A = 4 × 3.1416 × 6 = 75.39 square inches;
F, for a diameter of 4 inches, = 115 (see table of “[Pressure Factors]”). Then,
P = (75.39 × 0.0085 × 115)/2 = 37 tons, approximately.
Allowance for Given Pressure.—By transposing the preceding formula, the approximate allowance for a required ultimate tonnage can be determined. Thus, a = 2P ÷ AF. The average ultimate pressure in tons commonly used ranges from 7 to 10 times the diameter in inches. Assuming that the diameter of a machine steel shaft is 4 inches and an ultimate pressure of about 30 tons is desired for forcing it into a cast-iron hub having a length of 51/2 inches, what should be the allowance?
A = 4 × 3.1416 × 51/2 = 69 square inches,
F, for a diameter of 4 inches, = 115. Then,
| 2 × 30 | ||||
| a | = | ————— | = | 0.0075 inch. |
| 69 × 115 |
Shrinkage Fits.—When heat is applied to a piece of metal, such as iron or steel, as is commonly known, a certain amount of expansion takes place which increases as the temperature is increased, and also varies somewhat with different kinds of metal, copper and brass expanding more for a given increase in temperature than iron and steel. When any part which has been expanded by the application of heat is cooled, it contracts and resumes its original size. This expansive property of metals has been taken advantage of by mechanics in assembling various machine details. A cylindrical part which is to be held in position by a shrinkage fit is first turned a few thousandths of an inch larger than the hole; the diameter of the latter is then increased by heating, and after the part is inserted, the heated outer member is cooled, causing it to grip the pin or shaft with tremendous pressure.