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VariationalMethods

# VariationalMethods - MCE 561 Computational Methods in Solid...

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Unformatted text preview: MCE 561 Computational Methods in Solid Mechanics Variational Techniques Variational techniques will be used to cast governing differential equation(s) into integral variational form and determine approximate solutions using variational methods such as Rayleigh~ Ritz, Galerkin, Least—squares, Collocation, etc., processes that can be categorized as Weighted Residual Methods. This will lead to useful methods for developing ﬁnite element equations. Differential , Variational Approximate 9 Finite Element Equati0n(s) ‘ Formulation Solution Equation(s) Basic Definitions for Field Problems: Boundary—Initial Value Problems The boundary value problem concept involves the solution of some dependent ﬁeld variable u in a region of space Q enclosed by some boundary F. A governing differential equation exists in Q, and this models the physics of the problem. y Boundary F a a, Domain [2 Examples include, 2 2 V2 = 9—22 9—32 = 0 homogeneous type 6x 8y 2 2 ﬂ+ﬂ = [email protected] transient type 8x2 6y2 8t 2 2 S—Z+:—% = ltu eigenvalue type x y Typical boundary conditions on F would be “(x9y)lxygp :f(x3y)’f_ given while typical initial conditions which would exist in .Q at t = 0 would be “(959,0)li 8 Q = gen , g — given These conditions would normally be given or specified in the problem formulation. Continuity Definition A quantity is said to be of class C’“(Q) in domain £2 if all its partial derivatives up to and including m~th order exist and are continuous in Q. Variational Calculus Consider the following deﬁnite integral 1 : fabF(x,y(X),y/(X))dx = Icy) <1) For given constants a and b, the value of 1 depends on the function y(x). The function I, which depends on another function y(x), is called afanctz’onal. The fundamental problem in the calculus of variations (i.e. variational calculus) is concerned with ﬁnding the extremum (maximum or minimum) of functionals. I is said to be linear if1(ocy + (SW) = oc](y) + BI(w) for any scalars or and [3 and functions y(x) and w(x). A functional B(u,v) is said to be bilinear if it is linear in each of its two arguments u and V. If it is assumed that y(x) actually minimizes the integral in equation (1), any function in the neighborhood of y(x) can be represented by y(x) + €n(x) where n(x) is a continuously differentiable function that vanishes at x = a and x = b, and e is a small parameter. y WC) + 67706) In terms of this representation, we can write I = be(x,y+8n,y’+8n’)dx (2) a which assumes its minimum value when e = 0. Consequently from calculus, a necessary condition for an extremum at this point is that lgzo : 0 (3) Now / : fb( 8F d(y+en)+ 8F d0) +€n§)dx a 801+en) de (BO/+6115 (16 b 6F 8F — f [ n+ n/ldx a a(wen) (aw/+611) d] b 6F 8F / —— _ = —- +~—— dx = 0 dEiE—O fa (ayn ay/n) Integrating by parts yields (j b d] b 8F 6F b d 8F —-IE:O =f —~ndx+—-— I —f ——(——>ndx de a 6y 6y/ a d dx ay/ b 6F d 6F =f [—-——-—<-~]ndx = o a (9y dx 63/ For the integral to vanish for all n(x), 6F d 6F _______ __ z 0 ay way» (4) which is called the Euler equation associated with this paiticular problem. Hence the necessary condition to minimize (or maximize) the functional 1, is that F (x, y, y 9 must satisfy the corresponding Euler equation (4). Using the variation notation, let dy = en be the variation in y; then the ﬁrst order variation in F is 5F 2 Qan + 9553/ 8y 6y / So for an extremum of] we must have 61 = obe(x,y,y/)dx : o b b 6F 6F =f 5F(xw“)dx =f [—y + ———<5y’]dx a a 6y (93/ b 6F d 6F 8F b =f [—————(—->5y]dx +——- 6y! = 0 a 6y dx ay’ ay’ a andso 6F a’ 6F —-———~—(—~—) : 0 (9)) dx ay/ (5) b ﬂiéyl =0 By/ a The condition (5)2 is often referred to as the natural boundary condition. What we see from these results is that there exists a differential equation which corresponds to minimizing (or maximizing) a certain integral functional. Hence we can start with a given differential equation and retrace our steps to ﬁnd the corresponding variational problem. Certain approximate techniques can then be applied to the variational problem in order to ﬁnd a solution. This technique proves to be useful in formulating the ﬁnite element equations. Consider next the following functional in two dimensions, I : [f0F(x,y,u,ux,uy,um,uxy,uw)dxdy (6) where u = u(x,y) is a continuous function deﬁned in Q. It can be shown (see Energy Methods in Applied Mechanics, by H.L. Langhaar, John Wiley, 1962) that the Euler equation corresponding to the functional (6), is to the functional (6), is M a 61? 6 BF 82 817‘ 62 31? a2 61? - ( )- ( )+ ( )+ ( )+ ( ) — 0 (7) 82: 8x aux 8y Guy 8x2 Sum, axo’y 6qu 83/2 6a”) Consider the special cases of equation (7) for different types of F: F = é—(uxz + uyz) a Vzu = 0 Laplace Equation F : %(”x2 + uyz) + f(x,y)u => Va 2 ﬂx,y) Poisson Equation F = (u): + 2uxxuyy + uyzy) => _ 2 2 _ 4 _ . . . um“ + Zumy + uyyw — V V u — V u — 0 Bzharmonzc Equatzon Variational Formulation of Boundary-Value Problems Consider now the general partial differential equation in two dimensions, aFmaaF_aaF Bu 6x(6ux) 6y(6uy) HO 6 Q (8) where F = F (x, y, u, ux, u}) and with boundary conditions. (9) where I‘l + F2 = I‘ (total boundary) and 11K and ny are the components of the unit outward normal vector to F, ie. Domain 9 Now let us determine the variational form for the problem given by (8) and (9). First multiply by a so—called weight or test function w(x,y) and integrate over the domain Q 6F 8 GF 6 (BF w m~~—— ~~—— dd =0 [9 [Ga 8x(8ux) 6y(8uy)] x y Since w will be thought of as the variation in u, it is assumed to satisfy w = O on I‘. Next transfer differentiation from u to w by integrating by parts. This step actually employs the use of Green’s Theorem for multi-dimensional domains. 6F aw 6F 6w 6F 8F w———+———+—-——— dxd— w——~n +——n ds=0 fo[ Bu 6x aux 8y Guy] y ji‘ (aux x ) (10) where we have used the results [0 07¢)ng = frnfgdl‘ — f9(Vg)fdQ 01' f0 ggdxdy = frfgnxds—fnfggc: dxdy f0 ggdxdy = fpfgnyds jag? dxdy The development of the boundary terms leads us to consider now two fundamental types of boundary conditions: Natural Boundary Condition . Speciﬁcation of the coefﬁcient of w and its derivatives in the boundary integral, e.g., ——n +—n =q‘ on I‘ auxx auyy 1 Note that this quantity will be referred to as the secondary variable. Essential Boundary Condition Speciﬁcation of the dependent variable in the same form as the test function in the boundary integral, 6. g. u=ft on F2 This quantity will be referred to as the primary variable. 6 Note that the weight or test function w must satisfy the homogeneous form oft/16 essential boundary conditions. Rewriting relation (10) with w = 0 on F2 , yields 6F dw 8F aw 8F [ w + + Q du 8x aux 8y day a’xa’y—fF wcjds (11) which is the desired variational statement of the differential equation (8). Form ( l l) is also referred to as the weak form because the required continuity of the primary variable has been reduced. If F is linear in u, then (1 1) can be written as B(w,u) — 0(w) = O (12) BMW) = f (Elli +%£)dxdy Q ax aux By Buy A 6F t<w> = — [QWfdxdy + [P wqu, = a; =f(x,y> 1 It can be shown that whenever B is bilinear and symmetric i.e, B(u,w) = B(w,u), and if Q is linear, then a quadratic functional associated with B(u,w) - 0(w) is given by 1(a) = 1/2 Batu) - 0(a) (13) and minimization of 1(a) is equivalent to satisfying the weak form for most linear problems (see Shames and Dym, pp. 151-164)‘ Note that not all differential equations admit a functional formulation; however, most admit a weak form and this will be the basis of our ﬁnite element model. Example Determine the weak form and the associated quadratic functional (if it exists) of the following differential equation ddu ——(a—) — cu +x2 = 0 , O<x<l dx dx with boundary conditions, u(0) = 0 , a dug) =1 dx Forming the weak form 1 -1 ﬂ _ 2 __ f0w[dx(adx) cu+x]dx—O (a———— — cwu + wx2)dx — (waﬂﬂ = 0 [1 dw du 1 0 dx dx dx 0 a 33(1) =1 is the natural boundary condition while 11(0) = 0 is the essential boundary condition. So we have 1 dw du 1 (a—— 4 ~ cwu) dx + f we2 dx — w(1) = 0 0 6156 dx 0 or B(w,u)~ 0(w) = 0 since B is bilinear and symmetric, a quadratic functional exists for this problem and is given by 1 [(11) = % f0[a(%:—l)2 ~ c112 + Zux2]dx — u(1) Beam Example Consider the bending of an elastic beam. ’1‘ w ﬁx) “was...“ Bu —-:— -——-~ I. . a. A The transverse deﬂection is denoted by w(x), the slope is given by w'(x), the bending moment is M = E] w"(x), and the shear force is V = E] w’”(x). The governing equation is derived from equilibrium of a beam element, and the resulting equation is d 2 d 2w \$2M?) #06) = 0 where f(x) is the external distributed loading, E is modulus of elasticity, and 1 is the area moment of inertia of the cross—sectional area about the neutral axis. Note for this case w(x) and its derivatives up to fourth order must be continuous, i.e. w is of class C4 in the domain of interest. Recall that the strain energy in an elastic body is deﬁned to the stored internal energy due to the application of external forces doing work on the body. For an elastic body, this stored energy in completely recoverable when the external loads are removed. For the beam bending case including only bending effects, the only non—zero stress is 0x and this is given by the well—know relation 0" 2—D. A I The strain energy for an inﬁnitesimal element is dU = (—21— oxdydz)(8xdx) =%%[email protected]¢ The total strain energy for a beam of length l is then given by U z foil/4 5.0x ax dA dx: [07A if; dA dx 2 2 = uﬂ-dlengI(——dw)2dx 0 2E] 2 0 de Next consider the variational development for a beam problem with end conditions dw(0) w(0) = = 0 essential BC dx 2 Eli—33L = M , V(Q) = 0 natural BC 011:2 A4 0 First multiply the governing equation by a test function v(x) that is of class C2 and vanishes at x = 0 (satisﬁes homogeneous form of EBC), and integrate over the domain 2 dzw f0” v<x> [Q (E1 2;) — f(x)] dx = 0 Integrating by parts two times and incorporating the various boundary conditions yields the follow variational or weak form 2 2 ftEiiZf’ldx-ftvfax— o 6,le dx2 0 6MB) dx M020 The test function v will be viewed as the variation in w, consistent with the boundary conditions on the problem. The variational equation is equivalent to the differential equation. Also note that the continuity requirement on the variational form has been relaxed since it only requires w and v to have continuous derivatives up to order two. Following relation (13), a quadratic form can be obtained for this case by letting v = w, and multiplying the first term by 1/2; this yields I(w)=—;—f09EI(—23:—)2dx—fewfdx— (1142(0) d M0 air 0 dx Notice that this expression is exactly the total potential energy of this particular beam problem. Variational Methods of Approximation We now want to seek an approximate solution to our boundary value problems in the form of a linear combination of suitable approximation functions. An approximate solution is developed such that it satisﬁes the weak form or minimizes the quadratic functional of the problem under study. Several different methods exist which differ in the choice of the approximating functions that are used. Rayleigh-«Ritz Method Consider again the fundamental weak form B(w,u) — 0(w) = O (14) which is equivalent (with B symmetric) to minimizing the quadratic form. 1(a) = B(u,u)— 0(a) (15) l. 2 The Ritz method seeks an approximate solution to the problem in a series form N ”N = 261% + 4’0 (16) j=i where cj are the Ritz coefficients. The method is employed to the weak form (14) with w = (i)i (i=l,2,.,.N), so that 10 N 13(4),. 2;; cjcpj + (1)0) = 0(4),.) , i=1,2,...N J: and if B is bilinear, we get N ; B<¢p¢j> CJ. : 0(4),) _ B(¢ia¢0) (l7) Relation (17) represents a system of N linear algebraic equations for N unknown constants cj, i.e. N Z By. C]. = F], _] = l where F ,. = Q(d)i) - B(cl)i,¢0). The coefﬁcient matrix Bij = B(d)i,(j)j) must be non—singular in order to solve for the unknowns C}. With regard to the quadratic form (15), the Ritz method is employed to minimize [(uN) = 1(cj) = %B(uN,uN) — 0(uN). This is done by requiring that 01(cj) = 0 31(9) 2 0 61(0) 2 0 861 802 acN (18) and thus a system of N equations for N unknown cj's is developed. Equations (18) are identical to (17); however, set (17) is more general since B need not be symmetric for that case. The choice of approximating functions (bi and (ho are made as follows: 1. (I)o is selected to satisfy the speciﬁed non-homogeneous essential boundary conditions of the problem. 2. The set {(bi} (i=1, 2,...N) are selected to satisfy the homogeneous form of the essential boundary conditions (recall w = (bi). 3. {cbi} must produce a non-singular Bij = B(d)i,(l)j). 4. The set {(bi} is complete. These requirements guarantee, for linear problems, convergence of the Ritz method to the exact solution as N a co , i.e. 11 Rum) s [(un) s [(un_l) Wthh glves uN —» uemt as N -+ 00. Example Using the Ritz method, determine the approximate solution to the following beam problem 2 2 L(E1ﬂ)—Jg=o, Osst a’x2 air2 2 2 w(0)= ﬂ‘im) = O,E1d W(L) — M0, dam”! WW) — 0 dx dxl dx de Recall the variation form for this problem was B(v,w) “ 0(v) = 0 where L 2 2 L B(v,w) = f Eli—UM dx , 00)): f fovdx + Mom“ 0 abc2 air2 0 dx Select the approximation functions (bi to satisfy the homogeneous essential boundary conditions (Mo) = (MO) SO W6 try 4),. = xi*1,i=1,2,...N Since the speciﬁed essential boundary conditions were homogeneous we can set (1)0 = 0. With v = (bi, we get L By. = B((|)i,¢j) = f0E1(i+l)ixi”1j(i+1)xj'ldx : EIij(i+1)(j+1)L“j‘1 i+j —1 12 beHZ i+2 mm = + Macon" Considering the N=2 case, we get the two equations 2 12193 EI(4Lcl + 6L 02) = 3 + ZMOL L4 EI(6L2cl + 12L3c2) = f" + 3M0L2 for determining c1 and 02. Solving these yields 54142 + 12M0 fOL c1 — , c2 — — 24E] 12E] and so our approximate solution is given by lZMO + 5f0L2 2 f0 L 3 w2(x) = —-———— x — x 24E] 12E] Likewise if we pursue an N=3 case, the result is 2M +fL2 fL f o o x 2 __ o x 3 + o x 4 4E] 6E] 24E] W305) : which actually coincides with the exact solution. Example Find the Ritz solution of the following two—dimensional Poisson equation boundary value problem. Vsz—l in unitsquare: OSx,ySl T20 0nx=landy=l EAT—=0 on szandy=0 0% Recall for this case 13 1 awaT awar Bw, = —~———+—~—— dxd (T) f0f0(6x8x ayay) y Choose approximation forms for this problem to be (bi : cos (21—1)1tx cos (2i—1)Tty 2 2 trigonometric form (1),. = (1 ~x2)i (1 ~y2)‘ polynomial form which satisfy the essential conditions ((])0 = 0). Using the trigonometric form we get Tl=— cos ~— cos —— 71; 2 2 T2 = 0.3283988 cos 233 cos £23 + 0.001976 cos 2:35 cos 2:3 while for the polynomial form we get * T1 = {Eamon—yo Comparing these approximations at x = y = 1/2 (at the center of the square domain): T, = 0.1642557, T2 = 0.1651874, T]* = 0.1757813, and the and exact solution is 0.1810978. Weighted Residual Method (Galerkin Method) This is a generalization of the Ritz method in that the test functions, w, can be chosen from an independent set of functions, i.e, it will not be necessary for w = (bi. Consider a general differential equation A(u) =f in Q (19) 14 2 2 where A is some differential operator, e. g. A = V2 = i— + _8_ , u is the dependent variable and 6x2 dyz f is given. Similar to the Ritz method, we look for approximate solutions of the form N uN = (to + Z: cjcbj (20) j=l where (1)0 is required to satisfy all speciﬁed boundary conditions ((1)0 = 0 if all speciﬁed boundary conditions are homogeneous), and (I)i must satisfy the general requirements of the Ritz method and the speciﬁed homogeneous form of the boundary conditions. Deﬁne the residual R, by R = A(uN) —f¢ 0 (21) Note the residual depends on space and the coefﬁcients c}. Then the weighted residual is deﬁned by fella-(w) R(x,y,cj) dxdy (22) where it!i (x,y) are called the weighting functions, and are not necessarily the same as the approximating functions, d)i(x,y). The method employs the conditions that the weighted residual be zero, i.e. f 1p, chj) dxdy : 0 (2 OT fnwitAuN) — fidxdy = 0 (23) Now if A is linear, then (23) becomes N :1 [fawnup dxdyicj = fawn A<¢0>1dxdy J' and then WC can write 15 N Z Aijcj :fi j:l where Aij and fi are deﬁned appropriately. Note Aij is not symmetric, i.e. Aij at Aji. When the weighting functions are set equal to the trial or approximation functions, i.e, IIIi = (hi, the scheme is called the Galerkz'n method. For this case if A is linear and of even order, then the Galerkin method reduces to the Ritz method. Example Determine the Galerkin approximation of the following problem. dzu ——+u=x2,0<x<1 a’x2 with boundary conditions _ 313 : u(0)_o,dx(1) 1 The variational problem is given by dw du B w,u = —— - wu dx ( ) 0( dx dx ) 1 0(w) = _ wxzdx O The residual then becomes, R = —-[¢0 + Z 9.45.] +[<1>O + _1 cj¢j1 — x2 N ‘N ll .— K. Choose (1)() = K, (bl = x(2-X), (1)2 = x2(l — 2x/3)]. For the N = 2 case, the results are 16 d2 d2 2 R = c ——[x(2—x)] + c [x2(l ——x)] labc2 2dx2 3 + x + c x(2—x) + c x2(1—3x) — x2 l ' 2 3 =cl[x(2—x)—2] + 02[x2(1—%x) +2(1—2x)] + x — x2 with 111i = (1)1, the weighted residuals are 1 fx(2—x)R dx = o , i=1 0 1 2 fx2(1—_x)R dx = 0 , i=2 0 3 which yield the two equations 4 28 _ 7 —c1 + ——02 — ~— 5 45 60 17 29 1 —c1 + ——c2 : _ 90 315 36 Solving for the constants yields 01 = 623/4306 , c2 = 21/4306. So our approximate solution is uN = u2 = 1.2894x — 0.1398x2 + 0.00325x3 200s(1—x) — sinx cosl _ 2 ”exact + x 2' “2(5) = 0.6093 , u (5) = 0.6112 exact Applications to the Finite Element Method The classical variational techniques of Ritz, Galerkin, etc., which have been presented provide methods of ﬁnding approximate solutions to problems in engineering science. When compared with other techniques such as ﬁnite difference and ﬁnite element methods, the 17 formulational and computational efforts of a direct variational approach would be much less. Furthermore, the variational approximate solutions obtained are continuous functions of position in the domain under study; whereas most other approximate methods produce only piecewise continuous functions or functions known only at discrete points. However, even with these advantages, the variational methods suffer a very important drawback involved with the selection of the approximating functions. Apart from the properties the functions are required to satisfy, there exists no systematic procedure of constructing them. The selection process becomes more difficult when the domain is geometrically complex and/or the boundary conditions are complicated. If the functions are not selected from the domain space of the operator of the equation being solved (completeness property), the resulting solution could be either zero or wrong. For example, if one selects (i)l = l — cos(27tx/L) for a one—parameter Ritz solution of a cantilevered beam of length L, the resulting solution is incorrect because it vanishes at x = L where the deﬂection should be maximum. While this function satisﬁes the essential boundary conditions, w(0) = w'(0) = O of the problem, it cannot be used to represent a solution that is nonzero at x = L . An appropriate choice would be 1 - COS(Tle/L). Thus, without a judicious choice of the approximation functions, the resulting solution can be unacceptable. One cannot develop an automatic procedure for a given equation because the choice of approximation functions diﬂers with the boundary conditions. Even though variational techniques have such a limitation, it appears that these methods could provide powerful means of ﬁnding approximate solutions over domains of known simple shape with predetermined boundary conditions. This brings to mind the use of the ﬁnite element procedure in which a domain of complex shape is subdivided into a group of elements of simple geometry. On each element of the collection, the governing differential equation maybe formulated using any one of the variational methods. The approximation functions can then be systematically generated for each typical elemen...
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