Finite Element Method - Axisymmetric stress analysis _05

Finite Element Method - Axisymmetric stress analysis _05 The description of the laws of physics for space- and time-dependent problems are usually expressed in terms of partial differential equations (PDEs). For the vast majority of geometries and problems, these PDEs cannot be solved with analytical methods. Instead, an approximation of the equations can be constructed, typically based upon different types of discretizations. These discretization methods approximate the PDEs with numerical model equations, which can be solved using numerical methods. The solution to the numerical model equations are, in turn, an approximation of the real solution to the PDEs. The finite element method (FEM) is used to compute such approximations.

5

Axisymmetric stress analysis

5.1 Introduction

The problem of stress distribution in bodies of revolution (axisymmetric solids) under axisymmetric loading is of considerable practical interest The mathematical prob- lems presented are very similar to those of plane stress and plane strain as, once again, the situation is two dimensional 1,2 By symmetry, the two components of displacements in any plane section of the body along its axis of symmetry define completely the state of strain and, therefore, the state of stress Such a cross-section

is shown in Fig 5.1 If r and z denote respectively the radial and axial coordinates of a

point, with u and u being the corresponding displacements, it can readily be seen that

precisely the same displacement functions as those used in Chapter 4 can be used to

define the displacements within the triangular element i, j , m shown

The volume of material associated with an ‘element’ is now that of a body of revolution indicated in Fig 5.1, and all integrations have to be referred to this The triangular element is again used mainly for illustrative purposes, the principles developed being completely general

In plane stress or strain problems it was shown that internal work was associated with three strain components in the coordinate plane, the stress component normal to this plane not being involved due to zero values of either the stress or the strain

In the axisymmetrical situation any radial displacement automatically induces a strain

in the circumferential direction, and as the stresses in this direction are certainly non-zero, this fourth component of strain and of the associated stress has to be considered Here lies the essential difference in the treatment of the axisymmetric situation

The reader will find the algebra involved in this chapter somewhat more tedious than that in the previous one but, essentially, identical operations are once again

involved, following the general formulation of Chapter 2

5.2 Element characteristics

5.2.1 Displacement function

Using the triangular shape of element (Fig 5.1) with the nodes i, j , m numbered in the

Fig 5.1 Element of an axisymmetric solid

anticlockwise sense, we define the nodal displacement by its two components as

a i = { ; }

and the element displacements by the vector

ai

Obviously, as in Sec 4.2.1, a linear polynomial can be used to define uniquely the

displacements within the element As the algebra involved is identical to that of

Chapter 4 it will not be repeated here The displacement field is now given again by

Eq (4.7):

u = { :} = [ I N , , IN j , INm]ae

with

ai + bir + c,z

2A

N = , etc

and I a two-by-two identity matrix In the above

ai = rjzm - r z m J

bi = zj - Z ,

ci = r, - rj

(5.3)

(5.4)

etc., in cyclic order Once again A is the area of the element triangle

114 Axisyrnmetric stress analysis

5.2.2 Strain (total)

As already mentioned, four components of strain have now to be considered These

are, in fact, all the non-zero strain components possible in an axisymmetric deforma-

tion Figure 5.2 illustrates and defines these strains and the associated stresses

The strain vector defined below lists the strain components involved and defines them in terms of the displacements of a point The expressions involved are almost self-evident and will not be derived here The interested reader can consult a standard elasticity textbook3 for the full derivation We thus have

= s u

Using the displacement functions defined by Eqs (5.3) and (5.4) we have

E = Ba' = [Bi, B,, Bm]ae

in which

B =

Fig 5.2 Strains and stresses involved in the analysis of axisymmetric solids

With the B matrix now involving the coordinates r and z , the strains are no longer

constant within an element as in the plane stress or strain case This strain variation is

due to the E O term If the imposed nodal displacements are such that u is proportional

to r then indeed the strains will all be constant In addition, constant E, and -yrr strains

may be deduced from a linear displacement This is the only state of displacement

coincident with a constant strain condition and it is clear that the displacement

function satisfies the basic criterion of Chapter 2

5.2.3 Initial strain (thermal strain)

In general, four independent components of the initial strain vector can be

envisaged:

Eo = {

Trio

(5.7)

Although this can, in general, be variable within the element, it will be convenient to

take the initial strain as constant there

The most frequently encountered case of initial strain will be that due to thermal

expansion For an isotropic material we shall have then

where 8‘ is the average temperature rise in an element and a is the coefficient of

thermal expansion

The general case of anisotropy need not be considered since axial symmetry would

be impossible to achieve under such circumstances A case of some interest in practice

is that of a ‘stratified’ material, similar to the one discussed in Chapter 4, in which the

plane of isotropy is normal to the axis of symmetry (Fig 5.3) Here, two different

expansion coefficients are possible: one in the axial direction a, and another in the

plane normal to it, ar

Now the initial thermal strain becomes

Practical cases of such ‘stratified’ anisotropy often arise in laminated or fibreglass

construction of machine components

116 Axisymmetric stress analysis

Fig 5.3 Axisymmetrically stratified material

5.2.4 Elasticity matrix

The elasticity matrix D linking the strains E and the stresses (r in the standard form

[Eq (2.511,

(r = { ".) = D(E - E ~ ) + (ro

Trz

needs now to be derived

be simply presented as a special form

The anisotropic 'stratified' material will be considered first, as the isotropic case can

Anisotropic, stratified, material (Fig 5.3)

With the z-axis representing the normal to the planes of stratification we can rewrite Eqs (4.19) (again ignoring the initial strains and stresses for convenience) as

Or " 2 0 2 v100 v2ar a z u20R

& = - & = - - + - - -

(5.10)

vlur " 2 0 2 +2 Ti-,

Ti-z = -

G2

E # = - - - -

Writing again

we have on solving for the stresses, that

2

n(1 - n v q ) , n y ( 1 + VI), n(vl + n v $ ) ,

1 - V I , 2 n y ( 1 + v ) ,

I sym

D,!!?

Isotropic material

For an isotropic material we can obtain the D matrix by taking

E l = E 2 = E or n = l

and

VI = v2 = v

and using the well-known relationship between isotropic elastic constants

_ - _ = m = -

Substituting in Eq (5.1 1) we now have

(5.12)

0

0, ( 1 - 2v)/2

1 - v , v , v ,

E v , 1 - v, v,

D = ( l + v ) ( l - 2 v ) [ 6: v, 1 - v ,

0,

5.2.5 The stiffness matrix

The stiffness matrix of the element ijm can now be computed according to the general

relationship (2.13) Remembering that the volume integral has to be taken over the

whole ring of material we have

with B given by Eq (5.6) and D by either Eq (5.1 1) or Eq (5.12), depending on the

material

The integration cannot now be performed as simply as was the case in the plane

stress problem because the B matrix depends on the coordinates Two possibilities

exist: the first is that of numerical integration and the second of an explicit multiplica-

tion and term-by-term integration

The simplest numerical integration procedure is to evaluate all quantities for a cen-

troidal point

f

3 and ? =

3

r =

118 Axisymmetric stress analysis

In this case we have simply as a first approximation

with A being the triangle area and B the value of the strain-displacement matrix at the

centroidal point

More elaborate numerical integration schemes could be used by evaluating the inte- grand at several points of the triangle Such methods will be discussed in detail in Chapter 9 However, it can be shown that if the numerical integration is of such an order that the volume of the element is exactly determined by it, then in the limit

of subdivision, the solution will converge to the exact a n ~ w e r ~ The ‘one point’ inte- gration suggested here is of this type, as it is well known that the volume of a body of revolution is given exactly by the product of the area and the path swept around by its centroid With the simple triangular element used here a fairly fine subdivision is in any case needed for accuracy and most practical programs use the simple approxima- tion which, surprisingly perhaps, is in fact usually superior to exact integration (see Chapter lo) One reason for this is the occurrence of logarithmic terms in the exact formulation These involve ratios of the type ri/rm and, when the element is at a large distance from the axis, such terms tend to unity and evaluation of the logarithm

is inaccurate

5.2.6 External nodal forces

In the case of the two-dimensional problems of the previous chapter the question of assigning of the external loads was so obvious as not to need further comment In the present case, however, it is important to realize that the nodal forces represent a com- bined effect of the force acting along the whole circumference of the circle forming the element ‘node’ This point was already brought out in the integration of the expres- sions for the stiffness of an element, such integrations being conducted over the whole ring

Thus, if R represents the radial component of force per unit length of the circum-

ference of a node at a radius r, the external ‘force’ which will have to be introduced in the computation is

2rrR

In the axial direction we shall, similarly, have

2 r r Z

to represent the combined effect of axial forces

5.2.7 Nodal forces due to initial strain

Again, by Eq (2.13),

f e = -27r J BTDEor dr dz (5.15)

or noting that is constant,

The integration should be performed in a similar manner to that used in the determi-

nation of the stiffness

It will readily be seen that, again, an approximate expression using a centroidal

value is

Initial stress forces are treated in an identical manner

5.2.8 Distributed body forces

Distributed body forces, such as those due to gravity (if acting along the z-axis), cen-

trifugal force in rotating machine parts, or pore pressure, often occur in axisymmetric

problems

Let such forces be denoted by

per unit volume of material in the directions of r and z respectively By the general

equation (2.13) we have

Using a coordinate shift similar to that of Sec 4.2.7 it is easy to show that the first

approximation, if the body forces are constant, results in

(5.20)

Although this is not exact the error term will be found to decrease with reduction of

element size and, as it is also self-balancing, it will not introduce inaccuracies Indeed,

as will be shown in Chapter 10, the convergence rate is maintained

If the body forces are given by a potential similar to that defined in Sec 4.2.8, Le.,

(5.21) and if this potential is defined linearly by its nodal values, an expression equivalent to

Eq (4.42) can again be determined

In many problems the body forces vary proportionately to r For example in rotat-

ing machinery we have centrifugal forces

(5.22)

b, = w 2 pr where w is the angular velocity and p the density of the material

120 Axisymmetric stress analysis

Fig 5.4 Stresses in a sphere subject to an internal pressure (Poisson’s ratio v = 0.3: (a) triangular mesh - cen- troidal values; (b) triangular mesh - nodal averages; (c) quadrilateral mesh obtained by averaging adjacent triangles

5.2.9 Evaluation of stresses

The stresses now vary throughout the element, as will be appreciated from Eqs (4.5)

and (4.6) It is convenient now to evaluate the average stress at the centroid of the

element The stress matrix resulting from Eqs (5.6) and (2.3) gives there, as usual,

It will be found that a certain amount of oscillation of stress values between

elements occurs and better approximation can be achieved by averaging nodal

stresses or recovery procedures of Chapter 14

5.3 Some illustrative examples

Test problems such as those of a cylinder under constant axial or radial stress give, as

indeed would be expected, solutions which correspond to exact ones This is again an

obvious corollary of the ability of the displacement function to reproduce constant

strain conditions

A problem for which an exact solution is available and in which almost linear stress

gradients occur is that of a sphere subject to internal pressure Figure 5.4(a) shows the

centroidal stresses obtained using rather a coarse mesh, and the stress oscillation

around the exact values should be noted (This oscillation becomes even more pro-

nounced at larger values of Poisson’s ratio although the exact solution is independent

of it.) In Fig 5.4(b) the very much better approximation obtained by averaging the

stresses at nodal points is shown, and in Fig 5.4(c) a further improvement is given

by element averaging The close agreement with the exact solution even for the

very coarse subdivision used here shows the accuracy achievable The displacements

at nodes compared with the exact solution are given in Fig 5.5

- - - - - - -

Fig 5.5 Displacements of internal and external surfaces of sphere under loading of Fig 5.4

122 Axisymmetric stress analysis

Fig 5.6 Sphere subject to steady-state heat flow (100 "C internal temperature, 0°C external temperature):

(a) temperature and stress variation on radial section; (b) 'quadrilateral' averages

Fig 5.7 A reactor pressure vessel (a) 'Quadrilateral' mesh used in analysis; this was generated automatically

by a computer (b) Stresses due to a uniform internal pressure (automatic computer plot) Solution based on quadrilateral averages (Poisson's ratio v = 0.1 5)

In Fig 5.6 thermal stresses in the same sphere are computed for the steady-state

temperature variation shown Again, excellent accuracy is demonstrated by compar-

ison with the exact solution

5.4 Early practical applications

Two examples of practical applications of the programs available for axisymmetrical

stress distribution are given here

5.4.1 A prestressed concrete reactor pressure vessel

Figure 5.7 shows the stress distribution in a relatively simple prototype pressure

vessel Due to symmetry only one-half of the vessel is analysed, the results given

here referring to the components of stress due to internal pressure

In Fig 5.8 contours of equal major principal stresses caused by temperature are

shown The thermal state is due to steady-state heat conduction and was itself

found by the finite element method in a way described in Chapter 7

-

Fig 5.8 A reactor pressure vessel Thermal stresses due to steady-state heat conduction Contours of major

principal stress in pounds per square inch (interior temperature 400 "C, exterior temperature 0 "C,

a = 5 x 10-6/"C.E=2.58x 1061b/in2, v=O.15)