Vehicle Dynamics - Half Car Modelling

AIM: To solve the given conceptual numerical based on 'Half Car Model'-

Question: Owing to the wide separation of the natural frequency of the sprung parts from that of the unsprung parts, the bounce and pitch motions of the vehicle body and the wheel motions exist almost independently. The sprung parts of a vehicle weigh 9.79 kN (2200 lb), its center of gravity is 106.7 cm (42 in.) behind the front axle, and the wheelbase is 228.6 cm (90 in.). The combined stiffness of the springs of the front suspension is 24.52 kN/m (140 lb/in.) and that of the rear suspension is 26.27 kN/m (150 lb/in.). The radius of gyration of the sprung parts about a horizontal transverse axis through the center of gravity is 102.6 cm (40.4 in.).

OBJECTIVES:

1. Calculate the natural frequencies of the pitch and bounce motions of the vehicle body

2. Also determine the locations of the oscillation centers

GIVEN:

Sprung mass weight, `W_s = 9.79kN`

CG distance from front axle, `a = 106.7cm = 1.067m`

Wheelbase, `l = 228.6cm = 2.286m`

Front spring stiffness, `k_(s_f)=24.52(kN)/m`

Rear spring stiffness, `k_(s_r)= 26.27(kN)/m`

Radius of gyration about horizontal axis, `r_y = 10.26cm = 0.1026m`

SOLUTION:

Sprung mass, `m_S = W_s/9.81kg`

CG distance from rear axle, `b = l-a= 228.6-106.7`

`b = 121.9cm = 1.219m`

Longitudinal moment of inertia about lateral axis, `I_y = m_s*r_y^2`

Simplified terms,

`D_1 = 1/m_s(k_f + k_r)`

`D_2 = 1/m_s(b*k_r-a*k_f)`

`D_3 = 1/I_y(a^2*k_f+b^2*k_r)`

`D_4 = 1/(m_sI_y)(a+b)^2k_rk_f`

Natural Frequencies,

Bounce, `omega_(n_1)^2 = (D_1 +D_3-sqrt((D_1+D_3)^2-4D_4))/2`

Pitch, `omega_(n_2)^2 = (D_1 +D_3+sqrt((D_1+D_3)^2-4D_4))/2`

Oscillation centres,

Bounce, `l_(O_1) = D_2/(omega_(n_1)^2 - D_1)`

Pitch, `l_(O_2) = D_2/(omega_(n_2)^2 - D_1)`

MATLAB code:

clear all
close all
clc

%inputs
ws = 9.79*10^3;     %sprung mass weight (N)
a = 106.7*10^-2;        %CG distance from front axle (m)
l = 228.6*10^-2;        %wheelbase (m)
ks_f = 24.52*10^3;      %front spring stiffness (N/m)
ks_r = 26.27*10^3;      %rear spring stiffness (N/m)
ry = 102.6*10^-2;       %radius of gyration (m)

ms = ws/9.81;       %sprung mass (kg)
b = l-a;        %CG distance from rear axle (m)
Iy = ms*ry^2;       %moment of inertia

%simplified terms
D1 = (ks_f + ks_r)/ms;
D2 = (b*ks_r - a*ks_f)/ms;
D3 = (a^2*ks_f + b^2*ks_r)/Iy;
D4 = ((a+b)^2*ks_r*ks_f)/(ms*Iy);

%Natural Frequencies
Wn_1 = sqrt((D1 + D3 - sqrt((D1+D3)^2 - 4*D4))/2)
Wn_2 = sqrt((D1 + D3 + sqrt((D1+D3)^2 - 4*D4))/2)

%Oscillation centres
lo_1 = D2/(Wn_1^2 - D1)
lo_2 = D2/(Wn_2^2 - D1)

OUTPUT:

FINAL ANSWERS:

Bounce natural frequency, `omega_(n_1) = 6.9794(rad)/s`

Pitch natural frequency, `omega_(n_2) = 8.1187(rad)/s`

Bounce oscillation centre, `l_(O_1) = -2.6923m = -269.23cm`

Pitch oscillation centre, `l_(O_2) = 0.3910m = 39.10cm`