課程名稱：微機電系統
 Microelectromechanical Systems
授課教師：王東安
Lecture 5
1
Lecture Outline
• Reading Senturia: Chapter 9
• Today’
s lecture
–Axially loaded beams
–Bending of beams
–Bending of plates
–Residual stress
–Stress gradient
2
Prologue
• Compliant mechanical structures can be modeled as
simple linear springs with a given spring constant. Now
we will go into detail of how to calculate the spring
constant of different mechanical structures
3
Axially loaded beams
• First let’
s consider a solid beam with an axial load
which creates tensile stress in the beam.
• tensile stress leads to a tensile strain, which can be
expressed as
1
1 F
x  x 
E
E Wh
• beam is elongated by an amount equal to the product
of the strain and the length of the beam
• Solving for the force, we find the spring constant
1 FL
L x L 
E Wh
EWh
EWh
F
L  k 
L
L
4
Axially loaded beams
• For most MEMS materials reasonable geometries
(cross-section and length), this becomes a very stiff
spring. The more serious problem is that it doesn’
t allow
a very long deflection unless we make L very long. We
are interested in finding springs with more flexibility in
the spring constant, and a more compact geometry for a
given maximum deflection.
5
Axially loaded beams
• Variable cross sections: An axially loaded beam with a variable cross
section gets an elongation that can be expressed as
1 F
F L dx
L 
dx  
k 
0 E A
0
x
E A
x
L
E
L dx

0 A
x
6
Bending Beams
• We will develop equations to describe bending of transversally loaded
beams
• types of supports and loads we will consider
7
Bending Beams
• four types of supports provide different kinds of reaction forces to the beam
• Free ends do not provide any type of support, neither reaction forces nor
moments. The beam is not restricted in any way (position or angle) at a free
end
• Sliding supports can provide a vertical reaction force, but no horizontal
forces or moments. This means that vertical position of the beam end is
fixed, but its horizontal position and its angle at the support can both take
arbitrary values.
• Pinned supports provide reaction forces, but not moments. The position of
the beam end is therefore fixed, but the angle of the beam is not
• Fixed supports provide both reaction forces and moments, and the end is
fixed in position and angle
• Fixed and free supports are common in compliant MEMS structures, while
pinned and sliding (pinned-on-rollers) require sliding surfaces, which
introduce reliability and accuracy problems in MEMS
8
Forces and Moments in Equilibrium
• Consider the fixed beam
equilibrium requires that
FR F
M R FL
• At an arbitrary position, x,
V 
x F
M 
x M R x 
V 
x FL Fx F 
L x 
• sign conventions for load, shear
force and moments
9
Forces and Moments
• condition for static equilibrium for an element of a
transversally loaded beam. Consider the forces
acting on a beam element
• In equilibrium, the total force and moment on the
element must be zero
0 q 
dx 
V dV 
V 
dV
q
dx
dx
0 
M dM 
M 
V dV 

dx q 
dx  
2
dM
V
dx
10
Bending of beams
• We will now derive expressions for the deflection of beams under
transversal loading conditions. We define a beam as a mechanical
structure in which the loading forces and the reaction forces from the
supports can create axial stress, but not normal stresses perpendicularly
to the beam axis.
• Consider the differential section of a beam
• The beam segment has become curved as a result of transversal forces
acting on the beam. Our goal is to find the differential equation that
governs the deflection of a beam under transversal loading. The radius
of curvature of the beam equals the second derivative of the deflection,
as we will prove later. We therefore start by considering the radius of
curvature, ρ
11
Beam Bending
• length of the neutral axis is
• length of the section at an
arbitrary position is a function of z
• strain is the difference between
the length of the neutral axis and
the length at position z.
• axial strain is then given by
Hook’
s law
dx d
dL 
z 
d
 z
z
dL dx  dx 
1 
dx

 
dL dx
x 

dx
z
dx  dx dx
z

x 

dx

z
x Ex  E

12
Beam bending
• axial stress in the beam under pure bending is
• calculate the Internal Bending Moment of the differential beam segment by
integrating the first moment of the stress over the cross section of the beam
• where W is the width of the beam (W is in general a function of z, W(z)). Consider this
expression for a moment. This is the bending moment that the differential beam segment
is applying to its supports. The bending moment, M0, that the supports are applying to the
beam is equal in magnitude, but of opposite sign
H
2
H

2
M W 
zx dz
13
Beam bending
• There is clearly some danger that the internal bending
moment will be confused with the applied bending
moment. In practice we avoid this confusion by using
clear pictures that unambiguously show the loading
conditions and therefore the distribution of moments.
However, it is important to note that we define the
Internal Bending Moment as the moment the beam
applies to the supports
14
Moment of Inertia
• The equation we just derived for the bending moment leads us to the very
important concept of the moment of inertia of a bending beam. Inserting the
expression we found for the axial stress, we find
 Ez 
E 2
2
M W 
zx dz W 
z  
dz  
W

z
dz 
H

 2
 
EI
M 

H
2
H

2
H
2
H

2
H
where
H
2
H

2
I Wz 2 dz
is the Moment of Inertia of the bending beam. This can be written more
generally as
2
I
W 
z 
z dz

cross sec tion
15
Moment of Inertia
• Moment of Inertia does not only depend on the beam cross-section, but
also our choice of direction and origin of the integrating variable (z in
our calculation). Under most circumstances the Moment of Inertia is
calculated with respect to the center of the beam cross section, i.e. we
chose the origin of the integrating variable, z, such that
H
2
H

2
W zdz 0
• Moment of Inertia of a rectangular beam with respect to the center of the
beam in a direction perpendicular to one of the sides of the rectangle
becomes particularly simple. It is given by
H
2
H

2
H
2
H

2
3
WH
I Wz 2 dz W z 2 dz 
12
• This is a very important formula that it is well worth committing to
memory! On the other hand it is equally important to note the
restrictions (rectangular cross section, bending parallel to one of the
sides, neutral axis in the center) on the validity of this expression.
16
Differential equation for transversally loaded
beams
• consider the transversally loaded beam
• length of a differential beam segment
dx
ds 
d
cos
• where θ is the angle of the beam with
respect to the x-axis, and ρ is the
radius of curvature of the bending beam.
This leads to the following differential
equations relating angle to curvature
d 1

dx 
• From the figure we see that
dw
tan 
dx
17
Differential equation
• The two equations above can be combined to give
d 2 w d 1
 
2
dx
dx 
• Using the expression we found for the radius of
curvature of the bending beam, we arrive at the basic
form of the differential equation for a transversally
loaded beam
d 2w
M

2
dx
EI
• Combined with the equation for the derivative of the
moment dM V and the shear force dV q
dx
dx
• this equation can be written as
d 3w
V


dx 3
EI
d 4w q

4
dx
EI
18
Beam with a point load
• For a beam of length L loaded with a force F
at its end, the moment is given by
M 
x F 
L x 
• Substituting this into the differential equation
for the deflection of the beam, we find
d 2w F 
L x 

dx 2
EI
• Integrate twice and apply the boundary
condition w
0 0 and dw 0
dx x 0
• leads to the solution as
F 3 FL 2 FL 2  x 
w 
x 
x 
x 
1 
6EI
2 EI
2 EI  3L 
• maximum value of the deflection occurs as
expected at the end of the beam, and it has
the value
wmax
FL 2  L  FL3

L
1  
2 EI  3L  3EI
19
Spring constant of bending beams
• From the expression for the maximum deflection of the
bending beam we find the spring constant
FL 2  L  FL3
wmax 
L
1  

2 EI  3L  3EI
3EI
F  3 wmax 
L
3EI
ktransversal  3
L
• Assuming a rectangular cross section, this simplifies to
ktransversal
3E WH 3 EWH 3
 3

L 12
4 L3
• compare this to the expression we found for the spring
constant of an axially loaded beam
EWH
k axial 
L
20
Spring constant of bending beams
• ratio of the two is
k axial
ktransversal
4 L2
 2
H
• length of the beam is by definition much larger than the
thickness so
kaxial
1
ktransversal
• Bending beams are therefore much more useful than
axially strained beams as springs in MEMS devices.
21
A practical case
• To get a feeling for practical numerical values, consider a polysilicon
beam with a rectangular 2 by 2 µm cross section and a length of 100
µm. Assuming that the Young’
s modulus for polysilicon is 160 GPa,
we find a spring constant of 0.64 N/m, which means that a 1 µm
deflection requires a force of 0.64 N. Forces on this order of
magnitude are easily obtainable with electrostatic MEMS actuators.
• Gravitational forces, on the other hand, tend to be insignificant in
MEMS of this size scale. To get a gravitational force of 0.64 µN in a
1g gravitational field, we need a silicon proof-mass volume of
0.64 
106 N
11 3
7
3

2
.
8

10
m

2
.
8

10

m
9.8 N kg 
2330 kg m 3
• This corresponds to a silicon cube with a side of roughly 300 µm,
which is huge compared to the polysilicon beam itself. Only if we
work very hard can we make MEMS devices sensitive to
accelerations on the order of 1G!
22
Stress in bending beams
• An important parameter in any compliant structure is its
maximum stress under a given load. The bending beam
we are studying has a radius of curvature given by
1 d 2w F
 2  
L x 
 dx
EI
• This radius has a minimum value at the support (x=0),
which means that the strain and the stress are
maximized at the support
H 1 H FL LH
6L
max     
F
F
2  2 EI 2 EI
EWH 2
LH
6L
max Emax  F  2 F
2I
WH
23
Stress in bending beams
• Our 100 µm polysilicon beam with a 2 by 2 µm cross
section loaded by a force of 0.64 µN at its end to create
a 1 µm maximum deflection has a maximum stress of 48
MPa. This is substantially less than the yield stress of
polysilicon, which is on the order of 5GPa.
• This calculation illustrates an important point of MEMS
design. Most compliant MEMS devices are operated well
below their yield stress. This is one of the reasons for the
impressive reliability, longevity and accuracy of MEMS.
24
Anticlastic curve
• We have defined a beam as a structure in which loading
forces and reaction forces from the supports can create
only axial stress. This axial stress will lead to axial strain,
but also to strain in directions perpendicular to the axis
through the Poisson effect.
• axial strain is
z
x 

• This leads to strain in the two perpendicular directions
z
y 
x 

z
z 
x 

25
Anticlastic curve
• The strain in the z-direction is of little consequence, but the strain in the ydirection leads to an interesting deformation phenomenon called anticlastic
curvature.
• The bending moment in the +y-direction sets up axial strain in the x-direction
perpendicular to the plane, and creates a Poisson strain in the y-direction.
This Poisson strain is compressive where the bending strain is tensile and
tensile where the bending stress is compressive, leading to the anticlastic
curvature
• The anticlastic curvature will, in principle, change the value of the moment of
inertia of the bending beam, and will therefore influence the calculated
deflection. This effect is, however, quite small, and can be neglected in most
MEMS devices.
26
Plates
• A beam is a structure in which the supports cannot
create transversal stress, or, in other words, the supports
do not oppose Poisson contraction perpendicularly to the
axial strain.
• A plate has supports that preclude Poisson contraction.
That is not to say that there is no strain in directions
perpendicular to some well-defined primary stress axis.
What it means is that we must consider both bending
strain and Poisson strain when deriving expression for
the bending of plates
• Preclude: To make impossible, as by action taken in
advance
27
plates
• Let’
s assume that in some segment of a plate we have
bending-induced uniaxial stress,σx . The plate, by
definition, is constrained such that εy =0, which means
that there must be an in-plane stress, σy =νσx, to
offset the Poisson contraction. The strain/stress
relationship can then be expressed
 y
x  x

E
1 2
x 
x 
E
E
x 

2 x
1 
• The ratio E/(1- ν2) is called the plate modulus
28
Bending of Plates
• For small deflections we can write in the principal
coordinate system
2
1 w
 2
x x
1 2 w
 2
y y
2 w
0

x
y
• As in the case of the bending beam, the strains vary
linearly with z throughout the cross section
z
x 
x
z
y 
y
29
Bending of Plates
• in-plane stress and strains are related as
1

x  x  y
E
E

1
y  x  y
E
E
• substitute the expressions for the bending strains to get
zE 
1


x 
 
2 
1  x y 

zE 
1



y 


1 2 


x 
y
• integrate these expressions to find the moments
 zE 1




  
Mx  
W
zx dz  
W
z 
dz
 1 2 x y 

H 2
H 2




H 2
H 2
H 2 2
1
H 3
WE 
1

WE

   z dz 
   

2 
2 


1  x y 
1  x y 
H 2
12
1
Mx
1 EH 3 
1







D
 
2 



W
12 1  x y 
x y 
30
Bending of Plates
• Similarly
where
1
My
1 EH 3 
1







D
 
2 



W
12 1  y x 
y x 
1 EH 3
D
12 1 2
is called the flexural rigidity of the plate. We can think
of the flexural rigidity of a plate as similar to the moment
of inertia of a beam
• width, W, of the plate in the above equations can be
chosen arbitrarily, and we therefore prefer to give the
moments normalized to the width.
31
Bending of Plates
• In the special case where the curvature is the same in the two
principal directions (i.e. it is the same in all directions), the equations
simplify to
M
D
 
1 

W

1
1
M
 

x y WD 
1 
• This differential equation has the solution
2 w 2 w
M
 2 
2

x

y
WD
1 
M
x 2 y 2
w
WD 
1  2
• This is a paraboloid, which for small deflections can be
approximated as a sphere.
• Paraboloid: x2/a2+y2/b2=z
32
Beams with Stress Gradients
• Stress gradients in thin films can
lead to curvature of freestanding
structures.
• Before the beam is released, there is
both an average stress and a stress
gradient in the thin film. The average
stress must go to zero when the
structure is released (i.e. when the
sacrificial layer is removed), but the
stress gradient remains if the films is
constrained such that it is still flat.
• Without such constraints, the film will
bend until the stress is relieved.
33
plates
• Most thin-film MEMS technologies suffer from the curvature problem.
This is bothersome in many applications, but it is particularly
problematic for optical MEMS, in which flatness to within a small
fraction of a wavelength is often required for correct operation.
• We will now carry out a calculation of the radius of curvature as a
function of stress gradient. To get the correct results we must
consider the Poisson effect.
• Plates with uniform radius of curvature, we found the following
expression relating curvature and bending moment
WD
1  1 WE 
1 
H3
1 WEH 3



2
M
12 
1  
M
12 
1 
M
~
1 EWH 3

12 M
~ is the biaxial modulus
• where E
34
Plate
• write the stress as (assume that the stress gradient is uniform in the
plane)
1

z
H 2
• moment becomes
H 2
H 2
2W
WH 2
2
M 
Wzdz  1 
z dz 
1
H
6
H 2
H 2
• Substituting this in the equation for the radius of curvature gives
~
~
3
1 EWH
EH
x 
6

12 WH 21
21
• In this calculation we have correctly accounted for the Poisson effect
under the assumption that the stress gradient is uniform in the plane,
and the beam therefore has the same curvature in all directions.
This is a particularly simple case in which the treatment of a plate
exactly follows that of a beam, the only difference being that we
must use the biaxial modulus instead of the Young’
s modulus in our
expressions.
35
Composite Plates
• Composite beams and plates, e.g.
polysilicon films with films of another
materials on top, are common in
MEMS. Different stress levels in
different parts of the composite lead
to curvature of the structures
• Varying stress levels in composite
structures cause curvature similar to
stress gradient in homogeneous films.
Before release, the thin film on top of
the plate is in tension due to thermal
expansion mismatch, or some other
process related effect. After release,
the average stress is zero, but the
stress in the film leads to a stress
gradient in the plate
36
Composite Plates
• it is assumed that the film is in tensile
stress before release. After release some
of the stress in the film will be relieved,
and this relaxation will cause a stress in
the plate such that the total stress
integrated over the cross section is zero.
Now assume for a moment that the plate
is held flat after release (i.e. the plate is
kept flat by normal forces, but there is no
axial forces applied to the plate). The
strain and stress are then both uniform
across the thickness of the plate, and their
magnitude can be found from the fact that
the total stress across the beam cross
~
section must be zero.
E H
ax
• where εax is the axial strain of the beam
after release, As long as the beam is
straight, this strain is uniform
1


~
0 ax E0 h 
0 h
ax  ~
~
E1H E0h
37
Composite Plates
• The stress in the film is
h
~
~
film 0 ax E0 0 ~ 0 ~ E0
E1H E0h
~
~
~
~
E1H E0h E0h
E1H
film 0
~
~
~
~ 0
E1H E0h
E1H E0h
• and in the plate, the stress is
given by
h
~
~
plate ax E1 ~ 0 ~ E1
E1H E0h
~
E1h
plate ~
~ 0
E1H E0h
We can use these expressions to calculate the bending moment in the plate. To simplify
our work we will assume that the plate is much thicker than the film, so that we can
neglect the reduction in the film stress upon release. In other word, we assume that
the stress in the film is too low to significantly change the length of the plate.
38
Composite Plates
• The axial strain after release leads to purely axial stress
in both parts of the composite beam. The bending
moment can be expressed as
H 2
H 2 h
H 2
H 2
M 
W
zplatedz  
W
z0dz
M W0
Hh
2
• Here we have used the approximation that the stress in
the film is unchanged and the stress in the plate is
negligible. Obviously, this is only true if the film is much
thinner than the plate.
39
Composite Plates
• A free plate without applied forces cannot support a finite
bending moment. Once the forces that keep the plate flat
are removed, the plate must curve to relive the bending
moment. The radius of curvature of the beam will create
bending moment of equal but opposite magnitude of the
H 2 h
built-in moment.
M curve  
W
curve zdz
H 2
H 2 h
M curve
H 2 h
 z
~

W
E
z

zdz




 
H 2
~
 
WE 
z 
curve zdz 
H 2
H 2 h
H 2
~
~
M curve
~ z2
~ z2
 
WE1  dz  WE0  dz


H 2
H 2
M curve
WH 2 ~
~

E1H 3E0h
12 


where E1and E0 is the biaxial modulus of the plate and the film, respectively
40
Composite Plates
• We now add Mcurve to the calculated bending moment caused by the
thin film stress to find the radius of curvature.
M curve M 0 


1
WH 2 ~
~
0 HhW 
E1H 3E0h 
2
12 
H ~
~

E1H 3E0h
60h


• From this radius of curvature we can find expressions for the strain
and stress
• the radius becomes infinite if the stress is zero, the film has zero
thickness, the plate is infinitely thick, or the plate is infinitely stiff.
Giving the film zero stiffness does not lead to an infinite radius, but
this is a fallacy. If the film does not have a finite biaxial modulus, it
would not be able to support a finite stress. It is also interesting to
compare the radius to that of a plate bent by a stress gradient (p35)
H ~
 E
21
41
Composite Plates
• We see that the ratio of the radii of curvature resulting from
these two different stress conditions are roughly given by
H/3h.
M curve M 0 


1
WH 2 ~
~
0 HhW 
E1H 3E0h 
2
12 
H ~
~

E1H 3E0h
60h


H ~

E
21
42
Example of Composite Plate with Finite Radius
of Curvature
• Consider a 2 by 100 by 100 µm polysilicon plate, fixed at
one end, and covered on one side with a 10 nm film with
a tensile stress of 200 MPa. Assume that the Young’
s
modulus is 160 GPa for polysilicon and 250 GPa for the
thin film, and that both materials have Poisson ratios of
0.3.
• Using the formulas we have derived, we find that the
initial stress relaxation is small (2 MPa), and the induced
stress in the plate is only 1 MPa. Our simplifying
assumptions that there is negligible stress in the plate
and insignificant change in the stress of the film after
release therefore seem to be well justified!
• The radius of curvature, calculated using our expression,
is 7.9 cm. This is a reasonable number for most MEMS
structures, even for optical devices, which are among the
most sensitive to surface profiles.
43
Impact of radius of curvature on optical beams
• To see what impact radius of
curvature has on optical
beams, consider the situation
• The curvature of the plate
refocuses the beam to a
smaller beam radius, which
creates a larger diffraction
angle. The ratio of this larger
diffraction angle to the initial
“
diffraction limited”angle can
be used as a measure to
quantify the quality of the
optical of the optical surface
44
Thermal Actuators
• how the stress induced bending we just described can be used to
create a thermally actuated MEMS valve
• The plate in the figure is a composite structure consisting of two
films with different thermal expansion. The top film has a lower
thermal expansion than the lower film, so that the top film is in
compressive stress at the operating temperature. This
compressive stress will bend the plate toward the substrate. If the
displacement of the composite structure, in the absence of
opposing forces, would be larger than the separation of the plate
from the substrate, the plate will be clamped to the substrate with
a finite clamping force. This can be used to seal off a hole in the
substrate
• If the plate can be lifted from the substrate the composite
structure can function as a valve. Lifting the plate can be
accomplished by heating the lower part of the plate more than the
upper part, thereby reliving the stress that creates the curvature
of the plate
45
Clamped Beams
• So far we have considered structures that are free to curve under an
applied bending moment. In fact, the structures we have treated cannot
support a bending moment without curvature. If we apply an axial force
to the structure, this situation changes dramatically. Axial stress
becomes an important factor in doubly supported, or clamped
• beam can be in tensile or compressive stress. If the beam has tensile
axial stress, the stress tend to keep the beam straight, and we expect
that the stress makes the beam stiffer. Compressive stress, on the other
hand, tends to increase deflection, i.e. it makes the beams less stiff. In
fact, the spring constant for transversal motion of a compressively
stressed beam can go to zero if the load is sufficiently high. This
phenomenon is called beam buckling
46
Clamped Beams
• To understand the effect of axial stress on the bending of beams
consider the beam segment
• The axial stress acting on the two ends of the beam segment
balance each other only if the beam is straight. If the beam is curved,
we must postulate another load to provide balance in the vertical
direction. This is a differential length of the beam, so we use small
angle approximations throughout.
d
WP0 dx q0 dx 20WH sin 0WHd
2 
where W and H are the width and thickness,
respectively. P0 and q0 represent the load per unit q WH d
0
0
dx
area and load per unit length, respectively.
This load is balanced by the axial force, so
d 2w
47
q 0WH 2
it is not contributing to the change in shear force 0
dx
Clamped Beams
• Recall the differential equations we found for the transversely loaded
bending beam
d 2w
M

dx 2
EI
dM
d 3w
V
V 


dx
dx 3
EI
dV
d 4w q
q  4 
dx
dx
EI
• In this last equation we must subtract the load that is balanced by
the axial force, because the load cannot both be balancing the axial
force and the change in shear force. (Note that we have defined q0
opposite of how we earlier defined q, so the subtraction becomes
addition.)
d 4w
EI 4 q q0 
dx
d 4w
d 2w
EI 4 0WH 2 q
dx
dx
This is the famous Euler beam equation
48
Euler beam equation
• The Euler beam equation is a
linear differential equation, so
finding its solutions is
straightforward. We will not go
through the details here, but
instead take the solutions from
the text (see Senturia, chapter
9.6.2, 9.6.3, and 9.7 for the
derivations), and discuss their
implications.
• For the doubly supported
beam with a uniform load (Fig.
8.17), q, we have the boundary
conditions
w
0 w
L 0
dw
dw

0
dx x 0 dx x L
49
Euler beam
2
2
2

q
x
L

2
Lx

x
• Without tensile stress, the solution is w 
2 EWH 3
L4 q
wmax 

3
32 EWH
qL
32 EWH 3
k

wmax
L3
• With a tensile force, N=WH σ0, we find
qL L
cosh
k0 L 2 
1 

wmax  

2



4 N 2
k0 sinh
k0 L 2 
4N
k
L
cosh
k0 L 2 
1
2
2
k0 sinh
k0 L 2 
where
12 N
k0 
EWH 3
This spring constant is significantly higher than the one for the beam without
axial load, even when the axial stress is significantly lower than the Young’
s
modulus of the beam
50
Buckling
• When the axial load is compressive rather than tensile, the equations
become slightly more complicated. The solutions are, however, very
interesting.
• For the case of a doubly supported beam with a point load in the
center, Senturia finds the following spring constant
k
1


n 1, odd
L
2
2 EIk n4 Nk n2
where
2n
kn 
L
51
Buckling
• When the denominator of the first term of the sum is zero, the spring
constant is zero. This happens when the load exceeds the critical
value
2
3
 EWH
N 

2
3 L
2 EH 2
Euler 
3 L2
• When the axial load exceeds this critical value, called the Euler
buckling limit, the beam cannot support any transversal loads, but
will instead spontaneously deflect sideways. This phenomenon is
called buckling.
• From a MEMS device point of view, buckling is very interesting
because it can be utilized to create bi-stable devices. For example,
we can design a bi-stable valve which can be driven between two
stable positions, one in which the valve is closed and one in which it
is open
52