tailieunhanh - Ebook Numerical analysis (9th edition): Part 2
(BQ) Part 2 book "Numerical analysis" has content: Iterative Techniques in matrix algebra, approximation theory, approximation theory, numerical solutions of nonlinear systems of equations, boundary value problems for ordinary differential equations, numerical solutions to partial differential equations. | CHAPTER 7 Iterative Techniques in Matrix Algebra Introduction Trusses are lightweight structures capable of carrying heavy loads. In bridge design, the individual members of the truss are connected with rotatable pin joints that permit forces to be transferred from one member of the truss to another. The accompanying figure shows a truss that is held stationary at the lower left endpoint ①, is permitted to move horizontally at the lower right endpoint ④, and has pin joints at ①, ②, ③, and ④. A load of 10,000 newtons (N) is placed at joint ③, and the resulting forces on the joints are given by f1 , f2 , f3 , f4 , and f5 , as shown. When positive, these forces indicate tension on the truss elements, and when negative, compression. The stationary support member could have both a horizontal force component F1 and a vertical force component F2 , but the movable support member has only a vertical force component F3 . f1 2 f4 f3 f1 F1 1 f3 f2 f2 3 f4 l1 f5 f5 4 10,000 N F2 F3 If the truss is in static equilibrium, the forces at each joint must add to the zero vector, so the sum of the horizontal and vertical components at each joint must be 0. This produces the system of linear equations shown in the accompanying table. An 8×8 matrix describing this system has 47 zero entries and only 17 nonzero entries. Matrices with a high percentage of zero entries are called sparse and are often solved using iterative, rather than direct, techniques. The iterative solution to this system is considered in Exercise 18 of Section and Exercise 10 in Section . Joint ① ② ③ ④ Horizontal Component −F1 + √ 2 f + f2 = 0 2 1 √ √ − 22 f1 + 23 f4 = 0 −f2 + f5 = 0 − √ 3 f 2 4 − f5 = 0 Vertical Component √ 2 f − F2 = 0 2 1 √ 1 − 22 f1 − f3 − 2 f4 = 0 f3 − 10,000 = 0 1 f 2 4 − F3 = 0 431 Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be .
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