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

☺ Answer. wa 3800 lbs PS: 300 psi WF WR 51. llo in Icha kin=h TR = llin=8 = 49 RTC: 0:45.mt Kp=0.63 a) ABG= 3 ( Grit Gt) b) FREO + Frib aid:9 c) BE RB "For CP=425 psi d) Pemas 3 & TC = 0.28 e = wf 49. & cuft Cur3800 WR ST 49 WRt WR 3800 51 CUR 3800 x 51 • 1938 lbs = WR wf=1862 Too Fraction force mtx.co=0.45 x 3800 =1710lbs. Itro Frachon force on noors.enedimt XWR. FER=0.46 x 1938 = 872-12DS : FER, torqueon rear wheel (TR) = FERx radius 872:1X11'in TR= 9 593.1 Obs.in G C = 49 *110 5 5 3.9'' Bar 497 51 Erw i fw x = llo- 539 = 56'1' K split pressure = torquainneoraneel TR near axle brake gain Gr 1 r R 1 300 psi g593.1 lbsin Gr Gra9593.1 lbsin = 31.977 300 psi

porque in front azele. = Tt = ftr X Radius ftp = ME xwt = 0.45X 1862 = 837.9 lbs = fef Tf = 837.9 lbs all in = 9216.9 lbs.in=Tb & Ps = It Gt It = 9216igebs in Gt 300 Psi GL = 30.723 lebs in Psi Now, KPslope of AB = tan o. PS (P2) 06 -400 = 0 - tan-' (0.63) 32.219 as wet know PR = Pa- Pa<3oopsi DR = 300+ 0.63 CPU-300) pas3oopsi Fact 2 Gt PE & for=rGrip ४ PŁ = Pr = PO = Opplied Pressure the deceleration Core) = 2. brake torque fref threr (وات) the front & Rear axle loads, wtz Waterie ). 9 OR static - h (w Jone 4 9 = 32.2 ft /sec2 Braking coefficients mt=flt lwt , My=fariwr nb - Dve Max Cmt, mm) = braking to = braking en efficiency AS, pa 425 psi > 300 psi >> PL = 425 PS1 and. Pr - 300 +0.63(425-300) -> Pr=378.75 Bil fået = 2 (30-723) *49.5 = 2374.05 lbs in x Psi -ubo Psi fer=2(31.971) X 378.75 = 2202.05 lbs ebs (واضح deceleration Oxe = freft Frer Ft/32 (ebs/fl152) ose = 2374,05+2202.05 38.776 ft 152-DƏ (3800 32.2) 4 Asel - 1. 209 ( Duelg) felsa WF = Wt static + ( Dee' = 162 t 14 2) (38.776) = 18624532.413 (-9800 32.2

6 =2444.413 ebas (P3 ME FRE 2374.05 -0.9712 wf 2444.413 Ws = tor Static hope =1938 - 582.413 ) cor= 1355.587.0b5 Mr Far wr 2202.05 = 16244 1355.589 mare OR (Mti Mr) = mase (0.9712,116244) = 1.6244 nb = Doe' 11204 0.74133 = 74.133%. 126244 mimor

as FR = atb (FF) FER = a + b ftf of faer = utbl fret) 2202.05 atb (2374 .05) atb (837.9) 2 372,1 1829.95 - bc1536.15) =) b = 1.1912 putting in 2) 372.1 - at 998.1545 a=-626.05 300 Noul, For saction coefficient of ME = 0.28 as WR= 1938 PbS1 wf = 1862 ebs FER= MEXOR 0.28 x 1938 542-64 lbs fef=mex We 0-28 X 1862 521.36 lbs Tre ferxt - 1542.64 X 11 - 5969.04 ebs.in TE: 521.36% !! = 5734.96 Abs.in as Ps= TR Gr=TR = 5969.04 19.8968 Abs.in ar Ps 300 Psi 4952 To 5734.96 19.1165 lbs.in Ps Psi from above we known, P+= 425 psi 4 Pro 378.75 psi Ixt=296 Pb = 2X 19.1165 x 425 r Fara 298* Pr 2x1908968 x 378.75 r 1370.166 lbs il Dx = fatet for a 1477.189 + 1370.166 = 24.1275 (119) (380013262) lbs & Da = 24.1275 tys² lebs f+182) is XP/ 1477. 184 ebb A18.4
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