Module 1: Lecture 1 Classification of Optimization Problems ......Solving two problems concerning size optimization of beams for stiffness and flexibility with volume constraint. What

Lecture 16a

Size optimization of beams for stiffness and flexibility

ME 260 at the Indian Inst i tute of Sc ience , Bengaluru

Str uc tur a l O pt imiz a t io n : S iz e , Sha pe , a nd To po lo g y

G . K . Ana ntha s ur e s h

P r o f e s s o r , M e c h a n i c a l E n g i n e e r i n g , I n d i a n I n s t i t u t e o f S c i e n c e , B a n a g a l o r e

s u r e s h @ i i s c . a c . i n

1

Structural Optimization: Size, Shape, and TopologyG. K. Ananthasuresh, IISc

Outline of the lectureSolving two problems concerning size optimization of beams for stiffness and flexibility with volume constraint.

What we will learn:

How to apply the eight steps we had used for bars to the case of beams.

2


Problem 1

3

( ) ( )

( )0

*

0

2 *

( ) : 0 0

: 0

: ,q( ), , ,12

L

A x

L

Min MC q wdx

Subject to

x EIw q E Aw q

Adx V

tData L x E V

=

− = − =

−

=

We assume here that only b(x) is variable.

( )I x A=2

; 112

t

= =

Minimize the mean compliance of a beam for given volume of material.


Steps in the solution procedure

4

Step 1: Write the Lagrangian

Step 2: Take variation of the Lagrangian w.r.t. the design variable and equate to zero to get the design equation.

Step 3: Take variation of the Lagrangian w.r.t. state variable(s) and equate to zero to get the adjoint equation(s).

Step 4: Collect all the equations, including the governing equation(s), complementarity condition(s), resource constraints, etc.

Step 5: Obtain the optimality criterion by substituting adjoint and equilibrium equations into the design equation, when it is possible.

Step 6: Identify all boundary conditions.

Step 7: Solve the equations analytically as much as possible.

Step 8: Use the optimality criteria method to solve the equations numerically.


Solution

5

( ) ( )

( )0

*

0

2 *

( ) : 0 0

: 0

: ,q( ), , ,12

L

A x

L

Min MC q wdx

Subject to

x EIw q E Aw q

Adx V

tData L x E V

=

− = − =

−

=

Minimize the mean compliance of a beam for given volume of material.

( )( ) *0

L

L q w E Aw q A dx V = + − + −Step 1


Expand the Lagrangian

6

( )( ) *0

L

L q w E Aw q A dx V = + − + −Step 1

( )( )

( )

*

0

*

0

*

0

2

L

L

L

L q w E Aw q A dx V

qw E A w Aw q A dx V

qw E A w E A w E Aw q A dx V

= + − + −

= + + − + −

= + + + − + −


Design equation

7

*

0

2

L

L qw E A w E A w E Aw q A dx V = + + + − + −Step 2

( ) ( )

( ) ( ) ( )

( ) ( )

( ) ( ) ( ) ( )

0 0

2 0

2 0

0

0

A

F F FL

A A A

E w E w E w

E w E w E w

E

E

E

w

w

E w E

E

w w E w

Ew w E w

E

= − + =

+ − + =

+ − + + =

+ − + =

+ − − + +

=

+

( ) 0w =

2F qw E A w E A w E Aw q A = + + + − +


Adjoint equation

8

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( )

( )

0 0

0 2 0

2 0

0

w

F F F F FL

w w w w w

q E A E A E A

q E A E A E A E

E EA

A

q E A E A E A

q

= − + − + =

− + − + =

+ − + + =

+ − + =

+ − ( ) ( ) ( ) ( )

( )

0

0

A E EE A A

A

A

q E

− + + =

+ =

2F qw E A w E A w E Aw q A = + + + − + Step 3


Collect all equations

9

Step 4

( ) 0E w + =

Unknowns

Adjoint equation ( ) 0q E A + =

Governing equation ( ) 0E Aw q − =

Feasibility condition*

0

0

L

Adx V−

Complementarity condition

*

0

0, 0

L

Adx V

− =

w = −

2E w =

Strain energy density is uniform throughout the beam.

And, cannot be zero.

So, the volume constraint is active.

Step 5Design equation

( ), ( ), ( ),A x w x x

Three functions and one scalar variable.

We have three differential equations and one scalar equation.


Identify all boundary conditions

10

Step 6

( )

− =

=

0

0

0

0

and

L

A A

L

A

F F A

F A

Boundary conditions for ( )A x

2F qw E A w E A w E Aw q A = + + + − +

( )

0

0

0

2 0

0

0

L

L

L

E w E w A

E w E w A

E ww E w w A

− =

− =

− + = since = −w

0

0 0

0

0 0

L

L L

E w A

E ww A E ww A

=

− = =



11

Step 6 Boundary conditions for ( )x

2F qw E A w E A w E Aw q A = + + + − +

( ) ( ) ( )

( ) ( )

( )

− + − =

− + =

− =

=

0

0

0

0

0

0

0

0

L

L

w w w w

w w w

L

w w

L

w

F F F F w

F F F w

F F w

F w



12

Step 6 Boundary conditions for ( )x

2F qw E A w E A w E Aw q A = + + + − +

( ) ( ) ( )

( ) ( )

( )

− + − =

− + =

− =

=

0

0

0

0

0

0

0

0

L

L

w w w w

w w w

L

w w

L

w

F F F F w

F F F w

F F w

F w

( )0

0L

E A w =

( ) 0

2 0

L

E A E A w − =

( ) ( ) 0

2 0

L

E A E A E A w − + =

( ) ( ) ( ) 0

2 0

L

E A E A E A w − + − =


Simplify 1st adjoint boundary condition

13

( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

0

0

0

0

0 0

2 0

2 0

0

0

0 0

L

L

L

L

L L

E A E A E A w

E A E A E A E A w

E A E A E A w

E A E A E A E A w

E A E A w E A E A E A w

E

− + − =

− + − − =

− + − =

− + + − =

− = − − =

−( ) ( ) 00

00L

L

E AA w E A w + = =


Simplify 2nd-4th adjoint boundary conditions

14

( )0

0L

E A w =

( ) ( )0

002 0

LL

E A E A wE A E A w = −− =

( ) ( )

( ) ( ) ( )

( ) ( )

( ) ( ) ( )

0

0

0

00

2 0

2 0

0

0 0

L

L

L

L

L

E A E A E A w

E A E A E A E A w

E A E A E A w

E w E AE A wA

− + =

− + + =

− + =

− + = =


All four adjoint boundary conditions

15

( )0

0L

E A E A w + =

( )0

0L

E A w =

( )0

0L

E A w =

( )0

0L

E A E A w − =

PinnedFixed0w w = = 0w w = =

0A =

0A A − =

0A =

No BC for ( )x No BC for ( )x

No BC for ( )x

No BC for ( )x

0A =

Notice that BCs of the state variable transfer to adjoint variable (most often). But be sure to keep the BCs on the design variable in mind.


All four adjoint boundary conditions

16

( )0

0L

E A E A w + =

( )0

0L

E A w =

( )0

0L

E A w =

( )0

0L

E A E A w − =

Transversely guided Free

0w w = =

0w w = =

0A A − =

0A =No BC for ( )x

No BC for ( )xNo BC for ( )x

No BC for ( )x

0A A + = 0A A + =

Notice that BCs of the state variable transfer to adjoint variable (most often). But be sure to keep the BCs on the design variable in mind.


Solving for a particular beam BCs

17

Step 7Transversely guided

0w w = =

Fixed

0w w= =

( ) 0E Aw q − =*

0

0

L

Adx V−

2E w

wE

=

=

0A A − =0A A + =0A = 0A A − =& &

0

0L

E ww E w w A − + = 0

0L

E ww A = &

0( )q x q=

( )2

00 1 0

2

q xA E q A C x C

E

= = + +

Take 0 = = Take 0A = = =

Satisfied at x = 0 and x = L

Satisfied at x = 0 and x = L

Solve for

01

q LC

E= −

using


Reconciliation of BCs for A(x)

18

20 0

02

q x q LA x C

E E = − +

From the previous slide, we haveStep 7

We can use the active volume constraint to solve for

( )

3 3*0 0

0

30

*0

6 2

3

q L q LC L V

E E

q LE

V C L

− + =

= −−

for + sign( )

30

*0

2

3

q LE

V C L = −

−for - sign&

( ) ( )

( ) ( )

2 * *0 0

03 2

2 * *0 0

03 2

3 3

2( )

3 3

4 2

x V C L V C Lx C

L LA x

x V C L V C Lx C

L L

− −− − +

= − −

− +

How do we choose the two possibilities (+ or -) and determine C0?


Validating with the numerical solution…

19

( ) ( )

( ) ( )

2 * *0 0

03 2

2 * *0 0

03 2

3 3

2( )

3 3

4 2

x V C L V C LA x C

L LA x

x V C L V C LA x C

L L

+

−

− − = − − +

= − −

= − +

We need to see which part of the domain should have the “+ curve” and which part should have “-curve”.

+ curve

- curve

- curve+ curve

x̂x

L

( )A x

0C

We need to find using

0ˆ andx C

ˆ ˆ( ) ( ) 0A x A x+ −= =

*

min max 020; 1; 50; 210; 1 3; 10; 1L d V E A E A q= = = = = − = =

( )A x

x

20

Problem 12

( )

( )

( )0

*

0

2 *

0

2 *

( ) : 0

( ) : 0

: 0

1: 0

2

: ,12

L

A x

d d

L

L

*d

Min V Adx

Subject to

x E Aw q

x E Av q

E Aw v dx

E Aw dx SE

tData L,q(x),q (x), = ,E,Δ SE

=

− =

− =

− =

− =

Minimize the volume of material of a beam (statically determinate or indeterminate) for a deflection constraint in its span with an upper bound on the strain energy.


The end note

21

Thanks

Siz

e o

pti

miz

atio

n o

f b

eam

s

Iterative numerical solution, when it is needed, remains the same.

We follow essentially the same eight steps

Identifying the optimality criterion is the highlight.

See the correlation between BCs and optimal profiles

Boundary conditions for the adjoint variable need to be carefully done.

Analytical solution may be segmented with multiple possibilities because of + and – of constant strain.

Documents

Module 1: Lecture 1 Classification of Optimization Problems ......Solving two problems concerning size optimization of beams for stiffness and flexibility with volume constraint. What