Question 17.5: Select drive components for a 2:1 reduction, 90-hp input at ...

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.

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Function: Hnom=90hp,n1=300rev/min,C/p=25,Ks=1.3H_{ nom }=90 hp , n_{1}=300 rev / min , C / p=25, K_{s}=1.3 Design factor: nd=1.5n_{d}=1.5

Sprocket teeth: N1=17 teeth, N2=34 teeth, K1=1,K2=1,1.7,2.5,3.3N_{1}=17 \text { teeth, } N_{2}=34 \text { teeth, } K_{1}=1, K_{2}=1,1.7,2.5,3.3 Chain number of strands:

 

Htab=ndKsHnomK1K2=1.5(1.3)90(1)K2=176K2H_{ tab }=\frac{n_{d} K_{s} H_{ nom }}{K_{1} K_{2}}=\frac{1.5(1.3) 90}{(1) K_{2}}=\frac{176}{K_{2}}

 

Form a table:

 

Number of Strands 176/K2 (Table 17–23) Chain Number (Table 17–19) Lubrication Type
1 176/1 = 176 200 CC ^{\prime}
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, K2K _{2}
Number of Strands K2K _{2}
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 (HtabH_{ tab } is 72.4 hp).

Number of pitches in the chain:

 

Lp=2Cp+N1+N22+(N2N1)24π2C/p=2(25)+17+342+(3417)24π2(25)=75.79 pitches \begin{aligned}\frac{L}{p} &=\frac{2 C}{p}+\frac{N_{1}+N_{2}}{2}+\frac{\left(N_{2}-N_{1}\right)^{2}}{4 \pi^{2} C / p} \\&=2(25)+\frac{17+34}{2}+\frac{(34-17)^{2}}{4 \pi^{2}(25)}=75.79 \text { pitches }\end{aligned}

 

Use 76 pitches. Then L/p = 76.

Identify the center-to-center distance: From Eqs. (17–35) and (17–36),

 

C=p4[A+A28(N2N12π)2]C=\frac{p}{4}\left[-A+\sqrt{A^{2}-8\left(\frac{N_{2}-N_{1}}{2 \pi}\right)^{2}}\right] (17–35)

 

A=N1+N22LpA=\frac{N_{1}+N_{2}}{2}-\frac{L}{p} (17–36)

 

A=N1+N22Lp=17+34276=50.5A=\frac{N_{1}+N_{2}}{2}-\frac{L}{p}=\frac{17+34}{2}-76=-50.5

 

C=p4[A+A28(N2N12π)2]=p4[50.5+50.528(34172π)2]=25.104p\begin{aligned}C &=\frac{p}{4}\left[-A+\sqrt{A^{2}-8\left(\frac{N_{2}-N_{1}}{2 \pi}\right)^{2}}\right] \\&=\frac{p}{4}\left[50.5+\sqrt{50.5^{2}-8\left(\frac{34-17}{2 \pi}\right)^{2}}\right]=25.104 p\end{aligned}

 

For a 140 chain, p = 1.75 in. Thus,

 

C=25.104p=25.104(1.75)=43.93 in C=25.104 p=25.104(1.75)=43.93 \text { 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.

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