“Formulation and calculation of isoparametric finite …webarchiv.ethz.ch/ibk/emeritus/fa/education/Seminare/...“Formulation and calculation of isoparametric finite element matrixes”-Formulation

““Formulation and calculation of Formulation and calculation of isoparametricisoparametric finite element matrixesfinite element matrixes””

--Formulation of structural elements Formulation of structural elements ((plate and general shell elementsplate and general shell elements))

Andres Andres MenaMena (PhD student)(PhD student)Institute of Structural Engineering, ETHInstitute of Structural Engineering, ETH

Date: 13.12.2006Date: 13.12.2006

Plate and general shell elementPlate and general shell element

Shell element definition

τzz = 0 (stress perpendicular to the midsurface)


- Formulation of the plane stress element

Content

- Rotational stiffness perpendicular to the element surface

- Formulation of the plate element

- The general shell element

- The patch test and the incompatible displacement modes


Plane stress element (4 nodes)

4 u4

v4

x

y

u1

v1

12

3

u2

v2

u3

v4

K δ = F

Linear static analysis

K =

k11

..

k77k88

symm

Stiffness matrix (K)

k22

k12

k78

. .k17 k18k27 k28

..

.

...

k33

k13k23

δT = [u1, v1, u2, v2, u3, v3, u4, v4]

FT = [Fu1, Fv1, Fu2, Fv2, Fu3, Fv3, Fu4, Fv4]


Plane stress element (4 nodes)

4 u4

v4

x

y

r

su1

v1

12

3

u2

v2

u3

v4

r = 1, s = 1

-1 ≤ r ≤ 1

-1 ≤ s ≤ 1 u(r,s) = Σ hi (r,s) uii = 1

4

v(r,s) = Σ hi (r,s) vii = 1

4

Displacement and Coordinate Interpolation

h1 = (1+r) (1+s) / 4

h2 = (1-r) (1+s) / 4

h3 = (1-r) (1-s) / 4

h4 = (1+r) (1-s) / 4


Target:

where

by using numerical integration instead of explicit integration,

where tij = thickness at the integration pointαij = weighting factor

Integration points (ri, sj) to evaluate Fij of a 4-node plane stress element

Bij , Jij , C are unknowns ?????

-1 ≤ r ≤ 1

-1 ≤ s ≤ 1

Volume evaluated in natural coordinates


Matrixes Jij and Bij

Shape functions to calculate the Jacobian (Jij)

x(r,s) = Σ hi (r,s) xii = 1

4

Shape functions to interpolate displacements

u(r,s) = Σ hi (r,s) uii = 1

4


Bij evaluated at the integration points (ri,sj)

= Bij3x8

3x1

*8x1


=

Plane stress and isotropic material

3x8

3x3

3x38x38x8

v1v2

4 u4

v4

u1

12

3

u2

u3

v4

K =

k11

..

k77k88

symm


k22

k12

k78

. .k17 k18k27 k28

..

.

...

k33

k13k23

8x8


Calculation of stress

Strains and stresses are calculated at the integration points

= TT

This example correspond to a 9-node element with 3x3 Gauss points


Plate bending element (4 nodes)

K δ = F

K =

k11

..

k1111k1212

symm


k22

k12

k1112

. .k111 k112k211 k212

..

.

...

k33

k13k23

δT = [w1, θx1, θy1, w2, θx2, θy2, w3, θx3, θy3, w4, θx4, θy4]

FT = [Fz1, Mx1, My1, Fz2, Mx2, My2, Fz3, Mx3, My3, Fz4, Mx4, My4]

y

z

x

4

θy1

w1

1

2

3

θx1

θy2

w2 θx2

w3

θx3

θy3

44

4


-1 ≤ r ≤ 1

-1 ≤ s ≤ 1

w(r,s) = Σ hi (r,s) wii = 1

4

βx(r,s)= -Σ hi (r,s) θyii = 1

4

Displacement, Rotation and Coordinate Interpolation

h1 = (1+r) (1+s) / 4h2 …

Plate bending element (4 nodes)

βy(r,s) = Σ hi (r,s) θxii = 1

4


Area evaluated in natural coordinates

dA = det J dr ds

Bending momentShear force

k (shear correction factor for rectangular cross sections) = 5/6


4

θy1

w1

1

2

3

θx1

θy2

w2 θx2

w3

θx3

θy3

44

4

Fext = K * u12x12

12x13x123x1 2x1 12x12x12 12x1

Bk and Bγ are valuated at the integration points (ri,sj)

12x12 12x3 3x3 3x12 12x2 2x2 2x12

12x112x1

Fext =

Linear shape functions to interpolate area forces


Shear locking elimination

Shear strains

Incorrect interpolation at the corner nodes

Correct evaluation that eliminates the shear locking in the 4-node plate bending element

w(r,s) = Σ hi (r,s) wii = 1

4

βx(r,s)= -Σ hi (r,s) θyii = 1

4

h1 = (1+r) (1+s) / 4h2 …

βy(r,s) = Σ hi (r,s) θxii = 1

4


Shell element

Rotational stiffness perpendicular to the element surface is not defined

Easy solution

One elegant solutions is to add a “real” rotational stiffness for θz,


Patch elements to test the performance of the shell element

It shows incompatibility of displacements

To overcome this deficiency, incompatible displacement modes are added

Test of the plane stress element


Incompatible displacement modes

Correction of the incompatible strain-displacement matrixStiffness matrix (K)

12x12

12x1 12x1

Stiffness matrix (8x8) is obtained by applying static condensation(kCC – kCI kII

-1 kIC) u = R8x8 8x1 8x1


Higher order patch tests

For the plane stress element

For the plate bending element


General Shell element

Analysis of complex shell geometries

r = 1, s = 1

-1 ≤ r ≤ 1

-1 ≤ s ≤ 1

-1 ≤ t ≤ 1



Coordinate interpolation

r = 1, s = 1

x

y

z

in plane yz

-1 ≤ t ≤ 1

x = x1

y = y1 + 0.5 * 0.8√0.2 * (-1/√2)

For t = 1

z = z1 + 0.5 * 0.8√0.2 * (1/√2)

x = x1

y = y1 + -0.5 * 0.8√0.2 * (-1/√2)

For t = -1

z = z1 + -0.5 * 0.8√0.2 * (1/√2)



Displacement interpolation

Strain – displacement matrix B(r,s,t)

B(r,s,t)



Stress-strain law



Shear locking

9-node shell element


Boundary conditions

Solid element Shell element

CONCLUSIONS

- Evaluate the element performance of FEM programs or codes using the patch tests.

- Methodology to overcome shear locking.

- To use shell elements instead of 3D elements

- To use quadrilateral shell elements instead of triangular.

- Compatibility of element degrees of freedom.

- To select properly plane stress, bending plate or shell.

THANK YOU

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“Formulation and calculation of isoparametric finite …webarchiv.ethz.ch/ibk/emeritus/fa/education/Seminare/...“Formulation and calculation of isoparametric finite element matrixes”-Formulation