EE C245 – ME C218 Introduction to MEMS Design Fall 2007ee290g/fa08/lectures/Lec16bc.EnergyMethodsI.f08.pdfEE C245 – ME C218 Introduction to MEMS Design Fall 2007 Prof Clark TProf

EE C245 – ME C218Introduction to MEMS Design

Fall 2007Fall 2007

Prof Clark T C NguyenProf. Clark T.-C. Nguyen

Dept of Electrical Engineering & Computer SciencesDept. of Electrical Engineering & Computer SciencesUniversity of California at Berkeley

Berkeley, CA 94720y

L t 16 E M th d I

EE C245: Introduction to MEMS Design Lecture 16 C. Nguyen 10/21/08 1

Lecture 16: Energy Methods I

Lecture Outline

• Reading: Senturia, Chpt. 9, 10g p• Lecture Topics:

Bending of beamsCantilever beam under small deflectionsCantilever beam under small deflectionsCombining cantilevers in series and parallelFolded suspensionspDesign implications of residual stress and stress gradientsEnergy Methods

Virtual WorkVirtual WorkEnergy FormulationsTapered Beam ExampleDoubly Clamped Beam ExampleLarge Deflection Analysis


Deflection of Folded Flexures

This equivalent to two cantilevers of

l th L /2length Lc/2

Composite cantilever free ends attach here free ends attach here

Half of F absorbed in other half 4 sets of these pairs each of


other half (symmetrical)

4 sets of these pairs, each of which gets ¼ of the total force F

Constituent Cantilever Spring Constant

• From our previous analysis:

( )yFyLF ⎟⎞

⎜⎛ 2

2 ( )yLEI

yFLyy

EILFyx c

z

c

cz

cc −=⎟⎟⎠

⎞⎜⎜⎝

⎛−= 3

631

2)( 2

3EIF• From which the spring constant is:

33

)( c

z

c

cc L

EILx

Fk ==

• Inserting Lc = L/2

3324

)2/(3

LEI

LEI

k zzc ==


Overall Spring Constant • Four pairs of clamped-guided beams

In each pair, beams bend in series(A t i fl ibl )

Rigid Truss

(Assume trusses are inflexible)• Force is shared by each pair → Fpair = F/4

Leg

L

FFpair


Folded-Beam Stiffness Ratios

• In the x-direction:Folded-beam suspension 24EI

k z=

• In the z-direction:S m fl d b d

3Lkx =

Same flexure and boundary conditions

24EIk x=

• In the y-direction:Shuttle

3Lkz =

y

LEWhk y

8=[See Senturia, §9.2]

• Thus: Folding truss 2

4 ⎟⎞

⎜⎛=

Lk yMuch

stiffer in


Anchor4 ⎟

⎠⎜⎝Wkx y-direction!

Folded-Beam Suspensions Permeate MEMS

Gyroscope [Draper Labs.]Accelerometer [ADXL-05, Analog Devices]


Micromechanical Filter [K. Wang, Univ. of Michigan]

Folded-Beam Suspensions Permeate MEMS

• Below: Micro-Oven Controlled Folded-Beam Resonator


Stressed Folded-FlexuresStressed Folded Flexures


Clamped-Guided Beam Under Axial Load

• Important case for MEMS suspensions, since the thin films comprising them are often under residual stress

• Consider small deflection case: y(x) « L

Lxz

Ly

W

Governing differential equation: (Euler Beam Equation)


Unit impulse @ x=LAxial Load

The Euler Beam Equation

RAxial Stress

• Axial stresses produce no net horizontal force; but as soon as the beam is bent, there is a net downward force

For equilibrium must postulate some kind of upward load For equilibrium, must postulate some kind of upward load on the beam to counteract the axial stress-derived forceFor ease of analysis, assume the beam is bent to angle π


The Euler Beam Equation

Note: Use of the full Note Use of the full bend angle of π to

establish conditions for load balance; but this load balance; but this returns us to case of

small displacements and small anglessmall angles


Clamped-Guided Beam Under Axial Load

• Important case for MEMS suspensions, since the thin films comprising them are often under residual stress

• Consider small deflection case: y(x) « L

Lxz

Ly

W

Governing differential equation: (Euler Beam Equation)


Unit impulse @ x=LAxial Load

Solving the ODE

• Can solve the ODE using standard methodsSenturia, pp. 232-235: solves ODE for case of point load

l d l d b ( hi h d fi B C ’ )on a clamped-clamped beam (which defines B.C.’s)For solution to the clamped-guided case: see S. Timoshenko, Strength of Materials II: Advanced Theory Timoshenko, Strength of Materials II Advanced Theory and Problems, McGraw-Hill, New York, 3rd Ed., 1955

• Result from Timoshenko:


Design Implications

• Straight flexuresLarge tensile S means flexure behaves like a tensioned i (f hi h k 1 L/S)wire (for which k-1 = L/S)

Large compressive S can lead to buckling (k-1 → ∞)

• Folded flexuresResidual stress only partially y p yreleasedLength from truss to shuttle’s to shuttle s centerline differs by Ls for inner and outer legsand outer legs


Effect on Spring Constant

• Residual compression on outer legs with same magnitude of tension on inner legs:

⎞⎛ ⎞⎛⎟⎠⎞

⎜⎝⎛±=

LLs

rb εεBeam Strain: ; Stress Force: WhLLES s

r ⎟⎠⎞

⎜⎝⎛±= ε

• Spring constant becomes:

• Remedies:Reduce the shoulder width Ls to minimize stress in legsCompliance in the truss lowers the axial compression and


Compliance in the truss lowers the axial compression and tension and reduces its effect on the spring constant

Energy MethodsEnergy Methods


More General Geometries

• Euler-Bernoulli beam theory works well for simple geometries• But how can we handle more complicated ones?• Example: tapered cantilever beam•Objective: Find an expression for displacement as a function

f l ti d i t l d F li d t th ti f th of location x under a point load F applied at the tip of the free end of a cantilever with tapered width W(x)

50% tW

50% taper

xxy


Solution: Use Principle of Virtual Work

• In an energy-conserving system (i.e., elastic materials), the energy stored in a body due to the quasi-static (i.e., slow)

i f f d b d f i l h k d action of surface and body forces is equal to the work done by these forces …

• Implication: if we can formulate stored energy stored energy as a function Implication: if we can formulate stored energy stored energy as a function of the deformation of a mechanical object, then we can determine how an object responds to a force by determining the shape the object must take in order to minimizeminimize the differencedifference UU between the stored energy and the work done by the forces:by the forces:

U = Stored Energy - Work DoneU = Stored Energy - Work Done

• Key idea: we don’t have to reach U = 0 to produce a very


Key idea: we don t have to reach U = 0 to produce a very useful, approximate analytical result for load-deflection

More Visual Description …


Fundamentals: Energy Density

• Strain energy density: [J/m3]To find work done in straining material

• Total strain energy [J]:Total strain energy [J]:Integrate over all strains (normal and shear)


Bending Energy Density

y(x) = transverse displacement

Neutral Axis

y( ) pof neutral axis

xy

• First, find the bending energy dWbend in an infinitesimal length dx:

y


Energy Due to Axial Load

xy

• Strain due to axial load S contributes an energy dWstretchin length dx, since lengthening of the different element dx(to ds) results in a strain εx


Shear Strain Energy

Shear Modulus

• See W.C. Albert, “Vibrating Quartz Crystal Beam Accelerometer,” Proc. ISA Int. Instrumentation Symp., May 1982 pp 33-441982, pp. 33-44


Applying the Principle of Virtual Work

• Basic Procedure:Guess the form of the beam deflection under the applied loadsloadsVary the parameters in the beam deflection function in order to minimize:

Sum strain energiesAssumes point load

ii

ij

j uFWU ∑∑ −=ij

Displacement at point load

Find minima by simply setting derivatives to zero• S S t i 244 f l i ith

at point load


• See Senturia, pg. 244, for a general expression with distrubuted surface loads and body forces

Example: Tapered Cantilever Beam•Objective: Find an expression for displacement as a function of location x under a point load F applied at the tip of the free end of a cantilever with tapered width W(x)free end of a cantilever with tapered width W(x)

W

Adjustable

50% taperW

Adjustable parameters: minimize U

xy

• Start by guessing the solution3

32

2)( xcxcxy +=

y

Start by guessing the solutionIt should satisfy the boundary conditionsThe strain energy integrals shouldn’t be too tedious

Thi i ht t tt h th d th h i


This might not matter much these days, though, since one could just use matlab or mathematica

Strain Energy And Work By F

(Bending Energy)(Bending Energy)

(Using our guess)

Tip Deflectionp f

EWh3


Find c2 and c3 That Minimize U

•Minimize U → basically, find the c2 and c3 that brings U closest to zero (which is what it would be if we had guessed

l )correctly)• The c2 and c3 that minimize U are the ones for which the partial derivatives of U with respective to them are zero:partial derivatives of U with respective to them are zero:

• Proceed:First, evaluate the integral to get an expression for U:


Minimize U (cont)

• Evaluate the derivatives and set to zero:

• S l th i lt ti t t d • Solve the simultaneous equations to get c2 and c3:


The Virtual Work-Derived Solution

• And the solution:

• Solve for tip deflection and obtain the spring constant:

• Compare with previous solution for constant-width cantilever beam (using Euler theory):beam (using Euler theory):

13% smaller than tapered width case


tapered-width case

Comparison With Finite Element Simulation

• Below: ANSYS finite element model withL = 500 μm Wbase = 20 μm E = 170 GPaaseh = 2 μm Wtip = 10 μm

l ( • Result: (from static analysis)

k = 0 471 μN/mk 0.471 μN/m• This matches the result from energy minimization to 3 significant figures


Need a Better Approximation?

• Add more terms to the polynomial• Add other strain energy terms:gy

Shear: more significant as the beam gets shorterAxial: more significant as deflections become larger

• Both of the above remedies make the math more complex • Both of the above remedies make the math more complex, so encourage the use of math software, such as Mathematica, Matlab, or Maplep

• Finite element analysis is really just energy minimization• If this is the case, then why ever use energy minimization analytically (i.e., by hand)?

Analytical expressions, even approximate ones, give insight into parameter dependencies that FEA cannotinsight into parameter dependencies that FEA cannotCan compare the importance of different termsShould use in tandem with FEA for design


Documents

EE C245 – ME C218 Introduction to MEMS Design Fall 2007ee290g/fa08/lectures/Lec16bc.EnergyMethodsI.f08.pdfEE C245 – ME C218 Introduction to MEMS Design Fall 2007 Prof Clark TProf