Question

1) In the figure below, a truck is modeled as a 2-DOF system (DOFs: bounce, x(t) and pitch, 0(t) motion of the truck with respect to its center of gravity, c.g.). i) Determine the EOMs using the free-body diagram provided below (denote the mass of the truck as m and mass moment of inertia wrt to its c.g.as ) = mr? where r is the radius of gyration) ii) Assuming that the influence of unbalanced tires can be modeled as harmonic moment M(t) = 2 sin(3t) acting on pitch, 0(t) DOF, put the EOMs in matrix form. iii) Assume zero ICs and the following system parameters and solve for the steady-state response of the truck via modal analyses: r2 = 0.64 m² m = 4000 kg 6 = 62 = 2000 N-s/m kı = ky = 20,000 N/m 1 = 0.9 m 12 = 1.4 m iv) Plot the steady-state response in physical DOFs (bounce, x(t) and pitch, 0(t)). Bounce Pitch Good Humor c.g. le co ki k2 4 12 c. x(1)

Answer 1

ICC-100) 1462-110) sy lp nolie To l2 In the 7. kcantho) ↑ Elmi to) 20 260). By D'Alembents principle; Eftma o mütk (2-1,0) te, (n-118) + K₂ Cathe to crit b) =0 mü + (kitka) u (kelp-Kalz) 0 + (6 +62) si - (ali-Cab) co 70- Bay HIT + 10 (Acw=twe, cw-arej JÖ - Klecacatio) - Gli cei - 0) 4 Kz 12 (nrl20) + C l 2 (rithee) -O- ne Jö -kali-kal) x -Gli ogl) in + hili? thach? Jo + 6 1² th o =O

Writing equation (l (2) in matrin forom (p?-1852 - Trotz 10 b) 183-183) kitke - Chile - Ky lz ) + 123]07% - Chili kak) (Kl2 + kabe L'equation of ·m2 motion (EoH), J = my² = 40008 0.64 = 2560kg.m m = yooo kg a=4 = 2000 M/m Kp =k2 = 20000 Ym. lp=o.gm, l=1.4m [regos on Engine side) so xip 2 so e X El-ym Substituting in the Fom. youo O DOG looo ] a) + Bar 2560 looo 5540 entom 017-photo- -187 =[ Incefil conditions at too, id) ED, OCA) 20 at too ito, 8(t) = 0 lo O M = 2 sint for forced ribration case

we have. 4000 1000 LC = LG so 40,000 10,000 yooo 10,000 55,400 we will take Laplace transform both side, soo by following unctial conditions. wler. 52x05) 1034 at 340 1000 [ 2500 ш Now zen woth (5)X₂S = 0,- (@w) 5. C dtz (4.027 (5) @25 Leild)) = SXO, LCOH)=sos). 6 3 2 X LC2 sin3t) = zetas 5249 2 X (S) 4000 1000 IS CXE), 149,000 10,0001$) yooo + LOCS S2 Oce) 1000 8540 nas (9,000 1,400 Tear = s2tq

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Now, DIS (-10+5) XS) DE) 20 (9+52) 18805 +soy s4 -8.5695²410,580s +52900 (altstas E OVE DLS) SC2tq) (2,059 of goy s3 +85695²110,58os +52900) 334 simplifying XS = 36 11139700 (529) 12 X 3 4235 11139700 (52+9) 3*( 3523053_19686452 +11593755+2582000) + 11139.700 ( 28054t goy3+856952+10ssos +52900) OF and 00) = 25 {* 21 11439700 bras Sebass 5 botas Is bass (at182. SEEGA 25229* Ostal) + 65o8sal tzo basso ta shos tharo SS bass +52909) The numerator and denominato hay red as well as and complen root for denominator a case of Xcs), and Oo). makes it diffrow it to take to time domain. howeven pultey s= jw for different a have the frequency response of the system.. we com.
It will need another 1hour to solve fully