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# Design and assembly analysis of a gear train of a gear box report

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### Design and assembly analysis of a gear train of a gear box report

1. 1. DESIGN AND ASSEMBLY ANALYSIS OF A GEARTRAIN OF A GEAR BOX1. NOMENCLATUREP=powerdp=diameter of pinionα=helix angleBHN=brinel hardness numberE=youngs modulusEG=youngs modulus of gear materialEP=youngs modulus of pinion materialCompressive stress =Bending stress = aModule =mgear ratio = GRNo of teeth on gear =Diameter of gear =Diameter of pinion =Normal pitch =Normal pressure angle = фNpressure angle = фb=face widthEquivalent no of teeth on pinion =Equivalent no of teeth on gear =Y1=tooth form factorPeripheral speed = VDynamic tooth load =C = deformation factorStatic tooth load =K =load stress factorWw=wear loadX = over hangTorque =TEquivalent twisting moment =
2. 2. ABSTRACTDifferential is used when a vehicle takes a turn, the outer wheel on a longer radiusthan the inner wheel. The outer wheel turns faster than the inner wheel that is when there is arelative movement between the two rear wheels. If the two rear wheels are rigidly fixed to a rearaxle the inner wheel will slip which cause rapid tire wear, steering difficulties and poor loadholding.Differential is a part of inner axle housing assembly, which includes thedifferential rear axles, wheels and bearings. The differential consists of a system of gearsarranged in such a way that connects the propeller shaft with the rear axles.The following components consists the differential:1. Crown wheel and pinion.2. Sun gears3. Differential casingIn the present work all the parts of differential are designed under static condition andmodeled. The required data is taken from journal paper. Modeling and assembly is donein Pro/Engineer. The detailed drawings of all parts are to be furnished.The main aim of the project is to focus on the mechanical design and contactanalysis on assembly of gears in gear box when they transmit power at different speeds at2500 rpm, 5000 rpm. Presently used materials are Cast iron and Cast steel. For validatingdesign Structural Analysis is also conducted by varying the materials for gears, Cast Ironand of Aluminum Alloy.The analysis is conducted to verify the best material for the gears in the gear boxat higher speeds by analyzing stress, displacement and also by considering weightreduction.The analysis is done in Cosmos software. Modeling is done in the Pro/Engineer.
3. 3. 3. INTRODUCTION TO WHEELED VEHICLEDRIVE LINES3.1AXLES AND SUSPENSION SYSTEMSAutomotive drive lines and suspension systems have changed quite a bit since the firstautomobile was built. At first, automobile axles were attached directly to the main frame of thevehicle. This caused many problems. For example, the vehicle produced a very rough ride. Also,rigid construction did not work well on rough ground because sometimes one of the wheelswould not touch the ground. If the wheel off the ground was a drive wheel, the vehicle losttraction and stopped. This problem proved a need for a more flexible vehicle.The problem was corrected by using springs between the axles and the frame. The early springswere the same type used on the horse-drawn buggy. They allowed the wheels and axles to moveup and down separate from the body. The body moved very little compared to the wheels andaxles, and the ride was much smoother.Allowing the axles to move separate from the body also kept the wheels on the ground overrough roads, but this caused a new problem. The old drive train between the engine and the axlewould not work. The train had to be made to move more. This was done by adding movablejoints in the drive shaft known as universal joints. Some early vehicles used only one universaljoint on the drive shaft, while later vehicles used two universal joints on the drive shaft. Driveshafts are now usually called propeller shafts. Some long-wheel-base trucks now use as many asfour propeller shafts between the transmissionand the drive axle. These propeller shafts are connected by universal joints.Early automobiles were made up of a body, a power plant, and a running gear. The running gearwas made up of the wheels, axles, springs, drive shaft, and transmission. The transmission wasoften mounted midway between the engine and rear axle. It was connected to the engine and therear axle by drive shafts.The term "running gear" is not used any more. A new term, "chassis," is now used to identify theold running gear plus the power plant. In modern vehicles, the transmission is generally mountedon the engine and is part of the power plant.The chassis of modern vehicles, especially the frame, spring, and axles, must be very strong andyet quite flexible.
4. 4. 3.2 INTRODUCTION TO AXLE SHAFTAn axle is a central shaft for a rotating wheel or gear. On wheeled vehicles, the axle may befixed to the wheels, rotating with them, or fixed to its surroundings, with the wheels rotatingaround the axle. In the former case, bearings or bushings are provided at the mounting pointswhere the axle is supported. In the latter case, a bearing or bushing sits inside the hole in thewheel to allow the wheel or gear to rotate around the axle.Axles are an integral component of a wheeled vehicle. In a live-axle suspension system, the axlesserve to transmit driving torque to the wheel, as well as to maintain the position of the wheelsrelative to each other and to the vehicle body. The axles in this system must also bear the weightof the vehicle plus any cargo. A non-driving axle, such as the front beam axle in Heavy dutytrucks and some 2 wheel drive light trucks and vans, will have no shaft. It serves only as asuspension and steering component. Conversely, many front wheel drive cars have a solid rearbeam axle.In other types of suspension systems, the axles serve only to transmit driving torque to thewheels; The position and angle of the wheel hubs is a function of the suspension system. This istypical of the independent suspension found on most newer cars and SUVs, and on the front ofmany light trucks. These systems still have a differential, but it will not have attached axlehousing tubes. It may be attached to the vehicle frame or body, or integral in a transaxle. Theaxle shafts (usually C.V. type) then transmit driving torque to the wheels. Like a full floatingaxle system, the shafts in an independent suspension system do not support and vehicle weight."Axle" in reference to a vehicle also has a more ambiguous definition, meaning parallel wheelson opposing sides of the vehicle, regardless of their mechanical connection type to each otherand the vehicle frame or body.3.2.1 Types of Rear Axle ShaftsIn rear wheel drive vehicles, the rear wheels are the driving wheels, whereas in the vehicles withfront wheels drive the front wheels are the driving wheels. Almost all the rear axles in themodern cars are live axles, which means that these axles move with the wheels, or revolve withthe wheels and are known as live axles. Dead Axles are those axles which remain stationary anddo not move with the wheels.Rear axles / Live Axles are further classified into three types:1. Full Float axles2. Semi Float Axles3. Three quarter floating axles.
5. 5. 3.2.2 Semi float axle:The Semi float axle is used in light trucks and passenger vehicle / buses. In the vehicles equippedwith Semi Float axle the shaft as well as the differential housing supports the weight of thevehicle . The wheel hub is directly connected to the axle shaft or is an extension of the same, theinner end of the axle shaft is splined and it is supported by the final drive unit. The outer end issupported by a single bearing inside the axle casing / axle tube. The vehicle load is transmitted toeach of the axle shafts through the casing and the bearing, this causes a bending load and atendency to shear at a point. Besides the side forces also cause end thrust and bending moment inthe axle shafts, which have to take driving torque also. The semi float axle is the simplest and thecheapest of all types, because of which it is widely used on cars. However, since axle shafts haveto support all loads, they have to be of larger diameter for the same torque transmitted to theother types of axle supporting. The axle shafts take the stress caused by turning, skidding orwobbling of the wheels. The axle shafts are flanged or tapered on the ends. When the taperedaxles are used, the brake , the brake drum and hub are pressed onto the shafts, using keys toprevent the assemblies from turning on the shafts. In some cases, the outer ends of the shaftsmay have serrations or splines to correspond with those on the drum and hub assembly. Ifin case the axle shaft breaks on the vehicle using this type of arrangement, the wheel of thevehicle will get separated from the vehicle.3.2.3 Full float axle:Full Float Axle is considered as a robust one and is used for heavy vehicles / trucks meant tocarry heavy loads. The axle shaft has flanges at the outer ends, which are connected to theflanged sleeve by means of bolts. There are two taper roller bearings supporting the axle castingin the hub, which take up any side load. Thus in case of Full Float axles , the axle shafts carryonly the driving torque. The weight of the vehicle and the end thrust are not carried by them. Theweight of the vehicle is completely supported by the wheels and the axle casing. As the axleshafts carry only the driving torque, their failure or removal does not effect the wheels. Thus theaxle shafts can be taken out or replaced without jacking up the vehicle. For the same reasonvehicle can be towed even with a broken half shaft. We can say that the axle shaft takes thewhole weight of the vehicle and absorbs all types of stresses or end thrust caused by turning,skidding, and pulling. Full Float axle is considered as the most heavy and costly axle.3.2.4 Three quarter floating axle:This is a compromise between the more robust full float axle and the simplest semi float type ofaxle. In Semi Floating axle the bearing is located between the axle casing and the hub instead ofbeing between the axle casing and the shaft as in case of semi float axle. The axle shafts do nothave to withstand any shearing or bending actions due to the weight of the vehicle, which aretaken up by the axle casing through the hub and the bearing, provided the bearing lies in theplane of the road wheel. However, it has to take the end loads and the driving torque. Earlier,Three quarter floating axles were much popular for cases and lighter commercial vehicles, butwith the passage of time and with more improvements in the design, development, materials andfabrication techniques, preference is given to the Semi Float Axles, as these are simpler in designand cheaper to use.
6. 6. 4.LITERATURE SURVEY4.1 INTRODUCTION TO GEAR BOXA transmission or gearbox provides speed and torque conversions from a rotating power sourceto another device using gear ratios. In British English the term transmission refers to the wholedrive train, including gearbox, clutch, prop shaft (for rear-wheel drive), differential and finaldrive shafts. In American English, however, the distinction is made that a gearbox is any devicewhich converts speed and torque, whereas a transmission is a type of gearbox that can be"shifted" to dynamically change the speed: torque ratio, such as in a vehicle. The most commonuse is in motor vehicles, where the transmission adapts the output of the internal combustionengine to the drive wheels. Such engines need to operate at a relatively high rotational speed,which is inappropriate for starting, stopping, and slower travel. The transmission reduces thehigher engine speed to the slower wheel speed, increasing torque in the process. Transmissionsare also used on pedal bicycles, fixed machines, and anywhere else rotational speed and torqueneeds to be adapted.Often, a transmission will have multiple gear ratios (or simply "gears"), with the ability to switchbetween them as speed varies. This switching may be done manually (by the operator), orautomatically. Directional (forward and reverse) control may also be provided. Single-ratiotransmissions also exist, which simply change the speed and torque (and sometimes direction) ofmotor output.In motor vehicle applications, the transmission will generally be connected to the crankshaft ofthe engine. The output of the transmission is transmitted via driveshaft to one or moredifferentials, which in turn drive the wheels. While a differential may also provide gearreduction, its primary purpose is to change the direction of rotation.Conventional gear/belt transmissions are not the only mechanism for speed/torque adaptation.Alternative mechanisms include torque converters and power transformation (e.g., diesel-electrictransmission, hydraulic drive system, etc.). Hybrid configurations also exist.4.2 ExplanationEarly transmissions included the right-angle drives and other gearing in windmills, horse-powered devices, and steam engines, in support of pumping, milling, and hoisting.Most modern gearboxes are used to increase torque while reducing the speed of a prime moveroutput shaft (e.g. a motor crankshaft). This means that the output shaft of a gearbox will rotate atslower rate than the input shaft, and this reduction in speed will produce a mechanical advantage,causing an increase in torque. A gearbox can be setup to do the opposite and provide an increasein shaft speed with a reduction of torque. Some of the simplest gearboxes merely change thephysical direction in which power is transmitted.
7. 7. Many typical automobile transmissions include the ability to select one of several different gearratios. In this case, most of the gear ratios (often simply called "gears") are used to slow downthe output speed of the engine and increase torque. However, the highest gears may be"overdrive" types that increase the output speed.4.3 UsesGearboxes have found use in a wide variety of different—often stationary—applications, such aswind turbines.Transmissions are also used in agricultural, industrial, construction, mining and automotiveequipment. In addition to ordinary transmission equipped with gears, such equipment makesextensive use of the hydrostatic drive and electrical adjustable-speed drives.4.4 SimpleThe simplest transmissions, often called gearboxes to reflect their simplicity (although complexsystems are also called gearboxes in the vernacular), provide gear reduction (or, more rarely, anincrease in speed), sometimes in conjunction with a right-angle change in direction of the shaft(typically in helicopters). These are often used on PTO-powered agricultural equipment, sincethe axial PTO shaft is at odds with the usual need for the driven shaft, which is either vertical (aswith rotary mowers), or horizontally extending from one side of the implement to another (aswith manure spreaders, flail mowers, and forage wagons). More complex equipment, such assilage choppers and snow blowers, have drives with outputs in more than one direction.The gearbox in a wind turbine converts the slow, high-torque rotation of the turbine into muchfaster rotation of the electrical generator. These are much larger and more complicated than thePTO gearboxes in farm equipment. They weigh several tons and typically contain three stages toachieve an overall gear ratio from 40:1 to over 100:1, depending on the size of the turbine. (Foraerodynamic and structural reasons, larger turbines have to turn more slowly, but the generatorsall have to rotate at similar speeds of several thousand rpm.) The first stage of the gearbox isusually a planetary gear, for compactness, and to distribute the enormous torque of the turbineover more teeth of the low-speed shaft. Durability of these gearboxes has been a serious problemfor a long time.Regardless of where they are used, these simple transmissions all share an important feature: thegear ratio cannot be changed during use. It is fixed at the time the transmission is constructed.4.5 Multi-ratio systemsMany applications require the availability of multiple gear ratios. Often, this is to ease thestarting and stopping of a mechanical system, though another important need is that ofmaintaining good fuel efficiency.
8. 8. 4.6 Automotive basicsThe need for a transmission in an automobile is a consequence of the characteristics of theinternal combustion engine. Engines typically operate over a range of 600 to about 7000revolutions per minute (though this varies, and is typically less for diesel engines), while the carswheels rotate between 0 rpm and around 1800 rpm.Furthermore, the engine provides its highest torque outputs approximately in the middle of itsrange, while often the greatest torque is required when the vehicle is moving from rest ortraveling slowly. Therefore, a system that transforms the engines output so that it can supplyhigh torque at low speeds, but also operate at highway speeds with the motor still operatingwithin its limits, is required. Transmissions perform this transformation.Many transmissions and gears used in automotive and truck applications are contained in a castiron case, though more frequently aluminum is used for lower weight especially in cars. Thereare usually three shafts: a main shaft, a countershaft, and an idler shaft.The main shaft extends outside the case in both directions: the input shaft towards the engine,and the output shaft towards the rear axle (on rear wheel drive cars- front wheel drives generallyhave the engine and transmission mounted transversely, the differential being part of thetransmission assembly.) The shaft is suspended by the main bearings, and is split towards theinput end. At the point of the split, a pilot bearing holds the shafts together. The gears andclutches ride on the main shaft, the gears being free to turn relative to the main shaft except whenengaged by the clutches. Types of automobile transmissions include manual, automatic or semi-automatic transmission.4.7 Differential gear boxA differential is a device, usually but not necessarily employing gears, capable of transmittingtorque and rotation through three shafts, almost always used in one of two ways: in one way, itreceives one input and provides two outputs—this is found in most automobiles—and in theother way, it combines two inputs to create an output that is the sum, difference, or average, ofthe inputs.In automobiles and other wheeled vehicles, the differential allows each of the driving roadwheelsto rotate at different speeds, while for most vehicles supplying equal torque to each of them.4.7.1 PurposeA vehicles wheels rotate at different speeds, mainly when turning corners. The differential isdesigned to drive a pair of wheels with equal torque while allowing them to rotate at differentspeeds. In vehicles without a differential, such as karts, both driving wheels are forced to rotateat the same speed, usually on a common axle driven by a simple chain-drive mechanism. Whencornering, the inner wheel needs to travel a shorter distance than the outer wheel, so with nodifferential, the result is the inner wheel spinning and/or the outer wheel dragging, and this
9. 9. results in difficult and unpredictable handling, damage to tires and roads, and strain on (orpossible failure of) the entire drivetrain.4.7.2Functional descriptionThe following description of a differential applies to a "traditional" rear-wheel-drive car or truckwith an "open" or limited slip differential:Torque is supplied from the engine, via the transmission, to a drive shaft (British term: propellershaft, commonly and informally abbreviated to prop-shaft), which runs to the final drive unitthat contains the differential. A spiral bevel pinion gear takes its drive from the end of thepropeller shaft, and is encased within the housing of the final drive unit. This meshes with thelarge spiral bevel ring gear, known as the crown wheel. The crown wheel and pinion may meshin hypoid orientation, not shown. The crown wheel gear is attached to the differential carrier orcage, which contains the sun and planet wheels or gears, which are a cluster of four opposedbevel gears in perpendicular plane, so each bevel gear meshes with two neighbours, and rotatescounter to the third, that it faces and does not mesh with. The two sun wheel gears are aligned onthe same axis as the crown wheel gear, and drive the axle half shafts connected to the vehiclesdriven wheels. The other two planet gears are aligned on a perpendicular axis which changesorientation with the ring gears rotation. In the two figures shown above, only one planet gear(green) is illustrated, however, most automotive applications contain two opposing planet gears.Other differential designs employ different numbers of planet gears, depending on durabilityrequirements. As the differential carrier rotates, the changing axis orientation of the planet gearsimparts the motion of the ring gear to the motion of the sun gears by pushing on them rather thanturning against them (that is, the same teeth stay in the same mesh or contact position), butbecause the planet gears are not restricted from turning against each other, within that motion,the sun gears can counter-rotate relative to the ring gear and to each other under the same force(in which case the same teeth do not stay in contact).Thus, for example, if the car is making a turn to the right, the main crown wheel may make 10full rotations. During that time, the left wheel will make more rotations because it has further totravel, and the right wheel will make fewer rotations as it has less distance to travel. The sungears (which drive the axle half-shafts) will rotate in opposite directions relative to the ring gearby, say, 2 full turns each (4 full turns relative to each other), resulting in the left wheel making 12rotations, and the right wheel making 8 rotations.The rotation of the crown wheel gear is always the average of the rotations of the side sun gears.This is why, if the driven roadwheels are lifted clear of the ground with the engine off, and thedrive shaft is held (say leaving the transmission in gear, preventing the ring gear from turninginside the differential), manually rotating one driven roadwheel causes the opposite roadwheel torotate in the opposite direction by the same amount.When the vehicle is traveling in a straight line, there will be no differential movement of theplanetary system of gears other than the minute movements necessary to compensate for slightdifferences in wheel diameter, undulations in the road (which make for a longer or shorter wheelpath), etc.
11. 11. differential, it is better than a simple mechanical open differential with no electronic tractionassistance.A differential gearbox is a completely mechanical device that consist of an enclosure, some gearsand shafts. It has three input/output shafts (I will call them x, y and z shaft). I said input/outputbecause any of these three shafts can serve as either input or output shaft – but in any givenmoment at least one shaft must be output and at least one shaft must be input.Its main purpose is to sum or differentiate rotation rates (N) applied on its shafts, whilemaintaining constant torque (T) ratio between shafts at any given moment. In general we cansay:Where:- Nx, Ny and Nz are rotation rates [rotations per second] for axes x, y and z- Tx, Ty and Tz are torques [Newton-meter] on axes x, y and z- A, B, and C are constants – they depend on gear ratios inside the differential gearbox (notethat only two constants are actually needed, the third one is redundant)If two shafts are used as input shafts then the gearbox will add their rotation rates (usuallyweighted by some constants) and will deliver the result to third output shaft. On the other hand,if only single shaft is used as an input shaft, the gearbox can split its rotation rate to two othershafts (also usually weighted by some constants) – but note that the percentage of this ―rotationrate splitting‖ doesn’t have to have 50:50 or any other fixed ratio. Instead, this ratio is leveled tomaintain constant torque ratio between shafts (this works especially nice if applied loads havesuch characteristic that increased rotation rate increases the torque).The torque ratio between any two shafts is always maintained constant (a normal, two-shaftedgearbox behaves the same). The torque on all differential gearbox shafts can be increased ordecreased only simultaneously and proportionally. For example, if your car is driving along aroad and then one wheel steps into a mud, the increased drag will be simultaneously felt (in formof increased torque) also by both, the other wheel and the engine. But if the mud is so slipperythat your wheel loses its grip, then reduced torque will also be felt by the other wheel and by theengine. One wheel may spin in the mud, but the reduced torque on the other wheel will not beenough to pull you out.
12. 12. Three output shafts, together with belonging gears, are colored blue, yellow and green. Thesimplest is the blue (x) shaft . It only has one gear directly attached to it.Somewhat more complex is the yellow (y) shaft. The yellow gear attached to it drives (or isdriven by) another yellow double-gear (piggyback). The piggyback gear rotates around the greenshaft (it is not fixed to the green shaft). Otherwise there is no difference between blue gear andyellow piggyback gear – they both work symmetrically inside the differential gearbox.The most complex one is the green (z) shaft and its two green gears. These two gears aremounted to short stubs that go vertically from the shaft. As the shaft rotates both gears rotate allaround, carried by the stubs. In addition, every of these two gears can also rotate around the stubit is mounted on.The design shown below is invented to make things simpler. It is used sometimes in industry (forsmall gearboxes used in servo systems).There are (only) two gears, the blue one and the green one. The green one is special - it issomewhat flexible. Streched by the yellow axle from inside, the flexible green gear touches
13. 13. (becomes meshed with) the blue gear. The yellow axle can rotate inside the green ring-like gearcausing it to touch the blue gear at different positions, 360 degrees around.The thing is that the blue gear and the green gear dont have the same number of teeth. Forexample the green could have 100, while the blue could have 101 teeth. After the yellow axle isrotated for one revolution, the green and the blue will travel one tooth relative to each other.(The above picture is only 2-D crossection of such differential mechanism. In reality, the greengear must be very much extruded deeply into your monitor. It looks like a cup - the bottom endis non-deformable and its shaft is atteched there.)Sure, there are many disadvantages of such differential gearbox mechanism. Teeth on blue andgreen gears must be relatively small (there must be many of them), and the difference betweennumber of teeth must be small. Due to deformation of the green gear, teeth become non-idealy-shaped. Also, heat is disipated when material is deformed.4.7.4 USE OF DIFFERENTIALThe differential has three jobs:To aim the engine power at the wheelsTo act as the final gear reduction in the vehicle, slowing the rotational speed of thetransmission one final time before it hits the wheelsTo transmit the power to the wheels while allowing them to rotate at different speeds(This is the one that earned the differential its name.)4.7.4.1Automobile Differential Gear TrainThe gearing of an automobile differential is illustrated as following in final form.
14. 14. Automobile Differential Complete SchematicWithout the "square" set of four gears in the middle of the above diagram which yields to thefigure below, both wheels turn at the same angular velocity. This leads to problems when the carnegotiates a turn.Automobile Wheel Drive without DifferentialNow imagine the differential "square" alone, as illustrated in the following figure. It should beapparent that turning one wheel results in the opposite wheel turning in the opposite direction at
15. 15. the same rate.Automobile Differential AloneThis is how the automobile differential works. It only comes into play when one wheel needs torotate differentially with respect to its counterpart. When the car is moving in a straight line, thedifferential gears do not rotate with respect to their axes. When the car negotiates a turn,however, the differential allows the two wheels to rotate differentially with respect to each other.One problem with an automotive differential is that if one wheel is held stationary, thecounterpart wheel turns at twice its normal speed as can be seen by examining the completescheme of automobile differential. This can be problematic when one wheel does not haveenough traction, such as when it is in snow or mud. The wheel without traction will spin withoutproviding traction and the opposite wheel will stay still so that the car does not move. This is thereason for a device known as a "limited slip differential" or "traction control".4.7.4.2 GEAR BOX CASINGGearbox casing is the shell (metal casing) in which a train of gears is sealed. The gearboxcasing houses important transmission components like gears and shafts. Thus the strength of thegear box casing is an important parameter to be taken in to account while designing. The strengthof the gear box casing is important since the complete power is subjected to both static anddynamic loading.4.7.4.3 BEVEL GEARSBevel gears are gears where the axes of the two shafts intersect and the tooth-bearing faces of thegears themselves are conically shaped. Bevel gears are most often mounted on shafts that are 90degrees apart, but can be designed to work at other angles as well. The pitch surface of bevelgears is a cone.4.7.4.3.1 ApplicationsThe bevel gear has many diverse applications such as locomotives, marine applications,automobiles, printing presses, cooling towers, power plants, steel plants, railway track inspectionmachines, etc.
16. 16. For examples, see the following articles on:Bevel gears are used in differential drives, which can transmit power to two axlesspinning at different speeds, such as those on a cornering automobile.Bevel gears are used as the main mechanism for a hand drill. As the handle of the drill isturned in a vertical direction, the bevel gears change the rotation of the chuck to ahorizontal rotation. The bevel gears in a hand drill have the added advantage ofincreasing the speed of rotation of the chuck and this makes it possible to drill a range ofmaterials.The gears in a bevel gear planer permit minor adjustment during assembly and allow forsome displacement due to deflection under operating loads without concentrating the loadon the end of the tooth.Spiral bevel gears are important components on rotorcraft drive systems. Thesecomponents are required to operate at high speeds, high loads, and for a large number ofload cycles. In this application, spiral bevel gears are used to redirect the shaft from thehorizontal gas turbine engine to the vertical rotor.4.7.4.3.2 AdvantagesThis gear makes it possible to change the operating angle.Differing of the number of teeth (effectively diameter) on each wheel allows mechanicaladvantage to be changed. By increasing or decreasing the ratio of teeth between the driveand driven wheels one may change the ratio of rotations between the two, meaning thatthe rotational drive and torque of the second wheel can be changed in relation to the first,with speed increasing and torque decreasing, or speed decreasing and torque increasing.4.7.4.3.3 DisadvantagesOne wheel of such gear is designed to work with its complementary wheel and no other.Must be precisely mounted.The axes must be capable of supporting significant forces.
17. 17. 5. GEAR DIFFERENTIALA gear differential is a mechanism that is capable of adding and subtracting mechanically. Tobe more precise, we should say that it adds the total revolutions of two shafts. It alsosubtracts the total revolutions of one shaft from the total revolutions of anothershaft—and delivers the answer by a third shaft. The gear differential will continuously andaccurately add or subtract any number of revolutions. It will produce a continuous series ofanswers as the inputs change. Figure 11-8 is a cutaway drawing of a bevel gear differentialshowing all of its parts and how they relate to each other. Grouped around the center of themechanism are four bevel gears meshed together. The two bevel gears on either side are ―endgears.‖ The two bevel gears above and below are ―spider gears.‖ The long shaft running throughthe end gears and the three spur gears is the ―spider shaft.‖ The short shaft running through thespider gears together with the spider gears themselves make up the ―spider.‖ Each spidergear and end gear is bearing-mounted on its shaft and is free to rotate. The spider shaft connectswith the spider cross shaft at the center block where they intersect. The ends of the spider shaftare secured in flanges or hangers. The spider cross shaft and the spider shaft are also bearing-mounted and are free to rotate on their axis. Therefore, since the two shafts are rigidly connected,the spider (consisting of the spider cross shaft and the spider gears) must tumble, or spin, on theaxis of the spider shaft.5.1 Bevel Gear Differential
18. 18. 6. AIM OF THE PROJECTThe main aim of the project is to focus on the mechanical design and contactanalysis on assembly of gears in gear box when they transmit power at different speeds at2400 rpm, 5000 rpm. Analysis is also conducted by varying the materials for gears, CastIron, and Aluminum Alloy.The analysis is conducted to verify the best material for the gears in the gear boxat higher speeds by analyzing stress, displacement and also by considering weightreduction.Design calculations are done on the differential of Ashok leyland 2516M byvaring materials and speeds. Differential gear is modeled in Pro/Engineer. Analysis isdone on the differential by applying tangential and static loads. Freauency analysis is alsodone on the differential. Analysis is carried out using CosmosWorks.7.DESIGN CALCULATIONS OF DIFFERENTIAL7.2ALUMINUM ALLOY7475-T7617.2.1 2400 rpm7.2.1.1.1 Crown wheelDiameter of crown wheel = DG= 475mmNumber of teeth on gear = TG = 50Number of teeth on pinion = TP = 8Module = m=DG/TG=475/50=9.5=10(according to stds)Diameter of pinion = DP = m x TP=10x8=80mmMaterial used for both pinion and gear is aluminum alloy7475-T761Brinell hardness number(BHN)=140Pressure angle of teeth is 20° involute system Ø=20°P=162BHP = 162x745.7w=120803.4wWe know that velocity ratioV.R=TG/ TP= DG/DP= NP/NGV.R=TG/ TP=50/8=6.25V.R= NP/NG
19. 19. 6.25=2400/ NGNG=384rpmFor satisfactory operation of bevel gears the number of teeth in the pinion must not beLess than where v.r=velocity ratio= =7.5Since the shafts are at right angles therefore pitch angle for the pinionθp1=tan-1(1/v.r)= tan-1(1/6.25)=9.0Pitch angle of gear θp2=90°-9=81We know that formative number of teeth for pinionTEP= TPsec θp1=8sec9 =8And formative number of teeth for gearTEG= TGsec θp2=50sec81 =319.622Tooth form factor for the piniony1P=0.154-0.912/ TEP, for 20° full depth involute system=0.154-0.912/ 8=0.04And tooth form factor for geary1G=0.154-0.912/ TEG=0.154-0.912/ 319.622=0.151since the allowable static stresses(σO) for both pinion and gear is same (i.e σO=172.33 Mpa) andy1P is less than y1G, therefore the pinion is weaker. Thus the design should be based upon thepinionAllowable static stress(σO) =σu/3=517/3=172.33 Mpaσu=ultimate tensile strength=517 MpaTANGENTIAL TOOTH LOAD(WT)WT = ( σO x Cv).b.Π.m. y1P((L-b)/L)Cv=velocity factor =3/3+v, for teeth cut by form cuttersv=peripheral speed in m/sb=face widthm=module=10
20. 20. y1p=tooth form factorL=slant height of pitch coneDG= pitch diameter of gear =475DP= pitch diameter of gear =80V==10.048m/sCv==3/3+10.048=0.229L==240.844The factor (L-b/L) may be called as bevel factorFor satisfactory operation of the bevel gears the face width should be from 6.3m to 9.5mSo b is taken as 9.5mb= 9.5x10=95WT =(172.33x0.229)x95xΠx10x0.04( ) = 2922.51NDYNAMIC LOAD(WD)The dynamic load for bevel gears may be obtained in similar manner as discussed for spur gearsWD= WT+ WIWD =WT+V= pitch line velocityb=face widthc=deformation/dynamic factor in N/mmC=K=0.111 for 20° full depth involute systemEP=young’s modulus for material of pinion in N/mm2=70300 N/mm2EG=young’s modulus for material of gear in N/mm2= 70300N/mm2e=tooth error action in mme value for module=10 used in precision gears is e=0.023c== =90N/mmWD =WT+WD = 2922.51+
21. 21. WD =10532.23NSTATIC TOOTH LOAD (WS)The static tooth load or endurance strength of the tooth for bevel gear is given byWS= σe.b.Π.m. y1P( )(Flexible endurance limit) σe = 1.75XB.H.N = 1.75X140=245WS= ( )WS=18145NFor safety against tooth breakage the WS ≥1.25 Wd=13165.2875WEAR LOAD (WW)The maximum or limiting load for wear for bevel gears is given byWw=Dp,b,q,k have usual meanings as discussed in spur gears except that Q is based on formative orequivalent no.of teeth such that ratio factor Q= = =1.951K= load stress factor (also known as material combination factor )in N/mm2given byK= )σes= surface endurance limit in mpa or N/mm2σes=(2.8×B.H.N-70)N/mm2= (2.8×517-70)=322 N/mm2K= =0.72=cos 9 =0.987Ww= =10825.25NForces actingWT=WNCOSWN=normal load=WT/ COSWT=tangential forceWN= = =3110.070NRadial force WR=Mean radius (Rm)=(L-b/2)=(240.844-95/2)=32.111
22. 22. Axial force acting on the pinion shaftWRH=WT P1Tangential force acting at the mean radiusWT=T/RmT= torque on the pinionT ==T=480.661N-m=480661N-mmWT=480661/32.111=14968.733NAxial forceWRH=WT P1=14968.733=850.010NRadial force acting on the pinion shaftWRH=WT P1=14968.733=5366.752N7.2.1.1.2 SUN GEARDiameter of crown wheel = DG= 150mmDiameter of pinion = DP =70mmNumber of teeth on gear = TG =18Number of teeth on pinion = TP = 15D= DG+ DP=220T= TG+ TP = 33Module = m=D/T=220/33=6.66=7(according to stds)Material used for both pinion and gear is aluminum alloy7475-t761Brinell hardness number(BHN)=140Pressure angle of teeth is 20° involute system Ø=20°P=162BHP = 162x745.7w=120803.4wWe know that velocity ratioV.R=TG/ TP= DG/DP= NP/NGV.R= DG/DP =150/70=2.142V.R= NP/NG
23. 23. 2.142=2400/ NGNG=1120.448rpmSince the shafts are at right angles therefore pitch angle for the pinionθp1=tan-1(1/v.r)= tan-1(1/2.142)=25.025Pitch angle of gear θp2=90°-25.025=64.974We know that formative number of teeth for pinionTEP= TPsec θp1=15sec25.025 =16.554And formative number of teeth for gearTEG= TGsec θp2=18sec64.974 =42.55Tooth form factor for the piniony1P=0.154-0.912/ TEP, for 20° full depth involute system=0.154-0.912/ 16.554=0.099And tooth form factor for geary1G=0.154-0.912/ TEG=0.154-0.912/ 42.55=0.132Since the allowable static stresses(σO) for both pinion and gear is same (i.e σO=172.33 Mpa) andy1P is less than y1G, therefore the pinion is weaker. Thus the design should be based upon thepinionallowable static stress(σO) =σu/3=517/3=172.33 Mpaσu=ultimate tensile strength=517 MpaTANGENTIAL TOOTH LOAD(WT)WT =( σO x Cv).b.Π.m. y1P((L-b)/L)Cv=velocity factor =3/3+v, for teeth cut by form cuttersv=peripheral speed in m/sb=face widthm=module=7y1p=tooth form factorL=slant height of pitch coneDG= pitch diameter of gear =150mmDp= pitch diameter of gear =70mmV=
24. 24. ==8.792m/sCv==3/3+8.792=0.254L==82.764The factor (L-b/L) may be called as bevel factorFor satisfactory operation of the bevel gears the face width should be from 6.3m to 9.5mSo b is taken as 9.5mb= 9.5x7=66.5WT =(172.33x0.254)x66.5xΠx7x0.099( )=1244.7NDYNAMIC LOAD (WD)The dynamic load for bevel gears may be obtained in similar manner as discussed for spur gearsWD= WT+ WIWD =WT+V= pitch line velocityb=face widthc=deformation/dynamic factor in N/mmC=K=0.111 for 20° full depth involute systemEP=young’s modulus for material of pinion in N/mm2EG=young’s modulus for material of gear in N/mm2e=tooth error action in mme value for module=7 used in precision gears is e=0.0186c= = 72.57N/mmWD =WT+WD = 1244.7+WD = 5513.77STATIC TOOTH LOAD (WS)The static tooth load or endurance strength of the tooth for bevel gear is given byWS= σe.b.Π.m. y1P( )(Flexible endurance limit) σe = 1.75XB.H.N = 1.75X140 =245
25. 25. WS= ( )WS=6966.47NFor safety against tooth breakage the WS ≥1.25 Wd=6892.2125WS > WdWEAR LOAD (WW)The maximum or limiting load for wear for bevel gears is given byWw=Dp,b,q,k have usual meanings as discussed in spur gears except that Q is based on formative orequivalent no.of teeth such that ratio factor Q= = =1.439K= load stress factor (also known as material combination factor )in N/mm2given byK= )σes= surface endurance limit in mpa or N/mm2σes=(2.8×B.H.N-70)N/mm2= (2.8×140-70)=322 N/mm2K= =0.72=cos 25.025 =0.906Ww= =5322.62NForces actingWT=WNCOSWN=normal load=WT/ COSWT=tangential forceWN= = =1324.582NRadial force WR=WT tanMean radius (Rm)=(L-b/2)=(82.764-66.5/2)=20.944NAxial force acting on the pinion shaftWRH=WT P1
26. 26. Tangential force acting at the mean radiusWT=T/RmT= torque on the pinionT ==T=480.661N-m=480661N-mmWT=480661/20.944=22949.818NAxial forceWRH=WT P1=22949.818=3533.340NRadial force acting on the pinion shaftWRH=WT P1=22949.818=7567.863N7.2.2 5000 rpm7.2.2.1.1 Crown wheelDiameter of crown wheel = DG= 475mmNumber of teeth on gear = TG = 50Number of teeth on pinion = TP = 8Module = m=DG/TG=475/50=9.5=10(according to stds)Diameter of pinion = DP = mx TP= 10x8 =8 0mmMaterial used for both pinion and gear is aluminum alloy7475-T761Brinell hardness number(BHN)= 140Pressure angle of teeth is 20° involute system Ø=20°P=337.5BHP = 337.5x745.7w=251673.75wWe know that velocity ratioV.R=TG/ TP= DG/DP= NP/NGV.R=TG/ TP=50/8=6.25V.R= NP/NG6.25=5000/ NGNG=800rpm
27. 27. For satisfactory operation of bevel gears the number of teeth in the pinion must not beLess than where v.r=velocity ratio= =7.5Since the shafts are at right angles therefore pitch angle for the pinionθp1=tan-1(1/v.r)= tan-1(1/6.25)=9.0Pitch angle of gear θp2=90°-9=81We know that formative number of teeth for pinionTEP= TPsec θp1=8sec9 =8And formative number of teeth for gearTEG= TGsec θp2=50sec81 =319.622Tooth form factor for the piniony1P=0.154-0.912/ TEP, for 20° full depth involute system=0.154-0.912/ 8=0.04And tooth form factor for geary1G=0.154-0.912/ TEG=0.154-0.912/ 319.622=0.151Since the allowable static stresses(σO) for both pinion and gear is same (i.e σO=172.33 Mpa) andy1P is less than y1G, therefore the pinion is weaker. Thus the design should be based upon thepinionAllowable static stress(σO) =σu/3=517/3=172.33 Mpaσu=ultimate tensile strength=517 MpaTANGENTIAL TOOTH LOAD(WT)WT =( σO x Cv).b.Π.m. y1P((L-b)/L)Cv=velocity factor =3/3+v, for teeth cut by form cuttersv=peripheral speed in m/sb=face widthm=module=10y1p=tooth form factorL=slant height of pitch cone
28. 28. DG= pitch diameter of gear =475DP= pitch diameter of gear =80V==20.93m/sCv==3/3+20.93=0.125L==240.844The factor (L-b/L) may be called as bevel factorFor satisfactory operation of the bevel gears the face width should be from 6.3m to 9.5mSo b is taken as 9.5mb= 9.5x10=95WT =(172.33x0.125)x95xΠx10x0.04( ) = 1595.225NDYNAMIC LOAD (WD)WD= WT+ WIWD =WT+V= pitch line velocityb=face widthc=deformation/dynamic factor in N/mmC=K=0.111 for 20° full depth involute systemEP=young’s modulus for material of pinion in N/mm2EG=young’s modulus for material of gear in N/mm2e=tooth error action in mme value for module=10 used in precision gears is e=0.023c= =89.73WD =WT+=1595.225+WD =9830.39N
29. 29. STATIC TOOTH LOAD (WS)The static tooth load or endurance strength of the tooth for bevel gear is given byWS= σe.b.Π.m. y1P( )(Flexible endurance limit) σe = 1.75XB.H.N = 1.75X140=245=245 )WS=18145NFor safety against tooth breakage the WS ≥1.25 Wd=12287.9875ThereforeWS > WdWEAR LOAD (WW)The maximum or limiting load for wear for bevel gears is given byWw=Dp,b,q,k have usual meanings as discussed in spur gears except that Q is based on formative orequivalent no.of teeth such that ratio factor Q= = =1.951K= load stress factor (also known as material combination factor )in N/mm2given byK= )σes= surface endurance limit in mpa or N/mm2σes=(2.8×B.H.N-70)N/mm2= (2.8×140-70)=322 N/mm2K= =0.72=cos 9 =0.987WW= =10825.25NForces actingWT=WNCOSWN=normal load=WT/ COSWT=tangential forceWN=WN= =1697.602NRadial force WR=
30. 30. Mean radius (Rm)=(L-b/2)=(240.844-95/2)=32.111Axial force acting on the pinion shaftWRH=WT P1Tangential force acting at the mean radiusWT=T/RmT= torque on the pinionT ==T=480.905N-m=480905.254N-mmWT=480905.254/32.111=22961.480NAxial forceWRH=WT P1==3535.424NRadial force acting on the pinion shaftWRH=WT P1==7551.525N7.2.2.1.1 SUN GEARDiameter of crown wheel = DG= 150mmDiameter of pinion =TP=70mmNumber of teeth on gear =18Number of teeth on pinion=15D= DG+ DP=220T= TG+ TP = 33Module = m=D/T=220/33=6.66=7(according to stds)Material used for both pinion and gear is aluminum alloy7475-t761Brinell hardness number(BHN)=140
31. 31. Pressure angle of teeth is 20° involute system Ø=20°P=337.5BHP = 337.5x745.7w=251673.75wWe know that velocity ratioV.R=TG/ TP= DG/DP= NP/NGV.R= DG/DP =150/70=2.142V.R= NP/NG2.142=5000/ NGNG=2334.267rpmSince the shafts are at right angles therefore pitch angle for the pinionθp1=tan-1(1/v.r)= tan-1(1/2.142)=25.025Pitch angle of gear θp2=90°-25.025=64.974We know that formative number of teeth for pinionTEP= TPsec θp1=15sec25.025 =16.554And formative number of teeth for gearTEG= TGsec θp2=18sec64.974 =42.55Tooth form factor for the piniony1P=0.154-0.912/ TEP, for 20° full depth involute system=0.154-0.912/ 16.554=0.099And tooth form factor for geary1G=0.154-0.912/ TEG=0.154-0.912/ 42.55=0.132Since the allowable static stresses(σO) for both pinion and gear is same (i.e σO=172.33 Mpa) andy1P is less than y1G, therefore the pinion is weaker. Thus the design should be based upon thepinionAllowable static stress(σO) =σu/3=517/3=172.33 Mpaσu=ultimate tensile strength=517 MpaTANGENTIAL TOOTH LOAD(WT)WT=( σO x Cv).b.Π.m. y1P((L-b)/L)Cv=velocity factor =3/3+v,
32. 32. for teeth cut by form cuttersv=peripheral speed in m/sb=face widthm=module=7y1p=tooth form factorL=slant height of pitch coneDG= pitch diameter of gear =150mmDp= pitch diameter of gear =70mmV===18.316m/sCv==3/3+18.316=0.140L==82.764The factor (L-b/L) may be called as bevel factorFor satisfactory operation of the bevel gears the face width should be from 6.3m to 9.5mSo b is taken as 9.5mb= 9.5x7=66.5WT=(172.33x0.140)x66.5xΠx7x0.099( ) = 686NDYNAMIC LOAD (WD)WD= WT+ WIWD =WT+V= pitch line velocityb=face widthc=deformation/dynamic factor in N/mmC=K=0.111 for 20° full depth involute systemEP=young’s modulus for material of pinion in N/mm2=70300 N/mm2EG=young’s modulus for material of gear in N/mm2=70300 N/mm2e=tooth error action in mm
33. 33. e value for module=7 used in precision gears is e=0.0186c= = 72.57N/mmWD =WT+WD = 686+WD = 4620.13NSTATIC TOOTH LOAD (WS)The static tooth load or endurance strength of the tooth for bevel gear is given byWS= σe.b.Π.m. y1P( )(Flexible endurance limit) σe = 1.75XB.H.N = 1.75X140 =245WS= ( )WS=6966.47NFor safety against tooth breakage the WS ≥1.25 Wd=5775.162WS > WdWEAR LOAD (WW)The maximum or limiting load for wear for bevel gears is given byWw=Dp,b,q,k have usual meanings as discussed in spur gears except that Q is based on formative orequivalent no.of teeth such that ratio factor Q= = =1.439K= load stress factor (also known as material combination factor )in N/mm2given byK= )σes= surface endurance limit in mpa or N/mm2σes=(2.8×B.H.N-70)N/mm2= (2.8×140-70)=322K= =0.72=cos 25.025 =0.906Ww= =5323.347NForces acting
34. 34. WT=WNCOSWN=normal load=WT/ COSWT=tangential forceWN= = =730.025NRadial force WR=WT tanMean radius (Rm)=(L-b/2)=(82.764-66.5/2)=20.944NAxial force acting on the pinion shaftWRH=WT P1Tangential force acting at the mean radiusWT=T/RmT= torque on the pinionT ==T=480.905N-m=480905.254N-mmWT=480661/20.944=22961.480NAxial forceWRH=WT P1=22961.480=3535.424NRadial force acting on the pinion shaftWRH=WT P1==7551.525N
35. 35. 7.3 CAST IRON7.3.12400 rpm7.3.1.1.1Crown wheelDiameter of crown wheel = DG= 475mmNumber of teeth on gear = TG = 50Number of teeth on pinion = TP = 8Module = m=DG/TG=475/50=9.5=10(according to stds)Diameter of pinion = DP = mx TP= 10x8 = 80mmMaterial used for both pinion and gear is cast ironBrinell hardness number(BHN)=300Pressure angle of teeth is 20° involute system Ø=20°P=162BHP = 162x745.7w=120803.4wWe know that velocity ratioV.R=TG/ TP= DG/DP= NP/NGV.R=TG/ TP=50/8=6.25V.R= NP/NG6.25=2400/ NGNG=384rpmFor satisfactory operation of bevel gears the number odf teeth in the pinion must not beLess than where v.r=velocity ratio= =7.5Since the shafts are at right angles therefore pitch angle for the pinionθp1=tan-1(1/v.r)= tan-1(1/6.25)=9.0Pitch angle of gear θp2=90°-9=81We know that formative number of teeth for pinionTEP= TPsec θp1=8sec9 =8And formative number of teeth for gearTEG= TGsec θp2=50sec81 =319.622Tooth form factor for the piniony1P=0.154-0.912/ TEP, for 20° full depth involute system=0.154-0.912/ 8=0.04
36. 36. And tooth form factor for geary1G=0.154-0.912/ TEG=0.154-0.912/ 319.622=0.151Since the allowable static stresses(σO) for both pinion and gear is same (i.e σO=196 Mpa) andy1P is less than y1G, therefore the pinion is weaker. Thus the design should be based upon thepinionAllowable static stress(σO) =196 MpaTANGENTIAL TOOTH LOAD(WT)WT=( σO x Cv).b.Π.m. y1P((L-b)/L)Cv=velocity factor =3/3+v, for teeth cut by form cuttersv=peripheral speed in m/sb=face widthm=module=10y1p=tooth form factorL=slant height of pitch coneDG= pitch diameter of gear =475Dp= pitch diameter of gear =80V==10.048m/sCv==3/3+10.048=0.229L==240.844mmThe factor (L-b/L) may be called as bevel factorFor satisfactory operation of the bevel gears the face width should be from 6.3m to 9.5mSo b is taken as 9.5mb= 9.5x10=95mmWT=(196x0.229)x95xΠx10x0.04( )=3243.079N
37. 37. DYNAMIC LOAD (WD)WD= WT+ WIWD =WT+V= pitch line velocityb=face widthc=deformation/dynamic factor in N/mmC=K=0.111 for 20° full depth involute systemEP=young’s modulus for material of pinion in N/mm2=103000 N/mm2EG=young’s modulus for material of gear in N/mm2= 103000 N/mm2e=tooth error action in mme value for module=10 used in precision gears is e=0.023c==131.479N/mmWD =WT+WD = 3243.079+WD = 13110.818NSTATIC TOOTH LOAD (WS)The static tooth load or endurance strength of the tooth for bevel gear is given byWS= σe.b.Π.m. y1P( )(Flexible endurance limit) σe = 1.75XB.H.N = 1.75X300 =525N/mm2WS= )WS=37933.706NFor safety against tooth breakage the WS ≥1.25 Wd=16388.522NWEAR LOAD (WW)The maximum or limiting load for wear for bevel gears is given byWw=Dp,b,q,k have usual meanings as discussed in spur gears except that Q is based on formative orequivalent no.of teeth such that ratio factor Q= = =1.951K= load stress factor (also known as material combination factor )in N/mm2given byK= )
38. 38. σes= surface endurance limit in mpa or N/mm2σes=(2.8×B.H.N-70)N/mm2= (2.8×300-70)=770 N/mm2K==2.812=cos 9 =0.987Ww==42244.388NForces actingWT=WNCOSWN=normal load=WT/ COSWT=tangential forceWN= ==3451.212NWT=tangential force= WNCOS =2228.217cos20=2093.839NRadial force WR= =1180.384NMean radius (Rm)=(L-b/2)=(240.844-95/2)=32.111mmAxial force acting on the pinion shaftWRH=WT P1Tangential force acting at the mean radiusWRH=T/RmT= torque on the pinionT ==T=480.661N-m=480661N-mm
39. 39. WRH=480661/32.111=14968.733NAxial forceWRH=WT P1=14968.733=850.010NRadial force acting on the pinion shaftWRH=WT P1=14968.733=5366.752N7.3.1.1.2 SUN GEARDiameter of crown wheel = DG= 150mmDiameter of pinion = DP = 70mmNumber of teeth on gear = TG = 18Number of teeth on pinion = TP = 15D= DG+ DP=220T= TG+ TP = 33Module = m=D/T=220/33=6.66=7(according to stds)Material used for both pinion and gear is cast ironBrinell hardness number(BHN)=300Pressure angle of teeth is 20° involute system Ø=20°P=162BHP = 162x745.7w=120803.4wWe know that velocity ratioV.R=TG/ TP= DG/DP= NP/NGV.R= DG/DP =150/70=2.142V.R= NP/NG2.142=2400/ NGNG=1120.448rpmSince the shafts are at right angles therefore pitch angle for the pinionθp1=tan-1(1/v.r)= tan-1(1/2.142)=25.025Pitch angle of gear θp2=90°-25.025=64.974
40. 40. We know that formative number of teeth for pinionTEP= TPsec θp1=15sec25.025 =16.554And formative number of teeth for gearTEG= TGsec θp2=18sec64.974 =42.55Tooth form factor for the piniony1P=0.154-0.912/ TEP, for 20° full depth involute system=0.154-0.912/ 16.554=0.099And tooth form factor for geary1G=0.154-0.912/ TEG=0.154-0.912/ 42.55=0.132Since the allowable static stresses(σO) for both pinion and gear is same (i.e σO=126.66 Mpa) andy1P is less than y1G, therefore the pinion is weaker. Thus the design should be based upon thepinionAllowable static stress(σO) =196 MpaTANGENTIAL TOOTH LOAD(WT)WT=( σO x Cv).b.Π.m. y1P((L-b)/L)Cv=velocity factor =3/3+v, for teeth cut by form cuttersv=peripheral speed in m/sb=face widthm=module=7y1p=tooth form factorL=slant height of pitch coneDG= pitch diameter of gear =150mmDp= pitch diameter of gear =70mmV===8.792m/sCv==3/3+8.792=0.254L==82.764
41. 41. The factor (L-b/L) may be called as bevel factorFor satisfactory operation of the bevel gears the face width should be from 6.3m to 9.5mSo b is taken as 9.5mb= 9.5x7=66.5WT=(196x0.254)x66.5xΠx7x0.099( )=1415.587NDYNAMIC LOAD (WD)WD= WT+ WIWD =WT+V= pitch line velocityb=face widthc=deformation/dynamic factor in N/mmC=K=0.111 for 20° full depth involute systemEP=young’s modulus for material of pinion in N/mm2=103000 N/mm2EG=young’s modulus for material of gear in N/mm2=103000 N/mm2e=tooth error action in mme value for module=7 used in precision gears is e=0.0186c= = 106.3269N/mmWD =WT+WD = 1415.587+WD = 1883.579NSTATIC TOOTH LOAD (WS)The static tooth load or endurance strength of the tooth for bevel gear is given byWS= σe.b.Π.m. y1P( )(Flexible endurance limit) σe = 1.75XB.H.N = 1.75X300 =525WS= ( )WS=14928.16NFor safety against tooth breakage the WS ≥1.25 Wd=2354.39WS > Wd
42. 42. WEAR LOAD (WW)The maximum or limiting load for wear for bevel gears is given byWw=Dp,b,q,k have usual meanings as discussed in spur gears except that Q is based on formative orequivalent no.of teeth such that ratio factor Q= = =1.439K= load stress factor (also known as material combination factor )in N/mm2given byK= )σes= surface endurance limit in mpa or N/mm2σes=(2.8×B.H.N-70)N/mm2= (2.8×300-70)=770K= )=2.812=cos 25.025 =0.906Ww= =20790.62NForces actingWT=WNCOSWN=normal load=WT/ COSWT=tangential forceWN= = =1507.54NRadial force WR=WT tanMean radius (Rm)=(L-b/2)=(82.764-66.5/2)=20.944NAxial force acting on the pinion shaftWRH=WT P1Tangential force acting at the mean radiusWRH=T/Rm
43. 43. T= torque on the pinionT ==T=480.661N-m=480661N-mmWT=480661/20.944=22949.818NAxial forceWRH=WT P1=22949.818=3533.340NRadial force acting on the pinion shaftWRH=WT P1=22949.818=7567.863N7.3.2 5000 rpm7.3.2 .1.1Crown wheelDiameter of crown wheel = DG= 475mmNumber of teeth on gear = TG = 50Number of teeth on pinion = TP = 8Module = m=DG/TG=475/50=9.5=10(according to stds)Diameter of pinion = DP = mx TP = 10x8 = 80mmMaterial used for both pinion and gear is cast ironBrinell hardness number(BHN)=300Pressure angle of teeth is 20° involute system Ø=20°P=337.5BHP = 337.5x745.7w=251673.75wWe know that velocity ratioV.R=TG/ TP= DG/DP= NP/NGV.R=TG/ TP=50/8=6.25V.R= NP/NG6.25=5000/ NGNG=800rpm
44. 44. For satisfactory operation of bevel gears the number of teeth in the pinion must not beLess than where v.r=velocity ratio= =7.5Since the shafts are at right angles therefore pitch angle for the pinionθp1=tan-1(1/v.r)= tan-1(1/6.25)=9.0Pitch angle of gear θp2=90°-9=81We know that formative number of teeth for pinionTEP= TPsec θp1=8sec9 =8And formative number of teeth for gearTEG= TGsec θp2=50sec81 =319.622Tooth form factor for the piniony1P=0.154-0.912/ TEP, for 20° full depth involute system=0.154-0.912/ 8=0.04And tooth form factor for geary1G=0.154-0.912/ TEG=0.154-0.912/ 319.622=0.151Since the allowable static stresses(σO) for both pinion and gear is same (i.e σO=196 Mpa) andy1P is less than y1G, therefore the pinion is weaker. Thus the design should be based upon thepinionAllowable static stress(σO) =196 MpaTANGENTIAL TOOTH LOAD(WT)WT=( σO x Cv).b.Π.m. y1P((L-b)/L)Cv=velocity factor =3/3+v, for teeth cut by form cuttersv=peripheral speed in m/sb=face widthm=module=10y1p=tooth form factorL=slant height of pitch cone
45. 45. DG= pitch diameter of gear =475Dp= pitch diameter of gear =80V==20.93m/sCv=3/3+20.93=0.125L==240.844The factor (L-b/L) may be called as bevel factorFor satisfactory operation of the bevel gears the face width should be from 6.3m to 9.5mSo b is taken as 9.5mb= 9.5x10=95WT=(196x0.125)x95xΠx10x0.04( )=1770.24NDYNAMIC LOAD (WD)WD= WT+ WIWD =WT+V= pitch line velocityb=face widthc=deformation/dynamic factor in N/mmC=K=0.111 for 20° full depth involute systemEP=young’s modulus for material of pinion in N/mm2EG=young’s modulus for material of gear in N/mm2e=tooth error action in mme value for module=10 used in precision gears is e=0.023c==131.479N/mmWD =WT+=WD = 11214.311N
46. 46. STATIC LOAD (WS)The static tooth load or endurance strength of the tooth for bevel gear is given byWS= σe.b.Π.m. y1P( )(Flexible endurance limit) σe = 1.75XB.H.N = 1.75X300 =525=525 )WS=37933.706NFor safety against tooth breakage the WS ≥1.25 Wd=14017.88ThereforeWS > WdWEAR LOAD (WW)The maximum or limiting load for wear for bevel gears is given byWw=Dp,b,q,k have usual meanings as discussed in spur gears except that Q is based on formative orequivalent no.of teeth such that ratio factor Q= = =1.951K= load stress factor (also known as material combination factor )in N/mm2given byK= )σes= surface endurance limit in mpa or N/mm2σes=(2.8×B.H.N-70)N/mm2= (2.8×300-70)=770K= =2.812=cos 9 =0.987Ww= =42244.388NForces actingWT=WNCOSWN=normal load=WT/ COSWT=tangential forceWN= =WN=1883.850N
47. 47. Radial force WR= =1770.24 =644.314NMean radius (Rm)=(L-b/2)=(240.844-95/2)=32.111Axial force acting on the pinion shaftWRH=WT P1Tangential force acting at the mean radiusWRH=T/RmT= torque on the pinionT ==T=480.905N-m=480905.254N-mmWT=480905.254/32.111=22961.480NAxial forceWRH=WT P1==3535.424NRadial force acting on the pinion shaftWRH=WT P1==7551.525N
48. 48. 7.3.2 .1.2SUN GEARDiameter of crown wheel = DG= 150mmDiameter of pinion = DP = 70mmNumber of teeth on gear = TG = 18Number of teeth on pinion = TP = 15D= DG+ DP=220T= TG+ TP = 33Module = m=D/T=220/33=6.66=7(according to stds)Material used for both pinion and gear is cast ironBrinell hardness number(BHN)=300Pressure angle of teeth is 20° involute system Ø=20°P=337.5BHP = 337.5x745.7w=251673.75wWe know that velocity ratioV.R=TG/ TP= DG/DP= NP/NGV.R= DG/DP =150/70=2.142V.R= NP/NG2.142=5000/ NGNG=2334.267rpmSince the shafts are at right angles therefore pitch angle for the pinionθp1=tan-1(1/v.r)= tan-1(1/2.142)=25.025Pitch angle of gear θp2=90°-25.025=64.974We know that formative number of teeth for pinionTEP= TPsec θp1=15sec25.025 =16.554And formative number of teeth for gearTEG= TGsec θp2=18sec64.974 =42.55Tooth form factor for the piniony1P=0.154-0.912/ TEP, for 20° full depth involute system=0.154-0.912/ 16.554=0.099And tooth form factor for geary1G=0.154-0.912/ TEG=0.154-0.912/ 42.55=0.132
49. 49. Since the allowable static stresses(σO) for both pinion and gear is same (i.e σO=196 Mpa) andy1P is less than y1G, therefore the pinion is weaker. Thus the design should be based upon thepinionAllowable static stress(σO) =196 MpaTANGENTIAL TOOTH LOAD(WT)WT=( σO x Cv).b.Π.m. y1P((L-b)/L)Cv=velocity factor =3/3+v, for teeth cut by form cuttersv=peripheral speed in m/sb=face widthm=module=7y1p=tooth form factorL=slant height of pitch coneDG= pitch diameter of gear =150mmDp= pitch diameter of gear =70mmV===18.316m/sCv==3/3+18.316=0.140L==82.764The factor (L-b/L) may be called as bevel factorFor satisfactory operation of the bevel gears the face width should be from 6.3m to 9.5mSo b is taken as 9.5mb= 9.5x7=66.5WT=(196x0.140)x66.5xΠx7x0.099( )=780.245N
50. 50. DYNAMIC LOAD (WD)WD= WT+ WIWD =WT+V= pitch line velocityb=face widthc=deformation/dynamic factor in N/mmC=K=0.111 for 20° full depth involute systemEP=young’s modulus for material of pinion in N/mm2=103000 N/mm2EG=young’s modulus for material of gear in N/mm2=103000 N/mm2e=tooth error action in mme value for module=7 used in precision gears is e=0.0186c= = 106.326N/mmWD =WT+WD =780.245+WD = 7161.239NSTATIC LOAD (WS)The static tooth load or endurance strength of the tooth for bevel gear is given byWS= σe.b.Π.m. y1P( )(Flexible endurance limit) σe = 1.75XB.H.N = 1.75X300 =525WS= ( )WS=14928.16For safety against tooth breakage the WS ≥1.25 Wd=8951.548NWS > WdWEAR LOAD (WW)The maximum or limiting load for wear for bevel gears is given byWw=Dp,b,q,k have usual meanings as discussed in spur gears except that Q is based on formative orequivalent no.of teeth such that ratio factor Q= = =1.439K= load stress factor (also known as material combination factor )in N/mm2given byK= )σes= surface endurance limit in mpa or N/mm2
51. 51. σes=(2.8×B.H.N-70)N/mm2= (2.8×300-70)=770K= =2.812=cos 25.025 =0.906Ww= =20790.62NForces actingWT=WNCOSWN=normal load=WT/ COSWT=tangential forceWN= = =830.319NRadial force WR=WT tanMean radius (Rm)=(L-b/2)=(82.764-66.5/2)=20.944NAxial force acting on the pinion shaftWRH=WT P1Tangential force acting at the mean radiusWRH=T/RmT= torque on the pinionT ==T=480.905N-m=480905.254N-mmWT=480905.254/20.944=22961.480NAxial forceWRH=WT P1=22961.480=3535.424NRadial force acting on the pinion shaftWRH=WT P1==7551.525N
54. 54. 8.2 MODEL OF DIFFERENTIAL GEAR8.2.1CROWN8.2.2PINION
55. 55. 8.2.3 PLANET8.2.4 SUNGEAR
56. 56. 8.2.5 MAIN SHAFT_PINION8.2.6 MAIN SHAFT
57. 57. 8.2.7 CROWN ASSEMBLY8.2.7 CROWN ASSEMBLY EXPLODE VIEW
58. 58. 9.INTRODUCTION TO FEAFinite element analysis (FEA) is a fairly recent discipline crossing the boundaries ofmathematics, physics, engineering and computer science. The method has wide application andenjoys extensive utilization in the structural, thermal and fluid analysis areas. The finite elementmethod is comprised of three major phases: (1) pre-processing, in which the analyst develops afinite element mesh to divide the subject geometry into subdomains for mathematical analysis,and applies material properties and boundary conditions, (2) solution, during which the programderives the governing matrix equations from the model and solves for the primary quantities, and(3) post-processing, in which the analyst checks the validity of the solution, examines the valuesof primary quantities (such as displacements and stresses), and derives and examines additionalquantities (such as specialized stresses and error indicators).The advantages of FEA are numerous and important. A new design concept may be modeled todetermine its real world behavior under various load environments, and may therefore be refinedprior to the creation of drawings, when few dollars have been committed and changes areinexpensive. Once a detailed CAD model has been developed, FEA can analyze the design indetail, saving time and money by reducing the number of prototypes required. An existingproduct which is experiencing a field problem, or is simply being improved, can be analyzed tospeed an engineering change and reduce its cost. In addition, FEA can be performed onincreasingly affordable computer workstations and personal computers, and professionalassistance is available.It is also important to recognize the limitations of FEA. Commercial software packages and therequired hardware, which have seen substantial price reductions, still require a significantinvestment. The method can reduce product testing, but cannot totally replace it. Probably mostimportant, an inexperienced user can deliver incorrect answers, upon which expensive decisionswill be based. FEA is a demanding tool, in that the analyst must be proficient not only inelasticity or fluids, but also in mathematics, computer science, and especially the finite elementmethod itself.
59. 59. Which FEA package to use is a subject that cannot possibly be covered in this short discussion,and the choice involves personal preferences as well as package functionality. Where to run thepackage depends on the type of analyses being performed. A typical finite element solutionrequires a fast, modern disk subsystem for acceptable performance. Memory requirements are ofcourse dependent on the code, but in the interest of performance, the more the better, with arepresentative range measured in gigabytes per user. Processing power is the final link in theperformance chain, with clock speed, cache, pipelining and multi-processing all contributing tothe bottom line. These analyses can run for hours on the fastest systems, so computing power isof the essence.One aspect often overlooked when entering the finite element area is education. Withoutadequate training on the finite element method and the specific FEA package, a new user will notbe productive in a reasonable amount of time, and may in fact fail miserably. Expect to dedicateone to two weeks up front, and another one to two weeks over the first year, to either classroomor self-help education. It is also important that the user have a basic understanding of thecomputers operating system.
60. 60. 9.1 INTRODUCTION TO COSMOSWORKSCosmosworks is a useful software for design analysis in mechanical engineering. That’s anintroduction for you who would like to learn more about COSMOSWorks. COSMOSWorks is adesign analysis automation application fully integrated with SolidWorks.This software uses the Finite Element Method (FEM) to simulate the working conditions of yourdesigns and predict their behavior. FEM requires the solution of large systems of equations.Powered by fast solvers, COSMOSWorks makes it possible for designers to quickly check theintegrity of their designs and search for the optimum solution.A product development cycle typically includes the following steps:1 Build your model in the SolidWorks CAD system.2 Prototype the design.2 Test the prototype in the field.3 Evaluate the results of the field tests.4 Modify the design based on the field test results.Analysis Steps : You complete a study by performing the following steps:• Create a study defining its analysis type and options.• If needed, define parameters of your study. Parameters could be a model dimension, a materialproperty, a force value, or any other entity that you want to investigate its impact on the design.Analysis Background: Linear Static Analysis Frequency Analysis Linearized Buckling AnalysisThermal Analysis Optimization Studies, Material property, Material Models, Linear ElasticIsotropic.Plotting Results - Describes how to generate a result plot and result tools.Listing Results - Overview of the results that can be listed.Graphing Results - Shows you how to graph results.Results of Structural Studies - Lists results available from structural studies.Results of Thermal Studies - Lists results available from thermal studies.Reports - Explains the study report utility.Stress Check - Lists the basics of checking stress results and different criteria used in thechecking.
61. 61. 9.2 STRUCTURAL ANALYSIS OF DIFFERENTIAL GEAR9.2.1 2400 rpm9.2.1.2 ALUMINUM ALLOY9.2.1.2.1 TANGENTIAL LOAD9.2.1.2.2 Material PropertiesModel Reference Properties ComponentsName: al_ alloy7475-t761Model type: Linear Elastic IsotropicDefault failure criterion: Max von Mises StressYield strength: 1.65e+008 N/m^2Tensile strength: 3e+007 N/m^2Elastic modulus: 7e+010 N/m^2Poissons ratio: 0.33Mass density: 2600 kg/m^3Shear modulus: 3.189e+008 N/m^2Model9.2.1.1.4Units
62. 62. Unit system: SI (MKS)Length/Displacement mmTemperature KelvinAngular velocity Rad/secPressure/Stress N/m^2Unit system: SI (MKS)Length/Displacement mmTemperature KelvinAngular velocity Rad/secPressure/Stress N/m^2Mesh type Solid MeshMesher Used: Standard meshAutomatic Transition: OffInclude Mesh Auto Loops: OffJacobian points 4 PointsElement Size 22.7482 mmTolerance 1.13741 mmMesh Quality HighRemesh failed parts with incompatible mesh Off9.2.1.1.7 Mesh Information - DetailsTotal Nodes 39721
63. 63. 9.2.1.2.3 Loads and FixturesFixture name Fixture Image Fixture DetailsFixed-1Entities: 2 face(s)Type: Fixed GeometryResultant ForcesComponents X Y Z ResultantReaction force(N) 178.306 621.812 -91.4611 653.306Reaction Moment(N-m) 0 0 0 0Load name Load Image Load DetailsForce-1Entities: 9 face(s)Type: Apply normal forceValue: 2922.51 NTotal Elements 21842Maximum Aspect Ratio 250.8% of elements with Aspect Ratio < 3 63.3% of elements with Aspect Ratio > 10 1.11% of distorted elements(Jacobian) 0Time to complete mesh(hh;mm;ss): 00:01:25Computer name: WIN
64. 64. Force-2Entities: 11 face(s)Type: Apply normal forceValue: 1 N
65. 65. 9.2.1.2.4 Study ResultsName Type Min MaxStress1 VON: von Mises Stress 6.84248e-010 N/mm^2(MPa)Node: 386493.19018 N/mm^2(MPa)Node: 204752part_assm-2400_aluminiumally_tangential_load-Stress-Stress1Name Type Min MaxDisplacement1 URES: Resultant Displacement 0 mmNode: 31060.0241697 mmNode: 295072part_assm-2400_aluminiumally_tangential_load-Displacement-Displacement1
66. 66. Name Type Min MaxStrain1 ESTRN: Equivalent Strain 2.08727e-014Element: 208594.15934e-005Element: 10692part_assm-2400_aluminiumally_tangential_load-Strain-Strain1
67. 67. 9.2.1.2.5 STATIC LOAD9.2.1.2.6 Loads and FixturesFixture name Fixture Image Fixture DetailsFixed-1Entities: 2 face(s)Type: Fixed GeometryResultant ForcesComponents X Y Z ResultantReaction force(N) 1941.59 1881.96 -370.659 2729.27Reaction Moment(N-m) 0 0 0 0Load name Load Image Load DetailsForce-1Entities: 9 face(s)Type: Apply normal forceValue: 18145 NForce-2Entities: 11 face(s)Type: Apply normal forceValue: 6966.47 N
68. 68. 9.2.1.2.7 Study ResultsName Type Min MaxStress1 VON: von Mises Stress 8.11597e-007 N/mm^2(MPa)Node: 3357419.8068 N/mm^2(MPa)Node: 204752part_assm-2400_aluminiumally_static_load-Stress-Stress1Name Type Min MaxDisplacement1 URES: Resultant Displacement 0 mmNode: 31060.150063 mmNode: 29507
69. 69. 2part_assm-2400_aluminiumally_static_load-Displacement-Displacement1Name Type Min MaxStrain1 ESTRN: Equivalent Strain 1.52786e-011Element: 201070.000258239Element: 10692part_assm-2400_aluminiumally_static_load-Strain-Strain19.2.1.3 CAST IRON9.2.1.3.1 TANGENTIAL LOAD9.2.1.3.2 Material PropertiesModel Reference Properties ComponentName: Malleable Cast IronModel type: Linear Elastic IsotropicDefault failure criterion: Max von Mises StressYield strength: 2.75742e+008 N/m^2Tensile strength: 4.13613e+008 N/m^2Elastic modulus: 1.9e+011 N/m^2Poissons ratio: 0.27Mass density: 7300 kg/m^3Shear modulus: 8.6e+010 N/m^2Thermal expansioncoefficient:1.2e-005 /Kelvin
70. 70. 9.2.1.3.3 Loads and FixturesFixture name Fixture Image Fixture DetailsFixed-1Entities: 2 face(s)Type: Fixed GeometryResultant ForcesComponents X Y Z ResultantReaction force(N) 378.494 287.568 -36.4058 476.738Reaction Moment(N-m) 0 0 0 0Load name Load Image Load DetailsForce-1Entities: 9 face(s)Type: Apply normal forceValue: 3243.08 NForce-2Entities: 11 face(s)Type: Apply normal forceValue: 1415.59 N
71. 71. 9.2.1.3.4 Study ResultsName Type Min MaxStress1 VON: von Mises Stress 2.37779e-007N/mm^2 (MPa)Node: 382673.57544 N/mm^2(MPa)Node: 203652part_assm-2400_castiron_tangential_load-Stress-Stress1Name Type Min MaxDisplacement1 URES: Resultant Displacement 0 mmNode: 31060.0100566 mmNode: 303732part_assm-2400_castiron_tangential_load-Displacement-Displacement1
72. 72. Name Type Min MaxStrain1 ESTRN: Equivalent Strain 9.90157e-013Element: 212851.69558e-005Element: 99032part_assm-2400_castiron_tangential_load-Strain-Strain19.2.1.3.5 STATIC LOAD9.2.1.3.6 Loads and FixturesFixture name Fixture Image Fixture DetailsFixed-1Entities: 2 face(s)Type: Fixed GeometryResultant ForcesComponents X Y Z ResultantReaction force(N) 4230.77 3829.18 -471.957 5725.8Reaction Moment(N-m) 0 0 0 0
73. 73. Load name Load Image Load DetailsForce-1Entities: 9 face(s)Type: Apply normal forceValue: 37933.7 NForce-2Entities: 11 face(s)Type: Apply normal forceValue: 14928.2 N
74. 74. 9.2.1.3.7 Study ResultsName Type Min MaxStress1 VON: von Mises Stress 2.48589e-006 N/mm^2(MPa)Node: 3845841.8212 N/mm^2(MPa)Node: 203652part_assm-2400_castiron_static_load-Stress-Stress1Name Type Min MaxDisplacement1 URES: Resultant Displacement 0 mmNode: 31060.11763 mmNode: 303732part_assm-2400_castiron_static_load-Displacement-Displacement1
75. 75. Name Type Min MaxStrain1 ESTRN: Equivalent Strain 8.223e-012Element: 212850.000198329Element: 99032part_assm-2400_castiron_static_load-Strain-Strain1
76. 76. 9.2.2 5000 RPM9.2.2.2 ALUMINUM ALLOY9.2.2.2 .1 TANGENTIAL LOAD9.2.2.2 .2 Loads and FixturesFixture name Fixture Image Fixture DetailsFixed-1Entities: 2 face(s)Type: Fixed GeometryResultant ForcesComponents X Y Z ResultantReaction force(N) 196.301 117.513 -15.5613 229.316Reaction Moment(N-m) 0 0 0 0Load name Load Image Load DetailsForce-1Entities: 9 face(s)Type: Apply normal forceValue: 1595.22 NForce-2Entities: 11 face(s)Type: Apply normal forceValue: 780.245 N
77. 77. 9.2.2.2 .3 Study ResultsName Type Min MaxStress1 VON: von Mises Stress 1.68385e-007 N/mm^2(MPa)Node: 343261.70369 N/mm^2 (MPa)Node: 305162part_assm-5000_aluminiumally_tangential_load-Stress-Stress1Name Type Min MaxDisplacement1 URES: Resultant Displacement 0 mmNode: 31060.0131944 mmNode: 303732part_assm-5000_aluminiumally_tangential_load-Displacement-Displacement1
78. 78. Name Type Min MaxStrain1 ESTRN: Equivalent Strain 1.78999e-012Element: 215522.25582e-005Element: 99032part_assm-5000_aluminiumally_tangential_load-Strain-Strain1
79. 79. 9.2.2.2 .4 STATIC LOAD9.2.2.2 .5 Loads and FixturesFixture name Fixture Image Fixture DetailsFixed-1Entities: 2 face(s)Type: Fixed GeometryResultant ForcesComponents X Y Z ResultantReaction force(N) -1902.68 1875.88 446.924 2709.03Reaction Moment(N-m) 0 0 0 0Load name Load Image Load DetailsForce-1Entities: 9 face(s)Type: Apply normal forceValue: 18145 NForce-2Entities: 11 face(s)Type: Apply normal forceValue: 6966.47 N
80. 80. 9.2.2.2 .6 Study ResultsName Type Min MaxStress1 VON: von Mises Stress 7.79346e-007 N/mm^2(MPa)Node: 3846122.6949 N/mm^2(MPa)Node: 145842part_assm-5000_aluminiumally_static_load-Stress-Stress1Name Type Min MaxDisplacement1 URES: Resultant Displacement 0 mmNode: 31060.150036 mmNode: 317702part_assm-5000_aluminiumally_static_load-Displacement-Displacement1
81. 81. Name Type Min MaxStrain1 ESTRN: Equivalent Strain 1.90385e-011Element: 201870.000274774Element: 124102part_assm-5000_aluminiumally_static_load-Strain-Strain1
82. 82. 9.2.2.3 CAST IRON9.2.2.3 .1 TANGENTIAL LOAD9.2.2.3 .2 Loads and FixturesFixture name Fixture Image Fixture DetailsFixed-1Entities: 2 face(s)Type: Fixed GeometryResultant ForcesComponents X Y Z ResultantReaction force(N) 195.449 153.057 -14.6943 248.682Reaction Moment(N-m) 0 0 0 0Load name Load Image Load DetailsForce-1Entities: 9 face(s)Type: Apply normal forceValue: 1770.24 NForce-2Entities: 11 face(s)Type: Apply normal forceValue: 780.245 N
83. 83. 9.2.2.3 .3 Study ResultsName Type Min MaxStress1 VON: von Mises Stress 7.76622e-008 N/mm^2(MPa)Node: 384162.01579 N/mm^2(MPa)Node: 204752part_assm-5000_castiron_tangential_load-Stress-Stress1Name Type Min MaxDisplacement1 URES: Resultant Displacement 0 mmNode: 31060.00548866 mmNode: 295072part_assm-5000_castiron_tangential_load-Displacement-Displacement1
84. 84. Name Type Min MaxStrain1 ESTRN: Equivalent Strain 5.14296e-013Element: 184699.32532e-006Element: 10692part_assm-5000_castiron_tangential_load-Strain-Strain1
85. 85. 9.2.2.3 .4 STATIC LOAD9.2.2.3 .5 Loads and FixturesFixture name Fixture Image Fixture DetailsFixed-1Entities: 2 face(s)Type: Fixed GeometryResultant ForcesComponents X Y Z ResultantReaction force(N) 4093.63 3829.02 -1648.28 5842.6Reaction Moment(N-m) 0 0 0 0Load name Load Image Load DetailsForce-1Entities: 9 face(s)Type: Apply normal forceValue: 37933.7 NForce-2Entities: 11 face(s)Type: Apply normal forceValue: 14928.2 N
86. 86. 9.2.2.3 .6 Study ResultsName Type Min MaxStress1 VON: von Mises Stress 1.50476e-006 N/mm^2 (MPa)Node: 3864643.1949 N/mm^2 (MPa)Node: 204752part_assm-5000_castiron__static_load-Stress-Stress1Name Type Min MaxDisplacement1 URES: Resultant Displacement 0 mmNode: 31060.117614 mmNode: 295072part_assm-5000_castiron__static_load-Displacement-Displacement1
87. 87. Name Type Min MaxStrain1 ESTRN: Equivalent Strain 1.19677e-011Element: 184660.000199826Element: 10692part_assm-5000_castiron__static_load-Strain-Strain1
88. 88. 10. RESULTS TABLE10.1 2400 RPMTANGENTIAL Aluminum Alloy Cast IronLOAD (N) 2922.51 3243.08DISPLACEMENT(mm)0.0241696 0.0100566STRESS (N/mm2) 3.19018 3.57544STRAIN 4.1593e-51.69558 e-5STATICLOAD (N) 18143.3 37933.7DISPLACEMENT(mm)0.150063 0.11763STRESS (N/mm2) 19.8068 41.8212STRAIN 0.000258239 0.00019832910.2 5000 RPMTANGENTIAL Aluminum Alloy Cast IronLOAD (N) 1595.22 1770.24DISPLACEMENT(mm)0.0131944 0.00548866STRESS (N/mm2) 1.70369 2.01579STRAIN 2.2558e-59.32532 e-6STATICLOAD (N) 18143.3 37933.7DISPLACEMENT(mm)0.150036 0.117614STRESS (N/mm2) 22.6949 43.1949STRAIN 0.000274774 0.000199826
89. 89. 11. FREQUENCY ANALYSIS OF DIFFERENTIAL GEAR11.2 ALUMINUM ALLOY11.2.1 Study ResultsName Type Min MaxDisplacement1 URES: Resultant DisplacementPlot for Mode Shape: 1(Value =316.796 Hz)0 mmNode: 3106381.759 mmNode: 32802part_assm-aluminiumalloy_frequency-Displacement-Displacement1Name Type Min MaxDisplacement2 URES: Resultant Displacement Plotfor Mode Shape: 2(Value = 332.013Hz)0 mmNode: 3106396.691 mmNode: 44642part_assm-aluminiumalloy_frequency-Displacement-Displacement2
90. 90. Name Type Min MaxDisplacement3 URES: Resultant DisplacementPlot for Mode Shape: 3(Value =333.527 Hz)0 mmNode: 3106312.409 mmNode: 261192part_assm-aluminiumalloy_frequency-Displacement-Displacement3Name Type Min MaxDisplacement4 URES: Resultant DisplacementPlot for Mode Shape: 4(Value =865.487 Hz)0 mmNode: 3106723.674 mmNode: 2383
91. 91. 2part_assm-aluminiumalloy_frequency-Displacement-Displacement4Name Type Min MaxDisplacement5 URES: Resultant DisplacementPlot for Mode Shape: 5(Value =875.09 Hz)0 mmNode: 3106357.86 mmNode: 40342part_assm-aluminiumalloy_frequency-Displacement-Displacement511.2.2 Mode ListFrequency Number Rad/sec Hertz Seconds1 1990.5 316.8 0.00315662 2086.1 332.01 0.00301193 2095.6 333.53 0.00299834 5438 865.49 0.00115545 5498.4 875.09 0.001142711.2.3 Mass Participation (Normalized)Mode Number Frequency(Hertz) X direction Y direction Z direction1 316.8 9.8564e-005 5.8687e-007 0.0830342 332.01 0.087176 6.2493e-007 8.9872e-0053 333.53 0.0048507 8.7571e-009 6.3881e-0064 865.49 4.1886e-008 0.010812 1.4807e-0075 875.09 4.8546e-005 0.59579 3.4295e-007Sum X = 0.092174 Sum Y = 0.60661 Sum Z = 0.08313
92. 92. 11.3 CAST IRON11.3 .1 Study ResultsName Type Min MaxDisplacement1 URES: Resultant Displacement Plotfor Mode Shape: 1(Value = 316.612Hz)0 mmNode: 3106232.328 mmNode: 285412part_assm-c_i__frequency-Displacement-Displacement1Name Type Min MaxDisplacement2 URES: Resultant Displacement Plotfor Mode Shape: 2(Value = 331.666Hz)0 mmNode: 3106241.953 mmNode: 304472part_assm-c_i__frequency-Displacement-Displacement2
93. 93. Name Type Min MaxDisplacement3 URES: Resultant DisplacementPlot for Mode Shape: 3(Value =343.648 Hz)0 mmNode: 3106186.029 mmNode: 276132part_assm-c_i__frequency-Displacement-Displacement3Name Type Min MaxDisplacement4 URES: Resultant DisplacementPlot for Mode Shape: 4(Value =861.879 Hz)0 mmNode: 3106206.394 mmNode: 196322part_assm-c_i__frequency-Displacement-Displacement4
94. 94. Name Type Min MaxDisplacement5 URES: Resultant DisplacementPlot for Mode Shape: 5(Value =884.969 Hz)0 mmNode: 3106448.102 mmNode: 23842part_assm-c_i__frequency-Displacement-Displacement511.3 .2 Mode ListFrequency Number Rad/sec Hertz Seconds1 1989.3 316.61 0.00315842 2083.9 331.67 0.00301513 2159.2 343.65 0.002914 5415.3 861.88 0.00116035 5560.4 884.97 0.0011311.3 .3 Mass Participation (Normalized)Mode Number Frequency(Hertz) X direction Y direction Z direction1 316.61 5.4853e-005 3.5881e-007 0.0813922 331.67 0.090131 7.0582e-007 4.1476e-0053 343.65 8.998e-005 3.183e-009 4.3793e-0064 861.88 5.5549e-005 0.607 1.6037e-0075 884.97 6.3601e-008 0.0012827 8.1532e-007Sum X = 0.090331 Sum Y = 0.60829 Sum Z = 0.081439
95. 95. 12.CONCLUSIONIn our project we have designed a differential gear box for Ashok Leyland 2516M. Loads arecalculated when the gears are transmitting different speeds 2400rpm and 5000rpm and differentmaterials Aluminum Alloy and Cast Iron.Structural and Frequency analyses are done on the differential gear box to verify the bestmaterial by taking in to account stresses, displacements, weight etc.By observing the structural analysis results using Aluminum alloy the stress values are within thepermissible stress value. So using Aluminum Alloy is safe for differential gear. Whencomparing the stress values of the three materials for all speeds 2400rpm and 5000rpm, thevalues are less for Aluminum alloy than Cast Iron.By observing the frequency analysis, the vibrations are less for Aluminum Alloy than cast ironsince its natural frequency is less.And also weight of the Aluminum alloy reduces almost 3 times when compared with d Cast Ironsince its density is very less. Thereby mechanical efficiency will be increased.By observing analysis results, Aluminum Alloy is best material for Differential.
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