In Fig. 29, V might be the end load on a vertical shaft; and the two forces W might act in conjunction with it as in the case of Fig. 26, at the radius R. This case is not very often met with. It is usually possible to combine the moments, find an equivalent moment of a simple kind, and use the corresponding simple fiber stress. In the case in question we have a direct stress to be combined with a shearing stress, and mechanics gives us the following solution:

Fig. 30.

Let Ss = simple shearing stress (lbs. per sq in.). Let Sc = simple compressive stress (lbs. per sq. in.). Let Srs-- resultant shearing stress (lbs. per sq. in.). Let Src - resultant compressive stress (lb3. per sq. in.). We then have :

2WR = Snd3 /5.1; or, Ss = 5.1(2WR) / d3 (37)

Also, V = πd2Se / 4 ; pression, and shearing. For the material of which shafts are usually made, this is near enough to the truth to give safe and practical results. Using the expressions for internal moments of resistance as previously noted for circular sections, we then have:

Be = Sd3 / 10.2. (43)

Also, Te = Sd3 / 5.1. (44)

Either equation may be used ; the diameter d will result the same whichever equation is taken. For the sake of simplicity, equation 42 is generally preferred, equation 44 being taken in conjunction with it.

The expression √B2 + T2 is one that would be a long and tedious task to calculate. By inspection it is readily seen that this quantity can be graphically represented by means of a right-angled triangle having B and T as the sides. "We may then lay down on a piece of paper, to some convenient scale, the moments B and T as the sides of a right-angled triangle, when, upon measuring the hypothenuse, we can easily read off to the same scale √B2 + T2. Even if the drawing is made to a small scale, the accuracy of the reading will be sufficient to enable the value for d to be solved very closely. This graphical method is illustrated in Part I.