ECIV 520 A Structural Analysis II Stiffness Method – General Concepts

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 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

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

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

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

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

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

Case A Constitutive Equilibrium 11 1 P

Case B Constitutive Equilibrium 22

Case A+B A B

Stiffness Matrix 11 22 P1P1 P2P2

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

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

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

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

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

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

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

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

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

Case C For Both Springs (Superposition) Fix

Case A+B+C A B C

2-Springs

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

Use Energy Methods

2-Springs Compare to 1-Spring

Use Superposition 11 22 3

XX XX

XX XX

DOF not connected directly yield 0 in SM 0 0

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

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

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

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

Transformations Element stiffness equations in Local CS k = 1 1 11 22 P1P1 P2P2 Expand to 4 Local dof k 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

Transformations

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…

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

i k l j m Boundary Conditions PjPj PkPk PiPi

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 )

Derivation of Axial Force Element Fun!!!!!

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

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