Engineering Systems

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

Lumped Parameter Displacements of Joints fully describe solution

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

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

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

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

Derivation of Stiffness Matrix Using Basic Equations d1 d2 P1 P2

Derivation of Stiffness Matrix Using Basic Equations d2 + d1 = For each case write basic equations

Derivation of Stiffness Matrix Using Basic Equations P21 P11 Equilibrium X d2=0 d1 P11 Constitutive

Derivation of Stiffness Matrix Using Basic Equations P12 P22 Equilibrium d1=0 d2 Constitutive

Derivation of Stiffness Matrix Using Basic Equations P1 P2 Combined Action

Derivation of Stiffness Matrix Using Basic Equations In Matrix Form

Consider 2 Springs 1 2 3 k1 k2 2 elements 3 nodes 3 dof 1 Fix Fix 2 Fix Fix 3 Fix Fix

2-Springs Compare to 1-Spring

Use Superposition d1 d1 d2 d2 0 0 d3 d3 d1 d1 X X X X d2 d2 3 1 2 1 3 2 X X X X d3 d3 DOF not connected directly yield 0 in SM

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

Apply Boundary Conditions 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) uf Pf Kff Kfs Ksf Kss us Ps Kffuf+ Kfsus=Pf Ksfuf+ Kssus=Ps Ksfuf+ Kssus=Ps

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

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

Transformations P2y P2x Global CS 1 2 f P1x or y P1x -1 -1 P1y P2 P1 P1 P2 = = T T Similarly for u x or P2 P1 u1 u2 u2 u1 P1 P2 = = = = T T T T In General

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

SM in Global Coordinate System -1 R K R u P = [T] [0] [R]= K : Element SM in global CS [0] [T] Introduce the transformed variables… Both R and T Depend on Particular Element

Transformations l2 lm - l2 - lm m2 - lm - m2 K = AE/L l2 lm Symm. m2 For example for an axial element with k=AE/L l=cosf m=sinf

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

Engineering Systems