Variational Methods ME 6443
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Date Created: 11/02/15
ME6443 Variational Methods in Engineering Dr Ferri AxA02xL Elongation of an axial tapered rod A fixedfree axial rod of length L elastic modulus E and crosssectional area AX is subjected to a constant load P at its right end as shown On the left end the rod is xed Therefore there is a Geometric or Essential Boundary Condition at X l X 0 u00 2 5u00 lalb The strain energy of the rod is given by md jOL EAxu392dx 2 The virtual work is given by W P EuL 3 Using the static form of Hamilton39s Principle the weak form of the problem can be stated armd aww 55g EAxu392dx P5uLo 4 This is sometimes expressed as the Principle of Minimal Total Potential Energy V I1LEA Zld P 5 Total El0 xu J x L The objective is to nd the function uX having u0 0 essential BC such that I is minimized This is equivalent to the objective of nding a smooth function uX having u00 such that 4 is equal to zero for all smooth variations 5u also satisfying 5u00 Starting from 4 we can perform the variational operation 0L EAxu396u39dx P6uL 0 5 Integrating by parts we get L EAxu395u J0 Apply the essential BC to 5u0 and gather terms EAxu39 PIXL EuL JOL EAxu3939 5udx 0 Setting 5uL temporarily to zero we have loL EAOCW39 5udx 0 By the fundamental lemma of the calculus of variations we conclude that i EAx 0 dx dx Now the boundary term in 7 can be reconsidered EAxu39 PxL EuL 0 Since 5uL is arbitrary we obtain the Natural Boundary Condition EAx P xL EAmu39 6udx P6uL 0 6 7 8 9 10 11 Note that this is just a forcebalance on the right end of the rod The stress which is E times the strain dudX times the area is the tensile force at the rod tip This must equal the applied load P The Strong Formulation of the problem can be summarized EulerLagrange Equation 1 EAGQ 0 dx dx Natural BC EAx P dx xL Essential BC u0 0 12 13 14 The exact solution to this problem is available for the simple case of a linear variation in area with X Consider the function AxA02 xL 15 The integral of 12 gives du EAO2 xLEC1 16 The constant C1 is resolved by recognizing that at X L the lefthand side must equal P Thus C1 P Integrating again we have Jic 17 EA02 xL 2 PL x ux ln 2 C 18 EAO L 2 Applying essential BC we have PL PL u0 Eln2C2 0 C2 Em2 19 0 0 Thus the exact solution is ux ln2 ln2 ln 20 E40 L EA0 EA0 2 xL Weak Formulation Starting from Strong Formulation We will start with the Strong Formulation 12l4 but let39s pretend that we don t know or don t realize that a variational principle applies We can start by multiplying 12 by an arbitrary C0 function 5u having 5u0 0 Such a function is termed an admissible function see more below Then integrate over the domain L EAxu39 395u dx0 21 lo Integrate by parts E14xu396u IOL EAxu396u39dx 0 22 Now impose the requirement that u00 and therefore 5u00 and insert the natural BC 13 into the upperlimit of the boundary term in 22 PauL jOL EAxu396u39dx0 23 Comparing this with 5 we see that it is just the negative Equation 23 is termed the Weak Form of the governing equations for the system It can be stated in words as follows Of all functions ux which satis v the essential boundary condition u0 0 the actual solution is the one that satisfies 23 for all dupe also satis ving the essential boundary condition It can be shown that the strong and weak forms of the problem yield identical results The advantage of 23 is that it involves only the first derivative of u while 12 involves the second derivative of u This is one of the key reasons why 23 is more amenable to the development of approximate solutions The other feature of 23 which is so important for approximate solutions is that it automatically leads to the satisfaction of the natural boundary conditions Functions that satisfy the essential boundary conditions are termed admissible functions Here we will assume that all of our essential boundary conditions are homogeneous ie they require u to be zero as opposed to some specified nonzero value such as u0 1 It is also required that the admissible functions be smooth enough so that all necessary derivatives exist For this particular problem the space of admissible functions can be expressed as Au u00 ueCO 24 Note that admissible functions need only be continuous C0 the derivative must exist but can be discontinuous There is another important class of functions for approximate solutions comparison functions are admissible functions that in addition to satisfying the essential BC also satisfy the natural BC s For the particular problem under consideration Bu u00ueC0EAOu39LP 25 where B is the set of comparison functions A ObviouslyB C A ie all comparison functions are admissible functions but the converse is not true A third class of functions which pertains only to the approximate analyses of dynamic problems such as vibrating strings rods beams etc are eigenfunctions It may be noted that eigenfunctions are comparison functions which in addition also satisfy the governing differential equation As such they typically all functions eigenfunctions comparison functions must possess twice the number of derivatives as admissible functions Denoting the space of eigenfunctions as E it is clear that E CB C A This is displayed pictorially in the gure above Approximate Solutions The starting point for many approximate techniques is the weak formulation Assume that we can expand ux as a linear combination of Ritz functions IjX N um ch jx 26 jl The Ritz functions are admissible functions that derive their name from the Ritz approximation technique Ifthe functions IjX form a quotcomplete setquot then it can be shown that i gt u as N gt 00 In other words as we take more and more terms in the summation we come closer and closer to having the ability to approximate the true solution ux Let A denote the space of all functions it x that can be constructed using the Nterm series 26 Then it must be true that h h h 11 A1 CA2 CA3 CA100CA Another way of visualizing this is in the form ofa Venn diagram It is seen that as N gt oo the space spanned by the Ritz functions it x 3 mohong N 100 j lN becomes closer and closer to the entire N 10 setA of all admissible functions In the Ritz approximation technique we N 1 expand u and 5u using the same set of Ritz functions Thus along with 26 we have the related expressions N u39m jzlcj x 27 N Eu Z ci ix 28 il N Eu39 Z Eci i39x 29 il The coefficients Sci in 28 are arbitrary and independent allowing 5u to be a finitedimensional approximation of an arbitrary admissible function 5ux Substitution of 26 29 into 23 yields N L N N P Z 6ci iL 0 EAx Z cj J x Z gci i39xdx 0 il jl il 01 N N 360 P iLEcjj EAwwxxwxxMx 0 30 Since Eci i l 2 N are arbitrary and independent we must have N kcf i1N 31 g I J l where L I I kg J0 EAx x jxdx 32 f P L 33 In matrix form 3 1 can be written kc f 34 where k is an NxN symmetric quotstiffness matrixquot c is an le vector of unknowns and f is an le quotforcing vectorquot Once 34 is solved for c the approximate displacement field is given by expression 26 We have seen that the Ritz functions j x must be admissible functions Other than that there are two implied requirements The first is that the set of j x must be linearly independent If they are not then the stiffness matrix k will turn out to be singular thus preventing a unique determination of c The second consideration for functions j x is that they should be quotcompletequot in the sense that adding more terms refines the solution See further comments below Beyond the requirement of admissibility and linear independence there are a number of good choices for the functions j x and a multitude of bad choices A fair amount of skill and experience is needed to perform accurate approximations using quotglobalquot Ritz functions On the other hand finiteelement FE approximations which can be viewed as a Ritz approximations using a large number of spatiallysmall simple Ritz functions is less prone to this problem For the tapered rod under consideration here one attractive choice for Ritz functions is monomials of the form jxxLj j 12 N 35 First note that they are all admissible functions They have been normalized so that the tip displacement of each function is unity Also note that a Ritz summation of the form 26 with N 3 for example will encompass the entire space of cubic polynomials except for the constant term They are not comparison functions but they are linearly independent easy to differentiate and easy to integrate Substituting the expression for Ax and 35 into 32 we get L L EA x kg J0 EAX x jxdx to 732 3 w 2 dx 36 After much algebra we can obtain the following closedform expression k z39jz39jl 37 if L z39jz39j l Substitution of 35 into 33 gives fl P iL P for all i 38 ForN l we obtain kn 15 EAoL f1 P gt c1 2PL3EA0 Thus the N l approximation is simply 2P 39 u1x 3Eon ForN2 we obtain k EA0 32 43 f P PL 613 40 a gt c L 43 53 P EA0 313 or using 26 2 2x i EA0 13L 13 L These two approximations are compared in the gure below against the exact solution as given by 20 There is remarkable agreement between the exact solution and the Ritz approximation with N 2 Theoretically adding more terms would re ne the solution further It is also likely that had we used comparison functions rather than admissible functions the convergence would have been even better although there is no guarantee that this will be the case Finally had we omitted the first Ritz function xL from our summation it is unlikely that we would have obtained good results even with a high value of N Given the linear independence of the j x functions it is not possible to sum of monomials of order 2 and higher to get a linear polynomial In other words it is usually a good idea to keep the quotlowerorderquot functions and to retain all terms without skipping any This policy is in keeping with choosing quotcomplete setsquot of Ritz functions Tapered Rod Axial Displacement 07 l l l l uEA0PL 0 l l l l l l l l l 0 01 02 03 04 05 06 07 08 09 1 x over L It is instructive to examine a Ritz solution with N 5 Using Matlab the solution vector is given by c1 50030e001 c2 l2062e001 c3 61002e002 c4 l87l7e002 c5 29947e002 It is seen that the coefficients of the Ritz functions jx generally grow smaller as j increases This confirms our expectation that increasing N tends to make small refinements in the solution It should be noted that continuing to increase N could lead to some counterintuitive numerical results The reason for this is the relatively poor conditioning of the stiffness matrix as N gets larger For example with N 5 the condition number of k is l4l67e005 which is high but ME6443 Ritz Assumed Modes Method Dr Ferri Consider the eXural vibration of a uniform beam clamped at its left end such that the displacement and slope of the beam at X 0 is zero and having a point mass and spring at its right end The stiffness of the spring is speci ed to be k lOEUL3 and the point mass is speci ed to be m 05pAL The kinetic energy of the beam plus the point mass is given by l L 2 l 2 T EIO pAw dx3me where WL is the displacement of the beam at location X L The potential energy is expressed as l L 2 l 2 V EJ0 E w dx3kwL The displacement of the beam is approximated using a Ritz series of the form N Wxt Ell100 I 1 Due to the clamped boundary condition at X 0 the Ritz basis vectors must satisfy the essential boundary conditions llj 0 yj0 0 Here we choose to use the simple polynomial functions ljxxLj1 Substitution of the Ritz series into the kinetic energy eXpression gives T jijZI4qZw jjdm m wwallyjam 3 1 N N L x lj2 1 N N TEZ Zq39lq39j IO 1142 dxm or T Z Zq39lq39jmywhere 11 11 11 1 L lj2 x 1 1 2 m my m J0 1142 dxm pALJ0 d m Thus pAL gt 1 my 1 m I my 2 I lj3 pAL lj3 pAL mi Similarly the substitution of the Ritz series into the expression for potential energy gives 1 L N N 1 N N V 3I0 E1 ZMJ 214 q dx3kZl4La XIIILn 11 11 11 11 1 N n1 391 1 x H 1 N V Z quqj W10 EI dxk or V Z quqjkv 211 11 L L 2 11 11 where z39139z3911391 L EI k ky k TLEIZ dxk L3 yz11j0 116 Thus A k 3 ky wkgt kl W i L3 ij l J EjL3 ij l E Now use Lagrange s equations d3T 3T BVQi FLN dt 34139 39139 39139 This gives a set of N equations that can be written in matrix form m g k g 9 This constitutes a Ndegreeoffreedom NDOF free Vibration problem The free Vibration problem is characterized by natural 3 J 39 and 39 r J 39 r To nd them let qit ci sin0t i lN or 91 g sin03t Substitution into the matrix equation yields w2mkgsinwt9 2 w2mkg9 In terms of the quantities thJ and 13 this becomes wszLmHgv ij g9 pAL4 E De ning 12 w2 we obtain 12r132 9 For nontrivial solutions it is necessary that 12rh 0 There are N positive roots to this characteristic equation which can be ordered as 11lt g lt13 ltlt1N in IE Corresponding to each eigenvalue 7b there is a natural frequency wn Z a and an L eigenvector Q The elements of the eigenvector correspond to the values of the amplitudes of the generalized coordinates qit Let Din be the ithelement of the ntheigenvector Then when the system vibrates in mode n the generalized coordinates are given by ql r c n sinwnr in sinwnr The beam s vibratory shape modeshape can be calculated as N N I I Wx t 2 400910 Z 14 x SIHWJ WnxSlnwnt 11 11 where the nth modeshape is given by N wn x 2 Vi x in i1 Note that the eigenvectors like the modeshapes have arbitrary amplitude Often the g are scaled so that Qn mgn nlN Numerical Values 3 For N3 l 05 and CL 10 the nondimensional mass and stiffness matrices pAL E and are given by M 07000 06667 06429 06667 06429 06250 06429 06250 06111 K 140000 160000 180000 0000 220000 280000 180000 280000 388000 m The natural frequencies are given by natfreqs 41758 170180 798022 Dimensionally the natural frequencies in rads would be given as follows ml 41758 E 2 170180 E 3 798022 E L2 M L2 M L2 M The normalized eigenvectors are given by the columns of the matrix below PHI 22395 183953 371786 13996 320971 1009322 03101 131518 642020 Finally the eigenvectors can be used to nd approximations for the rst three modeshapes x 2 x 3 x 4 w1x22395 13996 03101 L L L 2 3 w2x 183953 320971 131518 2 3 4 mm 371786 1009322 642020 The modeshapes are plotted below Modeshape The Matlab Code used to produce these results is given by 96 9s Flexural vibration of a uniform beam fixed at its left end and d 96 having a spring and point mass at its right en 96 Define system matrices mrat05 96 ratio of point mass to beam mass krat 10 96 ratio of spring stiffness to EILA3 N3 96 Number of Ritz vectors MzerosNN KzerosNN PaJL l Mevmd A We wmiauw 9 th x gasquot qu quadc Jar 5m Fin51111 syshm errrdilcl a 441 fem Lw I Wm L I a damn verJar J m1 I r z 3 L quotm Mmlww 143 4cm m Wmul m at g M nu ma A v A w 2 AG EA Jx m 5441 w Jmma vi rmhu39a 1 L mam 52M L39w 51 31 m X4 91 AJJH mzr 44M m P m4 mam mm emu Qx z ssr I1 m M 6 Jr eronh39d nmlMJ WIIBJWIBI WI Bw A rem UP z r Vamfh m rm 6 and ma 4 A xa f hm nj a We 6 5mm k 4 xL 5 M Wce a lm EA w39Le kw o n5 2 A Juli swam w mum g Men 4 MM N Wx quotquotNU yam 2 Clt 39x 3 1 where mm mus mix0 14 M3 mam mm m and 1 39 M mu5 mini m mamas 26M Bc39s Th Weld Mime m M fwch Snj 3 mus3 5 Win ymt 3 mm m mm WKMQ z Janina 2 w W1 6 thova cm 7 A I4 3 is sq mmz 4 1 A will we Maw gutJig in Jamiquot mam MM 9 SmL L0H Wm RumN Lwf 4 In m mama 6 aqua JLJMI mama 61 m m h Such wwj hut H1 uullud i hunt 401m 3 n3 sme39wmayf Wu m we We 4w 7k quotwdljhle l resin1 a 3m f1 RMWN dvfx Ax 0 lt jw 5p gt 0 J where m m39m 7mm ltuvgt L awn b3 ltu gtfbuvdxx 7 was U ltu gto WAMdMLu u M r Newman mwmwui mm m Jmhr w 25 0JluN Wynn7 5rmj n 6M4 mallw m quotMM M Muwwoz M W A m govum l mama 1 Wm 3 mumam Weak517 1 44 than 95 560 sws omm39n 1ch L m 4 41 Jamal b Fr mm in smner sublime rt DJ Jule m wjumj me 961 w MM 14 15M 1 XMAquot JLN 9 o x m m 5439 m quot feknquot 42m RagaN i m 3mm 1 Lu 5m w a MugL my om yamk gamma ammang m N anms mdwemaa mmer quay 2 mm M w Merv an mum ma mam MJML5 3901 a rimm Iw m u m SA4 I 2qu use 3m a 5mm 71 is Wham h w gtmug WILL quotlaflnmhz m paint quot XJ wme 1mm 56 9073 jb v Wm up L A 1 la1 xumdimlhm h mm at 1 xi sumth 6 7 M s M RIJ39 AINXJ39O jLN 390
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