Allowance for Forced Fits.—The allowance per inch of diameter usually ranges from 0.001 inch to 0.0025 inch, 0.0015 being a fair average. Ordinarily, the allowance per inch decreases as the diameter increases; thus the total allowance for a diameter of 2 inches might be 0.004 inch, whereas for a diameter of 8 inches the total allowance might not be over 0.009 or 0.010 inch. In some shops the allowance is made practically the same for all diameters, the increased surface area of the larger sizes giving sufficient increase in pressure. The parts to be assembled by forced fits are usually made cylindrical, although sometimes they are slightly tapered. The advantages of the taper form are that the possibility of abrasion of the fitted surfaces is reduced; that less pressure is required in assembling; and that the parts are more readily separated when renewal is required. On the other hand, the taper fit is less reliable, because if it loosens, the entire fit is free with but little axial movement. Some lubricant, such as white lead and lard oil mixed to the consistency of paint, should be applied to the pin and bore before assembling, to reduce the tendency of abrasion.
Pressure for Forced Fits.—The pressure required for assembling cylindrical parts depends not only upon the allowance for the fit, but also upon the area of the fitted surfaces, the pressure increasing in proportion to the distance that the inner member is forced in. The approximate ultimate pressure in pounds can be determined by the use of the following formula in conjunction with the accompanying table of “[Pressure Factors].”
Pressure Factors
| Diameter, Inches | Pressure Factor | Diameter, Inches | Pressure Factor | Diameter, Inches | Pressure Factor | Diameter, Inches | Pressure Factor | Diameter, Inches | Pressure Factor |
| 1 | 500 | 31/2 | 132 | 6 | 75 | 9 | 48.7 | 14 | 30.5 |
| 11/4 | 395 | 33/4 | 123 | 61/4 | 72 | 91/2 | 46.0 | 141/2 | 29.4 |
| 11/2 | 325 | 4 | 115 | 61/2 | 69 | 10 | 43.5 | 15 | 28.3 |
| 13/4 | 276 | 41/4 | 108 | 63/4 | 66 | 101/2 | 41.3 | 151/2 | 27.4 |
| 2 | 240 | 41/2 | 101 | 7 | 64 | 11 | 39.3 | 16 | 26.5 |
| 21/4 | 212 | 43/4 | 96 | 71/4 | 61 | 111/2 | 37.5 | 161/2 | 25.6 |
| 21/2 | 189 | 5 | 91 | 71/2 | 59 | 12 | 35.9 | 17 | 24.8 |
| 23/4 | 171 | 51/4 | 86 | 73/4 | 57 | 121/2 | 34.4 | 171/2 | 24.1 |
| 3 | 156 | 51/2 | 82 | 8 | 55 | 13 | 33.0 | 18 | 23.4 |
| 31/4 | 143 | 53/4 | 78 | 81/2 | 52 | 131/2 | 31.7 | .... | .... |
Assuming that A = area of fitted surface; a = total allowance in inches; P = ultimate pressure required, in tons; F = pressure factor based upon assumption that the diameter of the hub is twice the diameter of the bore, that the shaft is of machine steel, and the hub of cast iron, then,
| A × a × F | ||
| P | = | ————— |
| 2 |
Example:—What will be the approximate pressure required for forcing a 4-inch machine steel shaft having an allowance of 0.0085 inch into a cast-iron hub 6 inches long?