A polyamide A-3 flat belt 6 in wide is used to transmit 15 hp under light shock conditions where Ks = 1.25, and a factor of safety equal to or greater than 1.1 is appropriate.The pulley rotational axes are parallel and in the horizontal plane. The shafts are 8 ft apart. The 6-in driving pulley

Question

A polyamide A-3 flat belt 6 in wide is used to transmit 15 hp under light shock conditions where [latex]K_{s}[/latex] = 1.25, and a factor of safety equal to or greater than 1.1 is appropriate.
The pulley rotational axes are parallel and in the horizontal plane. The shafts are 8 ft apart. The 6-in driving pulley rotates at 1750 rev/min in such a way that the loose side is on top. The driven pulley is 18 in in diameter. See Fig. 17–10. The factor of safety is for unquantifiable exigencies.
(a) Estimate the centrifugal tension [latex]F_{c}[/latex] and the torque T.
(b) Estimate the allowable [latex]F_{1}, F_{2}, F_{i}[/latex] and allowable power [latex]H_{a}[/latex] .
(c) Estimate the factor of safety. Is it satisfactory?

Accepted Answer

(a) Eq. (17–1):

[latex]θ_{d} = π − 2 sin^{−1} \frac{D − d}{2C}[/latex]
[latex]θ_{D} = π + 2 sin^{−1} \frac{D − d}{2C}[/latex] (17–1)

[latex]\phi= θ_{d} = π − 2 sin^{−1} \left[ \frac{18 − 6}{2(8)12}\right]= 3.0165[/latex] rad
[latex]exp( f \phi) = exp[0.8(3.0165)] = 11.17[/latex]
[latex]V = π(6)1750/12 = 2749[/latex] ft/min

Table 17–2:

[latex]w = 12 γ bt = 12(0.042)6(0.130) = 0.393[/latex] lbf/ft

Table 17–2
Properties of Some Flat- and Round-Belt Materials. (Diameter = d, thickness = t, width = w)

Coefficient of Friction	Specific Weight, lbf/[latex]in^{3}[/latex]	Allowable Tension per Unit Width at 600 ft/min, lbf/in	Minimum Pulley Diameter, in	Size, in	Specification	Material
0.4	0.035–0.045	30	3	t =[latex]\frac{11}{64}[/latex]	1 ply	Leather
0.4	0.035–0.045	33	[latex]3\frac{1}{2}[/latex]	t =[latex]\frac{13}{64}[/latex]
0.4	0.035–0.045	41	[latex]4\frac{1}{2}[/latex]	t =[latex]\frac{18}{64}[/latex]	2 ply
0.4	0.035–0.045	50	[latex]6^{a}[/latex]	t=[latex]\frac{20}{64}[/latex]
0.4	0.035–0.045	60	[latex]9^{a}[/latex]	t=[latex]\frac{23}{64}[/latex]
0.5	0.035	10	0.60	t= 0.03	F–[latex]0^{c}[/latex]	Polyamid[latex]e^{b}[/latex]
0.5	0.035	35	1.0	t = 0.05	F–[latex]1^{c}[/latex]
0.5	0.051	60	2.4	t = 0.07	F–[latex]2^{c}[/latex]
0.8	0.037	60	2.4	t = 0.11	A–[latex]2^{c}[/latex]
0.8	0.042	100	4.3	t = 0.13	A–[latex]3^{c}[/latex]
0.8	0.039	175	9.5	t = 0.20	A–[latex]4^{c}[/latex]
0.8	0.039	275	13.5	t = 0.25	A–[latex]5^{c}[/latex]
0.7	0.038–0.045	5.[latex]2^{e}[/latex]	See	t = 0.062	w = 0.50	Urethan[latex]e^{d}[/latex]
0.7	0.038–0.045	9.[latex]8^{e}[/latex]	Table	t = 0.078	w = 0.75
0.7	0.038–0.045	18.[latex]9^{e}[/latex]	17–3	t = 0.090	w = 1.25
0.7	0.038–0.045	8.[latex]3^{e}[/latex]	See	d=[latex]\frac{1}{4}[/latex]	Round
0.7	0.038–0.045	18.[latex]6^{e}[/latex]	Table	d=[latex]\frac{3}{8}[/latex]
0.7	0.038–0.045	33.[latex]0^{e}[/latex]	17–3	d=[latex]\frac{1}{2}[/latex]
0.7	0.038–0.045	74.[latex]3^{e}[/latex]		d=[latex]\frac{3}{4}[/latex]

a: Add 2 in to pulley size for belts 8 in wide or more.
b: Source: Habasit Engineering Manual, Habasit Belting, Inc., Chamblee (Atlanta), Ga.
c: Friction cover of acrylonitrile-butadiene rubber on both sides.
d: Source: Eagle Belting Co., Des Plaines, Ill.
e: At 6% elongation; 12% is maximum allowable value.

Table 17–3
Minimum Pulley Sizes for Flat and Round Urethane Belts. (Listed are the Pulley Diameters in Inches)
Source: Eagle Belting Co., Des Plaines, Ill.

Ratio of Pulley Speed to Belt Length, rev/(ft • min)			Belt Style	Belt Style
500 to 1000	250 to 499	Up to 250	Belt Style	Belt Style
0.50	0.44	0.38	0.50 × 0.062	Flat
0.75	0.63	0.50	0.75 × 0.078
0.75	0.63	0.50	1.25 × 0.090
2.00	1.75	1.50	[latex]\frac{1}{4}[/latex]	Round
3.00	2.62	2.25	[latex]\frac{3}{8}[/latex]
4.00	3.50	3.00	[latex]\frac{1}{2}[/latex]
7.00	6.00	5.00	[latex]\frac{3}{4}[/latex]

Eq. (e):

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

[latex]F_{c} =\frac{w}{g} \left(\frac{V}{60}\right)^{2}=\frac{0.393}{32.17} \left(\frac{2749}{60}\right)^{2}= 25.6[/latex] lbf

[latex]T =\frac{63 025H_{nom}K_{s}n_{d}}{n} =\frac{63 025(15)1.25(1.1)}{1750}[/latex]

= 742.8 lbf · in

(b) The necessary [latex](F_{1})_{a} − F_{2}[/latex] to transmit the torque T, from Eq. (h), is

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

[latex](F_{1})_{a} − F_{2} =\frac{2T}{d} =\frac{2(742.8)}{6} = 247.6[/latex] lbf

From Table 17–2 [latex]F_{a}[/latex] = 100 lbf. For polyamide belts [latex]C_{v}[/latex] = 1, and from Table 17–4 [latex]C_{p}[/latex] = 0.70. From Eq. (17–12) the allowable largest belt tension ([latex]F_{1}[/latex])a is

[latex](F_{1})_{a} = bF_{a}C_{p}C_{v} = 6(100)0.70(1) = 420[/latex] lbf

Table 17–4
Pulley Correction Factor [latex]C_{p}[/latex] for Flat Belts*

Small-Pulley Diameter, in						Material
Over 31.5	18 to 31.5	14, 16	9 to 12.5	4.5 to 8	1.6 to 4	Material
1.0	0.9	0.8	0.7	0.6	0.5	Leather
1.0	1.0	1.0	1.0	1.0	0.95	F–0	Polyamide,
1.0	1.0	1.0	0.95	0.92	0.70	F–1
1.0	1.0	1.0	0.96	0.86	0.73	F–2
1.0	1.0	1.0	0.96	0.86	0.73	A–2
1.0	0.96	0.94	0.87	0.70	__	A–3
0.92	0.85	0.80	0.71	__	__	A–4
0.91	0.77	0.72	__	__	__	A–5

*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.

then

[latex]F_{2} = (F_{1})_{a} − [(F_{1})_{a} − F_{2}] = 420 − 247.6 = 172.4[/latex] lbf

and from Eq. (i)

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

[latex]F_{i} =\frac{(F_{1})_{a} + F_{2}}{2} − F_{c} =\frac{420 + 172.4}{2} − 25.6 = 270.6[/latex] lbf

The combination [latex](F_{1})_{a}, F_{2}, and F_{i}[/latex] will transmit the design power of 15(1.25)(1.1) = 20.6 hp and protect the belt. We check the friction development by solving Eq. (17–7) for [latex]f ^{′}[/latex]:

[latex]\frac{F_{1} − mr^{2}ω^{2}}{F_{2} − mr^{2}ω^{2}} =\frac{F_{1} − F_{c}}{F_{2} − F_{c}} =exp( f \phi)[/latex] (17–7)

[latex]f ^{′} =\frac{1}{\phi} ln \frac{(F_{1})_{a} − F_{c}}{F_{2} − F_{c}} =\frac{1}{3.0165} ln \frac{420 − 25.6}{172.4 − 25.6} = 0.328[/latex]

From Table 17–2, f = 0.8. Since [latex]f ^{′} < f[/latex] , that is, 0.328 < 0.80, there is no danger of slipping.

(c)

[latex]n_{f s} =\frac{H}{H_{nom}K_{s}} =\frac{20.6}{15(1.25)} = 1.1[/latex] (as expected)

The belt is satisfactory and the maximum allowable belt tension exists. If the initial tension is maintained, the capacity is the design power of 20.6 hp.

Question 17.1: A polyamide A-3 flat belt 6 in wide is used to transmit 15 h...

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