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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

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Finite Element Method Chapter 9 Axisymmetric Elements

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