When a long rod of uniform cross-section is stretched by means of a suspended weight, a practically uniform tensile stress will act on horizontal surfaces, and vertical surfaces will be free from stress. The material is then said to be subjected to "simple longi tudinal stress." When a short pillar or block is compressed by opposite forces applied at its ends, we have, to a somewhat less close approximation, a state of simple longitudinal compressive stress.
We may imagine that a very small rectangular block of mate , rial, containing the point in question, has its faces parallel to the three principal planes. The theorem states that only normal stresses will act upon those faces, as shown in case D of fig. 1, which accordingly represents the most general state of compound stress that can occur. The three normal stresses pi, p2, P3 will in general all be different, and one or more may assume negative values (representing compressive stresses, as explained on p. 51).
gravitation or electrical attractions), p the density of the ma terial, and the acceleration in the direction Ox, at the point (x, y, z).
Fig. 2 shows a rectangular block of dimensions bx, by, I, in the directions of Ox, 0y, Oz respectively. The point A has co-ordinates x, y, and the stress on the face AD has a normal component which we denote (as above) by similarly, the stress on the face AB has a normal component which we denote by Y,. The tangen tial stress on AB has a component in the direction of Ox which we denote (as above) by and similarly, the tangential stress on Mohr's Circle Diagram for Compound Stress.—II. If, in fig. 3, the rectangular faces are principal planes of stress, X, will be zero. Writing and P2, in (4), for Y, and respectively, we have for this case: It is clear that p and q will be given, in terms of p2, by a circular diagram constructed as shown in fig. 4. If CA is drawn, at an angle 20 to OCN, to meet the circle at A, the co-ordinates ON, AN of A will represent p and q respectively.
Again, it is clear that the stress on AB (fig. 3) will not be af fected by the addition of a third principal stress acting on the triangular faces of the prism; so, in the general case, the circle BAX still gives the stresses on planes which are to the direction of In the same way, if OE (fig. 4) represents points on a circle having EB as diameter will relate the normal and tangential components of stress for all planes parallel to and points on a circle having EX as diameter will relate these components for all planes parallel to P2.
On planes which are inclined to all three of the principal planes, the stresses, in the general case, will depend upon all three of P2, p3. But it may be proved that points taken, in fig. 4, to relate the normal and tangential components of stress on such planes will in all cases lie within the shaded area of the diagram. Thus, if the normal stress on a plane is specified by ON, the intensity of the tangential stress lies between limits given by AN and A'N.
Case of Two Equal and Opposite Principal If, in equations (6), we make p, equal and opposite to the stress on planes which are equally inclined to the principal planes of stress (20 =9o°) will be given by p=0, q= pi. (7) Hence we see that a state of stress represented by equal and opposite principal stresses of intensity p and -p is equivalent to a state of simple shearing stress, of intensity p, on planes inclined at 45° to the principal planes of stress.