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The Problem

The Problem. Sines and Cosines. sin a 1 = (-12 - 0) / (20) = -0.6 cos a 1 = (16 - 0) / (20) = 0.8 sin a 2 = (12 - 0) / (15) = 0.8 cos a 2 = (9 - 0) / (15) = 0.6 . Element Matrices [S]. 3 4 1 2. 1 2 5 6. System Stiffness Matrix.

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The Problem

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  1. The Problem

  2. Sines and Cosines sina1 = (-12 - 0) / (20) = -0.6 cosa1 = (16 - 0) / (20) = 0.8 sina2 = (12 - 0) / (15) = 0.8 cosa2 = (9 - 0) / (15) = 0.6

  3. Element Matrices [S] 3 4 1 2 1 2 5 6

  4. System Stiffness Matrix

  5. Element Matrices [S] 3 4 1 2 1 2 5 6

  6. Summing Element Stiffnesses

  7. Summing Element Stiffnesses

  8. Two Matrix Contributions 1 2 3 4 5 6

  9. Final [K] 1 2 3 4 5 6

  10. Final Equation P=KX

  11. System Stiffness Matrices

  12. Solving the System of Equations

  13. Modify for Known Loads

  14. Modify for Boundary Conditions

  15. Modify to Ease Solution

  16. Return Symmetry

  17. Modified Equations

  18. Recap

  19. Initial Matrix

  20. Loads

  21. Boundary Conditions

  22. Symmetry

  23. Solution 10 = AE/L X1 100 = AE/L X2 0 = AE/L X3 0 = AE/L X4 0 = AE/L X5 0 = AE/L X6

  24. Force Calculation (f=sbX) {(X3i-X1i) cosai + (X4v-X2v) sinai} is simply the change in length t1 = AE/L {(10L/AE - 0)(0.8) + (100L/AE - 0)(-0.6)} + (0) t2 = AE/L {(0 - 10L/AE - 0)(0.6) + (0 - 100L/AE - 0)(0.8)} + (0) t1 = f2 = -f1 = -52 kips t2 = f4 = -f3 = -86 kips

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