Nonlinear FEM
11The TL Timoshenko
Beam Element:Formulation
NFEM Ch 11 – Slide 1
Nonlinear FEM
TL Beam Element Application
motion
current configuration
reference configuration
NFEM Ch 11 – Slide 2
Nonlinear FEM
Beam Terminology
motion
current configuration
reference configuration
Current cross section
Referencecross section
X
X
, x
Y, y
u X
uY
Z≡
(X)
(X)
θ (X) θ(X)
NFEM Ch 11 – Slide 3
Nonlinear FEM
Reduction To One Dimensional Model
motion
current configuration
reference configuration finite element idealizationof reference configuration
finite element idealizationof current configuration
NFEM Ch 11 – Slide 4
Nonlinear FEM
Beam Mathematical Models
uX1
uY 1
uX2
uY 2
θ1
θ2
uX1
uY 1
uX2
uY 2
θ1
θ2
1 2 1 2
C (Timoshenko) model
X; x
Y; y
C (BE) model10
NFEM Ch 11 – Slide 5
Nonlinear FEM
Timoshenko Beam Kinematics
normal to deformed beam axis
90
normal to referencebeam axis X
// X (X = X)
ds
θψ
ψdirection of deformedcross section
_ γ = ψ−θ
_
Note: in practice γ << θ; typically 0.1% or less. Magnitude of γ is grossly exaggerated in the figure for visualization convenience.
__
o
NFEM Ch 11 – Slide 6
Nonlinear FEM
Comparing The Two ModelsOver Individual Finite Element
2-node (cubic) elementfor Euler-Bernoulli beam model:plane sections remain plane andnormal to deformed longitudinal axis
2-node linear-displacement-and-rotations element for Timoshenko beam model:plane sections remain plane but notnormal to deformed longitudinal axis
C1
element with same DOFsC1
C0
NFEM Ch 11 – Slide 7
Nonlinear FEM
Element Kinematics:Reduction from 2D to 1D
1 2
P(x,y)
P (X,Y)0 C (X)0
XPuXCu
x ≡ xx
C(x ,y )C C
C (X,0)0
1
2
1 2
θ(X)
C(X+u ,u )X Y
XCXu (X) = u
1
2
C0
C
(a) (b)
X, xX, x
Y, y Y, y
0L
ψψ
ψSection rotation isθ=ψ+γ=ψ−γ
_
Py ≡y
P0Y ≡Y
P0 C0X≡X ≡X
CyC
P
X
reduction to 1D
Y YCu (X) = u
//Xψ
L
θ1 γ1
θ2 γ2
NFEM Ch 11 – Slide 8
Nonlinear FEM
Degrees of Freedom of Beam Element(they are model independent)
u =
u X1uY 1θ1
u X2uY 2θ2
f =
fX1fY 1fθ1fX2fY 2fθ2
NFEM Ch 11 – Slide 9
Nonlinear FEM
Element Formulation Path
X
X
Y
Y
Eqs (10.5), (10.8)
Eq. (10.9)
Node Displacements u
Element displacement field w = [u , u , θ]
Displacement gradients w' = [u' , u' , θ' ]
Generalized strains h = [ e , γ, κ ]
Stress resultants z = [ N ,V, M ]
Strain energy U
Eq. (10.15)
Eq. (10.29)
T
T
T
T
Tangent stiffness matrix K = K + K M G
Internal forces p
Eq. (10.34)
Eq. (10.40)
TT TδU = ∫ z B dX δu = p δuL 0vary U:
_
T Tδp = ∫ (B δz + δB z) dX δu = (K + K )L0
M Gvary p:_
NFEM Ch 11 – Slide 10
Nonlinear FEM
The Longest Trip Begins With The First Step
Element motion in Lagrangian description
Extended displacement vector
NFEM Ch 11 – Slide 11
Nonlinear FEM
Isoparametric Interpolation Of Extended Displacements
NFEM Ch 11 – Slide 12
Nonlinear FEM
Deformation Gradient andDisplacement Gradient Tensors
NFEM Ch 11 – Slide 13
Nonlinear FEM
Green-Lagrange Strains
Tensor form:
Tensor component form:
These are geometrically exact, but unwieldy. Simplifications are necessary - those are detailed in Chapter 10.
NFEM Ch 11 – Slide 14
Nonlinear FEM
Simplified Strain Field
We can characterize axial strains, shear strains and bending strains
Collected in the generalized strain vector
Axial strain is linearized with respect to displacements usinga technique called polar decomposition. The result is
NFEM Ch 11 – Slide 15
Nonlinear FEM
Geometrically Invariant Form ofAxial and Shear Strains
Using this invariant form we can pass to a beam element arbitrarily oriented in the reference configuration
NFEM Ch 11 – Slide 16
Nonlinear FEM
Arbitrary Reference Configuration
C0
C
ϕ
φ = ψ+ϕ
ϕ
ψ
uX1
1(X ,Y )1 1
1(x ,y )1 1
2(x ,y )2 2
2(X ,Y )2 2
uY1uX2
uY21θ
2θ
X, x
Y, y
γ// X
γ// X
γ// X_γ// Y
_
γ// Y_
γX_
γY_
M0 N 0V0
MN
V
C0
C
(a) (b)
NFEM Ch 11 – Slide 17
Nonlinear FEM
What Remains To Be Done
The remaining steps include:
PK2 stresses and stress resultants Strain energy Internal force Tangent stiffness: material + geometric Improving element performance
There is a lot of algebra left, even for this relatively simple element. A computer algebra system (CAS) can really help, especially to avoid algebra errors
NFEM Ch 11 – Slide 18
Nonlinear FEM
Flowcharted Steps (again)
X
X
Y
Y
Eqs (10.5), (10.8)
Eq. (10.9)
Node Displacements u
Element displacement field w = [u , u , θ]
Displacement gradients w' = [u' , u' , θ' ]
Generalized strains h = [ e , γ, κ ]
Stress resultants z = [ N ,V, M ]
Strain energy U
Eq. (10.15)
Eq. (10.29)
T
T
T
T
Tangent stiffness matrix K = K + K M G
Internal forces p
Eq. (10.34)
Eq. (10.40)
TT TδU = ∫ z B dX δu = p δuL 0vary U:
_
T Tδp = ∫ (B δz + δB z) dX δu = (K + K )L0
M Gvary p:_
So far we have worked up to here
The remaining isa lot of matrix algebra:1-2 months by hand,~1 week with a CAS
NFEM Ch 11 – Slide 19
Nonlinear FEMEnergy and Internal ForcesStrain energy of beam element
Vary to get internal force
Evaluate by one point Gauss quadrature:
NFEM Ch 11 – Slide 20
Nonlinear FEM
Tangent Stiffness: Material Component
Vary internal force to get tangent stiffness
Evaluate by one point Gauss quadrature:
Material stiffness matrix
NFEM Ch 11 – Slide 21
Nonlinear FEMMaterial Stiffness Subcomponents
in which
Material stiffness subcomponents due to axial, bending and shear:
NFEM Ch 11 – Slide 22
Nonlinear FEMMaterial Stiffness "Unlocking"
is formally replaced by in shear material stiffness
in which and
Eliminating "shear locking" by MacNeal's Residual Bending Flexibility (RBF) device:
Bending and shear components merge into a modified bending stiffness:
NFEM Ch 11 – Slide 23
Nonlinear FEMGeometric Stiffness
∫The geometric stiffness comes from the variation of strain-displacement matrix B while stress resultants (axial forces, shear forces and bending moments) are kept fixed:
After tons of algebra (see Notes) one arrives at
which evaluated by one point Gauss rule gives the closed form
NFEM Ch 11 – Slide 24