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

2. Lumped Parameter Displacements of Joints fully describe solution

3. 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}

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

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

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

7. Derivation of Stiffness Matrix Using Basic Equations 11 22 P1P1 P2P2

8. 22 = 11 + For each case write basic equations

9. Derivation of Stiffness Matrix Using Basic Equations  2 =0 11 P 11 P 21 X Equilibrium Constitutive P11

10. Derivation of Stiffness Matrix Using Basic Equations 22  1 =0 P 12 P 22 Equilibrium Constitutive

11. Derivation of Stiffness Matrix Using Basic Equations Combined Action P1P1 P2P2

12. Derivation of Stiffness Matrix Using Basic Equations In Matrix Form

13. Consider 2 Springs 2 elements 3 nodes 3 dof 1 Fix 2 3 123 k1k1 k2k2

14. 2-Springs Compare to 1-Spring

15. Use Superposition 11 22 33 123 1 2 3 XX XX 11 22 33 11 22 33 XX XX DOF not connected directly yield 0 in SM 0 0 11 22 33

16. 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

17. 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 )

18. 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

19. Transformations Consider P 1y P 1x 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 x y Global CS  P 1x P 2x 2 P 2y 1

20. Transformations In General P 1y P 1x P1P1 = T P1P1 x y Global CS  P 1x P 2x 2 P 2y 1 P2P2 = T P2P2 u2u2 = T u2u2 u1u1 = T u1u1 Similarly for u P1P1 = T P1P1 or P2P2 = T P2P2 or

21. 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

22. SM in Global Coordinate System Introduce the transformed variables… K u = P RR K : Element SM in global CS [T][0] [T] [R]= Both R and T Depend on Particular Element

23. Transformations For example for an axial element with k=AE/L AE/L l2l2 lm- l 2 - lm Symm. m2m2 - lm- m 2 l2l2 lm m2m2 K = l=cos  m=sin 

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