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. 𝐺𝑟 = 𝜋𝑟 2 (0.75 + 0.1)𝑧 = 0,85.5 = 4,25 𝑘𝑁 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 . 𝐺 = (1 + 2.5)4,75 = 14.785 𝑘𝑁 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 103 .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] 𝑟𝑘 = 10−3 ( 281 2 + 0,85.126,75) = 0.267 𝑚 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 𝑘𝜔 = p , м 837 785 = 1,07 , km - coefficient of adjustment of engine torque : kм 𝑘𝑀 = M д max Mp 150 146 , = 1,02 . 𝑎= 1−1,02.1,07(2−1,07) 𝑏= (1,07−1)2 = 0.4688 2(1 − 0.4688) = 1.155 2 − 1,07 𝑐= 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 : 𝒊ц = 𝟑. 𝟔 𝟕𝟐𝟔.𝟎,𝟐𝟕𝟔 𝟐𝟐𝟎 = 𝟑. 𝟑 , 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 𝑖𝑘1 = 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 𝑞 = √3,23 = 1,53 Gear ratios of the intermediate gears determine the following equations : For the second gear: 𝑖𝑘2 = 𝑖𝑘1 𝑞 = 3,23 1,53 = 2,11 For the third gear: 𝑖𝑘3 = 𝑖𝑘1 3,23 = = 1,38 𝑞 2 2,34 For the fourth gear: 𝑖𝑘4 = 𝑖𝑘1 3,23 = = 0.90 𝑞 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 : 𝒂𝒎𝒂𝒙 = 𝒈 б𝒂 [m/s2] Minimum stopping distance when braking to v = 0 [km / h]: 𝑺𝒎𝒊𝒏 = 𝒗𝟐 [m] 𝟐𝟔𝒂𝒎𝒂𝒙 Minimum stopping time v = 0: 𝒕𝒎𝒊𝒏 = 𝟕. 𝟐 𝑺𝒎𝒊𝒏 𝒗 Actual stopping distance when braking with maximum linear delay to v = 0 [km / h]: 𝑺𝒄 = (𝒕𝟏 + 𝒕𝟐 + 𝟎. 𝟓𝒕𝟑 ) 𝒗 𝟑.𝟔 + 𝒌𝒆 𝑺𝒎𝒊𝒏 Indeed brake stopping time with maximum linear delay to v = 0 [km / h]: 𝒕𝒄 = 𝒕𝟏 + 𝒕𝟐 + 𝒕𝟑 + 𝒕𝒎𝒊𝒏 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 ) 𝒈 ac = = 7,54 [m / s] maximum deceleration ( whichever is the dynamic calculations t.2.2.5 ) б𝒂 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= 𝑙2 +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 ) 𝑙2 = L / 2 = 1645 [mm]- distance from the center of gravity of the vehicle to the rear axle 𝜑0 = 0,8 - adhesion of the tire and the road ( whichever is the value for the dry time of dynamic calculations t.2.2.5 ) 𝛽0 = 1654 + 0,45.0,52 2720 = 0,78 3.3 Determination of the radius of the tire d rк 103 .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 ∑𝑺 = 𝑮 = 𝟎. 𝟎𝟔𝟔𝟑 [𝒎𝟐 ] 𝒒 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= 𝜷𝟎 ∑ 𝑺 𝟏𝟖𝟎𝒐 𝟒𝒓𝝑ср 𝝅 =0,0039 [m] For rear brakes b2 = wherein: (𝟏−𝜷𝟎 ) ∑ 𝑺 𝟏𝟖𝟎𝒐 𝟒𝒊𝒓𝝑ср 𝝅 = 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 𝑅3 −𝑟 3 𝑅тр = 3 𝑅2 −𝑟2 = 0,98 [𝑚𝑚] - 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 = 𝐹1 = Mc1 μ. R тр i 1145 = 14232 [𝑁] 0,41.0,098.2 3.11 Determination of clamping force of a rear wheel F2 = 𝐹2 = Mc2 μ.Rтр i 323 = 4146.34 [𝑁] 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 : 4𝐹2 𝑑𝑘2 = √ 𝜋𝑃𝑚𝑎𝑥 wherein: 𝑃𝑚𝑎𝑥 . = 12 ÷ 15 𝑀𝑃𝑎 - Using the amplifier For the front brakes : 4𝐹1 4.14232 𝑑𝑘1 = √ =√ = 0,0388 [𝑚] 𝜋𝑃𝑚𝑎𝑥 3,14.12 Rear brakes : 4𝐹2 4.4146 𝑑𝑘2 = √ =√ = 0,0209 [𝑚] 𝜋𝑃𝑚𝑎𝑥 3,14.12 Accept: 𝑑𝑘1 =39 [mm] и 𝑑𝑘2 =21 [mm] 3.13 Determination of the power ratio of the drive 𝑖𝑐 = Σ(𝐹1 + 𝐹2 ) = 36,54 𝐹𝑛 For treadle force Accepted: 𝐹𝑛 = 500 N (𝑖𝑐 for cars in the range 20 ÷ 40 , which is fulfilled) 3.14 Determination of the diameter of the master cylinder 2 (𝑑12 + 𝑑22 ) 𝐷=√ 𝑖𝑛 = 36 [𝑚𝑚] 𝑖𝑐 (𝑖𝑛 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 𝑆𝑛 ≤ (0.15 ÷ 0.18) m , depending on the design and clearance in the brake actuator . Accept 𝑆𝑛 = 0,16 [m] Travel of the treadle is obtained : 𝑆𝑛𝑝 = (0.4 ÷ 0.6) = 0.028 [m] 3.16 Determination of the force on the master silindar 𝐹сл.ц. = 𝑝𝛾 . 𝜋. 𝐷 2 2,5.3,14. 0,0362 = = 254 𝑘𝑁 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.𝐹гл.ц 𝐷𝑦 = √ 𝜋∆р 4.254,34 ≥ √ 3,14.0,5 ≥ 64,8 𝑚𝑚 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 𝑧= 0,294. 𝐺. 𝑚. 𝑑 4 𝑟23 − 𝑟13 wherein: d - is the diameter of the wire 𝑟23 - is the radius of the largest turn 𝑟13 - 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 𝑟 ln(𝑟2 ) 0,495 ln( ) 0,225 1 𝑚= = = 0,021 2. 𝜋. 𝑖 2.3,14.6 𝑧= 0,294. 𝐺. 𝑚. 𝑑4 0,294.0,7. 1011 . 0,21. 0,00464 𝑁 = = 26184 [ ] 3 3 3 3 0,0495 − 0,0225 𝑚 𝑟2 − 𝑟1 The spring force is given by addiction : 𝐹𝑛𝑝 = 𝑧. ∆ wherein: Δ is the displacement of the spring 𝐹𝑛𝑝 = 26184.0,002 = 52 [𝑁] 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