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GEARING
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TABLE 12.1 Efficiency of Worm-Gear Sets for = 0.05
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The pitch diameter of the gear is D = NG /P = 30/6 = 5 in. The center distance is thus C= The lead is L = px NW = 0.5236(2) = 1.0472 in From Eq. (12.3), = tan 1 L 1.0472 = tan 1 = 9.46 d 2 D+d 2+5 = = 3.5 in 2 2
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The pitch line velocity of the worm, in inches per minute, is VW = dnW = (2)(1200) = 7540 in/min The speed of the gear is nG = 1200(2)/30 = 80 r/min. The gear pitch line velocity is thus VG = DnG = (5)(80) = 1257 in/min The sliding velocity is the square root of the sum of the squares of VW and VG, or VS = 7540 VW = = 7644 in/min cos cos 9.46
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This result is the same as 637 feet per minute (ft/min); we enter Fig. 12.5 and find = 0.03. Proceeding now to the force analysis, we use the horsepower formula to find WW = (33 000)(12)(hp) (33 000)(12)(1) = = 52.5 lb VW 7540
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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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This force is the negative x direction. Using this value in the first of Eqs. (12.12) gives W= = Wx cos n sin + cos 52.5 = 278 lb cos 14.5 sin 9.46 + 0.03 cos 9.46
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From Eqs. (12.12) we find the other components of W to be Wy = W sin n = 278 sin 14.5 = 69.6 lb Wz = W(cos n cos sin ) = 278(cos 14.5 cos 9.46 0.03 sin 9.46 ) = 265 lb The components acting on the gear become WGa = Wx = 52.5 lb WGr = Wy = 69.6 lb WGt = Wz = 265 lb The torque can be obtained by summing moments about the x axis. This gives, in inch-pounds, T = 265(2.5) = 662.5 in lb It is because of the frictional loss that this output torque is less than the product of the gear ratio and the input torque (778 lb in).
12.5 STRENGTH AND POWER RATING
Because of the friction between the worm and the gear, power is consumed by the gear set, causing the input and output horsepower to differ by that amount and resulting in a necessity to provide for heat dissipation from the unit. Thus hp(in) = hp(out) + hp(friction loss) This expression can be translated to the gear parameters, resulting in hp(in) = VS Wf WGt DnW + 126 000mG 396 000 (12.17)
The force which can be transmitted WGt depends on tooth strength and is based on the gear, it being nearly always weaker than the worm (worm tooth strength can be computed by the methods used with screw threads, as in Chap. 13). Based on material strengths, an empirical relation is used. The equation is WGt = KsD0.8FeKmKv (12.18)
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.
WORM GEARING 12.10
GEARING
TABLE 12.2 Materials Factor Ks for Cylindrical Worm Gearing
where
Ks = materials and size correction factor, values for which are shown in Table 12.2 Fe = effective face width of gear; this is actual face width or two-thirds of worm pitch diameter, whichever is less Km = ratio correction factor; values in Table 12.3 Kv = velocity factor (Table 12.4)
Example 2. A gear catalog lists a 4-pitch, 141 2 pressure angle, single-thread hardened steel worm to mate with a 24-tooth sand-cast bronze gear. The gear has a 11 2-in face width. The worm has a 0.7854-in lead, 4.767 lead angle, 41 2-in face width, and a 3-in pitch diameter. Find the safe input horsepower. From Table 12.2, Ks = 700. The pitch diameter of the gear is D= NG 24 = = 6 in P 4
The pitch diameter of the worm is given as 3 in; two-thirds of this is 2 in. Since the face width of the gear is smaller (1.5 in), Fe = 1.5 in. Since mG = NG/NW = 24/1 =
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