**1. Introduction**

Mining vibrating screens are important equipment for mine washing and processing, and are widely used for mine grading, dehydration, and desliming in China [1,2], working as a forced vibration system under alternating loads [3–5]. The SLK3661W double-deck linear mining vibrating screen is shown in Figure 1, and its main structures include a screen box and four elastic supports, designed using principles of symmetry. As shown in Figure 2, the screen box is assembled from an exciter, a lateral plate, an exciting beam, reinforcing beams, upper-bearing beams, under-bearing beams, an upper-screen deck, and an under-screen deck. Additionally, each elastic support is composed of several metal helical springs. These springs are critical components in a mining vibrating screen's elastic supports, which directly affect the working performance of the mining vibrating screen [6,7]. However, long-term alternating loads are highly likely to lead to spring failure through spring stiffness decrease [8], causing a negative influence on the mining vibrating screen. On one hand, spring failures could lead to structural damages, such as beam fracture or lateral plate cracks [9–11]. On the other hand, spring failures could produce a loss of particle separation efficiency, thus hardly meeting practical process demands [12,13]. Therefore, it is necessary to diagnose the spring failures of a mining vibrating screen for routing maintenance, which can help to ensure safety and reliability.

**Figure 1.** The SLK3661W double-deck linear mining vibrating screen, unloading side view.

**Figure 2.** Structures of the SLK3661W double-deck linear mining vibrating screen.

Proper dynamic models provide a basis for diagnosing spring failures. In recent years, even though many studies have reported vibrating screen dynamic models on optimization [14–16], separation [17,18], and particle motion [19], there has been very little research reported on spring failure diagnosis. Aimed at spring failure diagnosis, Rodriguez et al. developed a two-dimensional, three-degree-of-freedom nonlinear model that considered one angular motion and damping, which allowed for the prediction of the behavior of a vibrating screen when there was a reduction in spring stiffness, and they used this model to determine a limit on spring failures before separation efficiency was affected [20]. Peng et al. presented a three-degree-of-freedom rigid plate structure to describe the isolation system, and they also proposed the method of stiffness identification by stiffness matrix disassembly; the numerical simulation results demonstrated the feasibility of the developed method [21]. However, each elastic support of a mining vibrating screen could have spring failures with spring stiffness decreases. The mining vibrating screen operating mode becomes spatial motion with very complicated dynamic characteristics, including multiple degrees of freedom.

The purpose of the present study is to explore the mining vibrating screen dynamic characteristics with spring failures, providing a theoretical basis for spring failure diagnosis. In this paper, a theoretical rigid body of a mining vibrating screen is proposed, the dynamic equation is established, and the steady-state solutions are obtained. Numerical simulations were carried out, and the results showed that the proposed model is feasible. In addition, spring failure simulations were also carried out, and the results indicated that the *x*, *y*, and *z* direction amplitude change rules for all the elastic supports were strongly related to spring failures. Hence, the key for spring failure diagnosis lies in obtaining the amplitude change rules, which can reveal the certain spring failures.

#### **2. Theoretical Rigid Body Model**

#### *2.1. The Model*

As shown in Figure 3, a six-degree-of-freedom theoretical rigid body model of spatial motion considering three rotations (Roll, Pitch, and Yaw) is proposed for exploring the mining vibrating screen dynamic characteristics with spring failures. The list of symbols is shown in abbreviations section.

**Figure 3.** Spatial motion dynamic model of the mining vibrating screen, including three translational degrees of freedom and three rotational degrees of freedom.

The screen box is simplified to a rigid body, and the four elastic support points are individually simplified as three mutually perpendicular springs *kix*, *kiy*, *kiz*, (where *i* is the elastic support point number, *i* = 1, 2, 3, 4). The movement of a rigid body is expressed by the position of a body frame *o'xyz* relative to the inertial frame *oxyz*. The *ox*-axis and *oz*-axis are mutually perpendicular and located in the horizontal plane, and the *oy*-axis is perpendicular to the horizontal plane. The origin *o'* of the body frame is located at the mass center of the rigid body, at all times. The *o'x*-axis and *o'z*-axis are mutually perpendicular and located in the rigid body plane, and the *o'y*-axis is perpendicular to the rigid body plane. Initially, the origin *o* of the inertial frame and the origin *o'* of the body frame are coincident. The distances between the mass center of the rigid body and the four elastic support points of spring are *r*1,*r*2,*r*3,*r*4 and, furthermore, the angles between them and *oz*-axis are *α*1, *α*2, *α*3, *α*4. Suppose that the rigid bodys mass is *m*, and the moments of inertia are *Jx*, *Jy*, *Jz*. Define *x*, *y*, *z* as the translation displacements of the rigid body and *γ*, *ϕ*, *θ* as the angular displacements in the inertial frame. The exciting force *f* is exerted on the rigid body as an alternating load, with included angles *β<sup>x</sup>*, *β<sup>y</sup>*, *βz* between exciting force and the *o'x*-axis, *o'y*-axis, and *o'z*-axis, respectively.

The dynamic equation is established by adopting the Lagrange method, and the processes are as follows. The three rotation angles are small, define cos *γ* = cos *ϕ* = cos *θ* = 1, sin *γ* = *γ*, sin *ϕ* = *ϕ*, and sin *θ* = *θ*.

#### *2.2. System Potential Energy*

The dynamic system includes three translation motions and three rotation motions. According to the Tait–Bryan angles in the literature [22], the rotation matrix between the body frame and the inertial frame was derived using the rotation system shown in Figure 4.

**Figure 4.** (**a**) Rotation of the inertial frame *oxyz* around the *ox*-axis by angle *γ*; (**b**) Rotation of the instantaneous system around the *oy*-axis by angle *ϕ*; (**c**) Rotation of the instantaneous system around the *oz*-axis by angle *θ*.

As the three rotation angles of the rigid body are small, they can be simplified as rotations around the *oxyz* axis. When rotating the rigid body around the *ox*-axis by the new angle of roll *γ*, the moment of inertia is *Jx*, and the rotation matrix is written as:

$$T\_{\mathcal{X}} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos \gamma & \sin \gamma \\ 0 & -\sin \gamma & \cos \gamma \end{bmatrix}. \tag{1}$$

When rotating the rigid body around the *oy*-axis by the new angle of yaw *ϕ*, the moment of inertia is *Jy*, and the rotation matrix is written as:

$$T\_Y = \begin{bmatrix} \cos \varrho & 0 & -\sin \varrho \\ 0 & 1 & 0 \\ \sin \varrho & 0 & \cos \varrho \end{bmatrix} . \tag{2}$$

When rotating the rigid body around the *oz*-axis by the new angle of pitch *θ*, the moment of inertia is *Jz*, and the rotation matrix is written as:

$$T\_z = \begin{bmatrix} \cos \theta & \sin \theta & 0 \\ -\sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{bmatrix} . \tag{3}$$

When rotating the rigid body in the sequence *oz-oy-ox*, the rotation matrix between the body frame and the inertial frame is obtained as:

$$R = T\_{\overline{z}} T\_{\overline{y}} T\_{\overline{x}} = \begin{bmatrix} \cos \theta \cos \varphi & \sin \theta \cos \gamma + \cos \theta \sin \varphi \sin \gamma & \sin \theta \sin \gamma - \cos \theta \sin \varphi \cos \gamma \\ -\sin \theta \cos \varphi & \cos \theta \cos \gamma - \sin \theta \sin \varphi \sin \gamma & \cos \theta \sin \gamma + \sin \theta \sin \varphi \cos \gamma \\ \sin \varphi & -\cos \varphi \sin \gamma & \cos \varphi \cos \gamma \end{bmatrix}. \tag{4}$$

Supposing that the coordinate of the mass center is (*<sup>x</sup>*, *y*, *z*) in an inertial frame, and any point of the rigid body is (*x*, *y*, *z*) in the body frame, the coordinate of any point of the rigid body in an inertial frame is written as:

$$
\begin{bmatrix} x\_d \\ y\_d \\ z\_d \end{bmatrix} = \begin{bmatrix} x \\ y \\ z \end{bmatrix} + \mathcal{R} \cdot \begin{bmatrix} x'\_d \\ y'\_d \\ z'\_d \end{bmatrix}. \tag{5}
$$

Moreover, the coordinates of the four spring support points in the body frame can be written as:

$$
\begin{bmatrix} x'\_1 \\ y'\_1 \\ z'\_1 \end{bmatrix} = \begin{bmatrix} -s\_1 \\ 0 \\ c\_1 \end{bmatrix}, \begin{bmatrix} x'\_2 \\ y'\_2 \\ z'\_2 \end{bmatrix} = \begin{bmatrix} s\_2 \\ 0 \\ c\_2 \end{bmatrix}, \begin{bmatrix} x'\_3 \\ y'\_3 \\ z'\_3 \end{bmatrix} = \begin{bmatrix} -s\_3 \\ 0 \\ -c\_3 \end{bmatrix}, \begin{bmatrix} x'\_4 \\ y'\_4 \\ z'\_4 \end{bmatrix} = \begin{bmatrix} s\_4 \\ 0 \\ -c\_4 \end{bmatrix}. \tag{6}
$$

In (6), *s*1 = *r*1 sin *α*1, *c*1 = *r*1 cos *α*1, *s*2 = *r*2 sin *α*2, *c*2 = *r*2 cos *α*2, *s*3 = *r*3 sin *α*3, *c*3 = *r*3 cos *α*3, *s*4 = *r*4 sin *α*4, and *c*4 = *r*4 cos *α*4.

In the initial state, the coordinate of the mass center is (0, 0, 0) in the inertial frame, and the rotation matrix is *R*0 = [1, 0, 0; 0, 1, 0; 0, 0, 1]T. Thus, the coordinates of the four spring support points in the inertial frame can be written as:

$$
\begin{bmatrix} x\_1 \\ y\_1 \\ z\_1 \end{bmatrix} = R\_0 \begin{bmatrix} -s\_1 \\ 0 \\ c\_1 \end{bmatrix}, \begin{bmatrix} x\_2 \\ y\_2 \\ z\_2 \end{bmatrix} = R\_0 \begin{bmatrix} s\_2 \\ 0 \\ c\_2 \end{bmatrix}, \begin{bmatrix} x\_3 \\ y\_3 \\ z\_3 \end{bmatrix} = R\_0 \begin{bmatrix} -s\_3 \\ 0 \\ -c\_3 \end{bmatrix}, \begin{bmatrix} x\_4 \\ y\_4 \\ z\_4 \end{bmatrix} = R\_0 \begin{bmatrix} s\_4 \\ 0 \\ -c\_4 \end{bmatrix}. \tag{7}
$$

In a motion state, the coordinate of the mass center is (*<sup>x</sup>*, *y*, *z*) in the inertial frame. Thus, the coordinates of the four spring support points in the inertial frame can be written as:

$$
\begin{bmatrix}
\Delta \mathbf{x}\_1 \\
\Delta y\_1 \\
\Delta z\_1
\end{bmatrix} = \begin{bmatrix}
\mathbf{x} \\
\mathbf{y} \\
\mathbf{z}
\end{bmatrix} + (\mathbf{R} - \mathbf{R}\_0) \begin{bmatrix}
\mathbf{0} \\
\mathbf{c}\_1
\end{bmatrix},
\begin{bmatrix}
\Delta \mathbf{x}\_2 \\
\Delta y\_2 \\
\Delta z\_2
\end{bmatrix} = \begin{bmatrix}
\mathbf{x} \\
\mathbf{y} \\
\mathbf{z}
\end{bmatrix} + (\mathbf{R} - \mathbf{R}\_0) \begin{bmatrix}
\mathbf{s}\_2 \\
\mathbf{0} \\
\mathbf{c}\_2
\end{bmatrix},
$$

$$
\begin{bmatrix}
\Delta \mathbf{x}\_3 \\
\Delta y\_3 \\
\Delta z\_3
\end{bmatrix} = \begin{bmatrix}
\mathbf{x} \\
\mathbf{y} \\
\mathbf{z}
\end{bmatrix} + (\mathbf{R} - \mathbf{R}\_0) \begin{bmatrix}
\mathbf{0} \\
\end{bmatrix},
\begin{bmatrix}
\Delta \mathbf{x}\_4 \\
\Delta y\_4 \\
\Delta z\_4
\end{bmatrix} = \begin{bmatrix}
\mathbf{x} \\
\mathbf{y} \\
\mathbf{z}
\end{bmatrix} + (\mathbf{R} - \mathbf{R}\_0) \begin{bmatrix}
\mathbf{s}\_4 \\
\mathbf{0} \\
\end{bmatrix}.
\tag{8}
$$

The results in (8) are equivalent to the spring compression and, therefore, the system potential energy is obtained as:

$$\begin{array}{ll} \mathcal{U} = \frac{1}{2}k\_{1x}\left[\mathbf{x} - c\_{1}(\boldsymbol{\varrho} - \boldsymbol{\gamma}\boldsymbol{\theta})\right]^{2} + \frac{1}{2}k\_{2x}\left[\mathbf{x} - c\_{2}(\boldsymbol{\varrho} - \boldsymbol{\gamma}\boldsymbol{\theta})\right]^{2} + \frac{1}{2}k\_{3x}\left[\mathbf{x} + c\_{3}(\boldsymbol{\varrho} - \boldsymbol{\gamma}\boldsymbol{\theta})\right]^{2} \\ \quad + \frac{1}{2}k\_{4x}\left[\mathbf{x} + c\_{4}(\boldsymbol{\varrho} - \boldsymbol{\gamma}\boldsymbol{\theta})\right]^{2} + \frac{1}{2}k\_{1y}\left[y + c\_{1}(\boldsymbol{\gamma} + q\boldsymbol{\theta}) + s\_{1}\boldsymbol{\theta}\right]^{2} \\ \quad + \frac{1}{2}k\_{2y}\left[y + c\_{2}(\boldsymbol{\gamma} + q\boldsymbol{\theta}) - s\_{2}\boldsymbol{\theta}\right]^{2} + \frac{1}{2}k\_{3y}\left[y - c\_{3}(\boldsymbol{\gamma} + q\boldsymbol{\theta}) + s\_{3}\boldsymbol{\theta}\right]^{2} \\ \quad + \frac{1}{2}k\_{4y}\left[-y + c\_{4}(\boldsymbol{\gamma} + q\boldsymbol{\theta}) + s\_{4}\boldsymbol{\theta}\right]^{2} + \frac{1}{2}k\_{1z}(z - s\_{1}\boldsymbol{\varrho})^{2} + \frac{1}{2}k\_{2z}(z + s\_{2}\boldsymbol{\varrho})^{2} \\ \quad + \frac{1}{2}k\_{3z}(z - s\_{3}\boldsymbol{\varrho})^{2} + \frac{1}{2}k\_{4z}(z + s\_{4}\boldsymbol{\varrho})^{2} \end{array} . \tag{9}$$

#### *2.3. System Kinetic Energy*

According to the literature [23,24], there is a relation expressing a rigid bodys spatial motion, which is written as: .

$$\begin{cases} \omega\_x = \dot{\gamma} - \theta \cos \varphi \tan \varphi\\ \omega\_y = \dot{\varphi} \cos \gamma + \dot{\theta} \cos \varphi \sin \gamma\\ \omega\_z = -\dot{\varphi} \sin \gamma + \dot{\theta} \cos \varphi \cos \gamma \end{cases} \tag{10}$$

Therefore, the system's kinetic energy is obtained as:

$$\begin{array}{ll} E &= \frac{1}{2}m\dot{\mathbf{x}}^2 + \frac{1}{2}m\dot{\mathbf{y}}^2 + \frac{1}{2}m\dot{\mathbf{z}}^2 + \frac{1}{2}I\_x\omega\_x^2 + \frac{1}{2}I\_y\omega\_y^2 + \frac{1}{2}I\_z\omega\_z^2\\ &= \frac{1}{2}m\dot{\mathbf{x}}^2 + \frac{1}{2}m\dot{\mathbf{y}}^2 + \frac{1}{2}m\dot{\mathbf{z}}^2 + \frac{1}{2}I\_x\left(\dot{\boldsymbol{\eta}} - \dot{\boldsymbol{\theta}}\boldsymbol{\eta}\right)^2 + \frac{1}{2}I\_y\left(\dot{\boldsymbol{\eta}} + \dot{\boldsymbol{\theta}}\boldsymbol{\gamma}\right)^2 + \frac{1}{2}I\_z\left(-\dot{\boldsymbol{\eta}}\boldsymbol{\gamma} + \dot{\boldsymbol{\theta}}\right)^2 \end{array} \tag{11}$$

#### *2.4. System Force Vector*

The mining vibrating screen in this study is equipped with two groups of counter-rotating vibrators, with each group having two pairs of eccentric blocks. Due to manufacturing errors and installation errors, the resultant force *f* typically does not pass through the center of mass of the screen box in practice. The resultant force *f* can be equivalent to a force vector. In the body frame, the force vector is written as: 

$$
\begin{bmatrix} f\_x' \\ f\_y' \\ f\_z' \end{bmatrix} = \begin{bmatrix} f \cos \beta\_x \sin \omega t \\ f \cos \beta\_y \sin \omega t \\ f \cos \beta\_z \sin \omega t \end{bmatrix}. \tag{12}
$$

On account of the force vector changing with the rigid body motion, the force vector in a body frame is written as: 

$$
\begin{bmatrix} f\_x \\ f\_y \\ f\_z \end{bmatrix} = \mathcal{R} \cdot \begin{bmatrix} f'\_x \\ f'\_y \\ f'\_z \end{bmatrix}. \tag{13}
$$

Meanwhile, supposing that the coordinate of the point exerting force is (*xf* , *yf* , *zf*) in a body frame, the coordinate of the point exerting force in an inertial frame can be written as:

$$
\begin{bmatrix} x\_f \\ y\_f \\ z\_f \end{bmatrix} = \begin{bmatrix} x \\ y \\ z \end{bmatrix} + \mathbb{R} \cdot \begin{bmatrix} x'\_f \\ y'\_f \\ z'\_f \end{bmatrix}. \tag{14}
$$

According to the literature [23,24], there is a relation expressing a rigid bodys spatial motion, which is written as: 

$$
\begin{bmatrix} M\_x \\ M\_y \\ M\_z \end{bmatrix} = \begin{bmatrix} 0 & -z\_f & y\_f \\ z\_f & 0 & -x\_f \\ -y\_f & x\_f & 0 \end{bmatrix} \cdot \begin{bmatrix} f\_x \\ f\_y \\ f\_z \end{bmatrix}. \tag{15}
$$

In an inertial frame, the system's force vector is obtained as:

$$\mathbf{F} = \begin{bmatrix} f\_x f\_y f\_z M\_x M\_y M\_z \end{bmatrix}^T. \tag{16}$$

*2.5. Dynamic Equation*

> After linearizing, the dynamic equation can be written as:

$$\mathbf{M}\ddot{\mathbf{x}} + \mathbf{K}\mathbf{x} = \mathbf{F}.\tag{17}$$

In (17), **x**¨ is the acceleration column vector:

$$
\ddot{\mathbf{x}} = \begin{bmatrix}
\ddot{x} & \ddot{y} & \ddot{z} & \ddot{\gamma} & \ddot{\varphi} & \ddot{\theta}
\end{bmatrix}^{\mathrm{T}}.\tag{18}
$$

**x** is the displacement column vector:

$$\mathbf{x} = \begin{bmatrix} \ \mathbf{x} & \ y & z & \gamma & \varrho & \theta \end{bmatrix}^{\mathrm{T}}.\tag{19}$$

**M** is the mass matrix:

$$\mathbf{M} = \begin{bmatrix} m & 0 & 0 & 0 & 0 & 0 \\ 0 & m & 0 & 0 & 0 & 0 \\ 0 & 0 & m & 0 & 0 & 0 \\ 0 & 0 & 0 & J\_x & 0 & 0 \\ 0 & 0 & 0 & 0 & J\_y & 0 \\ 0 & 0 & 0 & 0 & 0 & J\_z \end{bmatrix} \tag{20}$$

and **K** is the stiffness matrix:

$$\mathbf{K} = \begin{bmatrix} K\_{11} & 0 & 0 & 0 & K\_{15} & 0 \\ 0 & K\_{22} & 0 & K\_{24} & 0 & K\_{26} \\ 0 & 0 & K\_{33} & 0 & K\_{35} & 0 \\ 0 & K\_{42} & 0 & K\_{44} & 0 & K\_{46} \\ K\_{51} & 0 & K\_{53} & 0 & K\_{55} & 0 \\ 0 & K\_{62} & 0 & K\_{64} & 0 & K\_{66} \end{bmatrix} . \tag{21}$$

In (21), *K*11 = *k*1*x* + *k*2*x* + *k*3*x* + *k*4*x*, *K*15 = <sup>−</sup>*c*1*k*1*x* − *<sup>c</sup>*2*k*2*x* + *<sup>c</sup>*3*k*3*x* + *<sup>c</sup>*4*k*4*x*, *K*22= *<sup>k</sup>*1*y* + *<sup>k</sup>*2*y* + *<sup>k</sup>*3*y* − *<sup>k</sup>*4*y*, *K*24 = *<sup>c</sup>*1*k*1*y* + *<sup>c</sup>*2*k*2*y* − *<sup>c</sup>*3*k*3*y* + *<sup>c</sup>*4*k*4*y*, *K*26 = *<sup>s</sup>*1*k*1*y* − *<sup>s</sup>*2*k*2*y* + *<sup>s</sup>*3*k*3*y* + *<sup>s</sup>*4*k*4*y*, *K*33 = *k*1*z* + *k*2*z* + *k*3*z* + *k*4*z*, *K*35 = <sup>−</sup>*s*1*k*1*z* + *<sup>s</sup>*2*k*2*z* − *<sup>s</sup>*3*k*3*z* + *<sup>s</sup>*4*k*4*z*, *K*42 = *<sup>c</sup>*1*k*1*y* + *<sup>c</sup>*2*k*2*y* − *<sup>c</sup>*3*k*3*y* − *<sup>c</sup>*4*k*4*y*, *K*44 = *<sup>c</sup>*21*<sup>k</sup>*1*y* + *<sup>c</sup>*22*<sup>k</sup>*2*y* + *<sup>c</sup>*23*<sup>k</sup>*3*y* + *<sup>c</sup>*24*<sup>k</sup>*4*y*, *K*46 = *<sup>c</sup>*1*s*1*k*1*y* − *<sup>c</sup>*2*s*2*k*2*y* − *<sup>c</sup>*3*s*3*k*3*y* + *<sup>c</sup>*4*s*4*k*4*y*, *K*51 = <sup>−</sup>*c*1*k*1*x* − *<sup>c</sup>*2*k*2*x* + *<sup>c</sup>*3*k*3*x* + *<sup>c</sup>*4*k*4*x*, *K*53 = <sup>−</sup>*s*1*k*1*z* + *<sup>s</sup>*2*k*2*z* − *<sup>s</sup>*3*k*3*z* + *<sup>s</sup>*4*k*4*z*, *K*55 = *<sup>c</sup>*21*<sup>k</sup>*1*x* + *<sup>c</sup>*22*<sup>k</sup>*2*x* + *<sup>c</sup>*23*<sup>k</sup>*3*x* + *<sup>c</sup>*24*<sup>k</sup>*4*x* + *<sup>s</sup>*21*<sup>k</sup>*1*z* + *<sup>s</sup>*22*<sup>k</sup>*2*z* + *<sup>s</sup>*23*<sup>k</sup>*3*z* + *<sup>s</sup>*24*<sup>k</sup>*4*z*, *K*62 = *<sup>s</sup>*1*k*1*y* − *<sup>s</sup>*2*k*2*y* + *<sup>s</sup>*3*k*3*y* − *<sup>s</sup>*4*k*4*y*, *K*64 = *<sup>c</sup>*1*s*1*k*1*y* − *<sup>c</sup>*2*s*2*k*2*y* − *<sup>c</sup>*3*s*3*k*3*y* + *<sup>c</sup>*4*s*4*k*4*y*, *K*66 = *<sup>s</sup>*21*<sup>k</sup>*1*y* + *<sup>s</sup>*22*<sup>k</sup>*2*y* + *<sup>s</sup>*23*<sup>k</sup>*3*y* + *<sup>s</sup>*24*<sup>k</sup>*4*y*. Additionally, **F** is the force column vector:

$$\mathbf{F} = \begin{bmatrix} f\_x f\_y f\_z M\_x M\_y M\_z \end{bmatrix}^T. \tag{22}$$

According to the dynamic theory [20], the steady-state solutions of a forced vibration system can be written as:

$$\begin{cases} \begin{aligned} x &= X\_0 \sin \omega t \\ y &= Y\_0 \sin \omega t \\ z &= Z\_0 \sin \omega t \\ \gamma &= \Gamma\_0 \sin \omega t \\ \theta &= \Phi\_0 \sin \omega t \\ \theta &= \Theta\_0 \sin \omega t \end{aligned} \end{cases} \tag{23}$$

In (23), *ω* is the angular speed. It should be noted that the relations are valid for a constant rotational velocity of counter-rotating vibrators, for which there exists a resultant force acting along

a straight line towards the body of vibrating screen. Taking the derivative of both sides of (23), the acceleration can be written as:

$$\begin{cases} \ddot{\bar{x}} = -\omega^2 X\_0 \sin \omega t \\ \ddot{y} = -\omega^2 Y\_0 \sin \omega t \\ \ddot{z} = -\omega^2 Z\_0 \sin \omega t \\ \ddot{\gamma} = -\omega^2 \Gamma\_0 \sin \omega t \\ \ddot{\theta} = -\omega^2 \Phi\_0 \sin \omega t \\ \ddot{\theta} = -\omega^2 \Theta\_0 \sin \omega t \end{cases} \tag{24}$$

Bringing Equations (23) and (24) into Equation (22), the steady-state solutions can be obtained as:

$$
\begin{bmatrix} X\_0 \\ Y\_0 \\ Z\_0 \\ \Gamma\_0 \\ \Phi\_0 \\ \Theta\_0 \end{bmatrix} = \begin{bmatrix} K\_{11} - \omega^2 m & 0 & 0 & 0 & K\_{15} & 0 \\ 0 & K\_{22} - \omega^2 m & 0 & K\_{24} & 0 & K\_{26} \\ 0 & 0 & K\_{33} - \omega^2 m & 0 & K\_{35} & 0 \\ 0 & K\_{42} & 0 & K\_{44} - \omega^2 f\_x & 0 & K\_{46} \\ K\_{51} & 0 & K\_{53} & 0 & K\_{55} - \omega^2 f\_y & 0 \\ 0 & K\_{62} & 0 & K\_{64} & 0 & K\_{66} - \omega^2 f\_z \end{bmatrix}^{-1} \begin{bmatrix} f\_x \\ f\_y \\ f\_z \\ M\_x \\ M\_y \\ M\_z \end{bmatrix} \tag{25}
$$

The dynamic equation of a mining vibrating screen in spatial motion, shown above, gives the dynamic characteristics. In the following section, numerical simulations are carried out to verify the proposed model.
