Challenge

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 -

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