Dynamic model analysis of hydraulic support

Abstract

In view of the shortcomings of traditional hydraulic support system, such as poor mobility and limited range of adjustment, a new configuration with double parallelogram structure is proposed. Combining the kinematic diagram of the mechanism of the new support system and the relevant parameters of the balance moment, using the kinematic vector closed-loop modeling method and D’Alembert’s principle, the relationship equation between the generalized coordinates of the hydraulic support and the geometric position parameters is obtained, and the analysis model with plural-freedom is established. Dynamic response characteristics and natural frequencies of column hydraulic cylinder and balance hydraulic cylinder are analyzed. The results demonstrate that the natural frequencies of the improved balance hydraulic cylinder and column hydraulic cylinder have been greatly improved, and the safety performance of the hydraulic support has been greatly increased.

Introduction

As a key device for supporting the top beam and maintaining the safe production in the stope, the hydraulic support can be considered as a complex system that interacts with the surrounding rock.^1,2 The stress state of the hydraulic support depends on the relationship between the external mine pressure and the internal forces. ³ To understand the coupling relationship between the surrounding rock and the hydraulic support, a specific research of the dynamic characteristics is of great significance.

Many scholars have studied the hydraulic support system. Yang et al. ⁴ designed a new hydraulic support based on the kinematics modeling and simulation of the mechanism analysis of the bracket. Wang and colleagues^5–7 analyzed the mechanical characteristics of the elastic beam of the hydraulic support, the trajectory of parts, and the structural optimization design. Liang et al. ⁸ researched the force transfer characteristics of hydraulic support under impact loading. Using the AMESim software, the hydraulic system model of roadway rapid support equipment is built by Xu et al. ⁹ The curves of the displacement, speed, and acceleration characteristic of the piston rod in the working process are calculated. Oblak et al. ¹⁰ optimized the two sets of parameters of hydraulic support in mining industry based on mathematical programming method. Prebil et al. ¹¹ used the global optimization method and adaptive mesh refinement algorithm to determine the optimal connecting rod length in the hydraulic support mechanism.

The above research works mainly focus on the statics and kinematics analysis and do not involve dynamic analysis. As the mechanical vibration in hydraulic support device may cause the resonance and fatigue, the analysis of natural frequency will avoid unnecessary loss caused by the resonance.^12–14 Through the dynamic modeling of the hydraulic support, the dynamic response and frequency spectrum of hydraulic support under the external excitation are analyzed, which provides the theoretical basis for the design of the hydraulic support.

Working principle and simplification of hydraulic support

Working principle of hydraulic support

As the main safety support device for coal mining, the hydraulic support can not only support and control the roof beam, maintain the safe underground working space, but also enable the mechanization pass of other related equipment. The three-dimensional model of the traditional hydraulic support is shown in Figure 1(a), consisting of top beam, column hydraulic cylinder, shield beam, balancing hydraulic cylinder, and support floor. During the coal mining process, the piston’s movement of column hydraulic cylinder causes the top beam to be supported. When the balancing hydraulic cylinder is working, the adaptability of support to the unstable top beam is improved. In addition, when the support system is in unloading and semi-unloading state, the angle of the top beam is adjusted to make the support system in a proper working condition.

Figure 1.

Three-dimensional model: (a) traditional hydraulic support and (b) improved hydraulic support.

In the traditional hydraulic support system, the top beam is a kind of plural-freedom structure. Its mobility is poor and the adjustment range is limited. A new support configuration, as shown in Figure 1(b), is proposed by setting the concentrated force load mechanism above the top beam, using the configuration of the palletizing robot to change the top beam motion. The new hydraulic support is composed of two parallel quadrangles. Thus, the moving range of the column nest is more extensive. The driving mode is close to the palletizing plural-freedom robot. The piston of the column hydraulic cylinder moves and supports the top beam. The improved device improves the reliability of the top beam and also improves the ability of the support system to bear horizontal force. In addition, the shield beam can prevent the broken rock from entering the supporting working space.

Model of hydraulic support

To analyze the dynamic response characteristics of the improved hydraulic support, it is necessary to simplify the hydraulic support. The coupling model between improved hydraulic support and surrounding rock is shown in Figure 2. The simplified diagram of hydraulic support mechanism is shown in Figure 3. When the dynamic model is established, the length of balance hydraulic cylinder $l_4$ and column hydraulic cylinder $l_8$ is selected as the generalized coordinates. When the support is subjected to external impact (concentrated force f) and compared with other components, the balance hydraulic cylinder and column hydraulic cylinder show smaller stiffness. Therefore, it is simplified as a spring with a stiffness of $k_1$ , $k_2$ .

Figure 2.

The coupling model between improved hydraulic support and surrounding rock.

Figure 3.

The simplified diagram of hydraulic support mechanism.

The physical meaning of the other parameters is shown in Table 1. The related parameters at the equilibrium position are as follows: ${\theta }_0=0°$ , ${\theta }_1=89.42°$ , ${\theta }_2=59.89°$ , ${\theta }_3=113.57°$ , ${\theta }_4=119.98°$ , ${\theta }_5=0°$ , ${\theta }_6=0°$ ; $l_1=0.798\mathrm{m}$ , $l_2=0.094\mathrm{m}$ , $l_3=2.5\mathrm{m}$ , $l_4=1.89\mathrm{m}$ , $l_5=2.2\mathrm{m}$ , $l_6=3.82\mathrm{m}$ , $l_7=1.33\mathrm{m}$ , $l_8=5.45\mathrm{m}$ ; $H_1=0.22\mathrm{m}$ , $H_2=0.805\mathrm{m}$ , $H_3=0.375\mathrm{m}$ , $H_4=0.49\mathrm{m}$ , $H_5=0.553\mathrm{m}$ , $l_{s1}=3.112\mathrm{m}$ , $l_{s2}=1.279\mathrm{m}$ , $l_{s3}=1.045\mathrm{m}$ , $l_{s4}=1.78\mathrm{m}$ ; $m_1=1267.63\mathrm{kg}$ , $m_2=546\mathrm{kg}$ , $m_3=414.9\mathrm{kg}$ , $m_4=1066.11\mathrm{kg}$ , $m_5=4597.87\mathrm{kg}$ , $m_6=576.17\mathrm{kg}$ ; $k_1=7\times {10}^7\mathrm{N}/\mathrm{m}$ , $k_2=7\times {10}^7\mathrm{N}/\mathrm{m}$ ; $J_1=5146.73\mathrm{kg}{\mathrm{m}}^2$ , $J_2=1437.68\mathrm{kg}{\mathrm{m}}^2$ , $J_3=95.42\mathrm{kg}{\mathrm{m}}^2$ , $J_4=4195\mathrm{kg}{\mathrm{m}}^2$ .

Table 1.

Physical meaning of each parameter.

Symbol	Physical meaning	Symbol	Physical meaning	Symbol	Physical meaning
${\theta }_0$	Angle between support floor and x-axis	$l_7$	Length of DE	$m_1$	Mass of column hydraulic cylinder at balance position
${\theta }_1$	Angle between column hydraulic cylinder and x-axis	$l_8$	Length of column hydraulic cylinder	$m_2$	Mass of connecting rod
${\theta }_2$	Angle between connecting rod and x-axis	$H_1$	Distance from $A\prime$ to support floor	$m_3$	Mass of balance hydraulic cylinder
${\theta }_3$	Angle between balance hydraulic cylinder and x-axis	$H_2$	Distance from $B\prime$ to support floor	$m_4$	Mass of shield beam
${\theta }_4$	Angle between shield beam and x-axis	$H_3$	Distance from $C\prime$ to support floor	$m_5$	Mass of top beam
${\theta }_5$	Angle between top beam and x-axis	$H_4$	Distance from $D\prime$ to support floor	$m_6$	Mass of middle beam
${\theta }_6$	Angle between middle beam and x-axis	$H_5$	Distance from $E\prime$ to support floor	$k_1$	Stiffness of balance hydraulic cylinder
$l_1$	Length of AB	$l_{s1}$	Distance from $s_1$ to $A\prime$	$k_2$	Stiffness of column hydraulic cylinder
$l_2$	Length of BC	$l_{s2}$	Distance from $s_2$ to $C\prime$	$J_1$	Rotating inertia of column hydraulic cylinder
$l_3$	Length of connecting rod	$l_{s3}$	Distance from $s_3$ to $B\prime$	$J_2$	Rotating inertia of connecting rod
$l_4$	Length of balance hydraulic cylinder	$l_{s4}$	Distance from $s_4$ to $F\prime$	$J_3$	Rotating inertia of balance hydraulic cylinder
$l_5$	Length of middle beam	$l_{s5}$	Distance from $s_5$ to $G\prime$	$J_4$	Rotating inertia of shield beam
$l_6$	Length of shield beam	$l_{s6}$	Distance from $s_6$ to E

As shown in Figure 4, the shield beam of the hydraulic support can be considered as a parallelogram structure. Its equivalent mass can be calculated by center-of-mass motion theorem and torque equilibrium equation.

Figure 4.

The schematic of the parallelogram structure.

In the shield beam, $O\prime$ is the equivalent center of mass. The equivalent mass can be indicated as

$$m_{\mathrm{zs}}=m_{l_6}+m_{l_{06}}+m_{l_{16}}$$

(1)

In equation (1), $m_{\mathrm{zs}}$ is the equivalent mass. $m_{l_6}$ , $m_{l_{06}}$ , and $m_{l_{16}}$ mean the linkage mass. As $m_{\mathrm{zs}}$ is supported by points A₀ and B₀ in the vertical direction, the equivalent mass of the shield beam to point A₀ can be calculated by torque equilibrium equation

$$m_{\mathrm{zo}}=\frac{m_{\mathrm{zs}}{x}_4}{x_1}$$

(2)

In equation (2), $x_1$ is the distance between support points A₀ and B₀. $x_4$ is the distance between the equivalent center of mass $O\prime$ and the support point B₀ in X-axis direction. $x_{o\prime }$ is the distance between the equivalent center of mass $O\prime$ and the support point A₀ in X-axis direction

$$x_{o\prime }=\frac{m_{l_6}{x}_{l_6}+m_{l_{06}}{x}_{l_{06}}+m_{l_{16}}{x}_{l_{16}}}{m_{l_6}+m_{l_{06}}+m_{l_{16}}}$$

(3)

$$x_4=|x_{o\prime }-x_1|$$

(4)

In equations (3) and (4), $x_{l_6}$ , $x_{l_{06}}$ , and $x_{l_{16}}$ are the distance between the center of mass of the shield beam linkage and the support point A₀ in X-axis direction. As $x_{o\prime }$ is changing caused by the rotation of the parallelogram structure, $x_4$ is changing. As shown in equation (2), $m_{\mathrm{zs}}$ and $x_1$ are constant. Thus, $m_{\mathrm{zo}}$ is changing with $x_4$ and $m_{\mathrm{zo}}$ can be indicated as

$$m_{\mathrm{zo}}(t)={\overline{m}}_{\mathrm{zo}}+\Delta m_{\mathrm{zo}}(t)$$

(5)

In equation (5), ${\overline{m}}_{\mathrm{zo}}$ is the average value of $m_{\mathrm{zo}}$ ; $\Delta m_{\mathrm{zo}}(t)$ is the bias between the average value and the distributed value changing with time.

The equivalent moment of inertia of the parallelogram structure can be indicated as

$$J_{\mathrm{ze}}=m_{l_{06}}{\left(\frac{v_{06}}{\omega }\right)}^2+J_{l6}{\left(\frac{{\omega }_6}{\omega }\right)}^2+J_{l_{06}}{\left(\frac{{\omega }_{06}}{\omega }\right)}^2+J_{l_{16}}{\left(\frac{{\omega }_{16}}{\omega }\right)}^2$$

(6)

In equation (6), $\omega$ is the revolving speed of principal axis and $v_{06}$ is the center-of-mass speed of the linkage $l_{06}$ . $J_{l6}$ , $J_{l06}$ , and $J_{l16}$ mean the related moment of inertia. ${\omega }_6$ , ${\omega }_{06}$ , and ${\omega }_{16}$ mean the related revolving speed. As $v_{06}$ , ${\omega }_6$ , ${\omega }_{06}$ , and ${\omega }_{16}$ are changing with the posture of parallelogram structure, the equivalent moment of inertia $J_{\mathrm{ze}}$ is related with the posture of parallelogram structure, changing with time

$$J_{\mathrm{ze}}(t)={\overline{J}}_{\mathrm{ze}}+\Delta J_{\mathrm{zo}}(t)$$

(7)

In equation (7), ${\overline{J}}_{\mathrm{ze}}$ is the average value of $J_{\mathrm{ze}}$ and $\Delta J_{\mathrm{ze}}(t)$ is the bias between the average value and the distributed value changing with time.

As shown in Table 2, the equivalent mass of the shield beam is changing from 575.06 to 577.22 kg, and the average value is 576.17 kg; the equivalent moment of inertia is changing from 64.31 to 67.54 kg m², and the average value is 65.92 kg m². The equivalent simplification has little affection on frequency analysis.

Table 2.

Frequency analysis under equivalent simplification.

Equivalent mass (kg)	Equivalent moment of inertia (kg m2)	Balance hydraulic cylinder (Hz)	Column hydraulic cylinder (Hz)
576.17 (Average)	65.92 (Average)	28	160
577.55 (Max)	67.54 (Max)	28.4	161.2
575.06 (Min)	64.31 (Min)	27.8	159.7
577.55 (Max)	64.31 (Min)	28.1	160.8
575.06 (Min)	67.54 (Max)	27.9	160.4

Dynamics mathematical model

The relationship between the angular velocity and generalized coordinates

According to the kinematic vector closed-loop modeling method,^15,16 one can obtain

$$\{\begin{array}{c}{l}_2{e}^{i(\pi -{\theta }_0)}+H_2{e}^{i(\pi /2-{\theta }_0)}+l_4{e}^{i{\theta }_3}+l_5{e}^{i(\pi -{\theta }_0)}=H_3{e}^{i(\pi /2-{\theta }_0)}+l_3{e}^{i{\theta }_2}\hfill \\ \multicolumn{1}{c}{l_1{e}^{i(\pi -{\theta }_0)}+H_1{e}^{i(\pi /2-{\theta }_0)}+l_8{e}^{i{\theta }_1}+H_4{e}^{i(\pi /2+{\theta }_0)}+l_7{e}^{i{\theta }_0}}\\ \multicolumn{1}{c}{=H_2{e}^{i(\pi -{\theta }_0)}+l_4{e}^{i{\theta }_3}+l_5{e}^{i(\pi -{\theta }_0)}+l_6{e}^{i{\theta }_4}+H_5{e}^{i(\pi /2+{\theta }_0)}}\end{array}$$

(8)

Derivation of equation (8)

$$\begin{array}{c}\left[\begin{array}{cccc}0\hfill & l_3\sin {\theta }_2\hfill & -l_4\sin {\theta }_3\hfill & 0\hfill \\ \multicolumn{1}{c}{0}& l_3\cos {\theta }_2\hfill & l_4\cos {\theta }_3\hfill & 0\hfill \\ \multicolumn{1}{c}{l_8\sin {\theta }_1}& 0\hfill & -l_4\sin {\theta }_3\hfill & -l_6\sin {\theta }_4\hfill \\ \multicolumn{1}{c}{-l_8\cos {\theta }_1}& 0\hfill & l_4\cos {\theta }_3\hfill & l_6\cos {\theta }_4\hfill \end{array}\right]\left[\begin{array}{c}{\stackrel{·}{\theta }}_1\hfill \\ \multicolumn{1}{c}{{\stackrel{·}{\theta }}_2}\\ \multicolumn{1}{c}{{\stackrel{·}{\theta }}_3}\\ \multicolumn{1}{c}{{\stackrel{·}{\theta }}_4}\end{array}\right]\hfill \\ =\left[\begin{array}{ccc}(H_2-H_3-l_5-l_2)\sin {\theta }_0\hfill & -\cos {\theta }_3\hfill & 0\hfill \\ \multicolumn{1}{c}{(l_2+H_2+l_5-H_3)\cos {\theta }_0}& \sin {\theta }_3\hfill & 0\hfill \\ \multicolumn{1}{c}{(l_1-H_1+H_4-l_7-H_2-l_5-H_5)\sin {\theta }_0}& -\cos {\theta }_3\hfill & \cos {\theta }_1\hfill \\ \multicolumn{1}{c}{(l_1+H_1+H_4+l_7-H_2-l_5-H_5)\cos {\theta }_0}& -\sin {\theta }_3\hfill & \sin {\theta }_1\hfill \end{array}\right]\left[\begin{array}{c}{\stackrel{·}{\theta }}_0\hfill \\ \multicolumn{1}{c}{{\stackrel{·}{l}}_4}\\ \multicolumn{1}{c}{{\stackrel{·}{l}}_8}\end{array}\right]\hfill \end{array}$$

(9)

Substituting the value of each parameter into equation (9), it can be obtained

$$\left[\begin{array}{c}{\stackrel{·}{\theta }}_1\hfill \\ \multicolumn{1}{c}{{\stackrel{·}{\theta }}_2}\\ \multicolumn{1}{c}{{\stackrel{·}{\theta }}_3}\\ \multicolumn{1}{c}{{\stackrel{·}{\theta }}_4}\end{array}\right]=\left[\begin{array}{ccc}1.059\hfill & 0.568\hfill & -0.311\hfill \\ \multicolumn{1}{c}{8.770}& 2.389\hfill & 0\hfill \\ \multicolumn{1}{c}{10.949}& 2.752\hfill & 0\hfill \\ \multicolumn{1}{c}{-3.988}& 0.626\hfill & -0.515\hfill \end{array}\right]\left[\begin{array}{c}{\stackrel{·}{\theta }}_0\hfill \\ \multicolumn{1}{c}{{\stackrel{·}{l}}_4}\\ \multicolumn{1}{c}{{\stackrel{·}{l}}_8}\end{array}\right]$$

(10)

System kinetic energy

$$\left[\begin{array}{c}{x}_{s1}\hfill \\ \multicolumn{1}{c}{y_{s1}}\end{array}\right]=\left[\begin{array}{c}0\hfill \\ \multicolumn{1}{c}{H_1}\end{array}\right]+l_{s1}\left[\begin{array}{c}\cos {\theta }_1\hfill \\ \multicolumn{1}{c}{\sin {\theta }_1}\end{array}\right]$$

(11)

$$\left[\begin{array}{c}{x}_{s2}\hfill \\ \multicolumn{1}{c}{y_{s2}}\end{array}\right]=\left[\begin{array}{c}0\hfill \\ \multicolumn{1}{c}{H_3}\end{array}\right]+l_{s2}\left[\begin{array}{c}\cos {\theta }_2\hfill \\ \multicolumn{1}{c}{\sin {\theta }_2}\end{array}\right]+\left[\begin{array}{c}{l}_1+l_2\hfill \\ \multicolumn{1}{c}{0}\end{array}\right]$$

(12)

$$\left[\begin{array}{c}{x}_{s3}\hfill \\ \multicolumn{1}{c}{y_{s3}}\end{array}\right]=\left[\begin{array}{c}{l}_1\hfill \\ \multicolumn{1}{c}{0}\end{array}\right]+\left[\begin{array}{c}0\hfill \\ \multicolumn{1}{c}{H_2}\end{array}\right]+l_{s3}\left[\begin{array}{c}\cos {\theta }_3\hfill \\ \multicolumn{1}{c}{\sin {\theta }_3}\end{array}\right]$$

(13)

$$\left[\begin{array}{c}{x}_{s4}\hfill \\ \multicolumn{1}{c}{y_{s4}}\end{array}\right]=\left[\begin{array}{c}{l}_1+l_2\hfill \\ \multicolumn{1}{c}{0}\end{array}\right]+\left[\begin{array}{c}0\hfill \\ \multicolumn{1}{c}{H_3}\end{array}\right]+l_3\left[\begin{array}{c}\cos {\theta }_2\hfill \\ \multicolumn{1}{c}{\sin {\theta }_2}\end{array}\right]+l_{s4}\left[\begin{array}{c}\cos {\theta }_4\hfill \\ \multicolumn{1}{c}{\sin {\theta }_4}\end{array}\right]$$

(14)

$$\left[\begin{array}{c}{x}_{s5}\hfill \\ \multicolumn{1}{c}{y_{s5}}\end{array}\right]=\left[\begin{array}{c}{l}_1\hfill \\ \multicolumn{1}{c}{0}\end{array}\right]+\left[\begin{array}{c}0\hfill \\ \multicolumn{1}{c}{H_2}\end{array}\right]+l_4\left[\begin{array}{c}\cos {\theta }_3\hfill \\ \multicolumn{1}{c}{\sin {\theta }_3}\end{array}\right]+l_{s5}\left[\begin{array}{c}\cos {\theta }_0\hfill \\ \multicolumn{1}{c}{\sin {\theta }_0}\end{array}\right]$$

(15)

$$\left[\begin{array}{c}{x}_{s6}\hfill \\ \multicolumn{1}{c}{y_{s6}}\end{array}\right]=\left[\begin{array}{c}0\hfill \\ \multicolumn{1}{c}{H_1}\end{array}\right]+l_8\left[\begin{array}{c}\cos {\theta }_1\hfill \\ \multicolumn{1}{c}{\sin {\theta }_1}\end{array}\right]+\left[\begin{array}{c}0\hfill \\ \multicolumn{1}{c}{H_4}\end{array}\right]+l_{s6}\left[\begin{array}{c}\cos {\theta }_5\hfill \\ \multicolumn{1}{c}{\sin {\theta }_5}\end{array}\right]$$

(16)

where s_i is centroid of each component. By deriving the above formula and substituting the parameters, we can get

$$\left[\begin{array}{c}{\stackrel{·}{x}}_{s1}\hfill \\ \multicolumn{1}{c}{{\stackrel{·}{y}}_{s1}}\end{array}\right]=\left[\begin{array}{c}-1.768{\stackrel{·}{l}}_4+0.967{\stackrel{·}{l}}_8\hfill \\ \multicolumn{1}{c}{0.018{\stackrel{·}{l}}_4-0.010{\stackrel{·}{l}}_8}\end{array}\right]$$

(17)

$$\left[\begin{array}{c}{\stackrel{·}{x}}_{s2}\hfill \\ \multicolumn{1}{c}{{\stackrel{·}{y}}_{s2}}\end{array}\right]=\left[\begin{array}{c}-2.644{\stackrel{·}{l}}_4\hfill \\ \multicolumn{1}{c}{1.533{\stackrel{·}{l}}_4}\end{array}\right]$$

(18)

$$\left[\begin{array}{c}{\stackrel{·}{x}}_{s3}\hfill \\ \multicolumn{1}{c}{{\stackrel{·}{y}}_{s3}}\end{array}\right]=\left[\begin{array}{c}-2.636{\stackrel{·}{l}}_4\hfill \\ \multicolumn{1}{c}{-1.150{\stackrel{·}{l}}_4}\end{array}\right]$$

(19)

$$\left[\begin{array}{c}{\stackrel{·}{x}}_{s4}\hfill \\ \multicolumn{1}{c}{{\stackrel{·}{y}}_{s4}}\end{array}\right]=\left[\begin{array}{c}-4.202{\stackrel{·}{l}}_4+0.794{\stackrel{·}{l}}_8\hfill \\ \multicolumn{1}{c}{3.553{\stackrel{·}{l}}_4+0.458{\stackrel{·}{l}}_8}\end{array}\right]$$

(20)

$$\left[\begin{array}{c}{\stackrel{·}{x}}_{s5}\hfill \\ \multicolumn{1}{c}{{\stackrel{·}{y}}_{s5}}\end{array}\right]=\left[\begin{array}{c}-5.167{\stackrel{·}{l}}_4\hfill \\ \multicolumn{1}{c}{-1.163{\stackrel{·}{l}}_4}\end{array}\right]$$

(21)

$$\left[\begin{array}{c}{\stackrel{·}{x}}_{s6}\hfill \\ \multicolumn{1}{c}{{\stackrel{·}{y}}_{s6}}\end{array}\right]=\left[\begin{array}{c}-3.097{\stackrel{·}{l}}_4+1.704{\stackrel{·}{l}}_8\hfill \\ \multicolumn{1}{c}{0.031{\stackrel{·}{l}}_4+1.017{\stackrel{·}{l}}_8}\end{array}\right]$$

(22)

Now, the kinetic energy is

$$E=\frac{1}{2}\sum _{i=1}^6[m_i({\stackrel{·}{x}}_{\mathrm{si}}^2+{\stackrel{·}{y}}_{\mathrm{si}}^2)+J_i{\stackrel{·}{\theta }}_i^2]$$

(23)

Substituting equations (17)–(22) into equation (23), one can obtain

$$E=a_{11}{\stackrel{·}{l}}_4^2+a_{12}{\stackrel{·}{l}}_4{\stackrel{·}{l}}_8+a_{22}{\stackrel{·}{l}}_8^2$$

(24)

where $a_{11}=95760.3$ , $a_{12}=6569.07$ , and $a_{22}=2979.79$ .

System potential energy

The potential energy is

$$\begin{array}{c}\multicolumn{1}{c}{V=\frac{1}{2}{k}_1(l_4-l_4^{*}{)}^2+\frac{1}{2}{k}_2(l_8-l_8^{*}{)}^2}\\ =\frac{1}{2}{k}_1(l_4-1.89{)}^2+\frac{1}{2}{k}_2(l_8-5.45{)}^2\hfill \end{array}$$

(25)

where $l_4^{*}$ and $l_8^{*}$ are the length of balance hydraulic cylinder and column hydraulic cylinder, respectively, at the equilibrium position.

Generalized force of the system

The point coordinates of the external excitation force are as follows

$$\begin{array}{c}\multicolumn{1}{c}{\left[\begin{array}{c}{x}_h\hfill \\ \multicolumn{1}{c}{y_h}\end{array}\right]=\left[\begin{array}{c}0\hfill \\ \multicolumn{1}{c}{H_1}\end{array}\right]+l_8\left[\begin{array}{c}\cos {\theta }_1\hfill \\ \multicolumn{1}{c}{\sin {\theta }_1}\end{array}\right]}\\ +H_4\left[\begin{array}{c}\cos ({\theta }_5+\frac{\pi }{2})\hfill \\ \multicolumn{1}{c}{\sin ({\theta }_5+\frac{\pi }{2})}\end{array}\right]+h\left[\begin{array}{c}\cos {\theta }_5\hfill \\ \multicolumn{1}{c}{\sin {\theta }_5}\end{array}\right]\hfill \end{array}$$

(26)

Derivation of equation (26) and substitution of parameters can be obtained

$$\left[\begin{array}{c}{\stackrel{·}{x}}_h\hfill \\ \multicolumn{1}{c}{{\stackrel{·}{y}}_h}\end{array}\right]=\left[\begin{array}{cc}-3.096\hfill & 1.704\hfill \\ \multicolumn{1}{c}{0.031}& 0.982\hfill \end{array}\right]\left[\begin{array}{c}{\stackrel{·}{l}}_4\hfill \\ \multicolumn{1}{c}{{\stackrel{·}{l}}_8}\end{array}\right]$$

(27)

The instantaneous power of the generalized force is as follows

$$\begin{array}{c}\multicolumn{1}{c}{N=f{y}_h-{\stackrel{·}{l}}_4^2{C}_1(l_4)-{\stackrel{·}{l}}_8^2{C}_2(l_8)}\\ =(A_{21}f-C_1(l_4){\stackrel{·}{l}}_4){\stackrel{·}{l}}_4+(A_{22}f-C_2(l_8){\stackrel{·}{l}}_8){\stackrel{·}{l}}_8\hfill \\ =Q_1{\stackrel{·}{l}}_4+Q_2{\stackrel{·}{l}}_8\hfill \end{array}$$

(28)

where f is the magnitude of external load, and h is the distance from the external load to column nest. $Q_1=A_{21}f-C_1(l_4){\stackrel{·}{l}}_4$ , $Q_2=A_{22}f-C_2(l_8){\stackrel{·}{l}}_8$ , $C_1(l_4)$ and $C_2(l_8)$ are respectively damping of the balance hydraulic cylinder and column hydraulic cylinder. In general, the damping effect of the hydraulic cylinder is small when it is subjected to impact. Therefore, $Q_1=0.031f$ , $Q_2=0.983f$ . The peak value of external load acting on the column nest is 6000 N, and the bandwidth is 0.03 s. ¹⁷

Simulation solutions

According to D’Alembert’s principle, one can obtain

$$\frac{d}{\mathrm{dt}}\left(\frac{\partial E}{\partial {\stackrel{·}{l}}_i}\right)-\frac{\partial E}{\partial {\stackrel{·}{l}}_i}+\frac{\partial V}{\partial {\stackrel{·}{l}}_i}=Q_i$$

(29)

Substituting equations (25) and (27) into equation (29), one can obtain

$$2a_{11}{\stackrel{··}{l}}_4+a_{12}{\stackrel{··}{l}}_8+k_1(l_4-1.89)=A_{21}f-C_1(l_4){\stackrel{·}{l}}_4$$

(30)

$$2a_{22}{\stackrel{··}{l}}_8+a_{12}{\stackrel{··}{l}}_4+k_2(l_8-5.45)=A_{22}f-C_2(l_8){\stackrel{·}{l}}_8$$

(31)

Substituting the parameters into the above equations

$$1.915\times {10}^5{\stackrel{··}{l}}_4+6.569{\stackrel{··}{l}}_8+7\times {10}^7{l}_4=1.32\times {10}^8$$

(32)

$$5.959\times {10}^3{\stackrel{··}{l}}_8+6.569\times {10}^3{\stackrel{··}{l}}_4+7\times {10}^7{l}_8=3.87\times {10}^8$$

(33)

Combining equations (32) and (33)

$$\{\begin{array}{c}{\stackrel{··}{l}}_4=-0.034l_8-365.49l_4+691.77\hfill \\ \multicolumn{1}{c}{{\stackrel{··}{l}}_8=418.39l_4-1.22\times {10}^4{l}_8+3.83\times {10}^5}\end{array}$$

(34)

In MATLAB, equation (34) is solved using the fourth-order Runge–Kutta method. In the process of solving, auxiliary variables are introduced. $y_1=l_4$ , $y_2=d{l}_4/\mathrm{dt}$ , $y_3=l_8$ , $y_4=d{l}_8/\mathrm{dt}$ . Now, equation (34) is transformed into a first-order set of equations

$$\{\begin{array}{c}\frac{d{y}_1}{\mathrm{dt}}=y_2\hfill \\ \frac{d{y}_2}{\mathrm{dt}}=0.042y_3-378.64y_1-1592.99\hfill \\ \frac{d{y}_3}{\mathrm{dt}}=y_4\hfill \\ \frac{d{y}_4}{\mathrm{dt}}=418.39y_1-1.22\times {10}^4{y}_3+3.82\times {10}^5\hfill \end{array}$$

(35)

The displacement response curves of the balance hydraulic cylinder and column hydraulic cylinder are shown in Figures 5 and 6. In Figures 5 and 6, the maximum displacements of balance hydraulic cylinder and column hydraulic cylinder are 6.07 and 5.45 mm, respectively. The frequency spectrograms are shown in Figures 7 and 8. From the diagram, the natural frequencies of the balance hydraulic cylinder and column hydraulic cylinder are 28 and 160 Hz, respectively. In the traditional hydraulic support model, the natural frequency of the balance hydraulic cylinder is 10 Hz and the frequency of the column hydraulic cylinder is 17 Hz, which greatly improves the safety performance of the hydraulic support.

Figure 5.

Displacement response of balance hydraulic cylinder.

Figure 6.

Displacement response of column hydraulic cylinder.

Figure 7.

Frequency spectrograms of balance hydraulic cylinder.

Figure 8.

Frequency spectrograms of column hydraulic cylinder.

Conclusion

A novel hydraulic support with a configuration of double parallelogram structure is proposed in this article. The new design can decouple the support movement in X and Y directions. That allows the top beam to move along X or Y direction individually. On the contrary, the contact angle between the roof and the jack is determined by the coupling relationship of the traditional support structure. Thus, it is difficult to realize fully contact. Under unstructured roof conditions, the traditional support can only contact the roof fully in one state, while the new design can contact the roof fully in several states due to the separate movement along X and Y directions.

The relationship equation between the generalized coordinates of the hydraulic support and the position parameters of the mechanism is established using the kinematic vector closed-loop modeling method and D’Alembert’s principle. And the analytical model of the plural-freedom hydraulic support dynamics is carried out. The following results are obtained:

1. The maximum displacement response of the balance hydraulic cylinder is 6.07 mm, and the maximum displacement response of the column hydraulic cylinder is 5.45 mm.

2. The natural frequency of the balanced hydraulic cylinder and column hydraulic cylinder is 28 and 160 Hz, respectively. Compared with the 10 Hz of the balance hydraulic cylinder and 17 Hz of the column hydraulic cylinder in the traditional hydraulic support, the natural frequency is greatly increased, and the safety performance of the hydraulic support is greatly improved.