“Formulation and calculation of isoparametric finite element matrixes ...

Date: 13.12.2006 

“Formulation and calculation of 

isoparametric finite element matrixes” 

-Formulation of structural elements 

(plate and general shell elements) 

Andres Mena (PhD student) 

Institute of Structural Engineering, ETH

Plate and general shell element 

Shell element definition 

τ zz = 0 (stress perpendicular to the midsurface)


Content 

- Formulation of the plane stress element 

- Formulation of the plate element 

- Rotational stiffness perpendicular to the element surface 

- The patch test and the incompatible displacement modes 

- The general shell element


Plane stress element (4 nodes) 

u 1 

K δ = F 

2 

1 

v 4 

k 13 . 

u 3 3 

4 u 4 

k 22 k 23 

v k 33 

4 

y 

u 2 

v 2 

v 1 

K = 

Linear static analysis 

Stiffness matrix (K) 

k 11 k 12 . k 17 k 18 

. . k 27 k 28 

. . . 

. . . 

symm k 77 k 78 

k 88 

x 

δ T = [u 1 

, v 1 

, u 2 

, v 2 

, u 3 

, v 3 

, u 4 

, v 4 

] 

F T = [F u1 

, F v1 

, F u2 

, F v2 

, F u3 

, F v3 

, F u4 

, F v4 

]


Plane stress element (4 nodes) 

-1 ≤ r ≤ 1 

-1 ≤ s ≤ 1 

u 2 

v 2 

s 

v 1 

u 1 

2 

1 

r = 1, s = 1 

r 

v 4 

3 

4 u 4 

Displacement and Coordinate Interpolation 

4 

u(r,s) = Σ h i 

(r,s) u 

i = 1 

i 

4 

v(r,s) = Σ h i 

(r,s) v 

i = 1 

i 

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

h 2 = (1-r) (1+s) / 4 

h 3 = (1-r) (1-s) / 4 

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

y 

x


Target: 

Volume evaluated in natural coordinates 

where 

by using numerical integration instead of explicit integration, 

where t ij = thickness at the integration point 

α ij = weighting factor 

Integration points (r i , s j ) to evaluate F ij of a 4-node plane stress element 

-1 ≤ r ≤ 1 

-1 ≤ s ≤ 1 

B ij , J ij , C are unknowns ?????


Matrixes J ij and B ij 

Shape functions to calculate the Jacobian (J ij ) 

4 

x(r,s) = Σ h i 

(r,s) x 

i = 1 

i 

Shape functions to interpolate displacements 

4 

u(r,s) = Σ h i 

(r,s) u i 

i = 1


= B ij3x8 

* 

8x1 

3x1 

B ij evaluated at the integration points (r i ,s j )


Plane stress and isotropic material 

= 

3x3 

8x8 

8x3 

3x3 

3x8 

v 1 

u 2 

u 1 

2 

1 

3 

4 u 4 

v 2 

K = 

8x8 

u 3 

v 4 


k 11 k 12 k 13 . . k 17 k 18 

k 22 k 23 . . k 27 k 28 

k 33 . . . 

. . . 

symm k 77 k 78 

k 88


Calculation of stress 

T 

= 

T 

Strains and stresses are calculated at the integration points 

This example correspond to a 9-node 

element with 3x3 Gauss points


Plate bending element (4 nodes) 

θ x1 

w 1 

w 2 θx2 

1 

θ y2 

2 

4 

w 3 

4 

3 

θ y3 

4 

θ x3 

4 

θ y1 

K = 

K δ = F 


k 11 k 12 k 13 . . k 111 k 112 

k 22 k 23 . . k 211 k 212 

k 33 . . . 

. . . 

symm k 1111 k 1112 

k 1212 

z 

y 

δ T = [w 1 

, θ x1 

, θ y1 

, w 2 

, θ x2 

, θ y2 

, w 3 

, θ x3 

, θ y3 

, w 4 

, θ x4 

, θ y4 

] 

x 

F T = [F z1 

, M x1 

, M y1 

, F z2 

, M x2 

, M y2 

, F z3 

, M x3 

, M y3 

, F z4 

, M x4 

, M y4 

]


-1 ≤ r ≤ 1 

-1 ≤ s ≤ 1 

Plate bending element (4 nodes) 

Displacement, Rotation and Coordinate Interpolation 

4 

w(r,s) = Σ h i 

(r,s) w 

i = 1 

i 

4 

β x 

(r,s)= -Σ h i 

(r,s) θ 

i = 1 

yi 

4 

β y 

(r,s) = Σ h i 

(r,s) θ 

i = 1 

xi 

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

h 2 …


Area evaluated in natural coordinates 

dA = det J dr ds 

Bending moment 

Shear force 

k (shear correction factor for rectangular 

cross sections) = 5/6


3x1 3x12 12x1 2x1 2x12 12x1 

12x1 

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

B k and B γ are valuated at the integration points (r i ,s j ) 

θ x1 

θ y2 

w 2 θx2 

2 

w 1 

1 

θ y1 

F ext = K * u 

12x1 

12x12 

12x1 

4 

w 3 

4 

3 

θ y3 

4 

F ext = 

4 

Linear shape functions to interpolate area forces 

θ x3


Shear locking elimination 

4 

w(r,s) = Σ h i 

(r,s) w 

i = 1 

i 

Shear strains 

Incorrect interpolation 

at the corner nodes 

4 

β x 

(r,s)= -Σ h i 

(r,s) θ 

i = 1 

yi 

4 

β y 

(r,s) = Σ h i 

(r,s) θ 

i = 1 

xi 

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

h 2 … 

Correct evaluation that eliminates the shear locking 

in the 4-node plate bending element


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 θ z ,


Patch elements to test the performance of the shell element 

Test of the plane stress element 

It shows incompatibility of displacements 

To overcome this deficiency, incompatible displacement modes are added


Incompatible displacement modes 


Correction of the incompatible 

strain-displacement matrix 

12x1 

12x1 

12x12 

Stiffness matrix (8x8) is obtained by applying static condensation 

(k CC –k CI k II -1 k IC ) u = R 

8x8 8x1 8x1


Higher order patch tests 

For the plane stress element 

For the plate bending element


General Shell element 

Analysis of complex shell geometries 

-1 ≤ r ≤ 1 

-1 ≤ s ≤ 1 

-1 ≤ t ≤ 1 

r = 1, s = 1



Coordinate interpolation 

z 

y 

x 

in plane yz 

r = 1, s = 1 

-1 ≤ t ≤ 1 

For t = 1 

x = x 1 

y = y 1 + 0.5 * 0.8√0.2 * (-1/√2) 

z = z 1 + 0.5 * 0.8√0.2 * (1/√2) 

For t = -1 

x = x 1 

y = y 1 + -0.5 * 0.8√0.2 * (-1/√2) 

z = z 1 + -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 

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

- Compatibility of element degrees of freedom. 

- Methodology to overcome shear locking. 

- Evaluate the element performance of FEM programs or codes 

using the patch tests. 

- To use shell elements instead of 3D elements 

- To use quadrilateral shell elements instead of triangular.

THANK YOU

“Formulation and calculation of isoparametric finite element matrixes ...

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