ES240 Solid Mechanics Fall 2007

Transcription

1 ES4 Solid Mechanics Fall 7 Viscoelasiciy References NG McCrum, CP Buckley, and CB Bucknall, Principles of Polymer Engineering, nd ediion, Oxford Universiy Press, 997 A good balance of heory and applicaion for a firs inroducion John J Aklonis and WJ MacKnigh, Inroducion o Polymer Viscoelasiciy Wiley, New York, 983 JD Ferry, Viscoelasic properies of polymers 3 rd ediion, Wiley, New York, 98 RM Chrisensen, Theory of Viscoelasiciy, nd ediion Dover Publicaions, Inc New York, 98 This book describes many boundary value problems Newon s law of viscous deformaion Subjec o a load, a liquid flows The amoun of deformaion is ime-dependen and unlimied When he load is removed, he liquid does no recover is original shape The liquid has no memory To characerize such deformaion, we subjec he liquid o a sae of shear sress, and hen measure he shear srain as a funcion of ime The record, (, allows us o calculae he srain rae d / d For some liquids, we may be able o fi he experimenal daa o a linear relaion beween he srain rae and he sress If we can, we wrie d d = This linear fi is known as Newon s law of viscous deformaion (Some people s names simply sick whaever hey do The fiing parameer,, is known as he viscosiy of he liquid The viscosiy has he dimension of imesress, and is specific o each liquid; eg, waer (η ~ -3 spa and glass (η ~ spa Arrhenius equaion For a given liquid, he viscosiy decreases as he emperaure increases A liquid flows when molecules pass one anoher, a process ha is hermally acivaed Thus, he relaion beween he viscosiy and he emperaure is ofen fi o he Arrhenius equaion /9/7 Linear Elasiciy-

2 ES4 Solid Mechanics Fall 7 = exp q kt ' The pre-facor and he acivaion energy q are parameers used o fi experimenal daa kt is he emperaure in unis of energy If we would raher use he Kelvin (K as he uni for emperaure, he conversion facors beween he Kelvin and common unis of energy is k = 38 x -3 JK - = 863 x -5 evk - We may measure he viscosiy of a liquid a several emperaures, and hen plo he daa on he Arrhenius plo, namely, log vs / kt The acivaion energy is found from he slope of he plo, and he pre-facor he inercep 3 A liquid subjec o a muliaxial sae of sress Newon s law of viscosiy is analogous o Hooke s law of elasiciy (I don know who copied who in his case Following an analogous procedure, we may generalize Newon s law o a muli-axial sae of sress, ij Ofen a liquid is aken o be incompressible; he volumeric srain is ypically small compared o he shear srain Le v( x,x,x 3 be he velociy field The srain rae is a ensor d ij, given by d ij = v i + v j ' x j x i The componens of he sress relae o he componens of he srain rae as ij = v i + v ' j + x j x i ( 3 = d kk ij ij + 3 kk ij This is analogous o he generalized Hooke s law As a special case, under a uniaxial sress, he axial srain rae d / d relaes o he sress as = 3 d d The quaniy H = 3 defines he viscosiy under uniaxial ension /9/7 Linear Elasiciy-

3 ES4 Solid Mechanics Fall 7 4 Types of deformaion How a maerial deforms depends on he consiuion of he maerial We can ypically disinguish he following ypes of behavior: Elasic srain Under a load, he solid deforms insananeously Upon load removal, he solid recovers is original shape insananeously Deformaion occurs as a resul of bond sreching in crysals or chain sraighening in rubbers (cross-linked polymers Molecules do no change neighbors Picorial Represenaion: The spring The experimenal daa is fi o = / E Viscous srain Under a load, a maerial deforms indefiniely Molecules change neighbors Picorial Represenaion: The dashpo The experimenal daa is fi o = H d d Plasic srain The body does no recover is original shape when he load is removed Dislocaion moion Crysal rearranges Aoms change neighbors Srain is zero when < Y, indefinie when > Y Picorial Represenaion: A sliding block Saic fricion Elasic-plasic model Spring in series wih a fricional block Iso-sress The oal srain is he sum of he elasic srain (displacemen of he spring and he plasic srain (displacemen of he block Thermal srain The body expands or conracs when he emperaure changes Anharmonic energy well Picorial Represenaion: The spring changes lengh upon emperaure change Experimenal daa are ofen fi o = ( T T ref 5 Linear viscoelasiciy Creep es of a viscoelasic rod A rod of a polymer is held a a consan emperaure, sress free, for a long ime, so ha is lengh no longer changes wih ime The rod is in a sae of equilibrium A ime zero, we aach a weigh o he rod, and record is elongaion as a funcion of ime Thus, he sress in he rod is a /9/7 Linear Elasiciy-3

4 ES4 Solid Mechanics Fall 7 sep funcion of ime The rod is on is way from one equilibrium sae o anoher equilibrium sae The srain is an increasing funcion of ime, wih he following characerisics: Unrelaxed srain, U Insananeous afer we hang he weigh o he rod, he rod elongaes by a cerain amoun Relaxed srain, R Afer some ime, he molecules reach a new equilibrium configuraion, and he srain no long changes wih ime This is differen from a polymer mel or any liquid 3 Relaxaion ime, τ The ime scale o aain he relaxed srain The ime needed for he rod o change from he old o he new equilibrium sae Skech he funcion (, and mark he above quaniies To a firs approximaion, you may fi he record of he hisory of srain, (, o a funcion ( = U + ( R U, exp + ' - ( / We all do curve fiing, bu some of us are more sophisicaed (We are all in he guer, bu some of us are looking a he sars Oscar Wilde A molecular picure may look like his When a weigh is pu on, molecules change configuraions by wo processes A fas process occurs righ afer he weigh is pu on A slow process akes place over some ime Of course, a polymer may have more han wo processes, and he experimenal record of ( may no fi o he above formula Creep compliance The magniude of he srain should increase wih he weigh ha we hang o he rod Say a ime zero we load wo idenical rods, one wih weigh W, and he oher wih weigh W A any given ime, he elongaion of he second rod is wice ha of he firs rod This lineariy holds when he weigh is small We fi his experimenal observaion by ( = D(, /9/7 Linear Elasiciy-4

5 ES4 Solid Mechanics Fall 7 where is he suddenly applied sress, ( is he srain as a funcion of ime, and D ( is he compliance as a funcion of ime We can reinerpre he feaures of he funcion (, and denoe he unrelaxed compliance by D U, and he relaxed compliance by D R We may fi he experimenal record wih a formula: D, ( D + ( D - D - exp - = U R U ' + ( 3 Spring-dashpo models We nex ry o inerpre he experimenal observaions Your basic insinc migh ell you o look how molecules pass one anoher, and ry o relae various parameers o all he moving and jamming of he molecules This approach has been ried as long as he exisence of molecules was firs appreciaed, and has been dubbed as muli-scale modeling in recen years approach, however, is ime consuming, and ofen no very pracical If you suspend your basic insinc for he ime being, you may learn somehing from simple spring-dashpo models Maxwell model This model represens a maerial wih a spring in series wih a dashpo Because he wo elemens, he spring and he dashpo, are subjec o he same sress, he model is also known as an iso-sress model The oal srain is he sum of he elasic and he viscous srain, so ha d d = + d E d H This is an ODE For a prescribed hisory of sress, he ODE can deermine he hisory of srains For example, in he creep es we prescribe he hisory of sress as a sep funcion: The (, =, < The hisory of he srain will be ' (, < = +, E H We may also wrie he expression as /9/7 Linear Elasiciy-5

6 ES4 Solid Mechanics Fall 7 ( = E + ' (, where = H / E is a ime scale The model gives relaxaion modulus: ( = E + D ' The Maxwell model gives an unrelaxed srain, bu no a relaxed srain The dashpo is unconsrained: he rod will flow indefiniely under he weigh, and will never aain an equilibrium sae The long-ime behavior is liquid-like Kelvin model This model represens a maerial wih a spring in parallel wih a dashpo Because he wo elemens are subjec o he same srain, he model is also known as an iso-srain model The oal sress is he sum of he sress in he spring and he sress in he dashpo, so ha = E + H d d This equaion relaes a hisory of sress o a hisory of srain In he creep es, he sress is zero when <, and is held a a consan level when > We now ry o solve for he hisory of srain, ( The above equaion becomes an inhomogenous ODE wih consan coefficiens The general soluion is ( = E + Aexp ' (, where A is a consan of inegraion, and = H / E is he relaxaion ime A =, he dashpo allows no srain, ( = This iniial condiion deermines he consan A, so ha he hisory of srain is ( = - E exp ' +, ( / The srain is linear in he sress, and he creep compliance is /9/7 Linear Elasiciy-6

7 ES4 Solid Mechanics Fall 7 ( = + E exp D, ' ( - The Kelvin model gives a relaxed srain, bu no an unrelaxed srain In parallel, he spring canno elongae insananeously, and dashpo sops elongae evenually The long-ime behavior is solid like Zener model Neiher he Maxwell nor he Kelvin model is sufficien o describe he represenaive experimenal daa To describe he above observaions, we need a combinaion of he Maxwell and he Kelvin model In he Zener model, a spring of compliance D R D U and a dashpo of viscosiy Η are in parallel, and form a Kelvin uni The uni is hen in series wih anoher spring of compliance D U Le be he srain due o he Kelvin uni, so ha = + H D D R U The oal srain ( is he sum of he srain due o he Kelvin uni and he srain due o he spring in series: = + DU For he creep es, he oal srain is /, ( = DU + ( DR DU ' exp-, ( + where he relaxaion ime is = H ( D, ( D + ( D - D - exp - D R D U The creep compliance is = U R U ' + ( 4 Sress varies as a funcion of ime Now we load a rod wih a known hisory of sress, ( Wha will he hisory of srain, (, be? Of course, we can always measure he hisory of srain experimenally for each given hisory of sress, bu ha is a lo of measuremens We now describe a linear superposiion principle /9/7 Linear Elasiciy-7

8 ES4 Solid Mechanics Fall 7 Say we have conduced a creep es and measured he creep compliance, D ( In general he creep compliance is zero when <, and increases when > Remember, D ( is measured by subjec he solid o a sep sress We nex approximae a hisory of sress by a series of seps of sress: - A ime = u, he sress seps from zero o - A ime = u, he sress makes an addiional sep - A ime = u, he sress makes ye anoher sep The oal srain is he sum of he srains due o all he seps: ( = D( u + ( D( u + ( D( u As an example, consider a rod subjec o a sress a ime zero The sress is kep consan for a while and hen removed a ime The rod gradually conracs o is original lengh The sress is zero when <, held a a consan level beween < <, and dropped o zero again a > The hisory of srain will be ( = D( D( A ime zero, he srain seps up by D U, and hen increases wih ime A ime, he srain seps down by D U, and hen gradually decays o zero Bolzmann superposiion principle: A rod is held a a consan emperaure, sress free, for a long ime, and no longer changes is lengh This lengh is used as he reference o calculae he srain We hen apply a hisory of sress ( The hisory of srain is given by a superposiion: ( = D( u d du du In his equaion, is he ime a which he srain is measured, and u is he variable of inegraion (ie, a dummy variable The lower limi of he inegral is - because he complee sress hisory prior o conribues o he observed srain The upper limi of he inegral is, because he sress applied afer should have no effec on he srain measured a ime /9/7 Linear Elasiciy-8

9 ES4 Solid Mechanics Fall 7 5 The emperaure-dependence of creep Creep is a hermally acivaed process When he creep es is conduced a a higher fixed emperaure, he relaxaion ime is shorer This emperaure dependence may be undersood using any of he spring dashpo models For example, le us look a he Zener model Take ha he compliances of he wo springs are emperaure independen The viscosiy decreases as he emperaure increases, and may fi he Arrhenius equaion H = H exp q kt, where he pre-facor H and he acivaion energy q are parameers used o fi he experimenal daa Recall he relaxaion ime in he Zener model is given by = H ( D R D U Thus, he relaxaion ime fis he Arrhenius equaion, q ' = ' exp kt The shif facor: A real polymer may no fi wih he Zener model Raher, he emperaure effec is implemened by a shif facor Denoe he creep compliance a emperaure T by (, and he creep compliance a emperaure T by D T D T ( Plo D T and D T as funcion of log Each curve has wo plaeaus, one corresponding o he shor-ime unrelaxed compliance, he oher corresponding o he long-ime, relaxed compliance These plaeaus are approximaely independen of ime The wo curves are approximaely idenical afer a horizonal shif Tha is D T ( = D T a T where a T is called he shif facor: The curves D T and D T as funcion of log shif by log a T If we assume he Arrhenius equaion for he relaxaion ime, he shif facor is given by q, a T = exp - ' k + T T ( Creep of a real polymer may involve several acivaed processes, each wih is own acivaion energy Consequenly, he shif facor may show oher emperaure dependence /9/7 Linear Elasiciy-9

10 ES4 Solid Mechanics Fall 7 Acceleraed es: The above shifing relaion says ha he creep compliance a a long ime and a low emperaure is he same as he creep compliance a a shor ime and a high emperaure Relaions like his are known as he ime-emperaure correspondence, and form he basis for acceleraed ess, which are widely pracice in indusries You can shoren he ime of esing by increasing he emperaure Wih he correspondence, you can use he shor-ime response measured a a high emperaure o predic he long-ime response a a low emperaure 6 Thermal srain upon sudden change in emperaure A rod is held a a emperaure T ref for a long ime, so ha is lengh no longer changes wih ime A ime zero, he emperaure is changed from T ref o T, and is subsequenly held consan We measure he lengh of he rod as a funcion of ime Wrie ( = ( ( T T T ref We expec he coefficien of hermal expansion o be a funcion of boh ime and emperaure We can ge some feel for he CTE by using Zener s model again Le U be he CTE of he spring in series, and be he CTE of spring in parallel wih he dashpo The dashpo R U undergoes no hermal srain Le be he srain due o he Kelvin uni When he emperaure increases, he spring hermally expands, bu he dashpo canno respond immediaely Consequenly, he spring will be in compression, and he dashpo will be in ension Because no ne sress is applied, he sum of he sresses in he wo elemens mus vanish: ( ( T T = H R U Ref + DR DU The oal srain is he sum of he srain due o he Kelvin uni and he srain due o he spring in series: ( T = + U T Ref Solving his ODE, we obain he CTE as a funcion of ime / T, ( / + (/ - / - exp - where = H ( = U R U ', + ( D R D U This relaion is he same as ha of he creep compliance /9/7 Linear Elasiciy-

11 ES4 Solid Mechanics Fall 7 For a viscoelasic maerial, he CTE is sensiive o emperaure We can use he same shif facor a T o relae CTEs measured in wo experimens, one wih he emperaure changed from T ref o T, and he oher wih he emperaure changed from T ref o T Thus, ' T ( = ' T at Thermal srain subjec o ime-dependen emperaure: Now we combine he wo ideas: The Bolzmann superposiion principle and he shif facor A rod is held a a emperaure T ref for a long ime, so ha is lengh no longer changes wih ime This lengh is se o be he reference in calculaing he srain The rod is hen subjec o a hisory of emperaure, T ( hisory of srain is T dt du ( = ( du where he shifed ime is d' = a [ T( '] u T, The An algorihm o calculae he hisory of srain in a rod subjec o known hisories of sress and emperaure A rod is held a a emperaure T ref for a long ime, so ha he lengh of he rod no longer changes wih ime This lengh is se o be he reference in calculaing he srain The rod is hen subjec o hisories of emperaure and sress, ( srain is d, - ' T T du ( dt du ( = D ( + + ( du The shifed ime is d' = a [ T( '] u T The shif facor is a funcion of emperaure, a T ( T T and ( The hisory of /9/7 Linear Elasiciy-

12 ES4 Solid Mechanics Fall 7 In summary, for a give polymer, once he creep compliance is measured a one emperaure, D T (, he CTE is measured for one change in emperaure, (, and he shif facor is measured perhaps by creep ess a several emperaures, he above algorihm allows us o calculae he hisory of srain under any prescribed hisories of emperaure and sress 7 Relaxaion es A ime zero, apply a consan srain o a rod, and hen record he sress in he rod as a funcion of ime The sress decreases as a funcion of ime Main characerisics include U he unrelaxed sress is obained insananeously afer he applicaion of he sep-srain R he relaxed sress is obained afer a long ime of he applicaion of he sep-srain 3 is he characerisic ime for sress relaxaion For he same maerial, he characerisic ime for sress relaxaion es is differen from he characerisic ime for he creep es If he applied srain is sufficienly small, he sress is linear in he srain, namely, ( = E( The sress relaxaion modulus is a funcion of ime, E ( A suiable curve o fi he experimenal daa may be ' E( = E R + ( E U E exp R From physical consideraions, he compliance and he modulus a he wo limi saes are relaed: E U = / D U, E R =/ D R Zener s model relaes he relaxaion ime for he wo kinds of ess (sep-sress, and sep-srain: / = E U / E R T The effec of ime-dependen srain A ime -, boh sress and srain are zero Prescribe a hisory of srain ( u, where u represens ime, and he srain can be nonzero since u = Le E( be he relaxaion modulus The hisory of sress is given by a superposiion: /9/7 Linear Elasiciy-

13 ES4 Solid Mechanics Fall 7 ( = E( u d du du In his equaion, is he ime a which he sress is measured, and u is he inegraion variable The lower limi of he inegral is -, because he complee srain hisory prior o conribues o he observed sress The upper limi of he inegral is, because he srain applied afer should have no effec on he sress measured a ime A rod loaded a a consan srain rae As an example, consider ha case ha a esing machine is programmed o apply srain a a consan rae, ( = k, where k is he rae of srain Plo E(, ( and ( We should expec a nonlinear sress-srain relaion Use Bolzmann s superposiion principle o obain he sress hisory: ( = E( u d du du Noe d / du = k Change variable, v = u, and we obain ha Noe ha ( = k E( vdv E ( = E( vdv is he average modulus beween ime and We have ( = E ( ( Because he average modulus decreases wih ime, he sress-srain relaion is sub-linear 8 Cyclic sress and srain and he phase angle A a consan emperaure, program a loading machine o prescribe o a rod a cyclic hisory of srain: ( = sin /9/7 Linear Elasiciy-3

14 ES4 Solid Mechanics Fall 7 Here is he ampliude, and he frequency The srain repeas iself afer a period of ime equal o / Wha will be he sress as a funcion of ime? The answer depends on wha kind of maerials he rod is made of For a purely elasic solid, he sress is proporional o he srain a all ime, ( = E (, where Young s modulus E is a consan Consequenly, when he machine prescribes a sinusoidal srain ( = sin o he rod, he sress in he rod is ( = E sin The sress ampliude is linear in he srain ampliude, = E The sress response replicaes he srain inpu a all ime Tha is, he sress is in phase wih he srain For a purely viscous solid, he sress is proporional o he srain rae, ( = d / d The viscosiy η is a consan independen of ime Consequenly, when he machine is programmed o apply a sinusoidal srain ( = sin o he solid, he sress in he solid is ( = cos This expression can be rewrien as ( = ( sin ( + + ' The sress ampliude is linear in he srain ampliude, = The sress is ou of phase wih he srain; for a purely viscous solid, he sress leads he srain by a phase angle / Now consider a viscoelasic solid When he machine dicaes he solid o srain according o ( = sin, he sress as a funcion of ime looks complicaed in he firs few cycles Aferwards, he sress becomes cyclic, wih he same frequency, bu wih a phase shif This cyclic sress is wrien as ( = sin ( + The phase shif is beween and π/ Energy recovery and energy loss: Le L be he lengh of he polymer rod, and A he crosssecional area When he machine srain he solid according o ( = sin, he solid elongaes as ( = L sin /9/7 Linear Elasiciy-4

15 ES4 Solid Mechanics Fall 7 The sress in he rod is ( = sin ( + Recall an ideniy of rigonomery, ( = sin cos + cossin The force of he machine on he solid is sin + F ( = A ( sin cos + cos sin During a ime inerval d, he solid elongaes by d, and he machine does work Fd = F d d d o he solid In one full cycle he machine does work / F d d d = AL ( sin cos + cos sin (' cosd = AL ' sin / In a complee cycle, he elasic energy sored in he rod repeas is value Consequenly, he ne work done by he machine in a complee cycle mus all become hea The above expression is he energy dissipaed by he solid per cycle The power (energy per uni ime dissipaion per uni volume is sin For a purely elasic solid, he sress is in phase wih he srain, =, and no energy is dissipaed For a purely viscous solid, he sress is ou of phase wih srain, = /, and above formula gives he power dissipaed Sorage modulus and loss modulus When a linearly viscoelasic maerial is subjec o he cyclic srain, ( = sin, he hisory of sress is Define ( = sincos + cossin cos E ' =, sin E '' = The wo moduli are independen of he ampliude of he srain, When he rod is subjec o he cyclic srain ( = sin, he cyclic sress in he rod can be wrien as /9/7 Linear Elasiciy-5

16 ES4 Solid Mechanics Fall 7 ( = ( E'sin E'' cos + The sress is he sum of an in-phase response and ou-of-phase response We call E' sorage modulus, and E'' loss modulus The in-phase sress and srain resuls in elasic energy, which is compleely recoverable The π/ ou-of-phase sress and srain resuls in he dissipaed energy The loss angen is defined as an = E' ' E' Boh he sorage modulus and he loss modulus are funcions of he loading frequency Wrie he wo funcions as E '( and E ''( inuiion abou visco-elasiciy The following experimenal observaions conform o our When he loading frequency is low,, he sorage modulus is he same as he relaxed modulus, E '( = ER, and he loss modulus vanishes, E ''( = How low a frequency is low? The loading frequency should be compared wih he maerial ime scale, he relaxaion ime, The frequency is low when «When he loading frequency is high, >>, he sorage modulus is he same as he unrelaxed modulus, E '( = EU, and he loss modulus vanishes ''( = E 3 A inermediae some frequency, he sorage modulus increases rapidly, and he loss modulus peaks Complex variables Recall Euler s ideniy: exp ( i cos + i sin Wrie complex srain as = ( i = exp We can use his expression in algebra and in solving ODE A he end of he calculaion, we will jus ake he imaginary par, which gives ( = sin Le E be he complex modulus, defined as /9/7 Linear Elasiciy-6

17 ES4 Solid Mechanics Fall 7 E = E' + ie'' The complex modulus is a funcion of he loading frequency, E ( The complex sress is linear in he complex srain, E = Recall he definiions of he sorage modulus and he loss modulus, and we have ( i E = E exp, wih = / imaginary par, we recover he sress E Consequenly, he sress becomes = exp( i + i ( = sin + Compliance represenaion ( Taking he The cyclic experimen can also be represened in erms of compliance Le sress hisory be = sin The srain hisory mus be delayed by a phase facor δ Tha is, = sin ( Define wo kinds of compliances: D' = cos, D'' = sin and he complex compliance hisory is exp( i =, he srain hisory is D = Noe ha ( / D ( E = D = D' id' ' In he complex represenaion, when he sress The Zener model Compliance as a funcion of he loading frequency Recall he ODE for he Zener model: = + H, = + DU D D R U Le = exp( i and = D exp( i We find ha D D U = D R D U + i, where = H ( D R D U Taking he real and he imaginary par, we have /9/7 Linear Elasiciy-7

18 ES4 Solid Mechanics Fall 7 ( = D U + D D R U + (, D '' ( = D D R U + ( D ' ( When he loading frequency is low «, he sorage compliance is he same as he relaxed compliance, and he loss compliance vanishes When he loading frequency is high», he sorage compliance is he same as he unrelaxed compliance, and he loss compliance also vanishes 3 When ~, he sorage compliance decreases rapidly wih, and he loss compliance peaks 9 A general relaion beween modulus measured under various ess The Bolzmann superposiion principle allows us o relae compliance and modulus measured under various esing condiions Say we know he relaxaion modulus E ( ha we subjec he rod o a cyclic srain ( ( i = exp According o he Bolzmann superposiion principle, he sress is d ( = E( u du = i E( u exp( i udu du We now imagine Change he variable, v = u, and we obain ha ( i exp( i E( v exp( i vdv = Consequenly, he complex modulus is ( i E( v exp( i vdv E = Thus, he complex modulus relaes o he relaxaion modulus by he Fourier ransform The complex compliance is similarly relaed o he creep compliance Recall ha ( / D ( E = /9/7 Linear Elasiciy-8

19 ES4 Solid Mechanics Fall 7 Consequenly, all hese moduli and compliances can be convered from one o anoher They conain equivalen informaion Muliaxial sress sae When a solid is subjec o a hisory of muliaxial sress, ( pq ij ( = Dijpq( s ds, s where D ijpq ( is he ensor of creep compliance Example: Plane waves in a viscoelasic solid, he hisory of srain is Viscoelasiciy inroduces a ime scale ino he problem, so ha he waves mus be dispersive Here we limi our analysis o waves wih a sinusoidal profile To illusrae essenial feaures, we will consider waves propagaing in a rod Firs le us lis he governing equaions on he basis of he hree ingrediens of solid mechanics The balance of momenum requires ha u = x ij The srain-displacemen equaion is u = x These wo ingrediens are independen of he maerial The hird ingredien, he maerial law, will invoke visco-elasiciy We will use complex numbers: ( x, = u ( x, exp( i u ( x, = ( x, ( i exp ( x, ( x, ( i = exp /9/7 Linear Elasiciy-9

20 ES4 Solid Mechanics Fall 7 Using he complex ampliudes, he hree ingrediens of solid mechanics are wrien as follows The balance of momenum now requires ha x = u The srain-displacemen equaion now becomes u = x The maerials law is E = ( A combinaion of he hree ingrediens gives ha u E + u = x The soluion is u, ' ( ( x ( = Aexp + i ( x B exp i x E E A wave propagaing in he posiive direcion of x akes he form Wrie u, ( ( x = Aexp ' i ( x + i E ( i E = E exp The above soluion becomes /9/7 Linear Elasiciy-

21 ES4 Solid Mechanics Fall 7 /9/7 Linear Elasiciy- ( + ' ' = + ' ' = i x E i x E A i x i E i A x u ( ( ( ( ( cos exp sin exp exp exp, The wave propagaes a he velociy = sec ' ( E c The ampliude of wave decays exponenially as i propagaes The decay lengh is ( / an c l = This concludes our brief inroducion of viscoelasiciy The book by Ferry conains descripions of experimens and daa The book by Chrisensen conains many solved boundary value problems AQAQUS can be used o solve viscoelasic boundary value problems