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Because of production variations in washer parameters, both the foregoing statements carry cautionary notes. In the series stack, springs of the constant-load type (h/t = 1.41) may actually have a negative rate in some portion of their deflection range. When such a series stack is deflected, some washers will snap through, producing jumps in the load-deflection curve. To avoid this problem, the h/t ratio in a series stack design should not exceed 1.3. In the parallel stack, friction between the washers causes a hysteresis loop in the load-deflection curve (Fig. 6.34). The width of the loop increases with each washer added to the stack but may be reduced by adding lubrication as the washers burnish each other during use. Stacked washers normally require guide pins or sleeves to keep them in proper alignment. These guides should be hardened steel at HRC 48 minimum hardness. Clearance between the washer and the guide pin or sleeve should be about 1.5 percent of the appropriate diameter.
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6.7.3 Tolerances Load tolerances should be specified at test height. For carbon-steel washers with h/t < 0.25, use load tolerance of 15 percent. For washers with h/t > 0.25, use 10 percent. The recommended load tolerance for stainless steel and nonferrous washers is 15 percent. See Table 6.20 for outside- and inside-diameter tolerances.
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FIGURE 6.34 Hysteresis in stacked Belleville washers. (Associated Spring, Barnes Group Inc.)
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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.
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TABLE 6.20 Belleville Washer Diameter Tolerances
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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=
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OD2 (PF) = 0.054 in (1.37 mm) 19.2(107 )(h/t)
TABLE 6.21 Maximum Recommended Stress Levels for Belleville Washers in Static Applications
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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h = 1.41t = 1.41(0.054) = 0.076 in H = h + t = 0.076 + 0.054 = 0.130 in
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
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