Derive the equations of motion that model the four-DOF pitch-and-bounce model for the half-car shown below, and put the equations in vector-matrix form: M{x} +C{x} +K{x} = {f(t)} The model is developed for a half-car with one front wheel and one rear wheel. O is the pitch angle of the car, positive nose-up, and z, is the displacement of the center of gravity. z, and z, are the displacements of the front and rear wheels, respectively. z, (t) and z, (t) are inputs representing the base excitations at the front and rear tires, respectively. • The half-mass of the car, not including the wheels is m, and the mass moment of inertia of the car about a horizontal transverse axis through the car's center of mass is I,. The mass of each wheel and tire is m₂ = m₂. The distance from the center of gravity to the front axle is b and c is the distance from the center of gravity to the rear axle. The wheelbase is L=b+c. k and c, are the suspension stiffness and damping coefficient for the front suspension, respectively. •k and c, are the suspension stiffness and damping coefficient for the rear suspension, respectively. k and k, are, respectively, the stiffness of the front and rear tires.

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
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Derive the equations of motion that model the four-DOF pitch-and-bounce model for
the half-car shown below, and put the equations in vector-matrix form:
M{x} +C{x} + K{x} = {f(t)}
The model is developed for a half-car with one front wheel and one rear wheel. is the
pitch angle of the car, positive nose-up, and z, is the displacement of the center of
gravity. z, and z, are the displacements of the front and rear wheels, respectively.
z, (t) and z, (t) are inputs representing the base excitations at the front and rear tires,
respectively.
• The half-mass of the car, not including the wheels is m, and the mass moment of
inertia of the car about a horizontal transverse axis through the car's center of mass
is I,. The mass of each wheel and tire is m₂ = m₂.
• The distance from the center of gravity to the front axle is b and c is the distance
from the center of gravity to the rear axle. The wheelbase is L=b+c.
k and c, are the suspension stiffness and damping coefficient for the front
suspension, respectively.
•k and c, are the suspension stiffness and damping coefficient for the rear
●
suspension, respectively.
k and k, are, respectively, the stiffness
NO
z, (1)
kar
√
K₂
of the front and rear tires.
2
kaf
kif
72, (10)
Transcribed Image Text:Derive the equations of motion that model the four-DOF pitch-and-bounce model for the half-car shown below, and put the equations in vector-matrix form: M{x} +C{x} + K{x} = {f(t)} The model is developed for a half-car with one front wheel and one rear wheel. is the pitch angle of the car, positive nose-up, and z, is the displacement of the center of gravity. z, and z, are the displacements of the front and rear wheels, respectively. z, (t) and z, (t) are inputs representing the base excitations at the front and rear tires, respectively. • The half-mass of the car, not including the wheels is m, and the mass moment of inertia of the car about a horizontal transverse axis through the car's center of mass is I,. The mass of each wheel and tire is m₂ = m₂. • The distance from the center of gravity to the front axle is b and c is the distance from the center of gravity to the rear axle. The wheelbase is L=b+c. k and c, are the suspension stiffness and damping coefficient for the front suspension, respectively. •k and c, are the suspension stiffness and damping coefficient for the rear ● suspension, respectively. k and k, are, respectively, the stiffness NO z, (1) kar √ K₂ of the front and rear tires. 2 kaf kif 72, (10)
Answer:
{x}
M =
C =
K=
6
2₁
Z₂
cov
y
0
0
0
00 0
0 0
0
M3
m₂
0
0
m₂
0
b²c, +c²c,
-bc, + cc,
bcf
-cc,
-bc, + cc, bc, cc,
-C₁
-C₂
C₁ + C,
0
-C₁
-C,
[b²k+c²k-bkf+ck
-bkf+ckr
k₁f + k
bk
-kof
-ckgr
0
8]
0
{f(t)} =
k₁₁z, (t)
[k, z, (t))
- ST
Cf
0 C₂
bkof
__ksi
k + k
0
-ckgr
-k
0
k₂+k₁]
Transcribed Image Text:Answer: {x} M = C = K= 6 2₁ Z₂ cov y 0 0 0 00 0 0 0 0 M3 m₂ 0 0 m₂ 0 b²c, +c²c, -bc, + cc, bcf -cc, -bc, + cc, bc, cc, -C₁ -C₂ C₁ + C, 0 -C₁ -C, [b²k+c²k-bkf+ck -bkf+ckr k₁f + k bk -kof -ckgr 0 8] 0 {f(t)} = k₁₁z, (t) [k, z, (t)) - ST Cf 0 C₂ bkof __ksi k + k 0 -ckgr -k 0 k₂+k₁]
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