In the case of simple torsion the stress induced in the shaft is a shearing one. The external moment acts about the axis of the shaft, or is a polar moment; hence in the expression for the moment of the internal forces, the polar moment of inertia must be used. Now, from mechanics we have:
T = SI / c; and I /c = d2 /5.1 (for circular section of diameter d); therefore, T = Sd2 / 5.1 (30) from which the diameter for any given twisting moment and fiber stress can readily be found.
For a hollow shaft this expression becomes:
T = S(d04 - d14) /5.1 d0 (31)
The stresses induced in a pin or shaft under simple bending are compression and tension. The external moment in this case is transverse, or about an axis across the shaft; hence the direct moment of inertia is applicable to the equation of forces.
B = SI / c; and I / c = d2 / 10.2 (for circular section of diameter d); therefore, B = Sd2 / 10.2. (32)
For a hollow shaft or pin this expression becomes:
B = S(d04 - d14) / 10.2d0. (33)
Combined Stresses. In the greater number of cases met with in practice, we find two or more simple stresses acting at the same time, and, although the shaft may be strong enough for any one of them alone, it may fail under their combined action. The most common cases are discussed below.
In Fig. 28, the load W produces a tension acting over the whole area of d, due to its direct pull. It also produces a bending action due to the leverage R, which puts the fibers at B in tension and those at the opposite side in compression. It is evident, therefore, that by taking the algebraic sum of the stresses at either side we shall obtain the net stress. It is also evident that the greatest and controlling stress will occur on the side where the stresses add, or on the tension side. Hence, from mechanics:
W = πd2S / 4; or, S = 4W / πd2 (due to direct tension). (34)
Also, WR = Sd3 / 10.2; or, S = 10.2 WR / d3 (due to bending). (35)
Hence the combined tensional stress acting at the point B, or, in fact, at any point on the extreme outside of the vertical shaft toward the force W, is:
S = 4W / πd3 + 10.2 WR / d3 (36)
If Wasted in the opposite direction, the greatest stress would still be at the side B, but would bo a compression instead of a tension, of the same magnitude as before.