Question

HW #8 QUARTER CAR MODEL Sprunp Mass (14 of vehicle mass) m= 500kg ķ= 25000 ms Equesiti 0000 tor b: 3500 N m/sec Actua U HAU QUARTER OF m= 50 kg A CAR Unsprung Mass Egil position (mass of the wheel k = 200000 and associated compe hersts) U(U): Actuator force acting on Sprung and unsprung stiffness of tire X(t): Road heipht, m y (+): Sprung mass displacement, in (t): Unsprung -- --- Equilibrium position mass displacement, m. Determine the transfer functions G,G,G,G, GC y(s) = Gos) v(s) + GG) 06) UCS) = GC) y(s)+Gls)x)+ Gucs) mass, N.

Answer 1

equation of motion Fers sprung mass, mis roce body diagram velocity acceleralich misių 1 mns Ty ý = du ett d . I Lt) bely-v). va du Ksly-V) equaling Forus in vertical diredicin > 2, 3 + bột - )+ kg ( ) AC). ms it bsy + Ksy = Ult) + bs utks (ms.?t bos D + Ks )44) - uct)+ (bs D+ ks) VC) in laplace domain D5 Given mig. 500 kg ksa 25000 Nlm bs: 3500 NSim mua 50kg Kt: 2X10S Nim. = y(s) =- - bsstks N(E). u(s)__ Timis? +63 54 ks.) msis?+ best ks) => Yl5). Gics). VCS) + G2(5) .ucs) 1. bs Stks 3500s + 25000 whore G, (5) in s bs Saks: 500 5'+3500S + 25000 C12 (5) = mast bs Saks : 5005435005 A 95000 Scanned with CamScanner

Equation of motion for unsprung mass (mu) Fow body dicigram Ks (V-Y) uct) bs (4-4) mu ürl mu lir KT (V-XR) - Fvertical = 0 , , 7+ b (7-9) + KCV-4) + KCV-XR). - (4) my + bs V+(ks+K7)V :.Ult) + bs ý + ksy + kq XR [mu DP + bs: D + (Ks+ks) ] 264): - UCt).+ (bs: 17+ Ks ) y(t) + Ry Xplt) in Laplace domain D S -(5) (bsis + ks).4cs) KT Xes) · V(S) - - (mas+bg S4 Kotka) (mus+ bs 5 + kstkr) v(s). 2013 (5).y(s) 4 64(5) XX'S) + Gis. (s). Uls) bs: 5+ ks 3509 5 + 25000 5057+3500 5+25000+ 2x105 2 X 105 to- + mus+ best ks + kg) where G3 (5.) = nous?+bs S+kst kt . Kp 50+ 3,5005+ 2.256105 674(5) - MusPtb55+ Ks+kt As (S) = mesla bas+kst kr 505?+ 3500S +2:254105 Ics Scanned with CamScanner