13. Chapter 13 - Axial Symm _a4

8/11/2019 13. Chapter 13 - Axial Symm _a4

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______________________Basics of the Finite Element Method Applied in Civil Engineering

125

CHAPTER 3

AXIAL-SYMMETRIC STRESS ANALYSIS

The problem of stress distribution in bodies of revolution (or axial-

symmetric solids) under axial-symmetric loading condition has a significant

practical interest in civil engineering. Many structures accomplish such

symmetry properties, due to their shape and load pattern: spherical and

cylinder reservoirs, concrete pressure vessels, foundation piles, soil masses

subjected to circular footing loads, etc.

Because of the symmetry, the two components of displacement, u(or r) and

v (or z), define completely the strain and stress fields of such bodies. The

statement is true for any axial vertical cross section. Thus, the analysis can

be carried out in the 2D space, instead of performing a full 3D analysis.

An example of an axial-symmetric body that can be subjected to axial-

symmetric loads is shown in figure 13.1. The virtual three-dimensional

mesh may be thought as being made of ring-shape finite elements.

Considering the vertical cross section along the axis of symmetry

represented at the right hand side of the figure, the ring shape element

becomes the two-dimensional solid element [1, 2, 3, 4]. The behavior of theplanar element during a 2D analysis should reproduce precisely the results

of a full 3D analysis, regarding the radial and vertical displacements, the

stress field, etc.

13.1 The displacement field

Because of the symmetry, no displacements occur along the direction of

the cylindrical coordinate system. Thus, the displacement vector has only

in-plane components, uand v, along randzaxes respectively.

Nd =

= ),(

),(),(zrvzruzr , with [ ]4411 ... vuvuT = (13.1)


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Chapter 13 Axial-symmetric Stress Analysis_____________________________________

126

Using the isoparametric formulation - the element is quadrilateral, assumedlinear with 4 nodes - the shape functions in normalized coordinates are

)1)(1(4

1iii ttssN ++= (13.2)

and the displacement field has the already known expression

=

4

4

1

1

4321

4321...

0000

0000

),(

),(

v

u

v

u

NNNN

NNNN

tsv

tsu (13.3)

Fig. 13.1 Axisymmetric stress analysis. The 3D problem (a) andthe 2D approach (b)

z

r,

0=

)r(x

z

1 2

3

r,z- are the radial and axial coordinates

u, v-are the corresponding displacements

a. b.

4


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127

13.2The strain components

In common 2Dplane stressorplane strainproblems, either the stress or the

strain along the normal direction to the problems plane are zero. Thus, the

internal mechanical work associated with the third component < zT

z > is not

involved in the functionals expression. In axial-symmetric conditions, any

radial displacement r automatically induces a strain in circumferential

direction , hence the stresses in this direction are certainly non-zero.

Because of symmetry, all the derivatives with respect to vanish. The

displacement along the direction, the shear strains r and z, as well asthe corresponding shear stresses rand zare all zero. The circumferential

strain has the following expression (the reader may consult a standardelasticity text book):

r

u= . (13.4)

Thus, the fourth component is required for both strain and stress vectors*:

=

rz

z

r

,

=

rz

z

r

(13.5)

13.3 The strain - displacement relationship

The fourth term is added to the stain vector components:

r

ur

= ,

z

vz

= ,

r

v

z

urz

+

= andr

u= (13.6)

such that the strain vector becomes

* The x or r coordinates are in this case equivalent


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128

=

=

uz

vr

vz

uru

rrz

z

r

00110

10000

01000

00001

or B= (13.7)

To express the vector in terms of nodal values by using the shape

functions, the transformation from (r,z) to (s,t) coordinates is needed:

=

t

us

u

z

ur

u

1J ;

=

t

vs

v

z

vr

v

1J with

=

t

z

t

rs

z

s

r

J (13.8)

By substituting in :

44

33

22

11 u

s

Nu

s

Nu

s

Nu

s

N

s

u

+

+

+

=

(13.9)

The strain vector is expressed using the derivatives matrix B= B(s,t). This is

a 4 8 matrix, instead of a 3 8 one in the usual 2D plane stress or planestrain problems.

13.4 The stress field assessment and the elasticity matrix

The stress strain relationship E= written in full yields

=

rz

z

r

rz

z

r

E (13.10)


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129

For isotropic materials, the elasticity matrix (4 4) is:

+=

2

21000

01

01

01

)21)(1(

EE (13.11)

For stratified (or oriented) anisotropy, with material properties are 1, E1

along the radial direction and 2, E2along the vertical direction. Using the

notations2

1

E

E

n= and 2EG

m= , the elasticity matrix becomes

+

++

+=

A

nn

nnnn

nn

n

E

...

0)1(

0)()1(

0)1()1(1

)21)(1( 22

2

21

2

2

1212

2

1

2

211

2

E (13.12)

with )21)(1( 2211 nmA += .

13.5 The elemental stiffness matrix and the nodal forces

For the stiffness matrix evaluation the Gauss numerical integration is used:

dVtstseV

T ),(),( EBBk = (13.13)

with dV= r ddz dr= rddet J(s,t) ds dt

Evaluating the integral, the elemental stiffness matrix yields:

==

1

1

1

1

2

0

Jdet),(),(EB),(Bk

dsdt(s,t)dtsrtstsT


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= dsdt(s,t)tsrtstsd T

1

1

1

1

2

0

Jdet),(),(EB),(B

(13.14)

),t(stsrtstsHH jijijijiT

j

n

i

n

j

i JEBBk det),(),(),(21 1

=

= (13.15)

For the nodal body forces and the distributed loads

= VT dVfNr , = d

T

p pNr (13.16)

where dsdttsrdd ),(det J= with the following limits:

- for t= 1 dssrdd )1,(det = J ;- fors= 1 dttrdd ),1(det = J .

Thus, the load vectors yield:

= =

=n

i

n

j

jijiji

T

ji ),t(stsr),t(sHH1 1

det),(2 JfNr (13.17)

==n

i

ii

T

ip ssrNH1

)(det)(2 Jr or ==n

i

ii

T

ip ttrNH1

)(det)(2 Jr (13.18)

The method can be extended to deal with non-symmetrical loading if the

circumferential loading variation is expressed in circular harmonics. In this

case a tangential component w for the displacement vector must be

associated with the angular direction . However, it is usually easier to solve

such a problem as a full 3D model.

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13. Chapter 13 - Axial Symm _a4