two dimensional analysis – plane stress and plane strain · two dimensional analysis – plane...

Section 9: AXISYMMETRIC ELEMENTS

Washkewicz College of Engineering

Two Dimensional Analysis – Plane Stress and Plane StrainIn a large class of every day engineering problems certain approximations are made to simplify the structural analysis of three dimensional components. Recall that with the strong formulation there are 15 equations to solve in terms of 15 unknowns. One approach to reduce the computational effort in solving this system of equations is realizing that certain problems are really two dimensional. This reduces the number of equations to solve.

Three types of two dimensional idealizations are employed:1. Plane stress2. Plane strain3. Axisymmetry

For example, simplifying approximations can be made in analyzing deformations in a thin plate subjected to in-plane forces. Consider a prismatic structural member with a very short length or thickness (h). The mid plane contains the x and y coordinate axes and the thickness extends along the z axis to +h/2 and –h/2. If the member is not loaded on surfaces perpendicular to the z axis then

on these surfaces perpendicular the z axis and through the thickness. Assuming the remaining stress components are not dependent on z, then this is known as plane stress. 1

0 yzxzz



Here the stress matrix and the strain matrix take the following forms

For plane strain one dimension (say along the z axis) is exceedingly large relative to the other two dimensions. Applied forces act in the x – y plane and do not vary along the zdirection. Some practical examples include dams and tunnels as well as bars that are compressed along their length. Here

on these surfaces perpendicular the z axis. Assuming the remaining strain components are not dependent on z, then this is known as plane strain.

2

0 yzxzz

yxz

yxy

xyx

yxy

xyx

E

yxyxyxyx

yxyxyxyx

00

0,,0,,

0000,,0,,



Note that for plane strain the functional dependence of the displacements are

Thus

The axisymmetric formulation has the same geometric implication, i.e., a two dimensional problem with the additional wrinkle of posing the problem in a different coordinate system.

0

,

,

w

yxvv

yxuu

yxz

yxy

xyx

yxy

xyx

yxyxyxyx

yxyxyxyx

000,,0,,

0000,,0,,



AxisymmetryWe have considered “line” elements and two dimensional elements. We now turn our attention to axisymmetric elements. This is a special two dimensional element utilized when there is a specific type of geometric symmetry and load symmetry present in the problem. We use cylindrical coordinates (r, , z) to describe all aspects of the problem, i.e., displacements, strains and stresses. The z-axis is the axis of symmetry.

Examples of axisymmetric problems include, but are not limited to

• Pressure vessels

• Cylindrical shafts

• Hertzian contact between spheres

Use of axisymmetric elements provides computational efficiency relative to a full three dimensional analysis.

4



Quasi-Axisymmetric NotationSome fundamental concepts are presented first. An axisymmetric element is essentially a triangular (or quadrilateral) torus. Each node traces a circular line which are depicted as dashed lines in the figure below on the left:

The z-axis is the axis of symmetry. The component being modeled must have geometric as well as load symmetry with respect to this axis. These problems are best modeled in a polar coordinate system (r, , z). Corresponding displacements will be designated (u, v, w) and the two dimensional stress state is indicated in the figure on the right.



The following two load applications are acceptable for an axisymmetric analysis:

The next two are not because the load is functionally dependent on . The figure on the left represents the wind loads (windward and leeward) on a cooling tower.



As a previous figure indicates, we wish to use the notation, and to some extent the derivations, developed for the two dimensional constant strain element. That element will be “spun around” an axis of symmetry (the z-axis) and thereby generate an axisymmetric element.

In that context the “two dimensional” axisymmetric element presents itself in an r-z plane much the same way the constant strain element presents itself in the x-y plane. For the two dimensional constant strain and linear strain element we ignored displacements in the zdirection. Here we do not. We could wave our hands over the constant strain and linear strain elements and mumble something regarding plane stress or plane strain and somewhat ignore displacements in the z direction. We will rectify that when solid (tetrahedron) elements are developed. For axisymmetric elements the “out of plane” displacements are a bit more troubling.

The term “quasi-axisymmetric” refers to fact that in truly axisymmetric problems the circumferential displacements (v or u depending on the notation used) are identically zero. Where stresses and strains are not functionally dependent on and

this class of problems are labeled quasi-axisymmetric.

0v



The most familiar components that can be modeled in an axisymmetric fashion are thin walled pressure vessel (hot dog stresses). In fact any cylindrical pressure vessel, thin walled or otherwise, has the potential of being modeled as an axisymmetric problem. Consider the gun barrel for an Abrams M-1 tank:



A portion of the axisymmetric mesh used to model the tank barrel is depicted below:

Note that the mesh is predominantly quadrilateral axisymmetric elements, with a triangular element furnishing a transition as the throat of the barrel thins out. The point here is that there are a lot of elements in this mesh and they all trace out a torus when spun around the z-axis. If three dimensional elements had been used to model the full barrel the problem would have been too big to solve.

triangular transition element



Recall that in Cartesian coordinates the strain displacement relationships were

In cylindrical coordinates the strain displacement relationships are

Strain Displacement Relationships – Cylindrical Coordinates

yw

zv

zw

xw

zu

yv

xv

yu

yu

yzz

xzy

xyx

wrz

vzw

rw

zuv

rru

rv

rvu

rru

zz

rz

rrr

1

1

1



For plane strain in the z-direction

The stress and strain matrices take the following form

Any dependence upon z is suppressed for plane strain, and due to symmetry about the z-axis the strains in an axisymmetric component are independent of . Thus all derivatives with respect z and vanish

keeping in mind that w = 0 for plane strain.

0 zrzyzxzz

00

0

zz

rz

rrr

ru

rv

rv

ru

z

r

rr

0000

00000

r

rr



For plane stress in the z-direction

The stress and strain matrices take the following form

Due to symmetry about the z-axis the strains in an axisymmetric component are independent of . Thus all derivatives with respect vanish.

In addition, shear strains associated with the zero shear stresses are similarly zero

0 yzxzzrzz

rrzrrr Eru

rv

rv

ru

00000

r

rr

z

r

rr

0000

00 zrz



In axisymmetric problems radial displacements produce circumferential strains, which in turn generate circumferential stresses on face EBDF of the differential element shown below (this is not a depiction of a finite element).

Line AB becomes longer as it moves from its original position, marked in red, to its new position, marked with a solid black line. This lengthening produces strains in the direction.

y – axis into the pagez – axis out of the page



The strain in the radial direction is defined by the change in length in the radial direction of line segment BD in the previous figure divided by the original length (dr).

Looking at the change in length of line segment AB divided by its original length we would obtain from the previous figure

ru

udrruu

drr

1

ru

rdrddur



If we now look directly at face BDFE in the previous figure we would see the following (the original position of the face is marked with a dashed line)

Focusing on line segment BE, the change in its length divided by its original length would yield the strain in the z-direction, i.e.,

zw

wdzzww

dzz

1



m

m

j

j

i

i

wu

w

uwu

d

Notation – Three Node Triangular Axisymmetric ElementWe formulate the linear displacement function as follows

The nodal displacements are

zaraazrw

zaraazru

654

321

,,



The general displacement function can be expressed in matrix notation as

6

5

4

3

2

1

654

321

10000001

aaaaaa

zrzr

zaraazaraa

wu



To obtain the coefficients we substitute the coordinates of the nodes into the previous equations. This yields

We can solve for the first three coefficients and the last three coefficients from the following two systems of equations;

mmm

jjj

iii

mmm

jjj

iii

zaraaw

zaraawzaraawzaraau

zaraauzaraau

654

654

654

321

321

321

3

2

1

1

11

aaa

zr

zrzr

u

uu

mm

jj

ii

m

j

i

6

5

4

1

11

aaa

zr

zrzr

w

ww

mm

jj

ii

m

j

i



Inverting the first expression leads to

The method of cofactors is used to invert the 3 x 3 matrix, i.e.,

where

m

j

i

mm

jj

ii

u

uu

zr

zrzr

aaa

1

3

2

1

1

11

mji

mji

mji

mm

jj

ii

Azr

zrzr

21

1

11

1

mm

jj

ii

zr

zrzr

A

1

11

2



This determinant is

Note that A is the area of the triangular element. Thus (see appendix in a Calculus book – Cramer’s rule)

Now

jimimjmji zzrzzrzzrA 2

ijmmijjmi

jimimjmji

jijimmimijmjmji

rrrrrr

zzzzzz

rzzrzrrzrzzr

m

j

i

mji

mji

mji

u

uu

Aaaa

21

3

2

1

m

j

i

mji

mji

mji

w

ww

Aaaa

21

6

5

4



Utilizing the matrix expressions from the previous we once again formulate linear displacement function as follows

mmjjii

mmjjii

mmjjii

m

j

i

mji

mji

mji

uuu

uuu

uuu

Azr

u

uu

Azr

aaa

zr

zaraazru

211

211

1

,

3

2

1

321



This leads to

If we identify

then

mmmmjjjj

iiii

uzrA

uzrA

uzrA

zru

21

21

21,

zrA

N

zrA

N

zrA

N

mmmm

jjjj

iiii

21

21

21

mmjjii uNuNuNzru ,



Similarly

mmjjii

mmjjii

mmjjii

m

j

i

mji

mji

mji

vww

vww

www

Azr

w

ww

Azr

aaa

zr

zaraazrw

211

211

1

,

6

5

4

654



and

Once again

thus

mmmmjjjj

iiii

wzrA

wzrA

wzrA

zrw

21

21

21,

zrA

N

zrA

N

zrA

N

mmmm

jjjj

iiii

21

21

21

mmjjii wNwNwNzrw ,



Now

m

m

j

j

i

i

mji

mji

mmjjii

mmjjii

wu

w

uwu

NNN

NNN

wNwNwN

uNuNuN

zrwzru

000

000

,,



Strain Displacement Relationships The strains associated with a two dimensional element are

with

rw

zu

ruzwru

rz

z

r

mm

jj

ii

mmjjii

ur

Nur

Nu

rN

uNuNuNrr

u



this leads to

and

A

zrArr

N

Azr

ArrN

Azr

ArrN

mmmm

m

jjjj

j

iiii

i

221

221

221

mmjjii

mm

jj

ii

uuuA

uA

uA

uAr

u

21

222



For circumferential strain

mm

mm

jj

jj

ii

ii

mmmmjjjj

iiii

ur

zr

urz

r

urz

rA

uzrA

uzrA

uzrA

rru

21

21

21

21

1



Similar derivations lead to

In a matrix format the strain in a two dimensional element takes the following form

mmjjiimmjjii

mmjjii

uuuuuuAr

wzu

uuuAz

w

21

21

m

m

j

j

i

i

mmjjii

mm

mjj

jii

i

mji

mji

rz

z

r

wu

w

uwu

rz

rrz

rrz

rA

000

000

000

21



dB

rz

z

r

Note that the {B} matrix is a function of r and z coordinates. It is not dependent only on the nodal coordinate information. In general, the circumferential strain will not be constant within the element in the radial or z directions. However, the circumferential strain will be constant throughout the element in the circumferential direction.

mmjjii

mm

mjj

jii

i

mji

mji

rz

rrz

rrz

rA

B

000

000

000

21



Clayton Rencis (2000) Paper



One final note. The strain displacement relationships have singularities built into the expressions. As the edge of the element approaches the axis of revolution the (1/r)approaches infinity. How close the edge of the element is to the axis of revolution dictates how much influence this effect has on the numerical calculations.

This can have a pronounced effect on numerical computations for the stiffness matrix. For a three node triangular element the computations for the stiffness matrix have been made at the centroid of the element, which can be very close to the axis of revolution depending on the size of the axisymmetric element.

Clayton and Rencis (2000 and 1998) advocate numerical techniques to circumvent this issue. We will revisit the issue and their approach to the problem when the topic of integration points and quadrature rules are introduced.



Stress Strain Relationships

The constitutive relationship for axisymmtric elements is given by

where

rz

z

r

rz

z

r

D

221000

0101

01

211

ED



Element Stiffness Matrix

The stiffness matrix is determined once again with the following expression:

after integrating along the circumferential boundary. However, the [B] matrix is functionally dependent on r and z. Thus the stiffness matrix [k] is a function of r and z and is a 6x6 matrix. We can evaluate the integration above by one of three methods:

1. Numerical integration (Gaussian quadrature presented in a later section)2. Explicit multiplication and term-by-term multiplication3. Evaluate {B} at the centroid of the element. The centroid of an element is defined

by its coordinates, i.e.,

A

TT

V

TT

dzdrrBDB

dVBDBk

2

33mjimji zzz

zrrr

r



Now define

and as a first approximation take

If the triangular mesh is fine enough then acceptable results can be obtained using this third method. However, most commercial codes will utilize numerical integration, i.e., a quadrature rule.

zrBB ,

BDBArk TT2



In class example



Accounting for Body Forces and Surface Tractions

Gravity loads in the direction of the z-axis or centrifugal loads in the direction of the r-axis dominate axisymmetric analyses. Body forces at the nodes are defined through the expression

where

and

V

Tb dVXNf

b

b

ZR

X

rRb 2



Consider the body forces at node i

where

Thus

where Ni =1 at node i, and the origin of the coordinate axes has been located at the centroid of the element. Note that the body forces, especially the centrifugal body force, is evaluated at the centroid as well.

dzdrrZR

NfV b

bTibi

2

i

iTi N

NN

00

rAZR

fb

bbi

3

2



Similar considerations at nodes j and m lead to

where

The expression for { fb } is a first approximation relative to the radially directed body forces.

3

2 Ar

ZRZRZR

ff

f

fff

f

b

b

b

b

b

b

bmz

bmr

bjz

bjr

biz

bir

b

rRb 2



For surface tractions recall that

For radial and axial pressure loads, which are typical to the internal surface for an axisymmetric problem

The expression at the top can be evaluated at each node. For node j

S

Ts dSTNf

z

r

pp

T

dzr

pp

zrzr

A

dzrpp

NN

dzrpp

Nf

jz

rz

z iii

iii

jz

rz

z i

i

jz

rz

z

Tisj

m

j

m

j

m

j

20

021

20

0

2



Performing the integration at the other two nodes lead to

z

r

z

rjmj

smz

smr

sjz

sjr

siz

sir

s

ppppzzr

ff

f

fff

f

00

22



In class example



Basic Scorecard – Axisymmetric Element

zrA

N

zrA

N

zrA

N

mmmm

jjjj

iiii

21

21

21

jimimjmji zzrzzrzzrA 2

ijmmijjmi

jimimjmji

jijimmimijmjmji

rrrrrr

zzzzzz

rzzrzrrzrzzr



zrAxx

NA

zrAxx

NA

zrAxx

NA

mmmmm

jjjjj

iiiii

21

2

21

2

21

2

mmjjii

mm

mjj

jii

i

mji

mji

rz

rrz

rrz

rA

B

000

000

000

21



FKddKF 1

N

e

ekK1

AtBDBk TT

221000

0101

01

211

ED



m

m

j

j

i

i

mmjjii

mm

mjj

jii

i

mji

mji

rz

z

r

wu

w

uwu

rz

rrz

rrz

rA

000

000

000

21

rz

z

r

rz

z

r

D

m

m

j

j

i

i

wu

w

uwu

d

two dimensional analysis – plane stress and plane strain · two dimensional analysis – plane...

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