We use the basic theory of motion equations of an elastically supported rigid body when solving the vertical vibration of the vehicles, both road and rail. A rigid body of general shape can be elastically supported or suspended in an inertial orthonormal Cartesian space, which is defined by a stationary component system. In our case, it is a more complex system where we solve the vertical oscillation of the vehicle (model) including the effect of asymmetry based on various assumptions. As already mentioned, a quarter, half, or full 3D model of the mechanical system with several degrees of freedom, with different mass distributions and with different elastic and dissipative elements (bindings) is used. The kinetic excitation is considered, and the solution will be demonstrated analytically, numerically, and experimentally.
The time course of vertical displacement, i.e., the change in the position of the centre of gravity wT(t) and the rotation of the model about the central axes of inertia φx(t), φy(t), was chosen as the criterion for comparing the individual cases.
2.1. Individual Models Motion Equations
The solution was carried out for a spatial model (3D), a planar model, and a quartic model. The solution is found by determining the equations of motion of a mechanical system with three degrees of freedom.
2.1.1. Spatial Model
The equations of motion for the vibration of a spatial model of an elastically supported rigid body were determined from the Lagrange equations of the second kind
where
p—number of degrees of freedom,
qj—generalised coordinate,
Ek—kinetic energy,
Ep—potential energy,
Rd—Rayleigh dissipated energy,
Qj—excitation forces.
The solution procedure is known from the literature. After substituting for the individual variables and performing the appropriate derivations, we obtain a system of linear inhomogeneous equations, which can be written in matrix form
where
Mh—mass matrix,
Mb—damping matrix,
Mk—stiffness matrix,
Qj—vector of generalised excitation function (in our case kinematic excitation of the type
).
To solve Equation (5), we use Laplace integral transforms, modifying the elements of the matrix Mh by multiplying it from the left by the diagonal matrix D = (dij) of the third order, whose elements are equal to the inverse of the diagonal elements of the matrix Mh.
In the case of a symmetrical mechanical system, the mass matrix is unitary. If we multiply Equation (5) by the diagonal matrix
D from the left, we get
We rewrite this equation in the form
where
M—mass matrix,
B—damping matrix,
K—stiffness matrix,
Fj—vector of generalised kinematic excitation function.
The solution to Equation (7) is found by determining the elements of the damping matrix
bij and the stiffness matrix
aij. The calculation of these elements is given, e.g., in [
22].
If the vertical component of the force at the
m-th wheel location between the wheel and the ground is given by
where
km—spring stiffness of
m-th wheel,
hm(
t)—the height of the unevenness at the point of contact of
m-th wheel,
bm—damping coefficient of
m-th wheel, then the component of the generalised function
F1(
t)—is physically the acceleration component of the vertical displacement [ms
−2] and is given by the ratio of the product of the vertical components of the wheel forces and the mass of the vehicle model
The component of the generalised function
F2(
t)—the physical component of the angular acceleration with respect to the
x-axis [s
−2]—is given by the ratio of the product of the moments of the components of the wheel forces to the
x-axis and the moment of inertia of the vehicle model to this axis
The component of the generalised function
F3(
t)—physically the component of the angular acceleration with respect to the
y-axis [s
−2]—is given by the ratio of the product of the moments of the wheel force components to the
y-axis and the moment of inertia of the vehicle model to this axis
where
lxm a
lym for
m = 1, 2, 3, 4 are the distances of the support (elastic viscoelastic damping) from the axes
and
of the geometric symmetry with the origin at point
C,
Jx,
Jy—moments of inertia of the vehicle mass to the respective axes
lxm,
lym.
By defining the elements of the matrices M, B, K and the components of the vector F, a system of simultaneous linear inhomogeneous differential equations, the equations of motion of a spatial model of a vehicle with three degrees of freedom, with complete asymmetry: the mass distribution of the system () of the geometry and stiffness of the elastic support, the geometry and intensity of the viscous damping, and with the asymmetry of the kinematic excitation defined by the road roughness h(x) → h(t) is determined.
It should be emphasised that the investigation of vehicle oscillations assuming a planar vehicle model with a longitudinal axis of symmetry, and especially assuming a quarter vehicle model with both longitudinal and transverse axes of symmetry, will substantially change the definition of the elements of the matrices of the equations of motion (7) and thus the required solution. This can be documented by comparing the matrix elements of the single-individual vehicle models.
2.1.2. Planar Model
For spatial mode of vehicle with a longitudinal axis of symmetry, which has only two degrees of freedom, w(t) and φy(t), the following can be assumed:
Mass matrix M is diagonal and unit, and for mass distribution is valid Dxy = Dyx = 0, => S23 = S32 = 0, ey = 0.
The geometry of support is defined by dimensions ly1 = ly4, ly2 = ly3, lx1 = lx4, lx2 = lx3.
Damping intensity b1 = b4, b2 = b3.
Stiffness of elastic support is given by formulas k1 = k4, k2 = k3.
Then damping matrix
B elements are
Stiffness matrix
K elements are
Based on the above relations, we can determine the components of the vertical forces acting at the location of the
m-th wheel between the wheel and the road (rail) for crossing over unevenness of height
h1 =
h4,
h2 =
h3 2.1.3. Quarter Model
For quarter model of vehicle with a longitudinal and transverse axis of symmetry, i.e., for a system of one degree of freedom w(t) the following is assumed:
Mass distribution Dxy = Dyx = 0 → S23 = S32 = 0, ex = 0, ey = 0.
Mass matric M is unit, the centre of gravity of the system is identical to the centre of geometric gravity C, the main central axes of inertia are identical to axes of geometrical symmetry.
The support geometry is determined by dimensions lxj = lx, lyj = ly for j = 1, 2, 3, 4.
Damping intensity j = b, for j = 1, 2, 3, 4.
Stiffness of elastic support is given by kj = k.
Then, damping matrix
B elements are
Stiffness matrix
K elements are
Components of the generalised kinematic excitation function vector for
kj =
h 2.1.4. Models Discussion
From the brief analysis of the equations of motion discussed above, and the forces on the individual wheels, it follows that both the plane and quarter models require complete symmetry along the longitudinal axis (plane model) or along both the longitudinal and transverse axes (quarter model).
Due to the distribution of the different structural groups in the vehicle, no vehicle is geometrically symmetrical, it is clear that neither the quarter nor the half model are able to determine the dynamic properties of the vehicle with sufficient accuracy. In addition, the asymmetry in the loading of the vehicle (both cargo and possibly passengers) must be taken into account. For this reason, it is advisable to use a full spatial model for the further solution, which may be generally unsymmetrical (inertia and geometric axes are not parallel) or only unsymmetrical but with parallel inertia and geometric symmetry axes.
2.2. Simple Model of Four Axles Wagon
We will show the solution of this system on the chassis of a model of a railway vehicle, considering the effect of asymmetry. We assume a forced oscillation of the model, which is represented by a spatially elastically supported rigid plate with single and multiple primary linear suspension by coil springs (two-axle railway vehicle model). It is a kinematically excited system of three rigid bodies elastically supported and bounded, considering the effect of asymmetry [
1,
23,
24]. However, these authors did not address the issue of asymmetry.
A simple computational model of the vehicle chassis was chosen for the calculation. The model consists of two two-axle chassis with simple primary suspension and simple vehicle body suspension. The chassis are replaced by rigid plates of mass
m1 and
m2 with symmetric mass distribution (centre of gravity is identical to the kinematic centre, and the main central axes of inertia are identical to the axes of geometric symmetry). However, we consider the asymmetry of the spring stiffness parameter and their geometrical support. The model of the vehicle body is considered as a rigid plate of mass
m for which an asymmetric mass distribution is considered (the position of the centre of gravity is deflected by
ex and
ey from the geometric centre, and the main central axes of inertia are rotated with respect to the axes of geometric symmetry). The computational model is shown in
Figure 2.
The derivation of the following relationships is based on general assumptions, i.e., body stiffness, small displacements and rotations, and linear spring characteristics. We consider only the vertical displacements of arbitrary points of individual bodies. This vertical change in the position of a point of a body is determined by the displacement of the body’s centre of gravity w, w1 and w2, the rotations φx, φx1, φx2, φy, φy1, φy2 about the central axes of inertia of individual masses m, m1, m2 and its distance from these axes. Thus, we solve a system of bodies with nine degrees of freedom.
The equations of motion are derived from the Lagrange equations of the second kind (4). Kinetic energy of the system
where
Jx,
Jy—moments of inertia,
Dxy—deviation moment to central axes of inertia of chassis with mass
m,
Jx1,
Jy1,
Jx2 and
Jy2—inertia moments to the main central axes of bogies with mass
m1,
m2.
The potential energy of the system depends on the displacements of the individual masses in the places of their elastic support, i.e., in the places of their elastic bound. In our case, in the chassis model we have chosen the system of marking for points Ajki, their coordinates xjki, yjki, vertical displacements wjki and stiffness constants kjki. The individual indices correspond to masses j = 1, 2, quadrants k = 1, 2, 3, 4 and spring order i = 1, 2, ...., n (for the case of multiple support).
Similarly, the vertical change hjki is marked, which is at the point of contact of the axle (spring) with the rigid base, to which we relate the position of the body. This vertical change represents crossing over an unevenness in the road (rail)—kinematic excitation.
We can therefore determine the displacements of the individual points of the chassis, which are given by the relations for body
m1,
j = 1
Point | Vertical displacement | Stiffness constant | |
A111 | | | (19) |
A121 | | |
A131 | | |
A141 | | |
For body
m2,
j = 2
Point | Vertical displacement | Stiffness constant | |
A211 | | | (20) |
A221 | | |
A231 | | |
A241 | | |
In the case of a model of a body with a simple spring suspension in the axis of the chassis, the division into quadrants cannot be used to mark the individual points of action of the springs. Marking of the points, coordinates, and stiffness constants
Bjki,
xjki,
yjki,
kjki,
j = 0 body, mass
m,
k = 1, 2 marking of chassis 1 and 2, respectively,
i = 1, 4 belonging to the half-point
A111, respectively,
A141 follows from
Figure 2.
Displacement of individual point of body are given by
Point | Vertical displacement | Stiffness constant | |
B011 | | | (21) |
B014 | | |
B021 | | |
B024 | | |
Potential energy of bogies
m1,
m2 and body
m is
Kinetic energy from (18) and potential energy from (22) are substituted into the Equation (4) resulting in
where
j—degree of system freedom,
qj = (
φx,
φy,
w,
w1,
w2,
φx1,
φy1,
φx2,
φy2)
T and
Qj—generalised forces.
The solution (after derivation with respect to individual coordinates) is given by motion equations in matrix form
The individual elements of the stiffness matrix
κij are functions of stiffness constants
kjki, the dimensions of support
xjki,
yjki and the eccentricities
ex and
ey. The detailed solution of the individual elements is given, e.g., in [
22].
The generalised force functions
Qj ((23) and (24)) are again functions of the stiffness constants of the chassis springs
kjki and functions of the road unevenness (elevation) at time
h(
t) at the location of the wheel springs. Then the kinematic excitation functions in Equation (24) are given by
where
kjkihjki(
t)—the shape of the time function of kinematic excitation at individual axle points and their sequence.
Equation (24) can be rearranged to
Then the equation is multiplied by matrix
D (
dij) from the left, where
dij = 1/
αij—diagonal elements of mass matrix
Mh(
αij). After arrangement, the result is [
25,
26]
Mass matrix can be expressed in form
M =
E +
S and is substituted into (26)
where elements of matrix
S are
sij. These elements are given by equations
s11 =
s22 = 0;
sij = 0 for
i = 3, …, 9;
j = 3, …, 9;
s12 =
Dxy/
Jx;
s21 =
Dxy/
Jy. These quantities represent the effect of asymmetric distribution of mass, i.e., rotation of main central inertia axes to central axes parallel with geometric axes of the plate (vehicle). Similarly, the elements
aij of matrix
K are determined by division of
i-th row of stiffness matrix by
i-th element at diagonal of inertia matrix. Excitation function of time
Fj(
t) is determined by the same procedure with function
Qj(
t).
System of motion equations in case of symmetry is transformed into
where stiffness matrix is
Elements of stiffness matrix are function of spring stiffness
k12 and
k0 and function of displacement
x12,
y12,
x0,
y0. In the case of complete symmetry, the system of nine differentiated simultaneous equations is transformed to the system of seven simultaneous equations and two independent equations of harmonic motion for
q7 and
q9(
y9).
The solution of the system of Equation (26) can be performed by any of the numerical methods (MATLAB, MAPLE, etc.). Analytical solution is possible by applying matrix calculus, Lagrange’s method of variance of constants, or by Laplace transform. The Laplace transformation yields a system of linear algebraic equations for zero initial conditions [
27].
As indicated, the solution can also be done by applying the Laplace transformation, where Equation (27) after modifications becomes
where
and
are images of function
qj(
t) and
Fj(
t) for
j = 4 … 9,
p—parameter of Laplace transformation. By rewriting the Equation (32) into the matrix form, it is recommended to solve the system of linear algebraic equations by Cramer rule
where
D(
p)—determinant of matrix of above-mentioned system of equations, which is defined as
Equations for evaluation of real coefficients
A2(n − i) can be found in [
12]. These equations are generally valid for
n ≥ 2 and 0 ≤
i <
n.
Equations for evaluation of coefficients A2(n − i) for 2 < i < n are very complex and should be determined using MAPLE or MATLAB.
To determine the search required function qj(t) by reverse transformation of the images , it is convenient to modify the ratio of the determinants Dji(p) and D(p) so that the product of the expression and image obtained by the modification allows the application of the image convolution theorem.
The polynomial in Equation (34) with real coefficients can be replaced by the product of quadratic binomials
After performing the product on the right side, a polynomial is obtained
where real coefficients
are given by
By comparing the coefficients of the polynomials for the same powers of the parameter p2(n − i) on both sides of Equation (36) A2(n − i) = B2(n − i) a system of equations is obtained to determine ωi, ωj, ωl, ….
By comparing Equation (37) with the relations determining the coupling of the coefficients of the algebraic equation and its root factors, the algebraic equation is obtained
Determining the roots of the frequency Equation (38) is the most complicated part of the proposed solution, especially in terms of numerical accuracy, i.e.,
ω2—the circular frequencies of the functions
yj(
t), for
j = 1, 2, 3, …,
n. The next solution, after modifications and back-transformation, would give the convolution integral
where
ωk is the solution of Equation (38). From the known functions
qj(
t), i.e., the solution of the system of Equation (26), the required quantities of the system of Equation (24) are determined.
This method makes it possible to determine the vertical displacements of arbitrary points of the chassis or body of the vehicle, for example, the displacements of the points described by Equations (19)–(21). Thus, a program can be developed to calculate the vertical displacements of arbitrary points of railway vehicle chassis models with a large range of design variations.