Question

Figure 1 shows a basic model of an automobile, travelling over horizontal road. The car has a mass M, and a mass moment of inertia Jo= Ma(measured with respect to the center of gravity (c.g)). The vertical displacement of the center of gravity is e(t), and the rotation is e(t) (measured with respect to the position of static equilibrium). The total stiffness at the front wheels is equal to ki and the total stiffness at the rear wheels is ke. The distance between the front wheels and the center of gravity is a, and the distance between the rear wheels and the center of gravity is 2a. IfM= 1500 kg, ki = 2000 N/m, ka = 1000 Nm and a=0.5 m, determine the natural frequencies and mode shapes of the automobile. X(t) c.g 76(t) M, JoMap w w- k2 2a a Figure 1: A Two degree of freedom automobile system.

Answer 1

xt! M, :MA -F'k cxcutareck)! Answer: Ki FBD of the_system: J. öit) x 16) Torna 7 sa Í. Mårt) 20-0, ☺ Ki (xCHE 20 0041] considering the linear moment of mass, Applying Newtons grond law of motion Müet)+ Kilxatia 0111)+ Kg exlt !+a o(+1) KÝCH + Skit ky\x46) -/a 01 El to milt) +(K1+Ks ) 21t) + (Q Kg - 20 Kp) 014) = 0 considering the rolation of mass about its centre of gravity Torque due to spring 'k, & Kg getire TK, -ki (x1)-20 0641).(20) Tk, = + K3 (XLt)+a olt) (a) podu torque Torque due to inertia Note: - HVC Te Toöck)

Trött) - 9AK, (x (4) - 2010(t1)tak, cult)of noctl).. To öCE 1+x1)(ak-, K;) + oct] (Pak;+ a ki) = 0 © .: The equations of motion are Müellt (Kites ) x64) + Caks -20 Ki) olti To öll + latky-322.K;)x() + (QK3+20 ki) oct) = 0 put x: Asincet - Aw'sinust 0 : B sincet ö: -BW-sincet) o M(-Aw'sinwt) + (kit ka) (A SinW1) + ca kj- 20 ki) (B.Sincol :0 Al-Mu't Ki+k;) + Blak; -2aK) = 0 201ki-aki 1/B -MW't kitky > Jo-Bw'sinwef)+ (aks -- 20 K) (Asmut) + Cakarta KU (Bsinowe)ieu Bl-Iow't a K3+2a k) = Alzaki-akz ) - Iowa ks+20K, No - 6) akinak

w; : 7.4065, W, *= 1.2602 W mi 2.7215 rad/sec Win 6,1226 rad/sec Natural frequencies are Win = 2.7215 radises Win = 1.1926 rad/sai finding Miode shapes; © Wine 2.7215 rad(see substitue in eq@] AL B, 49-7215)+2 - 7.4065+) -5.4065 AL -0.1849 let Balradlec 2 B, AL-0.1849 9 Won= 1.1226 radlsee : 1.3517 A B2 1.260272 let Biz iradle Az i 1.3517 AL: 1:3515 B

Mude shapes اررر 18 A, nd موال 1)

To : Mia? : 1500 X 0.95 » 375 kq-m? Given M: 150019 KI: 2000 N/m Kg, 1000 N/m 0.5 m substitute the given (6) A kr 2000 - 500 - 1500 W + 3000 values in eq@ € 296 1 -wet 2 6. А - 37.500?+ 500+ 9000 sto 1500 -375w?+2500 1500 - 15w't 1000 1. 60 60 67 - 15W4100 -wr2 60 = [150+ 100) (-2+2) #15694 zowe?-loow +200 60: 12 a 3wt Gw? - 20W?+ 40 34426 642+28 = 0 let wap 3p? 262+28 = 0 P = 7.4065 , P, :1.2602