FINITE ELEMENT : TECHNOLOGY -...

FINITE ELEMENT : TECHNOLOGY

Georges CailletaudEcole des Mines de Paris, Centre des Materiaux

UMR CNRS 7633

Contents 1/41

Contents

Gauss integrationDefinition4-node quadrilateralQuality of the integration

Patch test

Rigid body mode

Locking

Spurious modes

Gauss integration

r -point Gauss integration on a [-1:+1] segment:∫ +1

−1

f (ξ)dξ ≈r∑1

wi f (ξi )

gives exact result for a (2r − 1) order polynomEvaluation at sampling points ξi , combined with weigthing coefficientswi

Example, order 2:f (ξ) = 1 : 2 = w1 + w2

f (ξ) = ξ : 0 = w1ξ1 + w2ξ2

f (ξ) = ξ2 : 2/3 = w1ξ21 + w2ξ

22

f (ξ) = ξ3 : 0 = w1ξ31 + w2ξ

32

then w1 = w2 = 1, and ξ1 = −ξ2 = 1/√

3

Gauss integration 4/41

Onedimensional Gauss integration

One integration point

-1

0

1

2

3

4

5

-1 -0.5 0 0.5 1

f(x)

x

f (x) = x5 + x3 + 3x2 − 1/2

Gauss approximation

∫ 1

−1

f (ξ)dξ ≈ 2 f (0 )



Two integration points

-1

0

1

2

3

4

5

-1 -0.5 0 0.5 1

f(x)

x

f (x) = x5 + x3 + 3x2 − 1/2

Gauss approximation

∫ 1

−1

f (ξ)dξ ≈ 1× f (−1/√

3) + 1× f (1/√

3)



Three integration points

-1

0

1

2

3

4

5

-1 -0.5 0 0.5 1

f(x)

x

f (x) = x5 + x3 + 3x2 − 1/2

Gauss approximation

∫ 1

−1

f (ξ)dξ =5

9f (−

√3/5) +

8

9f (0) +

5

9f (√

3/5)


Gauss integration in a N-dimensional space

∫ 1

−1

∫ 1

−1

∫ 1

−1

f (ξ, η, ζ)dξdηdζ =

∫ 1

−1

dξ

∫ 1

−1

dη

∫ 1

−1

f (ξ, η, ζ)dζ

Respectively r1, r2, r3 Gauss points in each direction, so that:

∫ 1

−1

∫ 1

−1

∫ 1

−1

f (ξ, η, ζ)dξdηdζ =r1∑

i=1

r2∑j=1

r3∑k=1

wiwjwk f (ξi , ηj , ζk)

Usually, r1 = r2 = r3

Special integration rules for triangles


Contents


Patch test

Rigid body mode

Locking

Spurious modes

4-node quadrilateral (1)

1

2

3

4

x

y

1 2

34

−1

−1

1

1 ξ

η

Bilinear interpolation of the geometry and of the unknown function

x = N1x1 + N2x2 + N3x3 + N4x4 y = N1y1 + N2y2 + N3y3 + N4y4

ux = N1qx1 + N2qx2 + N3qx3 + N4qx4 uy = N1qy1 + N2qy2 + N3qy3 + N4qy4

Shape functions

N1(ξ, η) = (1− ξ)(1− η)/4 N2(ξ, η) = (1 + ξ)(1− η)/4

N1(ξ, η) = (1 + ξ)(1 + η)/4 N2(ξ, η) = (1− ξ)(1 + η)/4

Jacobian matrix

[J] =1

4

„−x1 + x2 + x3 − x4 + η(x1 − x2 + x3 − x4) −y1 + y2 + y3 − y4 + η(y1 − y2 + y3 − y4)−x1 − x2 + x3 + x4 + ξ(x1 − x2 + x3 − x4) −y1 − y2 + y3 + y4 + ξ(y1 − y2 + y3 − y4)

«Gauss integration 10/41


Determinant of the Jacobian matrix: terms in ξ and η

8J =(y4 − y2)(x3 − x1)− (y3 − y1)(x4 − x2)

+ ((y3 − y4)(x2 − x1)− (y2 − y1)(x3 − x4)) ξ

+((y4 − y1)(x3 − x2)− (y3 − y2)(x4 − x1)) η

Inverse of the jacobian matrix: homographic function in ξ and η

[J]−1 =4

J

„−y1 − y2 + y3 + y4 + ξ(y1 − y2 + y3 − y4) −x1 − x2 + x3 + x4 + ξ(x1 − x2 + x3 − x4)+y1 − y2 − y3 + y4 − η(y1 − y2 + y3 − y4) +x1 − x2 − x3 + x4 − η(x1 − x2 + x3 − x4)

«

Derivative of the shape functions:„∂N1/∂x∂N1/∂y

«= [J]−1

„∂N1/∂ξ∂N1/∂η

«= [J]−1

„−(1− η)/4−(1− ξ)/4

«Terms in„

∂N1/∂x∂N1/∂y

«= 1

J

„(1, ξ) (1, ξ)(1, η) (1, η)

« „−(1− η)/4−(1− ξ)/4

«=

0BB@(1, ξ, η, ξη, ξ2)

(1, ξ, η)(1, ξ, η, ξη, η2)

(1, ξ, η)

1CCAGauss integration 11/41


The jacobian is an homographic function for a generic quad.

[K ] is obtained by Gauss integration, [K ] =RΩ[B]T [D][B]dΩ

[K ] =

Z 1

−1

Z 1

−1

[B]T [D][B]Jdξdη =

pXi=1

pXj=1

wiwjJ((ξi , ηj )[B]T (ξi , ηj )[D][B](ξi , ηj )

The stiffness matrix includes terms like(1, ξ, η, ξη, ξ2, ..., η4, ξ4, ξ2η2)

(1, ξ, η)For a generic quad, the integration of [K ] is never exact

The internal forces are computed as:

[Fint ] =

ZΩ[B]T [σ]dΩ =

pXi=1

pXj=1

wiwjJ((ξi , ηj )[B]T (ξi , ηj )[σ(ξi , ηj )]

The determinant at the denominator of [B] vanishes with thedeterminant due to elementary volumeThe internal forces (σ constant) include terms like(1, ξ, η, ξη, ξ2, ..., η2, ξ2, ξ2η, ξη2)Internal forces are integrated with a 2× 2 rule



If the ”real world” quad is a parallelogram, the relation (x , y)–ξ, η are linear andnot bilinear, so that the partial derivatives ∂x/∂ξ, etc. . . are constant. Thejacobian is also constant.

The following terms are present in the interpolation functions and theirderivatives:

[N] 1 ξ η ξη[∂N/∂ξ] 0 1 0 η[∂N/∂η] 0 0 1 ξ

[K ] is obtained by Gauss integration, using a constant J.

The product [B]T (ξi , ηi )[D][B](ξj , ηj), and also the stiffness matrix,include terms like ξiηj with i + j 6 2[K ] is exactly integrated with a 2× 2 rule

The internal forces present only linear terms ×σ

Only one Gauss point is needed for constant stress state, and 2× 2 for linear stresses


Contents


Patch test

Rigid body mode

Locking

Spurious modes

Number of Gauss points for an exact integration (1)


For generic geometries, the computation of [B] involves derivativesof the shape functions and partial derivative of the coordinate of thephysical space wrt the reference coordinates:

∂Ni

∂x=

∂Ni

∂ξ

∂ξ

∂x

A typical space derivative term is . . . . . . . . . . . . . . . . . . . . .∂ξ

∂x=

1

J

∂y

∂η

A typical term of [B] is then . . . . . . . . . . . . . . . . . . . . . . . . . . .1

J

∂Ni

∂ξ

∂y

∂η

A typical term of [K ] is then . . . . . . . . . . . . . . . . . . . . . .1

J

(∂Ni

∂ξ

∂y

∂η

)2

A typical term of [Fint ] is then . . . . . . . . . . . . . . . . . . . . . . . . . . . .∂Ni

∂ξ

∂y

∂η


Number of Gauss points for an exact integration (2)

For linear geometries, and constant jacobian matrix (introducing theconstant a):

∂Ni

∂x= a

∂Ni

∂ξ

A typical term of [B] is then . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .∂Ni

∂ξ

A typical term of [K ] is then . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(∂Ni

∂ξ

)2

A typical term of [Fint ] is then . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .∂Ni

∂ξ


Rectangular C2D8 element

Uniform jacobian


[N] 1 ξ η ξ2 ξη η2 ξ2η ξη2

[∂N/∂ξ] 0 1 0 2ξ η 0 2ξη η2

[∂N/∂η] 0 0 1 0 ξ 2η ξ2 2ξη

The product [B]T [D][B] includes terms like ξiηj with i + j ≤ 4

3× 3 points for a full integration (too much, exact until 5)

2× 2 points are for reduced integration


Number of Gauss points needed

Element Geometry Loading [K ] [Fint ]C2D4 Linear Constant 4 1C2D4 Bilinear Constant NO 1C2D4 Linear Linear 4 4C2D4 Bilinear Linear NO 4

C2D8 Linear Constant 4 4C2D8 Bilinear Constant NO 4C2D8 Bilinear Linear NO 4C2D8 Generic Constant NO 4

C3D8 Linear Constant 8 1C3D8 Trilinear Constant NO 8

C3D20 Linear Constant 27 8C3D20 Trilinear Constant NO 8C3D20 Trilinear Linear NO 27C3D20 Generic Constant NO 27


Precision of the Gauss integration method

a

a α

x

y

a

Compute I =

∫ +1

−1

∫ +1

−1

1

Jdξdη

Using a mapping on a square[0..1]:

x = (1 + (α− 1)η)aξ

y = aη

[J] =

(a(1 + (α− 1)η) a(α− 1)ξ

0 a

)

I =1

a2

∫ 1

0

dη

1 + (α− 1)η

Order α = 2 α = 5 α = 101 0.66666 0.33333 0.181822 0.69231 0.39130 0.234043 0.69312 0.40067 0.24962

exact 0.69315 0.40236 0.25584

Analytic expression:

I =log α

a2(α− 1)

(after Dhatt and Touzot)


Global algorithm

For each loading increment, do while ‖Riter‖ > EPSI :iter = 0; iter < ITERMAX ; iter + +

1 Update displacements: ∆uiter+1 = ∆uiter + δuiter

2 Compute ∆ε = [B].∆uiter+1 then ∆ε∼ for each Gauss point

3 Integrate the constitutive equation: ∆ε∼→ ∆σ∼, ∆αI ,∆σ∼∆ε∼

4 Compute int and ext forces: Fint(ut + ∆uiter+1) , Fe5 Compute the residual force: Riter+1 = Fint − Fe6 New displacement increment: δuiter+1 = −[K ]−1.Riter+1


Convergence

Value of the residual forces < Rε, e.g.

||R||n =

(∑i

Rni

)1/n

; ||R||∞ = maxi|Ri |

Relative values:

||Ri − Re ||||Re ||

< ε

Displacements ∣∣∣∣Uk+1 − Uk

∣∣∣∣n

< Uε

Energy [Uk+1 − Uk

]T. Rk < Wε


The concept of patch (engineering version)

Apply a given displacement field on the external (blue) nodes

Check the results in the internal (red) nodes

For instance, uniform strain; or shear, or bending

Check with a bending displacement field: ux = xy anduy = −0.5(x2 + νy 2), assuming bilinear geometry (with ξη term)

The resulting displacement for uy should have terms like (a + bξ + cη + dξη)2. A

nine node quad will pass, but not an eight-node quad (missing ξ2η2 term in thepolynomial base).The eight-node pass, provided the edges are straightThis demonstrates also the limitations of high order elements. For them, a complexshape (terms in ξ3η, ξη3 for a cubic interpolation introduces terms like ξ6η2,etc. . . for a correct patch test simulation. They are not in the polynomial basis...

Patch test 22/41

Rigid body mode (1)

Example of a 2D plane element

Zero strain:

u1,1 = 0 ; u2,2 = 0 ; u1,2 + u2,1 = 0

Possible displacement field:

u1 = A− Cx2 ; u2 = B + Cx1

3 rigid body modes:

2 translations

(10

)and

(01

); 1 rotation

(−x1

x2

)

Patch test 23/41

Rigid body mode (2)

Example of a 2D axisymmetric element

Zero strain:

εr = ur ,r = 0 ; εθ =ur

r= 0 ; uz,z = 0

Possible displacement field:

uz = A

Only 1 rigid body modes: 1 translation

(01

)

Rigid body mode 24/41

Rank sufficiency/deficiency

No zero-energy mode other than rigid body modes

r is the rank of the elementary stiffness matrix (number ofevaluations)

Check r with respect to nF − nR (nF is the number of element DOF,nR is the number of rigid body modes)

Rank sufficient element iff r > nF − nR

Rank deficiency, d in the case d = nF − nR − r > 0

Each Gauss point adds nE to the rank of the matrix (nE is the orderof the stress-strain matrix, nG the number of Gauss points),r = nEnG

RULE: nEnG > nF − nR


Rank-sufficient Gauss integration

Element n nF nF − nR Min ng rule3-node triangle 3 6 3 1 1-pt6-node triangle 6 12 9 3 3-pt4-node quadrilateral 4 8 5 2 2x28-node quadrilateral 8 16 13 5 3x39-node quadrilateral 9 18 15 5 3x38-node hexahedron 8 24 18 3 2x2x220-node hexahedron 20 60 54 9 3x3x3


Stiffness matrix of a rectangular element

rectangle [±a,±a/2]; E=96; ν=1/3

[K ] =

0BBBBBBBBBB@

42 18 −6 0 −21 −18 −15 018 78 0 30 −18 −39 0 −69−6 0 42 −18 −15 0 −21 180 30 −18 78 0 −69 18 −39

−21 −18 −15 0 42 18 −6 0−18 −39 0 −69 18 78 0 30−15 0 −21 18 −6 0 42 −180 −69 18 −39 0 30 −18 78

1CCCCCCCCCCA

Eigenvalues = 223.4 90 78 46.36 42 0 0 0 (three rigid body modes)

Carlos Felippa


Stiffness matrix of a trapezoidal element

a

a

2a1 2

34

E=4206384; ν=1/3

Rule Eigenvalues obtained with different Gauss rules (scaled by 10−6)

1x1 8.77276 3.68059 2.26900 0 0 0 0 02x2 8.90944 4.09769 3.18565 2.64523 1.54678 0 0 03x3 8.91237 4.11571 3.19925 2.66438 1.56155 0 0 04x4 8.91246 4.11627 3.19966 2.66496 1.56199 0 0 0

Three rigid body modes, but a rank deficiency by TWO is too few Gauss pointsare used

Carlos Felippa


Analysis of the locking penomenon

For bad reasons, the element becomes too stiff

Shear locking

Volumetric locking

Trapezoidal locking

Locking in fields

Locking 29/41

Alias functions

Function which tries to mimic a given function in one element

Basis for a C2D3: (1, ξ, η), for a C2D4: (1, ξ, η, ξη),for a C2D8: (1, ξ, η, ξ2, ξη, η2, ξ2η, ξη2).

Alias for various functions:

Function ξ2 ξη η2 ξ3 ξ2η ξη2 η3

C2D3 ξ 0 η ξ 0 0 ηC2D4 1 OK 1 ξ η ξ ηC2D8 OK OK OK ξ OK OK η

Locking 30/41

Shear locking

C2D4, −L ≤ x1 ≤ L, −1 ≤ x2 ≤ 1

Actual fieldu1 = x1x2

u2 = −x21/2

Aliased fieldu1 = x1x2

u2 = −L2/2 !!

Computed shear for the alias, ε12 = x/2 !! (actual solution: 0)Computed stored elastic energy We for the real field and Wa for the alias:

Wa

Wo= 1 +

1− ν

2L2

Solve the problem by computing shear on the middle of the element

Locking 31/41

Shear locking (2)

Analytic solution

ε11 = u1,1 = x2 σ11 =Ex2/(1− ν2)

ε22 = u2,2 = 0 σ22 =νEx2/(1− ν2)

2ε12 = u2,1 + u1,2 = 0 σ12 =0

Solution with the alias

ε11 = u1,1 = x2 σ11 =Ex2/(1− ν2)

ε22 = u2,2 = 0 σ22 =νEx2/(1− ν2)

2ε12 = u2,1 + u1,2 = x1 σ12 =(1/2)Ex1/(1 + ν)

Wo =1

2

∫Ω

σ∼ : ε∼dΩ

Locking 32/41

Dilatational locking

C2D4, −L ≤ x1 ≤ L, −1 ≤ x2 ≤ 1

Actual fieldu1 = x1x2

u2 = −x21

2− ν

2(1− ν)x22

Aliased fieldu1 = x1x2

u2 = −L2

2− ν

2(1− ν)!!

The computed stored elastic energy Wa for the alias tends to infinity if νtends to 0.5

Wa =E

2(1 + ν)

(1− ν

1− 2νx22 +

x21

2

)instead of: We =

E

2(1− ν2)x22

Solve the problem by adding a non conform. displacement (x21 − L2, x2

2 − 1)

Locking 33/41

Dilatational locking (2)

Analytic solution

ε11 = u1,1 = x2 σ11 =Ex2/(1− ν2)

ε22 = u2,2 = −νx2/(1− ν) σ22 =0

ε33 = u3,3 = 0 σ33 =

2ε12 = u2,1 + u1,2 = 0 σ12 =0

Solution with the alias

ε11 = u1,1 = x2 σ11 =E (1− ν)x2/(1 + ν)(1− 2ν)

ε22 = u2,2 = 0 σ22 =νEx2/(1 + ν)(1− 2ν)

2ε12 = u2,1 + u1,2 = x1 σ12 =(1/2)Ex1/(1 + ν)

Wo =1

2

∫Ω

σ∼ : ε∼dΩ

Locking 34/41

Locking of the 8-node rectangle

Consider a rectangle of length 2Λ and width 2 (with Λ > 1)

Displacement basis: 1, ξ, η, ξ2, ξη, η2, ξ2η, ξη2

Try u1 = x21 x2, u2 = −x3

1/3, so that ε11 = 2x1x2, ε22 = 0, ε12 = 0.

In fact, u2 represented by its alias, −x1Λ2/3, so that the shear is

x21 − Λ2/3

No locking if the shear is evaluated at the second order Gauss point(x1 = ±1/

√3)

Underintegration is a good remedy to locking

Locking 35/41

Locking of the 8-node rectangle (2)

Analytic solution

ε11 = u1,1 = 2x1x2 σ11 =2Ex1x2

ε22 = u2,2 = 0 σ22 =0

ε33 = u3,3 = 0 σ33 =

2ε12 = u2,1 + u1,2 = 0 σ12 =0

Solution with the alias (u1 = x21 x2 and u2 = −x1Λ

2/3)

ε11 = u1,1 = 2x1x2 σ11 =2Ex1x2

ε22 = u2,2 = 0 σ22 =0

ε33 = u3,3 = 0 σ33 =0

2ε12 = u2,1 + u1,2 = x21 − Λ2/3 σ12 =(1/2)(x2

1 − Λ2/3)/(1 + ν)

Wo =1

2

∫Ω

σ∼ : ε∼dΩ

Locking 36/41

Trapezoidal locking

2 (1+ ) Λ α

2 (1− ) αΛ

δ δx

y

Geometry: x1 = Λξ(1− αη), and x2 = η; ξ = x1/(1− αx2)Λ.

Displacement basis: 1, ξ, η, ξ2, ξη, η2, ξ2η, ξη2

Try u1 = x21 x2, u2 = −x2

1 /2, so that ε11 = x2, ε22 = 0, ε12 = 0.

In fact, the solution with the alias is:

ε11 =η − α

1− αηε22 = αΛ2 2ε12 = Λξ

„1 +

α(η − α)

1− αη

«All components are affected

Error on shear component suppressed if the evaluation is made at ξ = 0only.

Error on ε22 cannot be easily suppressed

Locking 37/41

Dilatational locking on triangles

triangle C2D3

Incompressible, BL → TR mesh.......x

y

z

Incompressible, TL → BR mesh.......x

y

z

Compressible, BL → TR mesh.........x

y

z

Locking 38/41

Spurious modes of a C2D4

Find a displacement field which does not produce any strain on the Gausspoints

u12 = u2

2 = u32 = u4

2 = 0

u11 = +a ; u2

1 = −a

u31 = +a ; u4

1 = −a 2

34

1

C2D4 — C2D4r — C2D8 — C2D8r

Spurious modes 39/41

Spurious modes

An element has ”internal degrees of freedom” which allow deformationprocess to occur in the element

Rigid body mode Φo such as: [K ] Φo = 0 everywhere

Spurious mode Φo such as: [K ] Φo = 0 in some places

The number of independent states is given by the total number of dof inthe element

Number of independent states - Rigid body mode - Number of evaluationof the strain components = Number of spurious modes

For an element of degree p, the number of strain evaluations is:3p2 (reduced integration, 2D); 3(p + 1)2 (full integration, 2D); 6p2

(reduced integration, 3D); 6p2 (reduced integration, 3D).

The number of strain states is:8p3 (Serendip, 2D); 2(p + 1)2 (Lagrange, 2D); 36p − 6 (Serendip, 3D);3(p + 1)3 − 6 Lagrange, 3D).


Spurious modes

Polynomial Serendip 2D Lagrange 2D Serendip 3D Lagrange 3Ddegree p

1 2 2 12 122 1 3 6 27

For the 8-node underintegrated element, the following is a spurious mode:

u1 = k1ξ(η2 − 1/3)

u2 = −k2η(ξ2 − 1/3)


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