Multi-Objective Optimization of a Two-Stage Helical Gearbox Using MARCOS Method

Abstract

In order to address the Multi-Objective Optimization Problem (MOOP) in building a two-stage helical gearbox, this work presents a novel application of the Multi-Criterion Decision-Making (MCDM) method. The aim of the study is to determine the optimal primary design factors that will increase gearbox efficiency while decreasing gearbox volume. Three main design parameters were chosen for assessment in this work: the first stage’s gear ratio, and the first and second stages’ Coefficients of Wheel Face Width (CWFW). In addition, the MOOP is divided into two phases: phase 1 solves the single-objective optimization problem to reduce the gap between variable levels, and phase 2 solves the MOOP to determine the optimal primary design factors. Furthermore, the Entropy approach was picked to compute the weight criteria, and the MARCOS method was chosen as an MCDM method to handle the multi-objective optimization issue. The following are important characteristics of the study: Firstly, the MCDM method (MARCOS technique) was successfully applied to solve a MOOP for the first time. Secondly, this work has looked into power losses during idle motion to calculate the efficiency of a two-stage helical gearbox. The results of the study were used in the design of a two-stage helical gearbox in order to identify the optimal values for three important design parameters.

1. Introduction

Industrial applications frequently utilize helical gearboxes because of their low cost, minimal complexity, and ease of design and manufacture. Therefore, many scientists have been interested in finding the best design for helical gearboxes.

Numerous studies on single- and multi-objective optimization for helical gearboxes have been conducted up to this point. The optimization problem for helical gearboxes has been approached for several single-objective functions, including minimum gear mass [1,2], minimum gear volume [2,3,4], minimum gearbox mass [1,5], minimum gearbox length [6,7], minimum gearbox across section area [8,9], minimum gearbox cost [10,11,12], and so on. The single-objective helical gearbox optimization problem has been solved by many different methods such as particle swarm optimization [4], Matlab optimization tool box [2], the direct search method [5,13,14] etc. Also, for helical gearboxes with several stages, such as one [4,14], two [6,9], three [12,15], and four stages [1,16], single-objective optimal design has also been addressed.

With various single objectives, the MOOP has been resolved. The lowest gearbox power loss overall and the lowest gearbox volume were examined in [17]. In order to lower both the gearing mass and the flank adhesive wear speed, the authors in [18] conducted optimization research. In [19], two single targets were selected: the maximum gear stress and the minimum gear mass. The multi-objective optimization solutions in [20] have enhanced the transmission error signal’s root mean square values as well as its maximal contact pressures. Implemented in [21] is the multi-objective optimization issue of choosing the best gear material for a helical gearbox to maximize surface fatigue and increase wear resistance. The optimal major design factor for maximizing gearbox efficiency and decreasing gearbox mass was identified in [13]. Besides this, MOOPs have also been solved using a variety of techniques, including the response surface methods [22], the PSO (particle swarm optimization) method [18], the NSGA-II method [17,18], the NSGA-II and the TOPSIS (Technique for Order of Preference by Similarity to Ideal Solution) methods [23], Grey Relation Analysis (GRA) and the Taguchi technique [13], etc. Recently, a genetic algorithm was used in [24] to solve a macro-geometry gear design optimization problem with two objective functions—gear mesh stiffness and dynamic behavior—in order to determine the optimal strategy for achieving a minimal amount of dynamic excitation. A computationally efficient and effective gear design optimization approach is provided based on the obtained results. A new framework for precise reliability analysis has been set out in [25]; it is predicated on employing metaheuristic algorithms to enhance the directional simulation. With the new improved version, the unit vector of direction is determined through the use of metaheuristic methods and expressed as a constrained optimization problem using the Harris Hawks Optimization technique. The suggested technique was compared to the performance of the first-order reliability method and six simulation-based reliability analysis methods. The outcomes demonstrate the enhanced directed simulation’s high performance capabilities in resolving extremely nonlinear engineering issues.

While helical gearbox multi-objective optimization has been extensively studied, the MCDM technique has not been used to find the optimal primary design parameters for these gearboxes. Moreover, the research indicated above did not account for the power loss that occurs when a gear is in an idle motion or when a gear is immersed in lubricant during bath lubrication. This paper presents the findings of a multi-objective optimization study conducted for a two-stage helical gearbox, with two specific objectives in mind: reducing gearbox volume and maximizing gearbox efficiency. The study looked at the first stage’s gear ratio and both stages’ CWFW as the three optimal primary design characteristics for the two-stage helical gearbox. Furthermore, the Entropy approach was utilized to determine the weights of the criteria, and the MARCOS method was chosen to handle the MOOP. One of the main conclusions of the research is the suggestion to apply an MCDM technique to tackle MOOPs in combination with two-step problem solving, tackling single- and multi-objective problems. In addition, the problem’s solutions are more effective than those of earlier studies. Moreover, the power losses incurred in idle motion have been added when calculating the efficiency of a two-stage helical gearbox.

2. Optimization Problem

In this part, the gearbox volume and efficiency are first calculated in order to build the optimization problem. Next, the specified objective functions and constraints are given. To facilitate calculations, the nomenclatures used in the optimization problem are presented in

Table 1. The nomenclatures for the optimization problem.

Parameters	Nomenclature	Units
Gearbox housing volume	Vgh	dm3
Gearbox width	B1	dm
Gearbox height	H	dm
Pitch diameter of the pinion of stage 1	dw11	mm
Pitch diameter of the gear of stage 2	dw21	mm
Pitch diameter of the pinion of stage 2	dw12	mm
Pitch diameter of the gear of stage 2	dw22	mm
Center distance of stage 1	aw1	mm
Center distance of stage 2	aw2	mm
Gear ratio of stage 1	u1	-
Gear ratio of stage 2	u2	-
Gearbox ratio	ugb	-
Gear width of stage 1	bw1	mm
Gear width of stage 2	bw2	mm
Wheel face width coefficient of stage 1	Xba1	-
Wheel face width coefficient of stage 2	Xba2	-
Material coefficient	ka	Mpa1/3
Allowable contact stress of stages 1	AS1	Mpa
Allowable contact stress of stages 2	AS2	Mpa
Contacting load ratio for pitting resistance	kHβ	-
Torque on the pinion of stage i	T1i	Nmm
Output torque	Tout	Nmm
Efficiency of a helical gear unit	ηhg	-
Efficiency of a rolling bearing pair	ηb	-
Length of shaft i	lsi	mm
Diameter of shaft i	dsi	mm
Allowable shear stress of shaft material	[τ]	MPa
Total power loss in the gearbox	Pl
Power loss in the gears	Plg	Kw
Power loss in the bearings	Plb	Kw
Power loss in the seals	Pls	Kw
Power loss in the idle motion	Pzo	Kw
Efficiency of a helical gearbox	ηhb	-
Efficiency of the i stage of the gearbox	ηgi	-
Friction coefficient	f	-
Friction coefficient of bearing	fb	-
Arc of approach on i stage	${\beta }_{\mathrm{a}\mathrm{i}}$
Arc of recess on i stage	${\beta }_{\mathrm{r}\mathrm{i}}$
Outside radius of the pinion	$R_{e1i}$	mm
Outside radius of the gear	$R_{e2i}$	mm
Base-circle radius of the pinion	$R_{01i}$	mm
Base-circle radius of the gear	$R_{02i}$	mm
Pressure angle	α	rad.
Sliding velocity of gear	v	m/s
Peripheral speed of bearing	vb	m/s
Load of bearing i	Fi	N
ISO Viscosity Grades number	VG40
Hydraulic moment of power losses	TH	Nm

.

2.1. Calculation of Gearbox Volume

The volume of the gearbox $V_{gb}$ can be calculated by (Figure 1):

$$V_{gb}=L·B_1·H$$

(1)

where L, B₁, and H are determined by [26]:

$$L={(d}_{w11}+d_{w21}/2+d_{w12}/2+d_{w22}/2+22.5)/0.975$$

(2)

$$H=max\left(d_{w21};d_{w22}\right)+8.5·S_G$$

(3)

$$B_1=b_{w1}+b_{w2}+6·S_G$$

(4)

$$S_G=0.005·L+4.5$$

(5)

In the above equations,

$$d_{w1i}=2·a_{wi}/(u_i+1)$$

(6)

$$d_{w2i}=2·a_{wi}·u_i·/(u_i+1)$$

(7)

$$b_{w1}=X_{ba1}{·a}_{w1}$$

(8)

$$b_{w2}=X_{ba2}{·a}_{w2}$$

(9)

In Equations (6) to (9), $a_{wi}$ is determined by [27]

$$a_{wi}=k_a·(u_i+1)·\sqrt[3]{T_{1i}·k_{H\beta }/({\left[{AS}_i\right]}^2·u_i·X_{bai})}$$

(10)

In which

$$T_{1i}=\frac{T_r}{\prod _{j=i}^3{(u}_i·{\eta }_{hg}^{3-i}·{\eta }_{be}^{4-i})}$$

(11)

2.2. Calculation of Gearbox Efficiency

The efficiency of a two-stage helical gearbox (%) is determined by

$${\eta }_{gb}=100-\frac{100·P_l}{P_{in}}$$

(12)

In which P_l can be found by [28]

$$P_l=P_{lg}+P_{lb}+P_{ls}+P_{Z0}$$

(13)

where P_lg, P_lb, P_ls, and P_zo are determined in the following way.

Determination of P_lg:

$$P_{lg}={\sum }_{i=1}^2{P}_{lgi}$$

(14)

in which

$$P_{lgi}=P_{gi}·\left(1-{\eta }_{gi}\right)$$

(15)

${\eta }_{gi}$ can be calculated by [29]

$${\eta }_{gi}=1-\left(\frac{1+1/{\mathrm{u}}_{\mathrm{i}}}{{\mathsf{\beta }}_{\mathrm{a}\mathrm{i}}+{\mathsf{\beta }}_{\mathrm{r}\mathrm{i}}}\right)·\frac{{\mathrm{f}}_{\mathrm{i}}}{2}·\left({\mathsf{\beta }}_{\mathrm{a}\mathrm{i}}^2+{\mathsf{\beta }}_{\mathrm{r}\mathrm{i}}^2\right)$$

(16)

while ${\mathsf{\beta }}_{\mathrm{a}\mathrm{i}}$ and ${\mathsf{\beta }}_{\mathrm{r}\mathrm{i}}$ can be found by [29]

$${\mathsf{\beta }}_{\mathrm{a}\mathrm{i}}=\frac{{\left({\mathrm{R}}_{\mathrm{e}2\mathrm{i}}^2-{\mathrm{R}}_{02\mathrm{i}}^2\right)}^{1/2}-{\mathrm{R}}_{2\mathrm{i}}·\mathrm{s}\mathrm{i}\mathrm{n}\mathsf{\alpha }}{{\mathrm{R}}_{01\mathrm{i}}}$$

(17)

$${\mathsf{\beta }}_{\mathrm{r}\mathrm{i}}=\frac{{\left({\mathrm{R}}_{\mathrm{e}1\mathrm{i}}^2-{\mathrm{R}}_{01\mathrm{i}}^2\right)}^{1/2}-{\mathrm{R}}_{1\mathrm{i}}·\mathrm{s}\mathrm{i}\mathrm{n}\mathsf{\alpha }}{{\mathrm{R}}_{01\mathrm{i}}}$$

(18)

In (16), f is calculated in the following way [13]:

-

If v ≤ 0.424 (m/s),

$$\mathrm{f}=-0.0877·v+0.0525$$

(19)

-

If v > 0.424 (m/s),

$$\mathrm{f}=0.0028·v+0.0104$$

(20)

Determination of P_lb [28]:

$$P_{lb}={\sum }_{i=1}^6{f}_b·F_i·v_i$$

(21)

where i = 1 ÷ 6 and $f_b=0.0011$ as the radical ball bearings with angular contact were used [28].

Determination of P_s [28]:

$$P_{\mathrm{s}}={\sum }_{i=1}^2{P}_{si}$$

(22)

In which i is the ordinal number of the seal (i = 1 ÷ 2) and $P_{si}$ is calculated by

$$P_{si}=\left[145-1.6·t_{oil}+350·\mathit{log}\mathit{log}\left({VG}_{40}+0.8\right)\right]·d_s^2·n·{10}^{-7}$$

(23)

Determination of P_zo [28]:

$$P_{Z0}={\sum }_{i=1}^k{T}_{Hi}·\frac{\pi ·n_i}{30}$$

(24)

In which k is the total number of gear pairs in the gearbox (k = 2); n is the number of revolutions of the driven gear; $T_{Hi}$ can be found by [28]

$$T_{Hi}=C_{Spi}·C_{1i}·e^{\frac{C_{2i}·v}{v_{t0}}}$$

(25)

In (25), $C_{Spi}=1$ for stage 1 under the circumstance wherein the involved oil must pass the mesh, and for stage 2, $C_{Spi}$ is computed using the following equation (Figure 2):

$$C_{Spi}={\left(\frac{4·e_{max}}{3·h_C}\right)}^{1.5}·\frac{2·h_{Ci}}{l_h}$$

(26)

where l_hi is determined by [28]

$$l_{hi}=\left(1.2÷2.0\right)·{\mathrm{d}}_{\mathrm{a}2\mathrm{i}}$$

(27)

In (25), C₁ and C₂ are calculated by [28]

$$C_{1\mathrm{i}}=0.063·\left(\frac{{\mathrm{e}}_{1\mathrm{i}}+{\mathrm{e}}_{2\mathrm{i}}}{{\mathrm{e}}_0}\right)+0.0128·\left(\frac{{\mathrm{b}}_{\mathrm{i}}}{{\mathrm{b}}_0}\right)$$

(28)

$$C_{2\mathrm{i}}=\frac{{\mathrm{e}}_{1\mathrm{i}}+{\mathrm{e}}_{2\mathrm{i}}}{80·{\mathrm{e}}_0}+0.2$$

(29)

In which e₀ = b₀ = 10 (mm).

2.3. Multi-Objective Optimization Problem

In this work, the MOOP can be express as follows:

$$\min y=F\left(X\right)={\left(f_1\left(X\right),f_2\left(X\right)\right)}^T$$

(30)

Here, two single objectives compose the MOOP:

-

Minimizing the gearbox volume—

$$\mathrm{m}\mathrm{i}\mathrm{n}{f}_1\left(X\right)=V_{gb}$$

(31)

-

Maximizing the gearbox efficiency—

$$\mathrm{m}\mathrm{i}\mathrm{n}{f}_2\left(X\right)={\mathsf{\eta }}_{\mathrm{g}\mathrm{b}}$$

(32)

where the vector representing the design variables is denoted by X. There are five primary design parameters for a two-stage helical gearbox: u_1, X_ba1, X_ba2, AS₁, and AS₂ [13]. Furthermore, it was shown that AS₁ and AS₂’s maximum values correspond to their ideal values [13]. Thus, three primary design factors, u₁, X_ba1 and X_ba2, were chosen as variables for the optimization problem in this work. As a result, we have:

$$X=\left\{u_1,{Xba}_1,{Xba}_2\right\}$$

(33)

The constraints that follow must be met by the multi-objective function

$$1\le u_1\le 9 \mathrm{and} 1\le u_2\le 9$$

(34)

$$0.25\le {Xba}_1\le 0.4 \mathrm{and} 0.25\le {Xba}_2\le 0.4$$

(35)

3. Methodology

3.1. Method to Solve the Multi-Objective Optimization

The multi-objective optimization issue with two objectives—minimum gearbox volume and highest gearbox efficiency—was described in Section 2. In addition, three primary design factors have been chosen to be variables in the optimization issue.

Table 2. Input parameters.

Variables	Symbol	Lower Bound	Upper Bound
Gearbox ratio of first stage	u1	1	9
CWFW of stage 1	Xba1	0.25	0.4
CWFW of stage 2	Xba2	0.25	0.4

lists these variables along with their minimum and maximum values. In fact, applying an MCDM approach to the MOO (multi-objective optimization) problem is challenging. The reason for this is that there are numerous options or possible solutions available for MOO problems. Each of the parameters in this study have limits, as indicated in Table 2, and the step between variables is 0.02 to ensure parameter accuracy and prevent missing the optimization problem’s solution. In this instance, there are $(9-1)/0.02·(0.4-0.25)/0.02·(0.4-0.25)/0.02=22.500$ (runs) options (or experimental runs) that need to be identified and compared. Because of the wide range of options, it is not feasible to directly handle the OMO problem using the MCDM approach. In this paper, the MCDM problem was solved using the MARCOS method, and the criterion weights were determined using the Entropy approach. A simulation experiment was constructed in order to provide the input data for the MOOP for a two-stage helical gearbox in the MCDM problem. Since this is a simulation experiment, there is no restriction on the number of experiments that can be conducted by utilizing the full factorial design. Because there are three experimental variables (as previously specified) and five levels for each variable, the total number of experiments will be 5³ = 125. However, u₁, which runs from 1 to 9 in Table 2, has the broadest distribution of the three variables mentioned. As a result, even with five levels, there was still a significant disparity between the levels of this variable (in this case, ((9−1)/4 = 2). To reduce this disparity, save time, and improve the accuracy of the outcomes, a technique for solving multi-objective problems was proposed (Figure 3). This procedure is broken down into two phases: phase 1 factors reduce the gap between levels by solving the single-objective optimization problem, and phase 2 factors find the optimal primary design by solving the MOOP. Additionally, in the process of addressing the multi-objective problem, the MARCOS issue will be rerun using the smaller distance between two levels of variables if the levels of a variable are not sufficiently close to one another (≤0.02) (or the best answer is not appropriate for the requirement) (see Figure 3).

3.2. Method to Solve MCDM Problem

In this work, the MARCOS method was selected for solving the MCDM problem. To apply the MARCOS approach, the following steps must be taken [30]:

-

Making initial decision-making matrix

$$X=\left[\begin{array}{c}\begin{array}{ccc}{x}_{11}& \cdots & x_{1n}\\ x_{21}& \cdots & x_{2n}\\ ⋮& \cdots & ⋮\end{array}\\ \begin{array}{ccc}{x}_{mn}& \cdots & x_{mn}\end{array}\end{array}\right]$$

(36)

where m and n are alternative and criterion numbers;

-

An extended initial matrix is produced by appending an ideal (AI) and anti-ideal solution (AAI) to the original decision-making matrix

$$X=\begin{array}{c}\begin{array}{c}AAI\\ A_1\end{array}\\ A_2\\ ⋮\\ A_m\\ AI\end{array}\left[\begin{array}{c}\begin{array}{ccc}{x}_{aa1}& \cdots & x_{aan}\\ x_{11}& \cdots & x_{1n}\\ x_{21}& \cdots & x_{2n}\end{array}\\ \begin{array}{ccc}⋮& ⋮& ⋮\\ x_{m1}& \cdots & x_{mn}\\ x_{ai1}& \cdots & x_{ain}\end{array}\end{array}\right]$$

(37)

Here, $AAI=min\left(x_{ij}\right)$ and $AI=max\left(x_{ij}\right)$ if the requirement set with criterion j is as large as possible; $AAI=max\left(x_{ij}\right)$ and $AI=min\left(x_{ij}\right)$ when criterion j is as small as possible; i = 1, 2, …, m; j = 1, 2,…, n;

-

One then normalizes the extended starting matrix (X). To calculate the normalized matrix $N={\left[n_{ij}\right]}_{m\times n}$, we use the following formula:

$$n_{ij}=x_{AI}/x_{ij}$$

(38)

$$n_{ij}=x_{ij}/x_{AI}$$

(39)

when the criterion j is as small as possible, use Equation (38), and when it is as large as possible, use Equation (39).

-

Determine the weighted normalized matrix by

$$C={\left[c_{ij}\right]}_{m\times n}$$

(40)

$$c_{ij}=n_{ij} · w_j$$

(41)

where w_j is the weight coefficient of criterion j.

-

Find the utility of alternatives K_i⁻ and K_i⁺ by

$$K_i^{-}=S_i/S_{AAI}$$

(42)

$$K_i^{+}=S_i/S_{AI}$$

(43)

In (42) and (43), S_i (i = 1, 2, …, m) is the sum of the elements of the weighted matrix C, and S_i is calculated by

$$S_i={\sum }_{i=1}^m{c}_{ij}$$

(44)

-

Find the utility function f(K_i) of alternatives by

$$f\left(K_i\right)=\frac{K_i^{+}+K_i^{-}}{1+\frac{1-f\left(K_i^{+}\right)}{f\left(K_i^{+}\right)}+\frac{1-f\left(K_i^{-}\right)}{f\left(K_i^{-}\right)}}$$

(45)

where the utility function linked to the anti-ideal solution is denoted by f(K_i⁻), while the utility function linked to the ideal solution is represented by f(K_i⁺). These functions can be found by

$$f\left(K_i^{-}\right)=K_i^{+}/(K_i^{+}+K_i^i)$$

(46)

$$f\left(K_i^{+}\right)=K_i^{-}/(K_i^{+}+K_i^i)$$

(47)

-

To determine which alternative has the highest utility function value, rank the options according to the final utility function values.

3.3. Method to Find the Weight of Criteria

In this paper, the entropy technique was used to establish the weights of the criteria. The actions listed below can be used to put this strategy into practice [31].

-

Finding indicator-normalized values,

$$p_{\mathrm{ij}}=\frac{x_{\mathrm{ij}}}{m+{\sum }_{i=1}^m{x}_{\mathrm{ij}}^2}$$

(48)

-

Calculating the Entropy for each indicator,

$$m{e}_j=-{\sum }_{i=1}^m\left[p_{ij}\times ln\left(p_{ij}\right)\right]-\left(1-{\sum }_{i=1}^m{p}_{ij}\right)\times ln\left(1-{\sum }_{i=1}^m{p}_{ij}\right)$$

(49)

-

Determining the weight of each indicator,

$${\mathrm{w}}_{\mathrm{j}}=\frac{1-{\mathrm{m}\mathrm{e}}_{\mathrm{j}}}{{\sum }_{\mathrm{j}=1}^{\mathrm{m}}\left(1-{\mathrm{m}\mathrm{e}}_{\mathrm{j}}\right)}$$

(50)

4. Single-Objective Optimization

In this study, the direct search strategy is used to solve the single-objective optimization problem. Moreover, a Matlab computer program has been created to address two single-objective issues: maximizing gearbox efficiency and minimizing gearbox volume. The following figures feature several of the program’s findings: In Figure 3, the connection between u₁ and V_gb is shown. When u₁ is at its optimal value, V_gb reaches its lowest value (Figure 4). Figure 5 shows the relationship between η_gb and u₁. Furthermore, the optimal value of u₁ at which η_gb achieves its maximum is depicted in Figure 5. Figure 6 and Figure 7 show the association of X_ba1 and X_ba2 with V_gb and η_gb, respectively. It can be seen that with an increase in X_ba1, V_gb will decrease (Figure 6a). In contrast, V_gb will fall as X_ba2 rises (Figure 7a). Additionally, as X_ba1 and X_ba2 rise, η_gb falls (Figure 6b and Figure 7b).

The values of optimal major design factors of two single-objective functions, V_gb and η_gb, are shown in

Table 3. Optimum main design factors for minimum V_gb and maximum η_gb.

Objective	Factor	ut
		10	15	20	25	30	35
	u1	3.83	5.02	6.09	7.06	7.98	8.84
Vgb	Xba1	0.25	0.25	0.25	0.25	0.25	0.25
	Xba2	0.4	0.4	0.4	0.4	0.4	0.4
	u1	2.49	2.98	3.49	3.98	4.42	4.79
ηgb	Xba1	0.25	0.25	0.25	0.25	0.25	0.25
	Xba2	0.25	0.25	0.25	0.25	0.25	0.25

. From the table, it is clear that for the minimum V_gb, the minimum values of X_ba1 and maximum values of X_ba2 (X_ba1 = 0.25 and X_ba2 = 0.4) are the ideal values for X_ba1 and X_ba2. The reason for this is that the cross-section dimension (determined by H*L) must be small in order for V_gb to be tiny. Approximately equal values for d_w21 and d_w22 are required for this [32]. Given that the second stage gets a significantly greater torque than the first, a_w2 will be significantly greater than a_w1. Consequently, to make d_w21 approximate, d_w22 requires an increase in X_ba2 to decrease a_w2 and a decrease in X_ba1 to increase a_w1 (Formula (9)).

Figure 8 illustrates the relationship between the optimum gear ratio for the first stage (u₁) and the overall gearbox ratio (u_t). Additionally,

Table 4. New constraints of u₁.

ut	u1
	Lower Limit	Upper Limit
10	2.39	2.93
15	3.08	5.12
20	3.59	6.19
25	4.08	7.16
30	4.52	8.08
35	4.89	8.94

displays newly calculated constraints for the variable u₁.

5. Multi-Objective Optimization

A computer program has been created to perform simulations. The gearbox ratios 10, 15, 20, 25, 30, and 35 were all considered for the analysis. For this problem, with u_t = 15, the solutions is displayed below. This overall gearbox ratio was used for 125 initial testing runs (as specified in Section 3). The experiment’s output values, the gearbox volume and efficiency, will be sent into MARCOS as input parameters in order to solve the MOOP. This procedure will be repeated until there is less than 0.02 separating two levels of u₁. The primary design parameters and output responses for u_t = 15 in the fifth and final run of the MARCOS experiment are shown in

Table 5. Main design parameters and output results for u_t = 15 in the 5th run of MARCOS.

Trial	u1	Xba1	Xba2	Vgb (dm3)	egb (%)
1	3.61	0.25	0.25	21.05	95.906
2	3.61	0.25	0.2875	20.41	95.963
3	3.61	0.25	0.325	19.91	95.947
4	3.61	0.25	0.3625	19.52	95.931
5	3.61	0.25	0.4	19.19	95.902
6	3.61	0.2875	0.25	21.35	95.407
…
25	3.61	0.4	0.4	19.95	93.52
26	3.63	0.25	0.25	21.01	95.888
27	3.63	0.25	0.2875	20.38	95.946
…
51	3.65	0.25	0.25	20.98	95.852
52	3.65	0.25	0.2875	20.34	95.929
53	3.65	0.25	0.325	19.84	95.913
…
75	3.67	0.4	0.4	19.86	93.42
76	3.67	0.25	0.25	20.94	95.837
77	3.67	0.25	0.2875	20.31	95.911
…
104	3.69	0.25	0.3625	19.39	95.852
105	3.69	0.25	0.4	19.07	95.836
106	3.69	0.2875	0.25	21.21	95.306
…
123	3.69	0.4	0.325	20.74	93.434
124	3.69	0.4	0.3625	20.24	93.405
125	3.69	0.4	0.4	19.82	93.387

The criteria’s weights have been established using the Entropy technique (see Section 3.3), as follows: First, use Equation (48) to derive the normalized values of p_ij. Equation (49) is used to determine each indicator mej’s entropy value. Finally, use Equation (50) to find the weight of the criteria w_j. The weights of V_gb and η_gb for the most recent MARCOS work run were determined to be 0.5565 and 0.4435, respectively. The MARCOS method’s multi-objective decision-making phases are outlined in Section 3.2. Specifically, they are as follows: Determine the ideal solution (AI) and the anti-ideal solution (AAI) using formula (36). The results show that, with AI, V_gb and η_gb were 17.21 (dm³) and 95.8 (%), and, with AAI, they were 20.1 (dm³) and 93.34 (%). The next step is to use Formulae (37) for V_gb and (39) for η_gb to derive the normalized values u_ij. The normalized values were then calculated using Formula (40), taking the weight c_ij into consideration. Moreover, the coefficients K_i^- and K_i⁺ were obtained from Equations (42) and (43). The values of f(Ki⁻) and f(Ki⁺) were determined using Equations (46) and (47). It was found that f(Ki⁻) = 0.501 and f(Ki⁺) = 0.499. Lastly, the values of f(K_i) were computed using Formula (44).

Table 6. Several calculated results and rankings of alternatives by MARCOS for u_t = 1.5.

Trial	K−	K+	f(K−)	f(K+)	f(Ki)	Rank
1	0.0084	0.0084	0.4990	0.5010	0.0056	87
2	0.0085	0.0085	0.4990	0.5010	0.0057	58
3	0.0087	0.0086	0.4990	0.5010	0.0058	30
4	0.0088	0.0087	0.4990	0.5010	0.0058	15
5	0.0088	0.0088	0.4990	0.5010	0.0059	5
6	0.0083	0.0083	0.4990	0.5010	0.0055	101
…
25	0.0085	0.0085	0.4990	0.5010	0.0057	56
26	0.0084	0.0084	0.4990	0.5010	0.0056	84
27	0.0085	0.0085	0.4990	0.5010	0.0057	55
…
51	0.0084	0.0084	0.4990	0.5010	0.0056	83
52	0.0085	0.0085	0.4990	0.5010	0.0057	52
53	0.0087	0.0086	0.4990	0.5010	0.0058	28
…
75	0.0085	0.0085	0.4990	0.5010	0.0057	49
76	0.0084	0.0084	0.4990	0.5010	0.0056	82
77	0.0086	0.0085	0.4990	0.5010	0.0057	50
…
104	0.0088	0.0087	0.4990	0.5010	0.0058	9
105	0.0088	0.0088	0.4990	0.5010	0.0059	1
106	0.0083	0.0083	0.4990	0.5010	0.0055	95
…
123	0.0083	0.0083	0.4990	0.5010	0.0055	91
124	0.0085	0.0084	0.4990	0.5010	0.0056	69
125	0.0085	0.0085	0.4990	0.5010	0.0057	46

displays the options’ ranking and the results of several parameters (for the last run of MARCOS work). Out of all the possibilities given, option 105 is the most ideal one, according to the table. Consequently, Table 5 shows the optimal values for the primary design features: u₁ = 3.69, X_ba1 = 0.25, and X_ba2 = 0.4.

Table 7. Optimum main design parameters.

No.	ut
	10	15	20	25	30	35
u1	3.55	3.69	4.18	4.47	5.06	5.39
Xba1	0.25	0.25	0.25	0.25	0.25	0.25
Xba2	0.4	0.4	0.4	0.4	0.4	0.4

shows the optimal values for the main design parameters that correspond to the remaining ut values of 10, 20, 25, 30, and 35, building on the previous discussion. The information in Table 6 permits the following deductions to be drawn.

Table 7 reveals that X_ba1 selects the lowest value (X_ba1 = 0.25), while X_ba2 selects the highest value (X_ba2 = 0.4). This is due to the fact that a small box’s cross-sectional area (LxH) is required to produce the smallest gearbox volume. In order to accomplish that, d_w21 and d_w22 must be about equal [32]. Additionally, because the second stage has a higher torque, a larger X_ba2 is required in order to decrease the diameter of d_w22. Conversely, a smaller X_ba1 must be chosen in order to raise d_w21 because the first stage has a lower torque.

The obtained values of the gearbox efficiency in this work were compared with the findings in [23] in order to evaluate the effectiveness in employing the formula for the power loss in the idle motion when calculating power loss in gears. Table 5 [23] shows that the gearbox efficiency, based on an input power of 10 kW, a maximum output power of 9.971 kW, and a minimum output power of 9.933 kW, will be between 99.33 and 99.71 percent when the overall gearbox ratio is 7.5. Furthermore, Table 4 indicates that the gearbox efficiency in this work will range from 93.258% to 95.963% when ut is increased from 10 to 35. Actually, a helical gear train’s efficiency is 0.93–0.98 (93–98%), whereas a pair of bearings’ efficiency is 0.99–0.995 (99–99.5%) [27]. The following formula will be used to calculate the efficiency of a two-stage helical gearbox based on these data:

$${\eta }_{gb}={\eta }_g^2·{\eta }_b^3={0.93}^2·{0.99}^3÷{0.98}^2·{0.995}^3=0.83÷0.95 \mathrm{or} {\eta }_{gb}=83÷95 \mathrm{or} (\%)$$

(51)

From the above analysis, it is clear that the gearbox efficiency of this work (93.258–95.963%) is reasonably near to reality (83–95%), when comparing with that in [23]. Furthermore, the gearbox efficiency shown in [23] (99.33–99.71%) is dramatically high and incompatible with reality. This demonstrates that the power loss in idle motion formula used in this work is a useful new tool that needs to be used.

Figure 9 shows that there is a definite first-order link between the ideal values of u₁ and u_t. Additionally, it was found that the following regression equation (with R² = 0.9804) may be used to calculate the optimal values of u₁:

$$u_1=0.0777·u_t+2.6414$$

(52)

Once u₁ has been determined, the optimal value of u₂ can be found using the following formula:

$$u_2=u_t/u_1$$

(53)

6. Conclusions

The MARCOS approach was utilized in this work to solve the MOOP related to the design of a two-stage helical gearbox. The study’s goal is to identify the best critical design factors that maximize gearbox efficiency while reducing gearbox volume. To do this, three essential design components were chosen: the CWFW for the first and second stages, and the first stage gear ratio. In addition, there are two steps in the MOOP solution process. Phase 1 is dedicated to solving the single-objective optimization problem of reducing the difference between variable values, whereas phase 2 is concerned with determining the optimal primary design factors. The following findings were drawn from this work:

-

The single-objective optimization problem speeds up and simplifies the resolution of the MOOP by bridging the gap between variable levels;

-

The three main design parameters for a two-stage helical gear gearbox, Equation (51) and Table 6, were recommended to have optimal values based on the study’s findings;

-

In regard to the important design characteristics, two single objectives—the minimal gearbox volume and the greatest gear-box efficiency—were assessed;

-

By using the MARCOS technique repeatedly until the desired results are reached, the MOOP can be solved more precisely (u₁ has an accuracy of less than 0.02);

-

The experimental data’s extraordinary degree of concordance with the proposed model of u₁ verifies their reliability;

-

Further research is required to determine how to apply the proposed approach for solving the MOOP for various domains and MCDM methods.