Introduction to MEMS Design
Introduction to MEMS Design MEC ENG C218
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This 16 page Class Notes was uploaded by Dayna Kessler on Thursday October 22, 2015. The Class Notes belongs to MEC ENG C218 at University of California - Berkeley taught by C. Nguyen in Fall. Since its upload, it has received 49 views. For similar materials see /class/226682/mec-eng-c218-university-of-california-berkeley in Mechanical Engineering at University of California - Berkeley.
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Date Created: 10/22/15
EE C245 ME C218 Introduction to MEMS Design Fall 2007 Prof Clark TC Nguyen Dept of Electrical Engineering amp Computer Sciences University of California at Berkeley Berkeley CA 94720 Lecture 16 Energy Methods Lecture 16 C Nguyen 101807 1 EE C245 Introduction to MEMS Design Announcements quotin ammm 39 Midterm Exam Date What39s best tl gtTuesday Oct 30 day before Halloween tl gtThursday Nov 1 day after Halloween 39 We chose Tuesday Oct 30 this is the date of our midterm exam 39 Correction to slide 19 of lecture 14 on next page and also lecture 14 has been reposted with the correction ti gtNote that this affects your homework5 Lecture 16 C Nguyen 101807 2 EE C245 Introduction to MEMS Design MEMS Stress Test Structure l AUC 39 Simple Approach use a clamped clamped beam ti gtCompressive stress causes buckling 5 Arrays with increasing length are used to determine the critical buckling load where h h39CkneSS 72 Eh2 acriticul 3 L2 H E Young39s modulus Pa I I 112Wh3 moment of inertia L W h indicated in the figure ti gtLimitationz Only compressive I stress is measurable I EE C245 Introduction to MEMS Design Lecture 16 C Nguyen 101807 3 Lecture Outline 39 Reading Senturia Chpts 9 10 39 Lecture Topics 5 Stressed FoldedFlexures lb Energy Methods 39Virtual Work 39Energy Formulations 39Tapered Beam Example 39Doubly Clamped Beam Example 39Large Deflection Analysis EE C245 Introduction to MEMS Design Lecture 16 C Nguyen 101807 4 Deflection of Folded Flexures to In This equivalent to two cantilevers of 5 length L2 Composite cantilever free ends attach here G a 3 4 Fa Lerb 39l l Half of F A 1 absorbed in A other half 4 sets of these pairs each of SYmme lf lcal which gets of the total force F 101007 EE C245 Introduction to MEMS Design Lectu e 7 C Nguyen 5 Constituent Cantilever Spring Constant mutingu 39 From our previous analysis FCLC 2E1 z Fcy2 6E1z xy y21 i 3L6 y 39 From which the spring constant is F 3E1 I kc c 3 z c xLc LC 39 Inserting Lc L2 3E1z 24EIz c L 23 L3 EE C245 Introduction to MEMS Design Lectu e 7 C Nguyen 101007 6 Overall Spring Constant 39 Four pairs of clamped guided beams m t gtIn each pair beams bend in series t gtAssume trusses are inflexible Force is shared by each pair gt Fpair F4 5 v H quot3 a Wyle v lwn 65 00 quot L flw 4 rrn39vr 09 fondsr 5 yr arlr ELL kc Erquot kgllkp 7 Sui5 Fm 36 kg kollkg Ayih howlto 7 F L F jmkq mm 9 5 l x orerwiw v kc I kc E kw AIIKAquotBquotquot ZVEI 0 t r 39 isquot 5245 Introduction to MEMS Design Lecture 2 C Nguyen 83007 7 FoldedBeam Stiffness Ratios Folded beam 39 In the x direction suspension 3 k x 24EIZ 3 39 In the z direction tsgtame flexure and boundary conditions 24EIx kz 3 L Shutle 39 In the y direction See Senturia 92 ky SEWh stiffer in y direction 2 39 Thus ky 4 L MUCh W x EE C245 Introduction to MEMS Design Lectu e 7 C Nguyen 101007 8 v my i Accelerometer ADXL 05 Ana H Aquot 39 p Micromechanical Filfer39 K Wong Univ of Michigan EE C245 Inmduc on to MEMS Design Lecture 7 C Nguyen 101007 9 I sWiiiiiiiiiiiiiiiiv quot 101007 10 EE C245 Inmduc on to MEMS Design Lecture 7 C Nguyen Stressed Folded Flexures EE C245 Introduction to MEMS Design Lecture 7 C Nguyen 101007 11 A clampedGuided Beam Under Axial Load i 397 9W 39 Important case for MEMS suspensions since the thin films comprising them are often under residual stress 39 Consider small deflection case yx L 4 Axial Load Unit impulse xL EE C245 Introduction to MEMS Design Lecture 7 C Nguyen 101007 12 E The Euler Beam Equation 39 ll um mare Thmbeam 45mm downher R 739 quotI 62 H Kioth W6P09 Axial Stress R O DWH GOWH 39 Axial stresses produce no net horizontal force but as soon as the beam is bent there is a net downward force b For equilibrium must postulate some kind of upward load on the beam to counteract the axial stressderived force 5 For ease of analysis assume the beam is bent to angle 1 DWIJ Fm lawH 11101 Panha upwd Equot M 3 g 2 a EelsD R P lgtvIRco6 lo zam 9 0 EE 6245 Introduction to MEMS Design Lecture 2 C Nguyen 83007 13 f mh lrml ZRWP 203wH 55 Morrile I dzu 39 J2 qr unit0194 h R 1 JV7 M L bemol39lwd N u f h fll M1 ball been ote seoteu 5 quot nial ml bend angle of n to d714 M dqor 12 F llth establish COI39IdI l39IOI39IS for 4quot61 Et 0 7 51 at load balance but this returns us to case of exleml hm small displacements adv 392 quot g 151 and small angles qwimlwamu y39 e QXI39J J l39rlr WLWM ol w ownmg rig My 1 EM 39quotquot Ewl TW AKWS E when EE 6245 Introduction to MEMS Design Lecture 2 C Nguyen 83007 14 ClampedGuided Beam Under Axial Load ElW ullT 39 Important case for MEMS suspensions since the thin films comprising them are often under residual stress 39 Consider small deflection case yx L z X quot lt L y 4 s S Axial Load Unit impulse xL EE C245 Introduction to MEMS Design Lecture 7 C Nguyen 101007 15 Solving the ODE ti gtenturia pp 232235 solves ODE for case of point load on a clampedclamped beam which defines BC39s tl gtFor solution to the clampedguided case see 5 Timoshenko Strength of Materials II Advanced Theory and Problems McGrawHill New York 3quotd Ed 1955 39 Result from Timoshenko 1 pL ZtamhprZ yx P3 F 3 gt a tension it 5 lt 0 compression if p 2tanpL2 yxL e PS F where p EE C245 Introduction to MEMS Design Lecture 7 C Nguyen 101007 16 Desngn Implications WWW 39 Straight flexures 5 Large tensile S means flexure behaves like a tensioned wire for which kquot1 US 5 Large compressive S can lead to buckling k391 oo 39 Folded flexures 5 Residual stress only partially released 5 Length from truss to shuttle39s centerline differs by Ls for inner and outer legs EE C245 Introduction to MEMS Design Lecture 7 C Nguyen 101007 17 Effect on Spring Constant 39 Residual compression on outer legs with same magnitude of tension on inner legs L L Beam Strain 8b i8rfs Stress Force S iEgrTslVh 39 Spring constant becomes k wequot kmquotr 34 PL2MDPL 2LPL Zt nMle2T 1451 plsl 39 Remedies 5Reduce the shoulder width Ls to minimize stress in legs 5Compliance in the truss lowers the axial compression and tension and reduces its effect on the spring constant EE C245 Introduction to MEMS Design Lecture 7 C Nguyen 101007 18 Energy Methods EE C245 Introduction to MEMS Design Lecture 7 C Nguyen 101007 19 Principle of Virtual Work 39 In an energyconserving system i e elastic materials the energy stored in a body due to the quasistatic i e slow action of surface and body forces is equal to the work done by these forces 39 mplication we can vary the surface and body forces to suit our convenience and find the minmum of the difference U between the stored energy and the work done by the forces U Stored Energy Work Done 39 Key idea we don39t have to reach U 0 to produce a very useful approximate anayfka result for loaddeflection EE C245 Introduction to MEMS Design Lecture 7 C Nguyen 101007 20 Energy Density l murmur 39 Strain energy density Jm3 tl gtFind work done in straining material w a39xdsx xaxis normal stress term 1 l 2 03 Sea we Beads E 39 Total strain energy J tl gtIntegrate over all strains normal and shear W Eh 8 r212 G793 y 7323W EE C245 Introduction to MEMS Design Lecture 7 C Nguyen 101007 21 Bending Energy Density l y Neutral l is Wrw um ludw yx transverse displacement of neutral axis 39 First find the bending energy deend in an infinitesimal length dx w 1 0M wail L56 9 in 39X 4 2 47 dz I 2 w He mm is 1 d 2 Pi d a dww war 45609 4 mega 3 EE 6245 Introduction to MEMS Design Lecture 15 C Nguyen 101607 22 my Energy Due to Axial Load rx Y 39 Strain due to axial load S contributes an energy dWsmmh in length dx since lengthening of the different element dx to ds results in a strain ex titanium diam 2 2 dwfwwrlquot H39 l 9 Mu i u ill 6w 5 WWW 9quot m dwawgew aliwle me xifliill EE 6245 Introduction to MEMS Design Lecture 15 C Nguyen 101607 23 Shear Strain Energy 5 3 m g fz 5 ax 46W 0 dx I Shear Modulus 39 See WC Albert Vibrating Quartz Crystal Beam Accelerometerquot Proc ISA Int Instrumentation Symp May 1982 pp 3344 EE C245 Introduction to MEMS Design Lecture 7 C Nguyen 101007 24 iyrtiirlifm Applying the Principle of Virtual Work Basic Procedure t34gtGuess the form of the beam deflection under the applied loads t34gtVary the parameters in the beam deflection function in order to minimize Assumes Sum strain energies point had UZWJZFI39 I39 j i Displacement at point load 39 See Senturia pg 244 for a general expression with distrubuted surface loads and body forces EE C245 Introduction to MEMS Design Lecture 7 C Nguyen 101007 25 Example Tapered Cantilever Beam 0 BTITRHIT 39 Ob iective 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 Wx Top view of wantile WW 1 V 1 WxW 2J W ll 50 taper 2quot x Lu AdJustable 1 1 x parameters f I yr mmlmlze U r4 x F x2 c3x3 39 x 39 Start by guessing the solution 39yx c2 t34gtIt should satisfy the boundary conditions t34gtThe strain energy integrals shouldn39t be too tedious 39This might not matter much these days though since one could just use matlab or mathematica EE C245 Introduction to MEMS Design Lecture 7 C Nguyen 101007 26 Strain Energy And Work By F UWbem F39yg J Bending Energy ally Wxk3 12 dxl fix 202 ich 3 Using our guess Wu W 2 50 Tip Deflection A F 2L62 GEES EE C245 Introduction to MEMS Design Lecture 7 C Nguyen 101007 27 Find c2 and c3 That Minimize U J 39 Minimize U basically find the c2 and c3 that brings U closest to zero which is what it would be if we had guessed correctly 39 The c2 and c3 that minimize U are the ones for which the partial derivatives of U with respective to them are zero av LU i1 3122 3153 39 Proceed 5 First evaluate the integral to get an expression for U 39 2 J53 3E263 3 3 2 f3 2 UEW32 L L liL FcL ML l 1 L 3 f 8 c 4 139 1 c EE C245 Introduction to MEMS Design Lecture 7 C Nguyen 101007 28 Minimize U cont Fm 39 Evaluate the derivatives and set to zero 3 3 ill0 ies F 5 m a L6 362 3 I 4 39 39 3 a U i m3c F 3 m c j 383 8 3 39 Solve the simultaneous equations to get c2 and c3 84 Ric 24 F 3993 3 3993 gt 3 13 m 13 BTW EE C245 Introduction to MEMS Design Lecture 7 C Nguyen 101007 29 The Virtual WorkDerived Solution HG Bh fllilll w A 39 And the solution A 24F 7 2 Jim l 1353 39 Solve for tip deflection and obtain the spring constant 24 5 3 135 L garL J c Cc c ag 3963 39 Compare with previous solution for constantwidth cantilever beam using Euler theory I 3 13 smaller than 3414 J2 taperedwidth case EE C245 Introduction to MEMS Design Lecture 7 C Nguyen 101007 30 Comparison With Finite Element Simulation 7mm quot quot 39 Below ANSYS finite element model with L 500 um wbm 20 um E 170 GPa h2pm W p10um 39 Result from static analysis bk 0471 uNm 39 This matches the result from energy minimization to 3 significant figures EE C245 Introduction to MEMS Design Lecture 7 C Nguyen 101007 31
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