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Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright 2004 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.

TABLE 27.7 Coefficients C for Use in Eq. (27.5)

Listing 2-3

where t and r designate the tangential and radial directions, respectively. For the inner member, the stresses at the contact surface are it = p b2 + a2 b2 a2 ir = p (27.12)

A stress-concentration factor may be needed for certain situations. A hub pressfitted to a shaft, for example, would be likely to have an increased pressure at the ends. So if either a brittle fracture or a fatigue failure is a possibility, then for such cases a stress-concentration factor in the range from 1.5 to 2 should be used. Example 2. A 11 2-in solid-steel shaft is fitted to a steel forging having an outside diameter of 21 2 in using a class FN3 fit. Determine the worst-condition stresses for each member.

Summary

Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright 2004 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.

FIGURE 27.3 A press-fitted assembly. Inner member has hole of radius a. Contact surface has radius b. Outer member has outside radius c.

CONTROL.BS2 program (Continued).

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Solution. The worst condition would occur when the hole is minimum and the shaft is maximum. From Table 27.7, we find C = 0 and C = +4.310 for the lower limit of the hole and upper limit of the shaft, respectively. Using Eq. (27.5), we find L = CD1/3 = +4.310(1.5)1/3 = 0.0049 in 1000

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Therefore, the maximum shaft has a diameter di = 1.5 + 0.0049 = 1.5049 in. Similarly, the minimum hole is do = 1.5000 in. The radial interference is = 0.5(0.0049) = 0.00245 in. For use in Eq. (27.10), we observe that b = 0.75 in and c = 1.25 in based on the nominal dimensions. Using E = 30 Mpsi, we find the contact pressure to be p= = E 2 (c b2 ) bc2 30(10)6(0.00245) [(1.25)2 (0.75)2] 0.75(1.25)2

= 62.7 kpsi Using Eq. (27.11) to get the stresses in the outer member gives ot = p c2 + b2 (1.25)2 + (0.75)2 = 133.2 kpsi 2 2 = 62.7 c b (1.25)2 (0.75)2

CONTENTS AT A GLANCE Dallas DS1302 Real-Time Clock Chip Dallas DS1620 Digital Thermometer/Thermostat Linear Technologies LTC1298 12-Bit Analog-to-Digital Converter Xicor X25640 8-kbyte EEPROM K1EL IBM AT Keyboard to RS-232 Serial Converter SONY Infrared Decoder Chip SVID Single-Chip Video Controller Multichannel Remote Serial I/O System SN7516 Differential Bus Transceiver HA7S ASCII to 1-Wire Host Adapter Eight-Channel Infrared Remote Control System Addressable Serial-to-Parallel Converter Addressable Parallel-to-Serial Encoder Addressable A/D and Digital Input/Output Control Device Fifty-six-Channel Radio Remote Control System Thirty-two-Channel Serial Input Scanner Alarm System

Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright 2004 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website.

For the inner member, the worst stress is given by it = and the result is ir = it = 62.7 kpsi pb2 = p = 62.7 kpsi b2 a2

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