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DESIGN AND ASSEMBLY ANALYSIS OF A GEAR

TRAIN OF A GEAR BOX

1. NOMENCLATURE

P=powerdG=diameter of geaardp=diameter of pinionα=helix angleBHN=brinel hardness numberE=youngs modulusEG=youngs modulus of gear materialEP=youngs modulus of pinion materialCompressive stress =σ c

Bending stress = σ baModule =mT p=no of teethon piniongear ratio = GRNo of teeth on gear = T G

Diameter of gear = DG

Diameter of pinion =DP

NG=speed of gearNormal pitch = PN

Pc=circuler pitchNormal pressure angle = фNpressure angle = фb=face widthEquivalent no of teeth on pinion = T EP

Equivalent no of teeth on gear = T EG

W T=tangential tooth loadσ 0=allowable static stressC v=velocity factorY1 =tooth form factorPeripheral speed = VDynamic tooth load = W DC = deformation factorStatic tooth load = W SK =load stress factorWw=wear loadX = over hangTorque =TEquivalent twisting moment = T e

τ=shear stress

ABSTRACT

Differential is used when a vehicle takes a turn, the outer wheel on a longer radius than the inner wheel. The outer wheel turns faster than the inner wheel that is when there is a relative movement between the two rear wheels. If the two rear wheels are rigidly fixed to a rear axle the inner wheel will slip which cause rapid tire wear, steering difficulties and poor load holding. Differential is a part of inner axle housing assembly, which includes the differential rear axles, wheels and bearings. The differential consists of a system of gears arranged 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 casing

In the present work all the parts of differential are designed under static condition and modeled. The required data is taken from journal paper. Modeling and assembly is done in 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 contact analysis on assembly of gears in gear box when they transmit power at different speeds at 2500 rpm, 5000 rpm. Presently used materials are Cast iron and Cast steel. For validating design Structural Analysis is also conducted by varying the materials for gears, Cast Iron and of Aluminum Alloy.

The analysis is conducted to verify the best material for the gears in the gear box at higher speeds by analyzing stress, displacement and also by considering weight reduction.

The analysis is done in Cosmos software. Modeling is done in the Pro/Engineer.

3. INTRODUCTION TO WHEELED VEHICLE DRIVE LINES

3.1AXLES AND SUSPENSION SYSTEMS

Automotive drive lines and suspension systems have changed quite a bit since the first automobile was built. At first, automobile axles were attached directly to the main frame of the vehicle. 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 wheels would not touch the ground. If the wheel off the ground was a drive wheel, the vehicle lost traction 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 springs were the same type used on the horse-drawn buggy. They allowed the wheels and axles to move up and down separate from the body. The body moved very little compared to the wheels and axles, and the ride was much smoother.

Allowing the axles to move separate from the body also kept the wheels on the ground over rough roads, but this caused a new problem. The old drive train between the engine and the axle would not work. The train had to be made to move more. This was done by adding movable joints in the drive shaft known as universal joints. Some early vehicles used only one universal joint on the drive shaft, while later vehicles used two universal joints on the drive shaft. Drive shafts are now usually called propeller shafts. Some long-wheel-base trucks now use as many as four 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 gear was made up of the wheels, axles, springs, drive shaft, and transmission. The transmission was often mounted midway between the engine and rear axle. It was connected to the engine and the rear axle by drive shafts.

The term "running gear" is not used any more. A new term, "chassis," is now used to identify the old running gear plus the power plant. In modern vehicles, the transmission is generally mounted on the engine and is part of the power plant.

The chassis of modern vehicles, especially the frame, spring, and axles, must be very strong and yet quite flexible.

3.2 INTRODUCTION TO AXLE SHAFT

An axle is a central shaft for a rotating wheel or gear. On wheeled vehicles, the axle may be fixed to the wheels, rotating with them, or fixed to its surroundings, with the wheels rotating around the axle. In the former case, bearings or bushings are provided at the mounting points where the axle is supported. In the latter case, a bearing or bushing sits inside the hole in the wheel 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 axles serve to transmit driving torque to the wheel, as well as to maintain the position of the wheels relative to each other and to the vehicle body. The axles in this system must also bear the weight of the vehicle plus any cargo. A non-driving axle, such as the front beam axle in Heavy duty trucks and some 2 wheel drive light trucks and vans, will have no shaft. It serves only as a suspension and steering component. Conversely, many front wheel drive cars have a solid rear beam axle.

In other types of suspension systems, the axles serve only to transmit driving torque to the wheels; The position and angle of the wheel hubs is a function of the suspension system. This is typical of the independent suspension found on most newer cars and SUV's, and on the front of many light trucks. These systems still have a differential, but it will not have attached axle housing tubes. It may be attached to the vehicle frame or body, or integral in a transaxle. The axle shafts (usually C.V. type) then transmit driving torque to the wheels. Like a full floating axle 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 wheels on opposing sides of the vehicle, regardless of their mechanical connection type to each other and the vehicle frame or body.

3.2.1 Types of Rear Axle Shafts

In rear wheel drive vehicles, the rear wheels are the driving wheels, whereas in the vehicles with front wheels drive the front wheels are the driving wheels. Almost all the rear axles in the modern cars are live axles, which means that these axles move with the wheels, or revolve with the wheels and are known as live axles. Dead Axles are those axles which remain stationary and do not move with the wheels.

Rear axles / Live Axles are further classified into three types:

1. Full Float axles

2. Semi Float Axles

3. Three quarter floating axles.

3.2.2 Semi float axle:

The Semi float axle is used in light trucks and passenger vehicle / buses. In the vehicles equipped with Semi Float axle the shaft as well as the differential housing supports the weight of the vehicle . The wheel hub is directly connected to the axle shaft or is an extension of the same, the inner end of the axle shaft is splined and it is supported by the final drive unit. The outer end is supported by a single bearing inside the axle casing / axle tube. The vehicle load is transmitted to each of the axle shafts through the casing and the bearing, this causes a bending load and a tendency to shear at a point. Besides the side forces also cause end thrust and bending moment in the axle shafts, which have to take driving torque also. The semi float axle is the simplest and the cheapest of all types, because of which it is widely used on cars. However, since axle shafts have to support all loads, they have to be of larger diameter for the same torque transmitted to the other types of axle supporting. The axle shafts take the stress caused by turning, skidding or wobbling of the wheels. The axle shafts are flanged or tapered on the ends. When the tapered axles are used, the brake , the brake drum and hub are pressed onto the shafts, using keys to prevent the assemblies from turning on the shafts. In some cases, the outer ends of the shafts may have serrations or splines to correspond with those on the drum and hub assembly. If in case the axle shaft breaks on the vehicle using this type of arrangement, the wheel of the vehicle 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 to carry heavy loads. The axle shaft has flanges at the outer ends, which are connected to the flanged sleeve by means of bolts. There are two taper roller bearings supporting the axle casting in the hub, which take up any side load. Thus in case of Full Float axles , the axle shafts carry only the driving torque. The weight of the vehicle and the end thrust are not carried by them. The weight of the vehicle is completely supported by the wheels and the axle casing. As the axle shafts carry only the driving torque, their failure or removal does not effect the wheels. Thus the axle shafts can be taken out or replaced without jacking up the vehicle. For the same reason vehicle can be towed even with a broken half shaft. We can say that the axle shaft takes the whole 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 of axle. In Semi Floating axle the bearing is located between the axle casing and the hub instead of being between the axle casing and the shaft as in case of semi float axle. The axle shafts do not have to withstand any shearing or bending actions due to the weight of the vehicle, which are taken up by the axle casing through the hub and the bearing, provided the bearing lies in the plane 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, but with the passage of time and with more improvements in the design, development, materials and

fabrication techniques, preference is given to the Semi Float Axles, as these are simpler in design and cheaper to use.

4.LITERATURE SURVEY

4.1 INTRODUCTION TO GEAR BOX

A transmission or gearbox provides speed and torque conversions from a rotating power source to another device using gear ratios. In British English the term transmission refers to the whole drive train, including gearbox, clutch, prop shaft (for rear-wheel drive), differential and final drive shafts. In American English, however, the distinction is made that a gearbox is any device which 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 common use is in motor vehicles, where the transmission adapts the output of the internal combustion engine 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 the higher engine speed to the slower wheel speed, increasing torque in the process. Transmissions are also used on pedal bicycles, fixed machines, and anywhere else rotational speed and torque needs to be adapted.

Often, a transmission will have multiple gear ratios (or simply "gears"), with the ability to switch between them as speed varies. This switching may be done manually (by the operator), or automatically. Directional (forward and reverse) control may also be provided. Single-ratio transmissions also exist, which simply change the speed and torque (and sometimes direction) of motor output.

In motor vehicle applications, the transmission will generally be connected to the crankshaft of the engine. The output of the transmission is transmitted via driveshaft to one or more differentials, which in turn drive the wheels. While a differential may also provide gear reduction, 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-electric transmission, hydraulic drive system, etc.). Hybrid configurations also exist.

4.2 Explanation

Early 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 mover output shaft (e.g. a motor crankshaft). This means that the output shaft of a gearbox will rotate at slower 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 increase

in shaft speed with a reduction of torque. Some of the simplest gearboxes merely change the physical direction in which power is transmitted.

Many typical automobile transmissions include the ability to select one of several different gear ratios. In this case, most of the gear ratios (often simply called "gears") are used to slow down the output speed of the engine and increase torque. However, the highest gears may be "overdrive" types that increase the output speed.

4.3 Uses

Gearboxes have found use in a wide variety of different—often stationary—applications, such as wind turbines.

Transmissions are also used in agricultural, industrial, construction, mining and automotive equipment. In addition to ordinary transmission equipped with gears, such equipment makes extensive use of the hydrostatic drive and electrical adjustable-speed drives.

4.4 Simple

The simplest transmissions, often called gearboxes to reflect their simplicity (although complex systems are also called gearboxes in the vernacular), provide gear reduction (or, more rarely, an increase 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, since the axial PTO shaft is at odds with the usual need for the driven shaft, which is either vertical (as with rotary mowers), or horizontally extending from one side of the implement to another (as with manure spreaders, flail mowers, and forage wagons). More complex equipment, such as silage 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 much faster rotation of the electrical generator. These are much larger and more complicated than the PTO gearboxes in farm equipment. They weigh several tons and typically contain three stages to achieve an overall gear ratio from 40:1 to over 100:1, depending on the size of the turbine. (For aerodynamic and structural reasons, larger turbines have to turn more slowly, but the generators all have to rotate at similar speeds of several thousand rpm.) The first stage of the gearbox is usually a planetary gear, for compactness, and to distribute the enormous torque of the turbine over more teeth of the low-speed shaft. Durability of these gearboxes has been a serious problem for a long time.

Regardless of where they are used, these simple transmissions all share an important feature: the gear ratio cannot be changed during use. It is fixed at the time the transmission is constructed.

4.5 Multi-ratio systems

Many applications require the availability of multiple gear ratios. Often, this is to ease the starting and stopping of a mechanical system, though another important need is that of maintaining good fuel efficiency.

4.6 Automotive basics

The need for a transmission in an automobile is a consequence of the characteristics of the internal combustion engine. Engines typically operate over a range of 600 to about 7000 revolutions per minute (though this varies, and is typically less for diesel engines), while the car's wheels rotate between 0 rpm and around 1800 rpm.

Furthermore, the engine provides its highest torque outputs approximately in the middle of its range, while often the greatest torque is required when the vehicle is moving from rest or traveling slowly. Therefore, a system that transforms the engine's output so that it can supply high torque at low speeds, but also operate at highway speeds with the motor still operating within its limits, is required. Transmissions perform this transformation.

Many transmissions and gears used in automotive and truck applications are contained in a cast iron case, though more frequently aluminum is used for lower weight especially in cars. There are 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 generally have the engine and transmission mounted transversely, the differential being part of the transmission assembly.) The shaft is suspended by the main bearings, and is split towards the input end. At the point of the split, a pilot bearing holds the shafts together. The gears and clutches ride on the main shaft, the gears being free to turn relative to the main shaft except when engaged by the clutches. Types of automobile transmissions include manual, automatic or semi-automatic transmission.

4.7 Differential gear box

A differential is a device, usually but not necessarily employing gears, capable of transmitting torque and rotation through three shafts, almost always used in one of two ways: in one way, it receives one input and provides two outputs—this is found in most automobiles—and in the other way, it combines two inputs to create an output that is the sum, difference, or average, of the inputs.

In automobiles and other wheeled vehicles, the differential allows each of the driving roadwheels to rotate at different speeds, while for most vehicles supplying equal torque to each of them.

4.7.1 Purpose

A vehicle's wheels rotate at different speeds, mainly when turning corners. The differential is designed to drive a pair of wheels with equal torque while allowing them to rotate at different speeds. In vehicles without a differential, such as karts, both driving wheels are forced to rotate at the same speed, usually on a common axle driven by a simple chain-drive mechanism. When cornering, the inner wheel needs to travel a shorter distance than the outer wheel, so with no differential, the result is the inner wheel spinning and/or the outer wheel dragging, and this

results in difficult and unpredictable handling, damage to tires and roads, and strain on (or possible failure of) the entire drivetrain.

4.7.2Functional description

The following description of a differential applies to a "traditional" rear-wheel-drive car or truck with an "open" or limited slip differential:

Torque is supplied from the engine, via the transmission, to a drive shaft (British term: 'propeller shaft', commonly and informally abbreviated to 'prop-shaft'), which runs to the final drive unit that contains the differential. A spiral bevel pinion gear takes its drive from the end of the propeller shaft, and is encased within the housing of the final drive unit. This meshes with the large spiral bevel ring gear, known as the crown wheel. The crown wheel and pinion may mesh in hypoid orientation, not shown. The crown wheel gear is attached to the differential carrier or cage, which contains the 'sun' and 'planet' wheels or gears, which are a cluster of four opposed bevel gears in perpendicular plane, so each bevel gear meshes with two neighbours, and rotates counter to the third, that it faces and does not mesh with. The two sun wheel gears are aligned on the same axis as the crown wheel gear, and drive the axle half shafts connected to the vehicle's driven wheels. The other two planet gears are aligned on a perpendicular axis which changes orientation with the ring gear's 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 durability requirements. As the differential carrier rotates, the changing axis orientation of the planet gears imparts the motion of the ring gear to the motion of the sun gears by pushing on them rather than turning against them (that is, the same teeth stay in the same mesh or contact position), but because 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 10 full rotations. During that time, the left wheel will make more rotations because it has further to travel, and the right wheel will make fewer rotations as it has less distance to travel. The sun gears (which drive the axle half-shafts) will rotate in opposite directions relative to the ring gear by, say, 2 full turns each (4 full turns relative to each other), resulting in the left wheel making 12 rotations, 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 the drive shaft is held (say leaving the transmission 'in gear', preventing the ring gear from turning inside the differential), manually rotating one driven roadwheel causes the opposite roadwheel to rotate in the opposite direction by the same amount.

When the vehicle is traveling in a straight line, there will be no differential movement of the planetary system of gears other than the minute movements necessary to compensate for slight differences in wheel diameter, undulations in the road (which make for a longer or shorter wheel path), etc.

4.7.3 Loss of traction

One undesirable side effect of a conventional differential is that it can limit traction under less than ideal conditions. The amount of traction required to propel the vehicle at any given moment depends on the load at that instant—how heavy the vehicle is, how much drag and friction there is, the gradient of the road, the vehicle's momentum, and so on.

The torque applied to each driving wheel is a result of the engine, transmission and drive axles applying a twisting force against the resistance of the traction at that roadwheel. Unless the load is exceptionally high, the drivetrain can usually supply as much torque as necessary, so the limiting factor is usually the traction under each wheel. It is therefore convenient to define traction as the amount of torque that can be generated between the tire and the road surface, before the wheel starts to slip. If the torque applied to drive wheels does not exceed the threshold of traction, the vehicle will be propelled in the desired direction; if not, then one or more wheels will simply spin.

To illustrate how a conventional "open" (non-locked or otherwise traction-aided) differential can limit torque applied to the driving wheels, imagine a simple rear-wheel drive vehicle, with one rear roadwheel on asphalt with good grip, and the other on a patch of slippery ice. Based on the load, gradient, etcetera, the vehicle requires a certain amount of torque applied to the drive wheels to move forward. If the two roadwheels were connected together without a differential, each roadwheel would be supplied with an equal amount of torque, and would push against the road surface as hard as possible. The roadwheel on ice would quickly reach the limit of traction, but would be unable to spin because the other roadwheel to which it is connected to has good traction. Therefore, when the good traction of the asphalt plus the poor traction from the ice together exceed the minimum required for forward propulsion, the vehicle accelerates.

With an open differential, however, where each tire is allowed to rotate at different speeds, as soon as the tire atop the ice patch exceeds the threshold of traction available to it, it will begin to spin or "slip". Additionally, once the traction threshold is broken and the tire experiences slip, the traction available will also decrease in accordance to the laws of kinetic friction. Since an open differential limits total torque applied to both drive wheels to the amount utilized by the lower traction wheel multiplied by a factor of 2, when one wheel is on a slippery surface, the total torque applied to the driving wheels will be lower than the minimum torque required for vehicle propulsion. Thus, the vehicle will not be propelled.

A proposed way to distribute the power to the wheels, is to use the concept of gearless differential, of which a review has been reported by Provatidis , but the various configurations seem to correspond either to the "sliding pins and cams" type, such as the ZF B-70 available for early VWs, or are a variation of the ball differential.

Many newer vehicles feature traction control, which partially mitigates the poor traction characteristics of an open differential by using the anti-lock braking system to limit or stop the slippage of the low traction wheel, thus transferring more torque to the wheel with good traction. While not as effective in propelling a vehicle under poor traction conditions as a traction-aided

differential, it is better than a simple mechanical open differential with no electronic traction assistance.

A differential gearbox is a completely mechanical device that consist of an enclosure, some gears and shafts. It has three input/output shafts (I will call them x, y and z shaft). I said input/output because any of these three shafts can serve as either input or output shaft – but in any given moment 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, while maintaining constant torque (T) ratio between shafts at any given moment. In general we can say:

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 (note that 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 (usually weighted 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 other shafts (also usually weighted by some constants) – but note that the percentage of this “rotation rate splitting” doesn’t have to have 50:50 or any other fixed ratio. Instead, this ratio is leveled to maintain constant torque ratio between shafts (this works especially nice if applied loads have such characteristic that increased rotation rate increases the torque).

The torque ratio between any two shafts is always maintained constant (a normal, two-shafted gearbox behaves the same). The torque on all differential gearbox shafts can be increased or decreased only simultaneously and proportionally. For example, if your car is driving along a road and then one wheel steps into a mud, the increased drag will be simultaneously felt (in form of increased torque) also by both, the other wheel and the engine. But if the mud is so slippery that your wheel loses its grip, then reduced torque will also be felt by the other wheel and by the engine. One wheel may spin in the mud, but the reduced torque on the other wheel will not be enough to pull you out.

Three output shafts, together with belonging gears, are colored blue, yellow and green. The simplest 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 is driven by) another yellow double-gear (piggyback). The piggyback gear rotates around the green shaft (it is not fixed to the green shaft). Otherwise there is no difference between blue gear and yellow 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 are mounted to short stubs that go vertically from the shaft. As the shaft rotates both gears rotate all around, carried by the stubs. In addition, every of these two gears can also rotate around the stub it is mounted on.

The design shown below is invented to make things simpler. It is used sometimes in industry (for small gearboxes used in servo systems).

There are (only) two gears, the blue one and the green one. The green one is special - it is somewhat flexible. Streched by the yellow axle from inside, the flexible green gear touches

(becomes meshed with) the blue gear. The yellow axle can rotate inside the green 'ring-like' gear causing it to touch the blue gear at different positions, 360 degrees around.

The thing is that the blue gear and the green gear don't have the same number of teeth. For example the green could have 100, while the blue could have 101 teeth. After the yellow axle is rotated 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 green gear must be very much extruded 'deeply into your monitor'. It looks like a cup - the bottom end is non-deformable and its shaft is atteched there.)

Sure, there are many disadvantages of such differential gearbox mechanism. Teeth on blue and green gears must be relatively small (there must be many of them), and the difference between number 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 DIFFERENTIAL

The differential has three jobs:

To aim the engine power at the wheels To act as the final gear reduction in the vehicle, slowing the rotational speed of the

transmission one final time before it hits the wheels To 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 Train

The gearing of an automobile differential is illustrated as following in final form.

Automobile Differential Complete Schematic

Without the "square" set of four gears in the middle of the above diagram which yields to the figure below, both wheels turn at the same angular velocity. This leads to problems when the car negotiates a turn.

Automobile Wheel Drive without Differential

Now imagine the differential "square" alone, as illustrated in the following figure. It should be apparent that turning one wheel results in the opposite wheel turning in the opposite direction at

the same rate.

Automobile Differential Alone This is how the automobile differential works. It only comes into play when one wheel needs to rotate differentially with respect to its counterpart. When the car is moving in a straight line, the differential 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, the counterpart wheel turns at twice its normal speed as can be seen by examining the complete scheme of automobile differential. This can be problematic when one wheel does not have enough traction, such as when it is in snow or mud. The wheel without traction will spin without providing traction and the opposite wheel will stay still so that the car does not move. This is the reason for a device known as a "limited slip differential" or "traction control".

4.7.4.2 GEAR BOX CASING

Gearbox casing is the shell (metal casing) in which a train of gears is sealed. The gearbox casing houses important transmission components like gears and shafts. Thus the strength of the gear box casing is an important parameter to be taken in to account while designing. The strength of the gear box casing is important since the complete power is subjected to both static and dynamic loading.

4.7.4.3 BEVEL GEARS

Bevel gears are gears where the axes of the two shafts intersect and the tooth-bearing faces of the gears themselves are conically shaped. Bevel gears are most often mounted on shafts that are 90 degrees apart, but can be designed to work at other angles as well. The pitch surface of bevel gears is a cone.

4.7.4.3.1 Applications

The bevel gear has many diverse applications such as locomotives, marine applications, automobiles, printing presses, cooling towers, power plants, steel plants, railway track inspection machines, etc.

For examples, see the following articles on:

Bevel gears are used in differential drives, which can transmit power to two axles spinning 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 is turned in a vertical direction, the bevel gears change the rotation of the chuck to a horizontal rotation. The bevel gears in a hand drill have the added advantage of increasing the speed of rotation of the chuck and this makes it possible to drill a range of materials.

The gears in a bevel gear planer permit minor adjustment during assembly and allow for some displacement due to deflection under operating loads without concentrating the load on the end of the tooth.

Spiral bevel gears are important components on rotorcraft drive systems. These components are required to operate at high speeds, high loads, and for a large number of load cycles. In this application, spiral bevel gears are used to redirect the shaft from the horizontal gas turbine engine to the vertical rotor.

4.7.4.3.2 Advantages

This gear makes it possible to change the operating angle. Differing of the number of teeth (effectively diameter) on each wheel allows mechanical

advantage to be changed. By increasing or decreasing the ratio of teeth between the drive and driven wheels one may change the ratio of rotations between the two, meaning that the 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 Disadvantages

One 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.

5. GEAR DIFFERENTIAL

A gear differential is a mechanism that is capable of adding and subtracting mechanically. To be more precise, we should say that it adds the total revolutions of two shafts. It also subtracts the total revolutions of one shaft from the total revolutions of another shaft—and delivers the answer by a third shaft. The gear differential will continuously and accurately add or subtract any number of revolutions. It will produce a continuous series of answers as the inputs change. Figure 11-8 is a cutaway drawing of a bevel gear differential showing all of its parts and how they relate to each other. Grouped around the center of the mechanism are four bevel gears meshed together. The two bevel gears on either side are “end gears.” The two bevel gears above and below are “spider gears.” The long shaft running through the end gears and the three spur gears is the “spider shaft.” The short shaft running through the spider gears together with the spider gears themselves make up the “spider.” Each spider gear and end gear is bearing-mounted on its shaft and is free to rotate. The spider shaft connects with the spider cross shaft at the center block where they intersect. The ends of the spider shaft are 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 the axis of the spider shaft.

5.1 Bevel Gear Differential

6. AIM OF THE PROJECT

The main aim of the project is to focus on the mechanical design and contact analysis on assembly of gears in gear box when they transmit power at different speeds at 2400 rpm, 5000 rpm. Analysis is also conducted by varying the materials for gears, Cast Iron, and Aluminum Alloy.

The analysis is conducted to verify the best material for the gears in the gear box at higher speeds by analyzing stress, displacement and also by considering weight reduction.

Design calculations are done on the differential of Ashok leyland 2516M by varing materials and speeds. Differential gear is modeled in Pro/Engineer. Analysis is done on the differential by applying tangential and static loads. Freauency analysis is also done on the differential. Analysis is carried out using CosmosWorks.

7.DESIGN CALCULATIONS OF DIFFERENTIAL

7.2ALUMINUM ALLOY7475-T7617.2.1 2400 rpm

7.2.1.1.1 Crown wheel

Diameter 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=80mm

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/NG

V.R=TG/ TP=50/8=6.25V.R= NP/NG

6.25=2400/ NG

NG=384rpm

For satisfactory operation of bevel gears the number of teeth in the pinion must not be

Less than 48

√1+( vr )2 where v.r=velocity ratio

=48

√1+(6.25)2=7.5

Since the shafts are at right angles therefore pitch angle for the pinionθp1=tan-1(1/v.r)= tan-1(1/6.25)=9.0

Pitch angle of gear θp2=90°-9=81

We know that formative number of teeth for pinionTEP= TPsec θp1=8sec9 =8And formative number of teeth for gearTEG= TGsec θp2=50sec81 =319.622

Tooth form factor for the pinion

y1P=0.154-0.912/ TEP, for 20° full depth involute system

=0.154-0.912/ 8=0.04

And tooth form factor for geary1

G=0.154-0.912/ TEG

=0.154-0.912/ 319.622=0.151

since the allowable static stresses(σO) for both pinion and gear is same (i.e σO=172.33 Mpa) and y1

P is less than y1G, therefore the pinion is weaker. Thus the design should be based upon the

pinion

Allowable static stress(σO) =σu/3=517/3=172.33 Mpaσu=ultimate tensile strength=517 Mpa

TANGENTIAL 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

¿√(DG

2)

2

+¿¿

DG= pitch diameter of gear =475DP= pitch diameter of gear =80

V=Π D p N P60 × 1000

=10.048m/s

Cv==3/3+10.048=0.229

L=√( 4752

)2

+¿¿

=240.844

The factor (L-b/L) may be called as bevel factor

For 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=95

WT =(172.33x0.229)x95xΠx10x0.04(240.844−95

240.844) = 2922.51N

DYNAMIC LOAD(WD)

The dynamic load for bevel gears may be obtained in similar manner as discussed for spur gearsWD= WT+ WI

WD =WT+21V (b .C+W T )21 v+√b . c+W t

V= pitch line velocity b=face widthc=deformation/dynamic factor in N/mm

C=K . e

1EP

+1

EG

K=0.111 for 20° full depth involute systemEP=young’s modulus for material of pinion in N/mm2=70300 N/mm2

EG=young’s modulus for material of gear in N/mm2= 70300N/mm2

e=tooth error action in mme value for module=10 used in precision gears is e=0.023

c==0.111× 0.023

170300

+1

70300=90N/mm

WD =WT+21V (b .C+W T )21 v+√b . c+W t

WD = 2922.51+21 ×10.048 (95 × 90+2922.51 )21× 10.048+√95 ×90+2922.5

WD =10532.23N

STATIC TOOTH LOAD (WS)

The static tooth load or endurance strength of the tooth for bevel gear is given by

WS= σe.b.Π.m. y1P(

L−bL

)

(Flexible endurance limit) σe = 1.75XB.H.N = 1.75X140=245

WS=245 × 95× π × 10× 0.041 (240.844−95

240.844)

WS=18145NFor safety against tooth breakage the WS ≥1.25 Wd=13165.2875

WEAR LOAD (WW)

The maximum or limiting load for wear for bevel gears is given by

Ww=DP× b ×q× k

cosθ p1

Dp,b,q,k have usual meanings as discussed in spur gears except that Q is based on formative or

equivalent no.of teeth such that ratio factor Q= 2 T EG

T EG+T EP

=2× 319.622319.622+8

=1.951

K= load stress factor (also known as material combination factor )in N/mm2 given by

K=σes

2×sin∅1.4

¿)

σes= surface endurance limit in mpa or N/mm2

∅=pressure angle σes=(2.8×B.H.N-70)N/mm2

= (2.8×517-70)=322 N/mm2

K=(3222 ) sin20

1.4 ( 170300

+ 170300 )=0.72

cosθ p 1=cos 9 =0.987

Ww=80 ×95 ×1.951 × 0.72

0.987=10825.25N

Forces acting

WT=WNCOS

WN=normal load=WT/ COSWT=tangential force

WN=tangential force

cos∅ =2922.51cos20

=3110.070N

Radial force WR=W T tan∅=3110.070 tan 20=1131.972

Mean radius (Rm)=(L-b/2)sin θp 1

=(240.844-95/2)sin 9 =32.111

Axial force acting on the pinion shaftWRH=WTtan∅ sin θP1

Tangential force acting at the mean radiusWT=T/Rm

T= torque on the pinion

T =p×60

2 π × N P

=120803.4 ×60

2 π × 2400T=480.661N-m=480661N-mm

WT=480661/32.111=14968.733N

Axial force WRH=WTtan∅ sin θP1

=14968.733tan20 sin 9 =850.010N

Radial force acting on the pinion shaftWRH=WTtan∅ cosθP1

=14968.733tan20 cos 9 =5366.752N

7.2.1.1.2 SUN GEAR

Diameter of crown wheel = DG= 150mmDiameter of pinion = DP =70mmNumber of teeth on gear = TG =18Number of teeth on pinion = TP = 15

D= DG+ DP=220T= TG+ TP = 33

Module = 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=120803.4w

We know that velocity ratioV.R=TG/ TP= DG/DP= NP/NG

V.R= DG/DP =150/70=2.142V.R= NP/NG

2.142=2400/ NG

NG=1120.448rpm

Since the shafts are at right angles therefore pitch angle for the pinionθp1=tan-1(1/v.r)= tan-1(1/2.142)=25.025

Pitch angle of gear θp2=90°-25.025=64.974

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 =42.55

Tooth form factor for the piniony1

P=0.154-0.912/ TEP, for 20° full depth involute system=0.154-0.912/ 16.554=0.099And tooth form factor for geary1

G=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) and y1

P is less than y1G, therefore the pinion is weaker. Thus the design should be based upon the

pinion

allowable static stress(σO) =σu/3=517/3=172.33 Mpaσu=ultimate tensile strength=517 Mpa

TANGENTIAL 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 cone

¿√(DG

2)

2

+¿¿

DG= pitch diameter of gear =150mmDp= pitch diameter of gear =70mm

V=Π D p N P60 × 1000

=Π × 70× 2400

60× 1000=8.792m/sCv==3/3+8.792=0.254

L=√( 1502

)2

+¿¿

=82.764

The factor (L-b/L) may be called as bevel factor For 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=66.5

WT =(172.33x0.254)x66.5xΠx7x0.099(82.764−66.5

82.764)

=1244.7N

DYNAMIC LOAD (WD)

The dynamic load for bevel gears may be obtained in similar manner as discussed for spur gearsWD= WT+ WI

WD =WT+21V (b .C+W T )21 v+√b . c+W t

V= pitch line velocity b=face widthc=deformation/dynamic factor in N/mm

C=K . e

1EP

+1

EG

K=0.111 for 20° full depth involute systemEP=young’s modulus for material of pinion in N/mm2

EG=young’s modulus for material of gear in N/mm2

e=tooth error action in mme value for module=7 used in precision gears is e=0.0186

c=0.111× 0.0186

170300

+1

70300= 72.57N/mm

WD =WT+21V (b .C+W T )21 v+√b . c+W t

WD = 1244.7+21× 8.792(66.5 ×72.57+1244.7)

21× 8.792+√66.5× 72.57+1244.7WD = 5513.77

STATIC TOOTH LOAD (WS)

The static tooth load or endurance strength of the tooth for bevel gear is given by

WS= σe.b.Π.m. y1P(

L−bL

)

(Flexible endurance limit) σe = 1.75XB.H.N = 1.75X140 =245

WS= 245 ×66.5 × π ×7× 0.099(82.764−66.5

82.764)

WS=6966.47NFor safety against tooth breakage the WS ≥1.25 Wd=6892.2125WS > Wd

WEAR LOAD (WW)

The maximum or limiting load for wear for bevel gears is given by

Ww=DP× b ×q× k

cosθ p1

Dp,b,q,k have usual meanings as discussed in spur gears except that Q is based on formative or

equivalent no.of teeth such that ratio factor Q= 2T EG

T EG+T EP

=2 × 42.55

42.55+16.554=1.439

K= load stress factor (also known as material combination factor )in N/mm2 given by

K=σes

2×sin∅1.4

¿)

σes= surface endurance limit in mpa or N/mm2

∅=pressure angle σes=(2.8×B.H.N-70)N/mm2

= (2.8×140-70)=322 N/mm2

K=(3222 ) sin20

1.4 ( 170300

+ 170300 )=0.72

cosθ p 1=cos 25.025 =0.906

Ww=70× 66.5 ×1.439 ×0.72

0.906=5322.62N

Forces acting

WT=WNCOS

WN=normal load=WT/ COSWT=tangential force

WN=tangential force

cos∅ =1244.70.939

=1324.582N

Radial force WR=WT tan∅=1324.582 tan 20=482.108 N

Mean radius (Rm)=(L-b/2)sin θp 1

=(82.764-66.5/2)sin 25.025

=20.944N

Axial force acting on the pinion shaftWRH=WTtan∅ sin θP1

Tangential force acting at the mean radiusWT=T/Rm

T= torque on the pinion

T =p ×60

2 π × N P

=120803.4 ×60

2 π × 2400T=480.661N-m=480661N-mm

WT=480661/20.944=22949.818N

Axial force WRH=WTtan∅ sin θP1

=22949.818tan20 sin 25.0259 =3533.340N

Radial force acting on the pinion shaftWRH=WTtan∅ cosθP1

=22949.818tan20 cos 25.025 =7567.863N

7.2.2 5000 rpm

7.2.2.1.1 Crown wheel

Diameter 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 0mm

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=337.5BHP = 337.5x745.7w=251673.75w

We know that velocity ratio

V.R=TG/ TP= DG/DP= NP/NG

V.R=TG/ TP=50/8=6.25V.R= NP/NG

6.25=5000/ NG

NG=800rpm

For satisfactory operation of bevel gears the number of teeth in the pinion must not be

Less than 48

√1+( vr )2 where v.r=velocity ratio

=48

√1+(6.25)2=7.5

Since the shafts are at right angles therefore pitch angle for the pinionθp1=tan-1(1/v.r)= tan-1(1/6.25)=9.0

Pitch angle of gear θp2=90°-9=81

We know that formative number of teeth for pinionTEP= TPsec θp1=8sec9 =8And formative number of teeth for gearTEG= TGsec θp2=50sec81 =319.622

Tooth form factor for the piniony1

P=0.154-0.912/ TEP, for 20° full depth involute system=0.154-0.912/ 8=0.04And tooth form factor for geary1

G=0.154-0.912/ TEG

=0.154-0.912/ 319.622=0.151

Since the allowable static stresses(σO) for both pinion and gear is same (i.e σO=172.33 Mpa) and y1

P is less than y1G, therefore the pinion is weaker. Thus the design should be based upon the

pinion

Allowable static stress(σO) =σu/3=517/3=172.33 Mpaσu=ultimate tensile strength=517 Mpa

TANGENTIAL 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/s

b=face widthm=module=10y1p=tooth form factorL=slant height of pitch cone

¿√(DG

2)

2

+¿¿

DG= pitch diameter of gear =475DP= pitch diameter of gear =80

V=Π D p N P60 × 1000

=20.93m/s

Cv==3/3+20.93=0.125

L=√( 4752

)2

+¿¿

=240.844

The factor (L-b/L) may be called as bevel factor For 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=95

WT =(172.33x0.125)x95xΠx10x0.04(240.844−95

240.844) = 1595.225N

DYNAMIC LOAD (WD)

WD= WT+ WI

WD =WT+21V (b .C+W T )21 v+√b . c+W t

V= pitch line velocity b=face widthc=deformation/dynamic factor in N/mm

C=K . e

1EP

+1

EG

K=0.111 for 20° full depth involute systemEP=young’s modulus for material of pinion in N/mm2

EG=young’s modulus for material of gear in N/mm2

e=tooth error action in mme value for module=10 used in precision gears is e=0.023

c=0.111× 0.023

170300

+1

70300=89.73

WD =WT+21V (b .C+W T )21 v+√b . c+W t

W D=1595.225+21 ×20.933 (95 × 90+1595.225 )

21× 20.933+√95 ×90+1595.225WD =9830.39N

STATIC TOOTH LOAD (WS)

The static tooth load or endurance strength of the tooth for bevel gear is given by

WS= σe.b.Π.m. y1P(

L−bL

)

(Flexible endurance limit) σe = 1.75XB.H.N = 1.75X140=245W S=245× 95× π × 10× 0.04¿)WS=18145NFor safety against tooth breakage the WS ≥1.25 Wd=12287.9875ThereforeWS > Wd

WEAR LOAD (WW)

The maximum or limiting load for wear for bevel gears is given by

Ww=DP× b ×q× k

cosθ p1

Dp,b,q,k have usual meanings as discussed in spur gears except that Q is based on formative or

equivalent no.of teeth such that ratio factor Q= 2 T EG

T EG+T EP

=2× 319.622319.622+8

=1.951

K= load stress factor (also known as material combination factor )in N/mm2 given by

K=σes

2×sin∅1.4

¿)

σes= surface endurance limit in mpa or N/mm2

∅=pressure angle σes=(2.8×B.H.N-70)N/mm2

= (2.8×140-70)=322 N/mm2

K=(3222 ) sin20

1.4 ( 170300

+ 170300 )=0.72

cosθ p 1=cos 9 =0.987

WW=80 ×95 ×1.951 × 0.72

0.987=10825.25N

Forces acting

WT=WNCOS

WN=normal load=WT/ COSWT=tangential force

WN=tangential force

cos∅

WN=1595.225

cos20=1697.602N

Radial force WR=W T tan∅=1595.225 tan 20=580.614 N

Mean radius (Rm)=(L-b/2)sin θp 1

=(240.844-95/2)sin 9 =32.111

Axial force acting on the pinion shaftWRH=WTtan∅ sin θP1

Tangential force acting at the mean radiusWT=T/Rm

T= torque on the pinion

T =p×60

2 π × N P

=251673.75× 60

2 π × 5000

T=480.905N-m=480905.254N-mm

WT=480905.254/32.111=22961.480NAxial force

WRH=WTtan∅ sin θP1

=22961.480 tan20 sin 25.025 =3535.424N

Radial force acting on the pinion shaft

WRH=WTtan∅ cosθP1

=22961.480 tan20 cos 25.025 =7551.525N

7.2.2.1.1 SUN GEARDiameter of crown wheel = DG= 150mmDiameter of pinion =TP=70mmNumber of teeth on gear =18Number of teeth on pinion=15

D= 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=337.5BHP = 337.5x745.7w=251673.75w

We know that velocity ratioV.R=TG/ TP= DG/DP= NP/NG

V.R= DG/DP =150/70=2.142V.R= NP/NG

2.142=5000/ NG

NG=2334.267rpm

Since the shafts are at right angles therefore pitch angle for the pinionθp1=tan-1(1/v.r)= tan-1(1/2.142)=25.025

Pitch angle of gear θp2=90°-25.025=64.974

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 =42.55

Tooth form factor for the piniony1

P=0.154-0.912/ TEP, for 20° full depth involute system=0.154-0.912/ 16.554

=0.099And tooth form factor for geary1

G=0.154-0.912/ TEG

=0.154-0.912/ 42.55=0.132

Since the allowable static stresses(σO) for both pinion and gear is same (i.e σO=172.33 Mpa) and y1

P is less than y1G, therefore the pinion is weaker. Thus the design should be based upon the

pinion

Allowable static stress(σO) =σu/3=517/3=172.33 Mpaσu=ultimate tensile strength=517 Mpa

TANGENTIAL 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=7

y1p=tooth form factor

L=slant height of pitch cone

¿√(DG

2)

2

+¿¿

DG= pitch diameter of gear =150mmDp= pitch diameter of gear =70mm

V=Π D p N P60 × 1000

=Π × 70× 5000

60× 1000=18.316m/sCv==3/3+18.316=0.140

L=√( 1502

)2

+¿¿

=82.764

The factor (L-b/L) may be called as bevel factor For 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=66.5

WT=(172.33x0.140)x66.5xΠx7x0.099(82.764−66.5

82.764) = 686N

DYNAMIC LOAD (WD)

WD= WT+ WI

WD =WT+21V (b .C+W T )21 v+√b . c+W t

V= pitch line velocity b=face widthc=deformation/dynamic factor in N/mm

C=K . e

1EP

+1

EG

K=0.111 for 20° full depth involute systemEP=young’s modulus for material of pinion in N/mm2=70300 N/mm2

EG=young’s modulus for material of gear in N/mm2=70300 N/mm2

e=tooth error action in mme value for module=7 used in precision gears is e=0.0186

c=0.111× 0.0186

170300

+1

70300= 72.57N/mm

WD =WT+21V (b .C+W T )21 v+√b . c+W t

WD = 686+21× 18.316(66.5 ×72.57+686)

21× 18.316+√66.5 ×72.57+686WD = 4620.13N

STATIC TOOTH LOAD (WS)

The static tooth load or endurance strength of the tooth for bevel gear is given by

WS= σe.b.Π.m. y1P(

L−bL

)

(Flexible endurance limit) σe = 1.75XB.H.N = 1.75X140 =245

WS= 245 ×66.5 × π ×7× 0.099(82.764−66.5

82.764)

WS=6966.47NFor safety against tooth breakage the WS ≥1.25 Wd=5775.162WS > Wd

WEAR LOAD (WW)

The maximum or limiting load for wear for bevel gears is given by

Ww=DP× b ×q× k

cosθ p1

Dp,b,q,k have usual meanings as discussed in spur gears except that Q is based on formative or

equivalent no.of teeth such that ratio factor Q= 2T EG

T EG+T EP

=2 × 42.55

42.55+16.554=1.439

K= load stress factor (also known as material combination factor )in N/mm2 given by

K=σes

2×sin∅1.4

¿)

σes= surface endurance limit in mpa or N/mm2

∅=pressure angle σes=(2.8×B.H.N-70)N/mm2

= (2.8×140-70)=322

K=(3222 ) sin20

1.4 ( 170300

+ 170300 )=0.72

cosθ p 1=cos 25.025 =0.906

Ww=70× 66.5 ×1.439 ×0.72

0.906=5323.347N

Forces acting

WT=WNCOS

WN=normal load=WT/ COSWT=tangential force

WN=tangential force

cos∅ =686

cos20=730.025N

Radial force WR=WT tan∅=686 tan 20=249.683 N

Mean radius (Rm)=(L-b/2)sin θp 1

=(82.764-66.5/2)sin 25.025 =20.944N

Axial force acting on the pinion shaft

WRH=WTtan∅ sin θP1

Tangential force acting at the mean radiusWT=T/Rm

T= torque on the pinion

T =p ×60

2 π × N P

=251673.75× 60

2 π × 5000T=480.905N-m=480905.254N-mm

WT=480661/20.944=22961.480N

Axial force WRH=WTtan∅ sin θP1

=22961.480tan20 sin 25.0259 =3535.424N

Radial force acting on the pinion shaftWRH=WTtan∅ cosθP1

=22961.480 tan20 cos 25.025 =7551.525N

7.3 CAST IRON

7.3.12400 rpm

7.3.1.1.1Crown wheel

Diameter 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 = 80mm

Material used for both pinion and gear is cast iron

Brinell hardness number(BHN)=300Pressure angle of teeth is 20° involute system Ø=20°P=162BHP = 162x745.7w=120803.4w

We know that velocity ratioV.R=TG/ TP= DG/DP= NP/NG

V.R=TG/ TP=50/8=6.25V.R= NP/NG

6.25=2400/ NG

NG=384rpm

For satisfactory operation of bevel gears the number odf teeth in the pinion must not be

Less than 48

√1+( vr )2 where v.r=velocity ratio

=48

√1+(6.25)2=7.5

Since 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=81

We know that formative number of teeth for pinionTEP= TPsec θp1=8sec9 =8And formative number of teeth for gearTEG= TGsec θp2=50sec81 =319.622

Tooth form factor for the piniony1

P=0.154-0.912/ TEP, for 20° full depth involute system=0.154-0.912/ 8=0.04

And tooth form factor for geary1

G=0.154-0.912/ TEG

=0.154-0.912/ 319.622=0.151

Since the allowable static stresses(σO) for both pinion and gear is same (i.e σO=196 Mpa) and y1P

is less than y1G, therefore the pinion is weaker. Thus the design should be based upon the pinion

Allowable static stress(σO) =196 Mpa

TANGENTIAL 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 cutters

v=peripheral speed in m/sb=face widthm=module=10y1p=tooth form factorL=slant height of pitch cone

¿√(DG

2)

2

+¿¿

DG= pitch diameter of gear =475Dp= pitch diameter of gear =80

V=Π D p N P60 × 1000

=10.048m/sCv==3/3+10.048=0.229

L=√( 4752

)2

+¿¿

=240.844mm

The factor (L-b/L) may be called as bevel factor For 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=95mm

WT=(196x0.229)x95xΠx10x0.04(240.844−95

240.844)

=3243.079N

DYNAMIC LOAD (WD)

WD= WT+ WI

WD =WT+21V (b .C+W T )21 v+√b . c+W t

V= pitch line velocity b=face widthc=deformation/dynamic factor in N/mm

C=K . e

1EP

+1

EG

K=0.111 for 20° full depth involute system

EP=young’s modulus for material of pinion in N/mm2=103000 N/mm2

EG=young’s modulus for material of gear in N/mm2= 103000 N/mm2

e=tooth error action in mm

e value for module=10 used in precision gears is e=0.023

c=0.111× 0.0231

103000+

1103000

=131.479N/mm

WD =WT+21V (b .C+W T )21 v+√b . c+W t

WD = 3243.079+21× 10.048(95 ×131.479+3243.079)

21× 10.048+√95 ×131.479+3243.079WD = 13110.818N

STATIC TOOTH LOAD (WS)

The static tooth load or endurance strength of the tooth for bevel gear is given by

WS= σe.b.Π.m. y1P(

L−bL

)

(Flexible endurance limit) σe = 1.75XB.H.N = 1.75X300 =525N/mm2

WS= 525 ×95 × π ×10 × 0.04¿)WS=37933.706NFor safety against tooth breakage the WS ≥1.25 Wd=16388.522N

WEAR LOAD (WW)

The maximum or limiting load for wear for bevel gears is given by

Ww=DP× b ×q× k

cosθ p1

Dp,b,q,k have usual meanings as discussed in spur gears except that Q is based on formative or

equivalent no.of teeth such that ratio factor Q= 2T EG

T EG+T EP

=2× 319.622319.622+8

=1.951

K= load stress factor (also known as material combination factor )in N/mm2 given by

K=σes

2×sin∅1.4

¿)

σes= surface endurance limit in mpa or N/mm2

∅=pressure angle σes=(2.8×B.H.N-70)N/mm2

= (2.8×300-70)=770 N/mm2

K=7702sin 20

1.4 ( 1103000

+ 1103000 )

=2.812

cosθ p 1=cos 9 =0.987

Ww=80 ×95 ×1.951 ×2.812

0.987=42244.388N

Forces acting

WT=WNCOS

WN=normal load=WT/ COSWT=tangential force

WN=tangential force

cos∅ =3243.079

cos20=3451.212N

WT=tangential force= WNCOS =2228.217cos20 =2093.839N

Radial force WR=3243.079 tan 20=1180.384N

Mean radius (Rm)=(L-b/2)sin θp 1

=(240.844-95/2)sin 9 =32.111mm

Axial force acting on the pinion shaftWRH=WTtan∅ sin θP1

Tangential force acting at the mean radiusWRH=T/Rm

T= torque on the pinion

T =p×60

2 π × N P

=120803.4 ×60

2 π × 2400T=480.661N-m=480661N-mm

WRH=480661/32.111=14968.733N

Axial force WRH=WTtan∅ sin θP1

=14968.733tan20 sin 9 =850.010N

Radial force acting on the pinion shaftWRH=WTtan∅ cosθP1

=14968.733tan20 cos 9 =5366.752N

7.3.1.1.2 SUN GEAR

Diameter of crown wheel = DG= 150mmDiameter of pinion = DP = 70mmNumber of teeth on gear = TG = 18Number of teeth on pinion = TP = 15

D= 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=120803.4w

We know that velocity ratioV.R=TG/ TP= DG/DP= NP/NG

V.R= DG/DP =150/70=2.142V.R= NP/NG

2.142=2400/ NG

NG=1120.448rpm

Since the shafts are at right angles therefore pitch angle for the pinionθp1=tan-1(1/v.r)= tan-1(1/2.142)=25.025

Pitch angle of gear θp2=90°-25.025=64.974

We know that formative number of teeth for pinion

TEP= TPsec θp1=15sec25.025 =16.554And formative number of teeth for gearTEG= TGsec θp2=18sec64.974 =42.55

Tooth form factor for the piniony1

P=0.154-0.912/ TEP, for 20° full depth involute system=0.154-0.912/ 16.554=0.099And tooth form factor for geary1

G=0.154-0.912/ TEG

=0.154-0.912/ 42.55=0.132

Since the allowable static stresses(σO) for both pinion and gear is same (i.e σO=126.66 Mpa) and y1

P is less than y1G, therefore the pinion is weaker. Thus the design should be based upon the

pinion

Allowable static stress(σO) =196 Mpa

TANGENTIAL 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 cone

¿√(DG

2)

2

+¿¿

DG= pitch diameter of gear =150mmDp= pitch diameter of gear =70mm

V=Π D p N P60 × 1000

=Π × 70× 2400

60× 1000=8.792m/s

Cv==3/3+8.792=0.254

L=√( 1502

)2

+¿¿

=82.764

The factor (L-b/L) may be called as bevel factor For 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=66.5

WT=(196x0.254)x66.5xΠx7x0.099(82.764−66.5

82.764)

=1415.587N

DYNAMIC LOAD (WD)

WD= WT+ WI

WD =WT+21V (b .C+W T )21 v+√b . c+W t

V= pitch line velocity b=face widthc=deformation/dynamic factor in N/mm

C=K . e

1EP

+1

EG

K=0.111 for 20° full depth involute systemEP=young’s modulus for material of pinion in N/mm2=103000 N/mm2

EG=young’s modulus for material of gear in N/mm2=103000 N/mm2

e=tooth error action in mm

e value for module=7 used in precision gears is e=0.0186

c=0.111× 0.0186

1103000

+1

103000= 106.3269N/mm

WD =WT+21V (b .C+W T )21 v+√b . c+W t

WD = 1415.587+21× 8.792(66.5 ×106.3269+1415.587)

21× 8.792+√66.5× 106.3269+1415.587WD = 1883.579N

STATIC TOOTH LOAD (WS)

The static tooth load or endurance strength of the tooth for bevel gear is given by

WS= σe.b.Π.m. y1P(

L−bL

)

(Flexible endurance limit) σe = 1.75XB.H.N = 1.75X300 =525

WS= 525 ×66.5 × π ×7× 0.099(82.764−66.5

82.764)

WS=14928.16NFor safety against tooth breakage the WS ≥1.25 Wd=2354.39WS > Wd

WEAR LOAD (WW)

The maximum or limiting load for wear for bevel gears is given by

Ww=DP× b ×q× k

cosθ p1

Dp,b,q,k have usual meanings as discussed in spur gears except that Q is based on formative or

equivalent no.of teeth such that ratio factor Q= 2T EG

T EG+T EP

=2 × 42.55

42.55+16.554=1.439

K= load stress factor (also known as material combination factor )in N/mm2 given by

K=σes

2×sin∅1.4

¿)

σes= surface endurance limit in mpa or N/mm2

∅=pressure angle σes=(2.8×B.H.N-70)N/mm2

= (2.8×300-70)=770

K=¿¿)=2.812cosθ p 1=cos 25.025 =0.906

Ww=70× 66.5×1.439 ×2.812

0.906=20790.62N

Forces acting

WT=WNCOS

WN=normal load=WT/ COSWT=tangential force

WN=tangential force

cos∅ =1415.587

0.939=1507.54N

Radial force WR=WT tan∅=1415.587 tan 20=515.231 N

Mean radius (Rm)=(L-b/2)sin θp 1

=(82.764-66.5/2)sin 25.025 =20.944N

Axial force acting on the pinion shaftWRH=WTtan∅ sin θP1

Tangential force acting at the mean radiusWRH=T/Rm

T= torque on the pinion

T =p ×60

2 π × N P

=120803.4 ×60

2 π × 2400T=480.661N-m=480661N-mm

WT=480661/20.944=22949.818N

Axial force WRH=WTtan∅ sin θP1

=22949.818tan20 sin 25.0259 =3533.340N

Radial force acting on the pinion shaftWRH=WTtan∅ cosθP1

=22949.818tan20 cos 25.025 =7567.863N

7.3.2 5000 rpm

7.3.2 .1.1Crown wheel

Diameter 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 = 80mm

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=251673.75w

We know that velocity ratio

V.R=TG/ TP= DG/DP= NP/NG

V.R=TG/ TP=50/8=6.25V.R= NP/NG

6.25=5000/ NG

NG=800rpm

For satisfactory operation of bevel gears the number of teeth in the pinion must not be

Less than 48

√1+( vr )2 where v.r=velocity ratio

=48

√1+(6.25)2=7.5

Since the shafts are at right angles therefore pitch angle for the pinionθp1=tan-1(1/v.r)= tan-1(1/6.25)=9.0

Pitch angle of gear θp2=90°-9=81

We know that formative number of teeth for pinionTEP= TPsec θp1=8sec9 =8And formative number of teeth for gearTEG= TGsec θp2=50sec81 =319.622

Tooth form factor for the piniony1

P=0.154-0.912/ TEP, for 20° full depth involute system=0.154-0.912/ 8=0.04And tooth form factor for geary1

G=0.154-0.912/ TEG

=0.154-0.912/ 319.622=0.151

Since the allowable static stresses(σO) for both pinion and gear is same (i.e σO=196 Mpa) and y1P

is less than y1G, therefore the pinion is weaker. Thus the design should be based upon the pinion

Allowable static stress(σO) =196 Mpa

TANGENTIAL 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

¿√(DG

2)

2

+¿¿

DG= pitch diameter of gear =475Dp= pitch diameter of gear =80

V=Π D p N P60 × 1000

=20.93m/s

Cv=3/3+20.93=0.125

L=√( 4752

)2

+¿¿

=240.844

The factor (L-b/L) may be called as bevel factor For 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=95

WT=(196x0.125)x95xΠx10x0.04(240.844−95

240.844)

=1770.24N

DYNAMIC LOAD (WD)

WD= WT+ WI

WD =WT+21V (b .C+W T )21 v+√b . c+W t

V= pitch line velocity b=face widthc=deformation/dynamic factor in N/mm

C=K . e

1EP

+1

EG

K=0.111 for 20° full depth involute system

EP=young’s modulus for material of pinion in N/mm2

EG=young’s modulus for material of gear in N/mm2

e=tooth error action in mme value for module=10 used in precision gears is e=0.023

c=0.111× 0.0231

103000+

1103000

=131.479N/mm

WD =WT+21V (b .C+W T )21 v+√b . c+W t

W D=1770.24+21× 20.933 (95 ×131.479+1770.24 )

21 ×20.933+√95 × 131.479+1770.24WD = 11214.311N

STATIC LOAD (WS)

The static tooth load or endurance strength of the tooth for bevel gear is given by

WS= σe.b.Π.m. y1P(

L−bL

)

(Flexible endurance limit) σe = 1.75XB.H.N = 1.75X300 =525W S=525× 95× π × 10× 0.04¿)WS=37933.706NFor safety against tooth breakage the WS ≥1.25 Wd=14017.88ThereforeWS > Wd

WEAR LOAD (WW)

The maximum or limiting load for wear for bevel gears is given by

Ww=DP× b ×q× k

cosθ p1

Dp,b,q,k have usual meanings as discussed in spur gears except that Q is based on formative or

equivalent no.of teeth such that ratio factor Q= 2 T EG

T EG+T EP

=2× 319.622319.622+8

=1.951

K= load stress factor (also known as material combination factor )in N/mm2 given by

K=σes

2×sin∅1.4

¿)

σes= surface endurance limit in mpa or N/mm2

∅=pressure angle σes=(2.8×B.H.N-70)N/mm2

= (2.8×300-70)=770

K=7702sin 20

1.4 ( 1103000

+ 1103000 )=2.812

cosθ p 1=cos 9 =0.987

Ww=80 ×95 ×1.951 ×2.812

0.987=42244.388N

Forces acting

WT=WNCOS

WN=normal load=WT/ COSWT=tangential force

WN=tangential force

cos∅ =1770.24cos20

WN=1883.850N

Radial force WR=W T tan 20=1770.24tan20=644.314N

Mean radius (Rm)=(L-b/2)sin θp 1

=(240.844-95/2)sin 9 =32.111

Axial force acting on the pinion shaftWRH=WTtan∅ sin θP1

Tangential force acting at the mean radiusWRH=T/Rm

T= torque on the pinion

T =p ×60

2 π × N P

=251673.75× 60

2 π × 5000T=480.905N-m=480905.254N-mm

WT=480905.254/32.111=22961.480N

Axial force

WRH=WTtan∅ sin θP1

=22961.480 tan20 sin 25.025 =3535.424N

Radial force acting on the pinion shaftWRH=WTtan∅ cosθP1

=22961.480 tan20 cos 25.025 =7551.525N

7.3.2 .1.2SUN GEAR

Diameter 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=251673.75w

We know that velocity ratioV.R=TG/ TP= DG/DP= NP/NG

V.R= DG/DP =150/70=2.142V.R= NP/NG

2.142=5000/ NG

NG=2334.267rpm

Since 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

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 =42.55

Tooth form factor for the piniony1

P=0.154-0.912/ TEP, for 20° full depth involute system=0.154-0.912/ 16.554=0.099And tooth form factor for geary1

G=0.154-0.912/ TEG

=0.154-0.912/ 42.55=0.132

Since the allowable static stresses(σO) for both pinion and gear is same (i.e σO=196 Mpa) and y1P

is less than y1G, therefore the pinion is weaker. Thus the design should be based upon the pinion

Allowable static stress(σO) =196 Mpa

TANGENTIAL 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 cutters

v=peripheral speed in m/sb=face widthm=module=7y1p=tooth form factorL=slant height of pitch cone

¿√(DG

2)

2

+¿¿

DG= pitch diameter of gear =150mmDp= pitch diameter of gear =70mm

V=Π D p N P60 × 1000

=Π × 70× 5000

60× 1000=18.316m/s

Cv==3/3+18.316=0.140

L=√( 1502

)2

+¿¿

=82.764

The factor (L-b/L) may be called as bevel factor For 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=66.5

WT=(196x0.140)x66.5xΠx7x0.099(82.764−66.5

82.764)

=780.245N

DYNAMIC LOAD (WD)

WD= WT+ WI

WD =WT+21V (b .C+W T )21 v+√b . c+W t

V= pitch line velocity b=face widthc=deformation/dynamic factor in N/mm

C=K . e

1EP

+1

EG

K=0.111 for 20° full depth involute systemEP=young’s modulus for material of pinion in N/mm2=103000 N/mm2

EG=young’s modulus for material of gear in N/mm2=103000 N/mm2

e=tooth error action in mme value for module=7 used in precision gears is e=0.0186

c=0.111× 0.0186

1103000

+1

103000= 106.326N/mm

WD =WT+21V (b .C+W T )21 v+√b . c+W t

WD =780.245+21× 18.316(66.5 ×106.326+780.245)

21× 18.316+√66.5 ×106.326+780.245WD = 7161.239N

STATIC LOAD (WS)

The static tooth load or endurance strength of the tooth for bevel gear is given by

WS= σe.b.Π.m. y1P(

L−bL

)

(Flexible endurance limit) σe = 1.75XB.H.N = 1.75X300 =525

WS= 525 ×66.5 × π ×7× 0.099(82.764−66.5

82.764)

WS=14928.16For safety against tooth breakage the WS ≥1.25 Wd=8951.548NWS > Wd

WEAR LOAD (WW)

The maximum or limiting load for wear for bevel gears is given by

Ww=DP× b ×q× k

cosθ p1

Dp,b,q,k have usual meanings as discussed in spur gears except that Q is based on formative or

equivalent no.of teeth such that ratio factor Q= 2 T EG

T EG+T EP

=2 × 42.55

42.55+16.554=1.439

K= load stress factor (also known as material combination factor )in N/mm2 given by

K=σes

2×sin∅1.4

¿)

σes= surface endurance limit in mpa or N/mm2

∅=pressure angle σes=(2.8×B.H.N-70)N/mm2

= (2.8×300-70)=770

K=7702sin 20

1.4 ( 1103000

+ 1103000 )=2.812

cosθ p 1=cos 25.025 =0.906

Ww=70× 66.5 ×1.439 ×2.812

0.906=20790.62N

Forces acting

WT=WNCOS

WN=normal load=WT/ COSWT=tangential force

WN=tangential force

cos∅ =780.245cos20

=830.319N

Radial force WR=WT tan∅=780.245 tan20=283.985 N

Mean radius (Rm)=(L-b/2)sin θp 1

=(82.764-66.5/2)sin 25.025 =20.944N

Axial force acting on the pinion shaftWRH=WTtan∅ sin θP1

Tangential force acting at the mean radiusWRH=T/Rm

T= torque on the pinion

T =p×60

2 π × N P

=251673.75× 60

2 π × 5000T=480.905N-m=480905.254N-mm

WT=480905.254/20.944=22961.480N

Axial force WRH=WTtan∅ sin θP1

=22961.480tan20 sin 25.0259 =3535.424N

Radial force acting on the pinion shaftWRH=WTtan∅ cosθP1

=22961.480 tan20 cos 25.025 =7551.525N

8. INTRODUCTION TO CAD

Computer-aided design (CAD), also known as computer-aided design and drafting (CADD), is the use of computer technology for the process of design and design-documentation. Computer Aided Drafting describes the process of drafting with a computer. CADD software, or environments, provide the user with input-tools for the purpose of streamlining design processes; drafting, documentation, and manufacturing processes. CADD output is often in the form of electronic files for print or machining operations. The development of CADD-based software is in direct correlation with the processes it seeks to economize; industry-based software

(construction, manufacturing, etc.) typically uses vector-based (linear) environments whereas graphic-based software utilizes raster-based (pixelated) environments.

CADD environments often involve more than just shapes. As in the manual drafting of technical and engineering drawings, the output of CAD must convey information, such as materials, processes, dimensions, and tolerances, according to application-specific conventions. CAD may be used to design curves and figures in two-dimensional (2D) space; or curves, surfaces, and solids in three-dimensional (3D) objects.

CAD is an important industrial art extensively used in many applications, including automotive, shipbuilding, and aerospace industries, industrial and architectural design, prosthetics, and many more. CAD is also widely used to produce computer animation for special effects in movies, advertising and technical manuals. The modern ubiquity and power of computers means that even perfume bottles and shampoo dispensers are designed using techniques unheard of by engineers of the 1960s. Because of its enormous economic importance, CAD has been a major driving force for research in computational geometry, computer graphics (both hardware and software), and discrete differential geometry.

Current computer-aided design software packages range from 2D vector-based drafting systems to 3D solid and surface modellers. Modern CAD packages can also frequently allow rotations in three dimensions, allowing viewing of a designed object from any desired angle, even from the inside looking out. Some CAD software is capable of dynamic mathematic modeling, in which case it may be marketed as CADD — computer-aided design and drafting.

CAD is used in the design of tools and machinery and in the drafting and design of all types of buildings, from small residential types (houses) to the largest commercial and industrial structures (hospitals and factories). CAD is mainly used for detailed engineering of 3D models and/or 2D drawings of physical components, but it is also used throughout the engineering process from conceptual design and layout of products, through strength and dynamic analysis of assemblies to definition of manufacturing methods of components. It can also be used to design objects.

CAD has become an especially important technology within the scope of computer-aided technologies, with benefits such as lower product development costs and a greatly shortened design cycle. CAD enables designers to lay out and develop work on screen, print it out and save it for future editing, saving time on their drawings.

8.1 INTRODUCTION TO PRO/ENGINEER

Pro/ENGINEER Wildfire is the standard in 3D product design, featuring industry-leading productivity tools that promote best practices in design while ensuring compliance with your industry and company standards. Integrated Pro/ENGINEER CAD/CAM/CAE solutions allow you to design faster than ever, while maximizing innovation and quality to ultimately create exceptional products.

Customer requirements may change and time pressures may continue to mount, but your product design needs remain the same - regardless of your project's scope, you need the powerful, easy-to-use, affordable solution that Pro/ENGINEER provides.

Pro/ENGINEER Wildfire Benefits:

•Unsurpassed geometry creation capabilities allow superior product differentiation and manufacturability•Fully integrated applications allow you to develop everything from concept to manufacturing within one application•Automatic propagation of design changes to all downstream deliverables allows you to design with confidence•Complete virtual simulation capabilities enable you to improve product performance and exceed product quality goals•Automated generation of associative tooling design, assembly instructions, and machine code allow for maximum production efficiencyPro ENGINEER can be packaged in different versions to suit your needs, from Pro/ENGINEER Foundation XE, to Advanced XE Package and Enterprise XE Package, Pro/ENGINEER Foundation XE Package brings together a broad base of functionality. From robust part modelling to advanced surfacing, powerful assembly modelling and simulation, your needs will be met with this scaleable solution. Flex3C and Flex Advantage Build on this base offering extended functionality of your choosing.

The main modules are

Part Design

Assembly

Drawing

Sheet Metal

8.2 MODEL OF DIFFERENTIAL GEAR

8.2.1CROWN

8.2.2PINION

8.2.3 PLANET

8.2.4 SUNGEAR

8.2.5 MAIN SHAFT_PINION

8.2.6 MAIN SHAFT

8.2.7 CROWN ASSEMBLY

8.2.7 CROWN ASSEMBLY EXPLODE VIEW

9.INTRODUCTION TO FEA

Finite element analysis (FEA) is a fairly recent discipline crossing the boundaries of

mathematics, physics, engineering and computer science. The method has wide application and

enjoys extensive utilization in the structural, thermal and fluid analysis areas. The finite element

method is comprised of three major phases: (1) pre-processing, in which the analyst develops a

finite element mesh to divide the subject geometry into subdomains for mathematical analysis,

and applies material properties and boundary conditions, (2) solution, during which the program

derives 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 values

of primary quantities (such as displacements and stresses), and derives and examines additional

quantities (such as specialized stresses and error indicators).

The advantages of FEA are numerous and important. A new design concept may be modeled to

determine its real world behavior under various load environments, and may therefore be refined

prior to the creation of drawings, when few dollars have been committed and changes are

inexpensive. Once a detailed CAD model has been developed, FEA can analyze the design in

detail, saving time and money by reducing the number of prototypes required. An existing

product which is experiencing a field problem, or is simply being improved, can be analyzed to

speed an engineering change and reduce its cost. In addition, FEA can be performed on

increasingly affordable computer workstations and personal computers, and professional

assistance is available.

It is also important to recognize the limitations of FEA. Commercial software packages and the

required hardware, which have seen substantial price reductions, still require a significant

investment. The method can reduce product testing, but cannot totally replace it. Probably most

important, an inexperienced user can deliver incorrect answers, upon which expensive decisions

will be based. FEA is a demanding tool, in that the analyst must be proficient not only in

elasticity or fluids, but also in mathematics, computer science, and especially the finite element

method itself.

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 the

package depends on the type of analyses being performed. A typical finite element solution

requires a fast, modern disk subsystem for acceptable performance. Memory requirements are of

course dependent on the code, but in the interest of performance, the more the better, with a

representative range measured in gigabytes per user. Processing power is the final link in the

performance chain, with clock speed, cache, pipelining and multi-processing all contributing to

the bottom line. These analyses can run for hours on the fastest systems, so computing power is

of the essence.

One aspect often overlooked when entering the finite element area is education. Without

adequate training on the finite element method and the specific FEA package, a new user will not

be productive in a reasonable amount of time, and may in fact fail miserably. Expect to dedicate

one to two weeks up front, and another one to two weeks over the first year, to either classroom

or self-help education. It is also important that the user have a basic understanding of the

computer's operating system.

9.1 INTRODUCTION TO COSMOSWORKS

Cosmosworks is a useful software for design analysis in mechanical engineering. That’s an

introduction for you who would like to learn more about COSMOSWorks. COSMOSWorks is a

design analysis automation application fully integrated with SolidWorks.

This software uses the Finite Element Method (FEM) to simulate the working conditions of your

designs 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 the

integrity 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 material

property, 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 Analysis

Thermal Analysis Optimization Studies, Material property, Material Models, Linear Elastic

Isotropic.

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 the

checking.

9.2 STRUCTURAL ANALYSIS OF DIFFERENTIAL GEAR

9.2.1 2400 rpm

9.2.1.2 ALUMINUM ALLOY

9.2.1.2.1 TANGENTIAL LOAD

9.2.1.2.2 Material Properties

Model Reference Properties Components

Name: al_ alloy7475-t761Model type: Linear Elastic Isotropic

Default failure criterion:

Max von Mises Stress

Yield strength: 1.65e+008 N/m^2Tensile strength: 3e+007 N/m^2Elastic modulus: 7e+010 N/m^2

Poisson's ratio: 0.33 Mass density: 2600 kg/m^3

Shear modulus: 3.189e+008 N/m^2

Model

9.2.1.1.4UnitsUnit system: SI (MKS)

Length/Displacement mm

Temperature Kelvin

Angular velocity Rad/sec

Pressure/Stress N/m^2

Unit system: SI (MKS)

Length/Displacement mm

Temperature Kelvin

Angular velocity Rad/sec

Pressure/Stress N/m^2

Mesh type Solid Mesh

Mesher Used: Standard mesh

Automatic Transition: Off

Include Mesh Auto Loops: Off

Jacobian points 4 Points

Element Size 22.7482 mm

Tolerance 1.13741 mm

Mesh Quality High

Remesh failed parts with incompatible mesh Off

9.2.1.1.7 Mesh Information - DetailsTotal Nodes 39721Total 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) 0

Time to complete mesh(hh;mm;ss): 00:01:25Computer name: WIN

9.2.1.2.3 Loads and Fixtures

Fixture name Fixture Image Fixture Details

Fixed-1

Entities: 2 face(s)Type: Fixed Geometry

Resultant ForcesComponents X Y Z Resultant

Reaction force(N) 178.306 621.812 -91.4611 653.306Reaction Moment(N-m) 0 0 0 0

Load name Load Image Load Details

Force-1

Entities: 9 face(s)Type: Apply normal force

Value: 2922.51 N

Force-2

Entities: 11 face(s)Type: Apply normal force

Value: 1 N

9.2.1.2.4 Study Results

Name Type Min MaxStress1 VON: von Mises Stress 6.84248e-010 N/mm^2

(MPa)Node: 38649

3.19018 N/mm^2 (MPa)Node: 20475

2part_assm-2400_aluminiumally_tangential_load-Stress-Stress1

Name Type Min MaxDisplacement1 URES: Resultant Displacement 0 mm

Node: 31060.0241697 mmNode: 29507

2part_assm-2400_aluminiumally_tangential_load-Displacement-Displacement1

Name Type Min MaxStrain1 ESTRN: Equivalent Strain 2.08727e-014

Element: 208594.15934e-005 Element: 1069

2part_assm-2400_aluminiumally_tangential_load-Strain-Strain1

9.2.1.2.5 STATIC LOAD

9.2.1.2.6 Loads and Fixtures

Fixture name Fixture Image Fixture Details

Fixed-1

Entities: 2 face(s)Type: Fixed Geometry

Resultant ForcesComponents X Y Z Resultant

Reaction force(N) 1941.59 1881.96 -370.659 2729.27Reaction Moment(N-m) 0 0 0 0

Load name Load Image Load Details

Force-1

Entities: 9 face(s)Type: Apply normal force

Value: 18145 N

Force-2

Entities: 11 face(s)Type: Apply normal force

Value: 6966.47 N

9.2.1.2.7 Study Results

Name Type Min MaxStress1 VON: von Mises Stress 8.11597e-007 N/mm^2

(MPa)Node: 33574

19.8068 N/mm^2 (MPa)Node: 20475

2part_assm-2400_aluminiumally_static_load-Stress-Stress1

Name Type Min MaxDisplacement1 URES: Resultant Displacement 0 mm

Node: 31060.150063 mmNode: 29507

Name Type Min Max2part_assm-2400_aluminiumally_static_load-Displacement-Displacement1

Name Type Min MaxStrain1 ESTRN: Equivalent Strain 1.52786e-011

Element: 201070.000258239 Element: 1069

2part_assm-2400_aluminiumally_static_load-Strain-Strain1

9.2.1.3 CAST IRON9.2.1.3.1 TANGENTIAL LOAD

9.2.1.3.2 Material Properties

Model Reference PropertiesComponen

t

Name: Malleable Cast IronModel type: Linear Elastic Isotropic

Default failure criterion:

Max von Mises Stress

Yield strength: 2.75742e+008 N/m^2Tensile strength: 4.13613e+008 N/m^2Elastic modulus: 1.9e+011 N/m^2

Poisson's ratio: 0.27 Mass density: 7300 kg/m^3

Shear modulus: 8.6e+010 N/m^2Thermal expansion

coefficient:1.2e-005 /Kelvin

9.2.1.3.3 Loads and Fixtures

Fixture name Fixture Image Fixture Details

Fixed-1

Entities: 2 face(s)Type: Fixed Geometry

Resultant ForcesComponents X Y Z Resultant

Reaction force(N) 378.494 287.568 -36.4058 476.738Reaction Moment(N-m) 0 0 0 0

Load name Load Image Load Details

Force-1

Entities: 9 face(s)Type: Apply normal force

Value: 3243.08 N

Force-2

Entities: 11 face(s)Type: Apply normal force

Value: 1415.59 N

9.2.1.3.4 Study Results

Name Type Min MaxStress1 VON: von Mises Stress 2.37779e-007

N/mm^2 (MPa)Node: 38267

3.57544 N/mm^2 (MPa)Node: 20365

2part_assm-2400_castiron_tangential_load-Stress-Stress1

Name Type Min MaxDisplacement1 URES: Resultant Displacement 0 mm

Node: 31060.0100566 mmNode: 30373

2part_assm-2400_castiron_tangential_load-Displacement-Displacement1

Name Type Min MaxStrain1 ESTRN: Equivalent Strain 9.90157e-013

Element: 212851.69558e-005 Element: 9903

2part_assm-2400_castiron_tangential_load-Strain-Strain1

9.2.1.3.5 STATIC LOAD

9.2.1.3.6 Loads and Fixtures

Fixture name Fixture Image Fixture Details

Fixed-1

Entities: 2 face(s)Type: Fixed Geometry

Resultant ForcesComponents X Y Z

Reaction force(N) 4230.77 3829.18 -471.957Reaction Moment(N-m) 0 0 0

Load name Load Image Load Details

Force-1

Entities: 9 face(s)Type: Apply normal force

Value: 37933.7 N

Force-2

Entities: 11 face(s)Type: Apply normal force

Value: 14928.2 N

9.2.1.3.7 Study Results

Name Type Min MaxStress1 VON: von Mises Stress 2.48589e-006 N/mm^2

(MPa)Node: 38458

41.8212 N/mm^2 (MPa)Node: 20365

2part_assm-2400_castiron_static_load-Stress-Stress1

Name Type Min MaxDisplacement1 URES: Resultant Displacement 0 mm

Node: 31060.11763 mmNode: 30373

2part_assm-2400_castiron_static_load-Displacement-Displacement1

Name Type Min MaxStrain1 ESTRN: Equivalent Strain 8.223e-012

Element: 212850.000198329 Element: 9903

2part_assm-2400_castiron_static_load-Strain-Strain1

9.2.2 5000 RPM

9.2.2.2 ALUMINUM ALLOY

9.2.2.2 .1 TANGENTIAL LOAD

9.2.2.2 .2 Loads and Fixtures

Fixture name Fixture Image Fixture Details

Fixed-1

Entities: 2 face(s)Type: Fixed Geometry

Resultant ForcesComponents X Y Z Resultant

Reaction force(N) 196.301 117.513 -15.5613 229.316Reaction Moment(N-m) 0 0 0 0

Load name Load Image Load Details

Force-1

Entities: 9 face(s)Type: Apply normal force

Value: 1595.22 N

Force-2

Entities: 11 face(s)Type: Apply normal force

Value: 780.245 N

9.2.2.2 .3 Study Results

Name Type Min MaxStress1 VON: von Mises Stress 1.68385e-007 N/mm^2

(MPa)Node: 34326

1.70369 N/mm^2 (MPa)Node: 30516

2part_assm-5000_aluminiumally_tangential_load-Stress-Stress1

Name Type Min MaxDisplacement1 URES: Resultant Displacement 0 mm

Node: 31060.0131944 mmNode: 30373

2part_assm-5000_aluminiumally_tangential_load-Displacement-Displacement1

Name Type Min MaxStrain1 ESTRN: Equivalent Strain 1.78999e-012

Element: 215522.25582e-005 Element: 9903

2part_assm-5000_aluminiumally_tangential_load-Strain-Strain1

9.2.2.2 .4 STATIC LOAD

9.2.2.2 .5 Loads and Fixtures

Fixture name Fixture Image Fixture Details

Fixed-1

Entities: 2 face(s)Type: Fixed Geometry

Resultant ForcesComponents X Y Z Resultant

Reaction force(N) -1902.68 1875.88 446.924 2709.03Reaction Moment(N-m) 0 0 0 0

Load name Load Image Load Details

Force-1

Entities: 9 face(s)Type: Apply normal force

Value: 18145 N

Force-2

Entities: 11 face(s)Type: Apply normal force

Value: 6966.47 N

9.2.2.2 .6 Study Results

Name Type Min MaxStress1 VON: von Mises Stress 7.79346e-007 N/mm^2

(MPa)Node: 38461

22.6949 N/mm^2 (MPa)Node: 14584

2part_assm-5000_aluminiumally_static_load-Stress-Stress1

Name Type Min MaxDisplacement1 URES: Resultant Displacement 0 mm

Node: 31060.150036 mmNode: 31770

2part_assm-5000_aluminiumally_static_load-Displacement-Displacement1

Name Type Min MaxStrain1 ESTRN: Equivalent Strain 1.90385e-011

Element: 201870.000274774 Element: 12410

2part_assm-5000_aluminiumally_static_load-Strain-Strain1

9.2.2.3 CAST IRON

9.2.2.3 .1 TANGENTIAL LOAD

9.2.2.3 .2 Loads and Fixtures

Fixture name Fixture Image Fixture Details

Fixed-1

Entities: 2 face(s)Type: Fixed Geometry

Resultant ForcesComponents X Y Z Resultant

Reaction force(N) 195.449 153.057 -14.6943 248.682Reaction Moment(N-m) 0 0 0 0

Load name Load Image Load Details

Force-1

Entities: 9 face(s)Type: Apply normal force

Value: 1770.24 N

Force-2

Entities: 11 face(s)Type: Apply normal force

Value: 780.245 N

9.2.2.3 .3 Study Results

Name Type Min MaxStress1 VON: von Mises Stress 7.76622e-008 N/mm^2

(MPa)Node: 38416

2.01579 N/mm^2 (MPa)Node: 20475

2part_assm-5000_castiron_tangential_load-Stress-Stress1

Name Type Min MaxDisplacement1 URES: Resultant Displacement 0 mm

Node: 31060.00548866 mmNode: 29507

2part_assm-5000_castiron_tangential_load-Displacement-Displacement1

Name Type Min MaxStrain1 ESTRN: Equivalent Strain 5.14296e-013

Element: 184699.32532e-006 Element: 1069

9.2.2.3 .4 STATIC LOAD

9.2.2.3 .5 Loads and Fixtures

Fixture name Fixture Image Fixture Details

Fixed-1

Entities: 2 face(s)Type: Fixed Geometry

Resultant ForcesComponents X Y Z Resultant

Reaction force(N) 4093.63 3829.02 -1648.28 5842.6Reaction Moment(N-m) 0 0 0 0

Load name Load Image Load Details

Force-1

Entities: 9 face(s)Type: Apply normal force

Value: 37933.7 N

Force-2

Entities: 11 face(s)Type: Apply normal force

Value: 14928.2 N

9.2.2.3 .6 Study Results

Name Type Min MaxStress1 VON: von Mises Stress 1.50476e-006 N/mm^2 (MPa)

Node: 3864643.1949 N/mm^2 (MPa)Node: 20475

2part_assm-5000_castiron__static_load-Stress-Stress1

Name Type Min MaxDisplacement1 URES: Resultant Displacement 0 mm

Node: 31060.117614 mmNode: 29507

2part_assm-5000_castiron__static_load-Displacement-Displacement1

Name Type Min MaxStrain1 ESTRN: Equivalent Strain 1.19677e-011

Element: 184660.000199826 Element: 1069

2part_assm-5000_castiron__static_load-Strain-Strain1

10. RESULTS TABLE

10.1 2400 RPM

TANGENTIAL Aluminum Alloy Cast Iron

LOAD (N) 2922.51 3243.08DISPLACEMENT

(mm)0.0241696 0.0100566

STRESS (N/mm2) 3.19018 3.57544STRAIN 4.1593e-5 1.69558 e-5

STATICLOAD (N) 18143.3 37933.7

DISPLACEMENT (mm)

0.150063 0.11763

STRESS (N/mm2) 19.8068 41.8212STRAIN 0.000258239 0.000198329

10.2 5000 RPM

TANGENTIAL Aluminum Alloy Cast Iron

LOAD (N) 1595.22 1770.24DISPLACEMENT

(mm)0.0131944 0.00548866

STRESS (N/mm2) 1.70369 2.01579STRAIN 2.2558e-5 9.32532 e-6

STATICLOAD (N) 18143.3 37933.7

DISPLACEMENT (mm)

0.150036 0.117614

STRESS (N/mm2) 22.6949 43.1949STRAIN 0.000274774 0.000199826

11. FREQUENCY ANALYSIS OF DIFFERENTIAL GEAR

11.2 ALUMINUM ALLOY

11.2.1 Study Results

Name Type Min MaxDisplacement1 URES: Resultant Displacement

Plot for Mode Shape: 1(Value = 316.796 Hz)

0 mmNode: 3106

381.759 mmNode: 3280

2part_assm-aluminiumalloy_frequency-Displacement-Displacement1

Name Type Min MaxDisplacement2 URES: Resultant Displacement Plot

for Mode Shape: 2(Value = 332.013 Hz)

0 mmNode: 3106

396.691 mmNode: 4464

2part_assm-aluminiumalloy_frequency-Displacement-Displacement2

Name Type Min MaxDisplacement3 URES: Resultant Displacement

Plot for Mode Shape: 3(Value = 333.527 Hz)

0 mmNode: 3106

312.409 mmNode: 26119

2part_assm-aluminiumalloy_frequency-Displacement-Displacement3

Name Type Min MaxDisplacement4 URES: Resultant Displacement

Plot for Mode Shape: 4(Value = 865.487 Hz)

0 mmNode: 3106

723.674 mmNode: 2383

Name Type Min Max2part_assm-aluminiumalloy_frequency-Displacement-Displacement4

Name Type Min MaxDisplacement5 URES: Resultant Displacement

Plot for Mode Shape: 5(Value = 875.09 Hz)

0 mmNode: 3106

357.86 mmNode: 4034

2part_assm-aluminiumalloy_frequency-Displacement-Displacement5

11.2.2 Mode List

Frequency Number Rad/sec Hertz Seconds

1 1990.5 316.8 0.0031566

2 2086.1 332.01 0.0030119

3 2095.6 333.53 0.0029983

4 5438 865.49 0.0011554

5 5498.4 875.09 0.0011427

11.2.3 Mass Participation (Normalized)

Mode NumberFrequency(Hertz

)X direction

Y direction Z direction

1 316.8 9.8564e-005 5.8687e-007 0.083034

2 332.01 0.087176 6.2493e-007 8.9872e-005

3 333.53 0.0048507 8.7571e-009 6.3881e-006

4 865.49 4.1886e-008 0.010812 1.4807e-007

5 875.09 4.8546e-005 0.59579 3.4295e-007

Sum X = 0.092174 Sum Y = 0.60661 Sum Z = 0.08313

11.3 CAST IRON

11.3 .1 Study Results

Name Type Min MaxDisplacement1 URES: Resultant Displacement Plot

for Mode Shape: 1(Value = 316.612 Hz)

0 mmNode: 3106

232.328 mmNode: 28541

2part_assm-c_i__frequency-Displacement-Displacement1

Name Type Min MaxDisplacement2 URES: Resultant Displacement Plot

for Mode Shape: 2(Value = 331.666 Hz)

0 mmNode: 3106

241.953 mmNode: 30447

2part_assm-c_i__frequency-Displacement-Displacement2

Name Type Min MaxDisplacement3 URES: Resultant Displacement

Plot for Mode Shape: 3(Value = 343.648 Hz)

0 mmNode: 3106

186.029 mmNode: 27613

2part_assm-c_i__frequency-Displacement-Displacement3

Name Type Min MaxDisplacement4 URES: Resultant Displacement

Plot for Mode Shape: 4(Value = 861.879 Hz)

0 mmNode: 3106

206.394 mmNode: 19632

2part_assm-c_i__frequency-Displacement-Displacement4

Name Type Min MaxDisplacement5 URES: Resultant Displacement

Plot for Mode Shape: 5(Value = 884.969 Hz)

0 mmNode: 3106

448.102 mmNode: 2384

2part_assm-c_i__frequency-Displacement-Displacement5

11.3 .2 Mode List

Frequency Number Rad/sec Hertz Seconds

1 1989.3 316.61 0.0031584

2 2083.9 331.67 0.0030151

3 2159.2 343.65 0.00291

4 5415.3 861.88 0.0011603

5 5560.4 884.97 0.00113

11.3 .3 Mass Participation (Normalized)

Mode NumberFrequency(Hertz

)X direction

Y direction Z direction

1 316.61 5.4853e-005 3.5881e-007 0.081392

2 331.67 0.090131 7.0582e-007 4.1476e-005

3 343.65 8.998e-005 3.183e-009 4.3793e-006

4 861.88 5.5549e-005 0.607 1.6037e-007

5 884.97 6.3601e-008 0.0012827 8.1532e-007

Sum X = 0.090331 Sum Y = 0.60829 Sum Z = 0.081439

12.CONCLUSION

In our project we have designed a differential gear box for Ashok Leyland 2516M. Loads are calculated when the gears are transmitting different speeds 2400rpm and 5000rpm and different materials Aluminum Alloy and Cast Iron.

Structural and Frequency analyses are done on the differential gear box to verify the best material by taking in to account stresses, displacements, weight etc.

By observing the structural analysis results using Aluminum alloy the stress values are within the permissible stress value. So using Aluminum Alloy is safe for differential gear. When comparing the stress values of the three materials for all speeds 2400rpm and 5000rpm, the values are less for Aluminum alloy than Cast Iron.

By observing the frequency analysis, the vibrations are less for Aluminum Alloy than cast iron since its natural frequency is less.

And also weight of the Aluminum alloy reduces almost 3 times when compared with d Cast Iron since its density is very less. Thereby mechanical efficiency will be increased.

By observing analysis results, Aluminum Alloy is best material for Differential.

13. REFERENCES

1. J.O.Nordiana, S.O.Ogbeide, N.N.Ehigiamusoe and F.I.Anyasi., 2007,”Computer aided designof a spur gear, “Journal of Engineering and Applied Sciences 2 (12); pp 17431747.2. Zeping Wei., 2004”Stresses and Deformations in Involute spur gears by Finite Elementmethod,” M.S, Thesis, College of Graduate Studies and research, University of Saskatchewan,Saskatchewan.3. Darle W.Dudley, 1954, Hand book of practical gear design4. Alec strokes, 1970, High performance of gear design5. Maitra, G.M, 2004, Hand Book of Gear Design, TataMcGrawHill,New Delhi..6. S.Md.Jalaluddin., 2006, “Machine Design, “Anuradha publications, Chennai.7. Thirupathi Chandrupatla, Ashok D.Belegundu, “Introduction to finite element in Engineering” ,20038. PSG, 2008.”Design data,” Kalaikathir Achchagam publishers, Coimbatore, India.9. S.Mahalingam, R.E.D Bishop, 1974, “Dynamic loading of Gear tooth”, Journal of sound andvibration, 36(2), pp17918910. S.H.Choi, J.Glienicke, D.C.Han, K.Urlichs, April 1999, “Dynamic Gear Loads due to coupledlateral, Torsional and Axial Vibrations in a helical Geared System” , Journal of vibration and

acoustics, Vol 121 /141.