eciv 520 a structural analysis ii

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ECIV 520 A Structural Analysis II Stiffness Method – General Concepts

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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. - PowerPoint PPT Presentation

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Page 1: ECIV 520 A  Structural Analysis II

ECIV 520 A Structural Analysis II

Stiffness Method – General Concepts

Page 2: ECIV 520 A  Structural Analysis II

Engineering Systems

Lumped Parameter

(Discrete)Continuous

• A finite number of state variables describe solution

• Algebraic Equations

• Differential Equations Govern Response

Page 3: ECIV 520 A  Structural Analysis II

Lumped Parameter

Displacements of Joints fully describe solutionDisplacements of Joints fully describe solution

Page 4: ECIV 520 A  Structural Analysis II

Matrix Structural Analysis - Objectives

UseEquations of EquilibriumConstitutive EquationsCompatibility Conditions

Basic Equations

Form

[A]{x}={b}

Solve for Unknown Displacements/Forces

{x}= [A]-1{b}

OR

Energy Principles

Page 5: ECIV 520 A  Structural Analysis II

TerminologyElement:Discrete Structural MemberNodes:Characteristic points that define elementD.O.F.:All possible directions of displacements @ a node

Page 6: ECIV 520 A  Structural Analysis II

Assumptions

• Equilibrium Pertains to Undeformed Configuration

• Linear Strain-Displacement Relationship

• Small Deformations

Page 7: ECIV 520 A  Structural Analysis II

The Stiffness MethodConsider a simple spring structural member

Undeformed Configuration

Deformed Configuration

Page 8: ECIV 520 A  Structural Analysis II

Derivation of Stiffness Matrix

1 2

P1

P2

Page 9: ECIV 520 A  Structural Analysis II

Derivation of Stiffness Matrix

=

+

For each case write basic equations

22P

1

11P

222P

21P

Page 10: ECIV 520 A  Structural Analysis II

Case A 0 ,0 21

1121 kkP Constitutive

Equilibrium1221 0 kPPP

2P

1

1P

Page 11: ECIV 520 A  Structural Analysis II

Case B 0,0 21

2121 kkP Constitutive

Equilibrium1221 0 kPPP

2

2P

1P

Page 12: ECIV 520 A  Structural Analysis II

Case A+B12 kP 11 kP A

21 kP 23 kP B

212 kkP 211 kkP

2

1

2

1

kk

kk

P

P

Page 13: ECIV 520 A  Structural Analysis II

Stiffness Matrix

2

1

2

1

kk

kk

P

PKδP

1 2

P1

P2

Page 14: ECIV 520 A  Structural Analysis II

Consider 2 Springs

2 elements 3 nodes 3 dof

Fix Fix

B

C

Fix Fix

1 2 3k1 k2

A

Fix Fix

Page 15: ECIV 520 A  Structural Analysis II

Case A – Spring 1

111211 kkP

Fix

P1 P2

1

Constitutive

Equilibrium11

12

121 0 kPPP

0,0 321

Page 16: ECIV 520 A  Structural Analysis II

Case A – Spring 2

02322

2 kP

Fix Fix

P2P3

1

Constitutive

Equilibrium 00 332

2 PPP

0,0 321

Page 17: ECIV 520 A  Structural Analysis II

Case AFix Fix

P1 P2 P3

1

112

21

22 kPPP

0,0 321

For Both Springs (Superposition)

111 kP

03 P

Page 18: ECIV 520 A  Structural Analysis II

Case B – Spring 1

211211 kkP Constitutive

Equilibrium21

12

121 0 kPPP

0,0 312

2P1

Page 19: ECIV 520 A  Structural Analysis II

P1P1

Case B – Spring 2

222322

2 kkP Constitutive

Equilibrium2233

22 0 kPPP

0,0 312

2

P2

P3

Page 20: ECIV 520 A  Structural Analysis II

Case B

P1 P2

P3

2

22212

21

22 kkPPP

For Both Springs (Superposition)

211 kP

223 kP

0,0 312

Page 21: ECIV 520 A  Structural Analysis II

Case C – Spring 1

01211 kPConstitutive

Equilibrium 00 12

121 PPP

0,0 213

P1 P2 3

Page 22: ECIV 520 A  Structural Analysis II

P1P1

Case C – Spring 2

322322

2 kkP Constitutive

Equilibrium3233

22 0 kPPP

0,0 213

P2 P3

3

Page 23: ECIV 520 A  Structural Analysis II

Case C

322

21

22 kPPP

For Both Springs (Superposition)

01 P

323 kP

0,0 213

Fix Fix

Page 24: ECIV 520 A  Structural Analysis II

Case A+B+C112 kP 111 kP 03 PA

22212 kkP 211 kP 223 kP B

322 kP 01 P 323 kP C

32

223

k

kP

21

111

k

kP

32

221

112

k

kk

kP

Page 25: ECIV 520 A  Structural Analysis II

2-Springs

3

2

1

3

2

1

22

22

11

11

0

0

P

P

P

kk

kk

kk

kk

Page 26: ECIV 520 A  Structural Analysis II

Use Energy Methods

Lets Have Fun !

Pick Up Pencil & Paper

Page 27: ECIV 520 A  Structural Analysis II

Use Energy Methods

3

2

1

321

3

2

1

22

22321

3

2

1

11

11

321

0

0

000

000

0

0

P

P

P

u

u

u

kk

kk

u

u

u

kk

kk

Page 28: ECIV 520 A  Structural Analysis II

2-Springs

3

2

1

3

2

1

22

22

11

11

0

0

P

P

P

kk

kk

kk

kk

Compare to 1-Spring

2

1

2

1

11

11

P

P

kk

kk

Page 29: ECIV 520 A  Structural Analysis II

Use Superposition1

2

3

x x x

x x x

x x x

x x x

x x x

x x x

1 2 3

1

2

3

Page 30: ECIV 520 A  Structural Analysis II

Use Superposition

x x x

x x x

x x x

x x x

x x x

x x x

1 2 3

1

2

3

1

2

3

X X

X X

11 22

11

22

1

2

3

1

2

3

X X

X X

11 22

11

22

X X

X X

11 22

11

22

1

2

3

X X

X X

11 22

11

22

1

2

3

1

2

3

X X

X X

11 22

11

22

X X

X X

11 22

11

22

Page 31: ECIV 520 A  Structural Analysis II

Use Superposition

x x x

x x x

x x x

x x x

x x x

x x x

1 2 3

1

2

3X X

X X

11 22

11

22

1

2

3

1

2

3

Page 32: ECIV 520 A  Structural Analysis II

1

2

3

1

2

3

Use Superposition

x x x

x x x

x x x

x x x

x x x

x x x

1 2 3

1

2

3X X

X X

22

22

33

33

Page 33: ECIV 520 A  Structural Analysis II

Use Superposition

x x x

x x x

x x x

x x x

x x x

x x x

1 2 3

1

2

3

DOF not connected directly yield 0 in SM

0

0

1

2

3

1

2

3

Page 34: ECIV 520 A  Structural Analysis II

Properties of Stiffness Matrix

SM is SymmetricBetti-Maxwell Law

SM is SingularNo Boundary Conditions Applied Yet

Main Diagonal of SM Positive

Necessary for Stability

Page 35: ECIV 520 A  Structural Analysis II

Transformations

P

k2k1

u1

u2

u3

u4

u3

u4

u5

u6x

yGlobal CS

xLocal CS

Objective: Transform State Variables from LCS to GCS

Page 36: ECIV 520 A  Structural Analysis II

Transformations

P1

y

P1x

x

yGlobal CS

P2xP2y

2

1

P1x

P1y P1x = P1xcosP1ysin

P1y = -P1xsinP1ycos

P1x

P1

y

=cos sin

-sin cos

P1x

P1

yP1 = T P1

Page 37: ECIV 520 A  Structural Analysis II

Transformations

In GeneralIn General

P1 = T P1

P2 = T P2

u2 = T u2

u1 = T u1

Similarly for uSimilarly for uP1 = T P1

-1

or

P2 = T P2

-1

or

P1y

P1x

P2xP2y

xy

xy

xy 2

1

P1x

P1y

P1y

P1x

P1y

P1x

P2xP2y P2xP2xP2yP2y

xy

xy

xy 2

1

xy

xy

xy 22

11

P1x

P1y

P1xP1x

P1yP1y

Page 38: ECIV 520 A  Structural Analysis II

TransformationsElement stiffness equations in Local CS

k =1 -1

-1 1

1

2

P1

P2

Expand to 4 Local dof

k

1 0 -1 00 0 0 0-1 0 1 00 0 0 0

u1x

u1y

u2x

u2y

=

P1x

P1y

P2x

P2y

P1x

P2xP2y

P1y

2

1

P1

P2

K u P

Page 39: ECIV 520 A  Structural Analysis II

Transformations

y

x

y

x

y

x

y

x

u

u

u

u

TT

TT

TT

TT

u

u

u

u

2

2

1

1

2221

1211

2221

1211

2

2

1

1

00

00

00

00

Ruu

P1x

P2xP2y

P1y

P1xP1x

P2xP2xP2yP2y

P1yP1y

xy

xy

xy 2

1

xy

xy

xy 22

11

Page 40: ECIV 520 A  Structural Analysis II

Transformations

y

x

y

x

y

x

y

x

P

P

P

P

TT

TT

TT

TT

P

P

P

P

2

2

1

1

2221

1211

2221

1211

2

2

1

1

00

00

00

00

RPP

P1x

P2xP2y

P1y

P1xP1x

P2xP2xP2yP2y

P1yP1y

xy

xy

xy 2

1

xy

xy

xy 22

11

Page 41: ECIV 520 A  Structural Analysis II

SM in Global Coordinate System

Introduce the transformed variables…

K u = PRR-1

K : Element SM in global CS

K u = PR R

K u = PLocal Coordinate System…

Page 42: ECIV 520 A  Structural Analysis II

Transformations

[T] [0]

[0] [T][R]=

Both R and TDepend on Particular Element

In this case (2D spring/axial element)

In General

Page 43: ECIV 520 A  Structural Analysis II

i

k

l

j

m

Boundary Conditions

Pj

Pk

Pi

Page 44: ECIV 520 A  Structural Analysis II

Apply Boundary Conditionskii kij kik kil kim ui

uj

uk

ul

kji kjj kjk kjl kjm

kki kkj kkk kkl kkm

kli klj klk kll klm

kli klj klk kll klm um

=

Pi

Pj

Pk

Pl

Pm

Kff

Kfs

Ksf Kss

uf

Pf

us Ps

Kffuf+ Kfsus=Pf

Ksfuf+ Kssus=Ps Ksfuf+ Kssus=Ps

uf = Kff (Pf + Kfsus)-1

Page 45: ECIV 520 A  Structural Analysis II

Derivation of Axial Force Element

Fun!!!!!

Page 46: ECIV 520 A  Structural Analysis II

Example

L2

3

L

1L

AE

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

P

da

Page 47: ECIV 520 A  Structural Analysis II

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