ES 240 Solid Mechanics Z. Suo. Vibration

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ES 4 Sold Mechancs Z Suo Vbraton Reference JP Den Hartog, Mechancal Vbratons, Dover Publcatons, New York hs eceptonal book, wrtten by a moshenko Medalst, s avalable on amazoncom at $6 When a bar s pulled, the stress feld n the bar s unform When the bar s bent, the stress feld s a functon of poston When the bar vbrates, the stress s a functon of poston and tme that s, the stress n the bar s a tme-dependent feld We wll see how the three elements of sold mechancs play together n ths new contet Before consderng structures, we ll frst recall a famlar problem: system of one degree of freedom A system of one degree of freedom Mass, sprng, dashpot Model a system by a mass, connected to a sprng and the dashpot n parallel he mass s on a frctonless ground he nput to the system s an eternal F t he output of the system s the dsplacement of force on the mass as a functon of tme, ( ) the mass as a functon of tme, () t Free-body dagram Let m be the mass Choose the orgn such that the sprng eerts no force when = Measure the dsplacement from ths poston he sgn conventon: pck one drecton as the postve drecton for the dsplacement When the dsplacement s ( t), the velocty s & () t, and the acceleraton s & & ( t) he sprng eerts a force k, and the dashpot eerts a force c & he sprng constant k and the dampng constant c are measured epermentally Newton s second law, Force = (Mass) (Acceleraton), gves the equaton of moton m & + c& + k = F t In wrtng the ODE, we hs s an ordnary dfferental equaton (ODE) for the functon ( ) put all the terms contanng the unknown functon ( t) terms on the rght-hand sde on the left-hand sde, and all the other Free vbraton ( F = ), and no dampng ( c = ) After vbraton s ntated, no eternal force s appled Wth no dampng, the mass wll vbrate forever he equaton of moton s m & + k = hs s a homogeneous ODE he general soluton to ths equaton has the form ( t) = Asn ωt + B cosωt, where A, B and ω are constants to be determned Insertng the soluton to the ODE, we fnd that ω = k / m he dsplacement repeats tself after a perod of tme equal to π / ω Per unt tme the mass vbrates ths many cycles: k f = ω π = π m We call ω the crcular frequency, and f the frequency Both are also called the natural frequency hs frequency s natural n that t s set by parameters of the system (e, the mass and the sprng constant), rather than the eternal force Determne the constants A and B by the ntal dsplacement and velocty For eample, suppose at tme zero the mass has a known dsplacement ( ) and s released at zero velocty, the subsequent dsplacement s t = cosω ( ) ( ) t //8 Vbraton -

ES 4 Sold Mechancs Z Suo Daly eperence suggests that to sustan a vbraton, one needs to apply a perodc force he above free vbraton lasts forever because our model s dealzed: the model has no dampng In practce, we can make dampng very small by a careful desgn So the model s an dealzaton of a system when the dampng s very small Forced vbraton, no dampng Now consder the eternal force as a functon of tme, F () t Accordng to Fourer, any perodc functon s a sum of many sne and cosne functons Consequently, we wll only consder an eternal force of form F ( t) = F sn Ωt he frequency of the force, Ω, beng determned by an eternal source, s unrelated to the natural frequency ω he equaton of moton now becomes m& + k = F sn Ωt hs s an nhomogeneous ODE It governs dsplacement as a functon of tme, partcular soluton, try the form () t C Ωt = sn, whch gves () t o fnd one F / k () t = sn Ωt ( Ω/ ω) he full soluton to the nhomogeneous, lnear ODE s the sum of all homogeneous soluton and one partcular soluton: F / k () t = A snω t + B cosω + sn Ωt ( Ω/ ω) In general, the drvng frequency Ω and the natural frequency ω are unrelated Consequently, ths moton s not perodc Wth dampng, the homogenous soluton wll de out, but the partcular soluton wll persst Consequently, the partcular soluton s the steady response of the system to the perodc force he steady response has the same frequency as the eternal force, Ω 5 k F 5-5 A B 3 4 5 C Ω /ω - -5 hree behavors of forced vbraton Let us focus on the steady soluton: F / k () t = sn Ωt ( Ω/ ω) Plot the ampltude of the dsplacement as a functon of the drvng frequency Dependng on the rato of the drvng frequency to the natural frequency, Ω / ω, we classfy three behavors as follows //8 Vbraton -

ES 4 Sold Mechancs Z Suo A Ω / ω «he frequency of the eternal force s low compared to the natural frequency he mass oscllates n phase wth the eternal force he behavor s smlar to the mass under a statc load, F / k B Ω / ω ~ he frequency of the eternal force s comparable to the natural frequency he system s sad to resonate wth the eternal force For ths reason, the natural frequency, ω, s also called the resonant frequency At resonance, the mass oscllates wth a large ampltude (he ampltude s fnte when dampng s ncluded) he load pushes at rght tme n the rght drecton C Ω / ω» he frequency of the eternal force s hgh compared to the natural frequency he mass oscllates ant-phase wth the eternal force As the frequency of the eternal force ncreases, the dsplacement ampltude dmnshes: the system s too slow to respond to the eternal force hs phenomenon s the bass for vbraton solaton Push the swng at the rght tme Here s an eample of resonance he length of the strng s l he acceleraton of gravty s g he natural frequency of the swng s ω = g/l When l = m, g = 98 m/s, the perod s = π / ω = 8 s In formulatng the equaton of moton, we normally neglect dampng, so that the mathematcal soluton says that an undamped swng wll persst forever In realty, the swng loses energy by moton of ar, by the frcton at varous onts, etc o sustan the swng, one needs to push from tme to tme If you push at a frequency Ω close to the natural frequency ω of the swng, n the rght drecton, the swng ampltude can even ncrease Vbraton solaton A vbratng machne s modeled by a mass m and a harmonc force, F sn Ωt For eample, f the machne contans a rotatng part, such as a fan, the harmonc force s due to mass mbalance When the machne s placed drectly on the ground, the force F sn Ωt s transmtted to the ground Of course, f you can, you should balance the fan to reduce the harmonc force tself Now suppose that you cannot balance the fan, and has to accept the harmonc force as gven, what can you do to reduce the vbraton transmtted to the ground? he soluton s to put the machne on a complant sprng, and then on the ground In ths case, the force transmtted to the ground s the same as the force n the sprng, k hus, the rato of the force transmtted to the ground to the force generated by the machne s Force transmtted to the ground k = = Force generated by the machne F Ω/ ω o make ths rato small, we need to make ω = ( ) Ω/ ω Recall that the natural frequency s k /m Use a complant sprng to reduce the natural frequency o solate the ground from vbraton, you don t need dampng Free vbraton wth dampng After vbraton s ntated, no eternal force s appled he vbraton decays over tme he effect of dampng s modeled by a dashpot he equaton of moton s m & + c& + k = hs s a homogeneous ODE wth constant coeffcents Q s for qualty factor he three parameters of the system, m, c and k, form a dmensonless group: mk Q = c //8 Vbraton -3

ES 4 Sold Mechancs Z Suo he three coeffcents, m, c and k, are all postve, so that < Q < + hs dmensonless rato s known as the qualty factor It quantfes the mportance of the dampng relatve to that of nerta and stffness he system s over-damped when Q, and undamped when Q = We wll be manly nterested n system where dampng s slght, so that the qualty factor far eceeds, eg n the range Q = 9 Soluton to the ODE Recall that the natural frequency for the undamped sprng-mass system s ω = k / m he above ODE can be rewrtten as ω & + & + ω = Q For any homogeneous ODE wth constant coeffcents, the soluton s of the form ( t) = ep( ρt) where ρ, known as the characterstc number, s to be determned Substtutng ths soluton nto the ODE, we obtan that ω ρ + ρ + ω = Q hs s a quadratc algebrac equaton for ρ he two roots are = ω ρ Q ± ω 4Q We are nterested n the stuaton where the dampng s small, so that the mass vbrates back and forward many tmes hat s, we assume that the qualty factor s a large number Denote q = ω 4Q hs s a real, postve quantty he two characterstc roots are = ω ρ Q ± q where = Recall the Euler equaton: ep ( q) = cosq + sn q he general soluton to the ODE s ωt = Q () t ep ( Asn qt + B cos qt) mk Q =, c k ω =, q = ω m 4Q hs soluton gves the dsplacement as a functon of tme he constants A and B are to be determned by the ntal condtons, namely the dsplacement and the velocty at tme zero Sgnfcant features of damped free vbraton We now eamne features of ths soluton that are mportant n applcaton he above soluton represents a damped wave, wth frequency q For slght dampng, Q >>, the frequency of the damped system s close to that of the undamped system, q ω = k / m //8 Vbraton -4

ES 4 Sold Mechancs Z Suo he vbraton ampltude decays wth tme In one cycle, the tme goes from t to t + π / q, ω the ampltude of the vbraton dmnshes from ep t to ep ω t π + Q Q q hus, π the rato of two consecutve ampltudes ep Q hs rato s the same for any two consecutve mama, ndependent of the ampltude of vbraton at the tme One can measure the rato of the consecutve ampltudes, and thereby determnes the qualty factor For the damped system to vbrate Q cycles, the tme needed s t = π Q / q πq / ω Consequently, the ampltude of the vbraton dmnshes by a factor ep ( π ) Roughly speakng, Q s the number of cycles for the vbraton to de out Forced vbraton, wth dampng he equaton of moton s m& + c& + k = F sn Ωt We have dscussed the homogeneous soluton We now need to fnd one partcular soluton o keep algebra clean, recall the defntons k mk ω = and Q = m c he equaton of moton becomes && & F + + = sn Ωt ω Qω k Now we use a standard trck to smplfy the algebra Wrte F ( t) = F ep( Ωt) Of course we known the eternal force s ust the real part of ths functon Wrte ( t) = ep( Ωt) Plug nto the equaton of moton, and we have F / k = Ω ω Ω + Qω Note that s a comple number he dsplacement of the mass should be = Re[ ep( Ωt) ] A useful nterpretaton of the soluton Wrte = ep φ where the ampltude of the dsplacement s ( ), F / k =, Ω Ω + ω Qω and the phase shft of the dsplacement relatve to the eternal force, φ, s gven by //8 Vbraton -5

ES 4 Sold Mechancs Z Suo Ω Qω tanφ = ( Ω / ω) Plot the dsplacement ampltude as a functon of the forcng frequency At the resonance, Ω / ω, we have = QF / k, namely, the vbraton ampltude s Q tmes of the statc dsplacement hs provdes another epermental way to determne the qualty factor he qualty factor determnes the sharpness of the resonance A large qualty factor corresponds to a sharp resonance hs s mportant to all resonance-based devces o dscuss the phase factor, let s consder three lmtng cases Very stff system Neglect the mass and the vscous terms n the equaton of moton, we have k = F sn( Ωt) Consequently, the dsplacement s n phase wth the eternal force c& = F sn Ωt he dsplacement s Very vscous system he equaton of moton s ( ) F F = cos cω cω force by a phase factor π / ( Ωt) = sn Ωt π hat s, the dsplacement lags behnd the eternal Very massve system he equaton of moton s m& = F sn( Ωt) he dsplacement s F F = sn( Ωt) = sn( Ωt π ) hat s, the dsplacement lags behnd the eternal mω mω force by a phase factor π Plot the phase factor as a functon of the forcng frequency for several values of the qualty factor he phase factor ranges between and π he trend s manly a competton between the three forces: elastc, vscous, and nertal When Ω /ω«, the elastc force prevals, and the dsplacement s n-phase wth the eternal force When Ω /ω», the nertal force prevals, and the dsplacement s π out-of-phase wth the eternal force When Ω / ω, the vscous force prevals, and the dsplacement s π / out-of-phase wth the eternal force Vbraton solaton wth a sprng and a dashpot n parallel he eternal force s F = F ep( Ωt) he dsplacement of the mass s = ep( Ωt), where s the same as above he force transmtted to the ground va the sprng and the dashpot s k + c& = ( k + Ωc) ep( Ωt) Consequently, we obtan that Forcetransmttedtotheground Forcegeneratedbythemachne = Ω + Qω Ω Ω + ω Qω Beam as a sprng Consder a cantlever and a lump of mass at ts end, ust lke a dver on a sprng board For the tme beng, we assume that the mass of the cantlever s neglgble compared to the lumped mass We can obtan the sprng constant as follows Apply a statc force F at the free end of the beam, and the beam end deflects by the dsplacement Accordng to the beam theory, the force-deflecton relaton s //8 Vbraton -6

ES 4 Sold Mechancs Z Suo 3EI F = 3 L Consequently, the sprng constant of the cantlever s 3EI k = 3 L For beams of other end condtons, the sprng constant takes the same form, but wth dfference numercal factors Consult standard tetbooks he beam s stff when Young s modulus s hgh, the moment of cross secton s hgh, and the beam s short If the lumped mass s M, the natural frequency s k 3EI ω = = 3 M L M Frequency meter A cantlever provdes elastcty A lump of mass m provdes nerta wo desgns: () several beams wth dfferent masses, and () a sngle beam wth adustable poston of the mass Place the devce on a vbratng machne, and watch for resonance Determne Young s modulus by measurng the resonant frequency You need to measure Young s modulus of a materal Make a beam out of the materal Make a devce so that you can change the forcng frequency Resonance eperment allows you to determne the natural frequency, and provdes a measurement of Young s modulus he advantage of the beam, as compared to other forms of sprngs, s that t can be made very small by usng the mcrofabrcaton technology Atomc force mcroscope (AFM) Mcro-electro-mechancal systems (MEMS) Accelerometer n an automoble ar bag Longtudnal vbraton of a rod When an elastc rod vbrates, each materal partcle provdes a degree of freedom hus, the rod has nfnte degrees of freedom We wll use the longtudnal vbraton of a rod to llustrate a fundamental phenomenon n structural vbraton: normal modes Consder a rod, cross-sectonal area A, length L, mass densty ρ, and Young s modulus E he rod s constraned to move along ts aal drecton, clamped at one end, and free to move at the other end We neglect dampng + d Reference confguraton u (, t) u ( + d, t) Confguraton at tme t he dsplacement s a tme-dependent feld ake the unstressed rod as the reference state Label each materal partcle by ts coordnate n the reference state o vsualze the moton of the materal partcles, place markers on the rod In the fgure, two markers ndcate two materal partcles, and + d When the rod s stressed, the markers move to new postons he dstance by whch each marker moves s the dsplacement of the u, t materal partcle Denote the dsplacement of the materal partcle at tme t by ( ) //8 Vbraton -7

ES 4 Sold Mechancs Z Suo hree ngredents of sold mechancs We now translate the three ngredents of sold mechancs nto equatons Stran-dsplacement relaton Denote the stran of the materal partcle at tme t by ε (, t) Look at a small pece of the rod between and + d At tme t, the dsplacement of materal partcle s u (, t), and the dsplacement of materal partcle + d s u ( + d, t) he stran of ths pece of the rod s elongaton u ( ) ( + d, t) u(, t) ε, t = = orgnallengh d We obtan the relaton between the stran feld and the dsplacement feld: ε = he partal dervatve s taken at a fed tme Materal law Denote the stress of the materal partcle at tme t by σ (, t) We assume that the rod s made of an elastc materal hat s, for every materal partcle and at any tme, Hooke s law relates the stress to the stran u σ = Eε ( t) A σ, + d ( + d t) A σ, u acceleraton t Newton s second law Draw the free body dagram of the small pece of the rod between A σ, t to the left, and the and + d At tme t, the stress at cross-secton gves a force ( ) stress at cross secton + d gves a force A ( + d, t) σ to the rght he pece of the rod has mass ρ Ad, and acceleraton u / t Apply Newton s law, Force = (Mass)(Acceleraton), to ths pece of the rod We obtan that Aσ ( + d, t) Aσ (, t) = ( ρad)( u / t ), or σ u = ρ t Put the three ngredents together A combnaton of the three boed equatons gves u u E = ρ t hs s a partal dfferental equaton that governs the dsplacement feld u (, t) It s known as the equaton of moton Separate spatal coordnates from tme As a frst dynamc phenomenon, consder free vbraton of the rod In a normal mode, each materal partcle vbrates wth ts ndvdual //8 Vbraton -8

ES 4 Sold Mechancs Z Suo ( ) ampltude U, but all materal partcles vbrate at the same natural crcular frequency ω (radan per unt tme) he dsplacement feld of such a normal mode takes the form u(, t) = U ( ) sn( ωt) he two varables and t are separated We net calculate the ampltude functon U ( ) and the natural frequency ω Substtutng the normal mode nto the equaton of moton, we obtan that E d U = ω U ρ d hs ODE for the ampltude functon U ( ) s homogeneous, and has constant coeffcents he general soluton to the ODE s ( ) ρ + ρ U = Asn ω B cos ω, E E where A and B are arbtrary constants he tme-dependent dsplacement feld s ρ ρ u(, t) = Asn ω + B cos ω snωt E E he three numbers, A, B, ω are yet to be determned Normal modes he above soluton satsfes the equaton of moton We now eamne the boundary condtons he rod s constraned at the left end, so that the dsplacement vanshes at = for all tme: u (, t) = he rod moves freely at the rght end, so that the stress vanshes at the rght = L for all tme Recall that σ = Eε = E u / he stress-free boundary condton means that u = L, alltme Apply the two boundary condtons, and we obtan that B =, ρ cos ρ A ω ωl = E E Possble solutons are as follows Frst, A = hs soluton makes the dsplacement vansh at all tme Second, ω = hs soluton makes the natural frequency vansh, so that the rod s statc he thrd possblty s of most nterest to us: cos ρ = E ωl hs requres that ρ π 3π 5π ωl =,, E or π E 3π E ω =, ω = L ρ L ρ he rod has nfnte many normal modes: st mode: π E ω = L ρ / / π u t A sn L = / 5π E, ω3 = L ρ, (, ) = sn ω t, σ (, t) = A E cos ω t / π π sn L L //8 Vbraton -9

ES 4 Sold Mechancs Z Suo / 3π E 3π 3π 3π nd mode: ω =, u(, t) = A sn snωt, σ (, t) = A E cos snωt L ρ L L L and so on so forth When ected, the moton of the rod s a superposton of all the normal modes 5 u/ A 8 6 4 σ EA / L 5 75 5 5 4 6 8 / L 4 6 8 / L 4 u A 5 / L 4 6 8 σ EA /L /L 4 6 8-5 - - -4 he frst mode s known as the fundamental mode It has the lowest frequency: ω E f = = π 4L ρ For steel, E = GPa, ρ = 78kg A rod of length L = m has a frequency about khz he audble range of an average human beng s between Hz to khz Vsualze engenmode Please take a look at a vdeo that shows the egenmodes of a plate (http://mechancaorg/node/4) Fnte element method for the dynamcs of an elastc sold Weak statement of momentum balance In three-dmensonal elastcty, momentum balance leads to three PDEs σ u + b = ρ t n the body, and the three stress-tracton relatons σ n = t on the surface of the body he momentum balance holds true f //8 Vbraton -

ES 4 Sold Mechancs Z Suo w dv holds true for every test functon w ( ) ρ u / t σ = ( b ρ u / t ) wdv + as the nerta force t w da It mght help you memorze the above by regardng Fnte element method Dvde a body nto many fnte elements Interpolate the dsplacement feld n an element as u = Nq, where u s the tme-dependent dsplacement feld nsde the element, and q s the tmedependent nodal dsplacement column he shape functon matr N s the same as that for statc problem he stran column s ε = Bq he stress column s σ = DBq hese steps are the same as n statc problems Insert these nterpolatons nto the weak statement Compared to the statc problem, the only new term s nerta term: ( ρ u / t ) δudv = q& mδq he sum s carred over all elements he mass matr for each element s m = ρ N NdV Let Q be the column of dsplacements of all the nodes n the body he global stffness matr and the global force column are assembled as before he global mass matr s assembled n a smlar way he weak statement takes the form ( MQ& + KQ F) δq =, whch must holds true for every varaton n the dsplacement column, δ Q We obtan that && M Q + KQ = F hs s a set of ODEs for the dsplacement column Q ( t) Normal mode analyss Consder free vbraton, where F = In a normal mode, all nodes vbrate at a sngle frequency, and each node vbrates wth ts ndvdual ampltude hat s, a normal mode takes the form Q ( t) = U snωt, where U s the column of the ampltude of the nodal dsplacements, and ω s a natural frequency Insert ths epresson nto M Q&& + KQ =, and we obtan that KU = ω MU hs s a generalzed egenvalue problem Both the mass matr and the stffness matr are postve defnte A n-dof system has n dstnct normal modes For eample, the frequency equaton for a DOF system s K K U M M U = ω K K U M M U or //8 Vbraton -

ES 4 Sold Mechancs Z Suo K K ω M ω M o obtan nonzero ampltude column [ U,U ] K ω M U = K ω M U, the determnant of the matr must vansh K ω M K ω M det = K ω M K ω M hs s quadratc equaton for ω, and has two solutons, each correspondng to a normal mode Appromate the rod as a -DOF system he computer readly assemble the stffness matr and mass matr, and performs normal mode analyss o gan some empathy for the computer, we now smulate the computer, and determne the fundamental frequency by usng the PVW We use the sngle element to represent the rod he dsplacement of the node on the left vanshes he dsplacement of the node on the rght, q ( t), s the degree of freedom Interpolate the dsplacement nsde the rod by a lnear functon u (, t) = q() t L From the eact soluton, we know the ampltude functon s snusodal, not lnear hus, our assumpton s wrong, resultng an appromate frequency Calculate the stran n the rod by ε = u /, gvng ε = q / L Usng Hooke s law, we obtan the stress σ = Eq / L he varaton n the dsplacement s δ u = δq L he varaton n the stran s δε = L δq Insertng the above nto the weak statement, we obtan that L L Eq δq q δq Ad = L L ρ & Ad L L Evaluate the ntegrals, and we obtan that ρ L q& + Eq δq = 3 he PVW requres that ths equaton hold true for every varaton δ q, so that hs ODE governs the functon wth the frequency ρ L q & + Eq = 3 q() t he soluton to ths ODE s snusodal, q t = constant snω, ( ) t //8 Vbraton -

ES 4 Sold Mechancs Z Suo 3 E ω = L ρ We make the followng comments hs appromate frequency takes the same form as the eact frequency, ecept for the numercal factor: 3 vs π / he appromate frequency s somewhat larger than the eact frequency hs trend s understood as follows In obtanng the appromate soluton, we have constraned the dsplacement feld to a small famly (e, the lnear dstrbuton) he constrant makes the rod appear to be more rgd, ncreasng the frequency By appromatng the rod wth a sngle degree of freedom, we can only fnd one normal mode If we want to fnd hgher modes, we must dvde the rod nto more elements Vew a partcular normal mode as a standng wave of some wavelength o resolve ths normal mode, the element sze should be smaller than a fracton of the wavelength If we dvde the rod nto many elements, the resultng dsplacement dstrbuton for the fundamental mode wll approach to the snusodal functon, and the frequency wll approach to the eact value In a homework problem, you wll apprecate these comments by dvdng the rod nto two lnear elements Propertes of the normal modes (IM Gel fand, Lectures on Lnear Algebra, Dover Publcatons) What can normal modes do for us? o answer ths queston, we need to learn a few more facts about the normal modes he normal mode analyss leads to an egenvalue problem: KU = λmu he natural frequency ω corresponds to the egenvalue, λ = ω he ampltude column U corresponds to the egenvector he egenvalues are roots to det [ K λ M] = hs s a polynomal of degree n for a system of n degrees of freedom Let the egenvalues be λ λ, λ,, and ther assocated egenvectors be, 3 λ n, U U, U,, U n, 3 o avod a certan subtle pont, we assume that the n egenvalues are dstnct Because the two matrces M and K are symmetrc and postve-defnte, ths egenvalue problem has several specfc propertes All egenvalues are real and postve numbers Let s say that an egenvalue λ mght a comple number, so that ts assocated egenvector U would be a comple column Denote the comple-conugate of U by U Multply U and KU = λmu, gvng U KU = λu MU he matr M s real and symmetrc, so that U MU = M UU + MU U + + M( UU + UU ) + Copnsequently, U MU s a real number Smlarly, U KU s a real number hs proves that λ s a real number he egenvector U must also be real Because M and K are postve defnte, U KU and U MU are postve numbers Consequently, λ s also postve //8 Vbraton -3

ES 4 Sold Mechancs Z Suo Egenvectors assocated wth dfferent egenvalues are orthogonal to one another Let λ and λ be two dfferent egenvalues, and U and U be ther assocated egenvectors Orthogonalty here means that If ths equaton holds, we also have U KU U MU o prove the orthoganalty, multply KU Smlarly, multply KU U U = = λ U MU KU = λ MU by, gvng U KU = = λ MU by, gvng = λ U = λ U Because M and K are symmetrc and real, U KU = U KU and U MU = U MU he dfference of the above two equatons gves that = λ λ hs proves that U MU = MU MU ( ) U MU Normalze each egenvector Each egenvector s determned up to a scalar We can choose the scalar so that the egenvector s normalzed, namely, U MU =, and U KU = λ U MU = λ Forced vbraton Ecte a system by a perodc force column, F t = sn Ω, ( ) t F where F s the ampltude of the force column, and Ω s the forcng frequency he equaton of moton becomes M Q & + KQ = F sn Ωt We want to determne Q () t Wrte Q ( t) as a lner superposton of the egenvectors: Q t = a t U + a t U + + a n t U U () ( ) ( ) ( ) n, Multplyng the equaton of moton by a partcular egenvector, a&& + ω a = b sn Ωt, U, we obtan that where b U = F hs equaton s dentcal to the equaton of moton of a -DOF system hus, the n-dof system s a superposton of n normal modes, each actng lke a -DOF system hs s an nhomogeneous ODE he full soluton s the sum of all homogeneous soluton and one partcular soluton b a () t = Asnω t + B cosωt + sn Ωt ω Ω Wth dampng (whch s neglected here), the homogenous soluton wll de out, but the partcular soluton wll persst he partcular soluton s //8 Vbraton -4

ES 4 Sold Mechancs Z Suo a b () t = sn Ωt ω Ω Intal value problem Normal modes are useful f you only care about a few modes, say to avod resonance or perform vbraton control If you are nterested n dynamc dsturbance of some wavelength much smaller than the overall structure sze, such as mpact and wave propagaton, you may want to evolve the dsplacement feld n tme Gven the ntal dsplacements and veloctes, as well as the eternal forces, M Q&& + KQ = F s a set of ODE that evolves the nodal dsplacements over tme hs s a standard numercal problem ABAQUS provdes ths opton //8 Vbraton -5