Nonlinear FEM
3
Residual
Force
Equations
NFEM Ch 3 – Slide 1
Nonlinear FEM
Total Force Residual Equation
Vector form
r(u,Λ) = 0
r = total force residual vector
u = state vector with displacement DOF
Λ = array of control parameters
Indicial form
ri (uj , Λ k ) = 0
NFEM Ch 3 – Slide 2
Nonlinear FEM
Black Box Interpretation
Control: Λ
State u
Residual
evaluation
NFEM Ch 3 – Slide 3
Residual r
Nonlinear FEM
Example
r1 = 4 u 1 − u 2 + u 2 u 3 − 6 1 = 0
r2 = 6 u 2 − u 1 + u 1 u 3 − 3 2 = 0
r 3 = 4 u 3 + u 1 u 2 − 3 2 = 0
Vector form:
r = r(u, Λ) =
r1 (u, Λ)
r2 (u, Λ)
r3 (u, Λ)
with
u=
=
u1
u2
u3
4 u 1 − u 2 + u 2 u 3 − 6 1
6 u 2 − u 1 + u 1 u 3 − 3 2
4 u 3 + u 1 u 2 − 3 2
1
Λ=
2
NFEM Ch 3 – Slide 4
=0
Nonlinear FEM
Correlation with Linear FEM
Master stiffness equations:
Ku = f
Transforming to force residual:
r= Ku − f =0
Control parameters are not needed in linear FEM.
NFEM Ch 3 – Slide 5
Nonlinear FEM
Conservative System: Derivation From Energy
total
potential
energy
r =
Π
=0
u
Physical meaning: force equilibrium is associated
with total potential energy Π being stationary
(its gradient of r w.r.t. state vector u vanishes)
NFEM Ch 3 – Slide 6
Nonlinear FEM
Balanced Force Residual Form
Split total
residual as
r= p−f=0
Move f to RHS
p(u) = f(u,Λ)
internal
force
vector
external
force
vector
NFEM Ch 3 – Slide 7
Nonlinear FEM
Balanced Force Residual Form: Black Box
Control: Λ
State u
Ext force
evaluation
f
−
Int force
evaluation
p
NFEM Ch 3 – Slide 8
Residual r
Nonlinear FEM
Previous Example
r1 = 4 u 1 − u 2 + u 2 u 3 − 6 1 = 0
r2 = 6 u 2 − u 1 + u 1 u 3 − 3 2 = 0
r 3 = 4 u 3 + u 1 u 2 − 3 2 = 0
r = r(u, Λ) =
p=
p1 (u)
p2 (u)
p3 (u)
r1 (u, Λ)
r2 (u, Λ)
r3 (u, Λ)
=
=
4 u 1 − u 2 + u 2 u 3 − 6 1
6 u 2 − u 1 + u 1 u 3 − 3 2
4 u 3 + u 1 u 2 − 3 2
4 u1 − u2 + u2 u3
6 u2 − u1 + u1 u3
4 u3 + u1 u2
,
f=
NFEM Ch 3 – Slide 9
f 1 (Λ)
f 2 (Λ)
f 3 (Λ)
=0
=3
21
2
2
Nonlinear FEM
Conservative System: Derivation from Energy
Π=U−W
total
potential
energy
p=
external
work
potential
internal
energy
∂U
,
∂u
f=
∂W
,
∂u
NFEM Ch 3 – Slide 10
Nonlinear FEM
(Tangent) Stiffness and Control Matrices
K=
∂r
∂u
with entries
Ki j =
Q=−
∂r
∂Λ
with entries
Qi j = −
An example next
NFEM Ch 3 – Slide 11
∂ri
∂u j
∂ri
∂ j
Nonlinear FEM
Residual Rate Forms
Define state and control in terms of pseudo time t :
u = u (t)
Λ = Λ(t)
The first two derivatives of r with respect to t are, using indicial form
ṙi =
∂ri
∂ri ˙
u̇ j +
j
∂u j
∂ j
∂ 2r
∂ri
∂ri ¨
∂ 2 ri ˙ i
r̈i =
ü j +
u̇ k +
k u̇ j +
j
∂u j
∂u j ∂u k
∂ u j ∂k
∂ j
∂ 2r
∂ 2 ri ˙ ˙
i
+
u̇ k +
k j
∂ j ∂u k
∂ j ∂k
in which a superposed dot denotes derivative wrt t, as in real dynamics
NFEM Ch 3 – Slide 12
Nonlinear FEM
Residual Rate ODEs in Matrix Form
Recall that
∂r
K=
∂u
Q=−
∂r
∂Λ
Using these definitions the previous ODEs in indicial form
can be rewritten in matrix form as
ṙ = Ku̇ − QΛ̇
first order rate form
r̈ = Kü + K̇u̇ − QΛ̈ − Q̇Λ̇
Examples in next slide
NFEM Ch 3 – Slide 13
second order rate form
Nonlinear FEM
Residual Computation Example
P = λ EA
E,A constant
A
C'
B
A
θ
θ
C
k
k = β EA
L
L
L
2L
Only one DOF: vertical motion of midspan joint C
Only one control parameter denoted by λ instead of Λ1
NFEM Ch 3 – Slide 14
B
u
Nonlinear FEM
Residual Computation Example (cnt'd)
P = λ EA
C'
A
θ
B
θ
u
k
P = λ EA
C'
State parameter
µ = u/L = tan θ
Control parameter
λ = P/EA
FAC
Fs
FBC
NFEM Ch 3 – Slide 15
Nonlinear FEM
Residual Computation Example (cnt'd)
2
r (µ, λ) = µ 2 + β − 1 + µ2
−λ=0
∂r
2(1 + µ2 ) 3/2 − 1
K =
=β+
∂µ
(1 + µ2 ) 3/2
NFEM Ch 3 – Slide 16
Nonlinear FEM
Residual Computation Example (cnt'd)
Load factor λ = P/EA
0.6
Response using
engineering
strain measure
0.4
0.2
0
β=1
β = 1/10
β = 1/100, 1/1000
(indistinguishable
at plot scale)
−0.2
−0.4
−0.6
−1
1
−0.5
0
0.5
Dimensionless displacement µ = u/L
NFEM Ch 3 – Slide 17
Nonlinear FEM
Stiffness coefficient K = dr/dµ
Residual Computation Example (cnt'd)
2.5
Stiffness using
engineering
strain measure
2
1.5
1
0.5
0
−1
1
−0.5
0
0.5
Dimensionless displacement µ = u/L
NFEM Ch 3 – Slide 18
Nonlinear FEM
Staging
Complicated nonlinear systems, such as structures,
are analyzed in stages because the superposition
principle no longer applies
Staging reduces multiple control parameters to
only one over each loading stage.
This single parameter is called the staging parameter
and will be usually denoted by λ
NFEM Ch 3 – Slide 19
Nonlinear FEM
Staging (cont'd)
Suppose that we have solved the residual equation for the control
parameters Λ A corresponding to point A in the control space.
We next want to advance the solution to Λ B corresponding to point B
Interpolate the control parameters linearly
Λ = (1 − λ)Λ A + λΛ B
where λ is the staging parameter that varies from 0 at A to 1 at B
The residual equation to ber solve over the stage is
r (u, λ) = 0
with the I.C. u = u A at λ = 0.
This equation has only one control parameter. It is studied in Chapter 4.
NFEM Ch 3 – Slide 20
Nonlinear FEM
To Illustrate Staging We Will Look
at a Suspension Bridge
NFEM Ch 3 – Slide 21