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

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

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

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

8. Derivation of Stiffness Matrix 11 22 P1P1 P2P2

9. = + For each case write basic equations 11 1 1 P 22

10. Case A Constitutive Equilibrium 11 1 P

11. Case B Constitutive Equilibrium 22

12. Case A+B A B

13. Stiffness Matrix 11 22 P1P1 P2P2

14. Consider 2 Springs 2 elements 3 nodes 3 dof Fix BC 123 k1k1 k2k2 A

15. Case A – Spring 1 Fix P1P1 P2P2 11 Constitutive Equilibrium

16. Case A – Spring 2 Fix P2P2 P3P3 11 Constitutive Equilibrium

17. Case A Fix P1P1 P2P2 P3P3 11 For Both Springs (Superposition)

18. Case B – Spring 1 Constitutive Equilibrium 22 P1P1

19. Case B – Spring 2 Constitutive Equilibrium 22 P2P2 P3P3

20. Case B P1P1 P2P2 P3P3 22 For Both Springs (Superposition)

21. Case C – Spring 1 Constitutive Equilibrium P1P1 P2P2 33

22. Case C – Spring 2 Constitutive Equilibrium P2P2 P3P3 33

23. Case C For Both Springs (Superposition) Fix

24. Case A+B+C 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. Use Superposition 11 22 33 123 1 2 3

30. 123 1 2 3

31. 123 1 2 3 XX XX

32. 123 1 2 3 XX XX

33. 123 1 2 3 DOF not connected directly yield 0 in SM 0 0

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. Transformations P k2k2 k1k1     u1u1 u2u2 u3u3 u4u4 u3u3 u4u4 u5u5 u6u6 x y Global CS x Local CS Objective: Transform State Variables from LCS to GCS

36. Transformations P 1y P 1x x y Global CS  P 2x P 2y 2 1 P 1x P 1y P 1x = P 1x cos  P 1y sin  P 1y = -P 1x sin  P 1y cos  P 1x P 1y = cos  sin  -sin  cos  P 1x P 1y P1P1 = T P1P1

37. Transformations In General P1P1 = T P1P1 P2P2 = T P2P2 u2u2 = T u2u2 u1u1 = T u1u1 Similarly for u P1P1 = T P1P1 or P2P2 = T P2P2 or

38. Transformations Element stiffness equations in Local CS k = 1 1 11 22 P1P1 P2P2 Expand to 4 Local dof k 100 0000 010 0000 u 1x u 1y u 2x u 2y = P 1x P 1y P 2x P 2y P 1x P 2x P 2y P 1y  2 1 P1P1 P2P2 K u P

39. Transformations

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

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

43. i k l j m Boundary Conditions PjPj PkPk PiPi

44. Apply Boundary Conditions k ii k ij k ik k il k im uiui ujuj ukuk ulul k ji k jj k jk k jl k jm k ki k kj k kk k kl k km k li k lj k lk k ll k lm k li k lj k lk k ll k lm umum = PiPi PjPj PkPk PlPl PmPm K ff K fs K sf K ss uf uf Pf Pf usus PsPs K ff u f + K fs u s =P f K sf u f + K ss u s =P s u f = K ff (P f + K fs u s )

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

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

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