Question

Prob. 2 (50): A vehicle has a total weight of 3800 bs and the front to rear weight distribution ratio is 49/51. The wheel base is 110 inches and the gravity center height is 14 inches. The tire radius is 11 inches. The rolling and air resistances are neglected in the brake design. The split point is selected at point A that corresponds to a value of road traction coefficient of 0.45. The split pressure p, and the proportioning factor k, of the proportioning valve are 300 psi and 0.63 respectively. a) Determine the axle brake gains. b) After the split point, the front and rear braking forces are related as Fr= a Ff+b. Determine the values of a and b. c) Determine the brake efficiency when the line pressure is 425 psi. d) Determine the maximum deceleration the vehicle can achieve when the vehicle is braked on a road surface with a traction coefficient of 0.28.

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Answer 1

p wftwR=3800 Sooms w=3800lbs RTC=0.45.=mt Ps= 3oopsi Kp=0.63 Wf WR a) ABG = 3 (Grit Gt) 51 D 110im by Fr=of Frib a, b = ? h=14 im C:>BE=mg for Cp=495 Pai 8= Ilim di) Poc mass 3 8 Tc=0.28 49 49 FORT WR=3800 51 R= 3800X51 1938 lbs = • WR wf-1862 L Fraction force santx w=0.45X3800=1710lbs ItR = fraction force on rear wheel = mtxwR .. FR=0.46x1938= 872.1 lbs =F+R torque on rear wheel (TR) = FtRx radius = 872 1Xllin TR=9593.1 obsin .9 c=49 X110= 53.911 49+51 90 = 110-53.9= 56.1" or Split pressure-tomque in meripheel rear ascle brake gain 300psi = 9593.10. beim Gr 31.977 - Gr=9593.1 lbsim 300 psi

Torque in front axle= Tt = Ft RX Radius Ftpe mt xwt = 0.45X1862- 837.9 lbs = Fif & Ps- If Gt= Tt = 9216.9ebsim. GL Ps 3oops GE 30.723 ebsim Psi Nowo, kp=slope of AB = tano 0.63=tamo = tan! (0.63) 92-21° as well known PR=Pa -pal 30opsi PR=300+ 0.63 (Pa-300) Pa> 300 psi Fact = 2 Gt. Pt & Focr=2 Gr.Pr. o Pt-Pop-pa-applied pressure The decceleration (Doc)=2 booke torque = Fooftfoor the front & Rear axle loads, wto wt .DOC static -g WR = wr h (w) Doc static & g=32:2 Pt/sec2 Braking coefficients My = Fathet, My=Foor/wr. braking efficiency max (MgM) As, Pa = 425 Psi > 300 psi => Pt=425 Psi n b = Ox -= boa

and, Por = 900+0.63(425-300) -Pr=37875 Psi Fiet=2(90.72232425 - 237605 @bobo xpoi albe Faer = 2(21977) X 27875 Psi 22.02-05 Obs =F+32 ebs deceleration Ox=FxFt For = (وادا) lebol (Falco) WF=2444.413 ebs = 0.9712 MF= FOCF = 2374.05 wf 2444,413 wga wystatic - h (w/g)boc = 1938-582-413 - wap=1355.587ebo جا 22202-05 = 1.6244 1355.587 Max of (M.2, Mr)= Max (0.912, 1.6244) -1.6244 1.204 -=0.74133 1.6244 Doc' Max (Mt, My b=74-133% as FR = atb (FF) FIR = atbfff & FOR=ar a+b(Ford) 2202-05 = a +b (23.7405) =_atb (837.9) 1829.95 = b(153615) 372.1

b=11912 putting im 2, 372.1= at 998.1545 a =-625.05 Gr 300 Now, for fraction coefficient of Me=0:28 AS WR=19380bs, Wf=1862.015 FER= MEX WR=0.28 x 1938-542.64 ebs Ft= MEXWf = 0.28% 1862=521,36 lbs TR= Ft Ror=54264 X11 = 5969.04 ebs. Im If=521.36x11=573496ebsim as Ps= TR Gro= TR-5964.04 – 19.8968 lbsim Po psi & Gf = Tg = 5734.96 =1901165 lbsim Po Psi From above we know, Pf=425Psi 8 fr=37875 Psi Focf = 2 Gif x PE 2X19.1165x 425 14770184 lbs Frem=2G X Po = 2*19.8968X378.75 = 1370-166 lbs 11 Doc = fact t Focs. 1477.189 +1370.166 24.1275 13. 800/32-2) ebs Ft/s2 روما 2) Doc 24-1275.ft/92
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