Slide 2b.1 Stiff Structures, Compliant Mechanisms, and MEMS: A short course offered at IISc,...

Stiff Structures, Compliant Mechanisms, and MEMS: A short course offered at IISc, Bangalore, India. Aug.-Sep., 2003. G. K. Ananthasuresh Slide 2b.1

Lecture 2bOptimal design of Regular Structures for Stiffness and FlexibilityPrincipal features of optimal design of stiff structures and compliant mechanisms via analytical solutions for simple structural forms.


Contents• Stiffest bar for a given volume of material

– Equilibrium equations in strong form– And in weak form– Some insights

• Lightest beam for given stiffness– Interchanging the objective and an integral constraint

• Lightest beam for given deflection– What is right and what is wrong?– How to fix the formulation?

• Compliant mechanism: optimal juxtaposition of flexibility and stiffness– Proper way to formulate a compliant mechanism problem


Stiffest bar for a given volume of material

C/s shape does not matter for axial load

)(xA

Lx

)(xp = load

= cross-section profile

Measure of stiffness = mean compliance = L

dxxuxp0

)()()(xu = axial deflection

At equilibrium, mean compliance = 2 * strain energy

Stiffest structure minimum mean compliance least strain energy


Optimal design problem for the stiffest bar

0

0)(

Minimize

*

0

0)(

VdxA

puEApdx

duEA

dx

d

toSubject

dxupJ

L

L

xA

Static equilibrium equation

Volume constraint

(Essential boundary conditions will also be there.)

*

000

)( VdxAdxpuEAdxupLLLL

Lagrangian

variable(spatial)tindependen

variablestate)(

variabledesign)(

x

xu

xA


Solution for the stiffest bar problem

uEuE

uEuELA

0

0)(0

*

000

)( VdxAdxpuEAuAEdxupLLLL

0)(0)(

0)()(0

pEAAEEAp

EAAEpLu

Recall the equilibrium equation: 0)( puEA

u

Self-adjointness

Strain energy density is constant across the bar 2uEuE

Insight: Optimal use of material makes every point work equally hard.

Necessary conditions:


Cross-section profile for the stiffest bar

E

uuE

2

0)( puEASubstitute the above result into

0 pAE Solve for given loading, .)(xp

Use the natural boundary condition (i.e., the internal axial force is zero at the free end) to find .C

0)(0)( LAuEA Lx

CxE

pA c

cpxp )(Example: = constant. Then,

)( xLE

pA c

Volume constraint gives .

)(2

** xLL

VA x)(xA

Now, we know why Egyptian pyramids and temple towers linearly taper!

This tells us why volume constraint must be active.Recall complementarity condition!


Highlights and some observations

• Self-adjointness• Uniform strain energy density• Lagrange multiplier has a physical

meaning (as it usually does)• Optimal shape for stiffest structures does

not depend on…– Material property– Actual magnitude of loading but depends only

on the profile of the loading (here, it was constant)

)(2

** xLL

VA u 2uE


Weak form of equilibrium conditions in the stiffest bar problem

0

0)(

Minimize

*

0

0

0)(

VdxA

dxpvvuEA

toSubject

dxupJ

L

L

L

xA

Weak form of the equilibrium equation

Volume constraint

(Essential boundary conditions will also be there.)

*

000

VdxAxdpvvuEAdxupLLLL

Lagrangian

A scalar-function Lagrange multiplier is now replaced with a scalar-variable multiplier.

An additional state variable ( ) arose but that can be easily dealt with.

)(xv


Solution with the weak form

00 vuELA

*

000

VdxAdxpvvuEAdxupLLLL

0)(0 vAEpLu

0)(0 puAELvuv

Eu

uE

/)( 2

Necessary conditions

Same result as before

In numerical optimization procedures, it is easier to handle integral (global) constraints. Hence, writing equilibrium equations in the weak form is advantageous.We can see that advantage even in manipulations involved in analytical solutions such as this.


The lightest beam for given stiffness

)(xbL

x)(xp = load

= width profile

)(xw = transverse deflection

t

)(12

)(

12)(

23

xAbttbt

xI

In general, )()( xAxI for other cross-sections.

Mean compliance = L

dxwp0

0)(

0)(0

pwEA

dxpvvwEAL

Equilibrium equation

Weak

Strong(And, essential boundary conditions)


The lightest beam problem statement

0

0)(

Minimize

*

0

0

0)(

Sdxwp

dxpvvwEA

toSubject

dxAJ

L

L

L

xA

*

00

0

SdxwpdxpvvwAE

dxAL

LL

L

010 vwELA

0)(0 pvEALw

0)(0 pwAELv

Necessary conditions

Lagrangian:

wv

Ew

wE

2

22

1

1)(

Again the same story… the strain energy density is the same.Other viewpoint: uniformly stressed beam.This viewpoint leads to the “optimality criteria” method.


The lightest beam for given deflection at a point: (towards flexibility)

L

x)(xp = load

t

x̂

Consider a simply-supported (statically determinate) boundary condition.

0

Minimize

0

0)(

dxAE

Mm

toSubject

dxAJ

L

L

xA

dx

AE

MmdxAL

LL

00

E

Mmdx

E

Mm

E

MmA

AE

MmL

L

A

0

*2

1

010

is to determined using the deflection constraint.

mM= bending moment due to )(xp= bending moment due to unit dummy load applied at A

A


A problem and its fix

EMm

A

*

MmWhat if is negative/zero somewhere in the span of the beam?

Let us try by imposing a lower limit on the area of c/s…

0

0

Minimize

0

0)(

AA

dxAE

Mm

toSubject

dxAJ

l

L

L

xA

L

l

LL

dxAAdxAE

MmdxAL

000

)(

E

MmA

AE

MmLA )1(

010 *2

)(x will be zero wherever .)(*lAxA

0))(()( xAAx l (Recall complementarity)

0)( x in other part(s) of the beam.


Is the problem really fixed?

lAl

A

xA

xE

MmA

if

if **

x

*A

lA

*

1

A

lA

dxE

Mm

AE

Mm

l

Observe that if area goes below the lower limit at some point, it has the effect of decreasing , which in turn, decreases and hence further. Finally, the entire span of the beam will reach the lower limit!

*A

*A

Conclusion: Design for given deflection needs infinitesimally small volume.

Therefore, including only flexibility (i.e., deflection) requirement is not a well posed problem.


Is this conclusion a surprise?

0

Minimize

0

0)(

dxAE

Mm

toSubject

dxAJ

L

L

xA

0

Minimize

0

0)(

dxAE

Mm

toSubject

dxAJ

L

L

xA

Conclusion: Including only flexibility (i.e., deflection) requirement is not a well posed problem.

A stiffness requirement, albeit only at one point.

Making area as small as possible makes the objective and the constraint very happy!They have the same monotonicities, which makes it improper.

Ill-pose

dWell-

pose

d

The objective and the constraint have opposite monotonicities with respect to the area—an indication of a properly posed optimization problem.

A flexibility requirement at one point.


A good fix: add stiffness requirement

0

0

Minimize

*

0

2

0

0)(

SdxAE

M

dxAE

Mm

toSubject

dxAJ

L

L

L

xA

*

0

2

00

SdxAE

Mdx

AE

MmdxAL

LLL

E

MMmA

AE

M

AE

MmLA

2*

2

2

2010

*

02

2

02

1

1

SdxMMm

M

E

dxMMm

Mm

EL

L

Use the two constraints to solve for and .

Now, can be prevented from becoming small by choosing as small as needed.

)(* xA

*S

Why?


Well posed flexibility (compliance) problem

• Adding a stiffness requirement prevents an overly flexible (and hence undesirable and indeed inappropriate, as we just saw) design solution.

• Insight: a flexible structure (and a compliant mechanism) should be:– As flexible as needed, but…– It should also be reasonably stiff.

• Practical viewpoint: some stiffness is needed to withstand the applied loads– Overly flexible structure has excessive motion but not the ability

to effectively support or transfer the loads.

• Let us recall:– Structures support and transmit loads.– Mechanisms transfer/transform motion AND support and transmit

loads.

Slide 2b.1 Stiff Structures, Compliant Mechanisms, and MEMS: A short course offered at IISc,...

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