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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Accepted Answer

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

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

[latex]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}}[/latex]

Form a table:

Number of Strands	176/K2 (Table 17–23)	Chain Number (Table 17–19)	Lubrication Type
1	176/1 = 176	200	[latex]C ^{\prime}[/latex]
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, [latex]K _{2}[/latex]
Number of Strands	[latex]K _{2}[/latex]
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 ([latex]H_{ tab }[/latex] is 72.4 hp).

Number of pitches in the chain:

[latex]\begin{aligned}\frac{L}{p} &=\frac{2 C}{p}+\frac{N_{1}+N_{2}}{2}+\frac{\left(N_{2}-N_{1} ight)^{2}}{4 \pi^{2} C / p} \&=2(25)+\frac{17+34}{2}+\frac{(34-17)^{2}}{4 \pi^{2}(25)}=75.79 ext { pitches }\end{aligned}[/latex]

Use 76 pitches. Then L/p = 76.

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

[latex]C=\frac{p}{4}\left[-A+\sqrt{A^{2}-8\left(\frac{N_{2}-N_{1}}{2 \pi} ight)^{2}} ight][/latex] (17–35)

[latex]A=\frac{N_{1}+N_{2}}{2}-\frac{L}{p}[/latex] (17–36)

[latex]A=\frac{N_{1}+N_{2}}{2}-\frac{L}{p}=\frac{17+34}{2}-76=-50.5[/latex]

[latex]\begin{aligned}C &=\frac{p}{4}\left[-A+\sqrt{A^{2}-8\left(\frac{N_{2}-N_{1}}{2 \pi} ight)^{2}} ight] \&=\frac{p}{4}\left[50.5+\sqrt{50.5^{2}-8\left(\frac{34-17}{2 \pi} ight)^{2}} ight]=25.104 p\end{aligned}[/latex]

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

[latex]C=25.104 p=25.104(1.75)=43.93 ext { in }[/latex]

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.

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

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