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• Finite Element Method

Chapter 9

Axisymmetric Elements

• Definition of an axisymmetric solid

An axisymmetric solid (or a thick-walled body) of revolution is defined as a 3-D body that is generated by rotating a plane and is most easily described in cylindrical coordinates. Where z is called the axis of symmetry.

If the geometry, support conditions, loads, and material properties are all axially symmetric (all are independent of ), then the problem can be idealized as a two-dimensional one.

• Problems such as soil masses subjected to circular footing loads, thick-walled pressure vessels, and a rocket nozzle subjected to thermal and pressure loading can often be analyzed using axisymmetric elements.

Examples of an axisymmetric solid

• Examples of an axisymmetric solid

Enclosed pressure vesselEngine valve stem

• axisymmetric problems can be analyzed by a finite element of revolution, called axisymmetric elements. Each element consists of a solid ring, the cross-section of which is the shape of the particular element chosen (triangular, rectangular, or quadrilateral elements). An axisymmetric element has nodal circles rather than nodal points

FE axisymmetric elements

• Equations of Equilibrium:

The three-dimensional elasticity equations in cylindrical coordinates can be summarized as follows

011

021

01

bzrzzzr

rzr

rrzrrr

Zrzrr

Yrzrr

Xrzrr

• Equations of Equilibrium:

The three-dimensional strain-displacement relationships of elasticity in cylindrical coordinates were u, v, w are the displacements in the r, ,dz , respectively, are:

w

rz

v

z

w

z

u

r

w

r

uv

r

r

v

r

vu

rr

u

zz

zr

rr

1,

,1

1,

• Equations of Equilibrium:

The three-dimensional stress-strain relationships for isotropic elasticity are:

z

zr

r

z

r

z

zr

r

z

r

E

2

210000

02

21000

002

21000

0001

0001

0001

)21()1(

• Axisymmetric Stresses and Strains

In axisymmetric problems, because of the symmetry about the z -axis, the stresses are independent of the coordinate. Therefore, all derivatives with respect to vanish and the circumferential (tangent to direction) displacement component is zero; therefore,

0 0

, , ,

r z r z

r z r z

and

u u w u w

r r z z r

• Axisymmetric Stresses and Strains

,

,

r

z r z

u u

r r

w u w

z z r

• The stress-strain relationship for isotropic axisymmetric problems

zr

z

r

zr

z

r

E

2

21000

01

01

01

)21()1(

}{][}{ D

• Derivation of the Stiffness Matrix and Equations

Typical slice through an axisymmetric solid Discretized into triangular elements

Step 1: Discretize and Select Element Type

• Step 1: Discretize and Select Element Type

m

m

j

j

i

i

m

j

i

w

u

w

u

w

u

d

d

d

d}{

(ui , wi ) displacement components of node i in the r and z directions,respectively.

• Step 2: Select Displacement Functions

zaraazrw

zaraazru

654

321

),(

),(

6

5

4

3

2

1

654

321

1000

0001

),(

),(}{

a

a

a

a

a

a

zr

zr

zaraa

zaraa

zrw

zru

• Step 2: Select Displacement Functions

mmm

jjj

iii

mmm

jjj

iii

zaraaw

zaraaw

zaraaw

zaraau

zaraau

zaraau

654

654

654

321

321

321

In Matrix Form

m

j

i

mm

jj

ii

u

u

u

zr

zr

zr

a

a

a1

3

2

1

1

1

1

m

j

i

mm

jj

ii

w

w

w

zr

zr

zr

a

a

a1

6

5

4

1

1

1

• mm

jj

ii

yx

yx

yx

A

1

1

1

2

mji

jimimjmji yyxyyxyyxA

)()()(2

Solving for the a’s

A is the area of the triangle

m

j

i

mji

mji

mji

u

u

u

A

a

a

a

2

1

3

2

1

m

j

i

mji

mji

mji

w

w

w

A

a

a

a

2

1

6

5

4

• ijmmijjmi

jimimjmji

jijimimimjmjmji

rrrrrr

zzzzzz

rzzrrzzrrzzr

3

2

1

1}{

a

a

a

zru

m

j

i

mji

mji

mji

u

u

u

zrA

u

12

1}{

• 1{ } 1

2

1( , ) ( ) ( ) ( )

2

1( , ) ( ) ( ) ( )

2

( , ){ }

( , )

i i j j m m

i i j j m m

i i j j m m

i i i i j j j j m m m m

i i i i j j j j m m m m

i i j

u u u

u r z u u uA

u u u

u r z r z u r z u r z uA

w r z r z w r z w r z wA

N u Nu r z

w r z

j m m

i i j j m m

u N u

N v N v N v

• ( , ){ }

( , )

1

2

1

2

1

2

i i j j m m

i i j j m m

i i i i

j j j j

m m m m

N u N u N uu r z

N v N v N vw r z

N r zA

N r zA

N r zA

• [ ]

0 0 0{ }

0 0 0

{ } { }

0 0 0

0 0 0

i

i

i j m j

i j m j

m

m

i j m

i j m

u

w

N N N u

N N N w

u

w

d

N N N

N

N

N NN

• Step 3: Define the Strain/Displacement and Stress/Strain Relationships

2

6

312

3 5

{ }

{ }

r

z

rz

r

z

rz

u

r

w

z

u

r

u w

z r

a

a

a zaa

r r

a a

• }{][}{ dB

0 0 0

0 0 0

0 0 0

0

0

0

i

i j m

i

i j m

j

j ji i m mji j m

mi i j j m m

m

i

ii

ij

i j m i i ij i

mi i

m

u

w

uzz z

wr r r r r r

u

w

u

w

uB B B B z

wr r

u

w

B is a function of r and z

• Stress Strain Relationship

}{][][}{ dBD

zr

z

r

zr

z

r

E

2

21000

01

01

01

)21()1(

• Step 4 :Derive the Element Stiffness Matrix and Equations

[ ] [ ] [ ][ ]

[ ] 2 [ ] [ ][ ]

T

V

T

A

k B D B dV

k B D B r dr dz

1) Numerical integration (Gaussian quadrature)2) Explicit multiplication and term-by-term integration.3) Evaluate [B] at a centroidal point of the element

3 3

[ ( , )] [ ]

[ ] 2 [ ] [ ][ ]

i j m i j m

T

r r r z z zr r z z

B r z B

k r A B D B

• }{}{][}{][}{ PdSTNdVXNf

S

T

V

T

Concentrated nodal forcesBody forces

Surface Tractions

Step 4 :Derive the Element Stiffness Matrix and Equations

• { } [ ] { } 2 [ ]bT T

b

bV A

Rf N X dV N r dr dz

Z

for a machine part rotating with:a constant angular velocity about the z axis (zero angular acceleration); : material mass density (kg/m3) and r is the radial coordinate.Zb= body force per unit volume due to the force of gravity

(weight density, kN/m3).

Distributed Body ForcesBody forces can either : Gravity force in the direction of z axis, or Centrifugal forces in rotating machine parts in the direction of the r axisThe nodal equivalent body forces can found using

2

bR r

• 0{ } 2

0

2{ }

3

i b

bi

i bA

bir b

bi

biz b

N Rf r dr dz

N Z

f RArf

f Z

b

b

b

b

b

b

bmz

bmr

bjz

bjr

biz

bir

b

Z

R

Z

R

Z

R

Ar

f

f

f

f

f

f

f3

2}{

rRb 2

Distributed Body Forces

• { } [ ] { } [ ]rT T

s

zS S

pf N T dS N dS

p

dzrp

p

zr

zr

Af j

z

rz

zjjj

jjj

jsm

j

2

0

0

2

1}{

For example: along the vertical face jm of an element, let uniform loads pr and pz be applied, as shown along r=rj.

Surface Forces

• dzrp

p

zr

zr

Af j

z

rz

zjjj

jjj

jsm

j

2

0

0

2

1}{

z

r

z

rjmj

smz

smr

sjz

sjr

siz

sir

s

p

p

p

pzzr

f

f

f

f

f

f

f

0

0

2

)(2}{

Surface Forces

evaluated at r = rj ; z = z

• Step 5: Assemble the Element Equations to Obtain the Global Equations and Introduce Boundary Conditions

N

e

ekK1

)( ][][

}{][}{ dKF

N

e

efF1

)( ][][

Step 6: Solve for the Nodal Displacements

Step 7: Solve for the Element Stresses

• Steps 5 through 7

Steps 5 through 7, are analogous to those of Chapter 6 for the CST element, except the stresses are not constant in each element.

They are usually determined by: Determine the centroidal element stresses, Determine the nodal stresses for the element

and then average them. The latter method has been shown to be more

accurate in some cases

• Example 1

For the element of an axisymmetric body rotating with a constant angular velocity = 100 rev/min.Evaluate the approximate body force matrix, include the weight of the material, where the weight density w is 0.283 lb/in

3.The coordinates of the element (in inches) are shown in the figure.

The body forces per unit volume evaluated at the centroid of the element are

Zb =0.283 lb/in3

• All r-directed and z-directed nodal

body forces are equal

Example 1 -Continue

• Example 2

For the long, thick-walled cylinder under internal pressure p equal to 1 psi shown, determine the displacements and stresses.

• Example 2

To illustrate the finite element solution for the cylinder, we first discretize the cylinderinto four triangular elements, as shown. A horizontal slice of the cylinderrepresents the total cylinder behavior.

• Example 2

Assemblage of the Stiffness Matrix

0.5, 0, 1.0, 0, 0.75, 0.25

( 1; 2,

For element

5 )

1

i i j j m mr z r z r and z

i j and m

• Example 2 – Stiffness Matrix for ELEMENT 1

• Example 2 – Stiffness Matrix for ELEMENT 1

• Example 2 – Stiffness Matrix for ELEMENT 1

• 1.0, 0, 1.0,

For element 2

0.5, 0.75, 0.25

( 2; 3, 5 )

i i j j m mr z r z r and z

i j and m

Example 2 – Stiffness Matrix for ELEMENT 1

• Example 2 – Stiffness Matrix for ELEMENT 2 & 3

• Example 2 – Stiffness Matrix for ELEMENT 1

• Example 2 – Total Stiffness Matrix

• HW:

xxxx