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The line shaft illustrated in the figure is used to
Chapter 17, Problem 17.9(choose chapter or problem)
The line shaft illustrated in the figure is used to transmit power from an electric motor by means of flat-belt drives to various machines. Pulley A is driven by a vertical belt from the motor pulley. A belt from pulley B drives a machine tool at an angle of \(70^{\circ}\) from the vertical and at a center-to-center distance of 9 ft. Another belt from pulley C drives a grinder at a center-to-center distance of 11 ft. Pulley C has a double width to permit belt shifting as shown in Fig. 17-4. The belt from pulley D drives a dust-extractor fan whose axis is located horizontally 8 ft from the axis of the lineshaft. Additional data are
\(\begin{array}{|c|c|c|c|c|} \hline \text { Machine } & \begin{array}{l} \text { Speed, } \\ \text { rev/min } \end{array} & \begin{array}{l} \text { Power, } \\ \text { hp } \end{array} & \begin{array}{l} \text { Lineshaft } \\ \text { Pulley } \end{array} & \begin{array}{l} \text { Diameter, } \\ \text { in } \end{array} \\ \hline \text { Machine tool } & 400 & 12.5 & B & 16 \\ \hline \text { Grinder } & 300 & 4.5 & \mathrm{C} & 14 \\ \hline \text { Dust extractor } & 500 & 8 & \text { D } & 18 \\ \hline \end{array}\)
The power requirements, listed above, account for the overall efficiencies of the equipment. The two line-shaft bearings are mounted on hangers suspended from two overhead wide-flange beams. Select the belt types and sizes for each of the four drives. Make provision for replacing belts from time to time because of wear or permanent stretch.
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QUESTION:
The line shaft illustrated in the figure is used to transmit power from an electric motor by means of flat-belt drives to various machines. Pulley A is driven by a vertical belt from the motor pulley. A belt from pulley B drives a machine tool at an angle of \(70^{\circ}\) from the vertical and at a center-to-center distance of 9 ft. Another belt from pulley C drives a grinder at a center-to-center distance of 11 ft. Pulley C has a double width to permit belt shifting as shown in Fig. 17-4. The belt from pulley D drives a dust-extractor fan whose axis is located horizontally 8 ft from the axis of the lineshaft. Additional data are
\(\begin{array}{|c|c|c|c|c|} \hline \text { Machine } & \begin{array}{l} \text { Speed, } \\ \text { rev/min } \end{array} & \begin{array}{l} \text { Power, } \\ \text { hp } \end{array} & \begin{array}{l} \text { Lineshaft } \\ \text { Pulley } \end{array} & \begin{array}{l} \text { Diameter, } \\ \text { in } \end{array} \\ \hline \text { Machine tool } & 400 & 12.5 & B & 16 \\ \hline \text { Grinder } & 300 & 4.5 & \mathrm{C} & 14 \\ \hline \text { Dust extractor } & 500 & 8 & \text { D } & 18 \\ \hline \end{array}\)
The power requirements, listed above, account for the overall efficiencies of the equipment. The two line-shaft bearings are mounted on hangers suspended from two overhead wide-flange beams. Select the belt types and sizes for each of the four drives. Make provision for replacing belts from time to time because of wear or permanent stretch.
ANSWER:Step 1 of 4
We are assuming the velocity ratio for pulley A as m = 1.5.
The diameter of the pulley A is calculated as,
\(m = \frac{{{d_A}}}{D}\)
For m = 1.5 and \(D = 12\;{\rm{in}}\)
\(1.5 = \frac{{{d_A}}}{{12\;{\rm{in}}}}\)
\({d_A} = 18\;{\rm{in}}\)
The speed of the shaft of the pulley A is calculated as,
\(m = \frac{N}{{{n_A}}}\)
For m = 1.5 and \(N = 900\;{\rm{rev}}/\min\)
\(1.5 = \frac{{900\;{\rm{rev}}/\min }}{{{N_A}}}\)
\({N_A} = 600\;{\rm{rev}}/\min\)
The length of the belt between motor pulley and pulley A is calculated as,
\({L_1} = \frac{{\pi \left( {D + {d_A}} \right)}}{2} + \frac{{\left( {{d_A} - D} \right)}}{{2{C_1}}} + 2{C_1} - \frac{{{{\left( {{d_A} - D} \right)}^2}}}{{4{C_1}}}\)
For \(D = 12\;{\rm{in}} = 1\;{\rm{ft}}, {d_A} = 18\;{\rm{in}} = 1.5\;{\rm{ft}}\), and \({C_1} = 10\;{\rm{ft}}\)
\({L_1} = \frac{{\pi \left( {1\;{\rm{ft}} + 1.5\;{\rm{ft}}} \right)}}{2} + \frac{{\left( {1.5\;{\rm{ft}} - 1\;{\rm{ft}}} \right)}}{{2\left( {10\;{\rm{ft}}} \right)}} + 2\left( {10\;{\rm{ft}}} \right) - \frac{{{{\left( {1.5\;{\rm{ft}} - 1\;{\rm{ft}}} \right)}^2}}}{{4\left( {10\;{\rm{ft}}} \right)}}\)
\({L_1} = 23.95\;{\rm{ft}}\)
\({L_1} \approx 24\;{\rm{ft}}\)
The rotational speed of \(600\;{\rm{rev}}/\min\) and \(18\;\,{\rm{in}}\) diameter of the pulley results the belt speed approximately \(15\;{\rm{m}}/{\rm{s}}\). The distance between the pulley A and motor pulley is also short \(10\;{\rm{ft}}\left( {3.048\;{\rm{m}}} \right)\), so V belt of length is sufficient for power transmission between motor pulley and pulley A.
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Review this written solution for 131937) viewed: 1270 isbn: 9780073398204 | Mechanical Engineering Design - 10 Edition - Chapter 17 - Problem 17.9
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