9. Structures. Overview. Axially Loaded Beams. Axially Loaded Beams (ctnd.)

Size: px

Start display at page:

Download "9. Structures. Overview. Axially Loaded Beams. Axially Loaded Beams (ctnd.)"

Chad Craig
5 years ago
Views:

1 Overview. Structures icroelectromechaical systems iclude a large umber of devices spaig various physical pheomea for the acquisitio, trasductio, ad commuicatio of iformatio. Oe class of these devices employs movig micromechaical structures such as plates, resoatig beams etc. i their desig This chapter presets a brief overview of the physics ad dyamics of mechaical structures. ore specifically, we will cover beams, catilevers, plates, ad membraes. ially oaded Beams ially oaded Beams (ctd. Beam with Uiform Cross-Sectio W, H, are width, height, ad legth ad is applied uiaial load. The aial force is uiformly applied across the cross-sectio. The resultig tesile stress is: WH The strai i the beam is defied as: δ ε ad is related to the stress through Youg s modulus E: Eε combiig: δ EWH Defiig a sprig costat : we get: δ EWH aial beam Eample: calculate sprig costat for a aially loaded silico beam of legth μm ad square cross sectio of μm o the side: use E 6 GPa EWH (6 ( 6 6 N m

2 ially oaded Beams (ctd. ially oaded Beams (ctd. Beam of Varyig Cross-Sectio The elemet is cosidered small eough to have a uiform cross sectio alog its legth Thus: Δ( Where ( is the cross-sectioal area at the positio of the elemet The total legth chage is give by: δ δ Δ( E( E( E( Statically Idetermiate Beams Cosider a fied beam subjected to a thermal strai ε +α ΔT thermal T α TΔT + E The source of heat would geerate thermal epasio of the material ad would geerate a thermal stress thermal Eα T ΔT (6 (.8 (. 8Pa where α T is coefficiet of thermal epasio thermal ially oaded Beams (ctd. ially oaded Beams (ctd. Statically Idetermiate Beams However, sice the beam is clamped the total strai must be zero. The clampig poits are therefore applyig a aial stress thermal i such a way to make the total strai equal to zero Eample: Suppose the beam of previous eample is fied at both eds ad heated by C. Calculate resultig compressive stress. Use α T.8-6 K - Stress o Iclied Sectios Cosider a uiform beam with aial force cut at a agle θ with respect to its ais We separate the aial force alog two compoets ormal ad parallel to the force. N is the ormal force ad V is the shear force, where: cosθ si θ N V

3 ially oaded Beams (ctd. Bedig of Beams we defie the ormal ad shear stresses as: N V θ τθ θ where θ is the area of the iclied cross-sectio ad is give by: θ cosθ combiig we get: ormal stress θ cos θ ad shear stress τ θ θ θ cos si θ Types of Support ree: o costraits applied to etremity ied: fully costraied i positio ad agle Pied: costraied i positio ot i agle Pied o rollers: costraied i oe directio, free i the other oe ad i agle Bedig of Beams (ctd. Bedig of Beams (ctd. Types of oad Poit load total poit force [N] ' orce per uit width such as ' W [N/m] Distributed load q force per uit legth [N/m] P pressure q P W Reactio orces ad omets The reactio forces ad momets are the force ad momet provided by support uder eteral load i order to fulfill the costraits of the support or eample: a poit load o a clamped free catilever Uder static equilibrium m at ay poit of the beam or istace, at clampig poit : m R thus R

4 Bedig of Beams Bedig of Beams Deformatio due to momets ad shear forces ets ow cosider hypothetical situatio of splittig the beam ito two parts at poit where,l, r ( ad V,l V, r The total force actig o left-had part is: V R V,l,l (sice The momet with respect to fied support is: m R,l V, l ( Coclusio: i static coditios ad m is applicable at ay part i the beam. R Bedig of Beams (ctd. Bedig of Beams (ctd. Pure bedig of a trasversely loaded beam Cosider all possible loads o a differetial elemet of legth. The total force T actig o this elemet is: T q + (V + dv V i order to be zero we get: dv q The total momet applied o elemet with respect to left-had edge is: q T ( + d (V + dv eglectig the product of differetials, i order to be zero we get: d V Note: z is positive dowward. is eterally applied momet is radius of curvature dθ is presumed to have a legth whe the elemet is ot bet

5 Bedig of Beams (ctd. Bedig of Beams (ctd. Because stress ad strai are proportioal, the uiaial stress is give by: ze Note: for z > is egative (compressio for z < is positive (tesio The total momet about z i this segmet is give by: H / Wzdz H / EWz dz E WH Thus uder static coditios - thus : Bedig of Beams (ctd. Bedig of Beams (ctd. Differetial equatio for catilever beam The icremet of beam legth ds alog eutral ais is related to by: ds cos θ ad the slope of the beam at ay poit is: dw ta θ θ or ay give radius of curvature at positio the relatio betwee ds ad the icremetal subteded agle dθ is: ds dθ or small agles cosθ, taθ θ, thus ds ad ad: Combiig: rom previous sectio: Where is the iteral momet dθ dw θ 5

6 Bedig of Beams (ctd. Bedig of Beams (ctd. Catilever beam with poit load at etremity urther derivig we get: V sice d V The iteral momet at ay poit i the beam is give by: ( q ad: sice dv q Thus The boudary coditios are: ( w( dw Trial solutio: w + B + C + D fter applicatio of trial solutio to differetial equatio ad boudary coditios we get: C D 6 Thus: w The deflectio at ed of catilever is: ssigig a sprig costat: Eample: calculate sprig costat of the bet catilever to that of the aially loaded beam situatio (sectio. times smaller!!! Bedig of Beams (ctd. w( catilever EWH (6 ( ( N.6 m What about the stresses i the catilever The curvature is give by: Bedig of Beams (ctd. ( The stress iside the catilever was give by: ze Where z was the distace alog the thickess of catilever The stress is therefore maimum whe / is maimum which occurs at ma The stress is also maimum at the surfaces of the catilever (i.e. z ± H/ where H is thickess of catilever. We get H ma I Give that for a square cross-sectio: I WH We get 6 ma H W 6

7 Bedig of Beams (ctd. Plate Stiffess odulus or our catilever of μm ad square cross-sectio HWμm we get 6 ma few questios o this catilever: ( Pa 7.5 ( N a what force is required at etremity to produce a deflectio of μm? sol :.6(.6μN b what is maimum stress produced by this deflectio? sol : ma 7.5 ( 7.5 (.6 7.7Pa Plate stiffess modulus Util ow we have focused o thi beams with trasverse dimesios small tha their legth ets ow cosider more geeral situatio where thickess is ot beig eglected Cosider a plate uder eteral stress The plate will be sufficietly wide to build up a trasverse stress y to offset the Poisso cotractio i the trasverse directio i other words the plate is thick eough that it will geerate a strai ε y H/H we assume ε y thus we have: υ y ε E where υ is the Poisso ratio of the material ad y is the lateral strai that respods to offset ad stress i lateral directio we also have: y υ εy E combiig we get: Plate Stiffess odulus (ctd. E ε υ this quatity is the plate modulus, ad is geerally % greater tha E P(, y Plate i Pure Bedig Etedig approach employed for thi beams we would derive: + + P(, y dy dy D Where P(,y is a two dimesioal distributive load ad: D EH υ 7

8 Effects of residual stresses ad stress gradiets Stress gradiets i catilevers Deposited films ca have built-i residual stresses that will affect the mechaical behavior of machied mechaical devices. or istace o-uiform residual stresses will cause catilevers to curl. Residual stresses ca also itroduce o-liear respose i the deflectio of doubly-supported beams. Compressive residual stresses ca also cause a doublysupported beam or membrae to spotaeously buckle out of plae. Bedig due to residual stress Cosider a catilever machied out of a material cotaiig residual compressive stresses The aial stress i the beam is approimated by: z (H / where is the stress gradiet. The iteral momet about the middle of the catilever is: H / Wz dz WH 6 Stress gradiets i catilevers (ctd. Stress gradiets i catilevers (ctd. Neglectig trasverse Poisso effects we ca calculate the resultig bedig by treatig the built-i momet as beig eterally applied ad usig the result for the catilever uder bedig momet Immediately after release: Oce the sacrificial layer is removed, the catilever is free to epad to relieve the compressive stress to brig its average to zero The gradiet will however remai This will therefore create a et {word} stress at top surface ad a et compressio stress at bottom surface The catilever will bed upwards to decrease both these stresses. EH EWH I order to take ito accout additioal stiffess due to lateral Poisso effects oe simply replaces E i above results υ 8

9 Stress gradiets i catilevers (ctd. Bedig due to thi overlayer Imagie a catilever with o residual stress upo which a thi stressed overlayer of thickess h is deposited Prior to release the thi overlayer possess a bi-aial tesile stress fter release but prior to bedig a biaial relaatio would be epected to occur the average stress is zero However, give that the film is very thi compared to the beam we assume that eve after release the stress i the films are approimated to remai the same Stress gradiets i catilevers (ctd. Total momet about the cetre of the beam (per uit volume zdz Total with beam ad film we obtai H / H H + Hh h Hh Give the {word} ature of film we eed to compute a effective product (also per uit width H / E ~ I E ~ z dz E ~ z dz + E ~ z dz where E ~ is biaial modulus of beam ad E ~ is biaial modulus of film. ( E ~ I ( E ~ H E ~ eff H + h The radius of curvature is the fied by usig Thickess beam zdz + H h film zdz ( eff Total Thickess ( E ~ I eff H h Numerical applicatio: Cosider a μm log by μm square silico catilever beam. m film with tesile stress of Pa ad Youg s modulus of 5 GPa is deposited o the catilever. Poisso ratio of υ. is assumed for both materials. Calculate the bedig radius of curvature. ESi 6 E ~ 8 Pa υ. Si Efilm E ~ υ film Stress gradiets i catilevers (ctd Pa. 6 Hh ( ( ( ( E ~ I eff H ( E ~ H + E ~ h 7 ( E ~ I eff.55 ( E ~ 7 I eff mm ( [(8 ( + (57 ( ] Residual stresses i doubly supported beams Cosider a doubly supported beam with aial stress that has also bee bet to a radius of curvature (lefthad figure et s reduce this problem to etreme case where beam has bee bet by (righthad figure

10 The geometry of problem allows to postulate that a effective uiform pressure load eist i this structure This pressure cacels out the dowward force that iduces aial tesio The dowward vertical force due to is WH Uder static equilibrium, the et force is zero: We therefore have: Sice: We get: Residual stresses i doubly supported beams P + P P WP lso, sice the distributed load associated to the pressure P is give by we get: H q WH q P W Residual stresses i doubly supported beams We substitute this ito our differetial equatio qtot to get q + q where q Tot q + q ad q is a eterally applied load substitutig q above, we get ( WH q The Euler Beam Equatio ets solve this equatio without the uiaial residual stress q with dw dw w ( w ( Trial solutio is: w C + D + Residual stresses i doubly supported beams Residual stresses i doubly supported beams Substitutig ad applyig boudary coditios we get ( + q w EWH aimum deflectio is at /: q q EWH w ma stress-free EWH wma ets ow solve i presece of aial tesio i beam N WH Trial solutio is: With: N q + C + Dcosh w N EWH C q N q coth( / N q D Nsih( / The maimum deflectio i this case is w ma q cosh( / N sih( / with-stress q w ma N cosh( / sih( /

11 Bucklig of beams We ow cosider the effects of compressive stress o doublyclamped beams. We show that for a sufficietly large compressive stress the equilibrium positio is o loger straight but buckled. The beam is subjected to a poit load i its middle. Bucklig of beams (ctd. Usig Euler beam equatio: N δ( where δ( is a uit impulse fuctio δ(, δ( elsewhere. N is a tesio ad equal to N WH, where is aial stress. We employ a eigefuctio approach to aalyze this system. We defie the eigefuctio ψ ( as the solutios to: N λ ψ The eige fuctios of this form are ψ( cos( ad λ + N The complete trial solutio cosist of a eige fuctio {word} plus a solutio of the homogeeous solutio: Due to symmetry of problem B ad above ca be re-writte as: w( + B + w( C ψ( ( + C ψ ( The boudary coditios impose: + C ψ ( / irst boudary coditio is solved usig Secod coditio forces: dψ Thus Thus Bucklig of beams (ctd. dψ C C / w( / To calculate C we substitute back ito differetial equatio multiply each side by a arbitrary ψ itegrate ad employ epasio of eigefutio to fid: cos( / π π C cos ( ( + N C si( / The solutio becomes, π w( cos ( + ( N The maimum deflectio occurs at The related sprig costat is: Bucklig of beams (ctd. ( + N wma, odd beam, odd ( + N The deomiator vaishes for first term ( π EWH N Correspodig to a critical value of stress called the Euler bucklig limit: π EH euler

Principle Of Superposition

Principle Of Superposition ecture 5: PREIMINRY CONCEP O RUCUR NYI Priciple Of uperpositio Mathematically, the priciple of superpositio is stated as ( a ) G( a ) G( ) G a a or for a liear structural system, the respose at a give