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

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

Table 17–2:

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

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

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{0.393}{32.17} \left(\frac{2749}{60}\right)^{2}= 25.6 lbf

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

= 742.8 lbf · in

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

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

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

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

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

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.

then

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

and from Eq. (i)

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

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

The combination (F_{1})_{a}, F_{2},  and  F_{i} 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 f ^{′}:

\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}{3.0165} ln \frac{420 − 25.6}{172.4 − 25.6} = 0.328

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

(c)

n_{f s} =\frac{H}{H_{nom}K_{s}} =\frac{20.6}{15(1.25)} = 1.1 (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.