1 / 24

BTE 1013 ENGINEERING SCIENCEs

BTE 1013 ENGINEERING SCIENCEs. 5. MACHINES. NAZARIN B. NORDIN nazarin@icam.edu.my. What you will learn:. Simple machines Mechanical advantage / force ratio Movement ratio / velocity ratio Machine efficiency Gears: gear box ratio, gear efficiency Directions of rotations . Machines.

keahi
Download Presentation

BTE 1013 ENGINEERING SCIENCEs

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. BTE 1013 ENGINEERING SCIENCEs 5. MACHINES NAZARIN B. NORDIN nazarin@icam.edu.my

  2. What you will learn: • Simple machines • Mechanical advantage / force ratio • Movement ratio / velocity ratio • Machine efficiency • Gears: gear box ratio, gear efficiency • Directions of rotations

  3. Machines Is a device that can change magnitude or line of action, or both magnitude and line of action of a force. A simple machine usually amplifies an input force, call the effort , to give a larger output force, called the load. Examples; pulley systems, screw jacks, gear systems and lever systems.

  4. Machines

  5. Machines A machine is a device that receives energy in some available form and converts it to a useable form. For example, a person may wish to lift a weight of 2 tonnes; this is not possible unaided but the use of a jack or a hoist permits the person to achieve their objective. A lever such as a pry bar is an example of a simple machine. A relatively small manual force can be converted into a large force in order to lift or move an object.

  6. Levers Leverage and the use of levers occurs in the use of tools such as spanners, pry bars, pliers etc., and in many vehicle mechanisms such as clutch and brake pedals, throttle linkages and suspension units.

  7. Mechanical advantage The mechanical advantage of a machine is the ratio of the load to the effort: Lifting machines such as jacks and cranes have a large mechanical advantage so that relatively small manual forces can be used to raise heavy weights.

  8. Velocity ratio (movement ratio) The velocity ratio, VR, of a machine, or movement ratio, MR, is the ratio of the distance moved by the effort to the distance moved by the load:

  9. Efficiency of a machine The efficiency of a machine = (energy output/energy input) in the same time and this also

  10. Example problem In the trolley jack example shown before an effort of 250 newtons is lifting a load of 2 tonnes. In lifting the load through a distance of 15 cm the operator performs 40 pumping strokes of the handle each of which is 50 cm long. Calculate the mechanical advantage, velocity ratio and efficiency of the jack. MA = 785 MR =1333 The efficiency = 59%

  11. Gears Leverage and gears The gear ratio in this case is 4:1

  12. Gearbox Gear trains Ratio of gear set

  13. Problem 1 A simple machine raises a load of 160 kg through a distance of 1.6m. The effort applied to the machine is 200N and moves through a distance of 16m. Taking g as 9.8m/s2, determine the force ratio, movement ratio and efficiency of the machine.

  14. Solution 1 • Force ratio = Load/effort = 160kg/200N = (160 x 9.8)N/200N = 7.84 • Movement ratio = distance moved by effort/distance moved by the load = 16m/1.6m = 10 • Efficiency = (force ratio/movement ratio) x 100% = (7.84/10) x 100 = 78.4%

  15. Problem 2 For the simple machine of Problem 1, determine: (a) the distance moved by the effort to move the load through a distance of 0.9m, (b) the effort which would be required to raise a load of 200kg, assuming the same efficiency, (c) the efficiency if, due to lubrication, the effort to raise the 160kg load is reduced to 180N.

  16. Solution 2 • Distance moved by the effort = 10 x distance moved by the load = 10 x 0.9 = 9m • Since the force ratio is 7.84 effort = load/7.84 = (200 x 9.8) / 7.84 = 250N

  17. Solution 2 • The new force ratio load/effort = (160 x 9.8)/180 = 8.711 • The new efficiency after lubrication = (8.711/10) x 100 = 87.11%

  18. Problem 3 • A driver gear on a shaft of a motor has 35 teeth and meshes with a follower having 98 teeth. If the speed of the motor is 1400 revolutions per minute, find the speed of rotation of the follower.

  19. Solution 3 Speed of driver teeth on follower speed of follower teeth on driver 1400/speed of follower = 98/35 Speed of follower = (1400 x 35)/98 = 500 rev/min =

  20. Problem 4 • A compound gear train similar to that shown in last figure consists of a driver gear A, having 40 teeth, engaging with gear B, having 160 teeth. Attached to the same shaft as B, gear C has 48 teeth and meshes with gear D on the output shaft, having 96 teeth. Determine; • (a) the movement ratio of this gear system • (b) the efficiency ratio when the force ratio is 6

  21. Solution 4 • The speed of D = speed of A x TA/TB x TC/TD • movement ratio = NA/ND = TB/TA x TD/TC • = 160/40 x 96/48 • = 8 • (b) The efficiency = (force ratio/movement ratio)100% • = 6/8 x 100 • = 75%

  22. THANK YOU

More Related