Fig. 12. The analysis of the forces acting on the gears has been given on page 28, 4,000 pounds being taken at the pitch line. Using this same value, and choosing a T-shaped arm as a good form for a heavily loaded gear like the present one, let us consider that the rim is stiff enough to distribute the load equally between all the arms, and that each acts as a beam loaded at the end with its proportion of the tooth load. Before we can determine the length of these arms, however, we must fix upon the size of the flange which is to carry the driving bolts. This is taken at 13 inches. It could be smaller if desired, but drawing the bolts in toward the center increases the load on them, and 13 inches seems reasonable until it is proved otherwise. This makes the maximum moment which can come on an arm 4,000X11.5/ 6 = 7,666 inch-pounds.

Gears 10017

Fig. 12.

Now it is evident that the base of the T arm section, which lies in the plane of rotation, is most effective for driving, and that the center leg of the T does not add much to the driving capacity of the arm, although it increases the lateral stiffness of the arm, as well as providing in casting a free flow of metal between the rim and the hub. Hence the simplest way of treating the section of the arm for strength is to consider the base of the T only, of rectangular section, breadth b, and depth A, for which the internal moment of resistance is S x b x h2./6

Also, it is simplest to assume one dimension, say the breadth, and the allowable fibre stress, and figure for the depth. Taking the breadth at 1 1/8 inches, which looks about right, and the fibre stress at 2,500, and equating the external moment to the internal, we have 7,666 = 2,500 X 1.125 X h2 / 6 h2 = (6 X 7,666)/( 2,500 X1,125) = 16.4 h = √ 16.4 = 4.05 (say 4) Drawing in this size, and tapering the arm to the rim as in the case of the pulleys, making the depth of the rim according to the suggested proportions given in Part II, "Gears," giving the center leg of the T a thickness of 7/8 inch tapering to 1 inch, and heavily filleting the arms to the rim and center flange, we have a fairly well proportioned gear.

The next thing to determine is the size of the driving bolts. The circle upon which their centers lie may be 11 inches in diameter, and there will naturally be six bolts, one between each arm. These bolts are in pure shear, and the material of which they are to be made ought to be good for at least 8,000 pounds per square inch fibre stress. The force acting at the circumference of an 11-inch circle would be 4,000 x 18/ 5.5 =13,091 pounds.

Gears 10018

Equating the load on each bolt to the resisting shear gives:

13,091/6 = 8,000 x A = 8,000 x 3.1416 x d2/4.

Let A=area resisting shear. Let A = dia. of bolt.

Then A = π d2 / 4 d2 = 4 X 13,091/6 x 8,000 x 3.1416 = .35 d = √ .35 (say .6) 5/8 -inch bolts would do.

But |-inch bolts are pretty small to use in connection with such heavy machinery. They look out of proportion to the adjacent parts. Hence 7/8-inch bolts have been substituted as being better suited to the place in spite of the fact that theoretically they are larger than necessary. The extra cost is a small matter. These bolts may crush in the flange as well as shear off, but as there is an area of 7/8 X 15/8 = 1.422 square inches to take - 13,091/6 - =2,182 pounds, the pressure per square inch of projected area is only 2,182/1.422 =1,534 pounds, which is very low.

This gear needs no key to the shaft because all the power comes down the arms and passes off to the drum through the bolts, thus putting no torsional stress in the shaft. The face of the flange is counterbored so as to center the gear upon the drum, without relying upon the fit of the gear upon the shaft to do this.

The pinion is solid and needs no discussion for its design.