Fuzzy Reliability Analysis and Optimization Design of Hydraulic System of Bale Carrier for Tie Arm Device

Figure 1. Bale tie arm device

1. Square steel frame; 2. Slide rail block; 3. Lower swing arm; 4. Transverse thrust bar; 5. Drive shaft; 6. Lower shift bar; 7. Seat bearing; 8. Movable splint; 9. Drive shaft; 10. Upper arm rod; 11. Hydraulic cylinder; 12. Upshift rod; 13. Hexagonal nut; 14. Spacer; 15. Bolt; 16. Slide block seat

3. Establishment of Hydraulic System Model of Tie Arm Device

3.1. Establishment of Hydraulic Model for Tie Arm Device

To achieve the tie arm device in different positions of the lock, a two-way hydraulic lock is utilized. The bi-directional hydraulic lock shares a valve body and a control spool, the pilot-operated check valve is located on both sides, and the control spool is located in the middle. As shown in Figure 2, the two-way hydraulic lock works in the following manner: when the pressure oil from the A cavity goes to the left of the one-way valve into the A1 cavity and then through the control valve and right one-way valve, the original package and B1 cavity of the hydraulic oil are discharged through the B cavity. When the A and B cavities have no pressure oil, the conical valve conical valve and valve body close contact make the A1 and B1 cavities seal, and the implementing element is the bi-directional lock [12].

Figure 2.

Figure 2. Bi-directional hydraulic lock schematic diagram

In this system, we use position feedback to control the tie arm device in different positions of the lock. The position sensor is used to change the position signal of the hydraulic cylinder. The position cycle is set by the position cycle sub-model. The working principle of the system is as follows: the specified position is compared with the feedback position of the sensor to generate the error value, and the error multiplied by the gain signal drives the servo valve to turn on or off the hydraulic oil supply of the actuator and change the oil feeding direction. The system can automatically adjust until the error is zero [13]. The system model is shown in Figure 3.

Figure 3.

Figure 3. Tie arm device hydraulic system mode l

3.2. Basic Parameters Set

The hydraulic system part of the important parameters set is shown in Table 1.

Table 1. Tie arm device subsystem parameters

Parameters	Unit	Value
Fuel flow	L/min	35
Safety valve opening pressure	MPa	10
Hydraulic oil temperature	Degc	40
Hydraulic oil density	kg/m³	850
Directional valve to open the current	Ma	200
Directional valve natural frequency	Hz	50
Gain signal 1 value		10
Gain signal 2 value		250
Control valve core displacement limit	M	±0.0045
Control spool damping	N/m/s	1000
Conical valve taper	Degree	60
One-way valve core quality	kg	0.068
Check valve limit displacement	M	±0.00628
One-way valve core spring stiffness	N/mm	130
Horizontal push cylinder piston rod diameter	mm	56
Horizontal push hydraulic cylinder diameter	mm	80
Displacement sensor output signal gain	1/m	10

3.3. System Model Verification

In this paper, the simulation time is 12 seconds, the simulation step is 0.01 seconds, and the simulation results are shown in Figure 4.

Figure 4.

Figure 4. Tie arm device subsystem simulation

As can be seen from Figure 4(a), the piston rod accelerates from 0.2m/s in 1 ~ 2 seconds, stays steady at 2 ~ 5 seconds, decelerates to 0m/s in 5 ~ 6 seconds, and then the speed change of the piston rod retracts in 6 ~ 12 seconds. The aforementioned process is similar to that of the extension; From Figure 4(b), the piston rod extending 0.8m meets the actual situation. From Figure 4(c), the acceleration fluctuations greatly in the 1st, 5th, 6th, and 10th seconds. The piston rods either start from 0m/s or slow down at a steady speed, so the image is also conformity. From Figure 4(d), the inlet flow curve of the rodless cavity can match Figure 4(a) closely, so the hydraulic subsystem model is generally correct.

4. Hengshui Device Hydraulic System Optimization

4.1. Control Parameter Optimization

In AMESim, the dynamic characteristics of the three bit four way hydraulic servo valve spool are represented by a two order oscillatory system. Based on the automatic control principle, in the closed loop control system composed of the two order systems, the amplification factor of the preamplifier of the two order system has a great impact on the dynamic performance of the system. In this hydraulic subsystem, the preamplifier is represented by the gain 4, and the value of the gain 4 is adjusted repeatedly. The displacements of actual output and the piston rod of the hydraulic cylinder are observed. Given the difference between the expected value and the actual value, it can be obtained that the smaller the difference, the better the performance. The optimal parameter [14], which makes the system performance the best state, is found. Here, a traditional method is used to set the value of gain 4 to 200, 300, 400, 500, 600, and 700. The simulation length is set to 12 seconds, the simulation step is 0.05, and the batch mode is entered.

The desired displacement is calculated with the result of the piston rod displacement obtained from the simulation. The results are shown in Figure 5.

Figure 5.

Figure 5. The difference between piston rod displacement and expected value under different K4 values

It can be observed from Figure 5(a) that when the K4 value is equal to 200, the dynamic error of the system is close to 0.06m, and it takes about 1s for the system to reach a stable state. When the K4 value is equal to 700, the dynamic error of the system is close to 0.02m, and the system only takes about 0.3s to reach the steady state. Based on Figures 5(a), 5(b), and 5(c), it can be concluded that when K4 takes a greater value, the response of the system is faster and the dynamic error of the system is smaller.

As K4 takes a bigger value, however, it will cause overshoot of the following curve. The system will obviously oscillate and affect its own stability. We combine the curves under these different values and then perform local amplification, and the obvious difference is shown in Figure 6.

Figure 6.

Figure 6. Local magnification of piston rod displacement and expected value difference under different K4 values

From Figure 6, we can see that when the K4 value is 500, the system starts to oscillate within 5 ~ 6 seconds. When the K4 takes the value of 700, the oscillation of the system within 5 ~ 6 seconds is very obvious. With respect to both system error and stability, the value of gain 4 is repeated, and the optimum value range of K4 is 440 ~ 480.

4.2. Optimization of Structural Parameters

Several key structural parameters of the bi-directional hydraulic lock (i.e., control valve core mass, one-way valve core mass, control valve core damping, and one-way valve core stiffness) are set and simulated. Compared with the control valve core and one-way valve core of the system, the control valve damping value and the size of the one-way valve stiffness of the entire system has a greater impact. Accordingly, the aforementioned two parameters are studied [15].

4.2.1 Control Spool Damping Optimization

By changing the damping value of the control valve core, the curve of piston rod extension time under different damping values can be simulated. The results are shown in Figure 7. The average pressure curve and flow curve of the hydraulic cylinder under different damping of the control valve core are shown in Figures 8 and 9, respectively.

Figure 7.

Figure 7. Different control spool damping hydraulic cylinder piston rod out of time

Figure 8.

Figure 8. Different valve cylinder damping under the average working pressure cylinder

Figure 9.

Figure 9. Different control spool damping cylinder average working flow

As can be seen from Figure 7, as the control spool damping increases, the hydraulic cylinder piston rod extension time becomes significantly larger and grows linearly, increasing from the initial 5.35 seconds to 8.72 seconds.

It can be seen from Figures 8 and 9 that the damping of the control valve core is between 1000 and 7000N $∙$ s/m, and the average working pressure of the hydraulic cylinder does not change much and stays at 5.9MPa with little fluctuation. The hydraulic cylinder average flow decreases approximately from 9.2L/min to 7L/min, which indicates that the increase in control valve damping results in the reduced speed of the hydraulic cylinder piston rod. The reason behind this is that the flow changes more smoothly, so the stability of the hydraulic cylinder piston rod speed is better.

After comprehensive analysis based on Figures 7-9, it can be obtained that when the control valve spool damping values stay within the range of 3000 ~ 4000N·s/m, this can not only guarantee the piston rod out timely, but also maintain the stability of pressure as well as speed.

4.2.2. One-Way Valve Core Spring Stiffness Optimization

By changing the stiffness of the unidirectional valve core, the curve of the piston cylinder rod extension time under different unidirectional valve core stiffness can be simulated, as shown in Figure 10. The average pressure curve and average flow curve of the cylinder under different unidirectional valve core stiffness are shown in Figures 11 and 12, respectively.

Figure 10.

Figure 10. Piston rod extension time with different spring stiffness of check valve

From Figure 10, we can see that when the stiffness of the one-way valve core is below 150N/mm, the extension time of the piston rod of the hydraulic cylinder basically remains at 5.3 seconds. When the stiffness is greater than 150N/mm, the piston rod extension time is obviously longer. In addition, when the stiffness value is 400N/mm, the piston extension time reaches 10.79 seconds. This indicates that the stiffness of the one-way valve spring has a great influence on the performance of the hydraulic cylinder.

It can be seen from Figure 11 that the mean working pressure increases when the spring rate is less than 150N/mm. When the spring rate is greater than 150N/mm, the average working pressure also increases. As can be seen from Figure 12, the flow rate of the hydraulic cylinder changes drastically with an increase in spring stiffness. Spring stiffness is in the range of 50 ~ 400N/mm, and the cylinder flow quickly drops from 10.1L/min to 6.9L/min. However, the cylinder flow changes more smoothly, which indicates that the piston rod protruding speed is relatively stable [16].

Figure 11.

Figure 11. Average working pressure of hydraulic cylinder under different unidirectional valve core spring stiffness

Figure 12.

Figure 12. Average working flow of hydraulic cylinder under different unidirectional valve core spring stiffness

When the value of the check valve spring stiffness is in the range of 50 ~ 400N/mm, select the range of one-way valve core stiffness as 100 ~ 150N/mm. This can ensure that the piston rod extends in time, and it can also better ensure the pressure stability and speed stability.

5. Fuzzy FMECA Method

The fuzzy FMECA method combines the fuzzy comprehensive evaluation method and the FMECA method. The danger of the failure can be clearly manifested to make the traditional FMECA method more scientific [17].

The basic steps of the fuzzy FMECA method are as follows:

(1) Define a set of factors

The set of factors affect the object of evaluation, and different elements represent different influencing factors. Factor sets are usually represented by U, that is,

(1)

U = \{u_{1}, u_{2}, \dots, u_{i}, \dots, u_{n}\}

Where $u_{i}$ indicates the i influence factor, $i = 1,2, \dots, n$ .

(2) Set up the evaluation set

The evaluation set is usually indicated by V:

(2)

V = \{v_{1}, v_{2}, \dots, v_{j}, \dots, v_{m}\}

Where $v_{j}$ indicates the j rating, $j = 1,2, \dots, m$ .

(3) Establish the fuzzy factor evaluation matrix

A fuzzy mapping f for U to V is given, $f : U \to F (V), u_{i} \to f (u_{i})$ , and $f (u_{i}) = r_{i} = (r_{i 1}, r_{i 2}, \dots, r_{ij}, \dots, r_{im})$ , the fuzzy vector of evaluating $u_{i}$ , $r_{ij}$ is the membership of the factor i factor $u_{i}$ at factor level $v_{j}$ .

This article uses statistical methods to determine membership, an expert evaluation team composed of h persons is established, each member evaluates one and only one evaluation grade $v_{j}$ for each influencing factor $u_{i}$ of a certain failure mode, and if the $h_{ij}$ member in the h team member evaluating $u_{i}$ belongs to $v_{j}$ , then the $u_{i}$ ’s evaluation set is obtained [18].

The n failure modes of each factor evaluation set written by the fuzzy factor level evaluation matrix R are:

(3)

R = {(R_{1}, R_{2}, \dots, R_{n})}^{T} = [\begin{matrix} r_{11} & r_{12} & \dots & r_{1 m} \\ r_{21} & r_{22} & \dots & r_{2 m} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ r_{n 1} & r_{n 2} & \dots & r_{nm} \end{matrix}]

(4) Determine the weight of each factor set

Because each factor has a different influence on the failure, it is necessary to give the importance of each factor in the total evaluation during the process of comprehensive evaluation. The steps of using the analytic hierarchy process are described as follows:

$a_{ij}$ is used to represent the relative importance value of factor $u_{i}$ to $u_{j}$ , and the standard reference Table 2 is taken to determine the matrix A [19]:

(4)

A = [\begin{matrix} a_{11} & a_{12} & \dots & a_{1 m} \\ a_{21} & a_{22} & \dots & a_{2 m} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ a_{n 1} & a_{n 2} & \dots & a_{nm} \end{matrix}]

Table 2. AHP judgment matrix scale and its meaning

$a_{ij}$	Meaning
1	$u_{i} is as important as u_{j}$
3	$u_{i} is slightly more important than u_{j}$
5	$u_{i} is obviously more important than u_{j}$
7	$u_{i} is more important than u_{j}$
9	$u_{i} is more important than u_{j}$
one of 2, 4, 6, 8	$T he importance of u_{i} to u_{j} is between the$ $equivalents$

According to the judgment matrix A, it calculates its maximum eigenvalue $λ_{\max}$ and its corresponding eigenvector $ξ = (x_{1}, x_{2}, \dots, x_{n})$ . After normalization of the eigenvector, it can be used as the weight set W $= {(w_{1}, w_{2}, \dots, w_{n})}^{T}$ . Before normalization, the consistency test of the judgment matrix A must be performed to determine whether the weight distribution is reasonable. The consistency test formula is shown as

(5)

R_{C} = \frac{I_{C}}{I_{R}}

(6)

I_{C} = \frac{λ_{\max} - n}{n - 1}

Where $R_{C}$ indicates the ratio of random consistency of the judgment matrix; $I_{C}$ indicates the random consistency index of the judgment matrix; $I_{R}$ is the average random consistency of the matrix indicators. Table 3 can be obtained:

Table 3. $I_{R}$ values of 1-13 order judgement matrix

n	$I_{R}$	n	$I_{R}$
1	0.00	8	1.41
2	0.00	9	1.45
3	0.58	10	1.49
4	0.90	11	1.52
5	1.12	12	1.54
6	1.24	13	1.56
7	1.32

When $R_{C} < 0.1$ , the judgment matrix A has a satisfactory consistency; otherwise, the elements in A should be modified to meet the requirements.

(5) A fuzzy comprehensive evaluation

Rewrite the set of failure mode factor weights as a vector mode.

(7)

B = W ∙ R = [w_{1}, w_{2}, \dots, w_{n}] ∙ [\begin{matrix} r_{11} & r_{12} & \dots & r_{1 m} \\ r_{21} & r_{22} & \dots & r_{2 m} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ r_{n 1} & r_{n 2} & \dots & r_{nm} \end{matrix}] = (b_{1}, b_{2}, \dots, b_{n})

Where $B$ is the fuzzy comprehensive evaluation vector to determine the failure mode.

(6) Determine the level of comprehensive hazard

In order to interpret the result more intuitively, B is processed by the weighted average method to obtain a simple numerical value C to represent the evaluation result, that is:

(8)

C = B ∙ V

(7) Multi-level fuzzy comprehensive evaluationThe actual complex system consists of multiple subsystems. When we use multi-level fuzzy comprehensive evaluation, the first level fuzzy comprehensive evaluation of each sub-system failure mode is carried out, and the fuzzy comprehensive evaluation vector and comprehensive hazard level score are obtained. Then, system failure modes as a second-level fuzzy comprehensive evaluation of the factors are expressed asU = {fault mode 1, fault mode 2, $\dots$ , fault mode k}The level of factor V is the same, and then the weight set of each influencing factor is calculated by AHP. The comprehensive evaluation of the system is performed using the fuzzy comprehensive evaluation method. Through this method, we can obtain the second-level fuzzy comprehensive evaluation, and then the multi-level fuzzy evaluation of the system can be obtained.

6. The Cross-Pushing Device Hydraulic System Fuzzy FMECA Analysis

The design of the hydraulic control system is analyzed, combined with relevant information, servo valves, hydraulic cylinders, check valves, reversing valves, and piping in a liquid-controlled system. As an example, the fuzzy FMECA method is used to compute its reliability [20]. The FMECA tables of the five components are shown in Table 4.

Table 4. Hydraulic control system five parts FMECA table

Product name	Sequence Number	Failure mode	Cause of issue	Local effects	High level impact	The final impact	Severity level	Take measures
Servo	1	Slow response, slow response	Reduced oil pressure	Cylinder instability	Corresponding push, dial action instability	Tie arm device stacking quality decline	Ⅲ level	Increase the pressure of oil supply
Servo	2	Vibration appears	The system gain is too high	Cylinder instability	Corresponding push, dial action instability	Tie arm device stacking quality decline	Ⅱ level	Reduce system gain
Hydraulic Cylinder	3	Leakage	Seal failure	Cylinder lack of thrust	Corresponding to push, dial performance decline	Tie arm device stacking quality decline	Ⅲ level	Replace the seal
	4	Internal leakage	Seals wear	Cylinder instability	Corresponding to push, dial performance decline	Tie arm device stacking quality decline	Ⅲ level	Replace the seal
	5	Crawl	Hydraulic cylinder accumulates air	Cylinder instability	Hydraulic cylinder life is reduced	Tie arm device stacking quality decline	Ⅲ level	Exhaust the air
	6	Lack of thrust	Frictional resistance is too large	Reduced effective traction cylinder	Corresponding push, dial action instability	Tie arm device stacking quality decline	Ⅲ level	Adjust the cylinder and piston with the gap
Check valve	7	Oil is not countercurrent	Check valve stuck	Valve does not work	Corresponding push, dial no corresponding action	Tie arm device to stop working	Ⅱ level	Demolition, cleaning valve
Directional valve	8	Commutation is not reliable	Spring deformation fracture	Cylinder instability	Corresponding push, dial action instability	Tie arm device stacking quality decline	Ⅱ level	Adjust the gap
Directional valve	9	Commutation moves slowly	Poor lubrication	Change the valve function	Corresponding push, dial action instability	Tie arm device stacking quality decline	Ⅲ level	Adjust the viscosity of the lubricant
Pipeline	10	Pipeline blocked	Oil pollution	Inadequate supply of oil	Cylinder instability	Tie arm device stacking quality decline	Ⅳ level	Filter oil impurities

The steps of the fuzzy FMECA analysis of hydraulic system are as follows:

(1) Set up a set of factors

For the failure of the hydraulic system for hazard assessment, the set of factors used are as follows:

U = {probability of failure, severity, ease of detection, ease of maintenance}

(2) Set up an evaluation set

The factors are divided into four levels: V = {1, 2, 3, 4}. The different factors for the level of division are shown in Table 5.

Table 5. Factor ranking table

Influencing factors	Grade
Influencing factors	1	2	3	4
Failure probability level ( $u_{1}$ )	rarely happens	occasionally	occasionally happens	happens very often
Severity level ( $u_{2}$ )	slight influence	medium effect	tremendous influence	deadly impact
Detection level of difficulty ( $u_{3}$ )	accurate detection	not easy to detect	hard to detect	unable to detect
Maintenance level of difficulty ( $u_{4}$ )	simple debugging	re-install	replace parts	unable to repair

(3) Establish fuzzy evaluation matrix of failure mode

Ten experts proficient in the hydraulic system send the questionnaire. We have $R_{1}^{1} = \{0.1, 0.5, 0.4, 0\}$ , $R_{2}^{1} = \{0.6, 0.4, 0, 0\}$ , detection of the level of difficulty level set $R_{3}^{1} = \{0.2, 0.8, 0, 0\}$ , and maintenance level of difficulty level set $R_{4}^{1} = \{0.9, 0.1, 0, 0\}$ . Therefore, the fuzzy factor evaluation matrix is

$R^{1} = {(R_{1}^{1}, R_{2}^{1}, R_{3}^{1}, R_{4}^{1})}^{T} = [\begin{matrix} 0.1 & 0.5 & 0.4 & 0 \\ 0.6 & 0.4 & 0 & 0 \\ 0.2 & 0.8 & 0 & 0 \\ 0.9 & 0.1 & 0 & 0 \end{matrix}]$

Similarly, we can get the fuzzy factor evaluation matrix of the remaining nine failure modes as

$R^{2} = [\begin{matrix} 0.2 & 0.7 & 0.1 & 0 \\ 0 & 0.6 & 0.4 & 0 \\ 0 & 0.8 & 0.2 & 0 \\ 0 & 0.5 & 0.5 & 0 \end{matrix}]$

$R^{3} = [\begin{matrix} 0.2 & 0.5 & 0.3 & 0 \\ 0.7 & 0.3 & 0 & 0 \\ 0.6 & 0.4 & 0 & 0 \\ 0 & 0.2 & 0.6 & 0.2 \end{matrix}]$

$R^{4} = [\begin{matrix} 0.1 & 0.5 & 0.4 & 0 \\ 0.6 & 0.4 & 0 & 0 \\ 0 & 0.8 & 0.2 & 0 \\ 0 & 0.3 & 0.4 & 0.3 \end{matrix}]$

$R^{5} = [\begin{matrix} 0.2 & 0.6 & 0.2 & 0 \\ 0 & 0.5 & 0.5 & 0 \\ 0 & 0.9 & 0.1 & 0 \\ 0.7 & 0.3 & 0 & 0 \end{matrix}]$

$R^{6} = [\begin{matrix} 0.1 & 0.7 & 0.2 & 0 \\ 0.6 & 0.4 & 0 & 0 \\ 0 & 0.6 & 0.4 & 0 \\ 0.4 & 0.6 & 0 & 0 \end{matrix}]$

$R^{7} = [\begin{matrix} 0.2 & 0.4 & 0.4 & 0 \\ 0 & 0.4 & 0.6 & 0 \\ 0.2 & 0.5 & 0.3 & 0 \\ 0.4 & 0.4 & 0.2 & 0 \end{matrix}]$

$R^{8} = [\begin{matrix} 0.3 & 0.3 & 0.4 & 0 \\ 0 & 0.8 & 0.2 & 0 \\ 0.2 & 0.6 & 0.2 & 0 \\ 0.1 & 0.8 & 0.2 & 0 \end{matrix}]$

$R^{9} = [\begin{matrix} 0.2 & 0.2 & 0.6 & 0 \\ 0 & 0.4 & 0.6 & 0 \\ 0 & 0.5 & 0.5 & 0 \\ 0.1 & 0.8 & 0.1 & 0 \end{matrix}]$

$R^{10} = [\begin{matrix} 0 & 0.1 & 0.4 & 0.5 \\ 0.6 & 0.4 & 0 & 0 \\ 0.3 & 0.5 & 0.2 & 0 \\ 0.9 & 0.1 & 0 & 0 \end{matrix}]$

(4) Determine the weight set

Firstly, the factor weight set of failure mode 1 is solved. According to Tables 2 and 4, the relative importance of the four influencing factors of failure mode 1 are determined. The data are shown in Table 6.

Table 6. Failure mode 1 the relative importance and weight of each influencing factor

Influencing factors	$u_{1}$	$u_{2}$	$u_{3}$	$u_{4}$
$u_{1}$	1	3	7	5
$u_{2}$	1/3	1	5	3
$u_{3}$	1/7	1/5	1	1/3
$u_{4}$	1/5	1/3	3	1

Table 6 can be obtained by the failure mode 1 judgment matrix $A^{1}$ :

$A^{1} = [\begin{matrix} 1 & 3 & 7 & 5 \\ 1 / 3 & 1 & 5 & 3 \\ 1 / 7 & 1 / 5 & 1 & 1 / 3 \\ 1 / 5 & 1 / 3 & 3 & 1 \end{matrix}]$

From the MATLAB software, the maximum eigenvalue of the 4 $\times$ 4 judgment matrix $A^{1}$ is 4.1170, and the corresponding eigenvector is ξ = [0.8880, 0.4121, 0.0869, 0.1847]. Before the normalization, the judgment matrix A must be consistent. The sex test is to determine whether the distribution of weights is reasonable. If it is not, it is necessary to consult related experts again and combine relevant data to modify the relative importance of each influencing factor until the results meet the requirements.

Consistency test:

$I_{C} = \frac{λ_{\max} - n}{n - 1} = \frac{4.1170 - 4}{4 - 1} = 0.039$

Since the judgment matrix $A^{1}$ is 4th order, the value of $I_{R}$ that can be found in Table 3 is 0.90.

R_{C} = \frac{I_{C}}{I_{R}} = \frac{0.039}{0.90} = 0.0433

Since $R_{C}$ = 0.0433 < 0.1, the judgment matrix $A^{1}$ has a satisfactory consistency. Therefore, the normalized feature vector ξ = [0.8880, 0.4121, 0.0869, 0.1847] corresponding to the maximum eigenvalue 4.1170 can be used as a weight set W for each factor. As can be seen from above, the sum of the eigenvectors is not 1, so the eigenvectors are divided by the sum of these vectors, and the re-derived number is the weight vector. That is:

$W_{1} = \frac{ξ_{1}}{S} = \frac{0.8880}{(0.8880 + 0.4121 + 0.0869 + 0.1847)} = \frac{0.8880}{1.5717} = 0.5650$

$W_{2} = \frac{ξ_{2}}{S} = \frac{0.4121}{(0.8880 + 0.4121 + 0.0869 + 0.1847)} = \frac{0.4121}{1.5717} = 0.2622$

$W_{3} = \frac{ξ_{3}}{S} = \frac{0.0869}{(0.8880 + 0.4121 + 0.0869 + 0.1847)} = \frac{0.0869}{1.5717} = 0.0533$

$W_{4} = \frac{ξ_{4}}{S} = \frac{0.1847}{(0.8880 + 0.4121 + 0.0869 + 0.1847)} = \frac{0.1847}{1.5717} = 0.1175$

Thus, the factor weight set for failure mode 1 is

$W^{1} = [W_{1}, W_{2}, W_{3}, W_{4}] = [0.5650, 0.2622, 0.0553, 0.1175]$

The same weight set is used for the influence factors of each failure mode:

$W^{1} {= W}^{2} = W^{3} = W^{4} = W^{5} = W^{6} = W^{7} = W^{8} = W^{9} = W^{10}$

(5) A fuzzy comprehensive evaluation

The first-level fuzzy comprehensive evaluation of failure mode 1 can be obtained by Equation (7):

$B^{1} = W^{1} R^{1} = [0.3306, 0.4434, 0.2260, 0]$

Empathy:

$B^{2} = W^{2} R^{2} = [0.1130, 0.6558, 0.2312, 0]$

$B^{3} = W^{3} R^{3} = [0.3297, 0.4068, 0.2400, 0.0235]$

$B^{4} = W^{4} R^{4} = [0.2138, 0.4669, 0.2841, 0.0353]$

$B^{5} = W^{5} R^{5} = [0.1953, 0.5551, 0.2496, 0]$

$B^{6} = W^{6} R^{6} = [0.2608, 0.6041, 0.1351, 0]$

$B^{7} = W^{7} R^{7} = [0.1711, 0.4055, 0.4234, 0]$

$B^{8} = W^{8} R^{8} = [0.1923, 0.5064, 0.3130, 0]$

$B^{9} = W^{9} R^{9} = [0.1248, 0.3395, 0.5357, 0]$

$B^{10} = W^{10} R^{10} = [0.2797, 0.2008, 0.2371, 0.2825]$

(6) A comprehensive level of hazard

From Equation (8), the comprehensive hazard level of fault mode 1 is $C^{1} = B^{1} ∙ V^{T} = 1.8954$ . Similarly, the remaining nine kinds of failure modes of the comprehensive degree of risk are $C^{2} = 2.1182$ , $C^{3} = 1.9573$ , $C^{4} = 2.1411$ , $C^{5} = 2.0543$ , $C^{6} = 1.8743$ , $C^{7} = 2.2523$ , $C^{8} = 2.1441$ , $C^{9} = 2.4109$ , and $C^{10} = 2.5226$ .

According to the level of the integrated hazard level, the fault modes 1 to 10 may be ranked according to their degree of danger as follows [5]: fault mode 10 > fault mode 9 > fault mode 7 > fault mode 8 > fault mode 4 > fault mode 2 > fault mode 5 > fault mode 3 > fault mode 1 > fault mode 6.

(7) Two-level fuzzy comprehensive evaluation

For the servo valve, the set of factors $U_{1}^{'}$ = {failure mode 1, failure mode 2}, the evaluation set $V^{'}$ = {1, 2, 3, 4}, and the evaluation matrix $R_{1}^{'}$ = $[B^{1}, B^{2}]$ . According to the importance of these two failure modes, AHP [14] can be used to draw the weight set $W_{1}^{'} = \{0.167, 0.833\}$ , and the result of level 2 comprehensive evaluation is $B_{1}^{'} = W_{1}^{'} ∙ R_{1}^{'} = [0.1493, 0.6203, 0.2303, 0]$ , and the comprehensive hazard level is $C_{1}^{'} = B_{1}^{'} ∙ V^{T} = 2.0808$ . For the hydraulic cylinder, the comprehensive evaluation result and the comprehensive hazard degree can be respectively calculated as $B_{2}^{'} = [0.2838, 0.4539, 0.2398, 0.0225] and C_{2}^{'} = 2.001$ . For the check valve, the comprehensive evaluation result and comprehensive hazard degree can be respectively calculated as $B_{3}^{'} = [0.1711, 0.4055, 0.4234, 0] and C_{3}^{'} = 2.2523$ . For the directional control valve, the comprehensive evaluation result and the comprehensive hazard degree can be respectively calculated as $B_{4}^{'} = [0.1839, 0.4855, 0.3409, 0]$ and $C_{4}^{'} = 2.1776$ . For the pipeline, the comprehensive evaluation results and the comprehensive hazard degree can be respectively calculated as $B_{5}^{'} = [0.2797, 0.2008, 0.2371, 0.2825] and C_{5}^{'} = 2.5226$ .

According to the size of the integrated hazard, it can be seen that the arrangement of the five components with higher hydraulic system failure rate is: Piping > Check valve > Directional valve > Servo valve > Hydraulic cylinder. The most dangerous failure mode in the pipeline is the pipeline blockage. The servo valve in the most dangerous mode of failure is the emergence of vibration that is followed by the slow reaction. For the hydraulic cylinder, the failure mode of danger from large to small arrangement is: internal leakage, crawling, leakage, and lack of thrust.

The same method can be used to calculate the breakdown of the other components of the hydraulic system.

7. Conclusions

(1) Through the working principle of the bale pusher device, the model and parameters of the hydraulic system are built and simulated in the AMESim software. The simulation results show that the model is consistent with the actual situation.

(2) The combination of fuzzy theory and FMECA method overcomes the shortcomings of FMECA, quantifies the ambiguity index, and comprehensively evaluates these indicators using the quantitative calculation method, which can provide evaluation results that are more suitable for practical applications and reference for the further improvement of the reliability of the hydraulic system of the horizontal push device.

(3) By changing the control parameters and structural parameters, the structural parameters of the system are successfully optimized. The simulation results show that the optimal range of the preamplifier K4 is 440-480, the damping coefficient of the control valve core is 3000 ~ 4000N $∙$ s/m, and the stiffness of the one-way valve core is 100-150 N/mm.

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