Calculation of the differential of the car. Example of differential gear calculation

The external circumferential module of bevel gears of differentials is recommended to be chosen by analogy with the design of differentials of modern transport vehicles. For these purposes, the following formulas are used

;

; (1)

;

where is the empirical coefficient, ,

The number of teeth of the satellite,

Estimated moment,

Number of satellites

where is the number of teeth of the side gear.

The relation to conical differentials is , and so on.

In all cases, the condition of assembly must be observed

,

where is an integer.

All differential gears are spur gears. Ring gear width

where is the outer cone distance

The parameters of the initial circuit are taken in accordance with GOST 13754-88. The following options are allowed: . The displacement coefficients and are assumed to be equal in absolute value, but for the satellite it is positive, and for the gear it is negative.

With the initial contour according to GOST, they accept:

, Then ,

at , then .

In the differentials, blocking takes place with the help of hydraulic friction clutches. If the friction clutch blocks the axle shaft of the differential, then the moment of friction of the clutch

where is the calculated radius of the drive wheel,

bevel gear efficiency.

According to formula (1)

The number of teeth of a flat wheel, and for resp. gears at 90º. Corresponds to the contour of the rack teeth.

Coefficient taking into account the influence of the two-sided application of the load,

Endurance limit of teeth in bending, corresponding to the basic number of stress cycles,

Factor taking into account the influence of the shape of the tooth and the concentration on the bend,

Coefficient taking into account the dynamic load,

Cog width factor.

To calculate the side gears and pinion gears, the greatest moment is selected for the adhesion of the drive wheels to the road surface.

where is the adhesion coefficient,

Gear ratio,

Efficiency of final bevel gear.

The moment acting on the satellite

The satellite cross is calculated for shear from circumferential force

where is the average radius of the circumferential force on the cross.

where is the average radius of the contact surface of the satellite and the spike of the cross with respect to the axis of the side gears,

Cross spike diameter

The length of the cylindrical surface of the satellite under the spike of the cross.

The collapse stress in the contact of the cross spike with the differential housing is also calculated.

where is the length of the cylindrical surface of the differential housing under the spike of the cross.

The gears of the cross and differential cracker are made of high-alloy steels used for the manufacture of transmission units, carburized to a depth of 1.5 ... 1.9 mm and hardened to HRC e from 58 to 63 with a core hardness of 30 to 40. ductile iron 35 ... 10 or steel.

The number of teeth of the satellite is determined by the following formula

,

where is the gear ratio from the satellite to the side gear.

Usually taken in calculations, based on the condition of placing the side gears of the splined end of the axle shaft of the desired diameter and limiting the size of the differential.

In the differential satellites from 2 to 4.

half shafts

The axle shafts are used to transmit torque from the center differential to the driving wheels of the machine and, in fact, are the drive shafts. With dependent wheel suspension, the axle shafts are located inside the crankcase and, as a rule, are connected to the axle gears of the differential with splines, and with the hubs of the drive wheels using splines or flanges that are integral with the axle shafts. All types of axle shafts are calculated for fatigue resistance and static strength, assuming that the beams do not deform. The following force factors acting on the axle shaft are taken into account in the calculation:

in the case of intensive acceleration or deceleration, the maximum torque and bending moments act along the axes;

when skidding the machine on a turn, the bending moment relative to the horizontal axis of the site is taken into account;

in case of crossing over an obstacle, the bending moment is taken into account relative to the horizontal axis to the platform of the dangerous section of the semi-axle.

Take into account the dynamism factor used for highly loaded Vehicle in the range from 2 to 2.5, and for off-road vehicles from 2.5 to 3.

When calculating the estimates of the static strength of the semiaxes, additional stresses are applied:

s w:ascii="Cambria Math" w:h-ansi="Cambria Math"/> RI"> .

In this case, the equivalent voltage, which is compared with the allowable one, is calculated according to the following formulas

,

where is the diameter of the semiaxis in the dangerous section.

For semi-floating and ¾ unloaded semi-axes during intensive acceleration or braking

,

where are the bending moments about the axes and .

When skidding a car around a corner

When crossing obstacles

In existing designs, the diameter of the axle shafts of load vehicles is taken mm.

planetary gears

Basic ratios of planetary mechanisms.

A planetary mechanism is a mechanism consisting of gears, in which the geometric axis of at least one wheel is movable. gear wheel with movable geometric axis called a satellite. The satellite may have one or more gear rims, or consist of several gears in engagement.

Classification of three-link planetary mechanisms

The link in which the axles of the satellites are installed is the carrier (h). The gear wheel, the geometric axis of which coincides with the main axis of the mechanism, is the central one (a, b, k). The main link of the planetary mechanism is called the link that perceives the external moment in the loaded gear, and is central.

a - sun gear,

h - carrier,

g - satellite,

b - crown gear (epicyclic).

A planetary mechanism in which all 3 main links rotate is called a differential. Planetary mechanisms are designated by the correspondence of the available satellites, gearings and parameter values. Planetary mechanisms, in which the main links are 2 central wheels and a carrier, are designated 2k-h. The planetary gearbox may consist of one planetary gear or several connected to each other. The classification of three-link planetary gears of type 2k-h is given in the classification of three-link planetary gears. Three-link planetary gears of type A and D are more common in planetary gearboxes, much less often than type B. The kinematic and power characteristics of three-link planetary gears are determined by its kinematic parameter r wsp:rsidR="00000000"> "> , equal to the gear ratio from link a to link b with the carrier h stopped.

where and are the angular velocity and frequency of rotation of the link, respectively.

Expressions for determining the parameter, taking into account the sign, are indicated in the classification table of three-link planetary mechanisms. Reduced equation of parameter r wsp:rsidR="00000000"> "> is known as the Willis formula and can be directly used for calculation in the analysis and synthesis of planetary gearboxes, but it is more convenient to use it in a transformed form:

This equation is often referred to as the basic equation of the kinematics of a three-link mechanism. In some cases, the parameter is used k. In this case, the main equation of kinematics takes the following form

Ministry of Education of the Russian Federation

South Ural State University

Department "Automobiles"

Explanatory note to the course project

On the course: "Design and calculation of the car"

On the topic: "Calculation of the car VAZ 2104"

AT - 434.00.00.00.00 PZ

Completed by: student of group AT-434

Ivanov I.I.

Checked by: Ulanov A.G.

Chelyabinsk 2010

1. Grip calculation

1.1 Evaluation of wear resistance of friction linings

1.2 Evaluation of the heat stress of adhesion

2. Calculation of the driveline

3. Differential calculation

4. Synchronizer calculation

1. Grip calculation

Clutch assignment. Clutch Requirements

The clutch is designed to smoothly start the car from a stop, short-term separation of the engine and transmission when shifting gears and to prevent the impact on the transmission of large dynamic loads that occur in transient conditions and when driving on poorly paved roads. When designing friction clutches, in addition to the basic requirements (minimum dead weight, simplicity of design, high reliability, etc.)

The following must be ensured:

Reliable transmission of torque from the engine to the transmission under any operating conditions;

smooth starting of the car and full engagement of the clutch;

complete disconnection of the engine from the transmission with a guaranteed gap between the friction surfaces;

Minimal moment of inertia of the driven clutch elements for easier gear shifting and reduced wear of the friction surface in the synchronizer;

Necessary removal of heat from the friction surface;

protection of transmission from dynamic overloads.

Selectable Options

We select the outer diameter of the driven disk from the condition that M d max \u003d 116NChm and the maximum speed of the engine crankshaft w max \u003d 5600 rpm \u003d 586.1 rad / s:

D n \u003d 204 mm - outer diameter of the lining,

D n \u003d 146 mm - inner diameter of the lining,

d=3.3 mm – friction lining thickness,

і=2 is the number of pairs of friction surfaces.

1.1 Clutch wear assessment

The degree of loading and wear resistance of clutch linings are usually evaluated by two main parameters:

specific pressure on friction surfaces;

specific work of clutch slipping.

Calculation of specific pressure on friction surfaces:

p 0 = ≤ , N/m 2 ,

where p pr is the force of normal compression of disks, N;

F is the area of ​​the working surface of one friction lining,

F \u003d \u003d 0.785 H (0.204 2 + 0.146 2) \u003d 0.049 m 2;

[p 0 ]=0.25 MPa - allowable pressure, providing the required service life of the linings.

Determining the force of normal compression:

where M d max is the maximum torque of the engine, LFm; =1.5 – clutch safety factor; =0.4 – coefficient of friction; R cf - the average radius of the friction lining,

Rav = 0.0875 m, p pr = 2.485 kN, a

p 0 = , 0,05 < 0,25 МПа –

the required resource of linings is provided.

Calculation of the specific work of clutch slipping:

where Lsp is the specific work of slipping; L d - the work of slipping when starting the car from a place, J; F sum - the total area of ​​the working surfaces of the overlays, m 2;

j,

where J a is the moment of inertia of the vehicle, reduced to the input shaft of the gearbox,

About f \u003d dCh (b f) H LFmb

where, m a \u003d 1445 kg - the total mass of the car; m n \u003d 0 kg - the total mass of the trailer; i k and i 0 are gear ratios of the gearbox and final drive, respectively (ik =3.67, i 0 =3.9); d=1.46 – coefficient for accounting for rotating masses.

J a \u003d 1.46H1400H \u003d 0.67 NChm 2;

Estimated angular frequency of rotation of the engine crankshaft, rad/s; for a car with a carburetor engine; = = 586.1 3 = 195.35 rad/s, where, M r is the moment of resistance to movement when starting off,

M m = g LFm,

where, w = 0.02 is the rolling resistance coefficient (on a horizontal road with an asphalt surface); c tr \u003d 0.82 - efficiency transmission.

M m = = 4.14 NFM.

L d = = 50652 J.

L beats \u003d \u003d 0.52 MJ / m 2

L beats \u003d 0.52 MJ / m 2 \u003d 4 MJ / m 2,

therefore, the required resource of the overlays is provided.

1.2 Evaluation of the heat stress of the bond

The heating of the clutch parts for one inclusion is determined by the formula:

where = 0.5 is the proportion of heat consumed to heat the part; c=0.48 kJ/(kgChK) – heat capacity of the part; m d - weight of the part kg; [Дt]=1015 .

m d \u003d CHN (R n - R int)

where \u003d 7200m 3 / kg - the density of cast iron, R n \u003d 102 mm - the outer radius of the pressure disk,

R ext \u003d 73mm - the inner radius of the pressure plate, m d \u003d 4.92 kg.

Dt = = 10.7 [Dt]

1.3 Diaphragm spring calculation

The calculation scheme for determining the parameters of the diaphragm spring is shown in fig. 1. The diaphragm spring is a Belvia spring modified for use in automotive clutches. The pressure of the spring is created by its section between the support rings mounted on rivets fixed to the clutch housing and the outer edge of the spring resting against the clutch pressure plate. The petals are at the same time switching off levers, their elasticity contributes to smooth start clutch.

E is the modulus of elasticity of the first kind;

0.25 - Poisson's ratio;

H is the height of the spring;

h is the thickness of the spring;

f pr - spring deflection;

We accept that: h=2mm, a=60mm, c=70mm, d=80mm, b=90mm, H=5mm.

Table 1

R press, kN	f,mm
4,29	1
5,0	2
3,66	3
1,82	4
1	5
2,73	6
5,03	6,5

Fig.1 Diaphragm spring

Fig. 2 Graph of the dependence of displacement on the force on the spring

car clutch differential synchronizer

2. Calculation of the driveline

Initial data:

Prototype: VAZ-2103 car

Max. frequent rotation: 5600 rpm \u003d 586.1 rad / s

Engine torque: 116 Nm

Gear ratio 1 gear: 3.67

Gear ratio 4 gears: 1.00

Shaft inner diameter: 66mm

Wall thickness: 2mm

Cardan shaft length:

“Gearbox - Intermediate Bearing”: 606mm

“Intermediate Support - Rear Axle”: 785 mm

Shaft material density: 7800 kg/m2

2.1 Determining the critical speed

,

Determining the maximum speed of the cardan shaft:

,

where = 1.1…1.2

Reduced moment of inertia:

Mass of cardan shaft

Then the critical angular velocity for the cardan shaft is:

Conditional check:

In this case, the condition is satisfied, because

2.2 Determination of torsional stress

Shaft torsion stress:

M cr \u003d M dv. max Ch i 1 Chz kp = 116Ch3.67Ch0.99 = 421

Nm - torque at the output shaft of the gearbox in low gear,

Torsional moment of resistance.

Hence,

The condition for the torsion stress of the cardan shaft is fulfilled.

2.3 Calculation of the universal joint shaft

Determination of the crushing stress of the spikes of the cross:

where r = 47.2 mm is the distance between the centers of the needle rollers,

Angle of installation of cardan shaft,

3 0 - for cars.

Therefore, the normal force

Fig.3 Cross joint of cardan shaft bearing stress:

Determination of the bending stress of the spikes of the cross:

Shear stress definition:

where d w is the diameter of the spike, d w \u003d 14.7 mm.

Therefore, shear stress:

Conclusion: In the calculation, the main parameters of the cardan shaft of the rear wheel drive VAZ - 2104 were determined. The results obtained satisfy all standards and assumptions.

3. Differential calculation

It is necessary to determine the load on the teeth of the satellites, side gears, the spider and the loads from the side of the satellites on the differential housing.

Assembly requirements: When analyzing and evaluating the design of a differential, as well as other mechanisms, one should be guided by the requirements for them:

Torque distribution between wheels and axles in a proportion that provides the best performance (maximum traction, good stability and controllability)

In addition, the differential, as well as all mechanisms of the car, is subject to such general requirements: ensuring minimum dimensions and weight, ease of installation and maintenance, manufacturability, maintainability.

Prototype: As a prototype, let's take the differential of a VAZ - 2104 car. The differential is conical, two-satellite.

3.1 Determining the load on the tooth of the satellite and side gears

The load on the tooth of the satellite and side gears is determined from the condition that the circumferential force is distributed equally between all the satellites, and each satellite transmits the force with two teeth. Circumferential force acting on one satellite:

where, r 1 is the radius of the application, r 1 = 0.025 m;

r 2 \u003d 0.036 m;

n s is the number of satellites, n s = 2;

M to max - the maximum moment developed by the engine, M to max = 116 NFm;

u KP1 - gear ratio of the first gear, u KP1 = 3.67;

u GP - gear ratio of the main gear, u GP = 3.9;

K Z = 1.7 - safety factor for the automotive industry;

The spike of the cross under the satellite is under shear stress

Fig.4 Satellite tooth

where [ = 100 MPa, based on this, you can find d;

The spike of the cross under the satellite is also subjected to crushing stress

where [ = 55 MPa, based on this, you can find l 1;

The thorn of the cross under the pinion experiences a crushing stress at the attachment point in the differential housing under the action of circumferential force

where [ \u003d 55 MPa, based on this, l 2 can be found;

3.2 Determining the pressure of the end of the satellite on the differential housing

The pressure of the end face of the satellite on the differential housing is determined by the collapse stress.

where [ = 15 MPa;

4. Synchronizer calculation

Assembly requirements: When analyzing and evaluating the design of a gearbox, as well as other mechanisms, one should be guided by the requirements for them:

ensuring optimal traction - high-speed and fuel - economic properties car with a given external characteristics of the engine;

noiseless operation and gear shifting;

ease of management;

high efficiency;

In addition, the gearbox, like all mechanisms of the car, has the following general requirements:

Ensuring minimum dimensions and weight;

simplicity of device and maintenance;

· manufacturability;

maintainability;

The gearbox is four-speed with synchronizers in all forward gears. The main gear is cylindrical, helical.

Gear ratios:

first gear - 3.75;

second gear - 2.30;

third gear - 1.349;

fourth gear - 1;

reverse - 3.53;

main gear - 3.9;

n is the maximum speed of the engine crankshaft,

n - 5600 rpm;

4.1 Determining the friction torque in the synchronizer

To equalize the angular velocities of the connected elements, it is necessary to create a friction moment M tr on the surfaces of the cones

where t is the synchronization time, t = 1 s;

J is the moment of inertia corresponding to the parts rotating together with the gear of the gear being engaged;

w e - angular velocity of the crankshaft,

- gear ratio of the included gear, \u003d 2.30,

- gear ratio of the switched off gear, = 3.75.

;

;

The moment of inertia of the drive shaft is determined from the ratio

The moment of friction created on the hull surfaces can be expressed in terms of the normal force P n on the synchronization cones:

(3)

where P n is the normal force on the friction surface;

µ - coefficient of friction, µ = 0.06;

r cf is the average radius of the cone.

In turn, the normal force can be expressed in terms of the force Q created by the driver when the gear is engaged.

Substituting equation (4) into equation (3) and expressing the average radius of the cone, we get the following

Q - the force created by the driver when the gear is engaged is determined by the formula

where P rych - the force applied to the gear shift knob; P roar =60 N;

5 drive gear ratio,

Q \u003d 60h5 \u003d 12 N,

The width of the synchronizer ring along the generatrix of the cone is determined by the formula

where = 1 MPa is the conditional allowable pressure.

Fig 1. Synchronizer circuit

The surfaces of the blocking elements are made at an angle that satisfies the condition

where µ is the coefficient of friction of blocking surfaces,

29 mm - the average radius on which the blocking elements are located

In order for the gear to not be engaged until the angular velocities are completely equalized, the force Q applied to the synchronizer clutch must be less

Fig 2. Diagram of the dynamic synchronizer system

Let a differential gear be given, in which the numbers of teeth of all wheels are known (Fig. 9):

Rice. 9. Differential transmission. Calculation example.

z 1 =80; z 2 =20; z 2" =30; z 3 =30; n 1 =300 rpm; n H=200 rpm.

It is required to determine the number of revolutions of all transmission wheels.

According to the Willis formula:

"-" sign in front of the value n 3 corresponds to the case when the direction of rotation of link 4 is opposite to the direction of rotation of links 1 and H.

n 2 =n 2' , because z 2 and z 2' are rigidly fastened on one shaft.

If in the differential gear the leading links are interconnected by an additional gear, then we get closed differential gear.

Differential Closed Gear

Closed differential gear has one leading link (mobility) and movable central wheels.

As an example, consider a differential transmission, (Fig. 10, A) in which two leading links 1 and H. If these links are closed next to the wheels 1 ` , 5` , 5, 4, then you get a closed differential gear (Fig. 10, b).

Rice. 10 Getting differential closed transmission

Usually, for a kinematic study of such gears, a system of two algebraic equations is compiled. One of them is an equation for determining the gear ratio from the driving link to the driven link of the differential part using the Willis formula. The second equation is the closure equation for determining the gear ratio of the ordinary part of the transmission.

As a result of solving the resulting system, the angular velocities of all links are determined, and, accordingly, the gear ratio of the mechanism.

For the case in Fig. 10, b we take 1 as the leading link. The system of equations is written as:

The numerator and denominator of the left side of equation (6) are divided by w 1:

,

using (7), we get

To determine the angular velocities of the satellites, we use the technique from the previous example:

planetary gears

A planetary mechanism, in which one of the central wheels is fixed motionless, is called planetary gear. The fixed center wheel is called pivotal. For example, if in a differential gear (Fig. 10) the central wheel 3 is rigidly connected to the rack, then a planetary gear with one degree of freedom will be obtained (Fig. 11).

Therefore, by setting the movement of the central wheel 1, the value of the angular velocity of the carrier is obtained H. If w is given H, then w 1 can be determined.

Planetary gears are used to obtain significant gear ratios, increased efficiency values ​​with dimensions smaller than those of ordinary gears.

Rice. 11. Planetary gear.

To derive the gear ratio formula in the planetary gear (Fig. 11), the Willis formula is used:

,

since w 3 =0.

Therefore, with the drive wheel 1. with lead leash H.

- gear ratio of the reverse movement with a fixed leash and a loosened wheel 3: .

In general, for planetary gears:

where is the gear ratio from the movable wheel 1 to the stationary central wheel n with the leash stopped H.

It is determined by relations (8) for ordinary transmissions.

Mixed transfers

Gears consisting of ordinary and planetary mechanisms are called mixed or combined. The procedure for calculating such transfers is as follows:

1. The entire transmission is divided into separate simplest types of known transmissions according to the principle: the output link of the previous one is the input for the next stage.

2. The gear ratios of the selected mechanisms are calculated.

3. The total gear ratio of the entire mixed connection is equal to the product of the individual gear ratios from paragraph 2.

4. The determination of the angular velocities of the central wheels and satellites is based on the methods described in the previous sections.

As an illustration, consider a number of examples.

Example 1 Determine the gear ratio of the gearbox (Fig. 12).

Rice. 12. Scheme of the gearbox.

Solution.

a) We divide the mixed connection into an ordinary gear with multiple gearing (1,2,2`,3) and a planetary gear (3`,4,4`,5, H);

b) ;

f) To find the angular velocity of the satellites:

Example 2 Determine the gear ratio of the gearbox (Fig. 13).

Rice. 13. Scheme of the gearbox.

Solution.

a) We select elementary transmissions: (1,2); (2`,3,3`,4, H 1); (H 2 , 4`,5, 5`,6);

b) ;

G) ;

e) ;

e) ;

g) To, for example, find the angular velocity of satellites 3 - 3`, we use the formula:

where can be determined from item d).

Example 3 Determine the gear ratio, w 4, w 5 of the gearbox (Fig. 14).

Rice. 14. Scheme of the gearbox.

Solution.

a) We distinguish the following steps: ordinary transmission 1,2,2`,3; planetary gear 3`,4,6, H; planetary gear H,5,7,4`,8; ordinary transmission 8`,9;

V) (the “–” sign is chosen in accordance with the rule of arrows);

G) ;

e) ;

and) ;

h) With the leading track 1 from points c) and d) we find:

; Further, ,

.

Example 4 Based on the initial data, determine the number of teeth of the 9th and 10th wheels of the mechanism (Fig. 15).

Rice. 15. Diagram of the gearbox

Given:z 1 =20; z 2 =60; z 3 =20; z 4 =15; z 5 =60; z 6 =65; z 7 =78; z 8 =24; n 1 =3200 rpm; n 10 =200 rpm.

Solution.

A) ;

;

V) ;

e) ,

;

e) ;

g) From the condition of coaxiality of the entire mechanism:

h) .

Work order

1. Draw up a kinematic diagram of the studied gear mechanism. If the scheme is known, then go to step 2.

2. Determine the degree of mobility and the type of mechanism.

3. Depending on the condition of the problem, form the values ​​of the initial data: the number of teeth of the wheels, the modulus, the angular velocities of the driving links, etc.

4. Create an algorithm for calculating the gear ratio of the connection.

5. Make calculations.

6. If necessary, determine the values ​​of the angular velocities of all links of the mechanism by setting the numerical value of the angular velocity of the leading link.

7. For a full-scale mechanism, check the correctness of the resulting gear ratio by marking the relative direction of rotation of the driving and driven links and measuring the number of revolutions.

8. Draw conclusions based on the results of the work.

5. Variants of calculation tasks

No. Varianta	Kinematic scheme	Conditions
		Given: z 0 =20, z 1 =30, z 2 =100, z 3 =100, z 4 =30, z 5 =90, z 6 =20, z 7 =30, z 8 \u003d 10, w 0 \u003d 55 s -1. Find: i 0-8 , w 1 , w 8 .
		Given: z 0 =20, z 1 =56, z 2 =22, z 3 =18, z 4 =68, z 5 =24, z 6 =24, z 7 =40, z 8 =44, z 9 =64, z 10 =22, z 11 =28, z 12 =40, z 13 =20, z 14 =18, z 15 =102, n 0 =900 rpm. Find: i 0-15 , n 15 , n 5 , n 9 .
		Given: z 0 =20, z 1 =40, z 2 =35, z 3 =70, z 4 =15, z 5 =30, n 5 =115 rpm. Find: n 1 , n 4 .
		Given: z 0 =20, z 1 =60, z 2 =20, z 3 =15, z 4 =60, z 5 =65, z 6 =78, z 7 =24, m 8-9 =6, n 0 =3200 rpm, n 9 =200 rpm. Find: center distance between 8 and 9 wheels.
		Given: z 0 =24, z 1 =24, z 2 =28, z 3 =80, z 4 =28, z 4 =26, z 5 =30, z 6 =12, z 7 =28, n 8 =250 rpm. Find: n 0 .
		Given: z 0 =20, z 1 =22, z 2 =80, z 3 =80, z 4 =18, z 5 =30, z 6 =30, z 7 =18, n 0 =650 rpm. Find: i 0-7 , n 4 .
		Given: z 0 =80, z 1 =30, z 2 =40, z 3 =28, z 4 =24, z 5 =42, z 6 =40, z 7 =80, z 8 =28, z 9 \u003d 40, w 0 \u003d 10 s -1. Find: i 0-9 , w 3 , w 5 .
		Given: z 0 =20, z 1 =60, z 2 =20, z 3 =15, z 4 =60, Z 5 =65, z 6 =78, z 7 =24, n 0 =3200 rpm, n 9 =200 rpm. Find: z 8 and z 9 .
		Given: z 0 =20, z 1 =17, z 2 =57, z 3 =80, z 4 =25, z 5 =20, z 6 =85, z 7 =90, z 8 =14, z 9 =61, n 0 =900 rpm. Find: i 0-9 , n 1 , n 5 .
		Given: z 0 =20, z 1 =40, z 2 =30, z 3 =34, z 4 =30, z 5 =34, z 6 =28, z 7 =40, z 8 =20, z 9 =70, n 0 =300 rpm. Find: i 0-9 , n 1 .

Literature

1. Theory of mechanisms and mechanics of machines: a textbook for universities / K.V. Frolov [and others]; MSTU im. N. E. Bauman; Ed. K.V. Frolova. - 5th ed., ster. - M .: Publishing house of MSTU im. N. E. Bauman, 2004 .- 662 p.

2. I. I. Artobolevsky. Theory of mechanisms and machines. M., 1988.

3. I. I. Artobolevsky, B. V. Edelshtein. Collection of tasks on the theory of mechanisms and machines. M., 1973.

Goal of the work:

Determine the load on the teeth of satellites, side gears,

crosspiece and loads from the satellites on the differential case.

Prototype:

Let's take the Kia Spectra differential as a prototype.

Differential conical, two-satellite

Determining the load on the tooth of the satellite and side gears

The load on the tooth of the satellite and side gears is determined from the condition that the circumferential force is distributed equally between all the satellites, and each satellite transmits the force with two teeth. Circumferential force acting on one satellite,

where r1 is the force application radius,

nc is the number of satellites, nc = 2;

Mmax - maximum moment,

developed by the engine

Mmax = 130 N.m;

iTR - transmission ratio,

iTR = iKP1* iGP = ;

Kd - coefficient of dynamism,

2.5 > Kd > 1.5, in the calculation we take Kd \u003d 2.

Figure 12 Calculation scheme of the differential

The spike of the cross under the satellite is under shear stress

Transforming the formulas, we get:

where we take τav = 120 MPa, and based on this we can find d:

The spike of the cross under the satellite is also subjected to crushing stress:

where we take σcm = 60 MPa, based on this we find l1;

The thorn of the cross under the pinion experiences a crushing stress at the attachment point in the differential housing under the action of circumferential force:

where is the force application radius m;

where we take σcm = 60 MPa, and based on this we find l2;

During the calculation, the load on the teeth of the satellites, side gears, the cross and the load from the side of the satellites on the differential housing was determined. The loads calculated taking into account all assumptions satisfy the accepted conditions.