Lecture 6 - National Tsing Hua University
Transcript of 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)
2
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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====
∂∂∂∂∂∂∂∂−−−−====
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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 ====
∂∂∂∂∂∂∂∂====
3
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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
4
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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−−−−
−−−−====−−−−
−−−−====
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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
====
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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
6
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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
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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
++++====
====
−−−−++++++++====
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������������������������������������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
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������������������������������������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
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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 )(
)(
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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
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
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
=
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:
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)
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
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:
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
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)(
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
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:
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
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
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
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
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
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
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:
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
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
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