EE C245 ME C218 Introduction to MEMS Design

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1 EE C245 ME C218 Introduction to MEMS Design Fall 2007 Prof. Clark T.-C. Nguyen Dept. of Electrical Engineering & Computer Sciences University of California at Berkeley Berkeley, CA Lecture 16: Energy Methods I EE C245: Introduction to MEMS Design Lecture 16 C. Nguyen 10/21/08 1

2 Lecture Outline Reading: Senturia, Chpt. 9, 10 Lecture Topics: Bending of beams Cantilever beam under small deflections Combining cantilevers in series and parallel Folded suspensions Design implications of residual stress and stress gradients Energy Methods Virtual Work Energy Formulations Tapered Beam Example Doubly Clamped Beam Example Large Deflection Analysis EE C245: Introduction to MEMS Design Lecture 16 C. Nguyen 10/21/08 2

3 Deflection of Folded Flexures This equivalent to two cantilevers of length L c /2 Composite cantilever free ends attach here Half of F absorbed in other half (symmetrical) 4 sets of these pairs, each of which gets ¼ of the total force F EE C245: Introduction to MEMS Design Lecture 16 C. Nguyen 10/21/08 3

4 Constituent Cantilever Spring Constant From our previous analysis: x( y) 2 Fc Lc y Fc y y L EI z L = 2 1 = c 6EI z 2 ( y ) From which the spring constant is: Fc k c = = x L ) Inserting L c = L/2 c 3EI 3 ( c Lc z k 3EI z c = = 3 ( L / 2) 24EI L 3 z EE C245: Introduction to MEMS Design Lecture 16 C. Nguyen 10/21/08 4

5 Overall Spring Constant Rigid Truss Four pairs of clamped-guided beams In each pair, beams bend in series (Assume trusses are inflexible) ibl Force is shared by each pair F pair = F/4 Leg L F pair EE C245: Introduction to MEMS Design Lecture 16 C. Nguyen 10/21/08 5

6 Folded-Beam Stiffness Ratios Folded-beam suspension Anchor Shuttle Folding truss In the x-direction: k x = 24EI 3 L In the z-direction: Same flexure and boundary conditions k z = In the y-direction: [See Senturia, 9.2] Thus: k y k x = L 4 W 24EI 3 L z x 8EWh k y = L 2 Much stiffer in y-direction! EE C245: Introduction to MEMS Design Lecture 16 C. Nguyen 10/21/08 6

7 Folded-Beam Suspensions Permeate MEMS Accelerometer [ADXL-05, Analog Devices] Gyroscope [Draper Labs.] Micromechanical Filter [K. Wang, Univ. of Michigan] EE C245: Introduction to MEMS Design Lecture 16 C. Nguyen 10/21/08 7

8 Folded-Beam Suspensions Permeate MEMS Below: Micro-Oven Controlled Folded-Beam Resonator EE C245: Introduction to MEMS Design Lecture 16 C. Nguyen 10/21/08 8

9 Stressed Folded-Flexures Flexures EE C245: Introduction to MEMS Design Lecture 16 C. Nguyen 10/21/08 9

10 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 z y x W L Governing differential equation: (Euler Beam Equation) Axial Load Unit x=l EE C245: Introduction to MEMS Design Lecture 16 C. Nguyen 10/21/08 10

11 The Euler Beam Equation Axial Stress R 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 on the beam to counteract the axial stress-derived force For ease of analysis, assume the beam is bent to angle π EE C245: Introduction to MEMS Design Lecture 16 C. Nguyen 10/21/08 11

12 The Euler Beam Equation Note: Use of the full bend angle of π to establish conditions for load balance; but this returns us to case of small displacements and small angles EE C245: Introduction to MEMS Design Lecture 16 C. Nguyen 10/21/08 12

13 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 z y x W L Governing differential equation: (Euler Beam Equation) Axial Load Unit x=l EE C245: Introduction to MEMS Design Lecture 16 C. Nguyen 10/21/08 13

14 Solving the ODE Can solve the ODE using standard methods Senturia, pp : solves ODE for case of point load on a clamped-clamped l d beam (which h defines B.C. s) For solution to the clamped-guided case: see S. Timoshenko, Strength of Materials II: Advanced Theory and Problems, McGraw-Hill, New York, 3 rd Ed., 1955 Result from Timoshenko: EE C245: Introduction to MEMS Design Lecture 16 C. Nguyen 10/21/08 14

15 Design Implications Straight flexures Large tensile S means flexure behaves like a tensioned wire (for which h k -1 = L/S) Large compressive S can lead to buckling (k -1 ) Folded flexures Residual stress only ypartially released Length from truss to shuttle s s centerline differs by L s for inner and outer legs EE C245: Introduction to MEMS Design Lecture 16 C. Nguyen 10/21/08 15

16 Effect on Spring Constant Residual compression on outer legs with same magnitude of tension on inner legs: ε b = ± ε r L s L = Eε r Beam Strain: ; Stress Force: ± Wh Spring constant becomes: S Ls L Remedies: Reduce the shoulder width L s to minimize stress in legs Compliance in the truss lowers the axial compression and tension and reduces its effect on the spring constant EE C245: Introduction to MEMS Design Lecture 16 C. Nguyen 10/21/08 16

17 Energy Methods EE C245: Introduction to MEMS Design Lecture 16 C. Nguyen 10/21/08 17

18 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 of location x under a point load F applied at the tip of the free end of a cantilever with tapered width W(x) W 50% taper y x EE C245: Introduction to MEMS Design Lecture 16 C. Nguyen 10/21/08 18

19 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) action of surface and body forces is equal to the work done by these forces Implication: if we can formulate 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 minimize the difference U between the stored energy and the work done by the forces: U = Stored Energy - Work Done Key idea: we don t have to reach U = 0 to produce a very useful, approximate analytical result for load-deflection EE C245: Introduction to MEMS Design Lecture 16 C. Nguyen 10/21/08 19

20 More Visual Description EE C245: Introduction to MEMS Design Lecture 16 C. Nguyen 10/21/08 20

21 Fundamentals: Energy Density Strain energy density: [J/m 3 ] To find work done in straining material Total strain energy [J]: Integrate over all strains (normal and shear) EE C245: Introduction to MEMS Design Lecture 16 C. Nguyen 10/21/08 21

22 Bending Energy Density Neutral Axis y y(x) = transverse displacement of neutral axis x First, find the bending energy dw bend in an infinitesimal length dx: EE C245: Introduction to MEMS Design Lecture 16 C. Nguyen 10/21/08 22

23 Energy Due to Axial Load y x Strain due to axial load S contributes an energy dw stretch in length dx, since lengthening of the different element dx (to ds) results in a strain ε x EE C245: Introduction to MEMS Design Lecture 16 C. Nguyen 10/21/08 23

24 Shear Strain Energy Shear Modulus See W.C. Albert, Vibrating Quartz Crystal Beam Accelerometer, Proc. ISA Int. Instrumentation Symp., May 1982, pp EE C245: Introduction to MEMS Design Lecture 16 C. Nguyen 10/21/08 24

25 Applying the Principle of Virtual Work Basic Procedure: Guess the form of the beam deflection under the applied loads Vary the parameters in the beam deflection function in order to minimize: Sum strain energies Assumes point load U = W j j i F i u i Displacement at point load Find minima by simply setting derivatives to zero See Senturia, pg. 244, for a general expression with distrubuted surface loads and body forces EE C245: Introduction to MEMS Design Lecture 16 C. Nguyen 10/21/08 25

26 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) W 50% taper y x Adjustable parameters: minimize U y ( x) = c x + c x Start by guessing the solution 2 3 It should satisfy the boundary conditions The strain energy integrals shouldn t be too tedious This might not matter much these days, though, h since one could just use matlab or mathematica EE C245: Introduction to MEMS Design Lecture 16 C. Nguyen 10/21/

27 Strain Energy And Work By F (Bending Energy) (Using our guess) Tip Deflection EWh 3 EE C245: Introduction to MEMS Design Lecture 16 C. Nguyen 10/21/08 27

28 Find c 2 and c 3 That Minimize U Minimize U basically, find the c 2 and c 3 that brings U closest to zero (which is what it would be if we had guessed correctly) The c 2 and c 3 that minimize U are the ones for which the partial derivatives of U with respective to them are zero: Proceed: First, evaluate the integral to get an expression for U: EE C245: Introduction to MEMS Design Lecture 16 C. Nguyen 10/21/08 28

29 Minimize U (cont) Evaluate the derivatives and set to zero: Solve the simultaneous equations to get c 2 and c 3 : EE C245: Introduction to MEMS Design Lecture 16 C. Nguyen 10/21/08 29

30 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): 13% smaller than tapered-width case EE C245: Introduction to MEMS Design Lecture 16 C. Nguyen 10/21/08 30

31 Comparison With Finite Element Simulation Below: ANSYS finite element model with L = 500 μm W base = 20 μm E = 170 GPa h = 2 μm W tip = 10 μm Result: (from static analysis) k = μn/m This matches the result from energy minimization to 3 significant figures EE C245: Introduction to MEMS Design Lecture 16 C. Nguyen 10/21/08 31

32 Need a Better Approximation? Add more terms to the polynomial Add other strain energy terms: Shear: more significant as the beam gets shorter Axial: more significant as deflections become larger Both of the above remedies make the math more complex, so encourage the use of math software, such as Mathematica, Matlab, or Maple 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 cannot Can compare the importance of different terms Should use in tandem with FEA for design EE C245: Introduction to MEMS Design Lecture 16 C. Nguyen 10/21/08 32

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