The centrifugal force developed by the rim and arms tends to pull the arms from the hub. On the belt side, this is balanced to some extent by the belt wrap, which tends to compress the arm and relieve the tension. On the side away from the belt, the centrifugal action has full play, but the arm is usually of such cross-section that the intensity of this stress is very low. It may safely be neglected.

The rim being very thin in most cases, its distributing effect cannot be depended on, hence the driving force of the belt may be taken entirely by the arms immediately under the portion of the belt in contact with the pulley face. For a wrap of 180° this means that only one-half of the pulley arms can be considered as effective in transmitting the turning effort to the hub. Each of these arms is a lever fixed at one end to the hub and loaded at the other. A lever of this description is called a "cantilever" beam, its maximum moment existing at its fixed end. The load that each of these beams may be subjected to is P/N, and therefore the maximum external moment at the hub is 2PL / N. From mechanics we know that the internal moment of resistance of any beam section is SI / c, and that equilibrium of the beam can be satisfied only when the external moment is equal to the internal moment of resistance of the beam section. Equating these two, we have: Fig. 21.

2PL / N = SI / c. (20)

The arms of a pulley are usually of the elliptical or segmental cross-section, and may be of the proportions shown in Fig. 21.

For either of these sections the fraction - is approximately equal to 0.0393/ 13. For convenience (the error caused being on the safe side), L may be taken as equal to the full radius of the pulley R, whence 2PR /N = 2 T,. - T2)R = 0.0393Sh3, (21) in which S may be from 2,000 to 2,250 for cast iron.

Taking moments about the center of the pulley, and solving for P2, the force acting at the circumference of the hub, we hare:

2PR/ N = P2D1 / 2; or p2 = 4PR / N D1 (22

The area of an elliptical section is π times the product of the half axes. With the proportions of Fig. 21, this becomes: a = π X 0.2h X 0.5h = πh2. (23)

Equating the external force to the internal shearing resistance, we have:

4PR / Nd = πh2S3, (24) in which the shearing stress Ss may run from 1,500 to 1,800 for cast iron.

Although both bending and shearing stresses as calculated Above exist at the base of the arms, the bending is, in practically every case, the controlling factor in the design of the arms. An arm-section large enough to resist bending would have a very low intensity of shear.

If the number of arms be increased indefinitely, we come to a continuous arm or web, in which the bending action is eliminated. It may still shear off at the hub, where the area of metal is the least, at minimum circumference. In this case the area under shearing stress is π D1T; and the force at the circumference of the hub, as before, is: p2 = 4PR / ND1.

Equating external force to internal shearing resistance, we have:

4PR / ND1 = π D1 TSs; or, Ss = 4PR /π D1TN. (25)