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

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

Sprocket teeth: N_{1} = 17  teeth, N_{2} = 34  teeth, K_{1} = 1, K_{2} = 1, 1.7, 2.5, 3.3 Chain number of strands:

H_{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:

Table 17–19
Dimensions of American Standard Roller Chains—Single Strand Source: Compiled from ANSI B29.1-1975.

Table 17–23
Multiple-Strand Factors K_{2}

3 strands of number 140 chain (H_{tab} is 72.4 hp).
Number of pitches in the chain:

\frac{L}{p} =\frac{2C}{p} +\frac{N_{1} + N_{2}}{2} + \frac{(N_{2} − N_{1})^{2}}{4π^{2}C/p}
= 2(25) + \frac{17 + 34}{2} +\frac{(34 − 17)^{2}}{4π^{2}(25)} = 75.79 pitches

Use 76 pitches. Then L/p = 76.
Identify the center-to-center distance: From Eqs. (17–35) and (17–36),

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

=\frac{17 + 34}{2} − 76 = −50.5
C =\frac{p}{4} \left[−A + \sqrt{A^{2} − 8\left(\frac{N_{2} − N_{1}}{2π}\right)^{2}}\right]                       (17-35)

=\frac{p}{4} \left[ 50.5 + \sqrt{50.5^{2} − 8 \left(\frac{34 − 17}{2π}\right)^{2}}\right] = 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.