Chapter 17, Problem 4P

A flat-belt drive is to consist of two 4-ft-diameter cast-iron pulleys spaced 16 ft apart. Select a belt type to transmit 60 hp at a pulley speed of 380 rev/min. Use a service factor of 1.1 and a design factor of 1.0.

To determine

The type of belt.

The design parameter of belt.

Answer to Problem 4P

The type of belt is Polyamide A-3.

The length of the belt is $534.8\text{in}$.

The belt dip is $0.5625\text{ft}$.

Explanation of Solution

Write the expression for weight of the foot of belt.

    $w=12\lambda bt$                                                                        (I)

Here, width of belt is $b$ the size of the belt is $t$, and the specific weight is $\gamma$.

Write the expression for velocity of the smaller pulley.

    $V=\frac{\pi dn}{12}$                                                                        (II)

Here, the rotational speed of the smaller pulley is $n$ and diameter of the smaller pulley is $d$.

Write the expression for centrifugal tension

    $F_c=\frac{w}{g}{v}_{driving}{}^2$                                                                  (III)

Here, weight of the belt per $\text{ft}$ is $w$, the velocity of the driving pulley is $v_{driving}$, the acceleration due to gravity is $g$.

Write the expression for torque

    $T=\frac{63025H_{driven}{K}_s{n}_d}{n}$                                                       (IV)

Here, the power delivered is $H_{driven}$, the service factor is $K_s$, the design factor is $n_d$ and the speed of the rotation is $n$.

Write the expression for difference in tension in tight side and slack side.

    $\Delta F=\frac{2T}{d_{driving}}$                                                                          (V)

Here, torque is $T$, the diameter of the smaller pulley is $d_{driving}$.

Write the expression for actuating force on tight side of the belt.

    ${\left(F_1\right)}_a=b{F}_a{C}_P{C}_V$                                                                    (VI)

Here, the width of the flat belt is $b$, the pulley correction factor is $C_P$, the velocity correction factor is $C_V$ and the net actuating force per inch of the belt width is $F_a$.

Write the expression for tension in slack side of belt.

    $F_2={\left(F_1\right)}_a-\Delta F$                                                                      (VII)

Write the expression for the transmitted horse power.

    $H_a=\frac{\Delta F\times v_{driving}}{33000}$                                                                  (VIII)

Here, the velocity of the driving pulley is $v$ and the difference in tension in tight side and slack side of belt is $\Delta F$.

Write the expression for coefficient of friction of belt material.

    $f=\frac{1}{{\theta }_d}\ln \left(\frac{F_1-F_c}{F_2-F_c}\right)$                                                               (IX)

Here, the tension in tight side is $F_1$, tension in slack side is $F_2$ and centrifugal tension is $F_c$, angle of overlap of smaller pulley is ${\theta }_d$.

Write the expression for the length of the belt.

    $L=\sqrt{4C^2-{\left(D-d\right)}^2}+\frac{1}{2}\left(D{\theta }_D+d{\theta }_d\right)$                                 (X)

Here, the center distance between pulley is $C$, diameter of bigger pulley is $D$, diameter of smaller pulley is $d$, angle of overlap of smaller pulley is ${\theta }_d$ and angle of overlap of bigger pulley is ${\theta }_D$.

Write the expression for tight side in belt.

    $F_1=F_c+\Delta F\left[\frac{e^{f\theta }}{e^{f\theta }-1}\right]$                                                                 (XI)

Here, the centrifugal tension is $F_c$, the angle of overlap is $\theta$, the coefficient of the friction is $f$, the difference between tight side and slack side of the belt is $\Delta F$.

Write the expression for tension in slack side of belt.

    $F_2=F_1-\Delta F$                                                                                (XII)

Write the expression for initial tension in belt.

    $F_i=\frac{F_1+F_2}{2}-F_c$                                                                       (XIII)

Here, the centrifugal tension is $F_c$, the tension in tight side is $F_1$, and tension in slack side is $F_2$.

Write the expression for belt dip.

    $\text{dip}=\frac{3C^{2}w}{2F_i}$                                                                                (XIV)

Here, the center distance between pulley is $C$, weight of the belt per $\text{ft}$ is $w$ and initial tension in belt is $F_i$.

Conclusion:

Refer to table 17-2 “Properties of some flat- and round –belt materials.” to obtain the belt material with different parameters as polyamide A-3 belt.

1. 1. The allowable tension is $F_a=100\text{lbf}/\text{in}$.
2. 2. The specific weight is $\gamma =0.0\text{42}\text{lbf}/{\text{in}}^{\text{3}}$.
3. 3. The coefficient of the friction $f=0.8$.

Substitute $0.042\text{lbf}/{\text{in}}^3$ for $\lambda$, $6\text{in}$ for $b$ and $0.13\text{in}$ for $t$ in Equation (I).

    $\begin{array}{c}w=12\times \left(0.042\text{lbf}/{\text{in}}^3\right)\times \left(6\text{in}\right)\times \left(0.13\text{in}\right)\\ =\left(12\times 0.03276\right)\text{lbf}/\text{ft}\\ =0.3913\text{lbf}/\text{ft}\end{array}$

Substitute $48\text{in}$ for $d$ and $380\text{rev}/\min$ for $n$ in Equation (II).

    $\begin{array}{c}V=\frac{\pi \times 48\text{in}\times 380\text{rev}/\min }{12}\\ =\frac{18240\pi }{12}\text{ft}/\min \\ =4775\text{ft}/\min \end{array}$

Substitute $0.3913\text{lbf}/\text{ft}$ for $w$, $4775\text{ft}/\min$ for $v_{driving}$, and $32.2\text{ft}/{\text{s}}^{\text{2}}$ for $g$ in Equation (III).

    $\begin{array}{c}{F}_c=\frac{0.3913\text{lbf}/\text{ft}}{32.2\text{ft}/{\text{s}}^{\text{2}}}{\left[4775\text{ft}/\min \times \frac{1\text{min}}{60\text{s}}\right]}^2\\ =\frac{8.92\times {10}^6}{115920}\text{lbf}\\ =77\text{lbf}\end{array}$

Substitute $60\text{hp}$ for $H_{nom}$, $1.1$ for $K_s$, $1$ for $n_d$ and $380\text{rpm}$ for $n$ in Equation (IV).

    $\begin{array}{c}T=\frac{63025\times \left(60\text{hp}\right)\times \left(1.1\right)\times \left(1\right)}{\left(380\text{rpm}\right)}\\ =\frac{4159650}{380}\text{lbf}\cdot \text{in}\\ =10946\text{lbf}\cdot \text{in}\end{array}$

Substitute $48\text{in}$ for $d_{driving}$, and $10946\text{lbf}\cdot \text{in}$ for $T$ in Equation.(V).

    $\begin{array}{c}\Delta F=\frac{2\times \left(10946\text{lbf}\cdot \text{in}\right)}{48\text{in}}\\ =\frac{21892\text{lbf}\cdot \text{in}}{48\text{in}}\\ =456.1\text{lbf}\end{array}$

Substitute $6\text{in}$ for $b$, $100\text{lbf}/\text{in}$ for $F_a$, $1$ for $C_P$, and $1$ for $C_V$ in Equation (VI).

    $\begin{array}{c}{\left(F_1\right)}_a=\left(6\text{in}\right)\left(100\text{lbf}/\text{in}\right)\times 1\times 1\\ =600\text{lbf}\end{array}$

Substitute $600\text{lbf}$ for ${\left(F_1\right)}_a$, and $456.1\text{lbf}$ for $\Delta F$ in Equation.(VII).

    $\begin{array}{c}{F}_2=600\text{lbf}-456.1\text{lbf}\\ =143.9\text{lbf}\end{array}$

Substitute $456.1\text{lbf}$ for $\Delta F$ and $4775\text{ft}/\min$ for $v_{driving}$ in Equation (VIII).

    $\begin{array}{c}{H}_a=\frac{456.1\text{lbf}\times 4775\text{ft}/\min }{33000}\\ =\frac{2.178\times {10}^6}{33000}\\ =66\text{hp}\end{array}$

Substitute $600\text{lbf}$ for ${\left(F_1\right)}_a$, $143.9\text{lbf}$ for $F_2$ and $77.4\text{lbf}$ for $F_c$ in Equation (XIII).

    $\begin{array}{c}{F}_i=\frac{600\text{lbf}+143.9\text{lbf}}{2}-77.4\text{lbf}\\ =371.95\text{lbf}-77.4\text{lbf}\\ =294.55\text{lbf}\end{array}$

Substitute $600\text{lbf}$ for $F_1$, $143.9\text{lbf}$ for $F_2$, $\pi \text{rad}$ for ${\theta }_d$, and $77.4\text{lbf}$ for $F_c$ in Equation (IX).

    $\begin{array}{c}f=\frac{1}{\pi \text{rad}}\ln \left(\frac{600\text{lbf}-77.4\text{lbf}}{143.9\text{lbf}-77.4\text{lbf}}\right)\\ =\frac{1}{\pi \text{rad}}\times 2.06\\ =0.656\end{array}$

Substitute $48\text{in}$ for $D$ and $48\text{in}$ for $d$ and $192\text{in}$ $C$, $\pi \text{rad}$ for ${\theta }_D$ and $\pi \text{rad}$ for ${\theta }_d$ in Equation (X).

    $\begin{array}{c}L=\sqrt{4{\left(192\text{in}\right)}^2-{\left(48\text{in}-48\text{in}\right)}^2}+\frac{1}{2}\left(48\text{in}\times \pi \text{rad}+48\text{in}\times \pi \text{rad}\right)\\ =384\text{in+150}\text{.8}\text{in}\\ =534.8\text{in}\end{array}$

Thus, the length of the belt is $534.8\text{in}$.

Friction is not fully developed, so $b_{\min }$ is just smaller than $6\text{in}$, so the narrowest belt available should be chosen. The design can be improved by reducing the initial tension, which reduces $F_1$ and $F_2$ , thereby increasing belt life. This will bring $f$ to $0.80$.

Substitute $\pi \text{rad}$ for $\theta$ and $456.1\text{lbf}$ for $\Delta F$, $0.80$ for $f$ and $77.4\text{lbf}$ for $F_c$ in Equation (XI).

    $\begin{array}{c}{F}_1=77.4\text{lbf}+456.1\text{lbf}\left[\frac{e^{0.80\times \pi \text{rad}}}{e^{0.80\times \pi \text{rad}}-1}\right]\\ =77.4\text{lbf}+496.3\text{lbf}\\ =573.7\text{lbf}\end{array}$

Substitute $573.7\text{lbf}$ for $F_1$, and $456.1\text{lbf}$ for $\Delta F$ in Equation.(XII).

    $\begin{array}{c}{F}_2=573.7\text{lbf}-456.1\text{lbf}\\ =117.6\text{lbf}\end{array}$

Substitute $573.7\text{lbf}$ for $F_1$, $117.6\text{lbf}$ for $F_2$ and $77.4\text{lbf}$ for $F_c$ in Equation (XIII).

    $\begin{array}{c}{F}_i=\frac{573.7\text{lbf}+117.6\text{lbf}}{2}-77.4\text{lbf}\\ =345.65\text{lbf}-77.4\text{lbf}\\ =268.3\text{lbf}\end{array}$

Substitute $16\text{ft}$ for $C$, $0.393\text{lbf}/\text{ft}$ for $w$ and $268.3\text{lbf}$ for $F_i$ in Equation (XIV).

    $\begin{array}{c}\text{dip}=\frac{3\times {\left(16\text{ft}\right)}^2\times \left(0.393\text{lbf}/\text{ft}\right)}{2\left(268.3\text{lbf}\right)}\\ =\frac{301.824}{536.6}\text{ft}\\ =0.5625\text{ft}\end{array}$

Thus, the belt dip $0.5625\text{ft}$.

Refer to table 17-2 “Properties of some flat- and round –belt materials.” to obtain the belt material with different parameters as polyamide A-3 belt.

Thus, the type of belt is Polyamide A-3.