Download PDF by J Reddy: An Introduction to The Finite Element Method[Solutions]

By J Reddy

ISBN-10: 0072466855

ISBN-13: 9780072466850

J.N. Reddy's, An creation to the Finite aspect approach, 3rd version is an replace of 1 of the most well-liked FEM textbooks to be had. The booklet keeps its robust conceptual procedure, essentially reading the mathematical underpinnings of FEM, and delivering a common procedure of engineering program areas.
Known for its designated, rigorously chosen instance difficulties and vast collection of homework difficulties, the writer has comprehensively coated quite a lot of engineering parts making the e-book approriate for all engineering majors, and underscores the wide variety of use FEM has within the specialist world.
A supplementary textual content website positioned at http://www.mhhe.com/reddy3e comprises password-protected ideas to end-of-chapter difficulties, normal textbook info, supplementary chapters at the FEM1D and FEM2D desktop courses, and extra!

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Extra info for An Introduction to The Finite Element Method[Solutions]

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All rights reserved. 13: Solve the problem described by the following equations − d2 u = cos πx, 0 < x < 1; dx2 u(0) = 0, u(1) = 0 Use the uniform mesh of three linear elements to solve the problem and compare against the exact solution u(x) = 1 (cos πx + 2x − 1) π2 Solution: The main part of the problem is to compute the source vector for an element. We have fie = Z xb x cos πx ψie dx Z xa b µ ¶ xb − x = cos πx dx he xb ∙ µ ¶¸xb 1 xb 1 x sin πx − = cos πx + sin πx he π π2 π xa 1 1 = − sin πxa − (cos πxb − cos πxa ) π he π 2 µ ¶ Z xb x − xa cos πx dx f2e = he xa 1 1 (cos πxb − cos πxa ) + sin πxb = 2 he π π f1e The element equations are ∙ 3 −3 −3 3 ¸½ e ¾ u 1 ue2 = ½ e¾ f 1 f2e + ½ Qe1 Qe2 ¾ with the element source terms are given as follows.

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.

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.

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An Introduction to The Finite Element Method[Solutions] by J Reddy


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