FINITE ELEMENT : TECHNOLOGY -...
Transcript of FINITE ELEMENT : TECHNOLOGY -...
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FINITE ELEMENT : TECHNOLOGY
Georges CailletaudEcole des Mines de Paris, Centre des Materiaux
UMR CNRS 7633
Contents 1/41
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Contents
Gauss integrationDefinition4-node quadrilateralQuality of the integration
Patch test
Rigid body mode
Locking
Spurious modes
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Contents
Gauss integrationDefinition4-node quadrilateralQuality of the integration
Patch test
Rigid body mode
Locking
Spurious modes
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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
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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 )
Gauss integration 5/41
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Onedimensional Gauss integration
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)
Gauss integration 6/41
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Onedimensional Gauss integration
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 7/41
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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
Gauss integration 8/41
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Contents
Gauss integrationDefinition4-node quadrilateralQuality of the integration
Patch test
Rigid body mode
Locking
Spurious modes
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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
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4-node quadrilateral (2)
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
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4-node quadrilateral (3)
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
Gauss integration 12/41
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4-node quadrilateral (4)
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
Gauss integration 13/41
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Contents
Gauss integrationDefinition4-node quadrilateralQuality of the integration
Patch test
Rigid body mode
Locking
Spurious modes
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Number of Gauss points for an exact integration (1)
The following terms are present in the interpolation functions and theirderivatives:
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
Gauss integration 15/41
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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
Gauss integration 16/41
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Rectangular C2D8 element
Uniform jacobian
The following terms are present in the interpolation functions and theirderivatives:
[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
Gauss integration 17/41
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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
Gauss integration 18/41
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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)
Gauss integration 19/41
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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
Gauss integration 20/41
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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
Gauss integration 21/41
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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
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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
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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
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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
Rigid body mode 25/41
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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
Rigid body mode 26/41
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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
Rigid body mode 27/41
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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
Rigid body mode 28/41
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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 40/41
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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)
Spurious modes 41/41