The strength of a shaft to resist torsion is the cube of its diameter in inches, multiplied by the strength of the material of which the shaft is composed, per square inch of cross-sectional area, giving the strength in statical foot-pounds. The application of this rule is to find the necessary strength of the shaft to convey power irrespective of the distance from its centre at which it delivers such power.
But since the point at which the power to produce torsion is applied is at the rim of the pulley, the amount of torsion produced upon a shaft by a given stress must be obtained by multiplying the given amount of stress by the radius of the pulley in inches and parts of an inch. Example: the static stress upon a pulley, 24 inches diameter, is 100 lbs., what static torsion does it exert upon the shaft?
Here, stress 100 × 12 (radius of the pulley) = 1200 = static torsional stress.
In the following rules for finding the necessary diameters and strengths of shafts, the margin of extra diameter for strength necessary for safety is included, so that the given sizes are working diameters.
To find the necessary diameter of shaft from a given torsional stress.—Rule, divide the torsional stress expressed in statical foot lbs., by 57.2 for steel, by 27.7 for wrought iron, or by 18.5 for cast iron, and the cube root of the quotient is the required working diameter of shaft expressed in inches.
To find the maximum amount of horse-power capable, within good working limits, of being transmitted by a shaft of a given diameter.—Rule, multiply the cube of the diameter of the shaft, in inches, by its revolutions per minute and divide by 92 for steel, by 190 for wrought-iron, or by 285 for cast-iron shafts, and the quotient is the amount of horse-power.
Since, in this rule, the horse-power is a given quantity, the diameter of the pulley is of no consequence, since with a given stress it must have been taken into account in obtaining the horse-power.
To find the revolutions per minute a shaft will require to make to transmit a given amount of horse-power.—Rule, multiply the given amount of horse-power by 92 for steel, by 190 for wrought-iron, or by 285 for cast-iron shafts, and divide the product by the cube of the diameter of the shaft expressed in inches, and the quotient is the required revolutions per minute for the shaft.
The rule adopted by William Sellers and Co. to determine the size of shafts to transmit a given horse-power is:—Rule, divide the cube root of the horse-power by the revolutions per minute and multiply the quotient by 125, the product is the diameter of shaft required.
This gives a shaft strong enough to resist flexure, if the bearings are not too far apart. The distance apart that the bearings should be placed is an important consideration. Modern millwrights differ slightly in opinion in this respect: some construct their mills with beams 9 feet 6 inches apart, and put one hanger under each of the beams; others say 8 feet apart gives a better result. We are clearly of opinion that with 8 feet distance, and shafting lighter in proportion, the best result is obtained.