e Pathology

8/13/2019 e Pathology

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Contents

Gauss integrationDefinition4-node quadrilateralQuality of the integration

Patch test

Rigid body mode

Locking

Spurious modes
http://goforward/http://find/http://goback/


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Contents


Patch test

Rigid body mode

Locking

Spurious modes
http://find/


4/41

Gauss integration

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

+11

f()dr1

wif(i)

gives exact result for a (2r

1) order polynom

Evaluation at sampling points i, combined with weigthing coefficientswiExample, order 2:

f() = 1 : 2 = w1+w2f() = : 0 = w11+w22

f() =2

: 2/3 = w121+w2

22

f() =3 : 0 = w131+w2

32

then w1 =w2 = 1, and 1 = 2 = 1/

3

Gauss integration 4/41
http://find/


5/41

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

11

f()d2 f(0)

http://find/


6/41


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

11

f()d 1 f(1/

3) + 1 f(1/

3)

http://goforward/http://find/http://goback/


7/41


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

11

f()d=5

9f(

3/5) +

8

9f(0) +

5

9f(

3/5)

http://find/


8/41

Gauss integration in a N-dimensional space

11

11

11

f( , , )ddd=

11

d

11

d

11

f( , , )d

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

11

11

11

f( , , )ddd=

r1i=1

r2j=1

r3k=1

wiwjwkf(i, j, k)

Usually, r1 =r2 =r3

Special integration rules for triangles

http://find/


9/41

Contents


Patch test

Rigid body mode

Locking

Spurious modes
http://find/http://goback/


10/41

4-node quadrilateral (1)

1

2

3

4

x

y

1 2

34

1

1

1

1

Bilinear interpolation of the geometry and of the unknown functionx=N1x1+ N2x2+ N3x3+ N4x4 y=N1y1+ N2y2+ N3y3+ N4y4

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

Shape functions

N1(, ) = (1)(1)/4 N2(, ) = (1 +)(1)/4N1(, ) = (1 +)(1 +)/4 N2(, ) = (1)(1 +)/4

Jacobian matrix

[J] = 1

4

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

http://find/


11/41


Determinant of the Jacobian matrix: terms in and

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

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

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

Inverse of the jacobian matrix: homographic function in and

[J]1 =4

J

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

Derivative of the shape functions:N1/xN1/y

= [J]1

N1/N1/

= [J]1

(1 )/4(1 )/4

Terms in

N1/xN1/y

= 1

J

(1, ) (1, )(1, ) (1, )

(1 )/4(1 )/4

=

0BB@

(1, , , , 2)

(1, , )(1, , , , 2)

(1, , )

1CCA

http://find/


12/41


The jacobian is an homographic function for a generic quad.

[K] is obtained by Gauss integration, [K] =R

[B]T[D][B]d

[K] =

Z 11

Z 11

[B]T

[D][B]Jdd =

pXi=1

pXj=1

wiwjJ((i, j)[B]T

(i, j)[D][B](i, j)

The stiffness matrix includes terms like (1, ,,,2,...,4, 4, 22)

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

The internal forces are computed as:

[Fint] =

Z

[B]T[]d =

pXi=1

pXj=1

wiwjJ((i, j)[B]T(i, j)[(i, j)]

The determinant at the denominator of [B] vanishes with thedeterminant due to elementary volumeThe internal forces ( constant) include terms like(1, ,,,2,...,2, 2, 2,2)Internal forces are integrated with a 2 2 rule

http://find/


13/41


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

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

[N] 1

[N/] 0 1 0 [N/] 0 0 1

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

The product [B]T(i, i)[D][B](j, j), and also the stiffness matrix,include terms like ij with i+j 2

[K] is exactly integrated with a 2

2 rule

The internal forces present only linear terms

Only one Gauss point is needed for constant stress state, and2 2 for linear stresses

http://find/


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Contents


Patch test

Rigid body mode

Locking

Spurious modes
http://find/


15/41

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

x

A typical space derivative term is .....................

x =

1

J

y

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

J

Ni

y

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

J

Ni

y

2

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

y


( )
http://find/


16/41

Number of Gauss points for an exact integration (2)

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

Nix

=aNi

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

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

Ni

2

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


R l C2D8 l
http://find/


17/41

Rectangular C2D8 element

Uniform jacobianThe following terms are present in the interpolation functions and theirderivatives:

[N] 1 2 2 2 2

[N/] 0 1 0 2 0 2 2

[N/] 0 0 1 0 2 2 2

The product [B]T[D][B] includes terms like ij with i+j 43

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

2 2 points are for reducedintegration


N b f G i d d
http://find/


18/41

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 4

C2D8 Bilinear Linear NO 4C2D8 Generic Constant NO 4

C3D8 Linear Constant 8 1C3D8 Trilinear Constant NO 8

C3D20 Linear Constant 27 8

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


P i i f th G i t ti th d
http://find/


19/41

Precision of the Gauss integration method

a

a

x

y

a

Compute I =

+11

+11

1

Jdd

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

x= (1 + ( 1))ay=a

[J] =a(1 + (

1)) a(

1)

0 a

I= 1

a2

10

d

1 + ( 1)

Order = 2 = 5 = 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

a2( 1)(afterDhattandTouzot)


Gl b l l ith
http://find/


20/41

Global algorithm

For each loading increment, do while{R}iter >EPSI:iter= 0; iter


21/41

Convergence

Value of the residual forces


22/41

The concept ofpatch (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(x

2 +y2), assuming bilinear geometry (with term)

The resulting displacement for uy should have terms like (a+b+ c+ d)2. Anine node quad will pass, but not an eight-node quad (missing 22 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 3, 3 for a cubic interpolation introduces terms like 62,etc. . . for a correct patch test simulation. They are not in the polynomial basis...

Patch test 22/41

Rigid body mode (1)
http://find/


23/41

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

x1x2

Patch test 23/41

Rigid body mode (2)
http://find/


24/41

Rigid body mode (2)

Example of a 2D axisymmetric element

Zero strain:

r =ur,r= 0 ; =

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
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25/41

Rank sufficiency/deficiency

No zero-energy mode other than rigid body modesris the rank of the elementary stiffness matrix (number ofevaluations)

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

Rank sufficient element iffr nF nRRank deficiency, d in the case d=nF nR r 0Each Gauss point adds nEto the rank of the matrix (nE is the orderof the stress-strain matrix, nGthe number of Gauss points),

r=nEnGRULE: nEnG nF nR


Rank-sufficient Gauss integration
http://find/


26/41

Rank-sufficient Gauss integration

Element n nF nF nR Minng rule3-node triangle 3 6 3 1 1-pt6-node triangle 6 12 9 3 3-pt4-node quadrilateral 4 8 5 2 2x2

8-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
http://find/


27/41

Stiffness matrix of a rectangular element

rectangle [a,a/2]; E=96; =1/3

[K] =

0

BBBBBBBBBB@

42 18 6 0 21 18 15 018 78 0 30 18 39 0 696 0 42 18 15 0 21 18

0 30 18 78 0 69 18 3921 18 15 0 42 18 6 018 39 0 69 18 78 0 3015 0 21 18 6 0 42 18

0 69 18 39 0 30 18 78

1

CCCCCCCCCCA

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

Carlos Felippa


Stiffness matrix of a trapezoidal element
http://find/


28/41

Stiffness matrix of a trapezoidal element

a

a

2a1 2

34

E=4206384; =1/3

Rule Eigenvalues obtained with different Gauss rules (scaled by 106)

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 0

4x4 8.91246 4.11627 3.19966 2.66496 1.56199 0 0 0Three rigid body modes, but a rank deficiency by TWO is too few Gauss points

are used

Carlos Felippa


Analysis of the locking penomenon


29/41

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
http://find/


30/41

Alias functions

Function which tries to mimic a given function in one element

Basis for a C2D3: (1, , ), for a C2D4: (1, , , ),for a C2D8: (1, , , 2, , 2, 2,2).

Alias for various functions:Function 2 2 3 2 2 3

C2D3 0 0 0 C2D4 1 OK 1 C2D8 OK OK OK OK OK

Locking 30/41

Shear locking
http://find/


31/41

g

C2D4,L x1 L,1 x2 1Actual field

u1 = x1x2u2 = x21 /2

Aliased field u1 = x1x2

u2 =

L2/2 !!

Computed shear for the alias, 12 =x/2 !! (actual solution: 0)Computed stored elastic energy Wefor the real field and Wa for the alias:

WaWo

= 1 +1

2 L2

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

Locking 31/41

Shear locking (2)
http://find/


32/41

g ( )

Analytic solution

11 =u1,1 =x2 11 =Ex2/(1 2)22 =u2,2 = 0 22 =Ex2/(1 2)

212 =u2,1+u1,2 = 0 12 =0

Solution with the alias

11 =u1,1 =x2 11 =Ex2/(1 2)22 =u2,2 = 0 22 =Ex2/(1 2)

212 =u2,1+u1,2 =x1 12 =(1/2)Ex1/(1 +)

Wo=1

2

:

d

Locking 32/41

Dilatational locking
http://find/


33/41

g

C2D4,L x1 L,1 x2 1

Actual fieldu1 = x1x2

u2 = x21

2

2(1 ) x22

Aliased fieldu1 = x1x2

u2 =

L2

2

2(1 ) !!

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

Wa = E2(1 +)

1 1 2x

22 + x

21

2

instead of: We= E

2(1 2)x22

Solve the problem by adding a non conform. displacement (x21 L2, x22 1)

Locking 33/41

Dilatational locking (2)
http://find/


34/41

g ( )

Analytic solution

11 =u1,1 =x2 11 =Ex2/(1 2

)22 =u2,2 = x2/(1 ) 22 =033 =u3,3 = 0 33 =

212 =u2,1+u1,2 = 0 12 =0

Solution with the alias

11 =u1,1 =x2 11 =E(1 )x2/(1 +)(1 2)22 =u2,2 = 0 22 =Ex2/(1 +)(1 2)

212 =u2,1+u1,2 =x1 12 =(1/2)Ex1/(1 +)

Wo=1

2

:

d

Locking 34/41

Locking of the 8-node rectangle


35/41

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

Displacement basis: 1, , , 2, , 2, 2, 2

Try u1 =x21 x2, u2 = x31 /3, so that 11 = 2x1x2, 22 = 0, 12 = 0.

In fact, u2 represented by its alias,

x1

2/3, so that the shear is

x21 2/3No lockingif 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)
http://find/


36/41

( )

Analytic solution

11 =u1,1 = 2x1x2 11 =2Ex1x2

22 =u2,2 = 0 22 =0

33 =u3,3 = 0 33 =

212 =u2,1+u1,2 = 0 12 =0

Solution with the alias(u1

=x21

x2

and u2

=

x1

2/3)

11 =u1,1 = 2x1x2 11 =2Ex1x2

22 =u2,2 = 0 22 =0

33 =u3,3 = 0 33 =0

212 =u2,1+u1,2 =x

2

1 2

/3 12 =(1/2)(x

2

1 2

/3)/(1 +)

Wo=1

2

:

d

Locking 36/41

Trapezoidal locking
http://find/


37/41

2 (1+ )

2 (1 )

x

y

Geometry: x1 = (1), and x2 =; =x1/(1x2).Displacement basis: 1, , , 2, , 2, 2, 2

Try u1 =x21 x2, u2 =x

21/2, so that 11=x2, 22 = 0, 12 = 0.

In fact, the solution with the alias is:

11=

1

22 =2 212 = 1 +

()

1

All components are affected

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

Error on 22 cannot be easily suppressed

Locking 37/41

Dilatational locking on triangles
http://find/


38/41

triangle C2D3

Incompressible,BL TR mesh....... xy

z

Incompressible,TL BR mesh....... xy

z

Compressible,BL TR mesh......... xy

z

Locking 38/41

Spurious modes of a C2D4
http://com/Zescalier-comphttp://com/Zescalier-comphttp://com/Zescalier-comphttp://com/Zescalier-comphttp://com/Zescalier2http://com/Zescalierhttp://find/


39/41

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

u12 =u22 =u

32 =u

42 = 0

u11 = +a ; u21 = a

u31 = +a ; u41 = a 2

34

1

C2D4C2D4rC2D8C2D8r

Spurious modes 39/41

Spurious modes
http://com/Zpoteau_linhttp://com/Zpoteau_lin_rhttp://com/Zpoteau_quadhttp://com/Zpoteau_quad_rhttp://com/Zpoteau_quad_rhttp://com/Zpoteau_quadhttp://com/Zpoteau_lin_rhttp://com/Zpoteau_linhttp://find/


40/41

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

Rigid body mode {o} such as: [K] {o}= 0 everywhere

Spurious mode {o} such as: [K] {o}= 0 in some places

The number of independent states is given by the total number of dof in

the elementNumber 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); 6p

2

(reduced integration, 3D).The number of strain states is:8p3 (Serendip, 2D); 2(p+ 1)2 (Lagrange, 2D); 36p6 (Serendip, 3D);3(p+ 1)3 6 Lagrange, 3D).


Spurious modes


41/41

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 =k1(2 1/3)

u2= k2(2 1/3)


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