Question 18.CS.10: Belt Design Consider the toothed belt of a high-speed cutter...

See Figure 18.10 and Table 13.5; Sections 13.3, 13.4, and 13.5.

a. The appropriate belt pitch length is determined using Equation 13.9:

L=2 c+\pi\left(r_1+r_2\right)+\frac{1}{c}\left(r_2-r_1\right)^2       (a)

Substitution of the given data yields

\begin{aligned} L &=2(17)+\pi(1.25+2.25)+\frac{1}{17}(2.25-1.25)^2 \\ &=45.05  in . \end{aligned}

Comment: A standard toothed or timing belt with maximum length of 45 in. is selected.

b. An estimate of the center distance is given by Equation 13.10:

c=\frac{1}{4}\left[b+\sqrt{b^2-8\left(r_2-r_1\right)^2}\right]        (b)

Here, from Equation 13.11,

b=L-\pi\left(r_2+r_1\right)       (c)

Carrying the given numerical values into Equation (c), we have

b=45-\pi(2.25+1.25)=34  in .

Equation (b) is then

c=\frac{1}{4}\left[34+\sqrt{(34)^2-8(2.25-1.25)^2}\right]=16.97  in .

Comment: The requirement that c < 17 in. is satisfied.

c. The contact angle \phi, from Equations 13.7 and 13.6, equals

\sin \alpha=\frac{r_2-r_1}{c}     (13.6)

\phi=\pi-2 \alpha      (13.7)

\begin{aligned} \phi &=\pi-2 \sin ^{-1}\left\lgroup \frac{r_2-r_1}{c} \right\rgroup \\ &=\pi-2 \sin ^{-1}\left\lgroup \frac{2.25-1.25}{16.97} \right\rgroup =3.024  rad \end{aligned}

The tight-side tension in the belt is calculated, through the use of Equation 13.20 in which, for a toothed (or flat) belt, sin β = 1 and

F_1=F_c+\left\lgroup \frac{\gamma}{\gamma-1} \right\rgroup \frac{T_1}{r_1}     (13.20)

\begin{aligned} F_c &=\frac{w}{g} V^2 \\ &=\frac{0.007}{386.4}\left[\frac{\pi(2.5) 1800}{60}\right]^2=1.006  lb \end{aligned}

\gamma= e ^{f \phi}= e ^{(1)(3.024)}=20.57

\begin{aligned} T_1 &=\frac{33,000 hp }{n_1} \\ &=\frac{33,000(2)}{1800}=36.67  lb \cdot in . \end{aligned}

Therefore,

F_1=1.006+\left\lgroup \frac{20.57}{20.57-1} \right\rgroup \frac{36.67}{1.25}=31.84  lb

It follows that

\begin{aligned} F_2 &=F_1-\frac{T_1}{r_1} \\ &=31.84-\frac{36.67}{1.25}=2.50  lb \end{aligned}

The service factor, from Table 13.5, K_{s} = 1.4. The maximum belt tensile load is then obtained by Equation 13.22 as

\begin{aligned} F_{\max } &=K_s F_1 \\ &=1.4(31.84)=44.58  lb \end{aligned}

Comment: Recall from Section 13.2 that toothed belts can provide safe operation at speeds up to at least 16,000 fpm. This is well above the belt velocity. V=\pi(2.5)(1800) / 12= 1178  fpm , of the cutting machine.