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 independently. The sprung parts of a vehicle weigh 98 kN, its center of gravity is 108 cm behind the front axle and the wheelbase is 230 cm The combined stiffness of the springs of the front suspension is 25 kN/m and that of the rear suspension is 27 kN/m. The radius of gyration of the sprung parts about a horizontal transverse axis through the center of gravity is 0.92 m.
a) Calculate the natural frequencies of pitch and bounce motions of the vehicle body. Also determine the location of the oscillation centers.
b) If the radius of gyration of the sprung parts of the vehicle can be varied, determine the conditions under which the oscillation centers of the vehicle body will be located at the points of attachment of the front and rear springs. Also calculate the natural frequencies of the sprung parts.
To calculate the natural frequencies and the location of the oscillation centers, we need to use the following formulas:
1. Natural frequency of bounce motion (ωb):
ωb = √(k / m)
where k is the combined stiffness of the springs and m is the mass of the sprung parts.
2. Natural frequency of pitch motion (ωp):
ωp = √[k / (m * r^2)]
where r is the radius of gyration of the sprung parts.
3. Location of the oscillation centers:
Front oscillation center = (L * m1) / (m1 + m2)
Rear oscillation center = L - Front oscillation center
where L is the wheelbase, m1 is the sprung mass in front of the front axle, and m2 is the sprung mass behind the front axle.
Let's calculate these values step by step:
a) Calculate the natural frequencies and the location of the oscillation centers:
Given:
Sprung mass (m) = 98 kN
Combined stiffness of the front springs (k1) = 25 kN/m
Combined stiffness of the rear springs (k2) = 27 kN/m
Radius of gyration (r) = 0.92 m
Center of gravity (CG) = 108 cm = 1.08 m
Wheelbase (L) = 230 cm = 2.3 m
First, let's calculate the masses:
m1 = m * (CG / L)
= 98 kN * (1.08 m / 2.3 m)
= 46.739 kN
m2 = m - m1
= 98 kN - 46.739 kN
= 51.261 kN
Now, let's calculate the natural frequencies:
ωb = √(k1 / m)
= √(25 kN/m / 98 kN)
≈ 0.1587 rad/s
ωp = √[k1 / (m * r^2)]
= √[25 kN/m / (98 kN * (0.92 m)^2)]
≈ 0.4393 rad/s
Next, let's calculate the location of the oscillation centers:
Front oscillation center = (L * m1) / (m1 + m2)
= (2.3 m * 46.739 kN) / (46.739 kN + 51.261 kN)
≈ 1.042 m
Rear oscillation center = L - Front oscillation center
= 2.3 m - 1.042 m
≈ 1.258 m
Therefore, the natural frequency of bounce motion (ωb) is approximately 0.1587 rad/s, the natural frequency of pitch motion (ωp) is approximately 0.4393 rad/s, the front oscillation center is approximately 1.042 m from the front axle, and the rear oscillation center is approximately 1.258 m from the front axle.
b) Now, let's determine the conditions under which the oscillation centers will be located at the points of attachment of the front and rear springs:
To have the oscillation centers at the points of attachment of the front and rear springs, the front oscillation center should be at the front spring attachment point and the rear oscillation center should be at the rear spring attachment point.
In this case, the front oscillation center is approximately 1.042 m from the front axle and the rear oscillation center is approximately 1.258 m from the front axle.
Therefore, for the oscillation centers to be located at the points of attachment of the front and rear springs, the following condition must be met:
Front oscillation center = 0.0 m
Rear oscillation center = L - Front oscillation center
Rear oscillation center = L
L = Rear oscillation center
So, for the oscillation centers to be at the points of attachment of the front and rear springs, the rear oscillation center must be equal to the wheelbase (L).
Now, let's calculate the natural frequencies of the sprung parts under these conditions:
ωb = √(k1 / m)
ωp = √[k1 / (m * r^2)]
Using the given values:
m = 98 kN (unchanged)
k1 = 25 kN/m (unchanged)
r = 0.92 m (unchanged)
ωb = √(25 kN/m / 98 kN) ≈ 0.1587 rad/s (unchanged)
ωp = √[25 kN/m / (98 kN * (0.92 m)^2)] ≈ 0.4393 rad/s (unchanged)
Therefore, the natural frequencies of the sprung parts under the conditions where the oscillation centers are located at the points of attachment of the front and rear springs remain the same as in part a: ωb ≈ 0.1587 rad/s and ωp ≈ 0.4393 rad/s.