Machinery's Handbook, 31st Edition
644
Shrinkage Fits
When both hub and shaft are of steel:
(2a) (2b) If the shaft is solid, the factor C is taken from Table 1; if it is hollow and the hub is of steel, factor C is taken from Table 2; if it is hollow and the hub is of cast iron, the factor is taken from Table 3. Table 1. Factors for Calculating Shrinkage Fit Allowances for Steel Shafts and Steel or Cast-Iron Hubs Ratio of Diameters , , A T C 30 000 000 1 US = + ^ h . T C 25 4 206 843 + ^ h . A × 10 1 metric 9 =
Ratio of Diameters D D 1 2
Steel Hub
Cast-iron Hub
Steel Hub
Cast-iron Hub
D D
1 2
C
C
1.5 1.6 1.8 2.0 2.2 2.4 2.6
0.227 0.255 0.299 0.333 0.359 0.380 0.397
0.234 0.263 0.311 0.348 0.377 0.399 0.417
2.8 3.0 3.2 3.4 3.6 3.8 4.0
0.410 0.421 0.430 0.438 0.444 0.450 0.455
0.432 0.444 0.455 0.463 0.471 0.477
0.482 Values of factor C for solid steel shafts of nominal diameter D 1 , and hubs of steel or cast iron of nominal external and internal diameters D 2 and D 1 , respectively. Example 1 : A steel crank web 375 mm outside diameter is to be shrunk on a 250 mm solid steel shaft. Required is the allowance per mm of shaft diameter to produce a maximum ten sile stress in the crank of 170 MPa, assuming the stresses in the crank to be equivalent to those in a ring of the diameter given. The ratio of the external to the internal diameters equals 375 250 = 1.5; T = 170 MPa; from Table 1, C = 0.227. Substituting in Formula (2b) : . . . A 254 206843 170 1 0 227 0 026 mm # = + = ^ h Example 2 : Find the allowance per mm of diameter for a 250 mm shaft having a 125 mm axial through hole, other conditions being the same as in Example 1. The ratio of external to internal diameters of the hub equals 375 4 250 = 1.5, as before, and the ratio of external to internal diameters of the shaft equals 250 4 125 = 2. From Table 2, we find that factor C = 0.455; T = 170 MPa. Substituting these values in Formula (2b) : . . . A 254 206843 = 170 1+0455 = 0030mm ^ h The allowance is increased, as compared with Example 1, because the hollow shaft is more compressible. Example 3 : If the crank web in Example 1 is of cast iron and 28 MPa is the maximum ten sile stress in the hub, what is the allowance per mm of diameter? 1 2 = 1.5 T = 28 In Table 1 , we find that C = 0.234. Substituting in Formula (1b), for cast-iron hubs, A = 0.0076 mm, which, owing to the lower tensile strength of cast iron, is about one-third the shrinkage allowance in Example 1, although the stress is two-thirds of the elastic limit. D D
Copyright 2020, Industrial Press, Inc.
ebooks.industrialpress.com
Made with FlippingBook - Share PDF online