April 20, 2017

# Download An Intro. to the Finite Element Method [SOLUTIONS] by J. Reddy PDF

By J. Reddy

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016887 The exact solution is the same as the finite element solution at the nodes. 13 using the mixed boundary conditions µ ¶¯ du ¯¯ u(0) = 0, =0 dx ¯x=1 Use the uniform mesh of three linear elements to solve the problem and compare against the exact solution 1 u(x) = 2 (cos πx − 1) π Solution: The boundary conditions require U1 = 0 and Q32 = 0. 15198 U4 PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. ° Hence, the All rights reserved. 20264 Again, the exact solution is the same as the finite element solution at the nodes.

E. condensed equations) for the unknown voltages and currents. 3 R = 30 Ω 2 R = 35 Ω 1 V1= 10 volts R = 7 R=5Ω Ω 4 R = 15 Ω R = 10 Ω 6 V6= 200 volts 5 R=5Ω Fig. 3 for the direct current electric network shown in Fig. 4. PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. ° All rights reserved. 52 AN INTRODUCTION TO THE FINITE ELEMENT METHOD R=5Ω 3 R = 20 Ω 6 R=0Ω 8 R = 10 Ω R=5Ω 2 R=5Ω 5 R = 15 Ω R = 10 Ω 1 R = 20 Ω 4 R = 50 Ω V1= 110 volts 7 V7 = 40 volts Fig. 4 Solution: The assembled coeﬃcient matrix is ⎡ 1 5 + 1 20 ⎢ −1 ⎢ 5 ⎢ ⎢ 0 ⎢ 1 [K] = ⎢ ⎢ − 20 ⎢ 0 ⎢ ⎢ ⎣ 0 1 5 0 − 15 + 15 + 1 − 20 0 − 15 0 0 1 − 20 0 0 1 1 + 20 10 + 1 − 10 0 1 − 50 ⎤ 0 ⎥ 0 ⎥ ⎥ 0 ⎥ 1 − 50 ⎥ ⎥ ⎥ ⎥ 0 ⎥ 1 − 15 ⎦ 1 1 15 + 50 0 1 − 20 1 1 20 + 5 0 0 − 15 0 0 0 − 15 0 1 − 10 1 1 + 10 + 15 1 − 15 1 20 1 5 1 50 1 5 0 − 15 0 1 − 10 1 + 10 + 1 − 10 0 1 10 The condensed equations are ⎡ 9 20 ⎢− 1 ⎢ 20 ⎢ ⎢ 0 ⎢ 1 ⎣− 5 0 1 − 20 0 0 0 0 − 15 17 100 1 − 10 1 4 I1 = 0 − 15 0 1 − 10 2 5 1 − 10 1 5 0 − 15 0 1 − 10 1 + 10 + V1 − V2 V1 − V4 + , 5 20 PROPRIETARY MATERIAL.

1 Solution: The assembled matrix is ⎡ ⎢ ⎢ [K] = ⎢ ⎢ ⎣ 1 k1 2 −k1 k1 + k2 + k3 + k4 3 0 −k2 − k3 k2 + k3 + k5 4 0 −k4 −k5 k4 + k5 + k6 symm. PROPRIETARY MATERIAL. c The McGraw-Hill Companies, Inc. ° 5 ⎤ 0 0 ⎥ ⎥ 0 ⎥ ⎥ −k6 ⎦ k6 All rights reserved. 1 2 3 4 5 50 AN INTRODUCTION TO THE FINITE ELEMENT METHOD ⎡ ⎤ 60 −60 0 0 0 ⎢ 340 −180 −150 0 ⎥ ⎢ ⎥ ⎢ =⎢ 300 −270 0 ⎥ ⎥ ⎣ 270 −180 ⎦ symm. 180 The condensed equations for the unknown primary variables are ⎫ ⎤⎧ ⎡ ⎧ ⎫ 340 −180 −150 ⎨ U2 ⎬ ⎨ 100 ⎬ ⎣ −180 300 −270 ⎦ U3 = 0 ⎭ ⎩ ⎭ ⎩ −150 −270 270 80 U4 and unknown secondary variables are Q11 = −k1 U2 and Q62 = −k6 U5 .