ECIV 520 A Structural Analysis II

1. ECIV 520 A Structural Analysis II Stiffness Method – General Concepts

2. Engineering Systems Lumped Parameter (Discrete) Continuous • A finite number of state variables describe solution • Algebraic Equations • Differential Equations Govern Response

3. Lumped Parameter Displacements of Joints fully describe solution

4. Basic Equations Matrix Structural Analysis - Objectives Use Equations of Equilibrium Constitutive Equations Compatibility Conditions OR Energy Principles Form [A]{x}={b} Solve for Unknown Displacements/Forces {x}= [A]-1{b}

5. Terminology Element: Discrete Structural Member Nodes: Characteristic points that define element D.O.F.: All possible directions of displacements @ a node

6. Linear Strain-Displacement Relationship • Small Deformations Assumptions • Equilibrium Pertains to Undeformed Configuration

7. The Stiffness Method Consider a simple spring structural member Undeformed Configuration Deformed Configuration

8. Derivation of Stiffness Matrix d1 d2 P1 P2

9. d2 1 P 1 d1 = Derivation of Stiffness Matrix + For each case write basic equations

10. P 1 d1 Equilibrium Case A Constitutive

11. d2 Equilibrium Case B Constitutive

12. A B Case A+B

13. Stiffness Matrix d1 d2 P1 P2

14. A Fix Fix B Fix Fix C Fix Fix Consider 2 Springs k1 k2 1 2 3 2 elements 3 nodes 3 dof

15. Equilibrium Case A – Spring 1 Fix d1 P1 P2 Constitutive

16. Case A – Spring 2 Fix Fix d1 P3 P2 Constitutive Equilibrium

17. Case A Fix Fix d1 P1 P2 P3 For Both Springs (Superposition)

18. Case B – Spring 1 P1 d2 Constitutive Equilibrium

19. Case B – Spring 2 P2 P3 d2 Constitutive Equilibrium

20. Case B P1 P3 P2 d2 For Both Springs (Superposition)

21. Case C – Spring 1 P1 P2 d3 Constitutive Equilibrium

22. Case C – Spring 2 P2 P3 d3 Constitutive Equilibrium

23. Case C Fix Fix For Both Springs (Superposition)

24. A B C Case A+B+C

25. 2-Springs

26. Use Energy Methods Lets Have Fun ! Pick Up Pencil & Paper

27. Use Energy Methods

28. 2-Springs Compare to 1-Spring

29. d1 d2 d3 1 2 3 1 2 3 Use Superposition

30. 1 2 3 1 2 3 Use Superposition

31. X X 1 2 3 1 2 3 X X Use Superposition

32. X X 1 2 3 1 2 3 X X Use Superposition

33. 0 0 1 2 3 1 2 3 DOF not connected directly yield 0 in SM Use Superposition

34. Properties of Stiffness Matrix • SM is Symmetric • Betti-Maxwell Law • SM is Singular • No Boundary Conditions Applied Yet • Main Diagonal of SM Positive • Necessary for Stability

35. Global CS k1 k2 d3 u2 u6 u4 u4 d1 u1 u3 u3 u5 y P x x Local CS d2 d2 Transformations Objective: Transform State Variables from LCS to GCS

36. P2x P2y Global CS 1 2 P1x y P1y = -P1xsinf + P1ycosf P1y P1x P1x = P1xcosf + P1ysinf P1y cosf sinf P1x x = P1x -sinf cosf P1y P1y P1 P1 = T Transformations f

37. or -1 -1 P2 P1 P2 P1 = = T T Similarly for u or P2 u1 u2 P1 u1 P2 P1 u2 = = = = T T T T Transformations In General

38. P1y P2x P2y d1 1 -1 P1 k = 2 1 P2 -1 1 d2 P2 P1x f 1 0 -1 0 u1x P1x K u P 0 0 0 0 u1y P1y P1 = k -1 0 1 0 u2x P2x 0 0 0 0 u2y P2y Transformations Element stiffness equations in Local CS Expand to 4 Local dof

39. Transformations

40. Transformations

41. K u P = K R R u P = -1 R K R u P = K : Element SM in global CS SM in Global Coordinate System Local Coordinate System… Introduce the transformed variables…

42. [T] [0] [R]= [0] [T] Transformations In this case (2D spring/axial element) In General Both R and T Depend on Particular Element

43. Pk Pj Pi m i k l j Boundary Conditions

44. kii kij kik kil kim ui Pi uj Pj kji kjj kjk kjl kjm uk = Pk kki kkj kkk kkl kkm ul Pl kli klj klk kll klm um Pm kli klj klk kll klm -1 uf= Kff (Pf + Kfsus) Apply Boundary Conditions uf Pf Kff Kfs Ksf Kss us Ps Kffuf+ Kfsus=Pf Ksfuf+ Kssus=Ps Ksfuf+ Kssus=Ps

45. Derivation of Axial Force Element Fun!!!!!

46. Example P da Calculate nodal displacements for (a) P=10 & (b) da=1

47. In Summary • Derivation of element SM – Basic Equations • Structural SM by Superposition • Local & Global CS • Transformation • Application of Boundary Conditions • Solution of Stiffness Equations – Partitioning