## N Je

cos L0d0

Closure

The stress function concept described above was developed over 100 years ago. Despite this, however, the ideas contained are still of relevance today in providing a series of classical solutions to otherwise intractable problems, particularly in the study of plates and shells.

Examples

Example 8.1

At a point in a material subjected to a three-dimensional stress system the cartesian stress coordinates are:

Determine the normal, shear and resultant stresses on a plane whose normal makes angles of 52° with the X axis and 68° with the Y axis.

Solution

The direction cosines for the plane are as follows:

and, since I2 + m2 + n2 = 1, n2 = 1 - (0.61572 +0.37462)

Now from eqns. (8.13-15) the components of the resultant stress on the plane in the X,Y and Z directions are given by

Pxn = Oxxl + °xym + Oxzn Pyn = Oyy.m + OyXl + (J yZH Pzn = ozzn + azrl + azym

Pxn = (100 x 0.6157) + (40 x 0.3746) + (50 x 0.6935) = 111.2 MN/m2 Pyn = (80 X 0.3746) + (40 x 0.6157) + (-30 x 0.6935) = 33.8 MN/m2 pzn = (150 x 0.6935) + (50 x 0.6157) + (-30 x 0.3746) = 123.6 MN/m2 Therefore from eqn. (8.4) the resultant stress pn is given by

The normal stress an is given by eqn. (8.5),

= (111.2 X 0.6157) + (33.8 x 0.3746) + (123.6 x 0.6935) = 166.8 MN/m2 and the shear stress r„ is found from eqn. (8.6),

Example 8.2

Show how the equation of equilibrium in the radial direction of a cylindrical coordinate system can be reduced to the form

9°yr (QVr - Qee) _ Q 9 r r for use in applications involving long cylinders of thin uniform wall thickness.

Hence show that for such a cylinder of internal radius R0, external radius R and wall thickness T (Fig. 8.42) the radial stress arr at any thickness t is given by

where p is the internal pressure, the external pressure being zero.

Fig. 8.42.

For thin-walled cylinders the circumferential stress Oee can be assumed to be independent of radius.

What will be the equivalent expression for the circumferential stress? Solution

The relevant equation of equilibrium is dffrr 1 dOrf d<Tn (orr - (Tm) ------------\- r r = U

dr rW dz r

Now for long cylinders plane strain conditions may be assumed,

By symmetry, the stress conditions are independent of 9,

w and, in the absence of body forces,

Thus the equilibrium equation reduces to torr + {<*rr ~ aee) _ 0 dr r ~

Since ogg is independent of r this equation can be conveniently rearranged as follows:

dorr

0 0