Automotive Brake System Project

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Technical University of Sofia , Plovdiv branch
Project
topic: Designing the braking system of a car
Prepared by: Zhelyazko Nikolaev Angelov
Head: Assist Eng C. Taneva
Plovdiv
2012
CHAPTER I. INTRODUCTION
In 1990 . global fleet exceeds half a billion andannually increased by about 15 to 17,000,000 cars.
Vehicle is particularly dangerous object. Every year more than 300,000 people die in road accidents ,
not counting and wounded . Increasing vehicle safety remains one of the main directions in the
development of road transport.
Today automotive companies create different automatic control systems whose functionality
significantly exceed the capabilities of the driver. Create more and more new electronic engine
management , clutch , gearbox , etc. to enhance the performance properties of the car - in the first
place, these are safety , economy and ekologichnosta .
Brake control is the most important element in ensuring the safety of the movement of the car. New
step in perfecting a long creating anti-blocking system ( ABS ) that provides active safety of the
vehicle. The principle in the construction of each ABS is in maintaining the relative slip braking wheels
in a narrow range , providing high value for the coefficient of grip to the wheels with the ground at
any coverage. This allows you to keep the robustness of the car under braking and reduce stopping
distance .
The use of new , even the most advanced and stable braking systems do not prevent failures in their
work. For this reason, they need systematic and careful inspections during operation . Prevention of
brake systems using the latest methods and equipment will avoid or dramatically reduce the number
of victims or road accidents and victims .
Continuous improvement and sophistication of technology increased requirements to specialists in
the repair and service. Before we proceed to the repair must first establish exactly the type of fault ,
find a reason for it , and this requires knowledge of the principle of operation of all components of
the braking system. Correct determination of the cause of the fault and its location becomes
essential , even more important than the actual repair .
CHAPTER II. DYNAMIC PROPERTIES OF CAR
2.1. coupling calculation
2.1.1. Selection of the vehicle weight
Load capacity of passenger cars in kN is determined by the formula:
GТ   0,75  0,1.z ,
where
z = 5e number of passengers including the driver ;
0.75 - the weight of a passenger , kN;
0.2 - weight of luggage per passenger , kN.
 =  2 (0.75 + 0.1) = 0,85.5 = 4,25 
After determining the total weight of the car is given by :
G  1  nт  Gт ,
as the coefficient of load accepted
nт  2,5  3,5 .
 = (1 + 2.5)4,75 = 14.785 
2.1.2 . Selection of car tires
Tires are selected from catalogs and other catalogs where where
indicated : name and size of the tire and rim and the corresponding
(or several Type size of the wheels , to which may be mounted tire )
outside diameter , width , static or dynamic range , rated load with a
corresponding air pressure , type or tread permissible speed and
more.
Of similar models made since accept tires:
Front : 195/65 R 15
Rear : 195/65 R 15
The radius of the wheel is defined in [m] according to equation:
d

rк  103   .h  ,
2

Where:
d is the diameter of the rim , [mm]
λ - coefficient of radial deformation of the wheel ( 0.85 to 0.9 ) .
h - height of the tire , [mm]
 = 10−3 (
281
2
+ 0,85.126,75) = 0.267 
The ratio of the radial deformation of the wheels taken λ = 0,85 0,9, as lower values for light and
higher - for trucks. For the normal tire section height them h  b , and for low-profile tires
h   0,7  0,85 b where the section width of the tire.
2.1.3 . Selection of the efficiency of the transmission of the vehicle .
In the absence of kinematic scheme of the transmission, the calculation value of mechanical
efficiency ηm are:
For cars:
 м = 0,92  0,93;
The calculations assume :  м =
0,92
2.1.4 . Selection coefficient of drag and frontal area of the car.
Coefficient of drag kv is determined experimentally by blowing a model of the car in a wind tunnel or
road test of constructive similar designed car.Top surface S is chosen from prototype design and plot
the external transverse contour of the car and measure its area. To estimate in m2 can be used
dependence :
S  B.H Г
Where:
B is the track of the vehicle , [m]
HГ - overall height of the vehicle , [m]
In the absence of accurate data and a prototype design for the coefficient kv and frontal area S
calculations can be used totals .
The selected prototype frontal area :
Sq = 2,5.10-4 [kNs2/m4]
S = V.Hr = 0.5 [m2]
2.1.5 . Selection of motor vehicles.
Engine designed car is chosen from the condition to ensure the specified maximum speed Vmax.
Necessary engine power Pv for the maximum speed of the car when the print job driving conditions
are defined as:
2

Vmax
Pv   v .G  kв .S
13

 Vmax

 3,6. м
Where:
ψv is the coefficient of resistance of the road when driving at maximum speed – acceptance ψv =
0,02 0,04 minor.
The estimated power is selected the required maximum power . The type and design of the engine is
taken :
For carburetor engines with excess angular momentum.
Pe max   0,95  1,05 Рv

2202  220
Pv   0,02.14,785  0,00025.1,7


13
3,6.0,92


  0,295  1,58  66,425  124,4kW
Pe max  0,95Рv  0,95.124  117,8kW
As power
Pe max is selected existing engine or slightly larger maximum net power . Details of the
construction of the engine characteristics of the selected vehicle into the technical data of the
vehicle. They are 118kW, for 7600min- 1 , and at 150Nm 7000min- 1. Check for the correct choice of
engine is graphically capacity balance of the car in motion to direct transmission. For this purpose it is
necessary to have a graphical form of the external characteristic of the selected gear motor. In case
no such feature , and we can build it based on the data of the selected engine empirical correlations:
2
3
 
 
  
Pe  Pe max  a
 b
  c 
 

 p
 p 
  p  

,
2

  

M д  M p a  b
 c
 

p

  p  

Where:
ω is the current rotational speed of the crankshaft ;
ωp - angular velocity of the crankshaft at maximum power ;
Mp - engine torque and maximum power .
Mp 
Pe max
p
The coefficients a, b, and c depend on the type of engine are determined by the following relations :
For carburetor engines limiting angular velocity
a
1  k м .k  2  k 
 k  1
2
,b
2 1  a 
2  k
b
, с  k
2
Where:
kω is the coefficient of adjustment of the engine rotational speed :
k 
 =
p
,
м
837
785
= 1,07 ,
km - coefficient of adjustment of engine torque :
kм 
 =
M д max
Mp
150
146
,
= 1,02 .
=
1−1,02.1,07(2−1,07)
=
(1,07−1)2
= 0.4688
2(1 − 0.4688)
= 1.155
2 − 1,07
=
1,155
1,07 = −0.624
2
External speed feature is built in the velocity range ωmin - ωv.
Minimum steady rotational speed of the engine ωmin taken:
- For carburetor engines ωmin = 100 to 150 [rad / s]
Maximum angular velocity of the motor ωv, where the car has a top speed Vmax is taken :
For carburetor engines limiting angular velocity:
v   0,9  1,00  p
The calculations assume 726 [rad / s]
The formulas for Pe and Md is built outside speed characteristic of the engine speed range ωmin - ωv
( Figure 2.1 ) . In order to apply this feature capacity balance of the car to direct gear transmission
ratio of the central gear is determined by the equation:
iц  3,6
v .rк
Vmax
.
The selected prototype gear ratio of the central gear is :
ц = . 
.,

= .  ,
For the calculation of the speed using the relationship :
V  3,6
.rк
iц
,
The calculated speed is applied to the x-axis in the construction of the external speed characteristic.
The magnitude of V is chosen so as to match the Vmax ωv.
The results of the calculations for the construction of outdoor gear characteristics are shown in
tabl.2.1 .
Table 2.1
W
rad/s
V
km/h
Mg
Pe
Nm
kW
104,66
25,9049 90,28571
9,47661
157
38,8598 99,53863
15,6727
209,33
51,8123 107,9894
22,6707
261,66
64,7647 115,6399
30,3458
314
77,7196 122,4914
38,5734
366,33
90,6721 128,5412
47,2246
418,66
103,625 133,7907
56,1747
471
116,579 138,2407
65,2995
523,33
129,532 141,8895
74,4696
575,66
142,484
144,738
83,5606
628
155,439 146,7865
92,4483
680,33
168,392 148,0343
101,003
732,66
181,344 148,4818
109,101
785
194,299 148,1289
116,617
837,33
207,252 146,9757
123,423
868,73
220,815 145,8996
127,114
Outdoor gear characteristics
160
140
140
120
120
100
Mg, Nm
100
80
80
60
60
Mg
Pe
Nm
kW
40
40
20
20
0
0
0
50
100
150
200
250
w, rad/s
Fig. 2.1
2.1.6. Determining the gear ratios in the transmission of the vehicle.
The gear ratio of the gear its center is determined based on the specified maximum speed Vmax in
selecting the engine car designed by its formula mentioned above.
The gear ratio in first gear in the gearbox is determined by the formula:
iк1 
D1max .G.rк
M д max .iц . м
,
Where:
D1 max = ψ1 max is the maximum dynamic factor in first gear .
When α1 is not assigned to D1 max = ψ1 max values are accepted :
-for cars ψ1 max = 0,3 - 0,5.
The calculations assume ψ1 max = 0,4
1 =
0,4.14,785.0,267
= 3,23
0,15.3,3.0,92
The most common way to determine the gear ratios of the intermediate gears in the gearbox is by
ranking them in order with the following geometric correction.
Quotient of the geometric progression in gear ratio of direct drive ik = 1 is defined by:
q  n 1 iк1
,
Where:
n is the number of gears in the gearbox - usually assumed n = 3-6 for cars and trucks .
4
 = √3,23 = 1,53
Gear ratios of the intermediate gears determine the following equations :
For the second gear:
2 =
1

=
3,23
1,53
= 2,11
For the third gear:
3 =
1 3,23
=
= 1,38
 2 2,34
For the fourth gear:
4 =
1 3,23
=
= 0.90
 3 3,58
2.2.3 . Dynamic properties of the vehicle
Dynamic properties of avtomobilite set by solving the equations of power and capacity balance :
Fк  Ff  Fi  Fв  Fa  F  Fв  Fa ;
Pк  Pe . м  P  Pв  Pa .
These are nonlinear differential equations as entering them power values Pe and torque Md on the
outside of the engine speed characteristics are nonlinear functions of the angular velocity ω, related
to the speed of movement V.
2.2.1. Power balance of the car
This graph is constructed based on the external characteristic of the engine speed - the dependence
Md - ω. For different values of the angular velocity in the range ωmin - ωv calculated driving force FK
different gears equality :
Fк 
M д .iц .iк м
rк
,
By setting values of the angular velocity ω is calculated speed of movement of the car in all gears
dependence :
V  3,6
.rк
iц .iк
The results of the calculations are done in tabl.2.3 . They are used to construct the graph of the
power balance of the car in Fig.2.2 as FK = f (V) - for each gear:
The strength of the movement Ff vehicle for horizontal path is calculated by the equation:
Ff  f .G
Coefficient of resistance movement f is assumed to be constant and is selected from a table .
Air resistance FB is given by :
V2
Fв  kв .S
13
.
Results of calculations of the forces Ff, and FB depending on the speed V are plotted in tabl.2.2 .
Results in response curves forces Ff, Ff + FB . Their point of intersection of the curves of FK set the
maximum speed of the vehicle is moving in a horizontal V0 max and Vmax on sloping road section.
Table .2.2
V
Ff
Fв
km/h
---kN
0 kN-- 0
Ff + Fв
kN
0
0
45,729818
0,2218
0,0684
0,2901
60,97212
0,2218
0,1215
0,3433
76,214422
0,2218
0,1899
0,4117
91,459636
0,2218
0,2735
0,4952
106,70194
0,2218
0,3722
0,594
121,94424
0,2218
0,4861
0,7079
137,18945
0,2218
0,6153
0,8371
152,33564
0,2218
0,7587
0,9804
167,67406
0,2218
0,9191
1,1409
182,91927
0,2218
1,0939
1,3156
198,16157
0,2218
1,2838
1,5055
213,40388
0,2218
1,4888
1,7106
228,64909
0,2218
1,7092
1,9309
243,89139
0,2218
1,9446
2,1664
253,03736
0,2218
2,0932
2,315
Table .2.3
Mg
Nm
W
Fk1
V
Fk2
V
kN
V
km/h
V
Fk4
kN
rad/s
kN
km/h
kN
104,66
3,352024
9,43796
2,1845236
14,482 1,51308 20,909 1,1488
99,53863
157
3,695556
14,1578
2,4084043
107,9894
209,33
4,009308
18,8768
2,6128769
115,6399
261,66
4,293347
23,5958
2,7979862
122,4914
314
4,54772
28,3157
2,963762
128,5412
366,33
4,77233
33,0347
3,1101408
133,7907
418,66
4,967228
37,7536
3,2371562
138,2407
471
5,132442
42,4735
3,344827
141,869
523
5,267149
47,1627
144,738
575,66
5,373667
51,9115
146,7865
628
5,449723
148,0343
680,33
5,496049
148,4818
732,66
148,1289
785
146,9757
145,8996
90,28571
km/h
Fk3
km/h
Fk5
kN
V
km/h
27,538 0,90806
34,839547
21,724 1,66815 31,365 1,2666
41,31 1,00112
52,262649
28,965 1,80977 41,819 1,3741
55,079 1,08611
69,682423
36,206 1,93799 52,273 1,4714
68,848 1,16306
87,102196
43,449 2,05281
62,73 1,5586
82,619 1,23197
104,5253
2,1542 73,184 1,6356
96,388 1,29281
121,94507
57,931 2,24217 83,638 1,7024
50,69
110,16 1,34561
139,36485
65,173 2,31675 94,094
1,759
123,93 1,39037
156,78795
3,4326159
72,368 2,37756 104,48 1,8052
137,61 1,42686
174,09787
3,5020335
79,655 2,42564
115 1,8417
151,47 1,45571
191,6275
56,6314
3,5515994
86,898 2,45997 125,46 1,8678
165,24 1,47632
209,0506
61,3503
3,5817905
94,139 2,48088 135,91 1,8836
179,01 1,48887
226,47037
5,512664
66,0693
3,5926182
101,38 2,48838 146,37 1,8893
192,78 1,49337
243,89014
5,499561
70,7892
3,5840791
108,62 2,48246 156,82 1,8848
206,55 1,48982
261,31325
837,33
5,456746
75,5082
3,5561764
115,86 2,46314 167,28 1,8702
220,32 1,47822
278,73302
868,73
5,416792
78,3397
3,5301384
120,21
228,58
289,18555
2,4451 173,55 1,8565
1,4674
balance of power
6
5
F, kN
4
Fk1
Fk2
3
Fk3
Fk4
2
Fk5
1
Ff+Fв
0
0
50
100
150
200
250
300
350
v, km/h
Fig. 2.2
2.2.2 . Power balance of the car
Capacity balance is a graph ( Fig.2.3 ) utilization of capacity balance equation of the car on speed . It
is drawn on the outside of the engine speed characteristic Pe - ω as follows:
For different values of the angular velocity in the range ωmin - ωv, using the above equation that the
speed dependence Pe - ω be rebuilt in coordinates Pe - V for each gear. Obtained by multiplying the
values of Pe with efficiency ηm transmission are obtained dependencies Pk - V for each gear.
The results of the calculations are recorded in tabl.2.4 .
Table 2.4
W
Pe
rad/s
Pk
V
Pe
Pk
kW
V
kW
km/h
Pe
kW
Pk
V
Pe
Pk
V
Pe
Pk
V
kW
km/h
kW
kW
km/h
kW
km/h
kW
kW
km/h
kW
105
9,4766
8,813
9,438
9,4766
8,813
14,48
9,4766
8,8132
20,91
9,4766
8,813
27,54
9,477
8,8132
34,84
157
15,673
14,58
14,16
15,673
14,58
21,72
15,673
14,576
31,36
15,673
14,58
41,31
15,67
14,576
52,26
209
22,671
21,08
18,88
22,671
21,08
28,97
22,671
21,084
41,82
22,671
21,08
55,08
22,67
21,084
69,68
262
30,346
28,22
23,6
30,346
28,22
36,21
30,346
28,222
52,27
30,346
28,22
68,85
30,35
28,222
87,1
314
38,573
35,87
28,32
38,573
35,87
43,45
38,573
35,873
62,73
38,573
35,87
82,62
38,57
35,873
104,5
366
47,225
43,92
33,03
47,225
43,92
50,69
47,225
43,919
73,18
47,225
43,92
96,39
47,22
43,919
121,9
419
56,175
52,24
37,75
56,175
52,24
57,93
56,175
52,242
83,64
56,175
52,24
110,2
56,17
52,242
139,4
471
65,3
60,73
42,47
65,3
60,73
65,17
65,3
60,729
94,09
65,3
60,73
123,9
65,3
60,729
156,8
523
74,47
69,26
47,19
74,47
69,26
72,41
74,47
69,257
104,5
74,47
69,26
137,7
74,47
69,257
174,2
576
83,561
77,71
51,91
83,561
77,71
79,66
83,561
77,711
115
83,561
77,71
151,5
83,56
77,711
191,6
628
92,448
85,98
56,63
92,448
85,98
86,9
92,448
85,977
125,5
92,448
85,98
165,2
92,45
85,977
209,1
680
101
93,93
61,35
101
93,93
94,14
101
93,933
135,9
101
93,93
179
101
93,933
226,5
733
109,1
101,5
66,07
109,1
101,5
101,4
109,1
101,46
146,4
109,1
101,5
192,8
109,1
101,46
243,9
785
116,62
108,5
70,79
116,62
108,5
108,6
116,62
108,45
156,8
116,62
108,5
206,5
116,6
108,45
261,3
837
123,42
114,8
75,51
123,42
114,8
115,9
123,42
114,78
167,3
123,42
114,8
220,3
123,4
114,78
278,7
869
127,11
118,2
78,34
127,11
118,2
120,2
127,11
118,22
173,6
127,11
118,2
228,6
127,1
118,22
289,2
Power from the right side of the equation of capacity balance is obtained by multiplying pochlenno
equation of force balance at speed V. The results are in tabl.2.5 .
Table 2.5
V
P†=F†*V
km/h
Pв=Fв*V
kW
Pf + Pв
kW
kW
0
0
0
0
44,53035
2,74325
0,845665
3,58892
59,37285
3,65762
2,004442
5,66206
74,21536
4,57198
3,914814
8,48679
89,0607
5,48651
6,765316
12,2518
103,9032
6,40087
10,74278
17,1436
118,7457
7,31523
16,03554
23,3508
133,591
8,22976
22,83294
31,0627
148,4335
9,14412
31,28082
40,4249
163,2761
10,0585
41,68668
51,7452
178,1214
10,973
54,12253
65,0955
192,9639
11,8874
68,81102
80,6984
207,8064
12,8017
85,94223
98,744
222,6517
13,7163
105,7081
119,424
237,4942
14,6306
128,2889
142,92
246,4003
15,1793
143,2695
158,449
The intersection of the total resistance power Pf + PB curve Pk - V defines the maximum possible
speed Vmax under certain driving conditions laid down by the factor ψ.
Power balance of the car
160
140
Pe1
120
Pk1
Pe2
P, kW
100
Pk2
Pe3
80
Pk3
60
Pe4
Pk4
40
Pe5
20
Pk5
Pe+Pв
0
0
50
100
150
200
250
300
V, km/h
Fig.2.3
2.2.3. Dynamic characteristic of the vehicle
In the dynamic response of vehicles for different values of the rotational speed of the engine in the
range ωmin - ωv response curves of the dynamic factor D (Fig. 2.4) of the vehicle is loaded to the
formula:
Fк  Fв  M д .iц .iк . м
V2  1
D

 kв .S

G
r
13  G
к

and for calculating the velocity V using the aforementioned formula .
Results of the calculations for D for each gear are plotted in tabl.2.6 .
Table 2.6
W
Mg
rad/s
Nm
V
km/h
Fk
Fв
D1
V
Fk
Fв
D2
kN
-------
------
km/h
kN
-------
-------
104,66
90,28570901
9,43795778
3,35202433
0,002912069
0,226520951
14,481997
2,184523596
0,0068565
0,147288948
157
99,53863061
14,1578384
3,695556199
0,00655299
0,249509855
21,7243792
2,408404272
0,015429091
0,161851551
209,33
107,989425
18,8768173
4,00930761
0,011649388
0,270386082
28,9653777
2,612876941
0,027428621
0,174869687
261,66
115,6399366
23,5957962
4,293347036
0,018201822
0,289154225
36,2063762
2,797986226
0,042856401
0,186346285
314
122,4913979
28,3156769
4,547720238
0,026211959
0,305817266
43,4487584
2,963761951
0,061716362
0,196283097
366,33
128,5411909
33,0346558
4,772330017
0,035676739
0,320368839
50,6897569
3,11014077
0,084001297
0,204676325
418,66
133,7907009
37,7536347
4,967227811
0,046597554
0,332812327
57,9307553
3,237156205
0,109714483
0,211528016
471
138,2407019
42,4735153
5,132442348
0,058976907
0,343149506
65,1731376
3,344826979
0,138861815
0,216839037
523,33
141,8894934
47,1924942
5,267910495
0,072810069
0,351376424
72,414136
3,433111948
0,171432155
0,220607358
575,66
144,7380019
51,9114731
5,373666657
0,088099265
0,357495258
79,6551345
3,502033533
0,207430746
0,222834142
628
146,7865426
56,6313538
5,449722527
0,104847834
0,361506574
86,8975167
3,551599356
0,246865448
0,223519371
680,33
148,0343326
61,3503327
5,496049042
0,123049378
0,363408838
94,1385152
3,581790475
0,289721194
0,222662785
732,66
148,4818397
66,0693116
5,512663572
0,142706955
0,363203018
101,379514
3,59261821
0,33600519
0,220264662
785
148,12892
70,7891922
5,499560776
0,163824741
0,36088847
108,621896
3,584079081
0,385727263
0,2163241
837,33
146,9757084
75,5081711
5,456745659
0,186394666
0,35646608
115,862894
3,556176351
0,438868414
0,210842606
868,73
145,899568
78,3397388
5,416791951
0,200636441
0,352800508
120,20777
3,530138408
0,472400839
0,206813498
V
Fk
Fв
D3
km/h
kN
-------
-------
20,9085073 1,513080952 0,142919548 0,092672398
31,3647587 1,668148897 0,032160995
V
Fk
Fв
D4
km/h
kN
-------
-------
27,538034 1,148820723 0,024791993 0,076024939
V
Fk
Fв
D5
kN
-------
-------
34,839547 0,908056127 0,039681728
0,05873347
0,11065187 41,30968219 1,266557496 0,055789091 0,081891674 52,2626494 1,001118165 0,089295263
0,06167216
41,81901235 1,809774147 0,057173282 0,118539118 55,07869919 1,374087778 0,099177448 0,086229985 69,6824229 1,086112743 0,158742079
0,06272375
52,27326599 1,937987609 0,089331546 0,125035919 68,84771619 1,471435037 0,154961801 0,089041139 87,1021964 1,163058408 0,248029758
0,06188898
62,7295174
2,05280994 0,128643981 0,130143115 82,61936438 1,558614955 0,223156364 0,090325234 104,525299 1,231967557 0,357181052
0,05916716
73,18377104 2,154197265 0,175095565 0,133858755 96,38838137
1,63559422 0,303735079 0,090081782 121,945072 1,292813859 0,486154252
0,05455932
83,63802469 2,242172802 0,228693127 0,136183948 110,1573984
1,70239046 0,396709791 0,088311171 139,364846 1,345611249 0,634968315
0,04806512
94,09427609 2,316749518 0,289448957 0,137118739 123,9290466 1,759013523 0,502101819 0,085012628 156,787948 1,390367509 0,803657368
0,03968279
104,5485297 2,377898917 0,357339839 0,136662772 137,6980636
0,99215695
0,02941552
115,0027834 2,425636528 0,432376698 0,134816356 151,4670806 1,841686994 0,750035962 0,073835038 191,627495 1,455714652 1,200497398
0,01726190
125,4590348
1,80544177 0,619870892 0,080187411 174,207722 1,427065536
2,45996763 0,514575923 0,131578742 165,2387288
1,8677532 0,892625456 0,065953855 209,050597 1,476318022 1,428724209
0,00321906
135,9132884 2,480879103 0,603906103 0,126951167 179,0077458
1,88363043 1,047584888 0,056546875 226,470371 1,488867774 1,676750176
-0,01270763
146,3675421 2,488378789
0,70038226 0,120933144 192,7767627 1,889324636 1,214940316 0,045612737 243,890144 1,493368615 1,944617006
-0,03052068
156,8237935 2,482464276
0,80402488 0,113523125 206,5484109 1,884833987 1,394727275 0,033148915 261,313247 1,489819096 2,232381577
-0,05022404
167,2780471 2,463137824 0,914794358 0,104723941 220,3174279
1,8701602 1,586877065 0,019160171
173,5509989 2,445102992 0,984690647 0,098776621 228,5793644 1,856467087 1,708124883 0,010033291
278,73302 1,478220573 2,539933927
-0,07181016
289,18555
-0,0856682
1,4673972
2,73400154
dynamic response
0.4
0.35
0.3
do1
0.25
do2
D
0.2
do3
0.15
do4
0.1
do5
f
0.05
0
0
50
100
150
200
250
300
350
V, km/h
fig.2.4
2.2.4 . Acceleration of the car
More power and capacity balance and dynamic response are determined dynamic properties even
when the vehicle is moving. Dynamic characteristics of the car in motion with variable speed is
determined by the graph of acceleration ( fig.2.5 ) . Evaluation indicators for vehicle dynamics when
accelerating movement have acceleration time and way to accelerate at a speed range . These
indicators are defined as follows :
The equation of balance of power and acceleration of the vehicle can write the following equation :
a
D 
a
.g
,
Where:
g is the gravitational acceleration , [m/s2]
ψ - coefficient of resistance of the road;
δa - coefficient of the influence of the rotating masses of the vehicle.
In the absence of data on the moments of inertia Jd and Jk to determine the coefficient δa using the
empirical relationship:
 a  1,04   0,04  0,05.iк2 .
As the acceleration and depends on the difference D-ψ, for the construction of a characteristic
diagram - V is necessary to know the load of the vehicle and road conditions , as determined by the
coefficient ψ. Usually vehicle dynamics when accelerating explore a full load when traveling on a
horizontal paved road . In this case ψ = f and is taken from the table for the road conditions . To
perform calculations assume ψ = f. The dynamic factor is determined by the formula used by G0 is
replaced by the weight of the loaded vehicle G.
Results of calculations and for each gear are recorded in tabl.2.7 .
Table 2.7
Mg
W
Nm
rad/s
104,7
157
209,3
261,7
314
366,3
418,7
471
523,3
575,7
628
680,3
732,7
785
837,3
868,7
V
Fk
km/h
kN
90,2857
09
99,5386
9,437957
78
14,15783
3,35202
43
3,69555
31
107,989
43
115,639
84
18,87681
73
23,59579
62
4,00930
76
4,29334
94
122,491
4
128,541
62
28,31567
69
33,03465
7
4,54772
02
4,77233
19
133,790
7
138,240
58
37,75363
47
42,47351
4,96722
78
5,13244
7
141,889
49
144,738
53
47,19249
42
51,91147
23
5,26791
05
5,37366
146,786
54
148,034
31
56,63135
38
61,35033
67
5,44972
25
5,49604
33
148,481
84
148,128
27
66,06931
16
70,78919
9
5,51266
36
5,49956
92
146,975
71
145,899
22
75,50817
11
78,33973
08
5,45674
57
5,41679
57
88
2
Table 2.7
Fв
D1
а1
V
-------
------
m/s^2
km/h
0,00291 0,2265209 1,3287402 14,481996
21 0,2495098
51 1,4731527
26 21,724379
98
0,00655
3 0,2703860
55 1,6042938
84 28,965377
18
0,01164
94
82
46
67
0,01820 0,2891542 1,7221922 36,206376
18 0,3058172
25 1,8268667
69 43,448758
16
0,02621
2
66
85
37
0,03567 0,3203688 1,9182773 50,689756
67 0,3328123
39 1,9964453
98 57,930755
86
0,04659
76 0,3431495
27 2,0613818
7 65,173137
34
0,05897
69 0,3513764
06 2,1130620
44 72,414136
55
0,07281
01 0,3574952
24 2,1514995
07 79,655134
04
0,08809
93
58
29
53
0,10484 0,3615065 2,1766979 86,897516
78 0,3634088
74 2,1886476
61 94,138515
74
0,12304
94 0,3632030
38 2,1873547
74 101,37951
22
0,14270
7 0,3608884
18 2,1728151
47 108,62189
37
0,16382
47
7
37
59
0,18639 0,3564660 2,1450344 115,86289
47 0,3528005
8 2,1220078 120,20777
44
0,20063
64
08
72
02
Fk
Fв
kN
-------
2,1845235 0,0068565
96 0,0154290
2,4084042
72 0,0274286
91
2,6128769
41 0,0428564
21
2,7979862
26 0,0617163
01
2,9637619
51
62
3,1101407 0,0840012
7 0,1097144
97
3,2371562
05 0,1388618
83
3,3448269
79
15
3,4331119 0,1714321
48 0,2074307
55
3,5020335
33
46
3,5515993 0,2468654
56 0,2897211
48
3,5817904
75 0,3360051
94
3,5926182
1 0,3857272
9
3,5840790
81
63
3,5561763 0,4388684
51 0,4724008
14
3,5301384
08
39
D2
а2
-------
m/s^2
0,14728
89
0,16185
1,02869
75
1,14193
16
0,17486
97
0,18634
83
1,24316
92
1,33241
63
0,19628
31
0,20467
28
1,40968
29
1,47494
63
0,21152
8
0,21683
98
1,52822
95
1,56952
9
0,22060
74
0,22283
87
1,59883
17
1,61614
41
0,22351
94
0,22266
75
1,62147
6
1,61481
28
0,22026
47
0,21632
5
1,59616
69
1,56552
41
0,21084
26
0,20681
45
1,52289
97
1,49156
35
87
V
Fk
Fв
D3
а3
V
km/h
kN
-------
20,908
5073
31,364
1,5130
8095
1,6681
0,01429
1955
0,03216
7587
41,819
0123
52,273
489
1,8097
7415
1,9379
0995
0,05717
3282
0,08933
------0,10137 m/s^2
0,7391 km/h
27,538
2269
7882
034
0,11065 0,8185 41,309
187 0,8860
9418 55,078
682
0,11853
9118
937
699
0,12503 0,9416 68,847
266
62,729
5174
73,183
8761
2,0528
0994
2,1541
1546
0,12864
3981
0,17509
5919
0,13014
3115
0,13385
771
83,638
0247
94,094
9726
2,2421
728
2,3167
5565
0,22869
3127
0,28944
8755
0,13618
3948
0,13711
2761
104,54
853
115,00
4952
2,3778
9892
2,4256
8957
0,35733
9839
0,43237
8739
0,13666
2772
0,13481
2783
125,45
9035
135,91
3653
2,4599
6763
2,4808
6698
0,51457
5923
0,60390
6356
0,13157
8742
0,12695
3288
146,36
7542
156,82
791
2,4883
7879
2,4824
6103
0,70038
226
0,80402
1167
0,12093
3144
0,11352
3793
167,27
8047
173,55
6428
2,4631
3782
2,4451
488
0,91479
4358
0,98469
3125
0,10472
3941
0,09877
0999
0299
0647
6621
9369 82,619
716
0,9854
0137
364
1,0172 96,388
0003 110,15
381
1,0370
9916 123,92
74
1,0450
9916 137,69
905
1,0411
9696 151,46
806
1,0253
9523 165,23
708
0,9976
8755 179,00
873
0,9580
8449
775
0,9065 192,77
819 206,54
676
0,8431
6654
841
0,7678 220,31
6262 228,57
743
0,7169
6512
936
Fk
Fв
D4
kN
-----1,1488 -0,0247
2072 0,0557
92
1,2665
575 0,0991
891
1,3740
8778 0,1549
774
1,4714
3504 0,2231
618
1,5586
1495
564
1,6355 0,3037
9422 0,3967
351
1,7023
9046
098
1,7590 0,5021
1352 0,6198
018
1,8054
4177 0,7500
709
1,8416
8699
36
1,8677 0,8926
532 1,0475
255
1,8836
3043 1,2149
849
1,8893
2464 1,3947
403
1,8848
3399
273
1,8701 1,5868
602 1,7081
771
1,8564
6709
249
a4
V
Fk
kN
Fв
-------
D5
-------
a5
-------
m/s^2
km/h
0,07602
4939
0,08189
0,54360
2681
0,59586
34,8395
4701
52,2626
0,90805
6127
1,00111
0,03501
329
0,07878
0,059049
228
0,062382
0,39734
2998
0,42741
1674
0,08622
9985
0,08904
2836
0,63450
7977
0,65954
4935
69,6824
2286
87,1021
8165
1,08611
2743
1,16305
9938
0,14006
654
0,21884
701
0,063986
892
0,063862
2355
0,44188
2855
0,44076
1139
0,09032
5234
0,09008
9388
0,67098
7955
0,66881
9636
104,525
2987
121,945
8408
1,23196
7557
1,29281
9787
0,31515
9752
0,42895
605
0,062009
321
0,058427
1733
0,42404
4307
0,39173
1782
0,08831
1171
0,08501
9307
0,65304
6929
0,62366
0722
139,364
8457
156,787
3859
1,34561
1249
1,39036
9634
0,56026
616
0,70910
746
0,053117
693
0,046077
6963
0,34383
7997
0,28033
2628
0,08018
7411
0,07383
3908
0,58068
1471
0,52409
9481
174,207
7216
191,627
7509
1,42706
5536
1,45571
9442
0,87543
2603
1,05926
651
0,037310
31
0,026814
3787
0,20124
8599
0,10657
5038
0,06595
3855
0,05654
5304
0,45389
0694
0,37009
4951
209,050
5974
226,470
4652
1,47631
8022
1,48886
241
1,26063
9008
1,47948
6875
0,04561
2737
0,03314
4467
0,27269
4511
0,16166
3709
243,890
1444
261,313
7774
1,49336
8615
1,48981
5449
1,71583
8535
1,96974
491 0,00371
179
0,014587
691 9203
0,000634
0,12958
584
2235
000,0150 0,27103
8915 0,03705
8313
0,01916
0171 0,04424
8297
0,01003
3291 2837
2468
278,733
0203
289,185
9096
1,47822
0573
1,46739
5501
72
47002
0,032460
845
558
2,24111 0,051599
8171 0,063913
432
2,41235
43 23
m/s^2
6889
0,42811
4665
0,60075
5541
0,71183
1299
Vehicle acceleration
2.5
а, m/s^2
2
1.5
а1
а2
а3
1
a4
a5
0.5
0
0
50
100
150
200
V, km/h
fig.2.5
2.2.5 . Braking characteristics of the car
250
300
350
Maximum possible line tripping delay :
 =

б
[m/s2]
Minimum stopping distance when braking to v = 0 [km / h]:
 =

[m]

Minimum stopping time v = 0:
 = . 


Actual stopping distance when braking with maximum linear
delay to v = 0 [km / h]:
 = ( +  + .  )

.
+  
Indeed brake stopping time with maximum linear delay to v = 0 [km / h]:
 =  +  +  + 
Table 2.8
Vmax-V0 Smin suh
225
t min suh
Vmax-V0 S min mok
258,027532 8,256881025
200 203,8736056
225
t min mok
412,844037 13,21100919
7,3394498
200 326,1977577 11,74311928
175 156,0907293 6,422018575
175 249,7451582 10,27522937
150 114,6789031
5,50458735
150 183,4862387 8,807339457
125 79,63812717 4,587156125
125 127,4209991 7,339449547
100 50,96840139
3,6697249
100 81,54943941 5,871559638
75 28,66972578 2,752293675
75 45,87155967 4,403669728
50 12,74210035
1,83486245
50 20,38735985 2,935779819
25 3,185525087 0,917431225
25 5,096839963 1,467889909
0
0
Adhesion:
ϕ dry
0.8
0
0
0
0
ϕ wet 0.5
a max
dry
7.54615385
a max
wet
4.71634615
Braking characteristics of the car
450
14
400
12
350
10
S, m
300
250
8
200
6
Smin suh
S min mok
t min suh
150
4
100
2
50
0
0
0
50
100
150
200
250
V, km/h
CHAPTER III. DESIGN OF BRAKE SYSTEM
t min mok
3.1 Determination of the correct value of the total braking force
Fc = m. Ac, max = 11008,4 [N]
where:
m = 1460 [kg] - full weight of the car (when a traction calculation of 2.2.1 )

ac = = 7,54 [m / s] maximum deceleration ( whichever is the dynamic calculations t.2.2.5 )
б
dk
Fc = 1460.7,54 = 11008,4 [N]
Fig.3.1 . Scheme for calculating the disk brake mechanism
adhesion :
ϕ dry
0.8
ϕ wet 0.5
3.2 Determination
wheels
β0=
2 +ᵠ0 hg
L
a max
dry
7.54615385
not optimal coefficient between front and rear
a max
wet
4.71634615
where:
hg = 0,2 L = 524 [mm]- height of center of gravity of the car
L = 2720 [mm]- wheelbase vehicle ( taken from Technical data of the car )
2 = L / 2 = 1645 [mm]- distance from the center of gravity of the vehicle to the rear axle
0 = 0,8 - adhesion of the tire and the road ( whichever is the value for the dry time of dynamic
calculations t.2.2.5 )
0 =
1654 + 0,45.0,52
2720
= 0,78
3.3 Determination of the radius of the tire
d

rк  103   .h  ,
2

wherein:
d- rim diameter [mm]
λ - coefficient of radial deformation of the wheel
h - height of the tire [mm]
The ratio of the radial deformation of the wheels taken λ = 0,85  0,9, as lower values for light and
higher - for trucks. For the normal tire section height them
h  b , and for low-profile tires
h   0,7  0,85 b where the section width of the tire.
3.4 Required torque for a front wheel
Мс1 =0,5 .β0 .Fc . rk
Мс1 =0,5 .0,78.11008,4.0,267 = 1146,26 [Nm]
3.5 Required torque to one rear wheel
Мс2 = 0,5 (1- β0) Fc. rk
Мс2 = 0,5 (1- 0.78)11008,4.0,267 = 323,30 [Nm]
3.6 Selection of the diameter of the brake disc
Diameter of the brake disc is chosen according to the diameter of the wheel rim so that it remains
adequate radial clearance between the disc and rim ohlazhdane.Vanshniyat radius R of the jaws is 1
÷ 2 mm smaller than the brake disc .
To accept the calculations :
D disk = 240 [mm]- diameter disk
R disk = 115 [mm]- outer radius of the friction pad
r = (0.62-0.65) R = 75 [mm] - inner radius of the friction pad
3.7 Determination of the total area of the pad
∑ =

= .  [ ]

G - full vehicle weight
q = 0,22 .106 [Pa]- relative load
3.8 Determination of the width of the friction linings
For front brakes:
b1=
 ∑  
ср

=0,0039 [m]
For rear brakes
b2 =
wherein:
(− ) ∑  
ср

= 0,0020 [m]
r = (0.62-0.65) R = 75 [mm] - inner radius of the friction pad
θsr = 470 - average angle of pads
i = 1 - number of powered axles
S- total area of pads
3.9 Determination of braking torque
Torque with disc brakes without self amplifying is determined by the expression:
Ms = F.μ.i.Rtp
wherein:
2 3 − 3
тр = 3 2 −2 = 0,98 [] - average radius of friction
F - clamping force on the pistons of hydraulic cylinders
μ - coefficient of friction between the pads
( for clutch nakaladki μ = 0,41)
i - number of friction surfaces
3.10 Determination of clamping force on a front wheel
F1 =
1 =
Mc1
μ. R тр i
1145
= 14232 []
0,41.0,098.2
3.11 Determination of clamping force of a rear wheel
F2 =
2 =
Mc2
μ.Rтр i
323
= 4146.34 []
0,41.0,098.2
3.12 Determination of the size of the executive mechanism of the brake cylinders
The diameters of the front and rear wheel brake cylinders are different depending on the forces F1
and F2 and are calculated as the dependencies :
42
2 = √

wherein:
 . = 12 ÷ 15  - Using the amplifier
For the front brakes :
41
4.14232
1 = √
=√
= 0,0388 []

3,14.12
Rear brakes :
42
4.4146
2 = √
=√
= 0,0209 []

3,14.12
Accept:
1 =39 [mm] и 2 =21 [mm]
3.13 Determination of the power ratio of the drive
 =
Σ(1 + 2 )
= 36,54

For treadle force Accepted:  = 500 N
( for cars in the range 20 ÷ 40 , which is fulfilled)
3.14 Determination of the diameter of the master cylinder
2 (12 + 22 )
=√
 = 36 []

( kinematics ratio which for cars in the range ( 4 ÷ 5 ) . Accept = 4 , and with him to design the
engagement of the brake pedal )
3.15 Determining the course of the brake treadle
For existing structures during the brake treadle for cars in boundaries  ≤ (0.15 ÷ 0.18) m ,
depending on the design and clearance in the brake actuator .
Accept  = 0,16 [m]
Travel of the treadle is obtained :
 = (0.4 ÷ 0.6) = 0.028 [m]
3.16 Determination of the force on the master silindar
сл.ц. =
 . .  2 2,5.3,14. 0,0362
=
= 254 
4
4
Pγ - pressure in the brake mechanism
Δp- atmospheres pressure difference between the pressure of the dilution in amplifier
3.17 Determination of the diameter of the amplifier
4.гл.ц
 = √
∆р
4.254,34
≥ √ 3,14.0,5 ≥ 64,8 
The diameter of the amplifier choose D = 150 mm
3.18 Determination of the stiffness of the conical spring with a constant angle of inclination
=
0,294. . .  4
23 − 13
wherein:
d - is the diameter of the wire
23 - is the radius of the largest turn
13 - is the radius of the smallest coil
i - is the number of turns
G - is an assembly of linear deformation of the material.
Steel G = 0,7.10 ^ 11 Pa

ln(2 )
0,495
ln(
)
0,225
1
=
=
= 0,021
2. . 
2.3,14.6
=
0,294. . . 4 0,294.0,7. 1011 . 0,21. 0,00464

=
= 26184 [ ]
3
3
3
3
0,0495 − 0,0225

2 − 1
The spring force is given by addiction :
 = . ∆
wherein:
Δ is the displacement of the spring
 = 26184.0,002 = 52 []
3.19. Verifying calculations
3D model of the brake disk is mapped using the software product "SOLID
WORKS", and static parameters were determined by the method of finite
elements "Simulation" in the following sequence:
1. Drawing 3D models brake disc and pads "SOLID WORK Part" in the
following sizes
The model of the brake disc (fig.3.3) is made in the dimensions shown in
the drawing of the brake disc (fig.3.2).
fig.3.2
fig.3.3
The model of the pads (Fig.3.5) are drawn with dimensions adapted to the size
of the brake disc and consist of two elements (brake dust and steel plate
(Fig.3.4).
Fig.3.4
Fig.3.5
3D model of the piston (Fig. 3.7) is drawn with the dimensions obtained in the
calculations given in (Fig. 3.6).
(Fig. 3.6)
(Fig. 3.7)
2. The assembled unit is shown in (Fig. 3.8).
фиг. 3.8
1. Material is set by "Apply Material"
The brake disc is made of steel AISI 4130, for making
lining materials have been selected: stainless steel (SS) for the plate and
graphite friction part, a working piston of steel AISI 4340.
4. Attachment of the mechanism is through "Fixture"
The brake disc is fixed and has one degree of freedom (rotation about the axis
or tsilindrichnata povarhnina is stated in blue in (Fig. 3.9).
Fig. 3.9
Brake linings are fixed, have a degree of freedom (Fig. 3.10), in its outer
cylindrical surface in the axial direction.
Fig. 3.10
The working cylinder is secured (Fig.3.11). It has a degree of freedom in
the axial direction on the outer cylindrical surface.
Fig.3.11
5. Application of forces and moments acting on the mechanism by "External
loads" (Fig. 3.12).
The working cylinder and the other pad is loaded with a force equal to
14232 [N].
The brake disk is loaded with a torque equal to 1145 [Nm] and centrifugal
force caused by rotation of the disc with angular velocity ω=2200 rad / s
Fig. 3.12
6. The networking is done by "Create Mesh" (Figure 3.13).
Fig. 3.13
7. The results are obtained by running the task button "Run" from the taskbar
"simulation"
7.1 Strains of downforce on orphan working cylinder are presented in Fig.3.14
a and b.
Fig. 3.14 а
Fig. 3.14.б
- 7.2 Displacements obtained from the action of braking torque are:
- - In the radial direction (fig.3.15)
Fig. 3.15
-in the axial direction (Fig. 3.16)
Fig.3.16
conclusions:
- Voltages of the clamping force of the brake cylinders acting on the brake disc
pads, respectively, will lead to significant deformation or destruction of the
brake disk.
-Torque generated by the clamping force of the pad is
sufficient to overcome the moment created by the movement of the vehicle, or
it will be possible to perform braking by engagement
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