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1 ENE 5400 , Spring 2004 1 Lecture 6 Electrostatic Actuation Derived from an energy perspective Parallel-plate actuator Comb-drive actuator Mechanics of Material Beam Deflection » Integration method » Energy method Bending of Plates Flexure Spring Constants Residual Stress and Stress Gradient ENE 5400 , Spring 2004 2 Energy Perspective: Charge Control Due to the conservation of the energy of the system: By partial differentiation: Calculate stored energy by integration along the path: = e dW + _ z F e q V const q e e z W F = Stored energy Input energy Work done by force q’ q z z’ dq’=0 dz’=0 = e W *Source: H.H. Woodson and J.R. Melcher, Electromechanical Dynamics, Part I: Discrete Systems, Chap. 3, John Wiley & Sons, New York, 1968. (z,q)
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Page 1: Lecture 6 - National Tsing Hua University

1

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 1

Lecture 6

� Electrostatic Actuation�Derived from an energy perspective�Parallel-plate actuator�Comb-drive actuator

� Mechanics of Material �Beam Deflection

» Integration method» Energy method

�Bending of Plates� Flexure Spring Constants� Residual Stress and Stress Gradient

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 2

Energy Perspective: Charge Control

� Due to the conservation of the energy of the system:

� By partial differentiation:

� Calculate stored energy by integration along the path:

=edW +_ z Fe

q

V

constq

ee z

WF

∂∂∂∂∂∂∂∂−−−−====

Stored energyInput energy Work done by force

q’

q

z z’

dq’=0

dz’=0=eW

*Source: H.H. Woodson and J.R. Melcher, Electromechanical Dynamics, Part I:Discrete Systems, Chap. 3, John Wiley & Sons, New York, 1968.

(z,q)

Page 2: Lecture 6 - National Tsing Hua University

2

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 3

Cont’d

� The amount of charge on conductor is:

� By substitution, the energy We is:

� The electrostatic force in the z direction is:

� Is the charge control easy to implement?

CVq ===='

(((( )))) (((( )))) (((( ))))∫ ========q

e zCq

dqzC

qqzW

0

2

2'

',

(((( ))))(((( ))))

dzzdC

zCq

zW

Fconstq

ee 2

2

2====

∂∂∂∂∂∂∂∂−−−−====

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 4

Energy Perspective: Voltage Control

� Introduce the co-energy expression:

� By partial differentiation:

� Integrate along the path as shown:

� The electrostatic force in the z direction is:

constv

ee z

WF

∂∂∂∂∂∂∂∂====

'v’

V

z z’

dv’=0

dz’=0='

eW

(((( )))) 2'

21

Vdz

zdCz

WF

constv

ee ====

∂∂∂∂∂∂∂∂====

Page 3: Lecture 6 - National Tsing Hua University

3

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 5

Parallel-Plate Actuators

� The parallel-plate electrostatic force is :

� The force is nonlinear with respect to the applied voltage and the displacement

� Pull-in would happen due to a positive-feedback mechanism

gz

m, b, k

V +_

Fixed plate A

(((( ))))(((( ))))

22

22

21

21

21

VzgA

Vdz

zgA

dV

dzdC

F o

o

e −−−−====

−−−−========

εεεεεεεε

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 6

Electrostatic Pull-in: the Transfer-Function Perspective

� The dynamic equation is:

� At any equilibrium point (Zo, Vo):

� Consider a small variation of ∆∆∆∆z and ∆∆∆∆varound the operating point (Zo, Vo):

(((( ))))2

221

VzgA

Fkzzbzm oe −−−−

========++++++++εεεε

&&&

(((( ))))e

eVZeoo

Fzkzbzm

FFzZkzbzmoo

∆∆∆∆====∆∆∆∆++++∆∆∆∆++++∆∆∆∆∴∴∴∴

∆∆∆∆++++====∆∆∆∆++++++++∆∆∆∆++++∆∆∆∆

&&&

&&&

,

( )( )

( ) )(kZAV

Zgor

)(A

ZgkZV

Zg

AVkZ

o

ooo

o

ooo

o

ooo

22

12

2

22

22

2

2

ε=−

ε−

=⇒

−ε

=

gz

m, b, k

V +_

Fixed plate A

Page 4: Lecture 6 - National Tsing Hua University

4

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 7

Cont’d

� Use Taylor-series expansion and substitute (1) and (2):

� Therefore,

� Define the “negative” electrical spring ke: � At Zo = g/3, the electrical spring completely negates the

mechanical spring (the actuator is marginally stable)

( ) ( )

( ) vVkZ

zZg

kZ

vZg

AVz

ZgkZ

vvF

zzF

F

o

o

o

o

o

oo

o

o

V,Z

e

V,Z

ee

oooo

∆+∆−

=

∆−

ε+∆

−=∆⋅

∂∂

+∆⋅∂∂

=∆

22

22

vVkZ

ZgkZ

kzzbzm

Fzkzbzm

o

o

o

o

e

∆∆∆∆====

−−−−−−−−∆∆∆∆++++∆∆∆∆++++∆∆∆∆

∆∆∆∆====∆∆∆∆++++∆∆∆∆++++∆∆∆∆

22&&&

&&&

(((( ))))(((( ))))k

gZgZ

kZg

Zk

o

o

o

oe /1

/22−−−−

−−−−====−−−−

−−−−====

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 8

Cont’d

� The small-signal transfer function is:

� For displacements larger than g/3, the sum of k + ke (will have right-half-plane pole) is negative ⇒ the system is unstable

� Substitute Zo = g/3 into (1), we obtain the corresponding pull-in voltage:

(((( ))))e

oo

kkbsmsVkZ

vz

++++++++++++====

∆∆∆∆∆∆∆∆

2

/2

Akg

Vo

pi εεεε278 3

====

Page 5: Lecture 6 - National Tsing Hua University

5

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 9

Electrostatic Pull-in: Force-Balancing Perspective

� Solve Fe = Fm (3rd-order algebraic equation); there are normally two intersections except at the pull-in� No intersection (equilibrium state) after V > Vpi

1/3

2/3

1

z/g

V/Vpi

z/g < 1/3: stable equilibrium

z/g > 1/3: unstable equilibrium

(Vpi, g/3)

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 10

Lateral Comb-Drive Actuator

� Invented by W.C. Tang (Sensors and Actuators A, vol. 21, no. 1-3, pp. 328-331, 1990)

� Use the fringing field lines to drive, eliminating the 1/3 gap problem� Force is linear with respect

to the displacement ∆∆∆∆x� However the attainable force

is smaller than parallel plates� The capacitance and force are:

2

)(2)(

Vgh

F

gxLh

xC

oe

o

εεεε

εεεε

====

++++⋅⋅⋅⋅====

stator

rotorFe

+_ V

g

g

x

h

L

k

Page 6: Lecture 6 - National Tsing Hua University

6

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 11

Lateral Comb Drive with a Ground Plane

� Three conductors: rotor (r), stator (s), and ground plane (p)

� The differential co-energy of the comb-drive is:

� By integration of dWe’ along the 3-dimensional path:

==

=

r

s

'e

q

q

dW

='eW

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 12

Cont’d

� The lateral force is:

(((( )))) (((( )))) (((( )))) (((( ))))

(((( )))) 2,

222,

21

:021

21

21

srsspxe

r

rsrs

rrp

ssp

xe

VCCdxd

F

Vfor

VVdx

xdCV

dx

xdCV

dx

xdCF

++++====

====

−−−−++++++++====

Page 7: Lecture 6 - National Tsing Hua University

7

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 13

Comb-Drive Levitation Force

� With a bottom ground plane, the levitation force can be introduced due to the asymmetrical electric-field line distribution on top and bottom of the fingers

� The expression is analogous to the lateral force expression:

(((( )))) (((( )))) (((( )))) (((( ))))222, 2

121

21

rsrs

rrp

ssp

ze VVdz

xdCV

dz

xdCV

dz

xdCF −−−−++++++++====

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 14

Mechanics of Material

Page 8: Lecture 6 - National Tsing Hua University

8

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 15

Axial Stress and Strain

� Axial stress is normal to the applied surface� σσσσ = F/A (typically in N/m2, or Pa)� F is positive if it is a tensile force, and negative if it is a

compressive stress� Strain εεεε = ∆∆∆∆L/Lo (dimensionless)

Lo

Lo + ∆L

FF

A (area)

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 16

Shear Stress and Strain

� Shear stress is parallel to the applied surface: ττττ = F/A� Shear strain γγγγ (in radian) is the angular deformation with

respect to original structural shape

F

A (area)

γγγγ

Page 9: Lecture 6 - National Tsing Hua University

9

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 17

Shear Stress and Torsion

T

T

Take any cross-section

ρdF

τmax

c ρ

τ

Assume that the shearing stress varieslinearly with the distance ρJ

TJ

Tcc

J

dAc

dA

dFT

ρρρρττττ

ττττ

ττττ

ρρρρττττρρρρ

ττττρρρρ

ρρρρ

====∴∴∴∴

====⇒

====

====

====

====

max

max

max )(

)(

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 18

Material Properties

� Modulus of Elasticity (Young’s Modulus): E = stress / strain, σσσσ/εεεε� Modulus of Rigidity: G = shear stress / shear strain, ττττ/γγγγ� Poisson’s Ratio: v = lateral strain/axial strain, (∆∆∆∆w/Wo)/(∆∆∆∆l/Lo)� Relation between E, G, and v: G = E / [2(1 + v)]

Lo

Lo + ∆l

Wo

Wo - ∆w

F F

Page 10: Lecture 6 - National Tsing Hua University

10

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 19

MEMS Material Properties

� Material property database� MEMS Clearinghouse, http://www.memsnet.org/material

Young’s Modulus Density Thermal conductivity Thermal expansion Yield strength (GPa) (kg/m3) (W/m⋅⋅⋅⋅K) coefficient (K-1) (GPa)

Diamond* SiC*Al2O3*Si3N4*Iron* SiO2 (fibers)Silicon*W Aluminum

1035700530385196

73190410

70

3500320040003100780025002300

193002700

2000350

501980.3

1.4157178236

1 ×××× 10-6

3.3 ×××× 10-6

5.4 ×××× 10-6

0.8 ×××× 10-6

12 ×××× 10-6

0.55 ×××× 10-6

2.33 ×××× 10-6

4.5 ×××× 10-6

25 ×××× 10-6

532115.41412.68.4740.17

* Asterisks mean for single-crystal material* From K. Peterson, Proc. IEEE, May, 1982

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 20

(Second) Moment of Inertia of an Area

∫=

=

dAyI

dAydI

x

x

2

2

dA = (a-x)⋅dy

x

y

y

x dy

a

y

∫=

=

dAxI

dAxdI

y

y

2

2

dA = y⋅dx

xy

x

dx x

dA

y

r

∫= dArJo2

= Ix + Iy

Moment of inertia w.r.t. the x axis Moment of inertia w.r.t. the y axis Polar moment of inertia

Page 11: Lecture 6 - National Tsing Hua University

11

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 21

Moment of Inertia Examples

x’

x

y y’

h

b

I x’ = 1/12 bh3

I y’ = 1/12 b3hI x = 1/3 bh3

I y = 1/3 b3hJ c = 1/12 bh(b2+h2)

c

ch/3

h

b

I x’ = 1/36 bh3

I x = 1/12 bh3x’

x

x

y

or

I x = I y = 1/4 πr4

J o = ½ π r2

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 22

Pure Bending

� Our goal is to understand how the bending moment M is related to the radius of curvature (and thus the deflection) of structures

� A member subjected to equal and opposite couples acting in the same longitudinal plane is said to be in pure bending� For example, you can imagine M is the applied moment and

M’ is the reaction moment with the same magnitude in the reversed direction

MM’

CB

A

x

y

z

EI

xM

dx

yd )(2

2

=

Page 12: Lecture 6 - National Tsing Hua University

12

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 23

A Beam Cross Section in Pure Bending

� Based on the free-body diagram, a counter-balancing moment exists in any cross section; the moment is a combinational effect of the normal stress and shear stress acting on the plane

x

y

z

τxydA

τxzdA

σxdAy

z

x

y

z

M M

M

=

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 24

How Do the Stresses Relate to the Moment?

� The total stresses contribute to a single moment M:

x

y

z

τxydA

τxzdA

σxdAy

z

x

y

z

M M

M

=

x components:

Moments about y axis:

Moments about z axis:

Page 13: Lecture 6 - National Tsing Hua University

13

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 25

How Does Longitudinal Strain Relate to the Radius of Curvature?

� The plane corresponding to the neutral axis AB has no deformation (εεεεx = 0)

ρ

Μ Μy

x

y

ρ - y

A BC D

z

neutral axisyc

y

mx

m

AB

ABCDx

cy

c

yyL

LL

εεεεεεεε

ρρρρεεεερρρρρθρθρθρθ

ρθρθρθρθθθθθρρρρεεεε

−−−−====

====

−−−−====−−−−−−−−====

−−−−====

/

)(Longitudinal strain of CD:

The max. absolute value of strain:

A linear relationship: (note the minus sign due to the direction of M)

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 26

Cont’d

� Remember previously we have:

MdAyc

dAcy

EydAEydAy

MdAy

m

mxx

x

========

−−−−−−−−====−−−−====−−−−

====−−−−

∫∫ ∫

43421

2

)]()[())(()(

)(

σσσσ

εεεεεεεεσσσσ

σσσσ

I

IMyI

Mc

x

m

−−−−====∴∴∴∴

====⇒

σσσσ

σσσσ (note that it is positive since εm is positive)

Page 14: Lecture 6 - National Tsing Hua University

14

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 27

Cont’d

� Finally, bending moment is related to the radius of curvature by:

� Equation of the elastic curve (from your basic Calculus):

� Assuming a small beam deflection for an elastic beam:

EIM

IMy

Ey

Exx =ρ

∴−=ρ

−=ε=σ1

2/32

2

2

1

1

+

=

dxdy

dx

yd

ρ

EI

xM

dx

yd )(2

2

=

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 28

Boundary Conditions for Supported Beam

Α Β

(1) Cantilever beam (yA = 0, (dy/dx)A = 0)

ΑΒ

(2) Beam with a guided-end (yA = 0, (dy/dx)B = 0)

(3) Simply supported beam (yA = 0, yB = 0)

x

y

Page 15: Lecture 6 - National Tsing Hua University

15

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 29

Deflection of Beams by Integration

� Task 1: Specify bending moment M(x)� Use of free-body diagram: ∑Force = 0, and ∑Moment = 0 at

equilibrium� Example of a cantilever beam subjected to a pointed load P

P

Free-body diagram

x

M

Α Β

P

x

∑Force = 0 ⇒ V = -P∑Moment = 0 ⇒M = -Px

V

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 30

Deflection of Beams by Integration

� Task 2: Integration of the 2nd-order differential equation and insert boundary conditions:

� Tip deflection = . So spring constant = 3EI / L3EI

PL

3

3

)LxLx(EIP

y

PLCy

CxPLPxEIy

PLCdxdy

CPxdxdy

EI

PxMdx

ydEI

Lx

Lx

323

32

223

21

12

2

2

236

3

10

2

1

6

1

2

10

2

1

−+−=

−=⇒=

++−=

=⇒=

+−=

−==

=

=

1st integration:

2nd integration:

B.C.:

B.C.:

Final beam curve:

Page 16: Lecture 6 - National Tsing Hua University

16

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 31

Deflections of Beams

Lx

y

x

y

x

y

x

y

x

y

P

q

L/2

q

P

P

y = -PL3/(3EI) at x = L

y = -qL4/(8EI) at x = L

y = -PL3/(48EI) at x = L/2

L/2

y = -PL3/(192EI) at x = L/2

y = -qL4/(384EI) at x = L/2

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 32

Energy Method – Strain Energy

� The strain energy stored in a structural body is the work done by the applied load P with a deformation x1 :

� Strain-energy density:

� So the strain energy is:

210 2

11kxPdxU

x======== ∫

==VU

u

x1

P1P = kx

x

P

ε1

σ1σ = εE

ε

σ

∫==== dVE

U2

21σσσσ

Page 17: Lecture 6 - National Tsing Hua University

17

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 33

Strain Energy under Axial Loading

EALP

AdxEAP

dVEAP

U

2

)(2

2

2

2

2

2

2

====

====

====

PP’

A

L

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 34

Strain Energy in Bending

∫∫

====

====

====

====

L

I

L

x

dxEI

M

dxdAyEIM

dVEI

yM

dVE

U

0

2

2

0 2

2

2

22

2

2

)(2

2

2

43421

σσσσ

Α Β

P

x

Example:

∫ ====−−−−====

L

EILP

dxEI

PxU

0

322

62)(

Page 18: Lecture 6 - National Tsing Hua University

18

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 35

Elastic Strain Energy for Shearing Stresses

� The strain-energy density is expressed by:

� The total strain energy

G

G

dG

du

xy

xy

xyxy

xyxy

xy

xy

2

21

2

2

0

0

ττττ

γγγγ

γγγγγγγγ

γγγγττττγγγγ

γγγγ

====

====

====

====

∫==== dVG

U xy

2

2ττττ γγγγxy

ττττxy

τxy

π/2 - τxy

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 36

Strain Energy in Torsion

� The total strain energy:

GJLT

dxGJT

dxdAGJT

dVGJT

dVG

U

L

L

J

xy

2

2

)(2

2)/(

2

2

0

2

0

22

2

2

2

====

====

====

====

====

∫ ∫

43421

ρρρρ

ρρρρ

ττττ

φγ

T

T

L

ρ

Page 19: Lecture 6 - National Tsing Hua University

19

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 37

Deflections by the Castigliano’s Theorem

� The most effective method to compute structural deflections (or angles of rotation) for complex structures (e.g. meandering springs) or structures under various loads

� By theorem, the deflection xi of a structure at the point of application of a load Pi is determined by:� Step 1: calculate the total strain energy U� Step 2: do the partial differentiation so xi = ∂∂∂∂U/∂∂∂∂Pi

� For a beam:

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 38

Example: Cantilever Beam under a Distributed Load

� Goal: find the deflection and slope at A

� Deflection at A: add a dummy force QA at A

=M

Α Β

w

x

L

QA (dummy)

∫ ∂∂∂∂∂∂∂∂====

∂∂∂∂∂∂∂∂====

L

AAA dx

QM

EIM

QU

y0

{

↓↓↓↓====−−−−−−−−==== ∫∂∂∂∂∂∂∂∂

EIwL

dxxwxEI

yL

QM

A

A8

)()21

(1 4

0/

2

A

QA

x

M

Moment along the x axis:

Deflection at A:

Let QA = 0 in M:

Page 20: Lecture 6 - National Tsing Hua University

20

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 39

Example: Cantilever Beam under a Distributed Load

� Find the slope at A: add a dummy moment MA

Α Β

w

x

L

2

21

wxMM A −−−−−−−−====

dxMM

EIM

dxEI

MMM

U L

A

L

AAA ∫∫ ∂∂∂∂

∂∂∂∂====∂∂∂∂

∂∂∂∂====∂∂∂∂∂∂∂∂====

00

2

2θθθθ

{ EIwL

dxwxEI

L

MM

A

A6

)1()21

(1 3

0/

2 ====−−−−−−−−==== ∫∂∂∂∂∂∂∂∂

θθθθ

Moment along the x axis:

Angle of rotation at A:

Let MA = 0 in M: A

x

M

MA

MA (dummy)

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 40

Bending of Plates

� The analysis is very useful for design of pressure sensors� The small-amplitude deflection w(x,y) of a membrane under a two-

dimensional pressure p is:

� Based on solved w(x,y), the bending moments can be calculated as:

(((( ))))

(((( )))) thicknesshEh

D

asdefinedregidityflexuraltheisD

yxpyw

yxw

xw

D

:112

:

,2

2

3

4

4

22

4

4

4

νννν−−−−====

====

∂∂∂∂∂∂∂∂++++

∂∂∂∂∂∂∂∂∂∂∂∂++++

∂∂∂∂∂∂∂∂

Ref: S.P. Timoshenko and S. Woinowsky-Krieger, Theory of Plates and Shells, Chap. 4, McGraw-Hill, New York, 1959

∂∂∂∂∂∂∂∂++++

∂∂∂∂∂∂∂∂−−−−====

++++−−−−====

∂∂∂∂∂∂∂∂++++

∂∂∂∂∂∂∂∂−−−−====

++++−−−−====

2

2

2

2

2

2

2

2

1

1

yw

xw

DDM

yw

xw

DDM

yxy

yxx

ννννρρρρρρρρ

νννν

ννννρρρρνννν

ρρρρ

Page 21: Lecture 6 - National Tsing Hua University

21

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 41

Cont’d

� Assume the stress profile in the z direction at any point is triangular (zero at neutral plane and changes linearly), the stress at each position (x,y) can be calculated

� A square membrane has been solved by S.P. Timoshenko� Maximum bending moments appear at the center of the sides

of the membrane, and decrease towards the corners and towards the center of the membrane

� Accordingly, the maximum surface stress in the middle of the sides is:

(((( )))) 2

2

max 31.0ha

px ⋅⋅⋅⋅====σσσσ

a

h

p

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 42

Flexure Spring Constants

Source: G.K. Fedder, Ph.D. dissertation, UC Berkeley, 1994

Page 22: Lecture 6 - National Tsing Hua University

22

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 43

Procedure

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 44

Clamped-Clamped Flexure

� Modeled as four guided-end beams� Apply a lateral force Fx to find the

displacement δδδδx, and thus the spring constant kx

� An external bending moment, Mo, constrains the angle in the analysis

� The beam bending moment is:

� The strain energy of the beam is:

ξξξξxo FMM −−−−====

12

23

0

2

twI

dEIM

U

z

L

z

=

ξ⋅∫= θθθθo = 0

Page 23: Lecture 6 - National Tsing Hua University

23

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 45

Cont’d

� Apply Castigliano’s theorem, and the constraint θθθθo = 0:

� Once more, apply the theorem:

(((( ))))

(((( ))))ξξξξ

ξξξξξξξξξξξξθθθθ

−−−−========

====−−−−====∂∂∂∂∂∂∂∂====

∂∂∂∂∂∂∂∂==== ∫∫

2/,2/

01

00

LFMandLFMso

dFMEI

dMM

EIM

MU

xxo

L

xoz

L

ozoo

3,

3,

3,

3

0

2

0

/484

/484

/12/

122

LEIkkalso

LEIkk

LEIFk

EILF

dL

EIF

dFM

EIM

FU

xbeamzz

zbeamxx

zxxbeamx

z

xL

z

x

x

L

zxx

====⋅⋅⋅⋅====

====⋅⋅⋅⋅====

========

====

−−−−====∂∂∂∂∂∂∂∂====

∂∂∂∂∂∂∂∂==== ∫∫

δδδδ

ξξξξξξξξξξξξδδδδ

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 46

Crab-leg Flexure: kx

� Apply Fx, Fy, and Mo at the end of the thigh

� Bending moments of the thigh (Ma) and shin (Mb) are:

� Apply the boundary conditions:

(((( )))) ξξξξξξξξξξξξ

xayoxbob

yoa

FLFMFMM

FMM

−−−−−−−−====−−−−====

−−−−====

)2(0

)1(0

0,

0,

0,

0,

====∂∂∂∂∂∂∂∂++++

∂∂∂∂∂∂∂∂====

∂∂∂∂∂∂∂∂====

====∂∂∂∂∂∂∂∂++++

∂∂∂∂∂∂∂∂====

∂∂∂∂∂∂∂∂====

∫∫

∫∫

ξξξξξξξξδδδδ

ξξξξξξξξθθθθ

dFM

EIM

dFM

EIM

FU

dMM

EIM

dMM

EIM

MU

ba

ba

L

y

b

bz

bL

y

a

az

a

yy

L

o

b

bz

bL

o

a

az

a

oo

θθθθo = 0δδδδy = 0

Page 24: Lecture 6 - National Tsing Hua University

24

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 47

Cont’d

� Mo is solved from (1), and Fy is solved from (2) and the solved Mo :

� Mo and Fy eventually are functions of Fx

� Finally,

(((( ))))(((( ))))(((( ))))aabb

yaabybaxbo LwwL

FLwwFLLFLM 3

232

/2

/2

++++++++++++

====

(((( ))))(((( ))))aabba

xby LwwLL

FLF 3

2

/43++++

−−−−====

(((( ))))(((( ))))(((( ))))(((( ))))

x

xx

aabbbz

xaabbbL

x

b

bz

bL

x

a

az

a

xx

Fk

LwwLEIFLwwLL

dFM

EIM

dFM

EIM

FU ba

δδδδ

ξξξξξξξξδδδδ

⋅⋅⋅⋅====

++++++++====

∂∂∂∂∂∂∂∂++++

∂∂∂∂∂∂∂∂====

∂∂∂∂∂∂∂∂==== ∫∫

4

/43/

3,

33

0,

0,

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 48

Crab-leg Flexure: kz

� Apply Fz, Mo, and To at the end of the thigh

� Strain energy from torsion must be included in the z spring constant calculation

Side view

Sid

e vi

ew

( )

ππ

−=

ν+=

=

= oddi,i twi

tanhiw

twtJ

EG

U

155

3

2

11921

3

12

Page 25: Lecture 6 - National Tsing Hua University

25

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 49

Cont’d

� From the free-body diagram, we get:

� By solving

azob

zob

oa

zoa

LFMMT

FTM

TT

FMM

−−−−========−−−−====

====−−−−====

1

ξξξξ

ξξξξ

ξξξξξξξξδδδδ

ξξξξξξξξφφφφ

ξξξξξξξξθθθθ

dFT

GJT

FM

EIM

dFT

GJT

FM

EIM

dTT

GJT

TM

EIM

dTT

GJT

TM

EIM

dMT

GJT

MM

EIM

dMT

GJT

MM

EIM

ba

ba

ba

L

z

b

b

b

z

b

bx

bL

z

a

a

a

z

a

ax

az

L

o

b

b

b

o

b

bx

bL

o

a

a

a

o

a

ax

ao

L

o

b

b

b

o

b

bx

bL

o

a

a

a

o

a

ax

ao

∫∫

∫∫

∫∫

∂∂∂∂∂∂∂∂++++

∂∂∂∂∂∂∂∂++++

∂∂∂∂∂∂∂∂++++

∂∂∂∂∂∂∂∂====

====

∂∂∂∂∂∂∂∂++++

∂∂∂∂∂∂∂∂++++

∂∂∂∂∂∂∂∂++++

∂∂∂∂∂∂∂∂====

====

∂∂∂∂∂∂∂∂++++

∂∂∂∂∂∂∂∂++++

∂∂∂∂∂∂∂∂++++

∂∂∂∂∂∂∂∂====

0,

0,

0,

0,

0,

0,

0

0

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 50

Cont’d

� The z-direction spring constant is:

(((( ))))(((( ))))

bgbagabxebaxea

bgaeabagbgaeabaebeabagbebea

bagaebeabagbgaebbaebeaagbeb

bgaaebbeaagbebea

z

zz

GJSandGJSEISEISwhere

LSSLLSSSLLSSLLSSS

LLSSSLLSSSLLSSLSS

LSLSLSLSSS

Fk

≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡

++++++++++++

++++++++++++++++++++++++

====

⋅⋅⋅⋅====

,,,

44

44

48

4

,,

5244232

2344252

δδδδ

Page 26: Lecture 6 - National Tsing Hua University

26

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 51

Serpentine Flexure

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 52

Cont’d

� The moment of each beam segment is deduced from the free-body diagram:

� The total strain energy in the serpentine spring is:

� Simultaneously solve the three equations:

Page 27: Lecture 6 - National Tsing Hua University

27

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 53

Cont’d

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 54

Stress

� Three main kinds of stress in materials� Externally applied stress� Thermal stress� Intrinsic stress

� Thermal stress arise from different temperature coefficients of expansion (TCE), a� e.g., cool a thin-film material on substrate from Td to Tr

� εεεεth = thermal strain = (αααα2 – αααα1)(Td – Tr)� Intrinsic stress arises from interstitial atoms, mechanical

annealing, microvoids, gas entrapment, grains� After deposition of thin film, thermal + intrinsic stress = residue

stress

Page 28: Lecture 6 - National Tsing Hua University

28

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 55

Stress in Thin Films

� Single crystal silicon has almost no residual stress� Polysilicon exhibits both compressive and tensile stresses,

depending on deposition conditions� Silicon nitride is highly tensile

� Low-tensile-stress silicon nitride is silicon-rich, excellent for membranes

� Silicon oxides are always relatively highly compressive� Most metal are tensile, but dependent on deposition conditions

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 56

Residual Stress and Stress Gradient

� In doubly-supported beams, residual stresses modify the bending stiffness, and possible buckling phenomenon if the stress is compressive

� Non-uniform residual stresses in cantilevers, due to either a gradient through the cantilever thickness, or to the deposition of a different material onto a structure, can cause the cantilever to curl

Page 29: Lecture 6 - National Tsing Hua University

29

������������������������������������ENE 5400 �� ���� ���� ���� ��, Spring 2004 57

Stress Gradient and Radius of Curvature

� Assume that the axial stress in the beam is:

� Therefore the induced bending moment is:

� We know that:

� The stress is released in the form of curl1

3

21

121

σσσσρρρρ

ρρρρEh

MEwh

EIM

========∴∴∴∴====

(((( )))) yho 2/

1σσσσσσσσσσσσ −−−−====

(((( )))) σσσσσσσσ 22/

2/ 61

whdyyyMh

h====⋅⋅⋅⋅−−−−==== ∫−−−−

hw σo – σ1

σo + σ1

σo x


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