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Example. In a clutch, a minimum pressure of 202 lb (900 N) is required. This pressure must be held nearly constant as the clutch facing wears down 0.31 in (7.9 mm). The washer OD is 2.99 in (76 mm). The material washer OD is 2.99 in (76 mm). The material selected for the application is spring steel HRC 47-50. Solution 1. Base the load on a value 10 percent above the minimum load, or 202 + 10 percent = 223 lb (998 N). Assume OD/ID = 2. From Fig. 6.31, select a load-deflection curve which gives approximately constant load between 50 and 100 percent of deflection to flat. Choose the h/t = 1.41 curve. 2. From Fig. 6.31, the load at 50 percent of deflection to flat is 88 percent of the flat load. 3. Flat load is PF = 223/0.88 = 252 lb (1125 N). 4. From Fig. 6.30 [follow line AB from 1125 N to h/t = 1.41 and line BC to approximately 76-mm (2.99-in) OD], the estimated stress is 1500 MPa [218 kilopounds per square inch (kpsi)]. 5. From Table 6.21 maximum stress without set removed is 120 percent of tensile strength. From Fig. 6.3, the tensile strength at HRC 48 will be approximately 239 kpsi (1650 MPa). Yield point without residual stress will be (239 kpsi)(1.20) = 287 kpsi. Therefore 218 kpsi stress is less than the maximum stress of 287 kpsi. 6. Stock thickness is t=

directory. On a terminal, navigate to the directory where you have everything, and issue a command like this:

BASIC STAMP 1 (BS1)

OD2 (PF) = 0.054 in (1.37 mm) 19.2(107 )(h/t)

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.

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8. Refer to Fig. 6.31. The load of 202 lb will be reached at f1 = 50 percent of maximum available deflection. And f1 = 0.50(0.076) = 0.038 in deflection, or the load of 223 lb will be reached at H1 = H f1 = 0.130 0.038 = 0.092 in height at load. To allow for wear, the spring should be preloaded at H2 = H1 f(wear) = 0.092 0.032 = 0.060 in height. This preload corresponds to a deflection f2 = H H2 = 0.130 0.060 = 0.070 in. Then f2/h = 0.070/0.076 = 0.92, or 92 percent of h. 9. Because 92 percent of h exceeds the recommended 85 percent (the loaddeflection curve is not reliable beyond 85 percent deflection when the washer is compressed between flat surfaces), increase the deflection range to 40 to 85 percent. From Fig. 6.31, the load at 40 percent deflection is 78.5 percent, and PF = 223/0.785 = 284 lb. Repeat previous procedures 4, 5, 6, 7, and 8, and find that 100(f2/h) = 81 percent of h. The final design is as follows: Material: AISI 1074 OD = 2.99 in (76 mm) ID = 1.50 in (38 mm) t = 0.055 in (1.40 mm) nominal h = 0.078 in (1.95 mm) nominal Tensile stress ST1 = 29.5 kpsi ( 203 MPa) at f2 = 85 percent of h Tensile stress ST2 = 103 kpsi (710 MPa) at f2 = 85 percent of h

Spring washers are being used increasingly in applications where there is a requirement for miniaturization and compactness of design. They are used to absorb vibrations and both side and end play, to distribute loads, and to control end pressure. Design equations have been developed for determining the spring characteristics of curved, wave, and Belleville washers. There are no special design equations for slotted and finger washers. They are approximated by using Belleville and cantilever equations and then are refined through sampling and testing. 6.8.1 Curved Washers These springs (Fig. 6.35) exert relatively light thrust loads and are often used to absorb axial end play. The designer must provide space for diametral expansion which occurs as the washer is compressed during loading. Bearing surfaces should be hard, since the washer edges tend to dig in. The spring rate is approximately linear up to 80 percent of the available deflection. Beyond that the rate will be much higher than calculated. Load tolerance should not be specified closer than 20 percent. Approximate equations are P= and S= 1.5KP t2 (6.45) 4fEt 3 OD2(K) (6.44)

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