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Mechanics of Materials • stress and strain fl lb b di l i d i l • flexuralbeam bending analysis under simple loading conditions EECE 300 ‐ 2011 1 Learning Objectives • By the end of this section the student should be able to: • By the end of this section, the student should be able to: – describe the stress‐strain relationship for elastic materials – calculate beam deflection for simple loading conditions – describe comb drive actuators and how they are used in silicon‐ based microsystems Further Reading • Chang Liu, Foundations of MEMS, Chapter 3: Review of Essential Electrical and Mechanical Concepts EECE 300 ‐ 2011 2
Transcript of thethe endend ofof thisthis sectionsection thethe studentstudent...
Mechanics of Materials
• stress and strain
fl l b b di l i d i l• flexural beam bending analysis under simple loading conditions
EECE 300 ‐ 2011 1
Learning Objectives• By the end of this section the student should be able to:• By the end of this section, the student should be able to:
– describe the stress‐strain relationship for elastic materials
– calculate beam deflection for simple loading conditionsp g
– describe comb drive actuators and how they are used in silicon‐based microsystems
Further Reading
• Chang Liu, Foundations of MEMS, Chapter 3: Review of Essential Electrical and Mechanical Concepts
EECE 300 ‐ 2011 2
Mechanical design issues for MEMS devices
Example: MEMS accelerometer‐uses a loaded cantilever beam (or other design) to sense inertial motion (acceleration)sense inertial motion (acceleration)‐electrical circuitry to transduce the proof mass vibration into an electrical signal for amplification
A MEMS accelerometer is a key component of many automobile airbag sensors. The device consists of a center region, the proof mass, which is connected by springs to the surrounding structure. The interdigitated fingers are capacitors used by on‐chip electronics (not shown) to sense acceleration(not shown) to sense acceleration.
http://www.analog.com
EECE 300 ‐ 2011 3
Movable electrodes in the form of interdigitated fingers are attached to the proof mass. The fi d d i l t d f b k f ll l t d it If l ti i
EECE 300 ‐ 2011
fixed and moving electrodes form a bank of parallel‐connected capacitors. If an acceleration is applied to the chip, the proof mass will move under an inertial force against the chip frame. This changes the finger distances and therefore the total capacitance, which is read using on‐chip signal processing electronics. 4
Mechanical design issues for MEMS devices
Example: nanoscale cantilever mass sensor‐small amount of mass binds to cantilever (single bacterium single virus particle)bacterium, single virus particle)‐change in mass changes the resonance frequency
2 μm
A small gold dot rests on a silicon cantilever. The dot is a test mass for studying how the cantilever can be used to measure the masses of tiny particles including viruses withof tiny particles, including viruses, with attogram precision.
EECE 300 ‐ 2011
ILIC, et al. (2004) Attogram detection using nanoelectromechanical oscillators. Journal of Applied Physics,95, 3694‐3703. 5
Stress and strain
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Normal stress and strain
Mechanical stresses are either normal stress or shear stress.If the stress acts in a direction perpendicular to the cross section, it is called a normal stress. The normal stress σ is defined as the force F applied to a given area A. The SI unit of
i N/ 2 Pstress is N/m2, or Pa.
A normal stress can be tensile or compressive. The unit elongation is the strain. For a rod with uniform cross‐sectional area subjected to axial loading the normal strain is given byaxial loading, the normal strain is given by
F
EECE 300 ‐ 2011
F
L + ΔL
7
Poisson’s ratio
In reality, the applied longitudinal stress along the x‐axis not only produces a longitudinal elongation in the direction of the stress, but a reduction of the cross‐sectional area as well.The material must try to maintain constant atomic spacing and bulk volume.P i ’ i i d fi d h i b d l i di l l iPoisson’s ratio is defined as the ratio between transverse and longitudinal elongations:
y z
x x
ε ενε ε
= = 0
0
lateral strain
axial strain
Ld
d Lν Δ
= − = −Δ
x x
F Fcross
L
crosssection
EECE 300 ‐ 2011
L + ΔL
8
Elastic behavior – Hooke’s Law and springs
When a material experiences a stress, the usual result is a deformation resulting in a strain on the material.A common example is a spring which is either stretched or compressed.
F
This is an example of Hooke’s Law, a simple elastic response to an external force F. When F = 0, the Δx goes back to zero as long as the force remained in the elastic limit. There is no permanent deformation in the elastic limit.Atoms are held together with atomic forces If we imagine interatomic force acting asAtoms are held together with atomic forces. If we imagine interatomic force acting as springs to provide restoring force when atoms are pulled apart or pushed together, the modulus of elasticity is the measure of the stiffness of the interatomic spring near the equilibrium point.
FF
slope = k
EECE 300 ‐ 2011
Δx
slope k
9
Young’s modulus
Under small deformation, the stress and the strain terms are proportional to each other according to Hooke’s law:
The proportion constant E is called the modulus of elasticity or Young’s modulus. It is an intrinsic property of a material It is a constant for a given material irrespective of theintrinsic property of a material. It is a constant for a given material, irrespective of the shape and dimensions of the mechanical element.The Young’s modulus is a measure of the stiffness of a material. It is measured in stress units (force/area). E [=] Pa
i‐Clicker:i Clicker: a. k ~ Lb. k ~ 1/L
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Longitudinal stress and strain
A cylindrical silicon rod is pulled on both ends with a force of 10 mN. The rod is 1mm long and 100 μm in diameter. Find the stress and strain in the longitudinal direction of the rod, given that E = 130 x 109 N/m2.
310452a ×= −εi‐Clicker:
3
6
1018.3 c.
1045.2 b.
1045.2a.
×=
×=
×=
−
−
ε
ε
ε
51018.3 d. ×=ε
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Elastic behavior of materials
elastic regime
stress σ(F/A)
silicon 160 GPa
gold 78 GPa(F/A)
slope Eσε
Δ= =
Δ
bones 18 GPa
silicone rubber ~ 500 kPaε silicone rubber 500 kPa
brain tissue ~ 3 kPa
strain ε(δL/L)
EECE 300 ‐ 2011
( / )
12
Elastic limit and plastic behavior of materials
At some point when applying a stress, the material will reach a maximum strength at which point a deformation is produced. This is the yield strength. This is the elastic limit of the material.
Beyond this point, the material is plastically deformed. If the load is
d th t i lremoved, the material retains this deformation at zero load.
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Shear stress and strain
The magnitude of the shear stress is defined as
F=τ
The unit of τ is N/m2. Shear stress has no tendency to elongate or shorten the element in the x y and z directions Instead the shear stresses produce a change in the shape of the
Aτ
the x, y, and z directions. Instead, the shear stresses produce a change in the shape of the element.Shear strain, defined as the extent of rotational displacement, is
XΔX
Lγ Δ
=
The shear strain is unitless It is the angular displacement expressed in radians
tanθ θ≈ for small θ
The shear strain is unitless. It is the angular displacement expressed in radians.
Fθ
EECE 300 ‐ 2011
F
14
Shear modulus of elasticity
The shear stress and strain are related to each other by a proportional constant, called the shear modulus of elasticity G.
Gτ
Gτγ
=
The unit of G is N/m2. The value of G depends on the material, not the shape and dimensions of an object.
For a given material, E, G, and Poisson’s ratio are related by:
( )2 1
EG
ν=
+
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Flexural beam bending analysis
Flexural beams are commonly encountered in MEMS as spring support elements.It is important to calculate the bending of a beam under simple loading conditions, analyze induced internal stress, and determine the resonant frequency associated with the lelement.
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Possible boundary conditions
A flexural beam can be classified according to the combination of the two mechanical boundary conditions associated with it.
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Beam segment in pure bending
The loads acting on a beam cause the beam to bend (or flex), deforming its axis
When a beam is loaded by force, stresses and strains are created throughout the interior of the beam.
beam to bend (or flex), deforming its axis into a curve.Consider a portion of a beam (A–B) in pure bending. The cross section of the b i i b h ibeam is symmetric about the y axis.It is assumed that the cross sections of the beam, such as sections mn and pq, remain plane and normal to the plongitudinal axis.The lower part of the beam is in tensionand the upper part is in compression.
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Beam segment in pure bending
Somewhere between the top and bottom of the beam is a surface in which
When a beam is loaded by force, stresses and strains are created throughout the interior of the beam.
of the beam is a surface in which longitudinal lines do not change in length. This surface, indicated by st, is called the neutral surface of the beam.Th i i b h lThe intersection between the neutral surface with any cross‐sectional plane such as line tu is the neutral axis of the cross section.
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Deflection of beam under pure bending
The general method for calculating the curvature of the beam under small displacement is to solve a second order differential equation of a beam:
2∂2
( )y
EI M xx
∂=
∂where M(x) represents the bending moment at the cross section at location xwhere M(x) represents the bending moment at the cross section at location x.y represents the displacement at location x.The x‐axis runs along the longitudinal direction of the cantilever.
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Cantilever beam under small deflections
Beam segment under pure bending:
top is in tension
y
b i i i
neutral axis ε = 0
bottom is in compression
εx
εma
y
εmax
y
‐εmax
t/2‐t/2
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Bending
To calculate the magnitude of stresses at any location in the beam:At any section, the distributed stress contributes to distributed force, which subsequently gives rise to a reaction moment with respect to the neutral axis.b i f f dibasic concept of moment: force × distance
( )2( ) ( )t
M dF h h h dA h∫∫ ∫ ∫ ( )2
2
( ) ( )tA w hM dF h h h dA hσ
=−= =∫∫ ∫ ∫
Assuming that the magnitude of stress is linearly related to h and is the highest at the surface (denoted by σ ) this equation can be rewritten as:
2max max2 2max / 2 / 2 / 2
t t
t tw wh h
hM dA h h dA I
t t t
σ σσ⎛ ⎞= = =⎜ ⎟⎝ ⎠∫ ∫ ∫ ∫
surface (denoted by σmax), this equation can be rewritten as:
2 2/ 2 / 2 / 2w wh ht t t=− =−⎜ ⎟⎝ ⎠∫ ∫ ∫ ∫
The term I is called the moment of inertia associated with a particular cross section.
EECE 300 ‐ 2011 22
Bending strain and beam curvature
Radius of curvature → geometric connec on to strain
dθ
RR
( )/ 2 / 2R t d Rd tθ θε
+ − 21 d y=
also, for small deflections
( )max Rd R
εθ
= = 2R dx=
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Curvature and strain
Combining the curvature and moment results:
/ 2t Emax
/ 2t
Rε =
2max max2 2
t thM dA h h dA I
σ σ⎛ ⎞⎜ ⎟∫ ∫ ∫ ∫
Eσ ε=
21 d y
2max max2 2max
2 2/ 2 / 2 / 2t tw wh hM dA h h dA I
t t tσ
=− =−
⎛ ⎞= = =⎜ ⎟⎝ ⎠∫ ∫ ∫ ∫
2
1 d y
R dx=
2
2( )
yEI M x
x
∂=
∂
where M(x) represents the bending moment at the cross section at location x.y represents the displacement at location x.The x‐axis runs along the longitudinal direction of the cantilever
EECE 300 ‐ 2011
The x axis runs along the longitudinal direction of the cantilever.
24
Cantilever beam
Goal: find relation between tip deflection y(x = L) and applied load F.
Clamped end:y = 0 dy/dx = 0 at x = 0
x x = L
y = 0, dy/dx = 0 at x = 0
F
Assume tip deflection is small compared to length of beam.Shear stresses negligible.
( )( )M x F L x= −2
2( )
yM x EI
x
∂=
∂
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Cantilever beam ‐ bending i‐Clicker:deflection at beam tip:
2
2
)(b
2)(a.
LF
Ly
LEI
FLy
=
=
3
3)( c.
3)( b.
LEI
FLy
LEI
Ly
=
=
i‐Clicker:slope at beam tip: p p
2
2
2)(a.
F
LEI
FLy =
3
2
3)( c.
3)( b.
LEI
FLy
LEI
FLy
=
=
EECE 300 ‐ 2011 26
3EI
Review
: fin
taken
from nding m
om
mE. Po
pov, E en
t of in
ert
Engineerin
g tia
g M
echaniccs o
f Solids
EECE 300 ‐ 2011
s
27
Finding the spring constant
Beams are the most frequently encountered spring element in MEMS. These microbeams serve as mechanical springs for sensing and actuation. The stiffness of these beams is a frequently encountered design concern.Th iff i h i d b h i ( f ) Th h i lThe stiffness is characterized by the spring constant (or force constant). The mechanical spring constant is the ratio of the applied force and the resultant displacement:
FFk
x=
For a cantilever with a point‐loading on the free end, the maximum deflection occurs at the free end. For a fixed‐fixed bridge with a loading force in the center of the span, the center has the largest deflection.
EECE 300 ‐ 2011 28
Spring constant for cantilever
A fixed‐free beam with a rectangular cross section is one of the most common scenarios encountered in MEMS.
The free end of the beam will reach a certain bent angle,
The resultant vertical displacement equals
EECE 300 ‐ 2011
The spring constant of the cantilever is:
Liu, Foundations of MEMS 29
Spring constant for cantileverThe spring constant of the cantilever is:The spring constant of the cantilever is:
3
3 3
3
4
F EI Ewtk
x l l= = =
The spring constant:• decreases with increasing length• proportional to the width• strongly influenced by change in thickness due to the term t3• strongly influenced by change in thickness due to the term t3.The stiffness of the cantilever depends on the direction of the bending. If the force is applied longitudinally, the constant would be very different. The beam provides compliance in one direction and resistance to movement in another.
EECE 300 ‐ 2011 30
Springsi‐Clicker:
3
3
4 a.
L
bEaka =
b3
3
4 b.
4
L
Eabk
L
a =
L
b
ai‐Clicker:
3
3
a.bEa
kb = L
Fb3
3
3
4 b.
4 a.
L
Eabk
Lk
b
b
=
Fa
EECE 300 ‐ 2011 31
Spring systems – parallel connected springs
In many applications, two or more springs may be connected to form a spring system.In the parallel case:Same displacement → load is shared and the spring constant is the sum of the individual
ispring constants.
a
b
F/2
F/2
Lc
i‐Clicker:
kk 2a =F
b F/2
a
a
kk
kk
2
1 b.
2a.
=
=
EECE 300 ‐ 2011 32
Spring systems – serially connected springs
In many applications, two or more springs may be connected to form a spring system.In the serial case:Same load → deflec ons add
Ltotal
Fa b
Lc Lc
total
i‐Clicker:
kk 2a =Lc
a
a
kk
kk
2
1 b.
2a.
=
=
EECE 300 ‐ 2011 33
Moments of inertia of two beams – example
Consider two cantilever beams of the same length and material: one has a cross section of 100 μm by 5 μm, and a second one has a cross section of 50 μm by 8 μm. Which one is more resistant to flexural bending (i.e., stiffer)?
F
w1 t1 i‐Clicker:
a. beam 1 is stiffer1
F
w2 t2
L1b. beam 2 is stiffer
2
L2
2
EECE 300 ‐ 2011 34
End constraints and loading conditions
EECE 300 ‐ 2011Liu, Foundations of MEMS 35
End constraints and loading conditions
EECE 300 ‐ 2011Liu, Foundations of MEMS 36
Vertical translational plates
Fixed‐guided springs are often used to support rigid plates and facilitate their translation. Often, a plate is supported by two or more such beams. In these cases, one end of the beam is fixed, with all degrees of freedom limited. Another end of the spring can move in h i l di i b l di l i ll d b i i dthe vertical direction, but no angular displacement is allowed because it is connected to the stiff translational plate, which remains parallel to the substrate under allowable plate movement.
EECE 300 ‐ 2011Liu, Foundations of MEMS 37
Vertical translational plate – example
Find the expression of the force constant associated with the plate.
Start with the basic formula for the spring constant of a single fixed‐guided beam under a transverse loading force F. The maximum displacement x occurs at the guided end of the beam:
EECE 300 ‐ 2011 38
Vertical translational plate – example
The expression for the spring constant (force constant) of each single fixed‐guided beam is:
If a force is applied to a plate supported by n cantilevers with equal dimensions and force constants, each spring shares 1/nth of the total force load. The total force constant experienced by the spring is nk.The force constant associated with each fixed‐guided beam is:
For the plate supported by two fixed‐ For the plate supported by four fixed‐For the plate supported by two fixedguided beams, the equivalent force constant is
For the plate supported by four fixedguided beams, the equivalent force constant is
EECE 300 ‐ 2011 39
Capacitance Sensor Response
A parallel capacitor with an area (A) of 100 × 100 μm2 is supported by four cantilever beams. The plate is made of polycrystalline silicon that is t = 2 μm thick. The distance between the bottom of the plate and the substrate is d = 1 μm. Each cantilever beam is l 400 20 0 1 Fi d h l i h f i dl = 400 μm, w = 20 μm, t = 0.1 μm. Find the relative change of capacitance under an acceleration of 1 g.
EECE 300 ‐ 2011 40
Capacitance Sensor Response
A parallel capacitor with an area (A) of 100 × 100 μm2 is supported by four cantilever beams. The plate is made of polycrystalline silicon that is t = 2 μm thick. The distance between the bottom of the plate and the substrate is d = 1 μm. Each cantilever beam is l 400 20 0 1 Fi d h l i h f i dl = 400 μm, w = 20 μm, t = 0.1 μm. Find the relative change of capacitance under an acceleration of 1 m/s2.
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Capacitance Sensor Response
EECE 300 ‐ 2011 42
Microneedles for painless drug delivery
EECE 300 ‐ 2011N.-T. Nguyen and S. T. Wereley, Fundamentals And Applications of Microfluidics, Second ed: Artech House, 2006. 43
Molded polysilicon microneedles
EECE 300 ‐ 2011D. V. McAllister, M. G. Allen, and M. R. Prausnitz, "Microfabricated Microneedles for Gene and Drug Delivery," Annual Review of Biomedical Engineering, vol. 2, pp. 289-313, 2000. 44
4
Solid square
1Solid rectangle
141
12I W= 31
12I WH=
H h
( )4 4
Hollow square
1
12I W w= − ( )3 3
Hollow rectangle
1
12I WH wh= −
12
Solid circle
( )12
Hollow circle
4
4I R
π= ( )4 4
4I R r
π= −
EECE 300 ‐ 20113
Thin annulus
I R tπ= 45
Buckling of a microneedle
• If the needle’s length is relatively long compared to its width the first failure mode iscompared to its width, the first failure mode is buckling.
• The critical b cklin force
Fb
• The critical buckling force:
2EIπ24b
EIF
L
π=
EECE 300 ‐ 2011 46
Parylene microneedle – buckling
• A microneedle made of parylene C has channel dimensions 200 μm × 200 μm × 2 mm. The parylene layer is deposited over 4 hours at a rate of 5 μm/hr (over a sacrificial material) The Young’s modulus is 3 2 GPaμm/hr (over a sacrificial material). The Young s modulus is 3.2 GPa.
• Calculate the critical buckling force for this needle.
EECE 300 ‐ 2011 47
Bending of a microneedle
• As a rough estimate, the bending stiffness of a needle can be modeled by the spring constant k:
• For a hollow square needle:
EECE 300 ‐ 2011 48
Parylene microneedle – bending• A microneedle made of parylene C has channel (inner) dimensions 200 μmA microneedle made of parylene C has channel (inner) dimensions 200 μm
× 200 μm × 2 mm. The parylene layer is deposited over 4 hours at a rate of 5 μm/hr (over a sacrificial material). The Young’s modulus is 3.2 GPa.
• Calculate the tip deflection under a force of 15 mN at the tipCalculate the tip deflection under a force of 15 mN at the tip.
EECE 300 ‐ 2011 49
Polysilicon microneedle• If the previous needle were made of polysilicon, what are the tip p p y , p
deflection and critical buckling force? Assume all other parameters are the same, and Young’s modulus of polysilicon is 150 GPa.
i‐Clicker:
a. the tip deflection is directly proportional to Young’s modulus.b. the tip deflection is inversely proportional to Young’s modulus.
i‐Clicker:
a. the critical buckling force is directly proportional to Young’s modulus.b. the critical buckling force is inversely proportional to Young’s modulus.b. the critical buckling force is inversely proportional to Young s modulus.
EECE 300 ‐ 2011 50
Microneedle – mechanical designInsertion force• The area A of contact surface at the needle tip is considered the determining
factor for the insertion force Fi.• The puncture toughness Gp which is the work per area needed to initiate a crack:
W
Insertion force (static case where the insertion speed is slow and the kinetic energy
p
WG
A=
Insertion force (static case, where the insertion speed is slow and the kinetic energy of the needle is negligible)
1F F G A= +,max 0i pF F G A
χ= +
F0 and χ determined experimentally.F initial force
EECE 300 ‐ 2011
F0 initial forceχ characteristic insertion length
51
Insertion force of a microneedle• The tip radius of a microneedle is 50 μm. Assume a puncture p μ p
toughness of the skin of 30 kJ/m2, a characteristic insertion length of 150 μm, and an initial force F0 = 0.1 N. Determine the force required for puncturing the skin.
EECE 300 ‐ 2011 52
Differential thermal expansion
If two materials are in close contact during temperature changes (heating or cooling) and their thermal expansion coefficients are not equal, then differential expansion (during heating) or differential contraction (during cooling) can occur.Thi h d i hi fil d i i Th fil i d i d f T bThis can happen during thin film deposition. The film is deposited stress‐free at Tdep, but then at room temperature the film is under a thermal mismatch strain.
( ) Tε γ γ Δ( )1 2 Tε γ γ= − Δ
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Beam bending – intrinsic stress
EECE 300 ‐ 2011 54
Thermal coefficient of expansion
At the temperature increases, most solids expand in volume. The lateral dimensions of the object increases in all directions.
The volumetric thermal expansion coefficient α is the
Th li i ffi i i h h f l
The volumetric thermal expansion coefficient α is the ratio between the relative change of volume to the degree of temperature variation,
The linear expansion coefficient γ is the change of only one dimension of an object due to temperature variation,
The percentage change in length in an unconstrained object (free to expand at both ends) due to the increase in temperature is given by
If the object is fixed at both ends, then the thermal expansion produces a stress given by
EECE 300 ‐ 2011
j , p p g y
55
Thermal Expansion
EECE 300 ‐ 2011 56
Thermal actuatorAssume that you have a silicon beam that is 100 μm long, and 1μm square. You heat it by 100K. How much force do you get if you constrain it? How much elongation if you allow it to expand? Linear thermal coefficient of expansion for silicon is 2.3 × 10‐6/K . ESi = 160 GPa.
Area=ε = γ ΔT =ε = γ ΔT =σ = Ε ε =F AF = A σ =δL = ε L=
EECE 300 ‐ 2011 57
Thermal Actuators
Use thermal expansion for actuation
Very effective and high force output per unit area
Actuator translatesin this direction
Cold arm
Very effective and high force output per unit area
Cold arm
Current output pad
Hot arm
Cascaded thermal actuators for high force
EECE 300 ‐ 2011
Current input pad
J. Judy, UCLA 58
Bimetallic beam actuatorradius of curvatureradius of curvature
Bi t lli t ti th diff i th l ffi i t f i f t b d dBimetallic actuation uses the difference in thermal coefficient of expansion of two bonded solids. This principle is often called thermal bimorph actuation. Bimetallic actuators offer an almost linear deflection dependence on heating power. Their disadvantages are high power consumption and slow response.The resulting force is proportional to the difference between the thermal expansion coefficients of the two materials and the temperature difference.
EECE 300 ‐ 2011 59
Bimetallic beam actuator2L
The displacement of the beam tip is calculated as:
h di f i
( )2
Ly L
R≈
The radius of curvature is:
( ) ( ) ( )( ) ( )
2 22 2 2 21 1 1 2 2 2 1 1 1 2 2 2 1 1 2 2
2 1 1 1 1 2 2 2 1 2
2 2 3 2
6
b E t b E t b E t b E t t t t t
Tb E t b E t t tρ
γ γ
+ + + +=
− Δ +
where b1 and b2 are the widths of the two material layers.
( ) ( )2 1 1 1 1 2 2 2 1 2γ γ
( ) ( )3 EI y LTh i l t f th ti f th b ( ) ( )3
3beam
EI y LF
L=The equivalent force on the tip of the beam
(cantilever beam with free end) is:
The flexural rigidity of the composite beam is:
( ) ( ) ( ) ( )( )
2 22 2 2 21 1 1 2 2 2 1 1 1 2 2 2 1 1 2 2
1 1 1 2 2 2
2 2 3 2
12beam
b E t b E t b E t b E t t t t tEI
b E t b E t
+ + + +=
+
EECE 300 ‐ 2011 60
Bimetallic beam actuator
Assuming that the two layers have the same width b, the radius of curvature and the equivalent tip force can be simplified to:
2 2⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞2 2
1 1 1 1 2 2
2 2 2 2 1 11 2
2
3 1 1
6
t t E t t E
t t E t t Et t
tρ
⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + + +⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠+ ⎣ ⎦=⎛ ⎞( ) 1
2 12
61
tT
tγ γ
⎛ ⎞− Δ +⎜ ⎟
⎝ ⎠
( )1 22 1
31 14
t tbF T
Lγ γ+
= − Δ+
1 1 2 2t E t E+
EECE 300 ‐ 2011 61
Design of a thermomechanical valve with bimetallic actuator
A thermomechanical microvalve has a rigid square seat of 500 × 500 μm. The valve seat is suspended on four flexures. Each flexure is 500 μm long and 200 μm wide. The flexure is made of 10‐μm silicon and 2‐μm aluminum. The silicon heater is integrated in the flexure. Th l i l i d 400°C If ll l d i l i bThe aluminum layer is evaporated at 400°C. If a normally closed microvalve is to be designed at 25°C, what is the maximum gap between valve opening and the surface of the valve seat wafer?
Si: t1 = 10 μm γ1 = 2 3 × 10‐6 K‐1 E1 = 170 GPa
EECE 300 ‐ 2011
Si: t1 10 μm, γ1 2.3 × 10 K , E1 170 GPaAl: t2 = 2 μm, γ2 = 23 × 10‐6 K‐1, E2 = 70 GPa
62
Thermomechanical valve
EECE 300 ‐ 2011 63
Thermomechanical valve
EECE 300 ‐ 2011 64
Thermomechanical valve: minimum opening/closing pressure
If the gap between valve opening and surface of the valve seat is 5 μm, how large shouldIf the gap between valve opening and surface of the valve seat is 5 μm, how large should the inlet pressure be to open the valve? What is the temperature difference for opening the valve at zero inlet pressure? The valve opening is 200 × 200 μm.
EECE 300 ‐ 2011 65
