Question 1.2: Camshaft Torque Requirement A rotating camshaft (Figure 1.6)......
Camshaft Torque Requirement
A rotating camshaft (Figure 1.6) of an intermittent motion mechanism moves the follower in a direction at right angles to the cam axis. For the position shown, the follower is being moved upward by the lobe of the cam with a force F. A rotation of \theta corresponds to a follower motion of s. Determine the average torque T required to turn the camshaft during this interval.
Given
F=0.2 lb , \theta=8^{\circ}=0.14 rad , s=0.05 in .
Assumptions
The torque can be considered to be constant during the rotation. The friction forces can be omitted.
The work done on the camshaft equals the work done by the follower. Therefore, by Equations (1.8) and (1.9), we write
W=F s (1.8)
W=T \theta (1.9)
T \theta=F s (a)
Substituting the given numerical values,
T(0.14)=0.2(0.05)=0.01 lb \cdot in
The foregoing gives T=0.071 lb \cdot \text { in } .
Comments: Using the conversion factor (Table A.1), in SI units, the answer is:
T=0.071 \frac{10^3 N \cdot mm }{(0.7376) 12}=8.02 N \cdot m
The stress and deflection caused by force F at the contact surface between the cam and follower are considered in Chapter 8.
| TABLE A.1 Conversion Factors: SI Units to US Customary Units |
||
| Quantity | SI Unit | US Equivalent |
| Acceleration | m/s^2 (meter per square second) | 3.2808 ft/s² |
| Area | m^2 (square meter) | 10.76 ft² |
| Force | N (newton) | 0.2248 lb |
| Intensity of force | N/m (newton per meter) | 0.0685 lb/ft |
| Length | m (meter) | 3.2808 ft |
| Mass | kg (kilogram) | 2.2051 lb |
| Moment of a force, torque | N · m (newton meter) | 0.7376 lb · ft |
| Moment of inertia of a plane area | m^4 (meter to fourth power) | 2.4025× 10^6 in.^4 |
| Moment of inertia of a mass | kg · m^2 (kilogram meter squared) | 0.7376 ft · s² |
| Power | W (watt) | 0.7376 ft · lb/s |
| kW (kilowatt) | 1.3410 hp | |
| Pressure or stress | Pa (pascal) | 0. 145× 10^{−3} psi |
| Specific weight | kN/m^3 (kilonewton per cubic meter) | 3.684×10^{−3}lb/in.^ 3 |
| Velocity | m/s (meter per second) | 3.2808 ft/s |
| Volume | m^3 (cubic meter) | 35.3147 ft³ |
| Work or energy | J (joule, newton meter) | 0.7376 ft · lb |
| Notes: 1 mile, mi=5280 ft=1609 m; 1 kilogram, kg=2.20946 lb=9.807 N; 1 joule, J=1 N · m; 1 inch, in.=25.4 mm; 1 foot, ft=12 in.=304.6 mm; 1 acceleration of gravity, g=9. 8066 m/ s²=32.174 ft/s². | ||