## R rdr rdp J Vdr r dr r

The stresses oyr, and r^ are related to the stress function 0 in a similar manner to ox and oVy. The resulting values are:

_ 1 3<{> 1 d2<f> "" ~ ~rJr + 7-W

TffO

The derivation of these from the corresponding cartesian coordinate values is a worthwhile exercise for a winter evening.

8.27.4. Forms of stress function in polar coordinates

In cylindrical polars the stress function is, in general, of the form:

where f(r) is a function of r alone and n is an integer.

In exploring the form of cp in polars one can avoid the somewhat tedious polynomial expression used for the cartesian coordinates, by considering the following three cases:

(a) The axi-symmetric case when n — 0 (independent of 9), (j> = f(r). Here the biharmonic eqn. (8.102) reduces to:

1 d<p d2(j) Orr = - —, Ogg — —, Trf = 0 (8.106)

r dr drz

Equation (8.105) has a general solution:

0 = /1 (r) sin 9 or <p — f\(r)cos0. Equation (8.102) has the solution for f \(r) — A\r3 + Bi/r + C\r + D\r\nr (8.108)

i.e. 4> = (Air3 + Bx/r + C\r + DxrIn r)sin9 (or cos9)

(p — f n{r)s\x\n9 or <p = fn(r) cos n9 fn(r) = Anr" +Bnr-n + Cnrn+2 + Dnr~n+2 (8.109)

i.e. 4> = (Anrn +Bnr-" + C„rn+2 + Dnr~"+2) sin n9 (orcos n9)

Other useful solutions are <p = Crsix\9 or 4> = Crcos9 (8.110)

In the above A, B, C and D are constants of integration which enable formulation of the various problems.

As in the case of the cartesian coordinate system these stress functions must satisfy the compatibility relation embodied in the biharmonic equation (8.102). Although the reader is assured that they are satisfactory functions, checking them is always a beneficial exercise.

In those cases when it is not possible to adequately represent the form of the applied loading by a single term, say cos29, then a Fourier series representation using eqn. (8.109) can be used. Details of this are given by Timoshenko and Goodier.^

In the presentation that follows examples of these cases are given. It will be appreciated that the scope of these are by no means exhaustive but a number of worthwhile solutions are given to problems that would otherwise be intractable. Only the stress values are presented for these cases, although the derivation of the displacements is a natural extension.

8.27.5. Case 2 - Axi-symmetric case: solid shaft and thick cylinder radially loaded with uniform pressure

This obvious case will be briefly discussed since the Lamé equations which govern this problem are so well known and do provide a familiar starting point. Substituting eqn. (8.107) into the stress equations (8.106) results in orr = A(1 + 2 In r) + 2B + C/r2

When a solid shaft is loaded on the external surface, the constants A and C must vanish to avoid the singularity condition at r = 0. Hence orr = — 2B. That is uniform tension, or compression over the cross section.

In the case of the thick cylinder, three constants, A, B, and C have to be determined. The constant A is found by examining the form of the tangential displacement v in the cylinder. The expression for this turns out to be a multi-valued expression in 9, thus predicting a different displacement every time 9 is increased to 9 + 2n. That is every time we scan one complete revolution and arrive at the same point again we get a different value for v. To avoid this difficulty we put A — 0. Equations (8.111) are thus identical in form to the Lamé eqns. (10.3 and 10.4).t The two unknown constants are determined from the applied load conditions at the surface.

8.27.6. Case 3 - The pure bending of a rectangular section curved beam

Consider a circular arc curved beam of narrow rectangular cross-section and unit width, bent in the plane of curvature by end couples M (Fig. 8.33). The beam has a constant cross-section and the bending moment is constant along the beam. In view of this one would expect that the stress distribution will be the same on each radial cross-section, that is, it will be independent of 9. The axi-symmetric form of 0, as given in eqn. (8.107), can thus be used:-

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