5.masat 2012-2013 ahmadshell

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AHMAD SHELL FEM FORMULATION

U n i v e r s i t y o f M i n h o

Department of Civil EngineeringGuimarães, Portugal

[email protected] ∗∗∗∗ www.civil.uminho.pt



2

1 - Introduction

s3 s3

s2

Top surface

Degenerationprocess of avolume finiteelement into an

Joaquim Barros

Modeling and Advanced Structural Analysis Techniques

University of Minho – Department of Civil Engineering

s2

s1

s3

s2 s1

1

Middle surface of

shell element normal

Degenerated shell

finite element

Volume finite element Bottom surface ma s e n teelement



3

2 – Basic hypothesis

The deflection orthogonal to the shell middle surface of any point is small

when compared to the shell thickness;

The stress orthogonal to the shell middle surface is negligible;

Joaquim Barros



deformed shell middle surface.



4

3 – Types of nodal points

Structures that have coplanar and kink nodes

Joaquim Barros



Kink nodes

Coplanar nodes

Kink edges



5

ɵnC

D

ɵnC

ɵn A

ɵn D

ɵ ɵn B

ɵn D

Schematic representation of the procedure to distinguish coplanar and kink nodes

Joaquim Barros



ɵn A D

C

B

A

element

element

k

A B

C

D

k

k

B

A

(b) The coplanar node k has five

degrees of freedom. (c) The kink nde k has six degrees

of freedom. (a) The k node is common to the

elements A, B, C and D.

B



6

i1

i3 δ = u

δ = u

s2

n3 ( )nk 3

x3

v2

v3

Degrees of freedom and coordinate systems for coplanar nodes

Joaquim Barros



2

δ = u

Superfície s3=0

s1

3

k

nk 2

( )nk 1

( )δ 1 k

( )δ 3 k

( ) ( )δ θ 4 1k nk

=

( ) ( )δ θ 5 2k nk

= x2

x1

n1

( )δ 2 k

k

v1



7

( )δ 3 k

δ 3 3= u

x3

Degrees of freedom and coordinate system for kink nodes

Joaquim Barros



k

( )δ 1 k

( )δ 2 k

( )δ 4 k

( )δ 5 k ( )δ 6 k

δ θ 6 3= x

δ θ 5 2= x

δ θ 4 1= x

δ 2 2= u

x2

δ 1 1= u

x1



8

4 – Coordinate systems

Middle surface ( s3 0= )

s3 s2

s1

ɵn3 ɵn

n3

v3

Coordinate systems for the Ahamdfinite element formulation: (a)global, (b) natural, (c) nodal and (d)

tangential

(a)(b)

ɵi3

ɵi2

ɵi1

x u x3 3 3, , θ

x u x2 2 2, , θ

x u x1 1 1, , θ

Joaquim Barros



Middle surface ( s30= )

θ n2 n2 v2

θ n1

v1

n1

ɵn1

s2

n k 2,'

n k 1,

'

s1

k

k

v3

t 2' t 3 t 2

t 1 s1

π s3

s2

Surface s3 = constant

Integration points

t 3

t 3t 1

t 3

t 1

t 1

(c)

(d)



9

a ua1 1,

g ug

2 2,

l ul

2 2, β +

Auxiliary coordinate system

Joaquim Barros



i

b

( ) ( ) ( )u u ug l a3 3 3≡ ≡

( ) ( ) ( )g l a3 3 3≡ ≡

l ul1 1,

g u g1 1,

a ua

2 2,

α +



10

4.1 - Nodal coordinate system

n x

s

x

s

x

sk

k

T

1

1

1

2

1

3

1,

'=

∂

∂

∂

∂

∂

∂ n

x

s

x

s

x

sk

k

T

2

1

2

2

2

3

2,

'=

∂

∂

∂

∂

∂

∂ ( )' '

3, 1, 2,' '

1, 2,

1ˆ

k k k

k k

n n nn n

= ××

If is parallel to global axis, the nodal coordinate system is parallel tothe global coordinate system ( = , = and = ). If hasopposite direction to , = , =- and =- .

nk 3, 3 x

1,k n 1 x 2,k n2 x n

k 3, 3 n

k 3,

3 x 1,k n

1 2,k n

2 x nk 3, 3

Joaquim Barros



If is not in these special conditionsn k 3,

( )1, 3 3,

3 3,

22, 1 13, 2 3

1 ˆˆ ˆˆ ˆ

ˆ ˆ ˆ0

k k

k

k k

n i ni n

n i n i i

= ××

= − + −

( )ɵ ɵ ɵ, , ,n n nk k k 2 3 1= ×



11

s1

s3

s2

The vector at k node is defined from the coordinates of the correspondingnodes at top and bottom element surfaces

3n̂

Joaquim Barros



k

k θ n2

θ n1

n3

n2

n1

V 2

V 3

V 1

Not necessarily orthogonal to the middle

surface of shell element at node k .



12

5 – Geometry definition

1 1 31

2 2 3 32

1 1

3 3 33

2

n nk

k k

k k

k k

x x nh

x N x N s n

x x n

= =

= +

∑ ∑ 1,1

2,1

3,1

31,1

132,1

1 1 3 3

33,11

1

0 0 0 0 ... 0 0 0 02 2

nn n

n

x

x

x

n

hh n N N s N N s

n xhh

Joaquim Barros



k h

k _2

2 1 1 3 3

1,31

1 1 3 32,

3,

31,

32,

33,

......2 2

0 0 0 0 ... 0 0 0 02 2

n n

nn

n nn

n

n

n

n

x x hh N N s N N s x

x

n

n

n

=

,11

,22

,33

x

x

x

x

x N

x N X

x N

N X

=

=



13

6 – Displacement field

θ n k 1,

n k 3,

n k 2,

1 1,

2 2,

13 3,

k n

k k

k k

u u

u N u

u u=

=

∑

Joaquim Barros



n k 2 ,

θ n k 1,

n k 1,

uk

hk

k

k



14

k

s3

hk V 1

θ n k 2,

θ n k 1,

ɵ,n k 1 n k 1,

n k 2,

v1,k P

n k 3,

v2, k

s3

hk k V 2

θ n k 1,

θ n k 2,

ɵ,n k 2n k 1, n k 2,

P

n k 3,

Contribution of the rotations for the displacement field

Joaquim Barros



1, 3 2,v

2

nk k k

hs θ =

1 , 1, 1,

3 2, 1,

ˆv v .

ˆ.2

n k k k

nk k k

n

hs nθ

=

=

2, 3 1,v

2

nk k k

hs θ = −

2 , 2, 2,

3 1, 2,

ˆv v .

ˆ.2

n k k k

nk k k

n

hs nθ

=

= −1,

3 2, 1,

2,

ˆ ˆv U2

n

g k k k k k k n

k

hs n n

θ

θ

= = −



15

1

1,

2 3 2, 1,

1 2,

3

ˆ ˆ2

nng k k k k k k n

k k

uh

U u N s n n

u

θ

θ =

= = −

∑

n ggnk k k U T U =

1

2

3

n

n

k

k

u

U u

u

=

1

2

3

g

g

k

k

u

U u

u

=

1 11 12 13

2 21 22 23

3 31 32 33

ˆ

ˆ

ˆ

T

gn T

k

T

k k

n n n n

T n n n n

n n n n

= =

Joaquim Barros



1,

11 12 131,

2,

21 22 232,

3,

1,

1,

2,

2,

3,

ˆ

ˆ

g

k n

k g

k n

k gk k

g

k T

k g

k T

k g

k

n n n

n n n

n

n

θ θ

θ θ

θ

θ

θ

θ

=

=



16

1 1,

1,

2 3 2, 1, 2,

1 2,

3 3,

ˆˆ ˆ

ˆ2

g

k T nk gk

k k k k T k k g

k

unh

u N s n nn

u

θ

θ

θ =

= −

∑

1 21 11 1,

2 3 22 12 2,

1

3 23 13 3,

0

02

0

n

k nnk

k k

k n

k k

u n nh

u N s n n

u n n

θ

θ

θ =

−

= − −

∑

Joaquim Barros



11 12 13 1,

3 21 22 23 2,

1

31 32 33 3,

2

n

k nnk

k k

k n

k k

P P Ph N s P P P

P P P

θ θ

θ =

=

∑

11 21 12 11 13

21 22 22 12 23

31 23 32 13 33

0

0

0

P n P n P

P n P n P

P n P n P

= − = =

= − = =

= − = =



17

1 21 11 1,

11 12 13

2 3 22 12 2,

1 21 22 23

3 23 13 3,

11 12 13 1,

3 21 22 23 2,

1

31 32 33 3,

2

2

g

k ngk

k k

k gk k k

g

k ngk

k k

k g

k k

u n nn n nh

u N s n nn n n

u n n

P P Ph N s P P P

P P P

θ

θ

θ

θ θ

θ

=

=

−

= − −

=

∑

∑

Joaquim Barros



11 12 21 12 11 22 13 21 13 11 23

21 22 11 12 21 22 23 22 13 12 23

31 23 11 13 21 32 23 12 13 22 33 23 13 13 23

0

0

P P n n n n P n n n n

P n n n n P P n n n n

P n n n n P n n n n P n n n n

= = − + = − +

= − + = = − +

= − + = − + = − +

1 1 11 12 13 1

2 2 3 21 22 23 2

1 1

3 3 31 32 33 3

2

g g

n nk

k k

k k

k k k

u u P P Ph

u N u N s P P P

u u P P P

θ

θ

θ = =

= +

∑ ∑



18

1 1

2 2

3 3

n

n

n

k k

θ θ

θ θ

θ θ

=

1 1

2 2

3 3

g

g

g

k k

θ θ

θ θ

θ θ

=

if k is a coplanar node: if k is a kink node:

1 1 11 1 3 11,1 1 3 12,1 1 3 13,1 3 11, 3 12, 3 13,

1

0 0 ... 0 02 2 2 2 2 2

n n nn n n n n n n

g

h h hh h h N N s P N s P N s P N N s P N s P N s P

u

1,1

2,1

3,1

1,1

2,1

3,1

u

uu

θ

θ

θ

Joaquim Barros



1 1 1

2 1 1 3 21,1 1 3 22,1 1 3 23,1 3 21, 3 22, 3 23,

31

1 1 3 31,1

0 0 ... 0 02 2 2 2 2 2

0 02

n n n

n n n n n n n

u N N s P N s P N s P N N s P N s P N s P

uh

N N s P

=

1,1 1

1 3 32,1 1 3 33,1 3 31, 3 32, 3 33,2,

3,

1,

2,

3,

...

... 0 02 2 2 2 2

nn n n

n n n n n n nn

n

n

n

n

uh h hh h

N s P N s P N N s P N s P N s P u

u

θ

θ

θ

,11

,22

3 ,3

g

u

u

u

u

N u

u N U

u N

N U

=

=



19

7 – Strain components

1

1

21

2

2

1 2

12

2 1

t

t

w

t

wt

w w

t t

∂

∂

∂ ε ∂

ε ∂ ∂

ε γ ∂ ∂

+ = =

1 1 1

1 2 3

2 2 2

1 2 3

t

w w w

t t t

w w w

t t t

∂ ∂ ∂

∂ ∂ ∂

∂ ∂ ∂ ε

∂ ∂ ∂

=

Joaquim Barros



23

3213

3 2

31

3 1

wwt t

ww

t t

∂ ∂ γ ∂ ∂

∂ ∂

∂ ∂

+

+

3 3 3

1 2 3

w w w

t t t ∂ ∂ ∂

1 1 1

1 2 3

2 2 2

1 2 3

3 3 3

1 2 3

g

u u u

x x

u u u x x

u u u

x x

∂ ∂ ∂

∂ ∂ ∂

∂ ∂ ∂ ε ∂ ∂ ∂

∂ ∂ ∂

∂ ∂ ∂

=

T gt gt t gT T ε ε =

1 11 12 13

2 21 22 23

3 31 32 33

ˆ

ˆ

ˆ

gt

t t t t

T t t t t

t t t t

= =



20

1g s J ε ε −=

1 1 1

1 2 3

2 2 2

1 2 3

3 3 3

s

u u u

s s s

u u u

s s s

u u u

∂ ∂ ∂

∂ ∂ ∂

∂ ∂ ∂ ε

∂ ∂ ∂

∂ ∂ ∂

=

Joaquim Barros



1 2 3s s s∂ ∂ ∂

1 1 1

1 2 3

2 2 2

1 2 3

3 3 3

1 2 3

x x x

s s s

x x J s s s

x x x

s s s

∂ ∂ ∂

∂ ∂ ∂

∂ ∂ ∂ ∂ ∂ ∂

∂ ∂ ∂

∂ ∂ ∂

=



21

1

1 1,1 1, 1 31,1 31, 1

2 2,1 2, 3 32,1 32,

3 3,1 3, 33,1 33,

02

0 2

n n

n n

n n n n n

h

x x x N n n N

x x x s n n

x x x N n n h N

= +

⋯ ⋯

⋯ ⋮ ⋯ ⋱ ⋮

⋯ ⋯

1 x

Joaquim Barros



2 3 3

3

T T

V n V

x X N s L H N

x

= +

1,1 1,

2,1 2,

3,1 3,

n

T

n

n

x x

X x x x x

=

⋯

⋯

⋯

1

V

n

N

N N

=

⋮

31,1 31,

3 32,1 32,

33,1 33,

n

T

n n

n

n n

L n n

n n

=

⋯

⋯

⋯

1 02

02

n

h

H

h

=

⋱



22

1 1 1 11,1 1, 3 31,1 3 31,

1 1 1 1 1

= + + + +2 2

n n nn n

N N h x N N h x x s n s n

s s s s s

∂ ∂∂ ∂ ∂+

∂ ∂ ∂ ∂ ∂⋯ ⋯

1 1 1 11,1 1, 3 31,1 3 31,

2 2 2 2 2

1 11 31,1 31,

3

=2 2

= + +2 2

n n nn n

nn n

N N h x N N h x x s n s ns s s s s

h x h N n N n

s

∂ ∂∂ ∂ ∂+ + + + +∂ ∂ ∂ ∂ ∂

∂+

∂

⋯ ⋯

⋯

Joaquim Barros



2 1 1 12,1 2, 3 32,1 3 32,

1 1 1 1 1

2 1 1 12,1 2, 3 32,1 3 32,

2 2 2 2 2

2 11 32,1

3

=2 2

=2 2

= +2

n n n

n n

n n nn n


s s s s s


s s s s s

x h N n

s

∂ ∂∂ ∂ ∂+ + + + +

∂ ∂ ∂ ∂ ∂

∂ ∂∂ ∂ ∂+ + + + +

∂ ∂ ∂ ∂ ∂

∂

∂

⋯ ⋯

⋯ ⋯

32,+2

nn n

h N n+⋯



23

3 1 1 13,1 3, 3 33,1 3 33,

1 1 1 1 1

3 1 1 13,1 3, 3 33,1 3 33,

2 2 2 2 2

3 11 33,1

3

=2 2

=2 2

= +2

n n n

n n

n n nn n

x N N h N N h x s n s n

s s s s s

x N N h N N h x s n s n

s s s s s

x h N n

s

∂ ∂ ∂∂ ∂+ + + + +

∂ ∂ ∂ ∂ ∂

∂ ∂ ∂∂ ∂+ + + + +

∂ ∂ ∂ ∂ ∂

∂

∂

⋯ ⋯

⋯ ⋯

33,+2

nn n

h N n+⋯

Joaquim Barros





24

1 1 11 1

1 2 31 21,1 1,

2 2 22,1 2,

1 2 3

3,1 3,

3 3 31 2

1 2 3

0

0

n

n

n n n

x x x N N

s s ss s x x

x x x J x x

s s s

x x N N x x xs s

s s s

∂ ∂ ∂

∂ ∂ ∂

∂ ∂ ∂

∂ ∂ ∂

∂ ∂ ∂

∂ ∂ ∂

∂ ∂

∂ ∂

= = + ∂ ∂ ∂ ∂

⋯

⋯ ⋮ ⋮ ⋮

⋯

1 1 1 1 1 N h N h hs s N

∂ ∂

Joaquim Barros



31,1 31,

32,1 32,

33,1 33,

n

n

n

n n

n n

n n

⋯

⋯

⋯

1 2

3 3

1 2

2 2 2

2 2 2

n n n n nn

s s

N h N h hs s N

s s

∂ ∂

∂ ∂ ∂ ∂

⋮ ⋮ ⋮



25

1 1 1

1 2 3

2 2 21 2

1 2 3

3 3 3

1 2 3

x x xs s s

x x x J J J

s s s

x x x

s s s

∂ ∂ ∂ ∂ ∂ ∂

∂ ∂ ∂

∂ ∂ ∂

∂ ∂ ∂ ∂ ∂ ∂

= =

1 1 x ∂ ∂ 1 x ∂

Joaquim Barros



1 2

2 21

1 2

3 3

1 2

s s

x J

s s

x

s s

∂ ∂=

∂ ∂ ∂ ∂

∂ ∂

3

22

3

3

3

s

J s

x

s

∂

∂=

∂ ∂

∂

31 3

T T V V n

N N J X s L H

s s

∂ ∂= +

∂ ∂

32

T

n V J L H N =



26

, 3 3 ,

1 2

ni k k k

ij i k i k

k j j j

x N N h J x s n

s s s=

∂ ∂ ∂= = + ∂ ∂ ∂

∑ for 1 i 3 1 j 2∧

3 3 ,13 2

ni k

i k i k k

x h

J N ns =

∂= =

∂ ∑ for 1 i 3

≤ ≤ ≤ ≤

≤ ≤

31 1 1 1 2 1 = xu u x u x u ∂∂ ∂ ∂ ∂ ∂ ∂

Joaquim Barros



1 1 1 2 1 3 1

31 1 1 1 2 1

2 1 2 2 2 3 2

31 1 1 1 2 1

3 1 3 2 3 3 3

= + +

= + +

s x s x s x s

xu u x u x u

s x s x s x s

u u x u x u

s x s x s x s

∂ ∂ ∂ ∂ ∂ ∂ ∂

∂∂ ∂ ∂ ∂ ∂ ∂

∂ ∂ ∂ ∂ ∂ ∂ ∂

∂∂ ∂ ∂ ∂ ∂ ∂

∂ ∂ ∂ ∂ ∂ ∂ ∂



27

32 2 1 2 2 2

1 1 1 2 1 3 1

32 2 1 2 2 2

2 1 2 2 2 3 2

32 2 1 2 2 2

3 1 3 2 3 3 3

= + +

= + +

= + +

u u x u x u

s x s x s x s

xu u x u x u

s x s x s x s

xu u x u x u

s x s x s x s

∂∂ ∂ ∂ ∂ ∂ ∂

∂ ∂ ∂ ∂ ∂ ∂ ∂

∂∂ ∂ ∂ ∂ ∂ ∂

∂ ∂ ∂ ∂ ∂ ∂ ∂

∂∂ ∂ ∂ ∂ ∂ ∂

∂ ∂ ∂ ∂ ∂ ∂ ∂

Joaquim Barros



3 3 3 3 31 2

1 1 1 2 1 3 1

3 3 3 3 31 2

2 1 2 2 2 3 2

3 3 3 3 31 2

3 1 3 2 3 3 3

= + +

= + +

= + +

u u u u x x x

s x s x s x s

u u u u x x x

s x s x s x s

u u u u x x x

s x s x s x s

∂ ∂ ∂ ∂ ∂ ∂ ∂

∂ ∂ ∂ ∂ ∂∂ ∂

∂ ∂ ∂ ∂ ∂ ∂ ∂

∂ ∂ ∂ ∂ ∂∂ ∂

∂ ∂ ∂ ∂ ∂ ∂ ∂



28

1 1 1 1 1 1 1 1 1

1 2 3 1 2 3 1 2 3

2 2 2 2 2 2 2 2 2

1 2 3 1 2 3 1 2 3

3 3 3 3 3 3 3 3 3

1 2 3 1 2 3 1 2 3

u u u u u u x x x

s s s x x x s s s

u u u u u u x x x

s s s x x x s s s

u u u u u u x x x

s s s x x x s s s

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

= ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂

Joaquim Barros



u u J s x

∂ ∂=∂ ∂

1g s

J ε ε

−=

1 T

gt gt t sT J T ε ε − =



29

1 1 1

1 2 3

2 2 2

1 2 3

3 3 3

1 2 3

s

u M

u u u

s s s

u u u

s s s

u u u

s s s

D U

ε

∂ ∂ ∂

∂ ∂ ∂ ∂ ∂ ∂

= ∂ ∂ ∂

∂ ∂ ∂

∂ ∂ ∂

=

Joaquim Barros



,1 ,1 ,1

1 2 3

,2 ,2 ,2

1 2 3

,3 ,3 ,3

1 2 3

u u u

u u u

u

u u u

N N N s s s

N N N D

s s s

N N N s s s

∂ ∂ ∂

∂ ∂ ∂ ∂ ∂ ∂

= ∂ ∂ ∂

∂ ∂ ∂ ∂ ∂ ∂

0 0

0 0

0 0

M

U

U U

U

=

1 T

gt gt t

u M T D U J T ε − =



30

1,1 1 1 1 1 1 13 11,1 3 12,1 3 13,1

11 1 1 1

3 11, 3 12, 3 13,

1 1 1 1

,1 1 1 13

2 2 2

N = 0 0

2 2 s 2

0 02 2 2

= 0 02

∂ ∂ ∂ ∂ ∂

∂ ∂ ∂ ∂ ∂

∂ ∂ ∂ ∂

∂ ∂ ∂ ∂

∂ ∂ ∂

∂ ∂ ∂

⋯

⋯

u

n n n n n n nn n n

u

N N N h N h hs P s P s P

s s s s

N N h N h N hs P s P s P

s s s s

N N N hs

s s s

1 1 1 111,1 3 12,1 3 13,1

2 2

2 2

∂ ∂

∂ ∂⋯

N h N hP s P s P

s s

Joaquim Barros



3 11, 3 12, 3 13,

2 2 2 2

,1 1 1 11 11,1 1 12,1 1 1

3

0 0

2 2 2

= 0 0 02 2 2

∂ ∂ ∂ ∂

∂

∂

⋯ n n n n n n nn n n

u

s P s P s P

s s s s

N h h h N P N P N P

s3,1

11, 12, 13,

0 0 0 2 2 2

⋯

⋯

n n n

n n n n n n

h h h

N P N P N P



31

,2 1 1 1 1 1 1 13 21,1 3 22,1 3 23,1

1 1 1 1 1

3 21, 3 22, 3 23,

1 1 1 1

,2 1 1 13 21,1

2 2 2

= 0 02 2 2

0 02 2 2

= 0 02

∂ ∂ ∂ ∂ ∂

∂ ∂ ∂ ∂ ∂

∂ ∂ ∂ ∂

∂ ∂ ∂ ∂

∂ ∂ ∂

∂ ∂ ∂

⋯

⋯

u

n n n n n n nn n n

u

N N N h N h N hs P s P s P

s s s s s


s s s s

N N N hs P

s s s

1 1 1 13 22,1 3 23,1

2 2

2 2

∂ ∂

∂ ∂

∂ ∂ ∂ ∂

⋯ N h N h

s P s Ps s

N N h N h N h

Joaquim Barros



3 21, 3 22, 3 23,

2 2 2 2

,2 1 1 11 21,1 1 22,1 1 23,1

3

2 2 2

= 0 0 02 2 2

∂ ∂ ∂ ∂

∂

∂

⋯

⋯

n n n

u

s s s

s s s s N h h h

N P N P N Ps

21, 22, 23, 0 0 0 2 2 2

⋯

n n n

n n n n n n

h h h

N P N P N P



32

,3 1 1 1 1 1 1 13 31,1 3 32,1 3 33,1

1 1 1 1 1

3 31, 3 32, 3 33,

1 1 1 1

,3 1 1 13 31,

2 2 2

= 0 02 2 2

0 02 2 2

= 0 02

∂ ∂ ∂ ∂ ∂

∂ ∂ ∂ ∂ ∂

∂ ∂ ∂ ∂

∂ ∂ ∂ ∂

∂ ∂ ∂

∂ ∂ ∂

⋯

⋯

u

n n n n n n nn n n

u

N N N h N h N hs P s P s P

s s s s s


s s s s

N N N hs P

s s s

1 1 1 11 3 32,1 3 33,1

2 2

2 2

∂ ∂

∂ ∂⋯

N h N hs P s P

s s

Joaquim Barros



3 31, 3 32, 3 33,

2 2 2 2

,3 1 1 11 31,1 1 32,1 1 33,1

3

0 0

2 2 2

= 0 0 02 2 2

∂ ∂ ∂ ∂

∂

∂

⋯ n n n n n n nn n n

u

s P s P s P

s s s s

N h h h N P N P N P

s

31, 32, 33, 0 0 0 2 2 2

⋯

⋯

n n n

n n n n n n

h h h

N P N P N P



33

,

, 1 3 1, 3 2, 3 3,

1 1 1 1 1

= = 0 02 2 2

u i k k k k k k k u i i k i k i k

N N N h N h N h D s P s P s P

s s s s s

∂ ∂ ∂ ∂ ∂

∂ ∂ ∂ ∂ ∂ ⋯ ⋯

,

, 2 3 1, 3 2, 3 3,= = 0 02 2 2

u i k k k k k k k u i i k i k i k

N N N h N h N h D s P s P s P

s s s s s

∂ ∂ ∂ ∂ ∂

∂ ∂ ∂ ∂ ∂⋯ ⋯

for 1 i 3 j =1∧≤ ≤

Joaquim Barros



,, 3 1, 2, 3,

3

= = 0 0 02 2 2

u i k k k u i k i k k i k k i k N h h h D N P N P N P

s

∂

∂

⋯ ⋯

for 1 i 3 j =2∧≤ ≤

for 1 i 3 j =3∧≤ ≤



34

1

2

12

23

13

t

t

mb

t t

s

B

U B U

B

ε

ε

ε γ

γ γ

= = − − =

Joaquim Barros





35

8 – Stress components

{ }1 2 12 23 13, , , ,T t t t t t t σ σ σ τ τ τ =

{ }1 2 12, ,T

t t t t mbσ σ σ τ =

{ 23 13,T t t t

sσ τ τ =

t I

3

Stresses in the tangential coordinate system

Joaquim Barros



{ },T

t t t

mb sσ σ σ =

π

t I

1

I

t I

2σ t I 1

τ t t

I

1 2

τ t t

I

1 2

Surface of the integration

point in analysis

Generic integration point

Middle surface

τ τ t t

I

t t

I

2 3 3 2=

τ τ t t

I

t t

I

1 3 3 1=

τ t t

I

2 3τ t t

I

1 3

σ t

I

2



36

T gt gt g t T T σ σ =

1 12 13

12 2 23

13 230

t t t

t t t t

t t

σ τ τ

σ τ σ τ

τ τ

=

Joaquim Barros



1 12 13

12 2 23

13 23 3

g g g

g g g g

g g g

σ τ τ

σ τ σ τ

τ τ σ

=

{ }1 2 3 12 23 13, , , , ,T g g g g g g gσ σ σ σ τ τ τ =



37

9 – Constitutive equations for linear elastic behaviourt t

Dσ ε =

0

0

t t

mbmb mbt t

ss s

D

D

σ ε

σ ε

=

Joaquim Barros



2

1 0

1 01

10 0

2

mb E D

υ

υ υ

υ

= − −

1 00 1

s D F G =

0

0u d

t t t t mb mb T T mb

t t ss s

D

D

ε ε ε σ

σ ε

∆ ∆ − −

=



38

10 – Stiffness matrix for linear elastic materials

int

T t t

V

W dV δ δ ε σ = ∫

int

T T t t

V

W U B D B dV U δ δ = ∫

T t t

K B D B dV =

Joaquim Barros



V

,, , ,

, , ,,

E f E ff E fp E f

E pf E pp E p E E p

QU K K

U K K Q R

= +

, ,, ,,

,

E ff E fp E f E p E f

E f

K U Q K U

Q

= −

=



39

11 – Load vector

Joaquim Barros





40

12 – Resultant stresses

{ } { }1 2 1 2 1 2 1 2

2

3

2

2

, , , ,

h

T T t m t t t t t t t t

h

h

N N N dt σ σ σ σ −

= = ∫ { } { }2 1 1 2 1 2 1 2

2

3 3

2

2

, , , ,

h

T T t b t t t t t t t t

h

h

M M M t dt σ σ σ σ −

= = ∫

Membrane forces:Bending moments:

Joaquim Barros

Modeling and Advanced Structural Analysis Techniques University of Minho – Department of Civil Engineering

3

2

t

mb

h

dt σ

−

=

3 3

2

t

mb

h

t dt σ

−

=

{ } { }2 3 1 3 2 3 1 3

2

3

2

2

3

2

, ,

h

T T t

s t t t t t t t t

h

h

T t

s

h

V V dt

dt

σ τ τ

σ

−

−

= =

=

∫

∫

Shear forces:



41

{ { }1 2 1 2 2 1 1 2 2 3 1 3, , , , , , , , ,

T T t t t t

m b s t t t t t t t t t t t t N N N M M M V V σ σ σ σ = =

V t t

I

2 3

M t t

I

1 2

M t

I

2

N t t

I

1 2

N t

I

2t

I

3

Joaquim Barros


I

Middle surface of

the shell element

Generic integration point

M t t I

1 2 M t

I 1

V t t

I

1 3 N t t

I

1 2

N t

I

1

t

I

2

t I

1



42

{ }

( )

( )

1 2 1 2

2

3

2

1

31

1

3

1

, ,

2

1

2

u d

u d

h

T T t t

m mbt t t t

h

t t t

mb mb T T

t t t

mb mb T T

N N N dt

h

D ds

D h ds

σ σ

ε ε ε

ε ε ε

−

+

∆ ∆−

+

∆ ∆

−

= =

= − −

= − −

∫

∫

∫

Joaquim Barros


{ }

( )

1 2 1 2

2

3

2

1

3

1

1

3

1

, ,

2

22

u d

u

h

T T t t

m mbt t t t

h

t t t mb mb T T

t t

mb mb T

N N N dt

h D B U ds

h D B U ds

σ σ

ε ε

ε

−

+

∆ ∆

−

+

∆

−

= =

= − −

= −

∫

∫

∫

1

3

1

0d

t

T dsε

+

∆

−

=∫



43

{ } ( )( )

( )( )

1 2 1 2

3

1 1

1

3

1 1

1

, , , ,2

, ,2

u

u

nGPT t t t

mb mb mbm T t t t t k k

k

nGPt t

mb mb T k k

k

h N N N D B s s P U W h D

h D B s s P U W N

σ ε ∆

=

∆

=

= = −

= −

∑

∑

u u

u

t t

mbT T

t

T

N h D

h

ε

σ

∆ ∆

∆

=

=

Joaquim BarrosModeling and Advanced Structural Analysis Techniques University of Minho – Department of Civil Engineering



44

{ }

( )( )

2 1 1 2

2

3 3

2

1

3 31

12

3 3

1

, ,

2 2

4

u d

u d

h

T T t t

b mbt t t t

h

t t t

mb mb T T

t t t

mb mb T T

M M M t dt

h h D s ds

h D s ds

σ σ

ε ε ε

ε ε ε

−

+

∆ ∆

−

+

∆ ∆

−

= =

= − −

= − −

∫

∫∫


( )

( )( )

12

3 3

1

2 3

1 1 3

1

2

1

4

2, ,

4 3

,4

u d

d

t t t

mb mb T T

nGPt t

mb mb T k k

k

t mb mb

h D B U s ds

h D B s s P U s W

h D B s

ε ε

ε

+

∆ ∆

−

∆

=

= − −

= −

=

∫

∑

( )( )

( )( )

23

1 3

1

2 3

1 1 3

1

,6

, ,4

d

d

nGP

t mb T k k

k

nGPt t

mb mb T k k

k

hs P U s W D

h D B s s P U s W M

ε ∆

=

∆

=

−

= −

∑

∑

1

3 31 0u

t

T s dsε

+

∆−

=

∫1 1

2

3 3 3 3

1 1

2

3d d d

t t t

T T T s ds s dsε ε ε + +

∆ ∆ ∆

− −

= =∫ ∫



45

{ }2 3 1 3

2

3

2

1

3

1

1

3

1

,

2

2

h

T T t t

s st t t t

h

t

s s

t

s s

V V dt

h

D ds

h D B U ds

σ σ

ε

−

+

−

+

−

= =

=

=

∫

∫

∫


( )( )3

1 1

1

, ,2

nGP

t s s k k

k

h D B s s P U W

=

= ∑



46




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