Vehicle ride 1
1. Given:
Wheelbase = 2700 mm = 2.7m
Speed = 60mph = 60*1.6*5/18
Speed = 26.67 m/s
Front sprung Mass natural frequency at design = 1.15 Hz
i.)Time lag between front and rear suspension due to wheel base separation:
T(lag) = wheelbase/speed
T(lag) = 2.7/26.67
T(lag) = 0.10 seconds
In order to get a flat ride, the nose and tail part of the vehicle should come to the same point after striking a bump in the shortest interval of time.
Therefore,
T(front) = T(rear) + T(lag)
Front frequency = 1.15Hz
Time period of front sprung mass, T(front) = 1/1.15
T(front) = 0.8695 s
Therefore,
0.8695 = T(rear) + 0.1
T(rear) = 0.8695 - 0.1
T(rear) = 0.7695 s
Hence,
Recomended Rear sprung mass natural frequency,
f(rear) = 1/T(rear) = 1/0.7695
f(rear) = 1.29 Hz.
ii.) Since it has been asked to keep Front sprung mass natural frequency fixed it is same as the given value.
i.e., f(front) = 1.15 Hz.
iii.) Considering half load condition for front spring rate:
Kt = Ks(f) * (MR)^2
Kt - Tire/wheel rate
Ks(f) - front spring rate
MR - motion ratio
Ks(f) = Kt ÷ (MR)^2
At half load condition,
Kt = 245 N/mm and MR = 0.97
Ks(f) = 245 ÷ (0.97)^2
Ks(f) = 260.38 N/mm
iv.) Similarly, considering half load condition for rear spring too,
Kt = Ks(r) * (MR)^2
Ks(r) = Kt ÷ (MR)^2
Kt = 245 N/mm and MR = 0.85
Ks(r) = 245 ÷ (0.85)^2
Ks(r) = 339.1 N/mm
v.) Front Sprung mass natural frequency/front ride frequency is given by,
fs(f) = (1/2pi) * sqrt(Kr/Ms)
Ride rate, Kr = Kt*Ks/(Kt + Ks)
Ks = 260.38 N/mm
Kr = 245 * 260.38/(245 + 260.38)
Kr = 126.227 N/mm = 126227.98 N/m
Now taking into consideration curb and GVW load condition,
1.) Curb
Total mass of front portion of vehicle, M = 0.6*1500
M=900 kgs
Front Sprung mass, Ms = M - Mus
Ms = 900 - 90
Ms = 810 kgs
Therefore,
fs(f) = 0.1591 * sqrt(126227.98/810)
fs(f) = 1.98 Hz
2.) Gross Vehicle Weight(GVW)
Total mass of front portion of vehicle,
M = 0.52 * 2000
M = 1040 kgs
Front sprung mass, Ms = M - Mus
Ms = 1040 - 90
Ms = 950 kgs
Therefore,
fs(f) = 0.1591 * sqrt (126227.98/950)
fs(f) = 1.83 Hz
vi.) Rear sprung mass natural frequency/rear ride frequency
fs(r) = (1/2pi) * sqrt(Kr/Ms)
Ride rate, Kr = Kt*Ks/(Kt + Ks)
Ks = 339.1 N/mm
Kr = 245 * 339.1/(245 + 339.1)
Kr = 142.235 N/mm = 142235.06 N/m
Now taking into consideration curb and GVW load condition,
1.) Curb
Total mass of rear portion of vehicle, M = (1 - 0.6) *1500 =0.4 *1500
M= 600 kgs
Rear Sprung mass, Ms = M - Mus
Ms = 600 - 60
Ms = 540 kgs
Therefore,
fs(r) = 0.1591 * sqrt(142235.06/540)
fs(r) = 2.58 Hz
2.) Gross Vehicle Weight(GVW)
Total mass of rear portion of vehicle,
M = (1 - 0.52) * 2000 = 0.48 * 2000
M = 960 kgs
Rear sprung mass, Ms = M - Mus
Ms = 960 - 60
Ms = 900 kgs
Therefore,
fs(r) = 0.1591 * sqrt (142235.06/900)
fs(r) = 2.00 Hz
vii.) Front unsprung mass natural frequency at all three load condition,
fus(f) = (1/2pi) * sqrt (Keq/Mus)
Equivalent stifness, Keq = Ks + Kt
Ks = 260.38 N/mm
Keq = 260.38 + 245
Keq = 505.38 N/mm
Keq = 505380 N/m
1.) Curb
Total Unsprung mass (front)
Mus = 90 kgs
corner unsprung mass,
Mus/2 = 45 Kgs
fus(f) = 0.1591 * sqrt (505380/45)
fus(f) = 16.86 Hz
2.) Design
Total Unsprung mass (front)
Mus = 90 kgs
corner unsprung mass,
Mus/2 = 45 Kgs
fus(f) = 0.1591 * sqrt (505380/45)
fus(f) = 16.86 Hz
3.) GVW
Total Unsprung mass (front)
Mus = 90 kgs
corner unsprung mass,
Mus/2 = 45 Kgs
fus(f) = 0.1591 * sqrt (505380/45)
fus(f) = 16.86 Hz
viii.) Rear unsprung mass natural frequency at all three load condition,
fus(r) = (1/2pi) * sqrt (Keq/Mus)
Equivalent stifness, Keq = Ks + Kt
Ks = 339.1 N/mm
Keq = 339.1 + 245
Keq = 584.1 N/mm
Keq = 584100 N/m
1.) Curb
Total Unsprung mass (rear)
Mus = 60 kgs
corner unsprung mass,
Mus/2 = 30 Kgs
fus(r) = 0.1591 * sqrt (584100/30)
fus(r) = 22.2 Hz.
2.) Design
Total Unsprung mass (rear)
Mus = 60 kgs
corner unsprung mass,
Mus/2 = 30 Kgs
fus(r) = 0.1591 * sqrt (584100/30)
fus(r) = 22.2 Hz.
3.) GVW
Total Unsprung mass (rear)
Mus = 60 kgs
corner unsprung mass,
Mus/2 = 30 Kgs
fus(r) = 0.1591 * sqrt (584100/30)
fus(r) = 22.2 Hz.
ix.) Rear to front frequency ratio for all three load condition is given below,
1.)Curb
The front and rear sprung mass natural frequencies at curb are as calculated for question (v)
and (vi)
fs(f) = 1.98 Hz
fs(r) = 2.58 Hz
Therefore, rear/front frequency ratio at curb condition is,
fs(r)/fs(f) = 2.58/1.98
fs(r)/fs(f) = 1.303
2.) Design
Since Front sprung mass natural frequency at Design should be kept fixed,
fs(f) = 1.15 Hz
Rear sprung mass natural frequency at Design based on calculation in (ii),
fs(r) = 1.29 Hz
Therefore,rear/front frequency ratio at design condition is,
fs(r)/fs(f) = 1.29/1.15
fs(r)/fs(f) = 1.12
3.) GVW
The front and rear sprung mass natural frequencies at GVW are as calculated for question (v)
and (vi)
fs(f) = 1.83 Hz
fs(r) = 2.00 Hz
Therefore,rear/front frequency ratio at GVW condition is,
fs(r)/fs(f) = 2.00/1.83
fs(r)/fs(f) = 1.09