Select drive components for a 2:1 reduction, 90-hp input at 300 rev/min, moderate shock, an abnormally long 18-hour day, poor lubrication, cold temperatures, dirty surroundings, short drive C/p = 25.
Select drive components for a 2:1 reduction, 90-hp input at 300 rev/min, moderate shock, an abnormally long 18-hour day, poor lubrication, cold temperatures, dirty surroundings, short drive C/p = 25.
Function: Hnom=90hp,n1=300rev/min,C/p=25,Ks=1.3 Design factor: nd=1.5
Sprocket teeth: N1=17 teeth, N2=34 teeth, K1=1,K2=1,1.7,2.5,3.3 Chain number of strands:
Htab=K1K2ndKsHnom=(1)K21.5(1.3)90=K2176
Form a table:
Number of Strands | 176/K2 (Table 17–23) | Chain Number (Table 17–19) | Lubrication Type |
1 | 176/1 = 176 | 200 | C′ |
2 | 176/1.7 = 104 | 160 | C |
3 | 176/2.5 = 70.4 | 140 | B |
4 | 176/3.3 = 53.3 | 140 | B |
Table 17–23 Multiple-Strand Factors, K2 | |
Number of Strands | K2 |
1 | 1 |
2 | 1.7 |
3 | 2.5 |
4 | 3.3 |
5 | 3.9 |
6 | 4.6 |
8 | 6 |
Table 17–19 Dimensions of American Standard Roller Chains—Single Strand Source: Compiled from ANSI B29.1-1975. | ||||||
ANSI Chain Number | Pitch, in (mm) | Width, in (mm) | Minimum Tensile Strength, lbf (N) | Average Weight, lbf/ft (N/m) | Roller Diameter, in (mm) | Multiple- Strand Spacing, in (mm) |
25 | 0.250 | 0.125 | 780 | 0.09 | 0.130 | 0.252 |
(6.35) | (318) | (3 470) | (1.31) | (3.30) | (6.40) | |
35 | 0.375 | 0.188 | 1 760 | 0.21 | 0.200 | 0.399 |
(9.52) | (4.76) | (7 830) | (3.06) | (5.08) | (10.13) | |
41 | 0.500 | 0.250 | 1 500 | 0.25 | 0.306 | — |
(12.70) | (6.35) | (6 670) | (3.65) | (7.77) | — | |
40 | 0.500 | 0.312 | 3 130 | 0.42 | 0.312 | 0.566 |
(12.70) | (7.94) | (13 920) | (6.13) | (7.92) | (14.38) | |
50 | 0.625 | 0.375 | 4 880 | 0.69 | 0.400 | 0.713 |
(15.88) | (9.52) | (21 700) | (10.1) | (10.16) | (18.11) | |
60 | 0.750 | 0.500 | 7 030 | 1.00 | 0.469 | 0.897 |
(19.05) | (12.7) | (31 300) | (14.6) | (11.91) | (22.78) | |
80 | 1.000 | 0.625 | 12 500 | 1.71 | 0.625 | 1.153 |
(25.40) | (15.88) | (55 600) | (25.0) | (15.87) | (29.29) | |
100 | 1.250 | 0.750 | 19 500 | 2.58 | 0.750 | 1.409 |
(31.75) | (19.05) | (86 700) | (37.7) | (19.05) | (35.76) | |
120 | 1.500 | 1.000 | 28 000 | 3.87 | 0.875 | 1.789 |
(38.10) | (25.40) | (124 500) | (56.5) | (22.22) | (45.44) | |
140 | 1.750 | 1.000 | 38 000 | 4.95 | 1.000 | 1.924 |
(44.45) | (25.40) | (169 000) | (72.2) | (25.40) | (48.87) | |
160 | 2.000 | 1.250 | 50 000 | 6.61 | 1.125 | 2.305 |
(50.80) | (31.75) | (222 000) | (96.5) | (25.57) | (58.55) | |
180 | 2.250 | 1.406 | 63 000 | 9.06 | 1.406 | 2.592 |
(57.15) | (35.71) | (280 000) | (132.2) | (35.71) | (65.84) | |
200 | 2.500 | 1.500 | 78 000 | 10.96 | 1.562 | 2.817 |
(63.50) | (38.10) | (347 000) | (159.9) | (39.67) | (71.55) | |
240 | 3.000 | 1.875 | 112 000 | 16.4 | 1.875 | 3.458 |
(76.70) | (47.63) | (498 000) | (239) | (47.62) | (87.83) |
3 strands of number 140 chain (Htab is 72.4 hp).
Number of pitches in the chain:
pL=p2C+2N1+N2+4π2C/p(N2−N1)2=2(25)+217+34+4π2(25)(34−17)2=75.79 pitches
Use 76 pitches. Then L/p = 76.
Identify the center-to-center distance: From Eqs. (17–35) and (17–36),
C=4p[−A+A2−8(2πN2−N1)2] (17–35)
A=2N1+N2−pL (17–36)
A=2N1+N2−pL=217+34−76=−50.5
C=4p⎣⎢⎡−A+A2−8(2πN2−N1)2⎦⎥⎤=4p⎣⎢⎡50.5+50.52−8(2π34−17)2⎦⎥⎤=25.104p
For a 140 chain, p = 1.75 in. Thus,
C=25.104p=25.104(1.75)=43.93 in
Lubrication: Type B
Comment: This is operating on the pre-extreme portion of the power, so durability estimates other than 15 000 h are not available. Given the poor operating conditions, life will be much shorter.