1. The Problem

3. sin  1 = (-12 - 0) / (20) = -0.6 cos  1 = (16 - 0) / (20) = 0.8 sin  2 = (12 - 0) / (15) = 0.8 cos  2 = (9 - 0) / (15) = 0.6 Sines and Cosines

4. 3 41 2 1 25 6 Element Matrices [S]

5. System Stiffness Matrix

6. 3 41 2 1 25 6 Element Matrices [S]

7. Summing Element Stiffnesses.64-.48-.64.48 -.48.36.48-.36 -.64.48.64-.48.48-.36-.48.36

8. Summing Element Stiffnesses.64+.36-.48+.48-.64.48-.36-.48 -.48+.48.36+.64.48-.36-.48-.64.48.64-.48.48-.36-.48.36 -.36-.48.36.48 -.48-.64.48.64

9. 1 2 3 4 5 6 Two Matrix Contributions

10. 1 2 3 4 5 6 Final [K]

11. Final Equation P=KX

12. System Stiffness Matrices

13. Solving the System of Equations

14. Modify for Known Loads

15. Modify for Boundary Conditions

16. Modify to Ease Solution

17. Return Symmetry

18. Modified Equations

19. Recap

20. Initial Matrix

21. Loads

22. Boundary Conditions

23. Symmetry

24. 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

25. Force Calculation (f=sbX) {(X3i-X1i) cos  i + (X4v-X2v) sin  i} 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