eciv 520 a structural analysis ii
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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
Displacements of Joints fully describe solutionDisplacements of Joints fully describe solution
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
TerminologyElement:Discrete Structural MemberNodes:Characteristic points that define elementD.O.F.:All possible directions of displacements @ a node
Assumptions
• Equilibrium Pertains to Undeformed Configuration
• Linear Strain-Displacement Relationship
• Small Deformations
The Stiffness MethodConsider a simple spring structural member
Undeformed Configuration
Deformed Configuration
Derivation of Stiffness Matrix
1 2
P1
P2
Derivation of Stiffness Matrix
=
+
For each case write basic equations
22P
1
11P
222P
21P
Case A 0 ,0 21
1121 kkP Constitutive
Equilibrium1221 0 kPPP
2P
1
1P
Case B 0,0 21
2121 kkP Constitutive
Equilibrium1221 0 kPPP
2
2P
1P
Case A+B12 kP 11 kP A
21 kP 23 kP B
212 kkP 211 kkP
2
1
2
1
kk
kk
P
P
Stiffness Matrix
2
1
2
1
kk
kk
P
PKδP
1 2
P1
P2
Consider 2 Springs
2 elements 3 nodes 3 dof
Fix Fix
B
C
Fix Fix
1 2 3k1 k2
A
Fix Fix
Case A – Spring 1
111211 kkP
Fix
P1 P2
1
Constitutive
Equilibrium11
12
121 0 kPPP
0,0 321
Case A – Spring 2
02322
2 kP
Fix Fix
P2P3
1
Constitutive
Equilibrium 00 332
2 PPP
0,0 321
Case AFix Fix
P1 P2 P3
1
112
21
22 kPPP
0,0 321
For Both Springs (Superposition)
111 kP
03 P
Case B – Spring 1
211211 kkP Constitutive
Equilibrium21
12
121 0 kPPP
0,0 312
2P1
P1P1
Case B – Spring 2
222322
2 kkP Constitutive
Equilibrium2233
22 0 kPPP
0,0 312
2
P2
P3
Case B
P1 P2
P3
2
22212
21
22 kkPPP
For Both Springs (Superposition)
211 kP
223 kP
0,0 312
Case C – Spring 1
01211 kPConstitutive
Equilibrium 00 12
121 PPP
0,0 213
P1 P2 3
P1P1
Case C – Spring 2
322322
2 kkP Constitutive
Equilibrium3233
22 0 kPPP
0,0 213
P2 P3
3
Case C
322
21
22 kPPP
For Both Springs (Superposition)
01 P
323 kP
0,0 213
Fix Fix
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
2-Springs
3
2
1
3
2
1
22
22
11
11
0
0
P
P
P
kk
kk
kk
kk
Use Energy Methods
Lets Have Fun !
Pick Up Pencil & Paper
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
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
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
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
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
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
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
Properties of Stiffness Matrix
SM is SymmetricBetti-Maxwell Law
SM is SingularNo Boundary Conditions Applied Yet
Main Diagonal of SM Positive
Necessary for Stability
Transformations
P
k2k1
u1
u2
u3
u4
u3
u4
u5
u6x
yGlobal CS
xLocal CS
Objective: Transform State Variables from LCS to GCS
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
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
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
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
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
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…
Transformations
[T] [0]
[0] [T][R]=
Both R and TDepend on Particular Element
In this case (2D spring/axial element)
In General
i
k
l
j
m
Boundary Conditions
Pj
Pk
Pi
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
Derivation of Axial Force Element
Fun!!!!!
Example
L2
3
L
1L
AE
Calculate nodal displacements for (a) P=10 & (b) da=1
P
da
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
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