Vehicle Ride 1
1)
Given :-
Wheel base = 2700 mm
Speed = 60 kmph = 60 * 1.6 * 5/18 = 26.67 m/s
Front sprung mass natural frequency target at design = 1.15 Hz
Solution :-
i)
Due to wheel base separation, the time lag between front and rear suspension is
T = wheelbase/speed = 2.7/26.67
Time lag = 0.101 s
For flat ride, after striking a bump, the front and rear should come to the same point in the shortest interval of time.
Therefore,
T(front) = T(rear) + T(lag)
T(front) = 1/1.15 = 0.869 s.
Therefore, 0.869 = T(rear) + 0.101
T(rear) = 0.768 s.
Thus, recommended rear sprung mass frequency is 1/0.768 = 1.3 Hz
f(rear) = 1.3 Hz
ii)
Front sprung mass natural frequency at design = 1.15 Hz (as given it needs to be kept same)
iii)
The relation between wheel rate and the spring rate is Kw = Ks * MR2
where,
Ks = spring rate, Kw = wheel rate, MR = Motion ratio
Therefore, As front spring rate is same in all loading conditions;
Front spring rate, Ks = Kw/MR2
Ks = 245/(0.97)^2
Front spring rate, Ks = 260.38 N/mm
iv)
The relation between wheel rate and the spring rate is Kw = Ks * MR2
where,
Ks = spring rate, Kw = wheel rate, MR = Motion ratio
Therefore, as rear spring rate is same in all loading conditions;
Rear spring rate, Ks = Kw/MR2
Ks = 245/(0.85)^2
Rear spring rate, Ks = 339.1 N/mm
v)
Front sprung mass natural frequency, f = (1/2pi) * sqrt (Kr/ms)
where, Kr = ride rate = (Ks * Kw)/(Ks + Kw)
Kr = (260.38 * 245)/(260.38 + 245)
Kr = 126.227 N/mm = 126227 N/m
At curb, the total front mass = 0.6 * 1500 = 900 kg
The sprung mass = 900 - 90 = 810 kg
Therefore, sprung mass at each wheel for a quarter car model = 810/2 = 405 kg
Front sprung mass natural frequency at curb, f = (1/2pi) * sqrt (126227/405) = 2.81 Hz
At GVW, the total front mass = 0.52 * 2000 = 1040 kg
The sprung mass = 1040 - 90 = 950 kg
Therefore, sprung mass at each wheel for a quarter car model = 950/2 = 475 kg
Front sprung mass natural frequency at GVW, f = (1/2pi) * sqrt (126227/475) = 2.59 Hz
vi)
Rear sprung mass natural frequency, f = (1/2pi) * sqrt (Kr/ms)
where, Kr = ride rate = (Ks * Kw)/(Ks + Kw)
Kr = (339.1 * 245)/(339.1 + 245)
Kr = 142.235 N/mm = 142235 N/m
At curb, the total rear mass = 0.4 * 1500 = 600 kg
The sprung mass = 600 - 60 = 540 kg
Therefore, sprung mass at each wheel for a quarter car model = 540/2 = 270 kg
Rear sprung mass natural frequency at curb, f = (1/2pi) * sqrt (142235/270) = 3.65 Hz
At GVW, the total rear mass = 0.48 * 2000 = 960 kg
The sprung mass = 960 - 60 = 900 kg
Therefore, sprung mass at each wheel for a quarter car model = 900/2 = 450 kg
Rear sprung mass natural frequency at GVW, f = (1/2pi) * sqrt (142235/450) = 2.83 Hz
vii)
Front unsprung mass natural frequency, f = (1/2pi) * sqrt (Keq/mus)
where, Keq = Ks + Kw
Keq = 260.38 + 245
Keq = 505.38 N/mm = 505380 N/m
In all three conditions, the front unsprung mass is same at each wheel = 90/2 = 45 kg
Front unsprung mass natural frequency, f = (1/2pi) * sqrt (505380/45) = 16.86 Hz
viii)
Rear unsprung mass natural frequency, f = (1/2pi) * sqrt (Keq/mus)
where, Keq = Ks + Kw
Keq = 339.1 + 245
Keq = 584.1 N/mm = 584100 N/m
In all three conditions, the rear unsprung mass is same at each wheel = 60/2 = 30 kg
Rear unsprung mass natural frequency, f = (1/2pi) * sqrt (584100/30) = 22.2 Hz
ix)
At curb,
Front sprung mass natural frequency = 2.81 Hz ..... From (v)
Rear sprung mass natural frequency = 3.65 Hz ..... From (vi)
Therefore, Rear/Front frequency ratio at curb = 3.65/2.81 = 1.30
At design load,
Front sprung mass natural frequency = 1.15 Hz ..... (It is fixed as given)
Rear sprung mass natural frequency = 1.3 Hz ..... From (i)
Therefore, Rear/Front frequency ratio at design = 1.3/1.15 = 1.13
At GVW,
Front sprung mass natural frequency = 2.59 Hz ..... From (v)
Rear sprung mass natural frequency = 2.83 Hz ..... From (vi)
Therefore, Rear/Front frequency ratio at GVW = 2.83/2.59 = 1.09
2)
Bump oscillation profile at Curb -
Bump oscillation profile at Design -
Bump oscillation profile at GVW -