By Hamid R. Berenji (auth.), Anca L. Ralescu (eds.)

ISBN-10: 3540584099

ISBN-13: 9783540584094

This quantity constitutes the lawsuits of the second one Fuzzy common sense in AI Workshop, held along with IJCAI '93 in Chambéry, France in August 1993.

The publication includes complete revised types of the papers offered on the workshop and covers numerous elements of fuzzy common sense in contributions from popular researchers. This quantity displays the purpose of the workshop, specifically to function a platform for fruitful contacts among fuzzy engineering researchers attracted to AI and AI researchers attracted to fuzzy common sense purposes in engineering: in terms of designing clever structures they're faced with a similar difficulties and questions and wish for a similar winning results.

**Read Online or Download Fuzzy Logic in Artificial Intelligence: IJCAI '93 Workshop Chamberry, France, August 28, 1993 Proceedings PDF**

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**Extra info for Fuzzy Logic in Artificial Intelligence: IJCAI '93 Workshop Chamberry, France, August 28, 1993 Proceedings**

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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.

### Fuzzy Logic in Artificial Intelligence: IJCAI '93 Workshop Chamberry, France, August 28, 1993 Proceedings by Hamid R. Berenji (auth.), Anca L. Ralescu (eds.)

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