Mathematical Description and Study of the Vibration Deck of a Grain Sorting Machine

The Froude number is the most critical functional estimating parameter of the sorted grain material's movement on the sieve surface. It is expressed by the ratio of forces acting on the grain material moving along the deck, the sieve vibrations amplitude, and gravity force [5, 6, 20].

The dynamic oscillation scheme of the grain sorting machine deck is called an idealized set of solid bodies (frame, sieve, frame) with moments of inertia (mass), connected by elastic elements. At the same time, under the action of driving forces, the bodies of the dynamic scheme make vibrations, and are considered as the relative mobility of bodies formed along the axis or in the plane along which the vibrations of the working body occur. The main requirement, as a rule, for the dynamic scheme of the machine in question is to provide the necessary law of vibrations of the working body (body weight). At the same time, a lot of important roles is played by ensuring stability, balance, and the gain of the driving force.

Let's consider a calculation scheme for determining the coordinates of the nodal points of the kinematics of the deck of a pneumatic sorting table (for example, MOS-9) to calculate their speed of movement, the forces of inertia acting on the particles of grain material lying on the surface of the deck (Figure 1).

1.png

Figure 1. Design model of the pneumatic picking disk deck (MOS-9)

Note: 1-crank, 2-connecting rod, 3-deck, 4-suspensions

The considered dynamic scheme of the deck of the grain sorting machine is a single-mass (mass of the sieve mill) with a forced eccentric crank drive (resonant mode of operation). The main advantage of this scheme is high stability during the operation of the machine, the vibration amplitude of the working body does not change. But at the same time, there is also a disadvantage – unbalance, especially with an uneven supply of grain material entering the sorting. In this regard, it is necessary to rigidly fix the base of the machine. In most cases, to eliminate this drawback, in order to partially balance the dynamic loads coming to the eccentric shaft, unbalanced counterweights (loads) are installed, and in some designs, it serves as a second working body. The gain of the driving force in such machines is low, which significantly leads to significant loads in the drive [8, 21, 22].

Let O_k be the control points coordinates, k = 0, 1, 2 (Figure 1). Point A at time t has coordinates:

$x_{0}=R \cdot \cos \omega t$

$y_{0}=R \cdot \sin \omega t$      (1)

where, R is the length of the crank rotating round the radial circle.

Find the coordinates of points B and C using the following algorithm:

a) First, find the coordinates of point B from the conditions:

$\left\{\begin{array}{l}A B=L_{1} \\ O_{1} B=L_{2}\end{array}\right.$      (2)

Note that in this case, the system can have either two, one, or no solutions depending on the parameters L₁, L₂, R, t. We assume that the numbers L₁, L₂, and R are chosen so that for any t, the system has a solution.

b) Find the coordinates of point C from the coordinates of the already found point B from the system:

$\left\{\begin{array}{c}B C=L_{3} \\ O_{2} C=L_{4}\end{array}\right.$    （3）

In such a case, the previous paragraph's remark regarding the existence of a solution and the choice of a single solution is fair enough.

Speed of point A. $V_{A}=\omega \cdot O A$.

where the following formula determines the crank angular rate (for the MOS-9 machine, after the belt drive from the asynchronous electric motor, the crank rotation speed is n=500 rpm (Figure 2)):

$\omega=\frac{\pi \cdot n}{30}=\frac{3.14 \cdot 500}{30}=52.35 \mathrm{~s}^{-1}$  

Speed of point B:

$\left\{\begin{array}{l}\overrightarrow{V_{B}}=\overrightarrow{V_{A}}+\overrightarrow{V_{A B}} \\ \overrightarrow{V_{B}}=\overrightarrow{V_{O 1}}+\overrightarrow{V_{B O 1}}\end{array}\right.$     (4)

Speed of point C:

$\left\{\begin{array}{l}\overrightarrow{V_{C}}=\overrightarrow{V_{B}}+\overrightarrow{V_{B C}} \\ \overrightarrow{V_{C}}=\overrightarrow{V_{O 2}}+\overrightarrow{V_{C O 2}}\end{array}\right.$     (5)

Acceleration of point A at a constant angular rate a_A=ω²∙OA.

Acceleration of point B:

$\left\{\begin{array}{c}\overrightarrow{a_{B}}=\overrightarrow{a_{A}}+\overrightarrow{a_{A B}^{n}}+\overrightarrow{a_{A B}^{\tau}} \\ \overrightarrow{a_{B}}=\overrightarrow{a_{O 1}}+\overrightarrow{a_{B O 1}^{n}}+\overrightarrow{a_{B O 1}^{\tau}}\end{array}\right.$     (6)

where, the normal acceleration is $a_{A B}^{n}$:

$a_{A B}^{n}=\frac{V_{A B}^{2}}{A B}$     (7)

Acceleration of point C:

$\left\{\begin{array}{c}\overrightarrow{a_{C}}=\overrightarrow{a_{B}}+\overrightarrow{a_{B C}^{n}}+\overrightarrow{a_{B C}^{\tau}} \\ \overrightarrow{a_{C}}=\overrightarrow{a_{O 2}}+\overrightarrow{a_{B O 2}^{n}}+\overrightarrow{a_{B O 2}^{\tau}}\end{array}\right.$     (8)

where, the normal accelerations are $a_{B C}^{n}$ and $a_{B O 2}^{n}$:

$a_{B C}^{n}=\frac{V_{B C}^{2}}{B C}$ and $a_{B O 2}^{n}=\frac{V_{B O 2}^{2}}{B O 2}$     (9)

Figure 2 shows the calculation scheme for calculating the motion speeds and accelerations, the inertial forces acting on the sorted material. The presented vibration system of the grain sorting machine deck is a "hybrid" of a mathematical and spring pendulum. The deck is a link of the parallel four-link mechanism O₃AVO4. The sieve pan is mounted on suspensions 6 with a length of l, and a deck 8 is attached to the sieve pan so that tilt angle 9 and cross angle of inclination can be adjusted. Elastic elements 10 (springs) are placed at an angle to the plane of the deck frame. In the equilibrium position, the suspensions are located at an angle of 15° from the vertical, and the springs are not deformed. The angle of inclination α determines deck plane setting AB to the horizontal (in this particular case α=7°).

2.png

Figure 2. Design model of the pneumatic picking disk deck (MOS-9)

Note: 1 – electric motor, 2 – belt drive, 3 – pulley kit, 4 – eccentric, 5 – connecting rod, 6 – suspension, 7 – deck frame, 8 – the deck of the pneumatic picking disk, 9 – adjustment of the longitudinal inclination angle, 10 – elastic elements

Let a_i,x and b_i,y be the coordinates of the fixed control points O_i, i = 1, 2, 3, 4 (Figure 2). The right-handed Cartesian coordinate system is taken as a reference system with the coordinate origin at the drive point O of the deck Fx(t). The deck nodal points are dotted to compose equations of geometric constants, the trajectories of which are known: А and В. These points move round radial circles l = О₂А = О₃В. Point A lies both in crank O₂A and connecting rod AB. Point B lies both in crank O₃B and connecting rod AB.

When deflected from the equilibrium position by an angle θ the deck is set into vibration, which is due to a positive horizontal direction of the exciting force F(t) created by an asynchronous electric motor 1 using a belt drive 2 and an eccentric shaft 4 through a connecting rod 5. If x is the linear displacement of the deck from the equilibrium along an arc with a radius equal to the length l of the suspensions l, its angular displacement is θ = x/l (Figure 2).

When the deck is deflected from the equilibrium, a rotational moment M occurs and tends to return it to the equilibrium position.

$M=J \varepsilon$    (10)

where, J=ml² is the inertial moment, $\varepsilon=\frac{d^{2} \theta}{d t^{2}}$ is the angular acceleration.

The rotational movement M is produced by the forces of elasticity, gravity, inertia and resistance.

Deck vibration dynamic equation at small deviations θ under the action of a disturbing force force $\mathrm{F}(\mathrm{t})=\mathrm{F} \cdot \cos \omega t$ created by an electric motor:

$\begin{aligned} m l^{2} \cdot \frac{\mathrm{d}^{2} \theta}{\mathrm{dt}^{2}}=-\mathrm{mgl} & \\ & \cdot \sin \theta-\mathrm{kl}^{2} \\ & \cdot \sin \theta \cdot \cos \theta-\mathrm{f} \cdot \frac{\mathrm{d} \theta}{\mathrm{dt}} \cdot \mathrm{l}^{2}+\mathrm{F}(\mathrm{t}) \end{aligned}$      (11)

where, k is the stiffness coefficient, N/m, g is the gravity acceleration, m/s², f is the resistance coefficient:

$\begin{aligned} \frac{d^{2} \theta}{d t^{2}}=-\frac{g}{l} \cdot \sin \theta &-\frac{k}{m} \\ & \cdot \sin \theta \cdot \cos \theta-\frac{f}{m} \cdot \frac{d \theta}{d t}+\frac{F(t)}{m l^{2}} \end{aligned}$     (12)

This second-order nonlinear differential equation is not integrated in ordinary functions. Taking into account small deviations θ and expanding sin θ by its Taylor series expansion:

$\sin \theta=\theta-\frac{\theta^{3}}{3 !}+\frac{\theta^{5}}{5 !}-\cdots$

as well as ignoring the first-order expansion terms, a second order nonhomogeneous differential equation with constant coefficients can be obtained:

$\frac{\mathrm{d}^{2} \theta}{\mathrm{dt}^{2}}+\left(\frac{\mathrm{g}}{\mathrm{l}}+\frac{\mathrm{k}}{\mathrm{m}}\right) \cdot \theta+\frac{\mathrm{f}}{\mathrm{m}} \cdot \frac{\mathrm{d} \theta}{\mathrm{dt}}=\frac{\mathrm{F}(\mathrm{t})}{\mathrm{ml}^{2}}$     (13)

If the angle θ is denoted by moving x, then

$\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}+\left(\frac{\mathrm{g}}{\mathrm{l}}+\frac{\mathrm{k}}{\mathrm{m}}\right) \cdot \mathrm{x}+\frac{\mathrm{f}}{\mathrm{m}} \cdot \frac{\mathrm{dx}}{\mathrm{dt}}=\frac{\mathrm{F}(\mathrm{t})}{\mathrm{ml}^{2}}$     (13*)

Symbol $\frac{\mathrm{f}}{\mathrm{m}}=2 \beta$, where β is the attenuation coefficient, $\omega_{0}=\sqrt{\frac{g}{l}+\frac{k}{m}}$ is the circular vibration frequency, can be used. Vibration period $T=2 \pi \sqrt{\frac{m l}{k l+m g}}$. The damped vibrations frequency $\omega_{z}=\sqrt{\omega_{0}{ }^{2}-\beta^{2}}$ has a physical meaning at $\omega_{0} \geq \beta$.

The left part of the expression (13*) is a second-order linear differential equation that describes damped free vibrations. The right expression part, which depends on time, is the driving force.

For finding the law of the deck motion, it is necessary to solve this equation taking into account certain initial conditions: t=0; х = х₀; $\frac{d x}{d t}=v_{0}$; The sum of two functions is the general solution of such an equation:

$x(t)=x_{1}(t)+x_{2}(t)$     (14)

where, x₁(t) is the general solution of the homogeneous equation (left part (13*)), x₂(t) is, respectively, the specific solution of the inhomogeneous equation (right part (13*)).

A general solution of the left-hand side (13*), which is a damped vibration that disappears over time, is:

$\mathrm{x}_{1}=\mathrm{C}_{1} \sin \omega \mathrm{t}+\mathrm{C}_{2} \cos \omega \mathrm{t} \equiv \mathrm{C} \sin (\omega \mathrm{t}+\gamma)$$\left(C_{1}=\right.$ const,$C_{2}=$ const,$C=$ const,$\gamma=$ const $)$      (15)

where, (C = const, γ =const) are the integration constants,  is the amplitude of the deck free vibrations, ( is the phase of the rolling mill vibrations, γ is the phase of the initial vibration. Or

$\left.x_{1}\right|_{t \rightarrow \infty}=\lim _{t \rightarrow \infty}\left(A \cdot e^{-\beta t} \sin \left(\sqrt{\omega_{0}^{2}-\beta^{2}} t+\gamma\right)\right)=0$     (16)

The steady-state deck vibrations  consist only of violent vibrations under the exciting force's action, which is determined in the right part of the expression (13*). The specific solution of this equation must be found in the form:

$x_{2}=B \sin (\delta t+\mu)$     (17)

where, B is the amplitude of the violent vibration, $B=\frac{F(t)}{m \sqrt{\left(\omega_{0}^{2}-\delta^{2}\right)^{2}+4 \beta^{2} \delta^{2}}} ; \quad \sin \mu=\frac{2 B m \beta \delta}{F(t)}$; $\cos \mu=\frac{\operatorname{Bm}\left(\omega_{0}^{2}-\delta^{2}\right)}{\mathrm{F}(\mathrm{t})}$;$\delta$ is the frequency of the violent vibrations; μ is the angle that determines the phase displacement of violent vibrations and the exciting force.

According to the standard method applying the Mathcad program, the practical implementation of the model is carried out using the numerical quartic Runge-Kutta method.