Trusses are lightweight structures capable of carrying heavy loads. In bridge design, the individual members of the trus

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Trusses Are Lightweight Structures Capable Of Carrying Heavy Loads In Bridge Design The Individual Members Of The Trus 1 (148.04 KiB) Viewed 17 times

Trusses Are Lightweight Structures Capable Of Carrying Heavy Loads In Bridge Design The Individual Members Of The Trus 2 (38.16 KiB) Viewed 17 times

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 fi, f2. f3, fa, and fs, as shown in figure 1. 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 Fi and a vertical force component F2, but the movable support member has only a vertical force component F3. 11 12 12 Js Js 10,000 N F2 F Figure 1: Diagrammatic Representation of the Truss 1 NUMERICAL ANALYSIS ASSIGNMENT 3 MATH 351 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 below: Joint 1 2 Horizontal Component Vertical Component -Fi + 2 f1 + f2 = 0 fi - F2 = 0 -fi + sua fa = 0 - { f1 – f3 – į f4 = 0 - $2 + fs = 0 f3 – 10,000 = 0 - - s = 0 2 f4 – F3 = 0 3 4 An 8x8 matrix describing this system has 47 zero entries and only 17 nonzero entries. This linear system can be placed in the matrix form as shown below -1 0 0 1 0 0 0 -1 0 0 0 0 0 0 0 0 - 1 Mis ons 0 0 0 F2 F3 0 0 0 - 0 -1 - 0 woon woo FEF片方片七巧 0 0 0 0 - 1 0 1 0 0 0 10,000 0 0 0 0 0 0 0 f2 f3 f4 fs 0 0 0 0 - 0 0 0 0 0 0 0 0 0 - - 1 (a) Explain why the system of equations was reordered. (b) Using the matrix approach: 2K = Trk-1 + c, perform three steps iterations that approximate the
(a) Explain why the system of equations was reordered. (b) Using the matrix approach: x* = Tzk-1 + c, perform three steps iterations that approximate the solution of the resulting linear system using as initial approximation the vector all of whose entries are ls with 1. the Jacobi method 2. the Gauss-Siedal method 3. the SOR method with w= 1.25 and after each step with each method, calculate the relative error within 10-2 in the lo norm.