Z

zx dx

zy dy

zz dz

+FZ =0

The above equations have been derived by consideration of equilibrium of forces only, and this does not represent a complete check on the equilibrium of the system. This can only be achieved by an additional consideration of the moments of the forces which must also be in balance.

Consider, therefore, the element shown in Fig. 8.17 which, again for simplicity, shows only the stresses which produce moments about the Y axis. For convenience the origin of the cartesian coordinates has in this case been chosen to coincide with the centroid of the element. In this way the direct stress and body-force stress terms will be eliminated since the forces produced by these will have no moment about axes through the centroid.

It has been assumed that shear stresses rxy, ryz and rxz act on the coordinate planes passing through G so that they will each increase and decrease on either side of these planes as described above.

Thus, for equilibrium of moments about the Y axis, a

dz 2

d dxdy dz

dx 2

Dividing through by (dx dy dz) and simplifying, this reduces to

. ,1a dx. ! (r,v + I— r,v -7T-) 1 v « |dX zx 2' | |

Fig. 8.17. Element showing only stresses which contribute to a moment about the ¥ axis.

Similarly, by consideration of the equilibrium of moments about the X and Z axes, rzy ~ Tyz *xy = *yx

Thus the shears and complementary shears on adjacent faces are equal as in the simple two-dimensional case. The nine cartesian stress components thus reduce to six independent values,

(b) In cylindrical coordinates

The equations of equilibrium derived above in cartesian components are very useful for components and stress systems which can easily be referred to a set of three mutually perpendicular axes. There are many cases, however, e.g. those components with axial symmetry, where other coordinate axes prove far more convenient. One such set of axes is the cylindrical coordinate system with variables r, 6 and z as shown in Fig. 8.18.

Consider, therefore, the equilibrium in a radial direction of the element shown in Fig. 8.19(a). Again, for simplicity, only those stresses which produce force components in this direction are indicated. It must be observed, however, that in this case the terms will also produce components in the radial direction as shown by Fig. 8.19(b). The body-force stress components are denoted by Fr, Fz and Fg.

Therefore, resolving forces radially, on- + ~((Trr)dr or

drdz cos d6

Fig. 8.18. Cylindrical coordinates.

Fig. 8.19. (a) Element showing stresses which contribute to equilibrium in the radial and circumferential directions, (b) Radial components of hoop stresses.

With COS

1 and sin de de

-, this equation reduces to

Similarly, in the 6 direction, the relevant equilibrium equation reduces to 3 1 #o<M) 3 2°re n and in the Z direction (Fig. 8.20)

These are, then, the stress equations of equilibrium in cylindrical coordinates and in their most general form. Clearly these are difficult to memorise and, fortunately, very few problems arise in which the equations in this form are required. In many cases axial symmetry exists and circular sections remain concentric and circular throughout loading, i.e. Ore = 0.

Thus for axial symmetry the equations reduce to or dz r

r 86 dz

Further simplification applies in cases where the coordinate axes can be selected to coincide with principal stress directions as in the case of thick cylinders subjected to uniform pressure or thermal gradients. In such cases there will be no shear, and in the absence of body forces the equations reduce to the relatively simple forms

8.13. Principal stresses in a three-dimensional cartesian stress system

As an alternative to the graphical Mohr's circle procedures the principal stresses in three-dimensional complex stress systems can be determined analytically as follows.

The equations for the state of stress at a point derived in §8.3 may be combined to give the equation a\ - ("xx + Oyv + O-z) a2n + («xfOvv + OyyOzz + OxxOz7 - x]y - T2, - X^) On

- (oxxoyyGzz - oxxx\ - oyyx2a - ozzx2y + 2xxyxyzxxz) = 0 (8.42)

With a knowledge of the cartesian stress components this cubic equation can be solved for o„ to produce the three principal stress values required. A general procedure for the solution of cubic equations is given below.

8.13.1. Solution of cubic equations Consider the cubic equation jc3 + ax2 + bx + c = 0 (1)

ab 2 a3

we obtain the modified equation y* + py + « = 0 (5)

rl rJ

Now consider the standard trigonometric identity cos 30 = 4 cos3 0 — 3 cos 9 (8)

Rearranging and substituting z = cos 0, (9)

0 0

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