5 q4 q3 0 0 q2 q1 0 0
0 0 q2 q1 0 0 0 0 0 0
0 0 0 q2 q1 0 0 0 0 0
q3 0 0 0 q2 q1 0 0 0 0
0 0 q5 q4 q3 0 0 q2 q1 0
q6 0 0 q5 q4 q3 0 0 q2 q1
82
3-S: 2nd approach, cont.
:= M41
p17 p16 p15 p14 p13 0 0 q12 q11 q10 q9
0 p17 p16 p15 p14 p13 0 0 q12 q11 q10
0 0 p17 p16 p15 p14 p13 0 0 q12 q11
0 0 0 0 0 0 0 p17 p16 p15 p14
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 p17 p16 p15
0 0 0 0 0 0 0 0 0 p17 p16
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
83
3-S: 2nd approach, cont.
:= M42
p8 0 0 p7 p6 p5 p4 0 0 0 p3
q9 p8 0 0 p7 p6 p5 p4 0 0 0
q10 q9 p8 0 0 p7 p6 p5 p4 0 0
p13 0 0 q12 q11 q10 q9 p8 0 0 p7
0 0 0 p17 p16 p15 p14 p13 0 0 q12
0 0 0 0 0 0 0 0 0 0 p17
p14 p13 0 0 q12 q11 q10 q9 p8 0 0
p15 p14 p13 0 0 q12 q11 q10 q9 p8 0
0 0 0 0 p17 p16 p15 p14 p13 0 0
0 0 0 0 0 p17 p16 p15 p14 p13 0
84
3-S: 2nd approach, cont.
:= M43
p2 p1 0 0 0 0 0 0 0 0 0
p3 p2 p1 0 0 0 0 0 0 0 0
0 p3 p2 p1 0 0 0 0 0 0 0
p6 p5 p4 0 0 0 0 p3 p2 p1 0
q11 q10 q9 p8 0 0 p7 p6 p5 p4 0
p16 p15 p14 p13 0 0 q12 q11 q10 q9 p8
p7 p6 p5 p4 0 0 0 p3 p2 p1 0
0 p7 p6 p5 p4 0 0 0 p3 p2 p1
q12 q11 q10 q9 p8 0 0 p7 p6 p5 p4
0 q12 q11 q10 q9 p8 0 0 p7 p6 p5
85
3-S: 2nd approach, cont.
:= M44
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 p3 p2 p1 0 0 0 0 0 0
0 p7 p6 p5 p4 0 p3 p2 p1 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 p3 p2 p1 0 0 0 0 0
p4 0 0 p3 p2 p1 0 0 0 0
86
3-S: 2nd approach, cont.
• Equating the determinant of the coefficient matrix Mij to zero will yield a polynomial in the single variable L23.
• A numerical example is presented next.
87
3-S: 2nd approach, cont.
• The following parameters were selected in analysis of numerical example 2:
• strut lengths:• a34 = 12.0 in. a12 = 14.0 in.• non-compliant tie lengths:• a41 = 10.0 in.• spring 1 free length & spring constant: • L024 = 8.0 in. k1 = 1.0 lbf/in.• spring 2 free length & spring constant:• L031 = 2.6865 in. k2 = 2.0 bf/in.• spring 3 free length & spring constant: • L023 = 3.4651 in. k2 = 2.5 lbf/in.
88
3-S: 2nd approach, cont.
Find:L3, c1, c4.
Solution:
The result was 136 solutions for L3 which 7 were real.
89
3-S: 2nd approach, cont.
• Case 1 is presented here (7 real cases):
• Cosine θ1: -0.453571523
• Cosine θ4: -0.749999925
• L3: 7.01658755
90
3-S: 2nd approach, cont.
• The results of Case 1 is used to calculate L1, L2, and L3 presented here:
• L1: 13.0000000
• L2: 7.99999999
• L3: 7.01658830
91
3-S: 2nd approach, cont.
• Force balance is applied to each case to verify if the structure is in equilibrium.
• The summation of forces at point 2 and 3 is expected to become zero if the structure is at equilibrium.
• The force in each spring is calculated using F = k (Lfinal – Linitial).
92
3-S: 2nd approach, cont.
93
3-S: 2nd approach, cont.
• Testing of real values in second differentiation of energy equation to determine if potential energy of the structure is minimum or maximum.
• The structure with real values, pass force balance test and minimum potential energy is considered an acceptable solution.
94
3-S: 2nd approach, cont.
In these numerical examples :
• Obtained same results using two different approaches
• Satisfied force balance equilibrium evaluation• Was at minimum potential energy • Presents a stable structure.
95
3-S: 2nd approach, cont.
End of 3 Spring.
96
Agenda
• Objective• Literature review• Analysis of planar tensegrity mechanisms
– 2-spring planar tensegrity– 3-spring planar tensegrity– 4-spring planar tensegrity (added to proposal)
• Future work, Summary and Conclusion
97
4 spring planar tensegrity
• Analysis of a four elastic ties planar tensegrity mechanism to determine equilibrium configurations when no external forces or moments are applied.
• The equilibrium position is the potential energy stored in four springs is a minimum.
• A polynomial expressed in terms of the length of one of the springs is developed.
98
4-S
• given: a12, a34lengths of struts,
• find: At equilibrium position
• L1, L2, L3, and L4 length of spring 1, 2, 3 and 4.
•
• k1, L01 spring between points 4 and 2,
• k2, L02 spring between points 3 and 1,
• k3, L03 spring between points 3 and 2,
• k4, L04 spring between points 4 and 1.
99
4-S: cont.
4
3
2
1
L1
L2
L3
L4
a12
a34
100
4-S: cont.
Geometry equation :
• G1 (L14 L2
4) + G2 (L14 L2
2) + G3 (L32 L1
2 L22) + G4 (L1
2 L22) + G5
(L32 L2
2) + G6 (L22) + G7( L3
2 L12 ) + G8( L1
2 ) + G9( L34 ) +
G10( L32 ) + G1
1 = 0 (4-1)
101
4-S: cont.
• The potential energy of the system can be evaluated as :
20444
20333
20222
20111 )(
2
1)(
2
1)(
2
1)(
2
1LLkLLkLLkLLkU
At equilibrium, the potential energy will be a minimum.
102
4-S: cont.
0)()(1
404440111
1
dL
dLLLkLLk
dL
dU
0)()(2
404440222
2
dL
dLLLkLLk
dL
dU
0)()(3
404440333
3
dL
dLLLkLLk
dL
dU
103
4-S: cont.
212
23
234
23
43
23
24
212
234
21
23
21
212
23
22
21
22
234
224
234
23
212
234
23
24
24
212
24
22
212
22
23
22
21
22
234
22
421
1
4
2(
)]2[
aLaLLLLaaLLLaLLLLaLL
aLaaLLLaLLaLLLLLaLLL
dL
dL
)2(
)]2[2
122
32
342
34
32
32
42
122
342
12
32
12
122
32
22
12
22
342
24
212
23
23
24
212
234
21
24
234
24
21
212
21
23
41
21
234
21
222
2
4
aLaLLLLaaLLLaLLLLaLL
aLLLaaLLaLLaLLLLaLLL
dL
dL
)2(
)]2[2
122
32
342
34
32
32
42
122
342
12
32
12
122
32
22
12
22
342
24
21
234
212
234
24
212
234
24
23
24
44
21
24
212
22
24
22
21
223
3
4
aLaLLLLaaLLLaLLLLaLL
LaaaLaaLLLLLLaLLLLLL
dL
dL
104
4-S: cont.
Energy equations :
• D1(L1 L24) + D2(L1
3 L22) + D3(L1
2 L22) + D4(L3
2 L1 L22) + D5(L1 L2
2) + D6(L3
2 L22) + D7(L2
2) + D8(L34) + D9(L3
2) + D10 + D11(L32 L1
3) + D12(L1
3) + D1
3(L32 L1
2) + D14(L1
2) + D15(L3
4 L1) + D16(L3
2 L1) + D17 (L1) = 0
(4-10)
• E1(L12L2
3) + E2(L32L2
3) + E3(L23) + E4(L1
2L22) + E5(L3
2L22) + E6(L2
2) + E7(L1
4L2)+ E8(L32L2) + E9(L1
2L2) + E10(L34L2) + E11(L3
2L12L2) + E12(L2) +
E13(L32)+ E14(L1
2) + E15(L34) + E16 = 0 (4-12)
• F1(L12L2
2) + F2(L3L12L2
2) + F3(L33L2
2) + F4(L32L2
2) + F5(L3L22) + F6(L2
2) + F7(L3
3L12) + F8(L3 L1
2) + F9(L12) + F10(L3
5) + F11(L34) + F12(L3
3) + F13(L3
2)+ F14(L3) + F15 = 0 (4-14)
105
4-S: cont.
• Table 4-1 presents coeffecient G, D, E, and F used in the matrix.
• A numerical example is presented next.
106
4-S: cont.
Numerical Example:• strut lengths:• a12 = 14 in. a34 = 12 in.
• spring 1 free length & spring constant:• L01 = 8 in. k1 = 1 lbf/in.• L02 = 2.68659245 in. k2 = 2.0 lbf/in.• L03 = 3.46513678 in. k3 = 2.5 lbf/in.• L04 = 7.3082878 in. k4 = 1.5 lbf/in.
107
4-S: cont.
• Eighteen real roots have been found using Polynomial Continuation Method.
• The eighteen real roots for this example are presented in Table (4-2) of Chapter 4.
• Values in Table (4-2) can be either for maximum or minimum energy. The second derivative of a polynomial determines its curvature at specified point in Cartesian coordinate system. These values are tested in second differentiation of energy equation (4-20) with respect to L1, L2, and L3.
108
4-S: cont.
• The values of L1, L2, L3, and L4 listed in Table (4-2) satisfy the geometric constraints defined by equation (4-1).
• Each case of Table (4-2) must be analyzed to determine if the mechanism is in a maximum or minimum potential energy state. This is readily accomplished by evaluating the value of the second derivative of potential energy taken with respect to a change in length of spring 1, 2, and 3.
• 2nd derivatives are next.
109
4-S: cont.
12
42
04442
1
4412
1
2
)()(dL
LdLLk
dL
dLkk
dL
Ud
22
42
04442
2
4422
2
2
)()(dL
LdLLk
dL
dLkk
dL
Ud
32
42
04442
3
4432
3
2
)()(dL
LdLLk
dL
dLkk
dL
Ud
110
4-S: cont.
• Table (4-3) of Chapter 4 shows the value of the potential energy and the second derivative for each of the eighteen real cases.
• The only acceptable cases are: 2, 3, 13, 15, and 17. These cases are shown in Figures 8 through 12, bellow:
111
4-S: cont.
• Each of these five configurations was evaluated to determine if it was in equilibrium by determining if forces in the two struts and four elastic ties could be calculated such that the sum of forces at each of the four node points was zero.
• Table (4-4) presents four equilibrium cases of a tensegrity structure consists of two struts and four elastic ties for the
given initial values.
• Acceptable geometry is next.
112
4-S: cont.
113
4-S: cont.
• End of Chapter 4.
114
Agenda
• Objective• Literature review• Analysis of planar tensegrity mechanisms
– 2-spring planar tensegrity– 3-spring planar tensegrity– 4-spring planar tensegrity (added to proposal)
• Future work, Summary and Conclusion
115
3-spring spatial tensegrity system
116
3-spring spatial tensegrity system
The problem statement is given as follows:
Given:The length of non-compliant ties, struts, and initial length of compliant ties.
Find: The final length of compliant ties.
117
3-spring spatial tensegrity system
• The following three approaches were considered:
• minimum potential energy analysis• force balance analysis• linear dependence of 6 connector lines.
• Mathematical solution became very complicated.
118
3-spring spatial tensegrity system
• 5.1.1: 3D Platform Tensegrity:
Future suggested research work is a position analysis of a general three dimensional parallel platform device consisting of 3 struts, 3 springs, and 6 ties.
• 5.1.2: 2D Tensegrity structures:• Cases of two dimensional tensegrity structures proved to be
mathematically challenging in spite of their simple configurations. Different methods were tried to solve these problems in presented research. Other mathematical methods might exist to solve these problems with less complexity.
119
Conclusion
• 5.2: Summery and Conclusion:
• In this research, all possible combination of two dimensional tensegrity structures consists of struts, springs, and non compliance members were reviewed.
• A closed form solution could is obtained for case of 2 spring, 2 struts, and 2 non compliance members.
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• Finding a simple formula to represents a closed form solution to the other 2 dimensional tensegrity structures consisting of either 3 or 4 springs were not possible due to the complexity of mathematical solutions.
• A combination of mathematical, numerical and use of engineering software namely Maple, AutoCAD are used to suggest a procedure to solve these problems and verify the solutions.
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• The benefit of this research will be to set a basic mathematical foundation to the closed form solutions of tensegrity structures.
•The contribution of this research will be to provide a mathematical solution and/or analytical procedure for analysis of more complex structures in tensegrity structures hence robotic.
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Regards
• The author expressly thanks his committee members for their valuable suggestions on proposal, their time and effort to review this dissertation.
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End of presentation
Thank you.