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MMJ1153 – COMPUTATIONAL METHOD IN SOLID MECHANICS

D – AXISYMMETRIC PROBLEM FORMULATIONS

AXISYMMETRIC PROBLEM FORMULATIONS M.N. Tamin, CSMLab, UTM 1


Course Content:

A – INTRODUCTION AND OVERVIEW

Numerical method and Computer-Aided Engineering; Physical

problems; Mathematical models; Finite element method.

B – REVIEW OF 1-D FORMULATIONS

Elements and nodes, natural coordinates, interpolation function, bar

elements, constitutive equations, stiffness matrix, boundary


elements, constitutive equations, stiffness matrix, boundary

conditions, applied loads, theory of minimum potential energy; Plane

truss elements; Examples.

C – PLANE ELASTICITY PROBLEM FORMULATIONS

Constant-strain triangular (CST) elements; Plane stress, plane

strain; Axisymmetric elements; Stress calculations; Programming

structure; Numerical examples.


Axisymmetric Problems

• 3D axisymmetric solids or solids of revolution

• Subjected to axisymmetric loading

- Deformation and stress are

independent of θ- Reduced to 2D in rz-plane


- Reduced to 2D in rz-plane

(cylindrical coordinate)


Displacement

{ }

=w

uu

{ }

∂∂

+∂∂

∂∂∂∂

=

=

u

r

w

z

u

z

w

r

u

rz

z

r

θε

γ

ε

ε

ε

Strain and strain-displacement


r

u

{ }

=

θσ

τ

σ

σ

σrz

z

r

Stress


Generalized Hooke’s Law

−−

−

−

Ε=

z

y

x

z

y

x

εεε

νννν

νννννν

σσσ

0021

000

0001

0001

0001


( )( )

−

−−+

Ε=

yz

xz

xy

z

yz

xz

xy

z

γγγε

ν

ννν

τττσ

2

2100000

02

210000

002

000211

{ } [ ]{ }[ ] matrix elasticityor material a is where D

D εσ =


Constitutive equations - elasticity

{ } [ ]{ }εσ D=

Elasticity matrix

−− 10

11

νν

νν


[ ] ( )( )( )

( )

−−

−−

−−

−−

−+−

=

1011

012

2100

101

1

11

211

1

νν

νν

νν

νν

νν

νν

νννE

D


Discretization of Solution Domainz


ro


Finite Element Formulation

– Triangular element for axisymmetric problems

{ }

=

=2

2

1

1

4

3

2

1

u

w

u

w

u

q

q

q

q

q

q


Compare with triangular element

developed earlier for plane stress

and plane strain problem

3

3

6

5

w

u

q

q


Displacement Field

634221

533211

qNqNqNw

qNqNqNu

++=

++=

1

q

qηξ

η

ξ

−−=

=

=

13

2

1

N

N

N

Alternate displacement functions

Isoparametric Representation


{ } [ ]{ }u N q=

{ }

=

6

5

4

3

2

321

321

000

000

q

q

q

q

q

NNN

NNNu

ηξ −−=13N

( ) 1,2,3i , 2

1=++

∆= ycxbaN iiii

Refer to detailed derivation of

interpolation functions, Ni in terms of

nodal coordinates, from previous topic.

( )( ) 642

531

1

1

qqqw

qqqu

ηξηξ

ηξηξ

−−++=

−−++=


Strain-displacement Relations

r

u

r

w

z

u

z

w

r

u,,,

∂∂

+∂∂

∂∂

∂∂

The strain-displacement terms are:

ηηη

ξξξ

∂∂

∂∂

+∂∂

∂∂

=∂∂

∂∂

∂∂

+∂∂

∂∂

=∂∂

z

z

ur

r

uu

z

z

ur

r

uu

[ ]

∂∂∂∂

=

∂∂∂∂

∂∂

∂∂

∂∂

∂∂

=

∂∂

∂∂

z

u

r

u

J

z

u

r

u

zr

zr

u

u

ηη

ξξ

η

ξ


ηηη ∂∂∂∂∂ zr

[ ]

∂∂

∂∂

=

∂∂∂∂

−

η

ξu

u

J

z

u

r

u

1

Similarly:

[ ]

∂∂

∂∂

=

∂∂∂∂

−

η

ξw

w

J

z

w

r

w

1


Jacobian matrix

[ ]

=

∂∂

∂∂

∂∂

∂∂

=2323

1313

zr

zr

zr

zr

J

ηη

ξξjiij

jiij

zzz

rrr

−=

−=

( )( )

321332211 1 rrrrNrNrNr ηξηξ −−++=++=


[ ][ ]

−

−=−

1323

13231

det

1

rr

zz

JJ

[ ] 13232313det zrzrJ −=

[ ] eAJ 2det =

( ) 321332211

321332211

1 zzzzNzNzNz ηξηξ −−++=++=


Strain-displacement Relations

{ }

( ) ( )[ ]

( ) ( )[ ]

( ) ( ) ( ) ( )[ ]

++

−−−+−+−−

−+−−

−−−

=

∂∂

+∂∂

∂∂∂∂

=

=

qNqNqN

J

qqzqqzqqrqqr

J

qqrqqr

J

qqzqqz

u

r

w

z

u

z

w

r

u

rz

z

r

533211

6413622353135123

64136223

53135123

det

det

det

θε

γ

ε

ε

ε


++

r

qNqNqNr

u533211

{ }

[ ] [ ] [ ]

[ ] [ ] [ ]

[ ] [ ] [ ] [ ] [ ] [ ]

=

6

5

4

3

2

1

321

122131132332

211332

123123

000

detdetdetdetdetdet

det0

det0

det0

0det

0det

0det

q

q

q

q

q

q

r

N

r

N

r

N

J

z

J

r

J

z

J

r

J

z

J

r

J

r

J

r

J

r

J

z

J

z

J

z

ε { } [ ]{ }qB=ε


Potential Energy

The functional in the variational principle is the potential energy of a 2-D elastic body acted by surface and body forces.

( ) ( ) ( )wuWwuUwu ,,, −=π


( ) [ ] [ ][ ]{ }[ ] { } { }dlqTdVqfdVqBDBqwulV

b

V

T

∫∫∫ −−=1

2

1,π

( ) ( )( )∑=

=M

e

e wuwu1

,, ππ


Elemental volume

∫ ∫ ∫ ∫==V A A

dArddArdV

π

πθ2

0

2



( )

( ) 0=

∂∂∂∂

ei

e

w

u

π

π

Minimum Potential Energy Theorem

Recall the minimum potential energy theorem applied to

triangular element for plane stress / plane strain case:


[ ]( )

[ ][ ] ( ){ } { }( )

( ) { }( ) ( )

( )

0222

1

)( =−−ΒΒ

∂

∫∫∫∫∫eee

l

eee

A

e

b

e

i

A

T

i

dlrTdArfqdArD

w

πππ

[ ] { } { } { } )()()( ee

b

eTFqk +=

[ ] [ ] [ ][ ] dArBDBkT

A e

∫∫=)(

2π

Element stiffness matrix


Notes on [B] for evaluation of the term r

Ni

For simple approximation, both terms are evaluated at

the centroid of the triangle and used as representative

values for the triangle:

3

1321 === NNN

321 rrrr

++= radius of the centroid


3

321 rrrr

++= radius of the centroid

(from origin of cylindrical coordinate)

332211 rNrNrNr ++=

Alternately we can integrate when using:

[ ] [ ] [ ][ ]BDBArkTe)(2π=


Body Force

Suppose body force components, fr and fz, act at

the centroid of a triangular element.

The work done by these forces is given by,

Element Force Vector

{ } { } ( )+=∫ ∫A A

zrb

TdArwfufdArfu

e e

22 ππ


( ) ( )[ ]

≈

+++++= ∫

∫ ∫

z

r

z

r

z

r

e

A

zr

A A

f

f

f

f

f

f

qqqqqqAr

dArfqNqNqNfqNqNqNe

e e

654321

634221533211

3

2

2

π

π

If body force is the primary

load, for better accuracy,

use:

332211 rNrNrNr ++=


Body Force due to gravity and radial centrifugal force (inertia)

Element Force Vector

{ }

−≈

−=

=g

r

g

r

f

ff

z

r

b ρωρ

ρωρ 22

ω

ρ is mass per unit volume

of the element.


If body force is the primary

load, for better accuracy,

use:

332211 rNrNrNr ++=fz

fr

ω


Traction Force

Suppose a linearly varying traction components act along edge 1-2

of a triangular axisymmetric element.

The potential energy due to the traction force is,

{ } { }

+

+

+

+

= −∫r

z

r

l

T

Trr

Trr

Trr

qqqqldlrTue

6

26

26

2

2221

21

21

432121ππ


+zT

rr

6

26

21


Element Stress Calculations(Computed at the centroid of the element section)

[ ] [ ]{ } [ ][ ]{ }qBDD == εσ

zσ


rσθσ


Discussion on modeling issues and boundary

conditions for axisymmetric problems



Example: A thick cylinder subjected to internal pressure

�Boundary conditions

�Axisymmetric problem

AXISYMMETRIC PROBLEM FORMULATIONS M.N. Tamin, CSMLab, UTM

�Boundary conditions

w = constant

w = 0

u = constant


�Mesh

�3-node triangular elements



Theoretical result are calculated

using Lame equations

2

2

r

BA

r

BA

r

H

−=

+=

σ

σ


at R2

Date post:	03-Jan-2021
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