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Fig. 8.2.

Fig. 8.3. The state of stress on an inclined plane through any given point in a three-dimensional cartesian stress system.

Fig. 8.3. The state of stress on an inclined plane through any given point in a three-dimensional cartesian stress system.

ABC and YZ are given by the angle between the normals to both planes n and x, etc.) For convenience, let the plane ABC initially be some perpendicular distance h from Q so that the cartesian stress components actually acting at Q can be shown on the sides of the tetrahdedron element ABCQ so formed (Fig. 8.3). In the derivation below the value of h will be reduced to zero so that the equations obtained will relate to the condition when ABC passes through Q.

In addition to the cartesian components, the unknown components of the stress on the plane ABC, i.e. pxn, p■„, and pzn, are also indicated, as are the body-force field stress components which act at the centre of gravity of the tetrahedron. (To improve clarity of the diagram they are shown displaced from the element.)

Since body-force stresses are defined as forces/unit volume, the components in the X, Y and Z directions are of the form

where AS7i/3 is the volume of the tetrahedron. If the area of the surface ABC, i.e. AS, is assumed small then all stresses can be taken to be uniform and the component of force in the X direction due to axx is given by a„A5 cos nx

Stress components in the other axial directions will be similar in form. Thus, for equilibrium of forces in the X direction, h px„A.S + FXAS- = axxAScosnx + txy AS cos ny + rxz AS cos nz

As h —> 0 (i.e. plane ABC passes through Q), the second term above becomes very small and can be neglected. The above equation then reduces to pxn = axx cos nx + xxy cos ny + txz cos nz (8.1)

Similarly, for equilibrium of forces in the y and z directions, pyn — Oyy cos ny + ryx cos nx + tyz cos nz 1 (8.2)

The resultant stress p„ on the plane ABC is then given by

Pn = yJ(p2x„+Pyn+P2Zn) (8-4) The normal stress cr„ is given by resolution perpendicular to the face ABC, i.e. tr„ = pxn cos nx + pyn cos ny + pzn cos nz (8.5)

and, by Pythagoras' theorem (Fig. 8.4), the shear stress r„ is given by

Fig. 8.4. Normal, shear and resultant stresses on the plane ABC.

It is often convenient and quicker to define the line of action of the resultant stress pn by the direction cosines

I' = COS (p„x) = pxn/pn m = COS (pny) = Pyn/Pn n - COS(p„z) = Pzn/Pn

The direction of the plane ABC being given by other direction cosines

/=cosrtx, m = cos ny, n = cos ny It can be shown by simple geometry that

I2 + m2 + n2 = 1 and (I')2 + (mf + (n')2 = 1 Equations (8.1), (8.2) and (8.3) may now be written in two alternative ways.

(a) Using the common symbol a for stress and relying on the double suffix notation to discriminate between shear and direct stresses:

pyn — oyx cos nx + Oyy cos ny + ayz cos nz (8.11)

In each of the above equations the first suffix is common throughout, the second suffix on the right-hand-side terms are in the order x, y, z throughout, and in each case the cosine term relates to the second suffix. These points should aid memorisation of the equations.

(b) Using the direction cosine form :

Memory is again aided by the notes above, but in this case it is the direction cosines, I, m and n which relate to the appropriate second suffices x, y and z.

Thus, provided that the direction cosines of a plane are known, together with the cartesian stress components at some point Q on the plane, the direct, normal and shear stresses on the plane at Q may be determined using, firstly, eqns. (8.13-15) and, subsequently, eqns. (8.4-6).

Alternatively the procedure may be carried out graphically as will be shown in §8.9.

8.4. Direct, shear and resultant stresses on an oblique plane

Consider again the oblique plane ABC having direction cosines /, m and n, i.e. these are the cosines of the angle between the normal to plane and the x, y, z directions.

In general, the resultant stress on the plane p„ will not be normal to the plane and it can therefore be resolved into two alternative sets of components.

(a) In the co-ordinate directions giving components pxn, pyn and pzn, as shown in Fig. 8.5, with values given by eqns. (8.13), (8.14) and (8.15).

(b) Normal and tangential to the plane as shown in Fig. 8.6, giving components, of an (normal or direct stress) and r„ (shear stress) with values given by eqns. (8.5) and (8.6).

The value of the resultant stress can thus be obtained from either of the following equations:

Fig. 8.6. Normal and tangential components of resultant stress on an inclined plane.

these being alternative forms of eqns. (8.6) and (8.4) respectively. From eqn. (8.5) the normal stress on the plane is given by:

Substituting into eqn (8.5) and using the relationships axv = ayx\ axz — a:x and avz — crzy which will be proved in §8.12

<rn = crxx • I2 + ayy • m2 + • n1 + 2axy • Im + 2ayz • ran + 2axz 'In. (8.18)

and from eqn. (8.6) the shear stress on the plane will be given by

In the particular case where plane ABC is a principal plane (i.e. no shear stress):

c.vy = axz = Oyz = 0 and nxx = (71, oyy — (72 and ozz = 03

the above equations reduce to:

tr„ = orj ■ I2 + a2 • m2 + <x3 • n2 (8.20)

and since pxn = o\ • I pyn = a2 • m and Pzn = 03 • n

8.4.1. Line of action of resultant stress

As stated above, the resultant stress pn is generally not normal to the plane ABC but inclined to the x, y and z axes at angles 6X, 6V and 0Z - see Fig. 8.7.

Fig. 8.7. Line of action of resultant stress.

The components of p„ in the x, y and z directions are then

and the direction cosines which define the line of actions of the resultant stress are

0 0

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