Design a flat-belt drive to connect horizontal shafts on 16-ft centers. The velocity ratio is to be 2.25:1. The angular speed of the small driving pulley is 860 rev/min, and the nominal power transmission is to be 60 hp under very light shock.

Question

Accepted Answer

• Function: [latex]H_{ nom }=60[/latex] hp, 860 rev/min, 2.25:1 ratio, [latex]K_{s}=1.15, C=16[/latex] ft

• Design factor: [latex]n_{d}=1.05[/latex]

• Initial tension maintenance: catenary

• Belt material: polyamide

• Drive geometry, d, D

• Belt thickness: t

• Belt width: b

[latex]d=16 ext { in. } D=2.25 d=2.25(16)=36 in .[/latex]

Use polyamide A-3 belt; therefore t = 0.13 in and [latex]C_{v}=1[/latex].

Now there is one design decision remaining to be made, the belt width b.

Table 17–2:

Table 17–2 Properties of Some Flat- and Round-Belt Materials. (Diameter = d, thickness = t, width = w)
Material	Specification	Size, in	Minimum Pulley Diameter, in	Allowable Tension per Unit Width at 600 ft/min, lbf/in	Specific Weight, [latex] ext { Ibf/in }{ }^{3}[/latex]	Coefficient of Friction
Leather	1 ply	[latex]t=\frac{11}{64}[/latex]	3	30	0.035–0.045	0.4
		[latex]t=\frac{13}{64}[/latex]	[latex]3 \frac{1}{2}[/latex]	33	0.035–0.045	0.4
	2 ply	[latex]t=\frac{18}{64}[/latex]	[latex]4 \frac{1}{2}[/latex]	41	0.035–0.045	0.4
		[latex]t=\frac{20}{64}[/latex]	[latex]6^{a}[/latex]	50	0.035–0.045	0.4
		[latex]t=\frac{23}{64}[/latex]	[latex]9^{a}[/latex]	60	0.035–0.045	0.4
[latex] ext { Polyamide }^{b}[/latex]	[latex]F -0^{c}[/latex]	t = 0.03	0.6	10	0.035	0.5
	[latex]F -1^{c}[/latex]	t = 0.05	1	35	0.035	0.5
	[latex]F -2^{c}[/latex]	t = 0.07	2.4	60	0.051	0.5
	[latex]A -2^{c}[/latex]	t = 0.11	2.4	60	0.037	0.8
	[latex]A -3^{c}[/latex]	t = 0.13	4.3	100	0.042	0.8
	[latex]A -4^{c}[/latex]	t = 0.20	9.5	175	0.039	0.8
	[latex]A -5^{c}[/latex]	t = 0.25	13.5	275	0.039	0.8
[latex] ext { Urethane }^{d}[/latex]	[latex]w=0.50[/latex]	t = 0.062	see	[latex]5.2^{e}[/latex]	0.038–0.045	0.7
	[latex]w=0.75[/latex]	t = 0.078	Table	[latex]9.8^{e}[/latex]	0.038–0.045	0.7
	[latex]w=1.25[/latex]	t = 0.090	17–3	[latex]18.9^{e}[/latex]	0.038–0.045	0.7
	Round	[latex]d=\frac{1}{4}[/latex]	See	[latex]8.3^{e}[/latex]	0.038–0.045	0.7
		[latex]d=\frac{3}{8}[/latex]	Table	[latex]18.6^{e}[/latex]	0.038–0.045	0.7
		[latex]d=\frac{1}{2}[/latex]	17–3	[latex]33.0^{e}[/latex]	0.038–0.045	0.7
		[latex]d=\frac{3}{4}[/latex]		[latex]74.3^{e}[/latex]	0.038–0.045	0.7

*Average values of [latex]C_{p}[/latex] for the given ranges were approximated from curves in the Habasit Engineering Manual, Habasit Belting, Inc., Chamblee (Atlanta), Ga.

[latex]{ }^{a} Add[/latex] 2 in to pulley size for belts 8 in wide or more.

[latex] ext { bSource: }[/latex] Habasit Engineering Manual, Habasit Belting, Inc., Chamblee (Atlanta), Ga.

[latex]{ }^{c} ext { Friction c }[/latex] cover of acrylonitrile-butadiene rubber on both sides.

[latex]{ }^{d} ext { Source: }[/latex]: Eagle Belting Co., Des Plaines, Ill.

[latex]{ }^{c} At[/latex] 6% elongation; 12% is maximum allowable value.

[latex]\gamma=0.042 lbf / ext { in }^{3} \quad f=0.8 \quad F_{a}=100 lbf / ext { in at } 600 rev / min[/latex]

Table 17–4:

Table 17–4 Pulley Correction Factor [latex]C_{P}[/latex] for Flat Belts*
Material	Small-Pulley Diameter, in
Material	1.6 to 4	4.5 to 8	9 to 12.5	14, 16	18 to 31.5	Over 31.5
Leather	0.5	0.6	0.7	0.8	0.9	1.0
Polyamide, F–0	0.95	1.0	1.0	1.0	1.0	1.0
F–1	0.70	0.92	0.95	1.0	1.0	1.0
F–2	0.73	0.86	0.96	1.0	1.0	1.0
A–2	0.73	0.86	0.96	1.0	1.0	1.0
1A–3	—	0.70	0.87	0.94	0.96	1.0
A–4	—	—	0.71	0.80	0.85	0.92
A–5	—	—	—	0.72	0.77	0.91

*Average values of [latex]C_{P}[/latex] for the given ranges were approximated from curves in the Habasit Engineering Manual, Habasit Belting, Inc., Chamblee (Atlanta), Ga.

[latex]C_{p}=0.94[/latex]

Eq. (17–12):

[latex]\left(F_{1} ight)_{a}=b F_{a} C_{p} C_{v}[/latex] (17–12)

[latex]F_{1 a}=b(100) 0.94(1)=94.0 b lbf[/latex] (1)

[latex]H_{d}=H_{ nom } K_{s} n_{d}=60(1.15) 1.05=72.5 hp[/latex]

[latex]T=\frac{63025 H_{d}}{n}=\frac{63025(72.5)}{860}=5310 lbf \cdot ext { in }[/latex]

Estimate exp[latex](f \phi)[/latex] for full friction development:

Eq. (17–1):

[latex]\begin{array}{l} heta_{d}=\pi-2 \sin ^{-1} \frac{D-d}{2 C} \ heta_{D}=\pi+2 \sin ^{-1} \frac{D-d}{2 C}\end{array}[/latex] (17–1)

[latex]\phi= heta_{d}=\pi-2 \sin ^{-1} \frac{36-16}{2(16) 12}=3.037 rad[/latex]

[latex]\exp (f \phi)=\exp [0.80(3.037)]=11.35[/latex]

Estimate centrifugal tension [latex]F_{c}[/latex] in terms of belt width b:

[latex]w=12 \gamma b t=12(0.042) b(0.13)=0.0655 b lbf / ft[/latex]

[latex]V=\pi d n / 12=\pi(16) 860 / 12=3602 ft / min[/latex]

Eq. (e):

[latex]F_{c}=\frac{w}{g}\left(\frac{V}{60} ight)^{2}=\frac{w}{32.17}\left(\frac{V}{60} ight)^{2}[/latex] (e)

[latex]F_{c}=\frac{w}{g}\left(\frac{V}{60} ight)^{2}=\frac{0.0655 b}{32.17}\left(\frac{3602}{60} ight)^{2}=7.34 b lbf[/latex] (2)

For design conditions, that is, at [latex]H_{d}[/latex] power level, using Eq. (h) gives

[latex]F_{1}-F_{2}=\frac{2 T}{d}[/latex] (h)

[latex]\left(F_{1} ight)_{a}-F_{2}=2 T / d=2(5310) / 16=664 lbf[/latex] (3)

[latex]F_{2}=\left(F_{1} ight)_{a}-\left[\left(F_{1} ight)_{a}-F_{2} ight]=94.0 b-664 lbf[/latex] (4)

Using Eq. (i) gives

[latex]F_{i}=\frac{F_{1}+F_{2}}{2}-F_{c}[/latex] (i)

[latex]F_{i}=\frac{\left(F_{1} ight)_{a}+F_{2}}{2}-F_{c}=\frac{94.0 b+94.0 b-664}{2}-7.34 b=86.7 b-332 lbf[/latex] (5)

Place friction development at its highest level, using Eq. (17–7):

[latex]\frac{F_{1}-m r^{2} \omega^{2}}{F_{2}-m r^{2} \omega^{2}}=\frac{F_{1}-F_{c}}{F_{2}-F_{c}}=\exp (f \phi)[/latex] (17–7)

[latex]f \phi=\ln \frac{\left(F_{1} ight)_{a}-F_{c}}{F_{2}-F_{c}}=\ln \frac{94.0 b-7.34 b}{94.0 b-664-7.34 b}=\ln \frac{86.7 b}{86.7 b-664}[/latex]

Solving the preceding equation for belt width b at which friction is fully developed gives

[latex]b=\frac{664}{86.7} \frac{\exp (f \phi)}{\exp (f \phi)-1}=\frac{664}{86.7} \frac{11.38}{11.38-1}=8.40 ext { in }[/latex]

A belt width greater than 8.40 in will develop friction less than f = 0.80. The manufacturer’s data indicate that the next available larger width is 10 in.

Use 10-in-wide belt.

It follows that for a 10-in-wide belt

Eq. (2):

[latex]F_{c}=7.34(10)=73.4 lbf[/latex]

Eq. (1):

[latex]\left(F_{1} ight)_{a}=94(10)=940 lbf[/latex]

Eq. (4):

[latex]F_{2}=94(10)-664=276 lbf[/latex]

Eq. (5):

[latex]F_{i}=86.7(10)-332=535 lbf[/latex]

The transmitted power, from Eq. (3), is

[latex]H_{t}=\frac{\left[\left(F_{1} ight)_{a}-F_{2} ight] V}{33000}=\frac{664(3602)}{33000}=72.5 hp[/latex]

and the level of friction development [latex]f^{\prime}[/latex], from Eq. (17–7) is

[latex]f^{\prime}=\frac{1}{\phi} \ln \frac{\left(F_{1} ight)_{a}-F_{c}}{F_{2}-F_{c}}=\frac{1}{3.037} \ln \frac{940-73.4}{276-73.4}=0.479[/latex]

which is less than f = 0.8, and thus is satisfactory. Had a 9-in belt width been available, the analysis would show [latex]\left(F_{1} ight)_{a}=846 lbf , F_{2}=182 lbf , F_{i}=448 lbf[/latex], and [latex]f^{\prime}=0.63[/latex]. With a figure of merit available reflecting cost, thicker belts (A-4 or A-5) could be examined to ascertain which of the satisfactory alternatives is best. From Eq. (17–13) the catenary dip is

[latex]d i p=\frac{12(C / 12)^{2} w}{8 F_{i}}=\frac{C^{2} w}{96 F_{i}}[/latex] (17–13)

[latex]\operatorname{dip}=\frac{C^{2} w}{96 F_{i}}=\frac{[16(12)]^{2} 0.0655(10)}{96(535)}=0.470 in[/latex]

Question 17.2: Design a flat-belt drive to connect horizontal shafts on 16-...

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