Interval Multi-Objectives Optimization of Electric Wheel Dump Truck Frame Based on Blind Number Theory

applied sciences

Article Interval Multi-Objectives Optimization of Electric Wheel Dump Truck Frame Based on Blind Number Theory

Chengji Mi 1, Jidong Liu 2, Xuewen Xiao 1,*, Jinhua Liu 1, Rui Ming 1,*, Wentai Li 1 and Qishui Yao 1

1 Department of Mechanical Engineering, Hunan University of Technology, Zhuzhou 412007, China 2 Department of Mechanical Engineering, Hunan Technician College of Industry and Commerce, Zhuzhou 412006, China * Correspondence: [email protected] (X.X.); [email protected] (R.M.)

Received: 30 July 2019; Accepted: 22 August 2019; Published: 10 October 2019

Abstract: With respect to heavy carrying tasks and uneven road surface in open pit mine, the mechanical performances of electric wheel dump truck frame regarded as the main bearing component are extremely important for its safe operation. For the problem of insufficient strength and stiffness, an interval multi-objectives optimization design method based on blind number theory is presented in this paper, taking into account uncertain factors caused by manufacturing process. The allowable interval strength of welding material and permissible interval bending and torsional stiffness are obtained by means of experiments and simulations, using the blind number theory. Based on Latin hypercube sampling technique, the strength and stiffness of frame under different conditions is estimated. It shows that in the case of braking interference between interval strength and interval dynamic stress appears. Strength and stiffness are considered as optimization objectives, while the thicknesses of top longitudinal beam, stiffener, lateral longitudinal beam and hanging ear in the frame are deeded as design variables, and Young’s modulus and Poisson’s ratio are taken as uncertain variables. The optimized results show that the stress and deformation of the frame due to dynamic loading decreased. This result enables to formulate the following sentence: the mechanical performance of frame is improved.

Keywords: blind number theory; electric wheel dump truck; interval strength; interval stiﬀness; interval multi-objectives optimization

1. Introduction The electric wheel dump truck is the main carrying equipment in open pit mine. It has 300 tons of load capacity, and runs on uneven road surface for 24 working hours every day [1–3]. The frame is considered as one of pivotal bearing components, and thence its mechanical properties connected with strength and stiffness have significant influence on vehicle’s safe operation [4]. In designing of frame structure, the mechanical parameters and geometrical features are considered as deterministic values. However, there is loading caused by manufacturing process. This influences parental material and creates measurement errors [5], so that the mechanical parameters of engineering frames appear different than nominal [6,7]. Therefore, determination of the mechanical features of frame based on interval method is required. The blind number theory can characterize the uncertainties of multiple variables, and has simple operation rules and high credibility, which was extensively utilized in engineering fields including numerous uncertain information. Zhang et al. [8] presented a comprehensive evaluation model on

Appl. Sci. 2019, 9, 4214; doi:10.3390/app9204214 www.mdpi.com/journal/applsci Appl. Sci. 2019, 9, 4214 2 of 16 groundwater resources carrying capacity with blind information. They showed a risk assessment model of surcharge about groundwater resources carrying capacity. Their conception was formulated by means of blind reliability theory. This theory was also applied to define the blind numbers of the parameters of environmental system according to the characteristics of several kinds of uncertainties coexisting in water environmental system [9]. The uncertain variables in time-dependent stochastic process of mechanical structure were expressed in blind number, and its time-dependent reliability calculation model based on blind number theory was established [10]. However, this theory was rarely used to estimate mechanical properties of engineering parts for obtaining the interval parameters. In fact, some difficult working conditions are not easy to be predicted during designing, so that the mechanical features of engineering structures can be insufficient. After determination of the interval values by means of the blind number theory, the redesigning of component at optimization rules is needed. For the uncertain optimization problem, there are usually three methodologies i.e., probabilistic analysis, fuzzy analysis and non-probabilistic interval analysis methods. In the first approach the uncertain mathematical characteristics are transformed into random variables. Their randomness could be characterized by probability distribution function [11,12]. An optimization method for controller parameters of a gas turbine based on probabilistic robustness was presented and some uncertainties because of errors in measurement and manufacturing tolerances [11,12]. The membership function is utilized to describe the uncertain parameters using fuzzy theory [13]. A unified method was developed for the uncertainty quantification of dis brake squeal for two fuzzy-interval cases, while the unified uncertain response was computed with the aid of the combination of level-cut strategy, Taylor series expansion, subinterval analysis and Monte Carlo simulation [14]. Both probability distribution function and membership function need a lot of data, however, the engineering problems have finite data [15]. The interval method could make up for the above shortcomings [16]. The reliability optimization design method for turbine disk strength base on interval uncertainty was developed, in which the uncertainty and reliability were represented with interval model and non-probabilistic reliability model [17]. The fatigue strength of the A-type frame was carried out by means of genetic algorithm taking into account uncertainties of the component in design and manufacture [18]. Nevertheless, these papers focus attention to single objective optimization problem, while the interval multi-objectives optimization problem is insufficiently presented. This paper combines the limited experimental data with simulated results; the interval allowable strength and stiffness of frame were obtained, based on the blind number theory. Then, according to the deterministic values, the working dynamic stress under three conditions and two kinds of stiffness were all assessed. Finally, based on the response surface method, the interval multi-objectives optimization function was constructed. The optimal design was performed by genetic algorithm.

2. Strength and Stiffness of the Frame The electric wheel dump truck frame is manufactured in welding technology using high-strength low alloy quenched and tempered steel. There are multi-source uncertainties from manufacturing related to many variants of welds and their different mechanical properties, as well as two materials, which requires different welding process to produce the frame. Therefore, in order to predict strength and stiffness of the frame, the interval allowable values are determined by the use of the blind number theory.

2.1. Blind Number Theory R is supposed to be real number set, and R is considered as unknown rational number set, and G is rational number set [5–10]. Suppose a G, T [0, 1], and f (x) is deﬁned as grey function like this [5–10]: i ∈ i ∈ ( T , x = a (i = 1, 2, ..., n) f (x) = i i (1) 0 , x < {a , a , ..., an x G 1 2 } ∈ Appl. Sci. 2019, 9, 4214 3 of 16

Pn when i , j, a , a , and T = t 1(i = 1, ..., n), f (x) is regarded as blind number, which could be i j i ≤ i=1 characterized as [a1, an], f (x) . Ti is the credibility of ai, and t is the total credibility of f (x), and n is the order of f (x). A and B are supposed as blind number, and x G, y G. Two blind numbers could be expressed i ∈ i ∈ as [5–10]: ( α , x = x (i = 1, 2, m) A = f (x) i i ··· (2) 0, other ( β , y = y (j = 1, 2, n) B = f (y) j j ··· (3) 0, other Then, the probable value in matrix A B could be achieved according to the following ∗ equation [5–10]:  x y x y x y   1 1 1 i 1 n   ∗. ··· ∗. ··· ∗.   ......   . . . . .    A B =  x y x y x yn  (4)  i 1 i j i  ∗  ∗. ··· ∗. ··· ∗.   ......   . . . . .    xm y xm y xm yn ∗ 1 ··· ∗ j ··· ∗ The corresponding credibility in matrix A B could be obtained from [5–10]: ∗  α β α β α β   1 1 1 i 1 n   . ··· . ··· .   ......   . . . . .    AB =  α β α β α βn  (5)  i 1 i j i   . ··· . ··· .   ......   . . . . .    αmβ αmβ αmβn 1 ··· j ··· 2.2. The Strength of Welded Joints at Intervals In general, the ultimate tensile strength of welding material could be evaluated by:

F F σ = = (6) t A a b × where F is the maximum value of force during monotonic tensile test, A is a cross-sectional area, a and b are width and thickness of specimen, respectively. h i σt, σt is employed to describe interval allowable tensile strength of welded frame, and could be estimated by: h i h i F, F σt, σt = h i (7) [a, a] b, b × h i h i where F, F stands for interval maximum load, [a, a] is interval width of specimen, and b, b is interval thickness of specimen. According to Equation (7) and the blind number algorithms, the ultimate tensile strength of welded joints could be obtained if maximum loading, width and thickness of specimen at considered intervals are known. Five monotonic tensile tests were conducted on specimens of welded joint. The results are listed in Table1. The ﬂat dog-bone shaped specimen was designed to characterize the mechanical properties of butt joint [1,2,4,19]. Then, the ultimate tensile strength in each test was calculated by Equation (6), Table1[19]. Appl. Sci. 2019, 9, x FOR PEER REVIEW 4 of 16

Appl.properties Sci. 2019 of, 9, butt 4214 joint [1,2,4,19]. Then, the ultimate tensile strength in each test was calculated4 of by 16 Equation (6), Table 1 [19].

Table 1.1. MonotonicMonotonic experimentalexperimental datadata andand results.results.

Maximum ForceMaximum Width Force of SpecimenWidth of ThicknessThickness of ofUltimateUltimate Tensile Tensile Strength NumberNumber (kN) (kN) (mm) Specimen (mm)SpecimenSpecimen (mm) (mm) Strength(MPa) (MPa) 1 122.14 22.146.13 6.135.97 5.97604.98 604.98 2 218.05 18.056.22 6.225.81 5.81499.47 499.47 3 324.15 24.156.05 6.056.10 6.10654.38 654.38 4 427.12 27.125.89 5.896.04 6.04762.32 762.32 5 516.66 16.666.00 6.005.89 5.89471.42 471.42

BasedBased on thethe aboveabove ﬁvefive parameters,parameters, thethe initialinitial intervalinterval ofof tensiletensile strength,strength, widthwidth andand thicknessthickness ofof specimen specimen and and maximum maximum loading loading couldcould be be deﬁned defined as as [471.42,471.42,762.32 762.32] MPa,MPa, [5.89,5.89,6.22 6.22] mm, [5.81,5.81,6.10 6.10] mmmm and and[16.66, 16.66,27.12 27.12] kN. kN They. They were were considered considered as blind as numbers. blind numbers. According According to Equation to (7)Equation and operation (7) and operation rule of blind rule number, of blind the number, total 125 the probabletotal 125 valuesprobable of tensilevalues strengthof tensile and strength their correspondingand their corresponding credibility credibility values were value obtained,s were obtained, and some and of some them of werethem listedwere listed in Table in 2Table[ 19 ].2 Then, the interval tensile strength of welding material was determined by [439.11, 792.52] MPa. [19]. Then, the interval tensile strength of welding material was determined by 439.11,792.52 MPa.

Table 2. Values of ultimate tensile strength and credibility obtained by means of the blind Table 2. Values of ultimate tensile strength and credibility obtained by means of the blind numbers numbers theory. theory. Number 1 2 3 123 124 125 Number 1 2 3 123 124 125 Tensile strength Tensile439.11 strength (MPa) 443.44 439.11 443.44 445.57 445.57 777.97777.97 781.78 781.78 792.52 792.52 (MPa) CredibilityCredibility 1.000 0.9921.000 0.992 0.984 0.984 0.0240.024 0.016 0.016 0.008 0.008

2.3.2.3. StiStiffnessffness of Welded Frame TheThe dimension dimension of of frame frame and and load load capacity capacity is huge, is so huge, that sodefining that defining the stiffness the of stiffness the component of the iscomponent difficult through is difficult testing. through Thus, testing this parameter. Thus, this was parameter assessed was by finiteassessed element by finite analysis. element analysis. TheThe finitefinite element element model model of frameof frame established established by software by software HYPERMESH HYPERMES(Altair,H(Altair, Boston, Boston, MA, USA) MA, is shownUSA) is in shown Figure 1in. TheFigure frame 1. The was frame meshed was with meshed triangular with andtriangular quadrilateral and quadrilateral shell elements. shell There elements. were 165,985There were elements 165,985 and elements 159,450 nodes.and 159,450 nodes.

FigureFigure 1.1. Finite element model of frame.frame.

BeforeBefore estimating stiffness stiffness of of the the frame frame at atintervals intervals,, the the finite finite element element model model was wastestified. testified. The Thestress stress due dueto dynamic to dynamic loading loading in five in fivemeasurement measurement points points at full at fullload load at the at thespeed speed of 30 of km/h 30 km on/h the on themine mine road road was was measure measured,d, Fig Figureure 2.2 The. The strain strain gauge gauge rosette rosette was was cemented cemented in the selectedselected zoneszones ofof thethe frame,frame, FigureFigure3 [3 20[20]].. Both Both the the results results from from the the simulation simulation and and the the road road test test are listedare listed in Table in Table3[ 20 ].3 For[20] point. For point 4, the 4 high, the relativehigh relative error waserror caused was caused by neglecting by neglecting the loads the loads at this at region, this region because, because these massesthese masses from coolingfrom cooling system system and cab and were cab supposedwere supp toosed have to littlehave elittleffect effect on bearing on bearing state. state This. highThis Appl.Appl. Sci. Sci. 20192019,, 99,, x 4214x FORFOR PEERPEER REVIEWREVIEW 55 of of 16 16 highhigh stress stress level level could could be be produced produced by by geometricgeometric mutation. mutation. The The relative relative error error of of remaining remaining measurementstressmeasurement level could pointspoints be produced allall stayedstayed by withinwithin geometric 10%,10%, mutation.whichwhich showedshowed The relative thatthat thethe error simulated simulated of remaining resultsresults measurement agreedagreed well well withpointswith testedtested all stayed ones.ones. within 10%, which showed that the simulated results agreed well with tested ones.

FigureFigure 2. 2. DiagramDiagram ofof thethe fivefive ﬁve measurementmeasurement points. points.

FigureFigure 3. 3. TheThe strainstrain gaugegauge rosetterosette on on the the component component.component..

TableTable 3. 3. ComparisonComparison of of simulated simulated results results and and experimental experimental data data.data.. Simulated Results NumberNumber RoadRoad TestTest Road DataData Test (MPa)(MPa) Data (MPa) SimulatedSimulated ResultsResults (MPa)(MPa) RelativeRelativeRelative ErrorError Error(%)(%) (%) (MPa) 11 125125 121121 3.23.2 122 145145 125143143 1211.41.4 3.2 2 145 143 1.4 33 196196 117777 99.7.7 3 196 177 9.7 44 169169 16977 795.695.6 95.6 55 179179 179181181 1811.11.1 1.1

TheThe bendingbending stiffnessstiffness ofof frameframe atat intervalinterval couldcould bebe estimatedestimated asas followsfollows:: The bending stiﬀness of frame at interval could be estimated as follows: FF,,,, F F a a3333 a a hbb b bih b b b b i  3 3 CCCCbbbb,, F , F a , a (8)(8) h i b 4848bffffb , , b Cb, Cb = hbbbbi (8) 48 fb, fb wherewhere FFFF,, expressesexpresses thethe intervalinterval concentratedconcentrated bendingbending loadload appliedapplied toto thethe centercenter ofof gravitygravity ofof bbbb h i where F , Fexpresses the interval concentrated bending load applied to the center of gravity of the thethe goods, goods,b baaaabbbb,, isis wheelbasewheelbase of of this this truck truck atat interval interval,, and and ffff,, isis maximummaximum deformationdeformation ofof h i h i bbbb goods, a , a is wheelbase of this truck at interval, and f , f is maximum deformation of frame in frameframe inin forcebforceb centercenter atat intervalinterval [19][19].. FiveFive finitefinite elementelement modelsmodelsb b andand concentratedconcentrated loadloadinging ofof frameframe force center at interval [19]. Five ﬁnite element models and concentrated loading of frame on bending onon bendingbending fromfrom thethe componentcomponent projectproject werewere determineddetermined.. TheThe degreesdegrees ofof freedomfreedom atat suspensionssuspensions from the component project were determined. The degrees of freedom at suspensions were constrained. werewere constrained.constrained. Then,Then, thethe deformationdeformation ofof thethe frameframe underunder bendingbending couldcould bebe calculated,calculated, asas listedlisted inin Then, the deformation of the frame under bending could be calculated, as listed in Table4. TableTable 4.4.

Appl. Sci. 2019, 9, 4214 6 of 16

Table 4. Initial interval values under bending.

Concentrated Force Maximum Deformation Number Wheelbase (m) (kN) of Frame (m) 1 2.500 106 7.026 1.988 10 3 × × − 2 2.926 106 7.332 4.363 10 3 × × − 3 3.350 106 7.638 3.069 10 3 × × − 4 4.200 106 7.717 4.014 10 3 × × − 5 3.775 106 7.205 5.513 10 3 × × −

According to Equation (7) and blind number algorithms, the total 125 probable values of bending stiffness and corresponding credibility values were obtained, and some of them are presented in Table5. h i Then, the interval bending stiffness of the frame was defined by 3.28 109, 20.28 109 N m2. × × ·

Table 5. Probable values of bending stiﬀness and credibility.

Number 1 2 3 123 124 125 Bending stiﬀness ( 3.28 3.53 3.72 18.18 19.51 20.28 × 109 N.m2) Credibility 1.000 0.992 0.984 0.024 0.016 0.008

The torsional stiffness of frame at interval could be evaluated by: h ih ih i 2 2 h i Ft, Ft Lt , Lt at, at Ct, Ct = h i (9) ht, ht h i where Ft, Ft represents concentrated torsional loading applied to the front suspension at interval, h i h i 2 2 ab, ab is wheelbase of this truck taking interval, Lt , Lt is wheel-track of the truck connected with h i interval, and ht, ht is deflection of the frame due to torsion at interval. The other torsional condition was simulated applying to a pair of opposite loading at left and right front suspension with full constraints at rear suspension. Similarly, the torsional stiffness of frame h i at interval was defined by 0.84 109, 2.91 109 N m2/rad. × × · 3. Evaluation of Stress and Stiffness by Means of Latin Hypercube Sampling Method

3.1. Estimation of Stress Distribution under Three Loading Conditions The stress state component of the frame was mainly caused by bending in horizontal plane and torsion. The stress value due to dynamic loading was evaluated using the dynamic loading coefficient [21]: K + K C n = 1 + 1 2 1 (10) 2 G 1 + C2/V where K1 and K2 represent stiffness of front and rear spring systems, respectively. G is self-weight, C1 is road coefficient, C2 is experience coefficient, V is speed of truck. This evaluation could be implemented by finite element analysis under different material parameters and dynamic coefficient. For the bending condition in horizontal plane, the weight of the full load was applied, Figure1, and the maximum dynamic loading coe fficient was calculated of 2.5 [21,22]. In order to cover all possibilities of variables, the Latin Hypercube sampling method was used. Thanks to this theory the 40 sample data were obtained. Some of them are listed in Table6. The variables x1, x2, x3, x4 represented for the thicknesses of top longitudinal beam, stiffener, lateral longitudinal beam and hanging ear in the frame, respectively. x5, x6, n and S were Appl. Sci. 2019, 9, x FOR PEER REVIEW 7 of 16

The variables x1, x2, x3, x4 represented for the thicknesses of top longitudinal beam, stiffener, lateral longitudinal beam and hanging ear in the frame, respectively. x5, x6, n and S were deeded as Appl. Sci. 2019, 9, 4214 7 of 16 Young’s modulus, Poisson’s ratio, dynamic loading coefficient and corresponding maximum equivalent stress. One of corresponding values is shown in Figure 4. The maximum equivalent deededstress is as located Young’s at modulus,the bottom Poisson’s of lateral ratio, longitudinal dynamic loadingbeam around coeﬃcient rear andsuspension. corresponding Then, maximumaccording equivalentto the simulated stress. One maximum of corresponding equivalent values stress, is shownthe interval in Figure stress4. The due maximum to dynamic equivalent loading stress was iscalculated located at to the be bottom211.11,357.10 of lateral MPa longitudinal. Then, these beam values around were rear compared suspension. with Then, the accordingultimate tensile to the simulatedstrength at maximum interval, Fig equivalenture 5. stress, the interval stress due to dynamic loading was calculated to be [211.11, 357.10] MPa. Then, these values were compared with the ultimate tensile strength at interval, FigureAppl. 5Sci.. 2019, 9, x FOR PEER REVIEW 7 of 16

The variables x1, x2, x3, x4Table represented 6. Sample for data the and thicknesses corresponding of top responses. longitudinal beam, stiffener, lateral longitudinal beam and hanging ear in the frame, respectively. x5, x6, n and S were deeded as Number x1 (mm) x2 (mm) x3 (mm) x4 (mm) x5 (MPa) x6 n S (MPa) Young’s modulus, Poisson’s ratio, dynamic loading coefficient and corresponding maximum 1 44.05 16.79 25.49 45.28 200.3 0.271 2.37 338.8 equivalent stress. One of corresponding values is shown in Figure 4. The maximum equivalent 2 45.08 15.21 24.56 45.95 202.6 0.283 2.40 340.9 stress3 is located 45.23 at the bottom 16.38 of lateral 25.03 longitudinal 44.97 beam 210.0 around rear 0.273 suspension. 2.50 Then, according 352.7 to 38the simulated 45.28 maximum 15.15 equivalent 24.15 stress, 45.08 the interval 192.9 stress 0.269due to dynamic 1.94 loading 279.2 was calculated39 to 44.31 be [211.11,357.10 15.51] MPa. 25.74 Then, these 45.49 values 209.5 were comp 0.282ared with 2.09the ultimate 293.8 tensile 40 45.74 16.08 24.36 45.90 198.0 0.274 1.63 230.8 strength at interval, Figure 5.

Figure 4. Contour of vonMises stress under bending due to horizontal loading (unit: MPa).

Table 6. Sample data and corresponding responses.

Number x1 (mm) x2 (mm) x3 (mm) x4 (mm) x5 (MPa) x6 n S (MPa) 1 44.05 16.79 25.49 45.28 200.3 0.271 2.37 338.8 2 45.08 15.21 24.56 45.95 202.6 0.283 2.40 340.9 3 45.23 16.38 25.03 44.97 210.0 0.273 2.50 352.7 38 45.28 15.15 24.15 45.08 192.9 0.269 1.94 279.2 39 44.31 15.51 25.74 45.49 209.5 0.282 2.09 293.8 40 45.74 16.08 24.36 45.90 198.0 0.274 1.63 230.8 Figure 4. Contour of vonMises stress under bending due to horizontal loading (unit: MPa). Figure 4. Contour of vonMises stress under bending due to horizontal loading (unit: MPa).

Table 6. Sample data and corresponding responses.

Number x1 (mm) x2 (mm) x3 (mm) x4 (mm) x5 (MPa) x6 n S (MPa) 1 44.05 16.79 25.49 45.28 200.3 0.271 2.37 338.8 2 45.08 15.21 24.56 45.95 202.6 0.283 2.40 340.9 3211.11 45.23 357.1016.38 25.03439.11 44.97 210.0 0.273 792.522.50 352.7 38 45.28 15.15 24.15 45.08 192.9 0.269 1.94 279.2 Figure 5. Comparison of stress due to dynamic bending loading and the ultimate tensile strength at 39 44.31 15.51 25.74 45.49 209.5 0.282 2.09 293.8 Figure 5. Comparison of stress due to dynamic bending loading and the ultimate tensile strength at interval40 (unit: MPa). 45.74 16.08 24.36 45.90 198.0 0.274 1.63 230.8 interval (unit: MPa). The second significant condition was formulated in the following manner: an unexpected braking (fracturing)The second of the significant truck at full condition load. The was frame formulated is taken in with the a following huge inertial manner force.: an Based unexpected on the distancebraking (fracturing) to the fracture of andthe timetruck at at the full speed load. of The 30 kmframe/h for is thistaken electric with a wheel huge dump inertial truck, force. the Bas brakinged on decelerationthe distance wasto the calculated fracture toand be time 3.816 at m the/s2 [speed21–23 ].of In30 this km/h condition, for this theelectric dynamic wheel loading dump coetruck,fficient the wasbraking equal deceleration to 1.5 [21–23 was]. The calculated constraint to be was 3.816 the samem/s2 [21 as– the23] horizontal. In this condition, bending the case. dynamic The maximum loading equivalentcoefficient stresswas equal appeared to 1.5 at [21 the–23] same. The position constraint as the was bending the same one. as Then, the horizontal the stress caused bending by case. dynamic The loadingmaximum at interval equivalent211.11 was stress defined appeared357.10 as follows at the[343.31, same439.11 461.20 position], MPa, as the Figure bending6. It was one.792.52 clearly Then, seen the stress that therecaused was by overlapdynamic between loading the at stressinterval and was ultimate defined tensile as follows strength. 343.31,461.20 Therefore, the, MPa optimization, Figure 6. designIt was Figure 5. Comparison of stress due to dynamic bending loading and the ultimate tensile strength at wasclearly needed. seen that there was overlap between the stress and ultimate tensile strength. Therefore, the interval (unit: MPa). optimization design was needed. The second significant condition was formulated in the following manner: an unexpected braking (fracturing) of the truck at full load. The frame is taken with a huge inertial force. Based on the distance to the fracture and time at the speed of 30 km/h for this electric wheel dump truck, the braking deceleration was calculated to be 3.816 m/s2 [21–23]. In this condition, the dynamic loading coefficient was equal to 1.5 [21–23]. The constraint was the same as the horizontal bending case. The maximum equivalent stress appeared at the same position as the bending one. Then, the stress caused by dynamic loading at interval was defined as follows [343.31,461.20] , MPa, Figure 6. It was clearly seen that there was overlap between the stress and ultimate tensile strength. Therefore, the optimization design was needed. Appl. Sci. 2019, 9, x FOR PEER REVIEW 8 of 16 Appl. Sci. 2019, 9, 4214 8 of 16 Appl. Sci. 2019, 9, x FOR PEER REVIEW 8 of 16

343.31 439.11 461.20 792.52

Figure 6. Comparison343.31 of stress due to 439.11dynamic loading461.2 at breaking and ultimate792.52 tensile strength at the interval (unit: MPa). Figure 6. Comparison of stress due to dynamic loading at breaking and ultimate tensile strength at the Figure 6. Comparison of stress due to dynamic loading at breaking and ultimate tensile strength at intervalWhen the (unit: truck MPa). passed pit pavement, the frame was under torsional loading. In this case, a pair the interval (unit: MPa). of forces in opposite direction with 3.2 × 106 N was applied to the left and the right front suspension [21,24]When. The the dynamic truck passed loading pit coefficient pavement, was the the frame same was as under in the torsional braking loading. case, while In this only case, the a pairWhen of forces the truck in opposite passed directionpit pavement, with the 3.2 frame106 wasN was under applied torsional to the loading. left and In this the rightcase, a front pair translational degrees of freedom in vertical6 and× longitudinal directions at rear suspensions were suspensionof forces in [opposite21,24]. The direction dynamic with loading 3.2 × coe10ﬃ Ncient was was applied the same to the as left inthe and braking the right case, front while suspension only the limited. Then, the stress due to dynamic torsion at interval was defined as 222.60,374.54 , MPa, translational[21,24]. The degreesdynamic of freedomloading incoefficient vertical and was longitudinal the same directionsas in the at braking rear suspensions case, while were only limited. the Then,Figtranslationalure the 7. stress degrees due to of dynamic freedom torsion in vertical at interval and lo wasngitudinal deﬁned asdirections[222.60, 374.54at rear], MPa,suspensions Figure7 .were limited. Then, the stress due to dynamic torsion at interval was defined as [222.60,374.54] , MPa, Figure 7.

222.60 374.54 439.11 792.52

Figure 7. Comparison of stress due to dynamic torsion and ultimate tensile strength at the interval Figure 7. Comparison of stress due to dynamic torsion and ultimate tensile strength at the interval (unit: MPa). 222.60 439.11 792.52 (unit: MPa). 374.54 3.2. EstimationFigure 7. Comparison of Stiffness underof stress Two due Conditions to dynamic at Intervaltorsion and ultimate tensile strength at the interval 3.2. Estimation of Stiffness under Two Conditions at Interval A(unit: deformation MPa). of the frame was the second aim of the analysis. It was reached basing on the frameA sti deformationffness. In order of the to estimateframe was the the component second aim resistance of the analysis on bending. It was deformation, reached basing the weight on the of 3.2. Estimation of Stiffness under Two Conditions at Interval goodsframe andstiffness. carriage In order was applied to estimate as a kindthe component of concentrated resistance force inon the bending frame, deformation, Figure1. The the translational weight of degreesgoodsA anddeformation of freedom carriage inof was verticalthe appliedframe and was as lateral the a kind second directions of concentratedaim atof frontthe analysis. suspensions force inIt was the and reached frame, the all Fig basing translationalure 1on. Thethe degreestranslationalframe stiffness. of freedom degrees In order at rearof tofreedom suspensions estimate in thevertical were component limited.and lateral resistance Some direction of the on 40bendings samplesat front defo datasuspensionsrmation, are listed the and inweight Tablethe all7of. translational degrees of freedom at rear suspensions were limited. Some of the 40 samples data are Fgoodsc and Dandare carriage the applied was force applied and the as maximuma kind of deformation concentrated after force loading, in the respectively. frame, Figure The contour1. The oflistedtranslational displacement in Table degrees 7. in F verticalc andof freedom D direction are thein for vertical applied one sampling and force lateral and data directionsthe is displayedmaximum at front in deformation Figure suspensions8. According after and loading, tothe the all simulatedrespectively.translational results Thedegrees contour of this of freedom bending of displacement at condition rear suspension in and vertical Equations directionwere (8),limited. thefor bendingone Some sampling of sti theffness 40data samples of is frame displayed data was are ofin hFigure 8. 9According to9i the simulated2 results of this bending condition and Equation (8), the bending listed9.65 in10 Table, 10.50 7. 10Fc andN mD ,are Figure the9 .applied They were force exceeding and the themaximum limit value deformation and therefore after the loading, frame × × · 992 shouldstiffnessrespectively. be of redesigned. frame The contourwas of of9.65 displacement 10 ,10.50 10in vertical N·m , directionFigure 9. forThey one were sampling exceeding data theis displayed limit value in andFigure therefore 8. According the frame to the should simulated be redesigned. results of this bending condition and Equation (8), the bending Table 7.××Sample99 data and the corresponding2 responses. stiffness of frame was of 9.65 10 ,10.50 10 N·m , Figure 9. They were exceeding the limit value

andNumber thereforex 1the(mm) frame xshould2 (mm) be redesigned.x3 (mm) x 4 (mm) x5 (MPa) x6 Fc (N) D (mm) 2.851 1 45.23 17.00 24.87 44.36 208.6 0.256 × 2.435 106 2.779 2 44.05 15.92 24.51 45.23 199.6 0.281 × 2.540 106 3.246 3 44.00 15.00 24.28 45.18 205.3 0.277 2.851 106 × 1.940 38 44.21 16.33 25.49 44.46 192.9 0.269 × 2.740 106 2.090 39 44.51 16.69 25.95 45.08 209.5 0.282 × 2.844 106 1.630 40 44.41 15.62 24.26 45.49 198.0 0.274 3.334 106 ×

Figure 8. Contour of displacement in vertical direction under bending (unit: m). Figure 8. Contour of displacement in vertical direction under bending (unit: m). Appl. Sci. 2019, 9, x FOR PEER REVIEW 8 of 16

343.31 439.11 461.2 792.52

Figure 6. Comparison of stress due to dynamic loading at breaking and ultimate tensile strength at the interval (unit: MPa).

When the truck passed pit pavement, the frame was under torsional loading. In this case, a pair of forces in opposite direction with 3.2 × 106 N was applied to the left and the right front suspension [21,24]. The dynamic loading coefficient was the same as in the braking case, while only the translational degrees of freedom in vertical and longitudinal directions at rear suspensions were limited. Then, the stress due to dynamic torsion at interval was defined as [222.60,374.54] , MPa, Figure 7.

222.60 439.11 792.52 374.54 Figure 7. Comparison of stress due to dynamic torsion and ultimate tensile strength at the interval (unit: MPa).

3.2. Estimation of Stiffness under Two Conditions at Interval A deformation of the frame was the second aim of the analysis. It was reached basing on the frame stiffness. In order to estimate the component resistance on bending deformation, the weight of goods and carriage was applied as a kind of concentrated force in the frame, Figure 1. The translational degrees of freedom in vertical and lateral directions at front suspensions and the all translational degrees of freedom at rear suspensions were limited. Some of the 40 samples data are listed in Table 7. Fc and D are the applied force and the maximum deformation after loading, respectively. The contour of displacement in vertical direction for one sampling data is displayed in Figure 8. According to the simulated results of this bending condition and Equation (8), the bending stiffness of frame was of 9.65×× 1099 ,10.50 10 N·m2, Figure 9. They were exceeding the limit value Appl. Sci. 2019, 9, 4214  9 of 16 and therefore the frame should be redesigned.

Appl. Sci. 2019, 9, x FOR PEER REVIEW 9 of 16

Appl. Sci. 2019, 9, x FOR PEER REVIEW 9 of 16 Appl. Sci. 2019, 9, x FOR PEER REVIEWTable 7. Sample data and the corresponding responses. 9 of 16

Number x1 (mm) Tablex2 (mm) 7. Samplex3 (mm) data andx4 (mm)the corresponding x5 (MPa) responses.x6 Fc (N) D (mm) 1 45.23 Table17.00 7. Sample24.87 data and44.36 the corresponding 208.6 responses.0.256 2.851 × 106 2.435 Number x1 (mm) x2 (mm) x3 (mm) x4 (mm) x5 (MPa) x6 Fc (N) D (mm) 2 44.05 15.92 24.51 45.23 199.6 0.281 2.779 × 106 2.540 Numbe1 r x1 45.23 (mm) x217.00 (mm) x324.87 (mm) x444.36 (mm) x5 208.6(MPa) 0.256x6 2.851Fc (N) × 10 6 D2.435 (mm) 3 44.00 15.00 24.28 45.18 205.3 0.277 3.246 × 106 2.851 21 44.0545.23 15.9217.00 24.5124.87 45.2344.36 199.6208.6 0.2810.256 2.7792.851 × 106 2.5402.435 38 44.21 16.33 25.49 44.46 192.9 0.269 1.940 × 106 2.740 32 44.0044.05 15.0015.92 24.2824.51 45.1845.23 205.3199.6 0.2770.281 3.2462.779 × × 10 1066 2.8512.540 39 44.51 16.69 25.95 45.08 209.5 0.282 2.090 × 106 2.844 383 44.2144.00 16.3315.00 25.4924.28 44.4645.18 192.9205.3 0.2690.277 3.2461.940 ×× 101066 2.7402.851 40 44.41 15.62 24.26 45.49 198.0 0.274 1.630 ×106 3.334 3938 44.5144.21 16.6916.33 25.9525.49 45.0844.46 209.5192.9 0.2820.269 2.0901.940 × 106 2.8442.740 4039 44.4144.51 15.6216.69 24.2625.95 45.4945.08 198.0209.5 0.2740.282 2.0901.630 ××10 1066 3.3342.844 40 44.41 15.62 24.26 45.49 198.0 0.274 1.630 ×106 3.334 Figure 8. Contour of displacement in vertical direction under bending (unit: m). Figure 8. Contour of displacement in vertical direction under bending (unit: m).

9 3.28×109 9.65×109 10.50×109 20.28×10 9 9 9 9 3.28×10 9.65×10 10.50×109 20.28×109 Figure 9.3.28×10 Comparison9 of9.65×10 bending stiffness9 10.50×10 and its ultimate value at20.28×10 interval (unit: N·m2). 2 FigureFigure 9. 9.Comparison Comparison ofof bendingbending stistiffnessffness and and its its ultimate ultimate value value at at interval interval (unit: (unit: N N·mm2).). Similarly,Figure the 9. torsional Comparison stiffness of bending assessment stiffness of and frame its ultimate was conducted value at interval as the (unit:approach N·m· 2 ).mentioned above. The 40 sample data and their responses were obtained by finite element analysis. The contour Similarly,Similarly, the the torsional torsional sti stiffnessffness assessment assessment of of frame frame was was conductedconducted asas thethe approachapproach mentionedmentioned of displacement in vertical direction for one sampling data is displayed in Figure 10. The working above.above.Similarly, TheThe 4040 samplesample the torsional datadata andand stiffness theirtheir responses responsesassessment werewere of frame obtainedobtained was by byconducted finitefinite elementelement as theanalysis. analysis. approach TheThe mentioned contourcontour torsional stiffness of frame was calculated as follows: 1.35×× 1099 ,1.64 10 N·m2/rad, and compared ofofabove. displacementdisplacement The 40 sample inin verticalvertical data and directiondirection their responses forfor oneone samplingsampling were obtained datadata isis by displayed displayed finite element in in Figure Figure analysis. 10 10.. TheThe The workingworking contour h i of displacement in vertical direction for one sampling data×× is9 displayed999 in Figure22 10. The working torsionalwithtorsional the sti stiffnessultimateffness of oftorsional frame frame was was stiffness calculated calculated in asan as follows: intefollows:rsection1.35 1.35 number10 10, 1.64 ,1.64 of 10 10axis NN·m, Figurem /rad,/rad, 11. andand They comparedcompared were torsional stiffness of frame was calculated as follows: 1.35× ×× 1099 ,1.64× 10 N·m· 2/rad, and compared withintersected,with the the ultimate ultimate and thus torsional torsional some sti imff nessprovementstiffness in an in intersection of an frame inte rsectionis numberneeded. number of axis, Figureof axis 11, .Figure They were 11. intersected,They were andintersected,with thus the some ultimate and improvement thus torsional some im ofstiffnessprovement frame is in needed. ofan frame inte rsectionis needed. number of axis, Figure 11. They were intersected, and thus some improvement of frame is needed.

Figure 10. Contour of displacement in vertical direction under torsion (unit: m). Figure 10. Contour of displacement in vertical direction under torsion (unit: m).

9 9 9 2.76×109 0.84×10 1.35×10 1.64×10 9 9 9 9 Figure 11. Comparison0.84×10 of torsional1.35×10 stiﬀness and ultimate1.64×10 torsional stiﬀness2.76×10 at interval (unit: N m2/rad). Figure 11. Comparison0.84×10 of 9torsional1.35×10 stiffness9 and1.64×10 ultimate9 torsional 2.76×10stiffness at9 interval · (unit: N·m2/rad). Figure 11. Comparison of torsional stiffness and ultimate torsional stiffness at interval (unit: N·mFigure2/rad). 11. Comparison of torsional stiffness and ultimate torsional stiffness at interval (unit: N·m2/rad).

Appl. Sci. 2019, 9, x FOR PEER REVIEW 10 of 16

Appl. Sci. 2019, 9, 4214 10 of 16 4. Multi-Objectives Optimization of the Frame Based on Interval Analysis Method An analysis of strength and stiffness of the frame under exploitation indicated that the 4. Multi-Objectives Optimization of the Frame Based on Interval Analysis Method mechanical responses under several conditions could be better. Taking uncertain factors related to manufacturingAn analysis andof strength working and stage stiffnesss, theof the uncertain frame under multi exploitation-objectives indicated optimization that the design mechanical was responsesconducted. under several conditions could be better. Taking uncertain factors related to manufacturing and working stages, the uncertain multi-objectives optimization design was conducted. 4.1. Interval Multi-Objectives Optimization Method 4.1. Interval Multi-Objectives Optimization Method The uncertain parameters could be characterized as interval value, and the interval multiThe-objectives uncertain optimization parameters couldfunction be characterizedcould be described as interval as follows value, and[25]: the interval multi-objectives optimization function could be described as follows [25]:  min fi ( X , U )   s. t . : g ( X , U ) bILR b , b   minif (X, U i)  i i   i h i  i 1,2,...,I l ; X Ln R  s.t.: gi(X, U) b = b , b (11)   i i i  ILRILR≤ n   UUUUUUUUi = 1, 2,,,, ..., l; Xi  iΩ   i i  (11)  h i ∈ h i  I L Riq1,2,..., I L R  U U = U , U , Ui U = U , U  ∈ ∈ i i i  i = 1, 2, ..., q where fi (,) X U is objective function, gi (,) X U represents constraint function. X describes n wheren-dimensionalfi(X, U) isdesign objective variable function, withingi( Xthe, U )rangerepresents of  constraint, and U is function. q-dimensionalX describes uncertain n-dimensional variable designwithin variable the range within of U theI . rangeL and of ΩRnwere, and usedUis q-dimensional to mark upper uncertain and lower variable bound within of the range interval of Uparameter,I. L and R respectively.were used to markIn order upper to ensure and lower optimization bound of the efficiency interval and parameter, convergence, respectively. the uncertain In order toobjective ensure could optimization be transformed efficiency to and the convergence, certain one and the expressed uncertain objectiveby [25]: could be transformed to the certain one and expressed by [25]:  min(fcw ( X ), f ( X ))  min(f ( X , U )) max( f ( X , U ))   fXc () ( c( ) w( ))   min f X , f 2 X (12)  min( f (X,U))+max( f (X,U))  c max ( f ( X , U )) min( f ( X , U ))  f (Xw) = (12)  fX ( )  2  w max( f (X,U)) 2min( f (X,U)) f (X) = −2 where fXc () is the midpoint value of objective function, fXw() is the radius value of objective where f c(X) is the midpoint value of objective function, f w(X) is the radius value of objective function. function. Optimizing the midpoint value could improve mean value of objective function, Optimizing the midpoint value f c(X) could improve mean value of objective function, and optimizing and optimizing the radius value could reduce the sensitivity on the uncertainties and the radius value f w(X) could reduce the sensitivity on the uncertainties and enhance the stability and enhance the stability and robustness for the objective function. The solution of multi-objectives robustness for the objective function. The solution of multi-objectives optimization function could be optimization function could be transformed to search for the optimal result in the single-objective transformed to search for the optimal result in the single-objective optimization problem, which was optimization problem, which was evaluated by [25]: evaluated by [25]: minmin()f (Xf) = Xβfc( fXcw ()(1) X +  (1  β) f w ()( X ) (13) − where βis weightis weight coe coefficient.fficient. Its Its range range was was within within0 01β 1.. InIn this paper, it was taken as as 0.5 0.5 [25] [25].. ≤ ≤ In this this paper, paper the, the interval interval analysis analysis method method was utilizedwas utilized to optimize to optimize the frame the structure frame forstructure improving for mechanicalimproving mechanical performances. performances. The optimization The optimization process is shown process in is Figure shown 12 in. Figure 12.

Uncertainty Design Variables Variables

Latin Hypercube Experimental Design

Finite Element Simulation Analysis

Frame Strength Responses Bending Stiffness And Under Horizontal Bending, Emergency Braking, Torsion Torsional Stiffness Responses

Analytic Constructing An Hierarchy Approximate Model Process

Weight Coefficient Constructing Interval Multi-objective Functions

Searching For Optimal Solution Based On NSGA-II

N Whether To Meet The Optimization Conditions

Y Optimization Results Figure 12. Interval multi-objectivemulti-objective optimization process.process. Appl. Sci. 2019, 9, 4214 11 of 16

4.2. Determination of Optimization Function and Parameters According to the estimated results mentioned above, the strength under braking and the stiffness related to bending and torsion need to be improved. In order to avoid the conflict between optimized mechanical properties, all of them were considered as optimization objectives. In full consideration of uncertain factors during manufacturing and design stage, the thicknesses of top longitudinal beam, stiffener, lateral longitudinal beam and hanging ear in the frame were considered as design variables, and Young’s modulus x5 and Poisson’s ratio were taken as uncertain variables [26,27]. Moreover, the constraint of design and uncertain variables was within the range of 0.8x X 1.2x , i ≤ i ≤ i while the range of variation for the optimization objective was ascertained from the interval allowable values. The relationship between design variables and optimization objectives could be acquired through the response surface method. The 30 sample data for the four design variables and the six sample data for the uncertain variables were obtained by the Latin Hypercube sampling method. Some of them and their corresponding responses are listed in Table8. The y1, y2, y3, y4 and y5 represent the stress under dynamic bending, related to dynamic braking, connected with dynamic torsion, bending stiffness and torsional stiffness, respectively.

Table 8. Sample data and corresponding responses for constructing approximate model.

Number 1 2 3 4 5 6

x1 (mm) 50.28 44.07 41.59 46.55 36.62 38.48 x2 (mm) 18.76 19.20 16.99 16.77 15.89 18.32 x3 (mm) 27.69 22.41 22.07 30.00 26.90 28.28 x4 (mm) 44.69 48.41 36.62 47.79 39.72 50.28 x5 (MPa) 200.8 196.7 213.2 204.9 217.4 209.1 x6 0.262 0.273 0.278 0.284 0.267 0.257 y1 (MPa) 220.5 272.1 288.3 222.5 283.9 267.5 y2 (MPa) 362.1 430.5 448.5 353.7 418.7 394.9 y3 (MPa) 278.6 306.3 322.9 287.6 315.8 301.2 2 10 9 9 10 10 10 y4 (N m ) 1.095 10 9.395 10 9.544 10 1.118 10 1.011 10 1.048 10 y · × × × × × × 5 1.683 109 1.463 109 1.478 109 1.742 109 1.630 109 1.700 109 (N m2/rad) × × × × × × ·

Then, according to the simulated results, the response surface model of optimization objective function was ﬁtted by quadratic polynomial equation. Due to limited space, only model of strength in braking case was shown here:

y = 592390176.322597 6440001960.7152 x 387018782.531165 x 11501567583.2294 x + 340714204.106319 x 2 − × 1 − × 2 − × 3 × 4 +28383524629.0277 x 2 19085871274.882 x 2 + 53697949817.2065 x 2 3305903983.20956 x 2 × 1 − × 2 × 3 − × 4 (14) +5395879013.34058 x x + 80316087305.8226 x x + 34976216.0475603 x x × 1 2 × 1 3 × 1 4 +18318993227.7762 x x 942114760.313514 x x 1111863944.0763 x x × 2 3 − × 2 4 − × 3 4

The Pareto diagram analyzed from Equation (14) is shown in Figure 13. The design variable x3 and x1 had the most effect on optimization objective y2, and their influence ratios came to be –49.2% and –31.9%, respectively, which implied that they had negative correlation. The source item x1 – x2 had the most positive influence ratio. The main effect on optimization objective y2 is displayed in Figure 14, which shows that the variation of design variable x3 and x1 have the most influence on optimization objective y2. Then, the three-dimensional response surface between some of the design variables and optimization objective y1 is presented in Figures 15 and 16. It is clearly seen that there is an approximately linear relationship between certain stress due to dynamic loading y1 and design variable x1 and x3, while there was a non-linear relationship between the stress y1 and design variables x2 and x4. Moreover, in order to testify the established response of the surface model, the other 10 sample data and their corresponding simulated results were selected to compare with the predicted Appl. Sci. 2019, 9, x FOR PEER REVIEW 12 of 16

Appl. Sci. 2019, 9, x FOR PEER REVIEW 12 of 16 predicted ones obtained from the model. One of comparison results are shown in Figures 17 and 18. MostAppl.predictedAppl. Sci.ofSci. the2019 2019 ones ,points,9 9,, 4214 xobtained FOR were PEER aroundfrom REVIEW the the model. equal One line, of whichcomparison meant results that the are estimatedshown in Figuresresults from17 and1212 the ofof 18. 16 16 Appl. Sci. 2019, 9, x FOR PEER REVIEW 12 of 16 approximateMost of the responsepoints were surface around agreed the well equal with line, the which simulated meant results. that the estimated results from the approximatepredicted ones response obtained surface from agreedthe model. well One with of the comparison simulated resultsresults. are shown in Figures 17 and 18. onespredicted obtained ones fromobtained the model.from the One model. of comparison One of comparison results are results shown are in shown Figures in 17 Figures and 18 17. Mostand 18. of Most of the points were aroundx -x the equal line, which meant that the estimated results from the 1 2

Most of the points were aroundx 2 the equal line, which meant that the estimated results from the theapproximate points were response around surface the equal agreed2 line, whichwell with meant the thatsimulated the estimated results. results from the approximate x -x x 2 1 2 approximate response surface agreed4 well with the simulated results. response surface agreed well with the2 simulated results. x2 x2-x3 x 2 x 4 2 x1-x2

2x -x x x2 -x32 3 1x2 2 2 x x x 22 2 1 2x4 Sources items Sources

2 x x -x 3 2 1 x3 x4 2-x3 x 2 xx1-xx Sources items Sources 1 2 32 x -x

2 x 1 x 3 3 x23

2 x xx1 2 3 1 -50 -40 -30 -20 -10 0 10 Sources items Sources x2 x 3 Total effect on y2 x1 1-x3 Sources items Sources x -x -50 -40 -30 -20 -10 0 10 1 x31 Total effect on y2 xx 2 Figure1 3 13. Pareto diagram on y . x 3 -50 -40 -30 -20 -10 0 10 Figure 13. ParetoTotal effect diagram on y2 on y2. -50 -40 -30 -20 -10 0 10 x Total effect on y2 1

x2 Figure 13. Pareto diagram on y2. x 3 204.6 x1 Figure 13. Pareto diagram on y . x Figure 13. Pareto diagram on y22. 4 x 201.5 2 x 204.6 3 198.4 x x4 1 201.5 x x2 195.3 1 x3 198.4204.6 x2

192.2 x x4 204.6 3 195.3201.5 x 189.1 4 201.5 192.2198.4 186.0 Stress response value (MPa) 198.4 189.1195.3 182.9 195.3 186.0192.2 Stress response value (MPa) 192.2 182.9189.1low levels high Main effect 189.1186.0 Stress response value (MPa) low levels high 186.0 Stress response value (MPa) 182.9 Main effect 182.9 Figure 14. Main effect on y2. low levels high low Main effect high Figure 14. Mainlevels effect on y2. Main effect Figure 14. Main eﬀect on y2. Figure 14. Main effect on y2. Figure 14. Main effect on y2.

Figure 15. Three-dimensional response model with x1 and x3. Figure 15. Three-dimensional response model with x1 and x3. Figure 15. Three-dimensional response model with x1 and x3.

Figure 15. Three-dimensional response model with x1 and x3. Figure 15. Three-dimensional response model with x1 and x3.

Figure 16. Three-dimensional response model with x2 and x4. Figure 16. Three-dimensional response model with x2 and x4.

Figure 16. Three-dimensional response model with x2 and x4.

Figure 16. Three-dimensional response model with x2 and x4. Figure 16. Three-dimensional response model with x2 and x4. Appl. Sci. 2019, 9, x FOR PEER REVIEW 13 of 16

Appl.Appl. Sci. Sci.2019 2019, 9, ,9 4214, x FOR PEER REVIEW 1313 ofof 16 16

6 ) 2.0x10 Pa ( 1.9x106

6 6 ) 2.0x101.8x10 Pa

( 6 1.9x101.7x106

6 1.8x101.6x106

6 1.7x101.5x106

6 1.6x101.4x106

6 1.5x101.3x106

6 1.4x101.2x106

6 1.3x101.1x106

6 1.2x101.0x106 Radius value of predicted braking dynamic stress stress dynamic braking predicted of value Radius 1.0x106 1.1x106 1.2x106 1.3x1061.4x1061.5x1061.6x1061.7x106 1.8x106 1.9x106 2.0x106 1.1x106 Radius value of simulted braking dynamic stress (Pa) 1.0x106 Radius value of predicted braking dynamic stress stress dynamic braking predicted of value Radius 1.0x106 1.1x106 1.2x106 1.3x1061.4x1061.5x1061.6x1061.7x106 1.8x106 1.9x106 2.0x106 Radius value of simulted braking dynamic stress (Pa) Figure 17. Radius results between prediction and simulation. Figure 17. Radius results between prediction and simulation. Figure 17. Radius results between prediction and simulation. ) Pa ( 4.8x108

) 4.6x108 Pa ( 4.8x108 4.4x108 4.6x108 4.2x108 4.4x108 4.0x108 4.2x108 3.8x108 4.0x108 3.6x108 3.8x108 3.4x108 8 8 8 8 8 8 8 8 Midpointvalue predictedof braking stressdynamic 3.6x108 3.4x10 3.6x10 3.8x10 4.0x10 4.2x10 4.4x10 4.6x10 4.8x10 Midpoint value of simulted braking dynamic stress (Pa) 3.4x108 8 8 8 8 8 8 8 8 Midpointvalue predictedof braking stressdynamic 3.4x10 3.6x10 3.8x10 4.0x10 4.2x10 4.4x10 4.6x10 4.8x10 Figure 18. Midpoint resultsMidpoint value between of simulted braking predictiondynamic stress (Pa) and simulation. Figure 18. Midpoint results between prediction and simulation. 4.3. Optimized Results Figure 18. Midpoint results between prediction and simulation. 4.3. Optimized Results There were five optimization objectives for this frame, and thus the non-dominated sorting genetic 4.3. Optimized Results algorithmThere was were utilized five to optimization search for the objectives optimal solution for this in frame, the all and sample thus space the [non-dominated28,29]. Before starting sorting optimizationgeneticThere algorithm were design, five was the optimization utilized weight coeto ffiobjectivessearchcient for of everyforthe thisoptimal optimization frame, solution and objectivethus in thethe all hadnon-dominated sample to ascertain space firstly.sorting[28,29]. ThegeneticBefore analytic algorithmstarting hierarchy optimization was process utilized method design, to search was the usedweightfor the to calculatecooptimalefficient thesolution of coe everyffi cient.in optimizationthe Its all judgment sample objective matrixspace [28,29]. couldhad to beBeforeascertain estimated starting firstly. by [30optimization The]: analytic design, hierarchy the weightprocess co meefficientthod was of everyused tooptimization calculate the objective coefficient. had toIts judgment matrix could be estimated by [30]:  ascertain firstly. The analytic hierarchy α11 processα12 αme13thodα14 wasα15 used to calculate the coefficient. Its   judgment matrix could be estimated by α [30]:21 ααααα α22 α23 α24 α25   11 12 13 14 15  A =  α αααααα α α α  (15)  31 ααααα2132 2233 2334 2435 25   11 12 13 14 15   α = αααααα α α α   A41ααααα3142 3243 3344 3445 35  (15)  21 22 23 24 25   α αααααα α α α  51= ααααα4152 4253 4354 4455 45 A 31 32 33 34 35 (15) ααααα ααααα51 52 53 54 55 where αij is the ratio of importance between two41 optimization 42 43 44 objectives. 45 ααααα Accordingα to the analyzed results mentioned51 above 52 53 and 54 actual 55 engineering problem, the matrix A where ij is the ratio of importance between two optimization objectives. could be assigned by: whereAccording α is the to ratio the ofanalyzed importance results between mentioned two optimization above and actual objectives. engineering problem, the matrix ij  1 3/5 1 3/7 3/8  A could be assigned by:   According to the analyzed results 5 /mentioned3 1 6 abov/5e 5 /and7 actual 4/7  engineering problem, the matrix   A could be assigned by:  13/513/73/8 A =  1 5/6 1 4/7 4/8  (16)    7/35/3 7/5 1 7/4 6/5 1 5/7 7 4/7/8   13/513/73/8  A =1 5/6 1 4/7 4/8  (16) 8/35/3 7/4 1 8/ 6/54 8 5/7/7 4/7 1 7/3 7/5 7/4 1 7/8 A = 1 5/6 1 4/7 4/8  (16) The matrix A was processed to be normalized8/3 7/4 vector: 8/4 8/7 1 7/3 7/5 7/4 1 7/8 h 8/3 7/4 8/4 8/7 1 i The matrix A was processedU = 0.117 to be normalized 0.178 0.142 vector: 0.261 0.302 (17) The matrix A was processedU to= be[0.117 normalized 0.1780.142 vector: 0.261 0.302] (17) Therefore, U1, U2, U3, U4 and U5 were successively weight coefficient of five optimization U = [0.117 0.178 0.142 0.261 0.302] (17) objective.Therefore, Then, the U optimization1 , U2 , U3 , U design4 and wasU5 implementedwere successively by software weight ISIGHT(Engineous, coefficient of five Los optimization Angeles, CA,objective. USA), based Then, on the the optimization non-dominated design sorting was genetic implemented algorithm. by The software number ofISIGHT(Engineous, populations was 200, Los Therefore, U1 , U2 , U3 , U4 and U5 were successively weight coefficient of five optimization and the number of iterations was equal to 1000. After calculating for 51 minutes, the final optimized objective.Angeles, Then,CA, USA), the optimization based on thedesign non-dominated was implemented sorting by genetic software algorithm. ISIGHT(Engineous, The number Los of results are listed in Table9. Furthermore, the optimized objective values were also compared with Angeles,populations CA, was USA), 200, based and theon numberthe non-dominated of iterations sortingwas equal genetic to 1000. algorithm. After calculatingThe number for of51 the initial ones, Table 10. According to the optimized results, the thickness of four design variables populationsminutes, the was final 200, optimized and the resultsnumber are of listediterations in Table was 9.equal Furthermore, to 1000. Afterthe optimized calculating objective for 51 minutes, the final optimized results are listed in Table 9. Furthermore, the optimized objective Appl. Sci. 2019, 9, 4214 14 of 16 increased, and the top longitudinal beam had the largest increase in thickness. Then, the optimization with respect to the stress related to dynamic horizontal bending, unexpected braking (fracturing) and torsion were decreased by 19.18%, 17.25% and 9.06%, respectively. The bending and torsional stiffness were improved by 21.53% and 15.88%. Obviously, the mechanical responses of this frame were enhanced with a slight increase in weight, which achieved the goal of optimization.

Table 9. Comparison of the initial and optimized design variables.

Design Variables Initial Value Optimized Value

x1 (mm) 45 54 x2 (mm) 16 19 x3 (mm) 25 30 x4 (mm) 45 50

Table 10. Comparison of the initial and optimized certain objective values.

2 2 Type y1 (MPa) y2 (MPa) y3 (MPa) y4 (N m ) y5 (N m /rad) · · Initial value 253.4 404.7 300.2 9.96 109 1.599 109 × × Optimized 204.8 334.9 273.0 1.210 1010 1.853 109 value × ×

5. Conclusions Based on the experimental data and finite element analysis results, the ultimate strength and stiffness of electric wheel dump truck frame at interval can be obtained through blind number theory. The stresses under dynamic horizontal bending, unexpected braking (fracturing) and torsion as well as the stiffness under bending and torsion were estimated. The thickness of main bearing beams in frame was considered as a design variable, and the material properties were regarded as uncertain variables caused by manufacturing process. Based on the response surface method, the interval multi-objectives optimization function was obtained. Finally, the non-dominated sorting genetic algorithm was used to search for the optimal solution. The specific conclusions are listed as follows: 1. Based on the blind number theory, the ultimate strength of electric wheel dump truck welded frame at interval was [439.11, 792.52] MPa. The bending and torsional stiffness h i h i were 3.28 109, 20.28 109 N m2 and 0.84 109, 2.91 109 N m2/rad. Using Latin Hypercube × × · × × · method to acquire sample data, the stresses due to dynamic horizontal bending under horizontal bending, braking and torsional conditions were [211.11, 357.10] MPa, [343.31, 461.20] MPa and h i [222.60, 374.54] MPa, The stiffness under bending and torsion was 9.65 109, 10.50 109 N m2 h i × × · and 1.35 109, 1.64 109 N m2/rad. × × · 2. In full consideration of the uncertain factors caused by the manufacturing process, Young’s modulus and Poisson’s ratio were considered as uncertain variables. The thickness of four main bearing beams was regarded as a design variable. Based on the non-dominated sorting genetic algorithm, the stresses due to dynamic horizontal bending in horizontal bending, unexpected braking (fracturing) and torsion were decreased by 19.18%, 17.25% and 9.06%, respectively. The bending and torsional stiffness were improved by 21.53% and 15.88%. In the meantime, all design variables were increased by approximately 15%.

Author Contributions: Conceptualization, C.M. and J.L. (Jinhua Liu); Methodology, C.M.; Software, W.L.; Validation, W.L., X.X. and J.L. (Jidong Liu); Formal analysis, C.M.; Investigation, C.M.; Resources, R.M.; Data Curation, X.X.; Writing—Original Draft Preparation, C.M.; Writing—Review & Editing, Q.Y.; Visualization, X.X.; Supervision, Q.Y.; Project Administration, C.M.; Funding Acquisition, C.M. and W.L. Funding: This research was funded by Excellent Youth Project of Hunan Education Department (Grant No.: 18B301) and Postgraduate Research Innovation Project of Hunan Education Department (Grant No.: CX20190846) and National Natural Science Foundation of China (Grant No.: 51975192). Appl. Sci. 2019, 9, 4214 15 of 16

Acknowledgments: The authors gratefully thank for the support of Excellent Youth Project of Hunan Education Department (Grant No.: 18B301) and Postgraduate Research Innovation Project of Hunan Education Department (Grant No.: CX20190846) and National Natural Science Foundation of China (Grant No.: 51975192). Conﬂicts of Interest: The authors declare no conﬂict of interest. The funders had no role in the design of the study; in the collection, analyses, or interpretation of data; in the writing of the manuscript, and in the decision to publish the results.

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