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2D Scalar Example

# 2D Scalar Example - Example Poisson equation,—-V2y=1...

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Unformatted text preview: Example Poisson equation ,—-V2y=1 in82==((x,y): 0<(x,y)<1} 5-;<o,y>mw<x,o>w u<1,y>'mu(x,1>=:o | Line ofsymmetry “=0 Domain used for the triangular—element meshes Torsion problem 0.0 , Solution by linear triangular elements e 1 e e e e e_fA KiS‘)=ZX(Bi()Bj()l—Yi()7j()l Jim-““3“ 1 1 m1 0 1 1 Q9 (1) = m. _ u. (1) = _ (1) [K ] 2 1 2 1 (F ) M 1+Q2 0 —1 1 1 Q9 QWEﬁqPMWUdY I‘e _ 0'5[ (I) (1) ] d 0'5 (1) (I) 1 2 3 _.. qn ‘l/il(x9 y) y=0 x + [qn ¢i (xay)]x=0_5dy B ___=, 5 3 2 0 o _ 0 e 2 4 5 1 " 1 +f05[q,s>11><x,y)]x=ydx 3 5 6 __. l 1 l = Ql1)+ le> + Ql3) 1 2. 3 4 5 6 1 K“) 1(sz , Kg) 0 0 0 2 169+ K91“ K1?) . K5? + K3? K8) K8) + K5?) . (5 [K]=3 K§§’+Ké%’+l<f?’ 0 K§%’+K1‘§’ K1“? 4 Symmetric ' 1 I K £32” K £3) 0 5 ' K1?) + KS) + Ké‘? K5? 6 Kg) F1“) 1 2 F2(1)+ 173(2) + 171(3) . 3 E”[email protected]”+H” (F) = 4 F20) 5 F5214. 173(3) + 172(4) F301) assemblsd system [KmkﬂKwkﬂKmhﬂle <F®>=<F®>=<F®>=<FW> 1-—-1 0} 0 0 1 4 2: «1 0 ‘0 -—2 4: 6 m1 0:"2 ml" 0 0 2.~=1 4 ~= 0 0 0: 0 —-1 . 2 (29> + Q?) + Q3” Q3” boundary conditions w=e=e=o Qf‘) :- Qf‘l) + Q? = specified to be zero = 0 [email protected][email protected]\$+Q%%MQQ+%%] = specified to be zero + [cancel to give zero] = 0 Qe+Qe+ange+Qe=o Q53): Q8) + Q92) == specified to be zero + unknown = Q1332), unknown Q9+Q9+Q9\$QQ+QQwﬂmWI Qg“): Qg‘? + Qg‘? = unknown + zero due to symmetry = Q53), unknown Solution 0.5 “—0.5 0 U1 1 1 ——0.5 2.0 —- 1.0 U = —— 2 24 3 0 m 1.0 2.0 U3 3 532) 1 1 0 —0.5 0 U1 Q§32>+Qgg> =¢§Z 3 + 0 0 —1 U2 99 1 o o o e Ul 1 3 1 0.5 1 1 7.5 0.31250 U2 = 271: l 1 0.5 3 = 23 5.5 = 0.22917 U; 0 5 0.5 0.75 3 4.25 0.17708 Q53; — 51; —- 0.5U2 —-0.197917 1 Q53 + Q5? = -- g —— U3, = —-0.302083 1 g) 2.. :23 m0.041667 Table 4.1 Comparison of the finitevelement solutions u(0, y) with the series solution and the Ritz solution of Eqs. (4.47) . Series Tnangular elements Rectangular elements Ritz, solution, y 4elements 16 elements 4elements 16 elements . Eq. (2.96) Eq. (2.95) 0.0 0.3125 0.3013 0.3107 0.2984 0.3125 0.2947 0.25 0.2709? 0.2805 0.27591“ 0.2824 , 0.2930 0.2789 0.50 0.2292 0.2292 0.241 1 0.2322 0.2344 0.2293 0.75 0.1146" 0.1393 0.1205? 0.1414 0.1367 0.1397 1.0 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 llnterpolated values. u(x. 0) = 14(0,y) 0.25 ' 0.20 0_15 c1 4 elements FEM triangles 0 16 elements .. - Ritz solution (2.96) 0.10 16 elements (rectangles) - 0.05 7 -u-.\o-II~< J [=- =-'=- <7 6 elements ,I (triangles) O 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Figure 4.10 Comparison of the finite-element solution with the two-parameter Ritz solution and series solution of Eqs. (4.47). Table 4.4 Comparison of the finite-element solutions obtained by various triangularnelem'ent meshes with the series solution of the problem in Example 4.1 Finitea‘element solution 22.2....” Series Node Mesh 1 Mesh 2 ‘ Mesh 3 solution 1 0.31250 0.29167 0.25000 0.29469 2 0.22917 0.20833 0.20833 0.22934 4 0.22917 0.20833 0.20833 0.22934 5 0.17708 0.18750 0.16667 0.18114 Table 4.5 Convergence of the finitenelement solution (with the mesh refinementl) of the problem in Example 4.1 Location Finite-element solution Series x y 2X2 4X4 6X6 8X8 solution 0.0 0.0 0.31071 0.29839 0.29641 0.29560 0.29469 0.125 0.0 — — 0.29248 0.29167 0.29077 0.250 0.0 — 0.28239 0.28055 0.27975 0.27888 0.375 0.0 —— —— 0.26022 0.24943 0.25863 0.50 0.0 0.24107 0.23220 0.23081 0.23005 0.22934 0.625 0.0 — ~— — 0.19067 0.19009 0.750 0.0 —— 0.14137 0.14064 0.14014 0.13973 0875 0.0 ~— —— — 0.07709 0.07687 0.125 0.125 ~—— —- 0.28862 0.28781 0.28692 0.250 0.250 —- 0.26752 0.26580 0.26498 0.26415 0.375 0375 —~ -—- 0.22960 0.22873 0.22799 0.50 0.50 0.19286 0.18381 0.18282 0.18179 0.18114 0.625 0.625 ~— — ~— 0.12813 0.12757 0.750 0.750 - 0.07506 0.07481 0.07332 0.07282 0.875 0.875 ~— —~ —— 0.02561 0.02510 (C) (d) Figure 4.22 Mesh refinement: meshes in (a), (b), and (d) are uniform; mesh in (c) is nonuniform. (a) 2 X 2 mesh. (b) 4 X 4 mesh. (c) 6 X 6 mesh. (0') 8 X 8 mesh. ...
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