07 Axisymmetric Finite Element Modeling

8/10/2019 07 Axisymmetric Finite Element Modeling

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CE4257: Linear Finite Element Analysi s:

Part C: Axisymmetric Elements

Somsak SwaddiwudhipongDepartment of Civil & Environmental Engineering

National University of Singapore

Tel: 6516-2173Room: E1A-07-10

Email: [email protected]

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Objectives

To demonstrate the formulation of axisymmetric finite elements

and their applications to solve axisymmetric problems

(involving axisymmetric geometries under axisymmetric loads).

Intended Outcome

You should be ab le to (i) develop axisymmetric finite elements,

(ii) adopt appr opr iate axis ymmetr ic element s to analyse

axis ymmetr ic problems and (iii) further study to solve

problems involving axisymmetric geometries under

nonsymmetrical loads.

Objectives and Intended Outcome


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Part C: Axisymmetric Elements

4.8 Axisymmetric Elements4.8.1 Elasticity of Axisymmetric Problems4.8.2 Triangular Axisymmetric Element, TA34.8.3 Quadrilateral Axisymmetric Element, QA44.8.4 Boundary Condit ions4.8.5 Concluding Remarks

4

Applications of Axisymmet ric Elements


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4.8.1 Elasticity of Axisymmetric Problems

(4.8.1)

(4.8.2)

Only displacements in the r-z plane exist and all variables are independentof . The strains are given by:

; ; ;

0

0

10

r z rz

r

z

rz

u w u w ur z r r z

r

u zw

r

z r

D

1 01 0

1 0(1 )(1 2 )

1 22

E

sym

D(4.8.3)

(4.8.4)

4 2

Similar derivations may be made for displacement interpolation functionw(r,z), in the z-direction.

(4.8.5)

( ) / det

det *

1

1 1 2 2 3 3 2

3

1 1

2 2

3 3

1 2 3

1 2 3 3 2 1 2 3 1 3 2

2 3 1 1 3 2 3 1 2 1 3

3 1 2 2 1 3 1 2 3 2 1

111

2

i i i i

uu u u u

u

r zr zr z

u N N N N N N

N a b r c za r z r z b z z c r r a r z r z b z z c r r a r z r z b z z c r r

area of triangle

A

A

(4.8.6)

(4.8.7)

Writing both displacements in vector form, 11

21 1 2 2 3 3

21 1 2 2 3 3

3

3

1 2 3

1 2 3

0 0 00 0 0

uwuu u u uww v v vuw

N N NN N NN N NN N N

iu Nu

4.8.2 Triangular Axisymmetric Element, TA3 :

(4.8.8)

r

z Similar to T3 element,

w1

w2

Fig 4.8.3 Triangularaxisymmetric element


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Strain-Displacement Relation

(4.8.9)

Formulation procedure for element properties is similar to that of T3element. We have to add row 3 in strain matrix for hoop strain, that is,

Strain matrix B is not a constant matrix.

31 2

1

131 2

2

21 2 3

3

1 1 2 2 3 3 3

00 0 0

0 0 0 0

10 0 0 0

r

z

rz

N N N r r r r

w N N N u z z z z

ww N N N r r r r

N N N N N N w z r z r z r z r

u

u

u

iBu

(4.8.10)

31 2

1 2 3

1 1 2

2

2 3

3

3

1

00 0

0 0 01

2 2 22

0 0 0

bb b

c c c

A A A A

c b c b c

N N

r r r

b

N

B

r r r

Stiffness Matrix

(4.8.11).2V A

dV r dr dz T TK B DB B DB

B is not a constant matrix and the evaluation of the integral in Eq (4.8.11)is no longer a simple procedure. We adopt either numerical in tegrationor t he following approximate method to evaluate the integral values.

is B evaluated at and . Eq (4.8.12) provides acceptable accurateresults if small elements are used in the regions of high stressgradients. Once the consistent vector F i is evaluated as shown in thenext few slides, we have got the required set of element stiffnessequations: iu

2

V VdV dV r A

T T TK K B DB B DB B DB

B r z

(4.8.12)

1 2 33 r r r r 1 2 33 z z z z

, r z (4.8.13)


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Consistent Force Vectors ::

F = P+P T+P b(a) Nodal Circumferential Line Forces, p i:

1 1 1 1 2 2 2 2 3 3 3 32T

r z r z r zr p r p r p r p r p r p P

(4.8.14)

(4.8.15)

2i i iP r p

We normally set a uniform line forces along the circular ring node i. Thecircumferential line load of p ir or p iz on the structure is equivalent to a pointload P ir or P iz on axisymmetric element at the same node on the elementscross sectional view, and

If only a line force is applied along a circular ring node, say node 2,then only r 2p2r and r 2p2z exist. All other terms in Eq (4.8.15) vanish.

2

r

z

1

3iP

Consistent Force Vector is derived from 3 actions:(i) Nodal circumferential line forces, p i;(ii) Surface tractions, T along an edge;(iii) Body Forces, b

Consistent Force Vectors

(4.8.17)

(b) Surface Tractions, T along Edge 1-2:

(4.8.16)

1 1

1 1

2 2

2 2

3

3

2

0 0

0 0

0 0 2 2

0 0

0 0 00 0 0

T A S

r r

z zS S

dA r dS

N rL

N rL

N T T rLrdS dS

N T T rL

N N

T TP N T N T S is the coordinatealong element edge.

The kernel of integral in Eq (4.8.16) involves LiL j that can be integrated inaccordance with Eq (4.1.31) and finally we have for any inclined edge, l12 :

1 2

1 2

1 212

1 2

(2 )

(2 )

( 2 )2( 2 )6

0

0

r

z

r T

z

r r T

r r T

r r T lr r T

P122

2

0

0

r

z

r T

z

T

T

T rlT

P (4.8.18)

For vertical edge,r 1=r 2=r and hence:

l12 is the length of edge 1-2.

1 1 2 2 3 3 r r L r L r L


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(4.8.20)

(4.8.19)

1 1

1 1

2 2

2 2

3 3

3 3

0 00 0

0 02 2 2

0 0

0 0

0 0

r r z zV A A A

N rL N rL

N rLb bdV r dA rdA dA

N rLb b

N rL

N rL

T TbP N b N b

(c) Body Forces, b :

The kernel of integral in Eq (4.8.19) involves LiL jthat can be integrated using Eq (4.1.30) and finallywe have, for constant b r and b z:

_

1

_

1

_ 2

_

2

_

3

_

3

( 3 )

( 3 )

( 3 )212

( 3 )

( 3 )

( 3 )

r

z

r b

z

r

z

r r b

r r b

r r b A

r r b

r r b

r r b

P

Even when b r and b z are constant in eachelement, P b does not distribute the valuesat each node equally. Why?

1 1 2 2 3 3 r r L r L r L

Ex 4.8.1 Triangular Axisymmetric Element, TA3 :The coordinates (in cm) of a triangular ringelement are as shown in Fig E4.8.1. Theelement is under the body forces of b r =0 andbz=-10 N/cm 3. Evaluate the approximatestrain matrix and the consistent force vector.

( )( ) ( )( )

( )( ) ( )( )( )( ) ( )( )

det1 1

2 2

3 3

1 2 3 3 2

2 3 1 1 3

3 1 2 2 1

1 2 3

2 3 1

3 1 2

1 3 2

2 1 3

3 2 1

4 6 3 4 12

4 6 26 4 2

4 4 0

3 4 1

2 3 1

4 2 2

1 1 2 41 1 4 41

3 4 2 6 02 4 4 4 8

r zr zr z

a r z r za r z r za r z r z

b z zb z zb z zc r r c r r c r r

A

( ) ( ) ( )

(

1 3 6

24 12 12 12 8 16 4

2 areaof triangle) (b)

Fig E4.8.1 Triangular axisymmetricelement

(c)

_ _ 14At ( , ) (3, )

3r z

( )2 2 2 214 4

0 29 9

2r r r

AN a zb c

( )3 3 3 38 28 4

03 9 9

2r r r

AN a zb c

( )1 1 1 112 1 14 4

23 3 3 9

2r r r

AN a zb c

[Area of triangle is (2)(2)/2=2].

( ) /2 i i

i ir r r AN a zb c Det AEq (4.8.10):

(a)

r

u22

(4,4)(2,4)

(3.6)

z

1

3

u1

w1 w2

Eq (4.8.6):

310 N cm zb


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Strain Matrix and Force Vector

Eq (4.8.9):

(d)

31 2

1 2 3

1 2 3

1 1 2 2 3 3

00 0

0 0 01

2 2 22

0 0 0

bb b

c c c

AN AN AN Ar r r

c b c b c b

B _

2 0 2 0 0 0

0 1 0 1 0 21

4 4 44

0 0 09 9 9

1 2 1 2 2 0

B

_

1

_

1

_ 2

_

2

_

3

_

3

( 3 )0 0

( 3 )(2 3(3))( 10) 110

( 3 ) 0 02 N

(4 3(3))( 10) 13012 3 3( 3 )

0 0( 3 ) (3 3(3))( 10) 120

( 3 )

r

z

r b

z

r

z

r r b

r r b

r r b A

r r b

r r b

r r b

P

Eq (4.8.20):

(e)

P b does not distribute thevalues at each node equally.

2

(4,4)(2,4)

(3.6)

z

1

3

u1

w1 w2

r

4.8.3 Quadrilateral Axisymmetric Element, QA4 :

1 11 24 4

1 13 44 4

(1 )(1 ) (1 )(1 )

(1 )(1 ) (1 )(1 )

N s t N s t

N s t N s t

1

1

2

1 2 3 4 2

1 2 3 4 3

3

4

4

0 0 0 0

0 0 0 0

r

z

r

N N N N zr

N N N N r z

z

r

z

As an example, we will consider a QA4 element. The shape functions aresimilar to those of Q4 element as follows:

Geometry:

(4.8.21)

(4.8.22)

Displacement:

1

1

2

1 2 3 4 2

1 2 3 4 3

3

4

4

0 0 0 0

0 0 0 0

u

w

u

N N N N wu

N N N N uw

w

u

w

(4.8.23)

Fig 4.8.4 AxisymmetricQA4 master element


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Strain-Displacement Relation :

(4.8.24)

Formulation procedure for element properties is similar to that of Q4element. We have to add row 3 in strain matrix for hoop strain, that is,

Strain matrix B is normally derived through iso-parametric concept usingJacobian transformation matrix.

31 2 4

31 2 4

1 2 3 4

1 1 2 2 4 43 3

00 0 00

0 0 0 0 0

10 0 0 00

r

z

rz

N N N N r r r r r

N N N N u z z z z zw N N N N

r r r r r N N N N N N N N

z r z r z r z r z r

1

1

2

2

3

3

4

4

u

w

u

w

u

w

u

w

iBu

1 1 2 2 3 3 4 4, , , , r N s t r N s t r N s t r N s t r (4.8.22)

Jacobian Transformation Matrix

Chain rule gives:

where J is the Jacobian transformation matrix.

(4.8.25)

(4.8.26)

u u r u zs r s z su u r u zt r t z t

[ ]

u r z u us s s r r J u r z u ut t t z z

1 1 1

u u z z u z u z ur s t s s t s s t J u u r r u r u r u J J z t t s t s t t s


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Jacobian Transformation Matrix :

(4.8.22)

(4.8.28)

1 2

43[ ]

r zJ Js sJJ Jr z

t t

(4.8.27)

3 31 2 4 1 2 44 41 1 2 3 2 1 2 3

3 31 2 4 1 2 44 4 43 1 2 3 1 2 3

N NN N N NN NJ r r r r J z z z z

s s s s s s s s

N NN N N NN NJ r r r r J z z z zt t t t t t t t

Geometry:

Jacobian:

The characteristics (qualities) of the transformation (Jacobian) J matrixaffect greatly the FE performances adopting iso-parametric concept.

1 1 2 2 3 3 4 4, , , , r N s t r N s t r N s t r N s t r

1 1 2 2 3 3 4 4, , , , z N s t z N s t z N s t z N s t z

Elements of Strain Matrix

(4.8.29)

(4.8.31)

(4.8.30)

4 2

1

1 3

1 1 1

u u z z u z u z u u u

J Jr s t s s t s s t s tJu u J r r u r u r u u u

J Jz t t s t s t t s t s

J J

1

2

3

4

3 31 1 2 2 4 44 4 4 42 2 2 2

3 31 1 2 2 4 41 3 1 3 1 3 1 3

41 2 3

1

1

r r r r

z

uu

uu

N NN N N N N NJ J J J J J J J

s t s t s t s tN NN N N N N NJ J J J J J J J J

t s t s t s t s

uB B B Br Bu

z

1

2

3

4

4 2

42 31 3

i iir

zz z i iiz

uuuu

N NB J J J

s tB B B N N

B J J Jt s

14 2

2

31 3

4

41 2 3

41 2 3

1 1 r r r r zz z z

ww z w z w w wJ J wr t s s t s t

ww r w r w w wJ J

z s t t s t s w

B B B BB B B BJ J

1 1 2 2 3 3 4 4 N u N u N u N uu


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F = P+P T+P b

(a) Nodal Circumferential Line Forces, p i:

1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 42 T

r z r z r z r zr p r p r p r p r p r p r p r pP

(4.8.34)

(4.8.35)

2 i iP rp

We normally set a uniform line forces along the circular ring node i. Theline load p ir or p iz on the structure is equivalent to a point load P ir or P iz onaxisymmetric element at the same node on the elements cross section,and

If only a line force is applied along a circular ring node, say 3, then onlyr 3p3r and r 3p3z exist. All other terms in Eq (4.8.35) vanish.

Consistent Force Vectors(b) Surface Traction, T:

(4.8.36)

1

1

21

2

31

3

4

4

0

0

0

02 2

0

0

0

0

ir T t t z A S

N

N

N

N T dA r dS r J ds

N T

N

N

N

T TP N T N T

1 1 2 2 3 3 4 4, , , , G G G G G G G Gr N s t r N s t r N s t r N s t r We normally adopt Gauss quadrature to evaluate the integrals.

2 T A S

dA r dS T TP N T N T

Consider surface along t=t i: it t dS J ds

2 dA rdS

Note that dS is the differential length,while s is a natural coordinate.


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(4.8.37)

1

1

21 1

2

31 1

3

4

4

0

0

0

02 2 .

0

0

0

0

r zV A

N

N

N

N bdV r dA r J ds dt

N b

N

N

N

T TbP N b N b

(c) Body Force, b :

1 1 2 2 3 3 4 4, , , , G G G G G G G Gr N s t r N s t r N s t r N s t r

We normally adopt Gauss quadrature to evaluate the integrals.

2 2 . dV rdA r J ds dt

4.8.4 Boundary Conditions(a) Displacement u = 0 at the points along the axis of symmetry (z-axis);(b) Remove rigid body motion in the z-direction by setting w = 0 (at leastat one node).

4.8.5 Concluding Remarks

1. If geometry, materials and forces are axisymmetric, variables areindependent of and hence only u(r,z) and w(r,z) exist. 3-Dproblems can be solved as 2-D with additional hoop strain andstress;

2. Formulation process follows that of corresponding plane elementswith hoop actions and appropriate boundary conditions;

3. If geometry is axisymmetric but the loads are not, express the loadvia Fourier series comprising both symmetric and antisymmetriccomponents and each part is handled separately. More details aregiven in many textbooks including Asghar Bhati (2006), 154-166.


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10 N/mm(4,3)

(8,4)

(11,10)

(2,8)

12

43

21

3

658

7

4

z

r

Length unitin mm.

Fig. E4.8.1: Applied forces in axisymmetri c element

Exercise 4.8: Axisymmetric Elements

Exercise E4.8.1: Establish the strain matrix B and the consistent forcevector for a quadrilateral ring element shown in Fig. E4.8.1.

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