(a) Eq. (17–1):
\begin{aligned}\theta_{d} &=\pi-2 \sin ^{-1} \frac{D-d}{2 C} \\\theta_{D} &=\pi+2 \sin ^{-1} \frac{D-d}{2 C}\end{aligned} (17–1)
| 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, \text { Ibf/in }{ }^{3} |
Coefficient of Friction |
| Leather |
1 ply |
t=\frac{11}{64} |
3 |
30 |
0.035–0.045 |
0.4 |
|
|
t=\frac{13}{64} |
3 \frac{1}{2} |
33 |
0.035–0.045 |
0.4 |
| 2 ply |
t=\frac{18}{64} |
4 \frac{1}{2} |
41 |
0.035–0.045 |
0.4 |
|
t=\frac{20}{64} |
6^{a} |
50 |
0.035–0.045 |
0.4 |
| t=\frac{23}{64} |
9^{a} |
60 |
0.035–0.045 |
0.4 |
| \text { Polyamide }^{b} |
F -0^{c} |
t = 0.03 |
0.6 |
10 |
0.035 |
0.5 |
|
F -1^{c} |
t = 0.05 |
1 |
35 |
0.035 |
0.5 |
| F -2^{c} |
t = 0.07 |
2.4 |
60 |
0.051 |
0.5 |
| A -2^{c} |
t = 0.11 |
2.4 |
60 |
0.037 |
0.8 |
| A -3^{c} |
t = 0.13 |
4.3 |
100 |
0.042 |
0.8 |
| A -4^{c} |
t = 0.20 |
9.5 |
175 |
0.039 |
0.8 |
| A -5^{c} |
t = 0.25 |
13.5 |
275 |
0.039 |
0.8 |
| \text { Urethane }^{d} |
w=0.50 |
t = 0.062 |
see |
5.2^{e} |
0.038–0.045 |
0.7 |
|
w=0.75 |
t = 0.078 |
Table |
9.8^{e} |
0.038–0.045 |
0.7 |
| w=1.25 |
t = 0.090 |
17–3 |
18.9^{e} |
0.038–0.045 |
0.7 |
| Round |
d=\frac{1}{4} |
See |
8.3^{e} |
0.038–0.045 |
0.7 |
|
d=\frac{3}{8} |
Table |
18.6^{e} |
0.038–0.045 |
0.7 |
| d=\frac{1}{2} |
17–3 |
33.0^{e} |
0.038–0.045 |
0.7 |
| d=\frac{3}{4} |
74.3^{e} |
0.038–0.045 |
0.7 |
*Average values of C_{p} for the given ranges were approximated from curves in the Habasit Engineering Manual, Habasit Belting, Inc., Chamblee (Atlanta), Ga.
{ }^{a} Add 2 in to pulley size for belts 8 in wide or more.
\text { bSource: } Habasit Engineering Manual, Habasit Belting, Inc., Chamblee (Atlanta), Ga.
{ }^{c} \text { Friction c } cover of acrylonitrile-butadiene rubber on both sides.
{ }^{d} \text { Source: }: Eagle Belting Co., Des Plaines, Ill.
{ }^{c} At 6% elongation; 12% is maximum allowable value.
\phi=\theta_{d}=\pi-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=\pi(6) 1750 / 12=2749 ft / min
Table 17–2:
w=12 \gamma b t=12(0.042) 6(0.130)=0.393 lbf / ft
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
\begin{aligned}T &=\frac{63025 H_{\text {nom }} K_{s} n_{d}}{n}=\frac{63025(15) 1.25(1.1)}{1750} \\&=742.8 lbf \cdot \text { in }\end{aligned}
(b) The necessary \left(F_{1}\right)_{a}-F_{2} to transmit the torque T, from Eq. (h), is
F_{1}-F_{2}=\frac{2 T}{d} (h)
\left(F_{1}\right)_{a}-F_{2}=\frac{2 T}{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 (F1)a is
| Table 17–4 Pulley Correction Factor C_{P} for Flat Belts* |
| Material |
Small-Pulley Diameter, in |
| 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 C_{P} for the given ranges were approximated from curves in the Habasit Engineering Manual, Habasit Belting, Inc., Chamblee (Atlanta), Ga.
\left(F_{1}\right)_{a}=b F_{a} C_{p} C_{v} (17–12)
\left(F_{1}\right)_{a}=b F_{a} C_{p} C_{v}=6(100) 0.70(1)=420 lbf
then
F_{2}=\left(F_{1}\right)_{a}-\left[\left(F_{1}\right)_{a}-F_{2}\right]=420-247.6=172.4 lbf
and from Eq. (i)
H=\frac{\left(F_{1}-F_{2}\right) V}{33000} (i)
F_{i}=\frac{\left(F_{1}\right)_{a}+F_{2}}{2}-F_{c}=\frac{420+172.4}{2}-25.6=270.6 lbf
The combination \left(F_{1}\right)_{a}, F_{2}, \text { 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^{\prime}:
\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) (17–7)
f^{\prime}=\frac{1}{\phi} \ln \frac{\left(F_{1}\right)_{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^{\prime} < 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.
