Design of deployable structures using limit analysis of partially rigid frames 

with quadratic yield functions

Makoto OHSAKI (Kyoto University)Yuji MIYAZU  (Hiroshima University)Seiji TOMODA  (Hiroshima University)Seita TSUDA  (Okayama Prefectural University)

Planar mechanism (deployable structure) with partially rigid joints

● : rotational hinge- - - : removed member

Generate mechanism with partiallyrigid joints using optimization method

Input disp.

Output disp.

Mechanism with partially rigid joints

• Optimization approaches for generating link mechanisms– Truss elements for planar mechanism– Ideal (three‐axis) pin joints are needed for 3D mechanism

• Ideal pin joints are difficult to manufacture→ partially rigid connections are preferred

• No systematic approach to design of 3D link mechanism with partially rigid connections

Definition of local coordinates and member‐end force

Release momentT Torsion around x-axis

My Bending around y-axisMz Bending around z-axis

Local coordinates

Node i

x Node i

y

z

Rotational hinge around y-axis

i

j

i

j

x

z

y

kyiM

kziM

kyjM

kzjM

kT

kTkN

kN

Definition of local coordinates and member‐end force

6 x 2 member-end forcesSix equilibrium equations→ Six independent components

( , , , , , )k k k k k k k Tyi zi yj zjN T M M M Mf

Limit analysis problem with two loading conditions

inin

in in1

out

,

, (

max.

s. t.

| | 1, , )

n

ii

i

i

iw i n

f

f

y

p ph equilibrium

yield condition

in

in

out

: load factor

: load vector at input node: load vector at output node

: t

: yield moment or yield member fo

h row of equilibrium matr

r

ix

ce

i

i

i

w

ph

p

Procedure for generating infinitesimal mechanism

• Step 1: Define equilibrium matrix of rigidly jointed frame.

• Step 2: Find release conditions solving limit analysis problem.

• Step 3: Output nodal displacements (mechanism) given as dual variables.

| | for bending moment rotational hinge| | for torsional moment torsional hinge| | for axial force remove member

i i

i i

i i

ww

ff

w f

Procedure for generating finite mechanism

• Step 1: Find release conditions solving limit analysis problem.

• Step 2:  Geometrical non‐linear analysis. – If there exist internal forces, release the corresponding member‐end force and continue this process.

• Step 3: Output the release conditions and nodal displacements of mechanism.

Planar model 1

in 0.31.0 for member extension0.0001for hinge rotat

unit size :1

i n

1

oi

i

ww

u

Planar model 1

● : rotational hinge- - - : removed member

0.42 ≤ ≤ 0.63 ≥ 0.64( ≤ 0.41 → objective function is not bounded below)

Global mechanism

Local mechanism

Planar model 2

in 0.31.0 for member extension0.0001for hinge rotat

unit size :1

i n

1

oi

i

ww

u

Planar model 2

● : rotational hinge- - - : removed member

( ≤ 0.41 → objective function is not bounded below)

0.42 0.42,10000.0 for extension of member Aiw

Global mechanism

Local mechanism

3D mechanism of grid model

Symmetric w.r.t. XZ- and YZ-planesNode 1: fix rotations around X- and Y-axesNodes 2 and 4: Fix displacements in Y- and Z-directionsNodes 3 and 5: Fix displacements in X- and Z-directions

Global coordinateX

YZ

1 2

3

4

5

67

8 9Global coordinate

X

YZ

1

2

34

5

6

7

8

9

1.01.0

1.0

1.0

1.0

Boundary condition Specified deformation

3D mechanism of grid model

- : rotational hinge around local y-axis (bending)● : rotational hinge around local z-axis (bending)

x : rotational hinge around local x-axis (torsion)

Infinitesimal mechanism Finite mechanism

Local axis

x

zy

3D mechanism of grid model

Deformation of 3D mechanismwithout external load

1 2 3

4 5 6 7

8 9 10

242322

21201918

171615

14131211

Global coordinateX

Y

Z

Boundary conditionfixed in Z-directionfixed in X-directionfixed in Y-direction

node 1, 2 and 3: fixed around Y-directionnode 4, 11 and 18: fixed around X-direction

A

Boundary condition

Connect9-units

3D model 1

Pull node 1 downward→ Node 10 moves upward

1.0 for member extension0

unit size :1 11

.1 for bending and to

.0

rsioni

i

ww

3D model 2

Partially rigid frame with diagonal hinges

1

2

3

1.73521.0

Diagonal hinge (arbitrary direction)→ More diverse deformation

Reduce number of hinges

kN

kN

12kM

13kM

23kM

22kM

kT

kT

12

3

Optimization problem for mechanism with diagonal hinges

in6

out in in1

2 2 2 b2 3

2

maximize

subject to

( ( )) ( ( )) ( ( )) ,

( 1, , ; 1, 2)

( ( )) , ( 1

m

i ii

k k kj j

k

f

T M M w

k m j

N w k

a

h p p

f f f

f , , )m

load factor

equilibrium

yield condition for moment

yield condition for axial force

quadratic yield condition conditions

Normalization of u

Bending moment

Torsional moment

Optimality conditions

in1 0T p u

2 ( ) 0,

( 1, , ; 1, 2; 2,3)

T k ki jp jM c

k m j p

h u f

1 22 ( )( ) 0, ( 1, , ; 1,2)

T k k ki T c c

k m j

h u f

: member-end: axis: rotation of

member-end : bending moment at

member-end around axix

: torsional moment

kj

kjp

k

jpc

jM

jp

T

Axial force

Complemenrarity condition

Optimality conditions

02 ( ) 0, ( 1, , )

T k ki N c

k m

h u f

2 2 2 b2 3[( ( )) ( ( )) ( ( )) ] 0,

0, ( 1, , ; 1, 2)

k k kj j kj

kj

T M M w c

c k m j

f f f

20

0

[( ( )) ] 0,

0, ( 1, , ; 1,2)

k k

k

N w c

c k m j

af

0

: member-end: axis: extension

: axial force

k

k

jpc

N

Rotation or extension isnon-zero only when yieldcondition is satisfiedwith equality

Optimality condition

2 , ( 1, , ; 1, 2; 1, 2)k k kjp jp jM c k m j p

1 1 1 1 22 ( ), ( 1, , ; 1, 2)k k k k k kj i T c c k m j

Bending moment

Torsional moment

1

1 12 1 12

13 13

k k

k k kk

k k

Tc M

M

R1

2 22 2 22

23 23

k k

k k kk

k k

Tc M

M

R

Rotation around axis is proportional to

bending moment

(norm of rotation does not depend on axis)

kjp

kjp

kj

p

M

c

Auxiliary problem for  determination of parameter alpha

6

out1

2 2 2 b2 3

2

maximize

subject to

( ( )) ( ( )) ( ( )) ,

( 1, , ; 1, 2)

( ( )) , ( 1, , )

m

i ii

k k kj j

k

f

T M M w

k m j

N w k m

a

h p

f f f

f

load factor

equilibrium

yield condition for moment

yield condition for axial force

lower bound of 1/

Example 1

Example 1

1

2

3

1.73521.0

1

2

3

Vertical members 1-3 and 1-5

Horizontal members 1-2 and 1-4

Example of 3D‐mechanism with partially rigid joints

Symmetric w.r.t. XZ- and YZ-planesNode 1: fix rotations around X- and Y-axesNodes 2 and 4: Fix displacements in Y- and Z-directionsNodes 3 and 5: Fix displacements in X- and Z-directions

Global coordinateX

YZ

1 2

3

4

5

67

8 9Global coordinate

X

YZ

1

2

34

5

6

7

8

9

1.01.0

1.0

1.0

1.0

Boundary condition Specified deformation

Example 2

Conclusions

• New method for generating a deployable structure by solving a quadratic programming problem.

• Limit analysis problem with a quadratic yield function of the member‐end moments.

• Mechanism with diagonal hinges. • By allowing a diagonal hinge, the number of hinges can 

be reduced compared with the previous study, where only the hinges around the local axes are allowed.

• Additional hinges may be needed to generate a finite mechanism.