1. Topics: Mechanical Advantage Basic Machines Torque Power Efficiency Classes of Loads Service Factor Run-out and Overhung Load Note: Refer to chapter 3 in the textbook Mechanical Principles and Systems for Industrial Maintenance, Richard Knotek and Jon Stenerson, Prentice Hall, 2006 Some images in this presentation are taken from Wikimedia Commons, and are in the public domain. 1 MECH1200

2. Mechanical Advantage 2 A B Answer: MECH1200

3. Basic Machines There are six basic (simple) machines: 1. The lever 2. The wheel and axle 3. The pulley 4. The inclined plane 5. The wedge 6. The screw 3 MECH1200

4. The Simple Machines 4 The screw The wedge The inclined plane The lever The wheel and axleThe pulley MECH1200

5. The Three Classes of Lever Machine 5 Effort resistance First-class lever fulcrum Effort resistance Second-class lever fulcrum Third-class lever Effort resistance fulcrum MECH1200

6. The Pulley Combined pulleys increase mechanical advantage. Example 2: 6 1) What is the force that needs to be applied at the end of the rope to balance the weight? Solution: The force should be half of the weight as shown in the figure. 2) What is the mechanical advantage? Solution: MECH1200

7. Example 3: What is the force that needs to be applied at the end of the rope to balance the 100 lb weight? 7 Solution: The force should be 25% of the weight as shown in the figure. 2) What is the mechanical advantage? Solution: MECH1200

8. Torque The twisting effort around an axis. A vector quantity: It has direction and magnitude. The direction can be described as CW, CCW, about x,y, or z axis..etc. Torque = Force × Radius Some common units of torque: N.m ft.lb inch.lb 8 MECH1200

9. Power 9 MECH1200

10. Classes of Loads (see figures on pp.74-78) 1.Constant Torque Loads: the torque stays constant, and the speed varies. Therefore for these loads, as the speed increases, power consumed increases linearly. Example: A conveyer 2.Constant Power Loads: The power stays constant as the speed increases. Therefore, the torque is inversely proportional to the speed, because: Power = Torque x speed Thus: constant = Torque x speed, Thus: Torque = constant / speed Example: Lathe and milling machines 3.Variable Torque Loads: Torque requirement varies with speed. 10 MECH1200

11. Service Factor Service Factor: a multiplier that reflects the overall operating parameter that a motor will periodically function in without being damaged. Example: A class B induction motor is to be used in a piston compressor. It will be operated for 20 hours/day. a)What is the proper service factor (SF) for this application (Refer to the table in figure 3-10 on page 49)? b)What is the power required by the driven equipment if the motor nameplate power is 12 HP? Solution: a) SF = 1.4 b) Power requirement = Motor power x SF thus power requirement = 12 x 1.4 = 16.8 HP 11 MECH1200

12. Run-out and Overhung Load Run-out: a measure of eccentricity or roundness of an object. It can be radial or axial. It is measured with a dial indicator. multiplier that reflects the overall operating parameter that a motor will periodically function in without being damaged. Overhung Load (OHL): A force that is imposed on a shaft perpendicular to its axis. Examples that cause or increase overhung loads: 1.Shaft misalignment 2.Over tightened belts and chains 3.Load Location Factor (LF): Location of the load from the supporting bearing causes the load moment to increase. 12 MECH1200

13. Questions 13 MECH1200

14. 5. What is the mechanical advantage for each of the following pulley systems ? 14 Solution: 2 3 4 5 6 MECH1200