Chapter 17, Problem 2P

A 6-in-widc polyamide F-l flat belt is used to connect a 2-in-diameter pulley to drive a larger pulley with an angular velocity ratio of 0.5. The center-to-center distance is 9 ft. The angular speed of the small pulley is 1750 rev/min as it delivers 2 hp. The service is such that a service factor K_s of 1.25 is appropriate.

1. (a) Find F_c, F_i, (F₁)_a,. and F₂, assuming operation at the maximum tension limit.
2. (b) Find H_a, n_fs, and belt length.
3. (c) Find the dip.

To determine

The value of $F_c$.

The value of $F_i$.

The value of $F_{1a}$.

The value of $F_2$.

Answer to Problem 2P

The value of $F_c$ is $0.913\text{lbf}$.

The value of $F_i$ is $101.1\text{lbf}$.

The value of $F_{1a}$ is $147\text{lbf}$.

The value of $F_2$ is $57\text{lbf}$.

Explanation of Solution

Refer the Table $17-2$, “Properties of some flat and round belt materials”, to obtain the size as $t=0.05\text{in}$, the minimum pulley diameter as $d_{\min }=1.0\text{in}$, the allowable tension per unit width as $F_a=35\text{lbf}/in$, the specific weight as $\gamma =0.035\text{lbf}/{\text{in}}^3$.and the coefficient of friction is $f=0.5$ for F-1 belt.

Refer the Table 17-4, “Pulley correction factor $c_P$ for belts”, to obtain the correction factor $c_P=0.70$.

Write the expression for weight of the foot of belt.

$w=\lambda \cdot b\cdot t$ (I)

Here, width of belt is $b$.

Write the expression for velocity ratio.

$\frac{v_{driven}}{v_{driving}}=\frac{d_{driven}}{d_{driving}}$ (II)

Here, the velocity of the driven pulley is $v_{driven}$, the velocity of the driving pulley is $v_{driving}$, the diameter of the driven pulley is $d_{driven}$ and the diameter is $d_{driving}$.

Write the expression for contact angle for smaller pulley.

${\theta }_d=180°-2{\sin }^{-1}\left[\frac{D-d}{2C}\right]$ (III)

Here, the center distance between pulley is $C$, diameter of bigger pulley is $D$, diameter of smaller pulley is $d$.

Write the expression for velocity of the driving pulley.

$v_{driving}=\pi d_{driving}\cdot n$ (IV)

Here, diameter of the driving pulley is $d_{driving}$ and the speed of the rotation is $n$.

Write the expression for centrifugal tension

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

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

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

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 actuating force on tight side of the belt.

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

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 difference in tension in tight side and slack side.

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

Write the expression for tension in slack side of belt.

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

Write the expression for initial tension in belt.

$F_i=\frac{{\left(F_1\right)}_a+F_2}{2}-F_c$ (X)

Conclusion:

Substitute $0.035\text{lbf}/{\text{in}}^3$ for $\lambda$, $6\text{in}$ for $b$ and $0.05\text{in}$ for $t$ in Equation (I).

$\begin{array}{c}w=\left(0.035\text{lbf}/{\text{in}}^3\right)\cdot \left(6\text{in}\right)\cdot \left(0.05\text{in}\right)\\ =0.0105\text{lbf}/\text{in}\times \left[\frac{12\text{in}}{1\text{ft}}\right]\\ =0.126\text{lbf}/\text{ft}\end{array}$

Substitute $0.5$ for $\frac{v_{driven}}{v_{driving}}$ and $2.0\text{in}$ for $d_{driven}$ in Equation (II).

$\begin{array}{l}0.5=\frac{2.0\text{in}}{d_{driving}}\\ d_{driving}=4.0\text{in}\end{array}$

Substitute $4.0\text{in}$ for $D$ and $2.0\text{in}$ for $d$ and $108\text{in}$ $C$ in Equation.(III).

$\begin{array}{c}{\theta }_d=180°-2{\sin }^{-1}\left(\frac{4\text{in}-2\text{in}}{2\times 108\text{in}}\right)\\ =180°-1.06\\ =178.94°\times \frac{\pi \text{rad}}{180°}\\ =3.123\text{rad}\end{array}$

Substitute $2\text{in}$ for $d_{driving}$, $1750\text{rpm}$ for $n$ in Equation (IV).

$\begin{array}{c}{v}_{driving}=\pi \times \left[2\text{in}\times \frac{1\text{ft}}{12\text{in}}\right]\times 1750\text{rpm}\\ =\frac{1750\pi }{6}\text{ft}/\min \\ =916.3\text{ft}/\min \end{array}$

Substitute $0.126\text{lbf}/\text{ft}$ for $w$, $916.3\text{ft}/\min$ for $v_{driving}$, and $32.2\text{ft}/{\text{s}}^{\text{2}}$ for $g$ in Equation (V).

$\begin{array}{c}{F}_c=\frac{0.126\text{lbf}/\text{ft}}{32.2\text{ft}/{\text{s}}^{\text{2}}}{\left[916.3\text{ft}/\min \times \frac{1\text{min}}{60\text{s}}\right]}^2\\ =\frac{105.8\times {10}^3}{115920}\text{lbf}\\ =0.913\text{lbf}\end{array}$

Thus, the value of $F_c$ is $0.913\text{lbf}$.

Substitute $2\text{hp}$ for $H_{driven}$, $1.25$ for $K_s$, $1$ for $n_d$ and $1750\text{rpm}$ for $n$ in Equation (VI).

$\begin{array}{c}T=\frac{63025\left(2\text{hp}\right)\left(1.25\right)\left(1\right)}{\left(1750\text{rpm}\right)}\\ =\frac{157562.5}{1750}\text{lbf}\cdot \text{in}\\ =90\text{lbf}\cdot \text{in}\end{array}$

Substitute $6\text{in}$ for $b$, $35\text{lbf}/\text{in}$ for $F_a$, $0.7$ for $C_P$, and $1$ for $C_V$ in Equation (VII).

$\begin{array}{c}{\left(F_1\right)}_a=\left(6\text{in}\right)\left(35\text{lbf}/\text{in}\right)\times 0.7\times 1\\ =147\text{lbf}\end{array}$

Thus, the value of $F_{1a}$ is $147\text{lbf}$.

Substitute $2\text{in}$ for $d_{driving}$, and $90\text{lbf}\cdot \text{in}$ for $T$ in Equation.(VIII).

$\begin{array}{c}\Delta F=\frac{2\left(90\text{lbf}\cdot \text{in}\right)}{2\text{in}}\\ =90\text{lbf}\end{array}$

Substitute $147\text{lbf}$ for ${\left(F_1\right)}_a$, and $90\text{lbf}$ for $\Delta F$ in Equation.(IX).

$\begin{array}{c}{F}_2=147\text{lbf}-90\text{lbf}\\ =57\text{lbf}\end{array}$

Thus, the value of $F_2$ is $57\text{lbf}$.

Substitute $147\text{lbf}$ for ${\left(F_1\right)}_a$, $57\text{lbf}$ for $F_2$ and $0.913\text{lbf}$ for $F_c$ in Equation (X).

$\begin{array}{c}{F}_i=\frac{147\text{lbf}+57\text{lbf}}{2}-0.913\text{lbf}\\ =102\text{lbf}-0.913\text{lbf}\\ =101.1\text{lbf}\end{array}$

Thus, the value of $F_i$ is $101.1\text{lbf}$.

To determine

The value of $H_a$.

The value of ${\eta }_{fs}$.

The belt length.

Answer to Problem 2P

The value of $H_a$ is $2.5\text{hp}$.

The value of ${\eta }_{fs}$ is $1.0$.

The belt length is $225.5\text{in}$.

Explanation of Solution

Write the expression for the transmitted horse power.

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

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 factor of the safety.

$n_{fs}=\frac{H_a}{H_{driven}\cdot K_s}$ (XII)

Here, the transmitted horse power is $H_a$, the delivered horse power of smaller pulley is $H_{driven}$ and the service factor is $K_s$.

Write the expression for contact angle for bigger pulley.

${\theta }_D=180°+2{\sin }^{-1}\left[\frac{D-d}{2C}\right]$ (XIII)

Here, the center distance between pulley is $C$, diameter of bigger pulley is $D$, diameter of smaller pulley is $d$.

Write the expression for the length of the belt.

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

Conclusion:

Substitute $90\text{lbf}$ for $\Delta F$ and $916.3\text{ft}/\min$ for $v_{driving}$ in Equation (XI).

$\begin{array}{c}{H}_a=\frac{90\text{lbf}\times 916.3\text{ft}/\min }{33000}\\ =2.5\text{hp}\end{array}$

Thus, the value of $H_a$ is $2.5\text{hp}$.

Substitute $2.5\text{hp}$ for $H_a$, $2.0\text{hp}$ for $H_{driven}$ and $1.25$ for $K_s$ in Equation (XII).

$\begin{array}{c}{n}_{fs}=\frac{2.5\text{hp}}{2.0\text{hp}\times 1.25}\\ =1.0\end{array}$

Thus, the value of ${\eta }_{fs}$ is $1.0$.

Substitute $4.0\text{in}$ for $D$ and $2.0\text{in}$ for $d$ and $108\text{in}$ $C$ in Equation (XIII).

$\begin{array}{c}{\theta }_D=180°+2{\sin }^{-1}\left(\frac{4\text{in}-2\text{in}}{2\times 108\text{in}}\right)\\ =180°+1.06\\ =181.06°\times \frac{\pi \text{rad}}{180°}\\ =3.16\text{rad}\end{array}$

Substitute $4.0\text{in}$ for $D$ and $2.0\text{in}$ for $d$ and $108\text{in}$ $C$, $3.18\text{rad}$ for ${\theta }_D$ and $3.123\text{rad}$ for ${\theta }_d$ in Equation (XIV).

$\begin{array}{c}L=\sqrt{4{\left(108\text{in}\right)}^2-{\left(4\text{in}-2\text{in}\right)}^2}+\frac{1}{2}\left(4\text{in}\times 3.16\text{rad}+2\text{in}\times 3.123\text{rad}\right)\\ =216\text{in+9}\text{.5}\text{in}\\ =225.5\text{in}\end{array}$

Thus, the belt length is $225.5\text{in}$.

To determine

The dip of the belt.

Answer to Problem 2P

The belt dip is $0.151\text{in}$.

Explanation of Solution

Write the expression for belt dip.

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

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

Conclusion:

Substitute $9\text{ft}$ $C$, $0.126\text{lbf}/\text{ft}$ for $w$ and $101.1\text{lbf}$ for $F_i$ in Equation (XIII).

$\begin{array}{c}\text{dip}=\frac{3{\left(9\text{ft}\right)}^2\left(0.126\text{lbf}/\text{ft}\right)}{2\left(101.1\text{lbf}\right)}\\ =\frac{30.618}{202.2}\text{in}\\ =0.151\text{in}\end{array}$

Thus, the belt dip is $0.151\text{in}$.