Finite Element Analysis - Thermomechanics of Solids Part 11 pps

153 Solution Methods for Linear Problems 11.1 NUMERICAL METHODS IN FEA 11.1.1 S OLVING THE F INITE -E LEMENT E QUATIONS : S TATIC P ROBLEMS Consider the numerical solution of the linear system Kγγ γγ = f , in which K is the positive-definite and symmetric stiffness matrix. In many problems, it has a large dimension, but is also banded. The matrix can be “triangularized”: K = LL T , in which L is a lower triangular, nonsingular matrix (zeroes in all entries above the diagonal). We can introduce z = L T γγ γγ and obtain z by solving Lz = f . Next, γγ γγ can be computed by solving L T γγ γγ = z . Now Lz = f can be conveniently solved by forward substitution. In particular, Lz = f can be expanded as (11.1) Assuming that the diagonal entries are not too small, this equation can be solved, starting from the upper-left entry, using simple arithmetic: z 1 = f 1 / l 11 , z 2 = [ f 2 − l 21 z 1 ] / l 22 , z 3 = [ f 3 − l 31 z 1 − l 32 z 2 ] / l 33 , … . Next, the equation L T γγ γγ = z can be solved using backward substitution. The equation is expanded as (11.2) 11 l ll lll ll l z z z z f f f f nn nnn n 11 21 22 31 32 33 12 1 2 3 1 2 3 00 0 . . . . .                                                 =                         . ll l l lll ll l n nn nn nn nn nn nn n 11 12 1 22 22 21 2 11 1 1 2 3 0 00 0 00 0 0 . . . . . . . . ,,, ,, −− −− − −− −                                   γ γ γ γ               =                         − − f f f f n n n 1 2 1 . . . 0749_Frame_C11 Page 153 Wednesday, February 19, 2003 5:13 PM © 2003 by CRC CRC Press LLC 154 Finite Element Analysis: Thermomechanics of Solids Starting from the lower-right entry, solution can be achieved using simple arith- metic as γ n = f n / l nn , In both procedures, only one unknown is encountered in each step (row). 11.1.2 M ATRIX T RIANGULARIZATION AND S OLUTION OF L INEAR S YSTEMS We next consider how to triangularize K . Suppose that the upper-left ( j − 1) × ( j − 1) block K j − 1 has been triangularized: In determining whether the j × j block K j can be triangularized, we consider (11.3) in which k j is a ( j − 1) × 1 array of the first j − 1 entries of the j th column of K j . Simple manipulation suffices to furnish k j and l jj . (11.4) Note that λλ λλ j can be conveniently computed using forward substitution. Also, note that l jj = The fact that K j > 0 implies that l jj is real. Obviously, the triangularization process proceeds to the ( j + 1) st block and on to the complete stiffness matrix. As an example, consider (11.5) Clearly, For the second block, (11.6) γγγ γγ n n n nnn n n nnn nn n nn fl l fl l l −−− −−− −− −−−−− =− =− − 1111112 22 21122 []/,[ ]/, . ,, , , , K KLL T jjj−−− = 111 . K Kk kk L0 L 0 T T T T j jj jjj j jjj jj jj l l =         =                 − − − 1 1 1 λλ λλ , k jjj jj jj j j lk = =− − L T 1 λλ λλλλ k jj j j j − − − kk T1 K 1 . A 3 =             1 1 2 1 3 1 2 1 3 1 4 1 3 1 4 1 5 . LL 11 T =→1. 10 1 0 1 222 2 22 1 2 1 2 1 3 λ λ l l                 =         , 0749_Frame_C11 Page 154 Wednesday, February 19, 2003 5:13 PM © 2003 by CRC CRC Press LLC Solution Methods for Linear Problems 155 from which λ 2 = 1/2 and Now (11.7) We now proceed to the full matrix: (11.8) We conclude that l 31 = 1/3, l 32 = = 1/5 − 1/9 − 1/12 = 11.1.3 TRIANGULARIZATION OF ASYMMETRIC MATRICES Asymmetric stiffness matrices arise in a number of finite-element problems, includ- ing problems with unsteady rotation and thermomechanical coupling. If the matrix is still nonsingular, it can be decomposed into the product of a lower-triangular and an upper-triangular matrix: (11.9) Now, the j th block of the stiffness matrix admits the decomposition (11.10) in which it is assumed that the ( j − 1) th block has been decomposed in the previous step. Now, u j is obtained by forward substitution using L j−1 u j = k 1j , and λλ λλ j can be obtained by forward substitution using Finally, for l 22 2 13 12 1 12=− =/(/) / . L 2 10 12 1 12 =         // . 1 100 112 0 1 0112 00 1 212 2 1 2 1 3 1 2 1 3 1 4 1 3 1 4 1 5 33 1 2 31 32 33 1 2 31 32 33 1 2 31 1 2 1 3 31 32 31 31             = =                         =+ + LL T // // / lll l l l l ll ll lllll 32 31 2 32 2 33 2 12/ ++             112/,l 33 2 1 180 . KLU= . K Kk k L0 U 0 LU L U T T T TT j jj jjj j jjj jj jj jj jj j j j j jj jj k l l =         =                 = +         − − − −− − − 11 2 1 1 11 1 1 λλ λλλλ u u u u u Uk T jj j− = 12 λλ . uu jj jj jj j j lk=− λλ T , 0749_Frame_C11 Page 155 Wednesday, February 19, 2003 5:13 PM © 2003 by CRC CRC Press LLC 156 Finite Element Analysis: Thermomechanics of Solids which purpose u jj can be arbitrarily set to unity. An equation of the form Kx = f can now be solved by forward substitution applied to Lz = f, followed by backward substitution applied to Ux = z. 11.2 TIME INTEGRATION: STABILITY AND ACCURACY Much insight can be gained from considering the model equation: (11.11) in which λ is complex. If Re( λ ) > 0, for the initial value y(0) = y 0 , y(t) = y 0 exp(− λ t), then clearly y(t) → 0. The system is called asymptotically stable in this case. We now consider whether numerical-integration schemes to integrate Equation 11.11 have stability properties corresponding to asymptotic stability. For this pur- pose, we apply the trapezoidal rule, the properties of which will be discussed in a subsequent section. Consider time steps of duration h, and suppose that the solution has been calculated through the n th time step, and we seek to compute the solution at the (n + 1) st time step. The trapezoidal rule is given by (11.12) Consequently, (11.13) Clearly, y n+1 → 0 if < 1, and y n+1 → ∞ if > 1, in which |·| implies the magnitude. If the first inequality is satisfied, the numerical method is called A-stable (see Dahlquist and Bjork, 1974). We next write λ = λ r + i λ i , and now A-stability requires that (11.14) A-stability implies that λ r > 0, which is precisely the condition for asymptotic stability. Consider the matrix-vector system arising in the finite-element method: (11.15) dy dt y=− λ , dy dt yy h yyy nn nn ≈ − −≈− + + + 1 1 2 ,[]. λ λ y h h y h h y nn n + = − + = − +       1 0 12 12 12 12 λ λ λ λ / / / / 12 12 − + λ λ h h / / 12 12 − + λ λ h h / / 1 1 1 22 22 2 2 2 2 −+ ++ <                     λ λ λ λ r h i h r h i h . MDK0 ˙˙ ˙ ,( ˙ ( ˙ ,γγγγγγγγγγγγγγ++= = = 0 , 0)) 00 0749_Frame_C11 Page 156 Wednesday, February 19, 2003 5:13 PM © 2003 by CRC CRC Press LLC Solution Methods for Linear Problems 157 in which M, D, and K are positive-definite. Elementary manipulation serves to derive that (11.16) It follows that and γγ γγ → 0. We conclude that the system is asymptotically stable. Introducing the vector the n-dimensional, second-order system is written in state form as the (2n)-dimensional, first-order system of ordinary differential equations: (11.17) We next apply the trapezoidal rule to the system: (11.18) From the equation in the lower row, p n+1 = [γγ γγ n+1 − γγ γγ n ] − p n . Eliminating p n+1 in the upper row furnishes a formula underlying the classical Newmark method: (11.19) and K D can be called the dynamic stiffness matrix. Equation 11.19 can be solved by triangularization of K D , followed by forward and backward substitution. 11.3 NEWMARK’S METHOD To fix the important notions, consider the model equation (11.20) Suppose this equation is modeled as (11.21) d dt 1 2 1 2 ˙˙ ˙˙ .γγγγγγγγγγγγ TT T MK D0+       =− < ˙ γγ→0 p = ˙ ,γγ M0 0I p DK I0 p f 0               + −               =       • γγγγ . M0 0I pp DK I0 pp ff 0         − −           + −         + +           = +         + + + + + 1 1 1 2 1 2 1 2 1 1 1 1 1 h h nn nn nn nn nn () () () () () . γγγγγγγγ 2 h KrKMDK rMDK MDp ff DD γγ nn nnnnn hh hh y hh h ++ + + ==++       =+ −       ++       ++ 11 2 1 22 1 24 24 22 4 , () dy dx fy= (). αβ γδ yyhff nnnn++ ++ += 11 0[]. 0749_Frame_C11 Page 157 Wednesday, February 19, 2003 5:13 PM © 2003 by CRC CRC Press LLC 158 Finite Element Analysis: Thermomechanics of Solids We now use the Taylor series to express y n+1 and f n+1 in terms of y n and f n . Noting that and we obtain (11.22) For exact agreement through h 2 , the coefficients must satisfy (11.23) We also introduce the convenient normalization γ + δ = 1. Simple manipulation serves to derive that α = −1, β = 1, γ = 1/2, δ = 1/2, thus furnishing (11.24) which can be recognized as the trapezoidal rule. The trapezoidal rule is unique and optimal in having the following three char- acteristics: It is a “one-step method” using only the values at the beginning of the current time step. It is second-order-accurate; it agrees exactly with the Taylor series through h 2 . Applied to dy/dt + λ y = 0, with initial condition y(0) = y 0 , it is A-stable whenever a system described by the equation is asymptotically stable. 11.4 INTEGRAL EVALUATION BY GAUSSIAN QUADRATURE There are many integrations in the finite-element method, the accuracy and efficiency of which is critical. Fortunately, a method that is optimal in an important sense, called Gaussian quadrature, has long been known. It is based on converting physical coordinates to natural coordinates. Consider Let ξ = [2x − (a + b)]. Clearly, ξ maps the interval [a, b] into the interval [−1,1]. The integral now becomes Now consider the power series (11.25) from which (11.26) The advantages illustrated for integration on a symmetric interval demonstrate that, with n function evaluations, an integral can be evaluated exactly through (2n − 1) st order. ′ =yf nn ′′ = ′ yf nn , 02 2 =+ ′ + ′′ ++ ′ + ′′ + ′ αβγδ [/][].yyhyh yhyyhhy nn n n nn n αβ αγδ α γ += ++= +=000 2/. yy h fy fy nn nn + + − =+ 1 1 1 2 [()()], ∫ a b fxdx() . 1 ba− 1 1 1 ba fd − ∫ − () . ξξ f () 01 2 2 3 3 4 45 ξααξαξαξαξαξ =+ + + + + + 5 L, fd(. ξξ α α α )20 2 3 0 2 5 0 1 1 13 5 − ∫ =++ ++ ++L 0749_Frame_C11 Page 158 Wednesday, February 19, 2003 5:13 PM © 2003 by CRC CRC Press LLC Solution Methods for Linear Problems 159 Consider the first 2n − 1 terms in a power-series representation for a function: (11.27) Assume that n integration (Gauss) points ξ i and n weights are used as follows: (11.28) Comparison with Equation 11.26 implies that (11.29) It is necessary to solve for n integration points, ξ i , and n weights, w i . These are universal quantities. To integrate a given function, g( ξ ), exactly through ξ 2n−i , it is necessary to perform n function evaluations, namely to compute g( ξ i ). As an example, we seek two Gauss points and two weights. For n = 2, (11.30) From (ii) and (iv), leading to ξ 2 = − ξ 1 . From (i) and (iii), it follows that − ξ 2 = ξ 1 = The normalization w 1 = 1 implies that w 2 = 1. 11.5 MODAL ANALYSIS BY FEA 11.5.1 M ODAL DECOMPOSITION In the absence of damping, the finite-element equation for a linear mechanical system, which is unforced but has nonzero initial values, is described by . (11.31) Assume a solution of the form which furnishes upon substitution (11.32) The j th eigenvalue, λ j , is obtained by solving and a corre- sponding eigenvector vector, γγ γγ j , can also be computed (see Sample Problem 2). g n n (. ξααξ αξ ) 12 2 21 =+ ++ − L gd g w w w w ii i n i i n ii i n nii n i n (( . ξξ ξ α α α α ξ )) 1 1 1 12 1 2 1 − === − = ∫ ∑∑∑ ∑ ==+++ 1 21 L ww w w n w i i n i i n ii i n ii n i n ii n i n == = − = − = ∑∑ ∑ ∑ ∑ == = = − = 11 2 1 22 1 21 1 10 23 2 21 0,, /,, , . ξξ ξ ξ K ww w w ww ww 12 1122 11 2 22 2 11 3 22 3 20 2 3 0 += + = += += (i), (ii) (iii), (iv) ξξ ξξ ξξ w 11 1 2 2 2 0 ξξ ξ [],−= 1/ 3. MK0 ˙˙ ,( , ˙ ( ˙ γγγγγγγγγγγγ+= = = 00 00 )) γγγγ = ˆ (),exp t λ [] ˆ .KM0+= λ 2 γγ det( ) ,KM+= λ j 2 0 0749_Frame_C11 Page 159 Wednesday, February 19, 2003 5:13 PM © 2003 by CRC CRC Press LLC 160 Finite Element Analysis: Thermomechanics of Solids For the sake of generality, suppose that λ j and γγ γγ j are complex. Let denote the complex conjugate (Hermitian) transpose of γγ γγ j . Now, satisfies . Since M and K are real and positive-definite, it follows that λ j is pure imaginary: λ j = i ω j . Without loss of generality, we can take γγ γγ j to be real and orthonormal. Sample Problem 1 As an example, consider (11.33) Now det[K + λ 2 I] = 0 reduces to (11.34) with the roots (11.35) so that both are negative (since k 11 and k 22 are positive). We now consider eigenvectors. The eigenvalue equations for the i th and j th eigenvectors are written as (11.36) It is easily seen that the eigenvectors have arbitrary magnitudes, and for conve- nience, we assume that they have unit magnitude Simple manipulation furnishes that (11.37) γγ j H λ j 2 λ j jj jj 2 =− γγγγ γγγγ H H K M MK=         =         10 01 11 12 12 22 kk kk . () [ ] [ ] , λλ 22 11 22 2 11 22 12 2 0++ + +=kk kk k λ +− =− + ± + − + [] =− + ± − − [] , [][][ ] [][] 2 11 22 11 22 2 11 22 12 2 11 22 11 22 2 12 2 1 2 4 1 2 4 kk kk kkk kk kk k λλ +− 22 and [[KM 0 KM 0+= += ωω jj kk 22 ]g g ] γγγγ jj T = 1. γγγγγγγγγγγγγγγγ k jj k j k j k j k TT T T KK M M−− − =[]. ωω 22 0 0749_Frame_C11 Page 160 Wednesday, February 19, 2003 5:13 PM © 2003 by CRC CRC Press LLC Solution Methods for Linear Problems 161 Symmetry of K and M implies that (11.38) Assuming for convenience that the eigenvalues are all distinct, it follows that (11.39) The eigenvectors are thus said to be orthogonal with respect to M and K. The quantities and are called the ( j th ) modal mass and ( j th ) modal stiffness. Sample Problem 2 Consider (11.40) Let For the determinant to vanish, Using the first eigenvector satisfies (11.41) implying that The corresponding procedures for the sec- ond eigenvalue furnish that It is readily verified that (11.42) The modal matrix X is now defined as (11.43) γγγγγγγγγγγγγγγγγγγγ k jj k j k j k j k j kk j TT T T T KK M M M−= − =− =00 22 22 ,[ ( ) . ] ωω ωω γγγγγγγγ j k j k jk TT MK==≠00,,. µ jjj =γγγγ T M κ jjj =γγγγ T K 20 01 21 11 0 0 1 2 1 2               + − −               =       •• γ γ γ γ k m . ζωωω 22 0 2 0 2 == j km/, / . 112 2 −=± ± ζ / . 112 2 −= ± ζ /, 21 112 0 0 1 1 1 2 1 1 1 2 2 1 2 − −                 =       [] + [] = / ,, () () () () γ γ γγ γ γ 1 1 2 1 13 2 3 () () /, /.== γγ 1 2 2 2 13 2 3 () () /, /.==− γγγγγγγγ γγγγγγγγ ( ) () () ( ) ( ) () () ( ) ,/, /, , [ /], [ /] 2112 12 2112 12 043 43 0 4 3 112 4 3 112 TT TT MM KK === = = = =− =+ µµ κκ X = [].γγγγγγγγ 123 L n 0749_Frame_C11 Page 161 Wednesday, February 19, 2003 5:13 PM © 2003 by CRC CRC Press LLC 162 Finite Element Analysis: Thermomechanics of Solids Since the jk th entries of X T MX and X T KX are and respectively, it follows that (11.44) The modal matrix is said to be orthogonal with respect to M and K, but it is not purely orthogonal since X −1 ≠ X T . The governing equation is now rewritten as (11.45) implying the uncoupled modes (11.46) Suppose that g j (t) = g j0 sin ( ω t). Neglecting transients, the steady-state solution for the j th mode is (11.47) It is evident that if (resonance), the response amplitude for the j th mode is much greater than for the other modes, so that the structural motion under this excitation frequency illustrates the mode. For this reason, the modes can easily be animated. 11.5.2 COMPUTATION OF EIGENVECTORS AND EIGENVALUES Consider with Many methods have been proposed to compute the eigenvalues and eigenvectors of a large system. Here, we describe a method that is easy to visualize, which we call the hypercircle method. The vectors Kγγ γγ j and Mγγ γγ j must be parallel to each other. Furthermore, the vectors and must terminate at the same point in a hypersphere in n- dimensional space. Suppose that is the ν th iterate and that the two vectors γγγγ j k T M γγγγ j k T K , XMX XKX TT =                     =                     µ µ µ κ κ κ 1 2 1 2 0 0 0 0 . . . . . nn X MX X KX g X g X f TT 1 T ˙˙ ,,,ξξξξξξγγ+== = − µξ κξ jj jj j gt ˙˙ ().+= ξ κωµ ω j j jj g t= − 0 2 sin( ). ωωκµ 22 ~/ jjj = KMγγγγ jjj = ω 2 , γγγγ jj T = 1. KK T γγ γ jj j / 2 MM T γγ γ jj j / 2 γγ j () ν 0749_Frame_C11 Page 162 Wednesday, February 19, 2003 5:13 PM © 2003 by CRC CRC Press LLC [...]... Press LLC 0749_Frame_C11 Page 164 Wednesday, February 19, 2003 5:13 PM 164 Finite Element Analysis: Thermomechanics of Solids using transformations involving the matrix X1 The eigenvalues and eigenvectors of the deflated system can be computed by, for example, the hypercircle method described previously 11. 6 EXERCISES 1 Verify that the triangular factors L3 and LT for A3 in Equation 11. 5 are correct 3... γ k = 0, k (11. 50) which decomposes to 2  ω1   0T  0  1 2  − ωk  ˜ 0 T K n−1    0   0   0    =   ˜ M n−1   ηn−1   0  (11. 51) This implies the “deflated” eigenvalue problem ˜ ˜ [K n−1 − ω k M n−1 ]ηn−1 = 0 (11. 52) The eigenvalues of the deflated system are also eigenvalues of the original system The eigenvector ηn−1 can be used to compute the eigenvectors of the original... +1) M2 γ (jν +1) (11. 48) Alternatively, note that C(γ j ) = γ γ TK j Mγ j γ TK 2 γ j j γ TM2 γ j j (11. 49) is the cosine of the angle between two unit vectors, and as such it assumes the maximum value of unity when the vectors coincide A search can be executed on γ the hypersphere in the vicinity of the current point, seeking the path along which C(γj) increases until the maximum is attained Once... Using A3 in Equation 11. 5, use forward substitution followed by backward substitution to solve 1   A 3 γ = 1   1 3 Triangularize the matrix 36  K = 30  18  30 41 23 18   23  14   4 For the model equation dy/dx = f(y), develop a two-step numerical-integration model: (αyn+1 + βyn + γ yn−1 ) + h[δ f ( yn+1 ) + ε f ( yn ) + ζ f ( yn−1 )] = 0 What is the order of the integration method... Solution Methods for Linear Problems 165 8 (a) Find the modal masses µ1, µ2 and the modal stiffnesses κ1 and κ2 of the system 1 3  0 •• 0  γ 1   1    + 27 2  γ 2    −1 −1 γ 1   10    =   sin(10t ) 2  γ 2   20  (b) Determine the steady-state response of the system (i.e., particular solution to the equation) 9 Triangularize 2 µ A/ L   0   A  0 4 µ AL/Y 2 2 µ A/ L −A ...0749_Frame_C11 Page 163 Wednesday, February 19, 2003 5:13 PM Solution Methods for Linear Problems 163 do not coincide in direction Another iterate can be attained by an interval-halving method: ˆ Mγ (jν +1) =  Mγ (ν ) Kγ (ν ) 1 j j + (ν ) T K 2 γ (ν ) 2  γ (ν ) T M 2 γ (ν ) γ j j j  j   ,    ˆ γ (jν +1) = γ (jν +1) T ˆ ˆ γ (jν +1) M2 γ (jν +1) (11. 48) Alternatively, note that... T Mγ j ) Now, an efficient scheme is needed to “deflate” the system j j 2 so that ω1 ,and γ1 no longer are part of the eigenstructure in order to ensure that the solution scheme does not converge to values that have already been calculated 2 Given γ1 and ω1 , we can construct a vector p2 that is M-orthogonal to γ1 by T T T ˆ ˆ ˆ ˆ using an intermediate vector p2 : p2 = p2 − γ 1 Mp2 γ 1 Clearly, γ 1 Mp2... linear-mechanical system ˙ M˙˙ + Dγ + Kγ = f (t ), γ th suppose that γ(t) = γn at the n time step Derive KD and rn+1 such that γ st at the (n + 1) time step satisfies K D γ n+1 = rn+1 7 For the linear system 36  30  24  30 41 32 24 γ 1   1     32  γ 2  =  2     27 γ 3   3   triangularize the matrix and solve for γ1, γ2, γ3 © 2003 by CRC CRC Press LLC 0749_Frame_C11 Page . consider (11. 5) Clearly, For the second block, (11. 6) γγγ γγ n n n nnn n n nnn nn n nn fl l fl l l −−− −−− −− −−−−− =− =− − 111 1112 22 2112 2 []/,[ ]/, . ,, , , , K KLL T jjj−−− = 111 . K Kk kk L0 L 0 T T T T j jj jjj j jjj jj jj l l =         =                 − − − 1 1 1 λλ λλ , k jjj jj. 0 TT1 11 2 111 − [] = − ωωγγ kk , ω ω 1 2 1 2 1 1 1 0 0 0 0K 0 0M 0 T T ˜ ˜ . n k n n − − −         −                       =       ηη [ ˜˜ ].KM 0 n k nn−−− −= 111 ω ηη 0749_Frame_C11 Page 163 Wednesday, February 19, 2003 5:13 PM © 2003 by CRC CRC Press LLC 164 Finite Element Analysis: Thermomechanics of Solids using transformations. ,KM+= λ j 2 0 0749_Frame_C11 Page 159 Wednesday, February 19, 2003 5:13 PM © 2003 by CRC CRC Press LLC 160 Finite Element Analysis: Thermomechanics of Solids For the sake of generality, suppose