the tl timoshenko beam element: formulation · pdf fileelement for timoshenko beam model:...

Nonlinear FEM

11The TL Timoshenko

Beam Element:Formulation

NFEM Ch 11 – Slide 1

Nonlinear FEM

TL Beam Element Application

motion

current configuration

reference configuration


Nonlinear FEM

Beam Terminology

motion


reference configuration

Current cross section

Referencecross section

X

X

, x

Y, y

u X

uY

Z≡

(X)

(X)

θ (X) θ(X)


Nonlinear FEM

Reduction To One Dimensional Model

motion


reference configuration finite element idealizationof reference configuration

finite element idealizationof current configuration


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


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


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


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


Nonlinear FEM

Degrees of Freedom of Beam Element(they are model independent)

u =

u X1uY 1θ1

u X2uY 2θ2

f =

fX1fY 1fθ1fX2fY 2fθ2


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:_


Nonlinear FEM

The Longest Trip Begins With The First Step

Element motion in Lagrangian description

Extended displacement vector


Nonlinear FEM

Isoparametric Interpolation Of Extended Displacements


Nonlinear FEM

Deformation Gradient andDisplacement Gradient Tensors


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.


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


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


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)


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


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


Nonlinear FEMEnergy and Internal ForcesStrain energy of beam element

Vary to get internal force

Evaluate by one point Gauss quadrature:


Nonlinear FEM

Tangent Stiffness: Material Component

Vary internal force to get tangent stiffness

Evaluate by one point Gauss quadrature:

Material stiffness matrix


Nonlinear FEMMaterial Stiffness Subcomponents

in which

Material stiffness subcomponents due to axial, bending and shear:


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:


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


the tl timoshenko beam element: formulation · pdf fileelement for timoshenko beam model:...

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