Keywords

1 Theoretical Mathematical Model of the Tower-Type Orchard Sprayer

The study of the [6, 8] nonlinear dynamics of an agricultural tower pulverizer, coupled with a vehicle suspension, that is subject to random excitations due to soil irregularities, modeled as an inverted double pendulum over a moving suspension, with three degrees of freedom (one translation and two rotations). To take into account the random loadings, a parametric probabilistic approach was employed, where the external force was assumed to be a harmonic random process with random amplitude and frequency. The probability distribution of these random [8] parameters was constructed based on the known information through the maximum entropy principle. The results of numerical simulation show that large discrepancies in the system response can be seen when one compares the mean of the stochastic model with the nominal (deterministic) model. It is also noted that these responses are subject to a high level of uncertainty. Furthermore, an analysis of the system response probability distributions [6, 8] shows that they present asymmetries with respect to mean and unimodal behavior.

Some considerations were taken to arrive at the theoretical model. Initially, the simplifications [14, 15] will be considered. Where, initially, all the masses (truck, axle, wheel and tower with its eight fans) are concentrated in their centers of gravity. One of the main simplifications to be considered is that the operation of the sprayer will take place in short periods of time, because if long periods are taken the mass of the reservoir tank will have its center of gravity shifted, due to the fact that the sprayer starts the work with the reservoir full of defensive liquid and it is discharging during the spraying operation, until it finishes empty. Therefore, the mass of the reservoir tank is considered constant.

Once this is done, the reservoir mass and the chassis mass are grouped into a single invariant center of gravity \({m}_{2}\). All the concentrated masses of the eight fans are replaced by a single mass which, added to the motor mass, results in \({m}_{3}\), and the unbalanced mass of the direct current electric motor is \({m}_{0}\), located at the top of the tower. The tower, in turn, is represented by a negligible mass element of length \({L}_{2}\). The pivot point of the tower is represented by a point P, located at a distance \({L}_{1}\) above the trailer's center of gravity. This same junction P is represented by an elastic element \({k}_{t}\) and damping \({c}_{t}\) torsional and linear.

Starting from the premise that, when the system rotates around the point of concentrated mass \({m}_{2}\), the left and right tires have the same displacement, in opposite directions; thus, it is considered that the system only presents a translational movement in the vertical direction. Therefore, the theoretical system can be considered as being of ¼ of a vehicle or quarter-car, taking into account the displacement of only one of the wheels. This wheel is then represented by an element of mass m1 with linear stiffness \({k}_{1}\). And a vehicle-type MR damper, represented by F, is coupled to the system, in parallel to the viscous spring-damper suspension \({k}_{2}\) and c, respectively, according to the Bounc-Wen model.

An X–Y coordinate system is adopted, with X at ground level and Y passing through the center of mass of truck \({m}_{2}\). The tower will then have an angular offset \({\phi }_{1}\). It is the angular displacement of the unbalanced mass is \({\phi }_{2}\). Excitements resulting from irregularities presented by the soil of the orchards cause displacements in the tires represented by \({ye}_{1}\); mathematically \({ye}_{1}\) represents an excitation source that can be of the harmonic, transient, etc. type.

However, it is observed that in the simplification presented the model presented has four degrees of freedom, with only the mass of the fans concentrated at the top of the tower, called \({m}_{2}\) in the figure, and without taking in considering the mass of the wheels. Finally, the next simplification will be to add the unbalanced mass engine at the top of the tower and also the mass of a wheel, thus transforming the model into a ¼ vehicle (Quarter-car). Finally, all the simplifications mentioned above are represented in a theoretical scheme presented in the following figure, Fig. 1.

Fig. 1
figure 1

Theoretical mathematical model of the turret sprayer. a trailer and tower coupled and b scheme of forces applied

Full size image

The differential equations of motion are obtained from the Method of Energies and Conservative Forces, which is employed using the so-called Euler–Lagrange Equations.

2 Euler–Lagrange Energy Method

Euler–Lagrange equation is defined as:

$$\frac{d}{dt}\left(\frac{\partial L}{\partial {\dot{q}}_{j}}\right)-\frac{\partial L}{\partial {q}_{j}}={Q}_{j}$$
(1)

with j = 1,2,3…. Where \(L={T}_{T}-{V}_{T}\), called Lagrangian where \({T}_{T}\) represents the total kinetic energy of the system, \({V}_{T}\) is the work of conservative forces (potential energy of the chassis, tower and unbalanced motor masses, and the potential energy of the elastic elements). And Q is the work of all non-conservative forces (such as the energy dissipated by the damping elements).

2.1 Total Kinetic Energy of the System

The total kinetic energy of the \({T}_{T}\) system is the sum of the kinetic energy portions of the trailer, the tower, the unbalanced mass of the engine, and the tire, which are represented by \(T_{c} , \;T_{t} , \;T_{o} \;and\;T_{p}\) respectively. \(T_{T}\) is then defined as:

$${T}_{T}={T}_{p}+{T}_{c}+{T}_{t}+{T}_{0}$$
(2)

Where:

$${T}_{p}=\frac{1}{2}{m}_{1}{V}_{1}^{2}=\frac{1}{2}{m}_{1}\left({\dot{x}}_{1}^{2}+{\dot{y}}_{1}^{2}\right)$$
(3)
$${T}_{c}=\frac{1}{2}{m}_{2}{V}_{2}^{2}=\frac{1}{2}{m}_{2}({\dot{x}}_{1}^{2}+{\dot{y}}_{2}^{2})$$
(4)
$${T}_{t}=\frac{1}{2}{(m}_{3}){V}_{3}^{2}=\frac{1}{2}{(m}_{3})\left({\dot{x}}_{2}^{2}+{\dot{y}}_{3}^{2}\right)+\frac{1}{2}{m}_{3}{L}_{2}^{2}{\dot{\phi }}_{1}^{2}$$
(5)

With \({m}_{3}=M+m\), where M represents the concentrated mass of the fans and m the mass of the motor. The term \({m}_{3}{L}_{2 }^{2}\) in Eq. 5 represents the moment of inertia of the tower.

$${T}_{o}=\frac{1}{2}{m}_{0}\left({\dot{x}}_{0}^{2}+{\dot{y}}_{0}^{2}\right)+\frac{1}{2}{m}_{0}{r}^{2}{\dot{\phi }}_{2}^{2}$$
(6)

The horizontal and vertical positions of the \({m}_{3}\) tower mass are given as follows, respectively: The horizontal position is:

$$\begin{aligned} x_{2} & = L_{2} sin\phi_{1} - x_{1} \\ x_{2} & = L_{2} sin\phi_{1} - x_{1} \\ \end{aligned}$$
(7)

and its first and second derivatives, respectively in time:

$$\begin{aligned} \dot{x}_{2} & = L_{2} \dot{\phi }_{1} cos\phi_{1} - \dot{x}_{1} \\ \ddot{x}_{2} & = \ddot{x}_{1} L_{2} \ddot{\phi }_{1} cos\phi_{1} - L_{2} \phi_{1}^{2} sin\phi_{1} \\ \end{aligned}$$

The vertical position is:

$${y}_{3}={y}_{2}+{L}_{1}+{L}_{2}cos{\phi }_{1}$$
(8)

and its first and second derivatives, respectively in time:

$$\begin{aligned} \dot{y}_{3} & = \dot{y}_{2} + L_{1} + L_{2} \dot{\phi }_{1} sin\phi_{1} \\ \ddot{y}_{3} & = \ddot{y}_{2} - L_{2} \ddot{\phi }_{1} sin\phi_{1} - L_{2} \dot{\phi }_{1}^{2} cos\phi_{1} \\ \end{aligned}$$

The horizontal and vertical positions of the unbalanced mass \({m}_{0}\) can be described as follows:

$${x}_{0}={L}_{2}sin{\phi }_{1}-rcos{\phi }_{2}$$
(9)
$${y}_{0}={L}_{2}cos{\phi }_{1}-rsin{\phi }_{2}+{L}_{1}+{y}_{2}$$
(10)

With its derivatives \(\dot{x}_{0}\) and \(\dot{y}_{0}\) in time given by:

$$\begin{aligned} \dot{x}_{0} & = \dot{\phi }_{1} L_{2} cos + \dot{\phi }_{2} rsin\phi_{2} \\ \dot{y}_{0} & = - \dot{\phi }_{1} L_{2} sin\phi_{1} + \dot{\phi }_{2} rcos\phi_{2} + \dot{y}_{2} \\ \end{aligned}$$

representing the coordinates of the angular velocity of the unbalanced mass m0.

In this way, substituting Eqs. 3–6 in Eq. 2, we obtain:

$$\begin{array}{*{20}l} {T_{t} = \frac{1}{2}m_{1} \left( {\dot{x}_{1}^{2} + \dot{y}_{1}^{2} } \right) + \frac{1}{2}m_{2} \left( {\dot{x}_{1}^{2} + \dot{y}_{2}^{2} } \right) + \frac{1}{2}m_{3} \left( {\dot{x}_{2}^{2} + \dot{y}_{3}^{2} } \right)} \hfill \\ {\quad \quad \quad + \frac{1}{2}m_{3} L_{2}^{2} \dot{\phi }_{1}^{2} + \frac{1}{2}m_{0} \left( {\dot{x}_{0}^{2} + \dot{y}_{0}^{2} } \right) + \frac{1}{2}m_{0} r^{2} \dot{\phi }_{2}^{2} } \hfill \\ \end{array}$$
(11)

with respect to time of Eqs. 7 and 8 in the previous Eq. 11, we have:

$$\begin{aligned} T_{t} & = \frac{1}{2}m_{1} \dot{x}_{1}^{2} + \frac{1}{2}m_{1} \dot{y}_{1}^{2} + ~\frac{1}{2}m_{2} \dot{x}_{1}^{2} + \frac{1}{2}m_{2} \dot{y}_{2}^{2} + \\ & \frac{1}{2}m_{0} \left( {\dot{x}_{0}^{2} + \dot{y}_{0}^{2} } \right) + \frac{1}{2}m_{3} L_{2}^{2} \dot{\phi }_{1}^{2} + \frac{1}{2}m_{0} r^{2} \dot{\phi }_{2}^{2} \\ & \frac{1}{2}m_{3} ~\left( {\dot{x}_{1} + L_{2} \dot{\phi }_{1} cos\phi _{1} } \right)^{2} + \frac{1}{2}m_{3} \left( {\dot{y}_{2} - L_{2} \dot{\phi }_{1} sin\phi _{1} } \right)^{2} + \\ \end{aligned}$$
(12)

And now the first time derivatives of Eqs. 9 and 10 in Eq. 12, you get:

$$\begin{aligned} T_{T} & = \frac{1}{2}m_{1} \dot{x}_{1}^{2} + \frac{1}{2}m_{1} \dot{y}_{1}^{2} + ~\frac{1}{2}m_{2} \dot{x}_{1}^{2} + \frac{1}{2}m_{2} \dot{y}_{2}^{2} \\ & \quad + \frac{1}{2}m_{3} \left( {\dot{x}_{1} + L_{2} \dot{\phi }_{1} cos\phi _{1} } \right)^{2} + ~\frac{1}{2}m_{3} \left( {\dot{y}_{2} - L_{2} \dot{\phi }_{1} sin\phi _{1} } \right)^{2} + \\ & \frac{1}{2}m_{0} \left( {\dot{\phi }_{1} L_{2} cos\phi _{1} + \dot{\phi }_{2} rsin\phi _{2} } \right)^{2} + \frac{1}{2}m_{3} L_{2}^{2} \dot{\phi }_{1}^{2} + \frac{1}{2}m_{0} r^{2} \dot{\phi }_{2}^{2} \\ & \frac{1}{2}m_{0} \left( { - \dot{\phi }_{1} L_{2} sin\phi _{1} + \dot{\phi }_{2} rcos\phi _{2} + \dot{y}_{2} } \right)^{2} \\ \end{aligned}$$
(13)
$$\begin{aligned} T_{T} & = \frac{1}{2}m_{3} \dot{x}_{1}^{2} + \frac{1}{2}m_{3} \dot{y}_{2}^{2} - m_{3} \dot{x}_{1} L_{2} \dot{\phi }_{1} cos\phi _{1} - m_{3} \dot{y}_{2} L_{2} \dot{\phi }_{1} sin\phi _{1} + \\ & \frac{1}{2}m_{3} L_{2}^{2} \dot{\phi }_{1}^{2} \left( {cos^{2} \phi _{1} + sin^{2} \phi _{1} } \right) + \frac{1}{2}m_{0} \dot{\phi }_{1}^{2} L_{2}^{2} \left( {cos^{2} \phi _{1} + sin^{2} \phi _{1} } \right) + \\ & \frac{1}{2}m_{0} \dot{\phi }_{2}^{2} r^{2} \left( {cos^{2} \phi _{2} + sin^{2} \phi _{2} } \right) - \\ & m_{0} \dot{\phi }_{1} \dot{\phi }_{2} L_{2} r\left( {cos\phi _{2} sin\phi _{1} - cos\phi _{1} sin\phi _{2} } \right) + \\ & \frac{1}{2}m_{0} \dot{y}_{2}^{2} - m_{0} \dot{y}_{2} \dot{\phi }_{1} L_{2} sin\phi _{1} + m_{0} \dot{y}_{2} \dot{\phi }_{2} rcos\phi _{2} + \\ & \frac{1}{2}m_{1} \dot{x}_{1}^{2} + \frac{1}{2}m_{1} \dot{y}_{1}^{2} + ~\frac{1}{2}m_{2} \dot{x}_{1}^{2} + \frac{1}{2}m_{2} \dot{y}_{2}^{2} + \frac{1}{2}m_{3} L_{2}^{2} \dot{\phi }_{1}^{2} + \frac{1}{2}m_{0} r^{2} \dot{\phi }_{2}^{2} \\ \end{aligned}$$
(13b)

The lateral displacement of the trailer's center of gravity, \({x}_{1}\), limited by the tires, is very small compared to the magnitudes of the other displacements. Thus, it will be assumed that \({x}_{1}\) is constant [13]. So: if constant, then: if \({x}_{1}\cong\) constant, then: \({\dot{x}}_{1}\cong {\ddot{x}}_{1}\cong 0\).

In this way, making use of the trigonometric identities below:

$$si{n}^{2}{\phi }_{2}+co{s}^{2}{\phi }_{2}=1$$
$$cos{\phi }_{2}cos{\phi }_{1}+{sin\phi }_{1}sin{\phi }_{2}=\mathrm{cos}({\phi }_{2}-{\phi }_{1})$$

we have the following equation for the total kinetic energy of the system:

$$\begin{aligned} T_{T} & = \frac{1}{2}m_{3} \dot{y}_{2}^{2} - m_{3} \dot{y}_{1} L_{2} \dot{\phi }_{1} sin\phi _{1} + \frac{1}{2}m_{3} L_{2}^{2} \dot{\phi }_{1}^{2} + \frac{1}{2}m_{0} \dot{\phi }_{1}^{2} L_{2}^{2} \\ & \quad + \frac{1}{2}m_{0} r^{2} \dot{\phi }_{2}^{2} - m_{0} \dot{\phi }_{1} \dot{\phi }_{2} L_{2} r\cos \left( {\phi _{2} - \phi _{1} } \right) + \frac{1}{2}m_{0} \dot{y}_{2}^{2} - \\ & m_{0} \dot{y}_{2} \dot{\phi }_{1} L_{2} sin\phi _{1} + m_{0} \dot{y}_{2} \dot{\phi }_{2} rcos\phi _{2} + \frac{1}{2}m_{1} \dot{y}_{1}^{2} + \frac{1}{2}m_{2} \dot{y}_{2}^{2} + \\ & \frac{1}{2}m_{3} L_{2}^{2} \dot{\phi }_{1}^{2} + \frac{1}{2}m_{0} r^{2} \dot{\phi }_{2}^{2} \\ \end{aligned}$$
(14)
$$\begin{aligned} T_{T} & = \frac{1}{2}\left( {m_{2} + m_{3} + m_{0} } \right)\dot{y}_{2}^{2} - \left( {m_{3} + m_{0} } \right)L_{2} \dot{y}_{2} \dot{\phi }_{1} sin\phi _{1} + \\ & m_{3} L_{2}^{2} \dot{\phi }_{1}^{2} + m_{0} r^{2} \dot{\phi }_{2}^{2} + m_{0} \dot{y}_{2} \dot{\phi }_{2} rcos\phi _{2} - \\ & m_{0} r\dot{\phi }_{1} \dot{\phi }_{2} L_{2} \cos \left( {\phi _{2} - \phi _{1} } \right) + \frac{1}{2}m_{1} \dot{y}_{1}^{2} + \frac{1}{2}m_{0} L_{2}^{2} \dot{\phi }_{1}^{2} \\ \end{aligned}$$
(14a)

2.2 Total Potential Energy of the System

The total potential energy of the system or the work of the conservative forces of the \({V}_{T}\) system is given by the sum of the potential energy portions of the elastic elements of the \({K}_{1}\), \({K}_{2}\) and \({K}_{T}\) system, as follows:

$${V}_{T}={E}_{Pk1}+{E}_{Pk2}+{E}_{PkT}$$
(15)

The previous potential energy equation can then be rewritten as follows:

$$V_{T} = \frac{1}{2}K_{1} \left( {\Delta y_{{K_{1} }} } \right)^{2} + \frac{1}{2}K_{2} \left( {\Delta y_{{K_{2} }} } \right)^{2} + \frac{1}{2}K_{T} \left( {\Delta y_{{K_{T} }} } \right)^{2}$$
(16)

Replacing the appropriate displacements, Eqs. 8 and 10, we obtain the following equation:

$${V}_{T}=\frac{1}{2}{K}_{1}{\left({y}_{1}-{y}_{e1}\right)}^{2}+\frac{1}{2}{K}_{2}{\left({y}_{2}-{y}_{1}\right)}^{2}+\frac{1}{2}{K}_{T}{\phi }_{1}^{2}$$
(17)
$${V}_{T}=\frac{1}{2}{K}_{1}{\left({y}_{1}^{2}-{2{y}_{1}y}_{e1}+{y}_{e1}^{2}\right)}^{2}+\frac{1}{2}{K}_{2}{\left({y}_{1}^{2}-{2y}_{1}{y}_{2}+{y}_{1}^{2}\right)}^{2}+\frac{1}{2}{K}_{T}{\phi }_{1}^{2}$$
(17a)

From the previous equation (Eq. 17a), the effect of gravity as a conservative force was disregarded, due to its little influence on the response of the system.

2.3 Work of Non-Conserved Forces

The work of the non-conserved forces or total damping of the Q system represents the total energy dissipated by the damping elements, and is given by the sum of the energy dissipated by the suspension damper (\({F}_{c}\)) and by the junction damper torsional (\({F}_{{c}_{T}}\)), in addition to the energy dissipated by the damper with MR (\(F\)), as follows:

$$Q={F}_{c}+{F}_{{c}_{T}}+F$$
(18)

So the previous equation is rewritten as follows:

$$Q=C(\Delta {\dot{y}}_{C})+{C}_{T}(\Delta {\dot{y}}_{{C}_{T}})+F$$
(19)

where the terms \(\Delta {\dot{y}}_{C}\) and \(\Delta {\dot{y}}_{{C}_{T}}\) represent the deformation velocities of the damping elements, respectively. And they are given as follows:

$$\Delta {\dot{y}}_{C}={\dot{y}}_{2}-{\dot{y}}_{1}$$
(20)
$$\Delta {\dot{y}}_{{C}_{T}}={\dot{\phi }}_{1}$$
(21)

The strength of the MR damper is mathematically represented by the Bounce-Wen model. And then, replacing Eqs. 20 and 21 in 19, you can rewrite the expression for Q as follows:

$$Q=C\left({\dot{y}}_{2}-{\dot{y}}_{1}\right)+{C}_{T}\left({\dot{\phi }}_{1}\right)+\alpha z$$
(22)

where Va is the supply voltage, \({i}_{a}\) is the supply current, Φ is the magnetic flux and φ is the angular position of the motor [12].

$${L}_{a}\frac{{di}_{a}}{dt}+{R}_{a}{i}_{a}+{E}_{b}={V}_{a}$$
(23)

where Eb is the counter-electromotive force between the motor armature terminals given by:

\({E}_{b}={k}_{E}\frac{d{\phi }_{2}}{dt}\) where \({k}_{E}\) is the motor voltage constant and \({\phi }_{2}\) the motor angular position. Also according to [12], to study the interaction between the tower and the motor, the direct current motor is considered in a simplified way, and therefore the torque generated by the motor can be expressed as follows:

$${M}_{m}=\widehat{a}-\widehat{b}{\dot{\phi }}_{2}$$
(24)

where parameter â is related to the electrical voltage applied to the direct current motor and \(\widehat{b}\) related to the type of motor used, both defined as follows, respectively:

$$\widehat{a}=\frac{{k}_{m}{V}_{a}}{{R}_{a}}$$
(25)
$$\widehat{b}=\frac{{k}_{m}{k}_{b}}{{R}_{a}}$$
(26)

where \({R}_{a}\) is the motor electrical resistance, \({k}_{b}\) the motor voltage constant, \({V}_{a}\) the input voltage applied to the motor armature, \({k}_{m}\) is the motor torque constant.

Then, replacing Eq. 24, rewrite Eq. 22 as follows:

$$Q=C\left({\dot{y}}_{2}-{\dot{y}}_{1}\right)+{C}_{T}\left({\dot{\phi }}_{1}\right)+\alpha z+\widehat{a}-\widehat{b}{\dot{\phi }}_{2}$$
(27)

2.4 Application of the Euler–Lagrange Equation

The Lagrangian of the system under study is calculated by the difference between the total kinetic energy, the total potential energy and the dissipation energy of the system, is \(L={T}_{T}-{V}_{T}\). The Lagrangian is then obtained through the difference between Eqs. 7, 10 which results in [5]

$$\begin{aligned} L & = \frac{1}{2}\left( {m_{2} + m_{3} + m_{0} } \right)\dot{y}_{2}^{2} - \left( {m_{3} + m_{0} } \right)L_{2} \dot{y}_{2} \dot{\phi }_{1} sin\phi _{1} \\ & \quad + m_{3} L_{2}^{2} \dot{\phi }_{1}^{2} + m_{0} r^{2} \dot{\phi }_{2}^{2} - m_{0} \dot{y}_{2} \dot{\phi }_{2} rcos\phi _{2} - \\ & m_{0} r\dot{\phi }_{1} \dot{\phi }_{2} L_{2} \cos \left( {\phi _{2} - \phi _{1} } \right) + \frac{1}{2}m_{1} \dot{y}_{1}^{2} + \frac{1}{2}m_{0} L_{2}^{2} \dot{\phi }_{1}^{2} - \\ & \frac{1}{2}K_{1} \left( {y_{1}^{2} - 2y_{1} y_{{e1}} + y_{{e1}}^{2} } \right)^{2} + \frac{1}{2}K_{2} \left( {y_{1}^{2} - 2y_{1} y_{2} + y_{1}^{2} } \right)^{2} + \frac{1}{2}K_{T} \phi _{1}^{2} \\ \end{aligned}$$
(28)

For the application of the Lagrange Equation, the generalized coordinates of the system in question must be determined. For the system of the present work, the following generalized coordinates are defined: \({y}_{1}\), \({y}_{2}\), \({\phi }_{1}\) and \({\phi }_{2}\), so according to the Lagrange Equation, the Lagrangian (Eq. 16) must be derived in relation to these generalized coordinates.

But, once the equations of kinetic, potential and dissipation energies are deduced, the Hamilton Principle can be used for each one of the generalized coordinates. As will be shown below: For the \({y}_{1}\) coordinate, vertical displacement of the tire:

$$\begin{array}{*{20}l} {\frac{d}{dt}\left( {\frac{\partial L}{{\partial \dot{y}_{1} }}} \right) - \left( {\frac{\partial L}{{\partial y_{1} }}} \right) = Q_{1} } \hfill \\ {\frac{d}{dt}\left( {\frac{{\partial T_{T} }}{{\partial \dot{y}_{1} }} - \frac{{\partial V_{T} }}{{\partial \dot{y}_{1} }}} \right) - \left( {\frac{{\partial T_{T} }}{{\partial y_{1} }} - \frac{{\partial V_{T} }}{{\partial y_{1} }}} \right) = Q_{1} } \hfill \\ \end{array}$$
(29)

But: \(\frac{\partial {V}_{T}}{\partial {\dot{y}}_{1}}=0\) e \(\frac{\partial {T}_{T}}{\partial {y}_{1}}=0\) so the previous equation looks like this:

$$\frac{d}{dt}\left(\frac{\partial {T}_{T}}{\partial {\dot{y}}_{1}}\right)+\frac{\partial {V}_{T}}{\partial {y}_{1}}={Q}_{1}$$
$${m}_{1}{\ddot{y}}_{1}=-{K}_{1}{(y}_{1}-{y}_{e1})+{K}_{2}({y}_{2}-{y}_{1})+C\left({\dot{y}}_{2}-{\dot{y}}_{1}\right)-\delta z$$
(30)

where z is the evolutionary variable given by:

$$\dot{z}=-\gamma {\left|{\dot{y}}_{2}-{\dot{y}}_{1}\left|z\right|z\right|}^{n-1}z-\beta \left({\dot{y}}_{2}-{\dot{y}}_{1}\right){\left|z\right|}^{n}+\lambda ({\dot{y}}_{2}-{\dot{y}}_{1})$$

More details on the evolutionary variable z [5].

And ye1 is considered to be an excitation of the harmonic type and given as follows \({y}_{e1}=Acos(wt)\), where A is the amplitude that represents the irregularities of the ground surface.

For the \({y}_{2}\) coordinate, vertical displacement of the chassis:

$$\begin{array}{*{20}l} {\frac{d}{dt}\left( {\frac{\partial L}{{\partial \dot{y}_{2} }}} \right) - \left( {\frac{\partial L}{{\partial y_{2} }}} \right) = Q_{2} } \hfill \\ {\frac{d}{dt}\left( {\frac{{\partial T_{T} }}{{\partial \dot{y}_{2} }} - \frac{{\partial V_{T} }}{{\partial \dot{y}_{2} }}} \right) - \left( {\frac{{\partial T_{T} }}{{\partial y_{2} }} - \frac{{\partial V_{T} }}{{\partial y_{2} }}} \right) = Q_{2} } \hfill \\ \end{array}$$
(31)

But: \(\frac{\partial {V}_{T}}{\partial {\dot{y}}_{2}}=0\) e \(\frac{\partial {T}_{T}}{\partial {y}_{2}}=0\) so the previous equation looks like this:

$$\frac{d}{dt}\left(\frac{\partial {T}_{T}}{\partial {\dot{y}}_{2}}\right)+\frac{\partial {V}_{T}}{\partial {y}_{2}}={Q}_{2}$$
$$\begin{array}{*{20}l} {\left( {m_{2} + m_{3} + m_{0} } \right)\ddot{y}_{2} - \left( {m_{3} + m_{0} } \right)L_{2} \ddot{\phi }_{1} sin\phi_{1} + m_{0} r\ddot{\phi }_{2} cos\phi_{2} } \hfill \\ { = \left( {m_{3} + m_{0} } \right)L_{2} \dot{\phi }_{1}^{2} cos\phi_{1} + m_{0} r\dot{\phi }_{2}^{2} sin\phi_{2} + K_{2} \left( {y_{2} - y_{1} } \right) + } \hfill \\ {C\left( {\dot{y}_{1} - \dot{y}_{2} } \right) - \delta z} \hfill \\ \end{array}$$
(32)

Similarly, the Hamilton Principle is used for the \(\phi_{1}\) coordinate, angular displacement of the tower:

$$\begin{array}{*{20}l} {\frac{d}{dt}\left( {\frac{\partial L}{{\partial \dot{\phi }_{1} }}} \right) - \left( {\frac{\partial L}{{\partial \phi_{1} }}} \right) = Q_{3} } \hfill \\ {\frac{d}{dt}\left( {\frac{{\partial T_{T} }}{{\partial \dot{\phi }_{1} }} - \frac{{\partial V_{T} }}{{\partial \dot{\phi }_{1} }}} \right) - \left( {\frac{{\partial T_{T} }}{{\partial \phi_{1} }} - \frac{{\partial V_{T} }}{{\partial \phi_{1} }}} \right) = Q_{3} } \hfill \\ \end{array}$$
(33)

But: \(\frac{\partial {V}_{T}}{\partial {\dot{\phi }}_{1}}=0\) so the previous equation looks like this:

$$\frac{d}{dt}\left(\frac{\partial {T}_{T}}{\partial {\dot{\phi }}_{1}}\right)-\frac{\partial {T}_{T}}{\partial {\phi }_{1}}+\frac{\partial {V}_{T}}{\partial {\phi }_{1}}={Q}_{3}$$
$$\begin{array}{*{20}l} { - \left( {m_{3} + m_{0} } \right)\ddot{y}_{2} L_{2} sin\phi_{1} + 2m_{3} L_{2}^{2} \ddot{\phi }_{1} + m_{0} L_{2}^{2} \ddot{\phi }_{1} - } \hfill \\ {m_{0} rL_{2} \ddot{\phi }_{2} \cos \left( {\phi_{2} - \phi_{1} } \right) = - m_{0} rL_{2} \dot{\phi }_{2} \sin \left( {\phi_{2} - \phi_{1} } \right)\left( {\dot{\phi }_{2} - \dot{\phi }_{1} } \right) - } \hfill \\ {m_{0} rL_{2} \dot{\phi }_{2} \dot{\phi }_{1} \sin \left( {\phi_{2} - \phi_{1} } \right) - K_{T} \phi_{1} - C_{T} \dot{\phi }_{1} } \hfill \\ \end{array}$$
(34)

And now, using Hamilton's Principle for the \(\phi_{2}\) coordinate, angular displacement of the unbalanced mass, we have:

$$\begin{array}{*{20}l} {\frac{d}{dt}\left( {\frac{\partial L}{{\partial \dot{\phi }_{2} }}} \right) - \left( {\frac{\partial L}{{\partial \phi_{2} }}} \right) = Q_{4} } \hfill \\ {\frac{d}{dt}\left( {\frac{{\partial T_{T} }}{{\partial \dot{\phi }_{2} }} - \frac{{\partial V_{T} }}{{\partial \dot{\phi }_{2} }}} \right) - \left( {\frac{{\partial T_{T} }}{{\partial \phi_{2} }} - \frac{{\partial V_{T} }}{{\partial \phi_{2} }}} \right) = Q_{4} } \hfill \\ \end{array}$$
(35)

But: \(\frac{{\partial V_{T} }}{{\partial \dot{\phi }_{2} }} = 0\) so the previous equation looks like this:

$$\begin{array}{*{20}l} {\frac{d}{dt}\left( {\frac{{\partial T_{T} }}{{\partial \dot{\phi }_{2} }}} \right) - \frac{{\partial T_{T} }}{{\partial \phi_{2} }} + \frac{{\partial V_{T} }}{{\partial \phi_{2} }} = Q_{4} } \hfill \\ {2m_{0} r^{2} \ddot{\phi }_{2} + m_{0} r\ddot{y}_{2} cos\phi_{2} - m_{0} rL_{2} \ddot{\phi }_{1} cos\left( {\phi_{2} - \phi_{1} } \right)} \hfill \\ { = - m_{0} rL_{2} \dot{\phi }_{1} \sin \left( {\phi_{2} - \phi_{1} } \right)\left( {\dot{\phi }_{2} - \dot{\phi }_{1} } \right) + } \hfill \\ {m_{0} rL_{2} \dot{\phi }_{1} \dot{\phi }_{2} \sin \left( {\phi_{2} - \phi_{1} } \right) + \left( {\hat{a} - \hat{b}\dot{\phi }_{2} } \right)} \hfill \\ \end{array}$$
(36)

So the system of differential equations that define the movements of the system under

$$\begin{aligned} m_{1} \ddot{y}_{1} & = - K_{1} (y_{1} - y_{{e1}} ) + K_{2} \left( {y_{2} - y_{1} } \right) + C\left( {\dot{y}_{2} - \dot{y}_{1} } \right) - \delta z \\ & \left( {m_{2} + m_{3} + m_{0} } \right)\ddot{y}_{2} - \left( {m_{3} + m_{0} } \right)L_{2} \ddot{\phi }_{1} sin\phi _{1} + m_{0} r\ddot{\phi }_{2} cos\phi _{2} \\ & = \left( {m_{3} + m_{0} } \right)L_{2} \dot{\phi }_{1}^{2} cos\phi _{1} + \\ & m_{0} r\dot{\phi }_{2}^{2} sin\phi _{2} + K_{2} \left( {y_{2} - y_{1} } \right) + C\left( {\dot{y}_{1} - \dot{y}_{2} } \right) - \delta z \\ & \quad - \left( {m_{3} + m_{0} } \right)\ddot{y}_{2} L_{2} sin\phi _{1} + 2m_{3} L_{2}^{2} \ddot{\phi }_{1} + m_{0} L_{2}^{2} \ddot{\phi }_{1} \\ & \quad - m_{0} rL_{2} \ddot{\phi }_{2} \cos \left( {\phi _{2} - \phi _{1} } \right) = - \\ & m_{0} rL_{2} \dot{\phi }_{2} \sin \left( {\phi _{2} - \phi _{1} } \right)\left( {\dot{\phi }_{2} - \dot{\phi }_{1} } \right) \\ & \quad - m_{0} rL_{2} \dot{\phi }_{2} \dot{\phi }_{1} \sin \left( {\phi _{2} - \phi _{1} } \right) - K_{T} \phi _{1} - C_{T} \dot{\phi }_{1} \\ & 2m_{0} r^{2} \ddot{\phi }_{2} + m_{0} r\ddot{y}_{2} cos\phi _{2} - m_{0} rL_{2} \ddot{\phi }_{1} cos\left( {\phi _{2} - \phi _{1} } \right) \\ & = - m_{0} rL_{2} \dot{\phi }_{1} \sin \left( {\phi _{2} - \phi _{1} } \right)\left( {\dot{\phi }_{2} - \dot{\phi }_{1} } \right) + \\ & m_{0} rL_{2} \dot{\phi }_{1} \dot{\phi }_{2} \sin \left( {\phi _{2} - \phi _{1} } \right) + \left( {\hat{a} - \hat{b}\dot{\phi }_{2} } \right) \\ \end{aligned}$$
(37)

When the turret plus the non-ideal unbalanced motor are considered as a simple pendulum, then the angle \(\phi_{1}\) is considered small compared to \(\phi_{2}\). Therefore, the following relation is valid [13].

$$\left\{ {\begin{array}{*{20}l} {sin\phi_{1} \cong \phi_{1} } \hfill \\ {cos\phi_{1} \cong 1} \hfill \\ \end{array} } \right.$$

Thus, the system of equations presented above (Eq. 37) is rewritten as follows:

$$\begin{aligned} m_{1} \ddot{y}_{1} \\ & = - K_{1} (y_{1} - y_{{e1}} ) + K_{2} \left( {y_{2} - y_{1} } \right) + C\left( {\dot{y}_{2} - \dot{y}_{1} } \right) - \delta z \\ & \left( {m_{2} + m_{3} + m_{0} } \right)\ddot{y}_{2} - \left( {m_{3} + m_{0} } \right)L_{2} \ddot{\phi }_{1} \phi _{1} + m_{0} r\ddot{\phi }_{2} cos\phi _{2} \\ & = \left( {m_{3} + m_{0} } \right)L_{2} \dot{\phi }_{1}^{2} + \\ & m_{0} r\dot{\phi }_{2}^{2} sin\phi _{2} + K_{2} \left( {y_{2} - y_{1} } \right) + C\left( {\dot{y}_{1} - \dot{y}_{2} } \right) - \delta z \\ & \quad - \left( {m_{3} + m_{0} } \right)\ddot{y}_{2} L_{2} \phi _{1} + 2m_{3} L_{2}^{2} \ddot{\phi }_{1} + m_{0} L_{2}^{2} \ddot{\phi }_{1} \\ & \quad - m_{0} rL_{2} \ddot{\phi }_{2} \cos \left( {\phi _{2} - \phi _{1} } \right) = - \\ & m_{0} rL_{2} \dot{\phi }_{2} \sin \left( {\phi _{2} - \phi _{1} } \right)\left( {\dot{\phi }_{2} - \dot{\phi }_{1} } \right) \\ & \quad - m_{0} rL_{2} \dot{\phi }_{2} \dot{\phi }_{1} \sin \left( {\phi _{2} - \phi _{1} } \right) - K_{T} \phi _{1} - C_{T} \dot{\phi }_{1} \\ & 2m_{0} r^{2} \ddot{\phi }_{2} + m_{0} r\ddot{y}_{2} cos\phi _{2} - m_{0} rL_{2} \ddot{\phi }_{1} cos\left( {\phi _{2} - \phi _{1} } \right) \\ & = - m_{0} rL_{2} \dot{\phi }_{1} \sin \left( {\phi _{2} - \phi _{1} } \right)\left( {\dot{\phi }_{2} - \dot{\phi }_{1} } \right) + \\ & m_{0} rL_{2} \dot{\phi }_{1} \dot{\phi }_{2} \sin \left( {\phi _{2} - \phi _{1} } \right) + \left( {\hat{a} - \hat{b}\dot{\phi }_{2} } \right) \\ \end{aligned}$$
(38)

The previous system of equations (Eq. 38) is rewritten as follows:

$$\begin{aligned} \ddot{y}_{1} & = - q_{1} (y_{1} - y_{{e1}} ) + q_{2} \left( {y_{2} - y_{1} } \right) + p_{1} \left( {\dot{y}_{2} - \dot{y}_{1} } \right) - \mu z \\ & \left( {1 + \alpha _{1} + \alpha } \right)\ddot{y}_{2} - \left( {\alpha _{1} + \alpha } \right)\ddot{\phi }_{1} \phi _{1} + \alpha r\ddot{\phi }_{2} cos\phi _{2} \\ & = \left( {\alpha _{1} + \alpha } \right)L_{2} \dot{\phi }_{1}^{2} + \\ & \alpha r\dot{\phi }_{2}^{2} sin\phi _{2} + q_{3} \left( {y_{2} - y_{1} } \right) + p_{2} \left( {y_{1}^{'} - y_{2}^{'} } \right) - \mu _{1} z \\ & \quad - \frac{{m_{3} + m_{0} }}{{2m_{3} L_{2} }}\ddot{y}_{2} \phi _{1} + \left( {1 + \frac{{m_{0} }}{{2m_{3} }}} \right)\ddot{\phi }_{1} - \frac{{m_{0} r}}{{2m_{3} L_{2} }}\ddot{\phi }_{2} \cos \left( {\phi _{2} - \phi _{1} } \right) = - \\ & \frac{{m_{0} r}}{{2m_{3} L_{2} }}\dot{\phi }_{2} \sin \left( {\phi _{2} - \phi _{1} } \right)\left( {\dot{\phi }_{2} - \dot{\phi }_{1} } \right) \\ & \quad - \frac{{m_{0} r}}{{2m_{3} L_{2} }}\dot{\phi }_{2} \dot{\phi }_{1} \sin \left( {\phi _{2} - \phi _{1} } \right) - q_{4} \phi _{1} - p_{3} \dot{\phi }_{1} \\ & \frac{1}{{2r}}\ddot{y}_{2} cos\phi _{2} + \ddot{\phi }_{2} - \frac{{m_{0} L_{2} }}{{2r}}\ddot{\phi }_{1} cos\left( {\phi _{2} - \phi _{1} } \right) \\ & = \frac{{L_{2} }}{{2r}}\dot{\phi }_{1} \sin \left( {\phi _{2} - \phi _{1} } \right)\left( {\dot{\phi }_{2} - \dot{\phi }_{1} } \right) + \\ & \frac{{L_{2} }}{{2r}}\dot{\phi }_{1} \dot{\phi }_{2} \sin \left( {\phi _{2} - \phi _{1} } \right) + a - b\dot{\phi }_{2} \\ \end{aligned}$$
(39)

The parameters considered in the previous system are listed below:

$$\begin{array}{*{20}l} {q_{1} = \frac{{K_{1} }}{{m_{1} }},q_{2} = \frac{{K_{2} }}{{m_{1} }} ,q_{3} = \frac{{K_{2} }}{{m_{1} }},q_{3} = \frac{{K_{2} }}{{m_{2} }},p_{1} = \frac{C}{{m_{1} }},p_{2} = \frac{C}{{m_{1} }},p_{3} = \frac{{C_{T} }}{{m_{3} L_{2}^{2} }},\alpha = \frac{{m_{0} }}{{m_{2} }}\alpha_{1} = \frac{{m_{3} }}{{m_{2} }},} \hfill \\ {\quad \quad \quad \alpha_{2} = \frac{{m_{0} }}{{m_{3} }},\zeta = \frac{r}{{L_{2} }} ,\mu = \frac{\delta }{{m_{1} L_{2} }}, \mu_{1} = \frac{\delta }{{m_{2} L_{2} }}, a = \frac{{\hat{a}}}{{m_{0} r^{2} }},b = \frac{{\hat{b}}}{{m_{0} r^{2} }}.} \hfill \\ \end{array}$$

The system of equations is then rewritten, making use of a change of variables employing the following state variables:

$$u_1 = y_1 ,u_2 = y_1^{\prime} ,u_3 = y_2 , u_4 = y_2^{\prime} ,u_5 = \phi_1 ,u_6 = \phi_2^{\prime} ,u_7 = \phi_2 ,u_8 = \phi_2^{\prime} , \;{\text{e}}\,u_9 = z.$$

Which results in a new system of equations:

$$\left\{ {\begin{array}{*{20}l} {E_{1} y_{2} - E_{2} \phi_{1} + E_{3} \phi_{2} = f_{2} } \hfill \\ { - E_{4} y_{2} + E_{5} \phi_{1} + E_{6} \phi_{2} = f_{3} } \hfill \\ {E_{7} y_{2} - E_{8} \phi_{1} + E_{9} \phi_{2} = f_{4} } \hfill \\ \end{array} } \right.$$
(40)

Where do you have:

$$\begin{array}{*{20}l} {f_{1} = - q_{1} \left[ {u_{1} - A\cos (wt)} \right] + q_{2} \left( {u_{3} - u_{1} } \right) + p_{1} \left( {u_{4} - u_{2} } \right) - \mu_{1} z - u_{5} ) - q_{4} u_{5} - p_{3} u_{6} } \hfill \\ {f_{2} = \left( {\alpha_{1} + \alpha } \right)L_{2} u_{6}^{2} + \alpha ru_{8}^{2} \sin u_{7} - q_{3} \left( {u_{3} - u_{1} } \right) - p_{2} \left( {u_{4} - u_{2} } \right) + \mu_{1} z} \hfill \\ {f_{3} = - \frac{{m_{0} r}}{{2m_{3} L_{2} }}u_{8} \sin (u_{7} - u_{5} )\left( {u_{8} - u_{6} } \right) - \frac{{m_{0} r}}{{2m_{3} L_{2} }}u_{6} u_{8} \sin (u_{7} } \hfill \\ {f_{4} = - \frac{{L_{2} }}{2r}u_{6} \sin (u_{7} - u_{5} )\left( {u_{8} - u_{6} } \right) + \frac{{L_{2} }}{2r}u_{6} u_{8} \sin (u_{7} - u_{5} ) + \alpha - bu_{8} } \hfill \\ \end{array}$$

And also:

$$\begin{array}{*{20}l} {E_{1} = 1 + \alpha_{1} + \alpha E_{2} = \left( {\alpha_{1} + \alpha_{2} } \right)L_{2} u_{5} E_{3} = ar\cos (\alpha_{7} )E_{4} = \frac{{m_{3} + m_{0} }}{{2m_{3} L_{2} }}u_{5} E_{5} = 1 + \frac{{m_{0} }}{{2m_{3} }}} \hfill \\ {E_{6} = \frac{{m_{0} r}}{{2m_{3} L_{2} }}\cos (u_{7} - u_{5} )E_{7} = \frac{1}{2r}\cos (u_{7} )E_{8} = \frac{{m_{0} L_{2} }}{2r}\cos (u_{7} - u_{5} )E_{9} = 1} \hfill \\ \end{array}$$

So, rewriting the system in matrix form, you have:

$$\left[\begin{array}{ccc}{E}_{1}& {-E}_{2}& {E}_{3}\\ {-E}_{4}& {E}_{5}& {-E}_{6}\\ {E}_{7}& {-E}_{8}& {E}_{9}\end{array}\right]\left[\begin{array}{c}{y}_{2}^{"}\\ {\phi }_{1}^{"}\\ {\phi }_{2}^{"}\end{array}\right]=\left[\begin{array}{c}{f}_{2}\\ {f}_{3}\\ {f}_{4}\end{array}\right]$$
$$\left[\begin{array}{c}{y}_{2}^{"}\\ {\phi }_{1}^{"}\\ {\phi }_{2}^{"}\end{array}\right]=\left[\begin{array}{c}{f}_{2}\\ {f}_{3}\\ {f}_{4}\end{array}\right]{\left[\begin{array}{ccc}{E}_{1}& {-E}_{2}& {E}_{3}\\ {-E}_{4}& {E}_{5}& {-E}_{6}\\ {E}_{7}& {-E}_{8}& {E}_{9}\end{array}\right]}^{-1}$$
(41)

Finally, solving the inverse matrix and making the necessary multiplication of this answer in the system of equations above, we have the system of second order linear differential equations that govern the dynamics of the system under study, presented below:

$$\begin{array}{*{20}l} {y_{1} = f_{1} } \hfill \\ {y_{2} = 1/\Delta \left[ {\left( {E_{5} E_{9} - E_{6} E_{8} } \right)f_{2} + \left( {E_{2} E_{9} - E_{3} E_{8} } \right)f_{3} + \left( {E_{2} E_{6} - E_{3} E_{5} } \right)f_{4} } \right]} \hfill \\ {\emptyset_{1} = 1/\Delta \left[ {\left( {E_{4} E_{9} - E_{6} E_{7} } \right)f_{2} + \left( {E_{1} E_{9} - E_{3} E_{7} } \right)f_{3} + \left( {E_{1} E_{6} - E_{3} E_{4} } \right)f_{4} } \right]} \hfill \\ {\emptyset_{2} = 1/\Delta \left[ {\left( {E_{4} E_{8} - E_{5} E_{7} } \right)f_{2} + \left( {E_{1} E_{8} - E_{2} E_{7} } \right)f_{3} + \left( {E_{1} E_{5} - E_{2} E_{4} } \right)f_{4} } \right]} \hfill \\ {z^{\prime} = - \gamma \left| {{\text{u}}_{4} - {\text{u}}_{2} \left| {{\text{u}}_{9} } \right|{\text{u}}_{9} } \right|^{n - 1} - \beta \left( {{\text{u}}_{4} - {\text{u}}_{2} } \right)\left| {{\text{u}}_{9} } \right|^{n} + \lambda \left( {{\text{u}}_{4} - {\text{u}}_{2} } \right)} \hfill \\ \end{array}$$
(42)

The parameter Δ of the previous equation (Eq. 42) is given by:

$$\Delta = E_{1} E_{5} E_{9} - E_{1} E_{6} E_{8} - E_{2} E_{4} E_{9} + E_{2} E_{6} E_{7} + E_{3} E_{4} E_{8} - E_{3} E_{5} E_{7}$$

Making now, the right sides of the equations of y2″, ϕ1″ and ϕ2″ of the previous system (Eq. 42) equal to \(f_{5}\), \(f_{6}\) and \(f_{7}\), respectively, and also the equation of z equal to \(f_{8}\) as follows:

$$f_{8} = - \gamma \left| {{\text{u}}_{4} - {\text{u}}_{2} \left| {{\text{u}}_{9} } \right|{\text{u}}_{9} } \right|^{n - 1} - \beta \left( {{\text{u}}_{4} - {\text{u}}_{2} } \right)\left| {{\text{u}}_{9} } \right|^{n} + \lambda \left( {{\text{u}}_{4} - {\text{u}}_{2} } \right)$$
(43)

You can write the derivatives of system 42 as follows:

$$\begin{array}{*{20}l} {u_1^{\prime} = u_2 } \hfill \\ {u_2^{\prime} = f_1 } \hfill \\ {u_3^{\prime} = u_4 } \hfill \\ {u_4^{\prime} = f_5 } \hfill \\ {u_5^{\prime} = u_6 } \hfill \\ {u_6^{\prime} = f_6 } \hfill \\ {u_7^{\prime} = u_8 } \hfill \\ {u_8^{\prime} = f_7 } \hfill \\ {u_9^{\prime} = f_8 } \hfill \\ \end{array}$$
(44)

3 Results

3.1 System Responses in Time Domain

Time and phase domain responses will be presented for the following system components: wheel mass, trailer mass, mass concentrated at the top of the tower and unbalanced engine mass, respectively. The values considered for the dimensionless parameters of the system are those shown in Table 1, for a time interval of 0 ≤ t ≤ 100 and the initial conditions were considered as being null.

Table 1 Values for dimensionless parameters for the non-suspension system with MR
Full size table

The following table (Table 2) shows the values considered for the MR damper parameters.

Table 2 Values assigned to the parameters related to the MR damper
Full size table

Figure 2 shows the behavior of the displacements of the wheel in (a), the trailer in (b), the turret in (c) and the angular velocity of the unbalanced mass of the engine in (d), under suspension action with MR, in red compared to the MR shockless response in black. It can be seen that with the addition of the suspension with MR it significantly reduces the range of motion of each component of the system, having a lesser influence, in relation to the other components, on the displacement of the wheel, as expected. It is verified that the greatest influence is on the displacement of the trailer, which satisfies the proposed objectives, since the proposal is the reduction of the amplitudes of the trailer and the tower.

Fig. 2
figure 2figure 2

displacements of the wheel in (a), the trailer in (b), the turret in (c) and the angular velocity of the unbalanced mass of the engine in (d)

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3.2 Analyze the Influence of the External Force Applied on the Structure Described

We analyze the influence of the external force applied on the structure described by Eq. (42) considering amplitude A in the interval [0.1: 0.3] and frequency ω in the interval [1, 2]. For this we use the 4th order Runge–Kutta method with an integration step h = 0.001 and a total integration time t = 106[s]. We also consider a transient time of 40% of the total time and considering the initial conditions × 0 = [0, 0, 0, 0, 0, 0, 0, 0, 0]. Figure 3a represents the maximum amplitude of the displacement of the trailer and Fig. 3b of the tower that supports the motor for spraying, both are described by the set of Eq. 42. Thus, the yellow regions represent the maximum amplitude and the light gray region represents the minimum amplitude for the structure.

Fig. 3
figure 3

Representation of the maximum amplitude considering Eq. (42). a trailer displacement and b turret displacement

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Another analysis performed was the oscillation frequency and thus we obtain the fast Fourier transforms (FFT) with the variation of the amplitude of the external force (ω), in which we can observe a characteristic frequency of the Eq. (42) referring to the trailer and the tower. Figure 4. (a) represents the FFT sweeping the parameter ω in the interval [1, 2] of the trailer displacement, in which we can observe a natural frequency of the system in yellow and in gray scale are very low secondary frequencies. In Fig. 4b represents the variation ω for the same interval, however, for the tower displacement, in yellow it represents the dominant frequency of the system and in gray, low amplitude secondary frequencies.

Fig. 4
figure 4

Representation of the FFT for the parameter ω as Eq. (42). a trailer displacement and b turret displacement

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Therefore, we delimited the maximum amplitude regions for the displacement of the sprayer system truck and tower considering the external force applied to the system. This external force in Eq. (42) represents possible irregularities in the terrain where the spraying vehicle moves, as high displacements in the structure compromise the spraying application. The analysis of the frequency of for the parameter ω showed that there is a natural frequency for both the truck and the tower vibration, which showed a behavior of possible periodicity, such frequencies are between 0.1 and 0.5 [Hz]. Thus, Fig. 5a–c represent the time series of the trailer displacement for ω = 2.936 [Hz] and considering A = [0.0, 0.16, 0.5], respectively.

Fig. 5
figure 5

Time series of the trailer displacement a A = 0.0, b A = 1.6 and c A = 0.5

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Thus, Figs. 6a–c represent the time series of the trailer displacement for ω = 2.936 [Hz] and considering A = [0.0, 0.16, 0.5], respectively.

Fig. 6
figure 6

Time series of tower displacement a A = 0.0, b A = 1.6 and c A = 0.5

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3.3 Calculation of Resulting Vibrations with and Without MR

Thus, we calculated the vibrations resulting from the absence of the MR actuation in Eq. (42), considering the same constants of the previous case, that is, the same constants of the MR system actuation in the vehicle system. Fig. 7a represent the maximum amplitude of the tower and (7b) represent the maximum amplitude of the trailer with the variation of the parameters of the external force of the system. Colors from white to black represent the minimum to medium span and from black to yellow the medium to maximum span.

Fig. 7
figure 7

Representation of maximum amplitude as a function of external force parameters. a turret and b cart

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We also calculated the variation of the control parameter (a) for the system with and without MR application. Thus, Fig. 8a represents the behavior of the tower vibrations with the MR and Fig. 8b without the MR acting on the system with the variation of the parameter a = [3, 7]. It can be observed that there are peaks of maximum amplitude a = 5.329 and a = 7.691 for the system containing the MR and for the system with the MR there is a decay of the amplitude, however, it remains in a range from 0.04625 to 0.04616.

Fig. 8
figure 8

Representation of the maximum amplitude with the variation of the tower control parameter. a with the application of the MR and b without the application of the MR

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And figs. a and fig. B. Represents the behavior of the control parameter (a) for the trailer with and without the MR application, respectively (Fig. 9).

Fig. 9
figure 9

Representation of the maximum amplitude with the variation of the trailer control parameter. a with the application of the MR and b without the application of the MR.

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We can observe that there is a peak with maximum amplitude for the value of a = 4.2 in both cases with and without the MR, however, there is the appearance of a peak with amplitude at a = 8.8.

4 Conclusion

In this work, a vehicular suspension with Magneto Rheological (MR) was used, plus the addition of an unbalanced electric motor on top of the tower of a Quarter-car model that represents the structure of an orchard sprayer, in order to reduce vertical movements. mainly of the trailer or chassis, and also the angular movements of the turret. The use of the MR damper proved to be efficient for the main purpose for which it was used: to reduce the oscillation amplitudes of both the trailer or chassis (\({m}_{2}\)) and the tower (\({m}_{3}\)). It is clearly seen that there were reductions in amplitude, however in some components of the system, this reduction was minimal, as can be seen, for the wheel, and also for the tower. This fact can be justified due to the shock absorber with MR being strongly influenced by an external electric current, which was considered to be constant in this work. The analysis of the unbalanced electric motor showed that the \({m}_{3}\) mass amplitudes are strongly influenced by a possible unbalance of the sprayer fan blades.