prev

next

out of 21

View

3.823Download

13

Embed Size (px)

MECHANICS OF MACHINES DDA 3043POWER TRANSMISSION SYSTEM: GEAR SYSTEM1.1 Introduction to Gear SystemMechanical power transmission between shafts can be done in several ways. In most engineering practice, major power transmission used is gear system, belt drive, chain or rope drive. However, among these, gear system is the most efficient. The efficiency can go up to 98%. Gear system is efficient because it can perform high consistency of connection to produce high speed and load transfer with minimal noise of operation.It is analogical with belt drive system because belt drive is a flexible, easy to install way of power transmission mode. However, belt drives efficiency depends on the distance between the driver and driven pulley. The efficiency can be affected by belt drive slip, centrifugal effect and creep factor. Gears are used in many machines such as metal cutting machine tools, automobiles, hoists, rolling mill and so on.1.2 Types of Gear SystemThe function of gear is to transmit mechanical power from one shaft to another shaft with a certain speed ratio. Gear system parts include at least a set of gear that consist of Driver Gear and Driven Gear. Driver Gear is the gear that actuates power while Driven Gear is the gear that receives the power. A series of gear set is called Gear Train. Gear can be classified according to the relative position of the axes of mating gears. a) Parallel Axes ShaftThe shaft axes between driver and driven gear is parallel to each other. Example of this type of gear is Spur Gears and Helical Gears.1 Spur Gear Double Helical Gearb) Intersecting Axes ShaftThe shaft axes between driver and driven gear is perpendicular to each other. Example of gear is bevel gears. Bevel Gearc) Perpendicular Axes haftThe shaft axes between driver and driven gear are perpendicular to each other and do not intersect to each other. Example of gear is Worm Gear and Rack and Pinion Gear.2 Worm Gear Rack and Pinion Gear1.3 Relationship between Pitch Diameter and Pitch Circle Some of the important terminology of gear system is:Pitch Circle : An imaginary circle which by pure rolling action, would produce the same motion as the toothed gear wheel.Circular Pitch : The distance measured along the circumference of the pitch circle from a point on one tooth to the corresponding point on the adjacent tooth.Addendum Circle : Circle that limits the top of the teeth.Tooth Thickness : The width of the tooth measured along the pitch circle 3For both gear to mate efficiently, the pitch circle of both gear must be the same. Thus;2211NDNDp Thus 1212NNDD Where 1D = Diameter of driver gear2D = Diameter of driven gear1N = Number of teeth of driver gear2N = Number of teeth of driven gear1.4 Gear RatioConsider a gear set below; When two gear mate efficiently at point A, the velocity, vof both gear are the same. Thus;2 1 v v v with 2wDv Then from 2 22 2 1 1 D w D w will produce 1221DDww Where 1w = speed of driver gear2w =speed of driven gear4Driver gearDriven gearGear ratio nis defined as ratio of speed of driven gear with the speed of driver gear.12212112 NNDDwwnWhere 1 = angular acceleration of driver gear2 =angular acceleration of driven gear1.5 Gear Train Combination of gear wheels by means of which motion is transmitted from one shaft to another shaft is called Gear Train. In simple gear train, each shaft carries one gear only. Some gear trains consist of three gear that is driver gear, idler gear and driven gear. In Compound Gear Train, each shaft carries two wheels, except the first and the last. Simple gear train Compound gear train The idler gear doesnt affect the Gear Ratio of a gear system, but only affect the rotation of the driven gear. When the gear train is complex (consist of many gear sets), it is important for the designer to identify the rotation of the driver and the final driven gear respectively. However, there is a simple formula to determine the rotation of each successive gear in a gear train. 5For an ODD number of mating gears, the rotation of Driven gear is the SAME as Driver Gear.For an EVEN number of mating gears, the rotation of Driven gear is REVERSE of Driver Gear.Another classification of gear train is called Reverted Gear Train and Epicyclic Gear Train.1.6 Gear EfficiencyGear efficiency is defined as the ratio of Output Power from Driven Gear to the Input Power from Driver Gear. Gear efficiency measures how efficient a gear system is to transmit power. High value of gear efficiency reflects a more efficient gear system. Power loss in a gear system may come from sources like friction, slip, backlash and so on. From Power, T P , thenGear Efficiency, nTTTTPPG121 12 2122 1 , Where 1P = Input power from driver gear2P = Output power from driven gearn = Gear ratioIf the0 . 1 G, thus the torque at driver gear 1T is;2 1 ,21GnTT1.7 Power Transmission in a Gear Train SystemIn a gear train system, power loss normally happen in the bearing and gear due to friction and loading imposed on it and also power loss in overcoming shaft inertia. Consider a gear train consists of two sets of gear reducing arrangement. A motor is attached to the system with mIis 6the moment of inertia of motor shaft, TI is moment of inertia of middle shaft and GI is the moment of inertia of hoist which acts as the load of the system. Gear ratio and gear efficiency of gear set 1-2 is 2 / 1nand2 / 1 G, between gear set 3-4 is 4 / 3n and 4 / 3 G respectively. Let;mT = Torque of motorGT = Torque of hoistXT = Friction torque at bearing XDraw free body diagram and using Newton Second Law, I T Assume clockwise direction as positive value.For (A)m m m I T T 1 (1)For (B)T TI T T 3 2. (2)Since there is gear mating between gear 1 and 2, thus, must include in the analysis its own gear ratio and gear efficiency, and relate it to the inertia of middle shaft, TI .7HoistPreviously,2 / 1122 / 1 nTTG , thus it follows that 2 / 11 2 / 12n TT G.. (3)For (C)G G G X I T T T 4.. (4)also 4 / 34 / 3 34nTT G..(5)Using power, , T P power transfer to each gear component is;a) Power transfer by the motorm m m T P b) Power at gear 1( ) m m m m m I T T P 1 1c) Power at gear 22 / 1 1 2 GP P d) Power at gear 3( ) T T T T I T T P 2 3 3e) Power at gear 44 / 3 3 4 GP P f) Power at hoist( ) G G G X G G G I T T T P 4g) Overall power transfer efficiency, OmGOPP Thus if friction torque, XT effect is neglected, 8This concludes that 4 / 3 2 / 1 G GmGmTmGOPPPPPP

,`

.|

,`

.| Also; 4 / 3 2 / 1 n nTTmGO 1.8 Equivalent Moment of Inertia, equivIConsider a simple gear system as below Figure. In order for the driver gear A to start rotate, it must have enough torque to overcome its own inertia, AI first, and then another additional torque to start accelerate the driver gear B. However, to relate torque with the gear parameter, inertia term will be taken into account. For a simple gear system, the solution is straightforward, but when it comes to complex gear train design, it is useful to simplify / group together all inertia term in the system into a single compact inertia expression. The inertia term of each moving gear parts will be referred to a single part in the system, normally at motor side. 1. Torque at B to overcome BIB B B I T Refer B to gear A side. Use gear ratio,ABABn Thus, A B B n I T 2. Gear efficiency is related to power and thus torque of the mating gears, thus9Driven gear Driver gearIBIAABABGT n TPP 3. Therefore, torque at A, to accelerate BI( ) ( )GA BGA BGB BGBA n I n n InI n TT2 4. Therefore total torque at A to accelerate AI and BI isA A A TOTAL T I T + AGBA TOTAL n II T ]]]

+ 2, Or in general form, A equiv TOTAL I T (referred to motor side)Thus

,`

.|+ GBA equiv n II I2The derivation of equivI of this simple gear system can be extended to a double set of gear reducing problem as in section 1.7. By neglecting the friction torque effect, XT , thus,) )( () ( ) ( ) (4 / 3 , 2 / 1 ,24 / 322 / 12 / 1 ,22 / 1G GGGTm equiv n n I n II I + + 1.9 Gear Train Applications (Solved Problem)Example 1A motor is accelerating a 250 kg load with acceleration of 1.2 m/s2 through a gear system as shown below. The rope that carries the load are encircled on a hoist with diameter 1.2m.Gear for the hoists shaft has 200 teeth, gear for motor shaft has 20 teeth. Gear efficiency is 90%. Mass and radius of gyration of each shaft is as below;Mass (kg) Radius of gyration (mm)Motor shaft 250 100Hoist shaft 1100 50010Calculate the torque of the motor needed to bring up the load with acceleration 1.2 m/s2. Neglect friction effect. SolutionTotal torque at motor to bring up load2 1 M M total T T T + Where 1 MT = Torque to overcome equivalent inertia (refer to motor side).2 MT =Torque to accelerate the load through gear systema) Consider for 1 MTFrom GGM equiv n II I2+ Thus MI = Motor shaft inertia( ) 5 . 2 1 . 0 2502 2 mr IMkgm2GI= Hoist shaft inertia( ) 275 5 . 0 11002 GI kgm2Gear ratio, 1 . 02002021 NNnPut into ( )55 . 59 . 01 . 0 2755 . 22]]]

+ equivI kgm2 11Dia = 1.2 mHoistAcceleration of hoist, G G G r a Thus 26 . 02 . 1 G rad/sFrom the gear ratio, angular acceleration of motor, 201 . 02 nGm rad/sNow torque due to equivalent inertia, m equiv M I T 11 . 111 ) 20 ( 55 . 51 MT Nm.Then total torque referred to motor side is;5 . 183 1 . 1112 1+ + totalM M totalT T T T6 . 294 totalT NmExample 2Figure below shows a motor used to accelerate a hoist through two sets of gear reducing system. Moment of inertia for the motor shaft is 5 kgm2, middle shaft is 40 kgm2 and hoist shaft is 500 k