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

• Function: H_{nom} = 60 hp, 860 rev/min, 2.25:1 ratio, K_{s} = 1.15, C = 16 ft
• Design factor: n_{d} = 1.05
• Initial tension maintenance: catenary
• Belt material: polyamide
• Drive geometry, d, D
• Belt thickness: t
• Belt width: b

The last four could be design variables. Let’s make a few more a priori decisions.

d = 16 in, D = 2.25d = 2.25(16) = 36 in.

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

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

Table 17–2:      γ = 0.042 lbf/in^{3}      f = 0.8     F_{a} = 100  lbf/in  at  600  rev/min

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

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 17–4:       C_{p} = 0.94

Table 17–4
Pulley Correction Factor C_{p} for Flat Belts*

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

Eq. (17–12):

(F_{1})_{a} = bF_{a}C_{p}C_{v}            (17–12)

F_{1a} = b(100)0.94(1) = 94.0b lbf               (1)

H_{d} = H_{nom}K_{s}n_{d} = 60(1.15)1.05 = 72.5 hp
T =\frac{63  025H_{d}}{n} =\frac{63  025(72.5)}{860} = 5310 lbf · in

Estimate exp( f \phi) for full friction development:

Eq. (17–1):

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

\phi = θ_{d} = π − 2 sin^{−1} \frac{36 − 16}{2(16)12} = 3.037 rad
exp( f  \phi) = exp[0.80(3.037)] = 11.35

Estimate centrifugal tension F_{c} in terms of belt width b:

w = 12 γ  bt = 12(0.042)b(0.13) = 0.0655b lbf/ft
V = πdn/12 = π(16)860/12 = 3602 ft/min

Eq. (e):

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

F_{c} =\frac{w}{g} \left(\frac{V}{60}\right)^{2}=\frac{0.0655b}{32.17} \left(\frac{3602}{60}\right)^{2}= 7.34b lbf                  (2)

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

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

(F_{1})_{a} − F_{2} = 2T/d = 2(5310)/16 = 664 lbf                    (3)
F_{2} = (F_{1})_{a} − [(F_{1})_{a} − F_{2}] = 94.0b − 664 lbf                      (4)

Using Eq. (i) gives

F_{i} =\frac{F_{1} + F_{2}}{2} − F_{c}                            (i )

F_{i} =\frac{(F_{1})_{a} + F_{2}}{2} − F_{c} =\frac{94.0b + 94.0b − 664}{2} − 7.34b = 86.7b − 332 lbf                        (5)

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

\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)                (17–7)

f  \phi = ln \frac{(F_{1})_{a} − F_{c}}{F_{2} − F_{c}} = ln \frac{94.0b − 7.34b}{94.0b − 664 − 7.34b} =ln \frac{86.7b}{86.7b − 664}

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

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 in

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):      F_{c} = 7.34(10) = 73.4 lbf
Eq. (1):       (F_{1})_{a} = 94(10) = 940 lbf
Eq. (4):      F_{2} = 94(10) − 664 = 276 lbf
Eq. (5):      F_{i} = 86.7(10) − 332 = 535 lbf

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

H_{t} =\frac{[(F_{1})_{a} − F_{2}]V}{33  000} =\frac{664(3602)}{33  000} = 72.5 hp

and the level of friction development f^{′}, from Eq. (17–7) is

\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)                            (17–7)

f^{′}=\frac{1}{\phi} ln \frac{(F_{1})_{a} − F_{c}}{F_{2} − F_{c}} =\frac{1}{3.037} ln \frac{940 − 73.4}{276 − 73.4} = 0.479

which is less than f = 0.8, and thus is satisfactory. Had a 9-in belt width been available, the analysis would show (F_{1})_{a} = 846  lbf, F_{2} = 182  lbf, F_{i} = 448  lbf,  and  f^{′} = 0.63. 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

d =\frac{12L^{2}w}{8F_{i}} =\frac{3L^{2}w}{2F_{i}}                      (17–13)

d =\frac{3L^{2}w}{2F_{i}}=\frac{3(15^{2})0.0655(10)}{2(535)} = 0.413 in

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.