Axisymmetric stress analysis

Transcription

1 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 problems 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 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 Element characteristics Displacement function Using the triangular shape of element (Fig. 5.1) with the nodes i, j, m numbered in the

2 Element characteristics 1 13 Fig. 5.1 Element of an axisymmetric solid. anticlockwise sense, we define the nodal displacement by its two components as ai={;}, and the element displacements by the vector { ai ae = aj am (5.1) Obviously, as in Sec , 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 = { :} = [IN,, INj, INm]ae (5.3) with ai + bir + c,z N. =, etc. 2A and I a two-by-two identity matrix. In the above ai = rjzm - r z. mj bi = zj - Z, ci = r, - rj etc., in cyclic order. Once again A is the area of the element triangle. (5.4)

3 114 Axisyrnmetric stress analysis 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 deformation. 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 = su 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.

4 Element characteristics 1 15 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 EO 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 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.

5 116 Axisymmetric stress analysis Fig. 5.3 Axisymmetrically stratified material 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. The anisotropic 'stratified' material will be considered first, as the isotropic case can be simply presented as a special form. 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 "202 v100 &= v2ar az u20r r El E2 El E2 E2 E2 E#=---- vlur " Ti-, Ti-z = - El E2 El G2 (5.10)

6 Element characteristics 1 17 Writing again we have on solving for the stresses, that I 2 n(1 - nvq), ny(1 + VI), n(vl +nv$), VI, D,!!? ny(1 +v), d n(1 -nv;,, 0 sym. Isotropic material For an isotropic material we can obtain the D matrix by taking and El=E2=E or n=l VI = v2 = v and using the well-known relationship between isotropic elastic constants Substituting in Eq. (5.1 1) we now have _ G2 -_ =m=- 1 E2 E 2( 1 + v) 1 - v, v, v, E v, 1 - v, v, D= (l+v)(l-2v)[ 6: v, 1-v, 0, 0 0, (1-2v)/2 (5.11) (5.12) 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 f KG = 27r BTDBjr dr dz (5.13) with B given by Eq. (5.6) and D by either Eq. (5.11) 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 multiplication and term-by-term integration. The simplest numerical integration procedure is to evaluate all quantities for a centroidal point r= ri +rj +r, zi + zj + z, 3 and?= 3

7 118 Axisymmetric stress analysis In this case we have simply as a first approximation K; = 27rBTDB,TA (5.14) 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 integrand 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 an~wer.~ The one point integration 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 approximation 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 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 combined 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 expressions 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 circumference 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 2rrZ to represent the combined effect of axial forces Nodal forces due to initial strain Again, by Eq. (2.13), fe = -27r BTDEor dr dz (5.15) J

8 Element characteristics 1 19 or noting that is constant, fp = -2II( [BTrdrdz)D&o (5.16) The integration should be performed in a similar manner to that used in the determination 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. fp = -27rBTD&oFA (5.17) Distributed body forces Distributed body forces, such as those due to gravity (if acting along the z-axis), centrifugal force in rotating machine parts, or pore pressure, often occur in axisymmetric problems. Let such forces be denoted by b= { i:} (5.18) per unit volume of material in the directions of r and z respectively. By the general equation (2.13) we have fp = -2n[IN,( ::}rdrdr (5.19) Using a coordinate shift similar to that of Sec 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 , 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 rotating machinery we have centrifugal forces b, = w 2 pr where w is the angular velocity and p the density of the material. (5.22)

9 120 Axisymmetric stress analysis Fig. 5.4 Stresses in a sphere subject to an internal pressure (Poisson s ratio v = 0.3: (a) triangular mesh - centroidal values; (b) triangular mesh - nodal averages; (c) quadrilateral mesh obtained by averaging adjacent triangles.

10 Some illustrative examples 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, 8 = DBae - DE^ + oo (5.23) 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 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 pronounced 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 Fig. 5.5 Displacements of internal and external surfaces of sphere under loading of Fig. 5.4.

11 122 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).

12 Early practical applications 123 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 comparison 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 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=5x 10-6/"C.E=2.58x 1061b/in2, v=o.15). -

13 124 Axisymmetric stress analysis Foundation pile Figure 5.9 shows the stress distribution around a foundation pile penetrating two different strata. This non-homogeneous problem presents no difficulties and is treated by the standard approach given in this chapter in which the quadrilateral elements shown are assemblies of two triangles and the results are averaged. 5.5 Non-symmetrical loading The method described in the present chapter can be extended to deal with nonsymmetrical loading. If the circumferential loading variation is expressed in circular harmonics then it is still possible to focus attention on one axial section although the nodal degrees of freedom are now increased to three. Details of this process are described in references 5 and 6 and in Chapter 9 of Volume Axisymmetry - plane strain and plane stress In the previous chapter we noted that plane stress and strain analysis was done in terms of three stress and strain components and, indeed, both cases could be generally incorporated in a single program with an indicator changing appropriate constants in the matrix D. Doing this loses track of the a, component in the plane strain case which has to be separately evaluated. Further, special expressions [viz. Eq had to be used to introduce initial strains. This is inconvenient (especially when non-linear constitutive laws are used), and an alternative of writing the plane strain case in terms of four stress-strain components as a special case of axisymmetric analysis is highly recommended. If the axisymmetric strain definition of Eq. (5.5) is examined, we note that r = 03 gives E~ 0 and plane strain conditions are obtained. Thus, if we ignore the terms in B associated with E ~ replace, the coordinates randz by xandy and further change the volume of integration 27rr to 1 the plane strain formulation becomes available from the axisymmetric plane strain directly. Plane stress conditions can similarly be incorporated, requiring in addition substitution of the axisymmetric D matrix by Eqs (4.13) or (4.19) augmented by an appropriate zero row and column. Thus, at the cost of additional storage of the fourth stress and strain component, all the cases discussed can be incorporated in a single format.

14 Axisymmetry - plane strain and plane stress 125 Fig. 5.9 (a) A pile in stratified soil. Irregular mesh and data for the problem. (b) A pile in stratified soil. Plot of vertical stresses on horizontal sections. Solution also plotted for Boussinesq problem obtained by making E, = E2 = EP,l,, and this is compared with exact values.

15 126 Axisymmetric stress analysis References 1. R.W. Clough. Chapter 7 of Stress Analysis (eds O.C. Zienkiewicz and G.S. Holister), Wiley, R.W. Clough and Y.R. Rashid. Finite element analysis of axi-symmetric solids. Proc. ASCE, 91, EM.l, 71, S. Timoshenko and J.N. Goodier. Theory of Elasticity. 2nd ed., McGraw-Hill, B.M. Irons. Comment on Stiffness matrices for section element by I.R. Raju and A.K. Rao. JAZAA, 7, 156-7, E.L. Wilson. Structural analysis of axisymmetric solids. JAZAA, 3, , O.C. Zienkiewicz. The Finite Element Method. 3rd ed., McGraw-Hill, 1977.