外文翻譯--電梯傳動(dòng)軸的事故分析 英文版

Faculty of Mechanical Engineering, Istanbul Technical University, Gmssuyu 34437, Taksim, Istanbul, TurkeyNotch effectFinite element methodture occurred due to faulty design or manufacturing of the keyway (low radius of curvatureinfloortheworm gear is 28.6. The shaft rotates the pulley by a key. The four main ropes of the elevator are placed on the pulley and byrevolving in both direction of the drive shaft and pulley, the elevator moves up and down. Service speed of the elevator is0.6 m/s. The shaft is supported in three points in form of journal bearings (Fig. 2).1350-6307/$ - see front matter C211 2008 Elsevier Ltd. All rights reserved.* Corresponding author.E-mail addresses: .tr (A. Gksenli), .tr (I.B. Eryrek).Engineering Failure Analysis 16 (2009) 10111019Contents lists available at ScienceDirectEngineering Failure Analysisjournal homepage: /locate/engfailanaldoi:10.1016/j.engfailanal.2008.05.014and electric engine was broken, which lead the break system of the engine to fail. Due to the weight of balance weight, ele-vator lifted upwards with a huge speed. During the accident nobody was injured.2. Analysis of elevator drive systemDuring analysis of elevator drive system, it is concluded that torque, which is produced by an electric motor, is transmit-ted by a worm gear to the shaft. The electric engine, which produces 6.5 HP, is rotating 1500 rpm and reduction ratio of the1. IntroductionThe elevator drive shaft investigatedbottom of the building (Fig. 1). It is reportedis used in a building consisting of eightthe accident, two persons were leavingat keyway corner, causing high notch effect). In conclusion effect of change in radius of cur-vature on stress distribution is explained by using FEM and precautions which have to betaken to prevent a similar failure is clarified.C211 2008 Elsevier Ltd. All rights reserved.this paper was in service for 30 years. Elevator drive system is mounted at thethat no maintenance was applied on the shaft during operation life. The elevatorand 16 apartments. The elevator has a four person (320 kg) capacity. Just beforeelevator and because of sudden fracture of the shaft, connection between pulleyarticle infoArticle history:Received 23 May 2008Accepted 25 May 2008Available online 11 June 2008Keywords:Drive shaftElevatorFailure analysisabstractIn this study failure analysis of an elevator drive shaft is analyzed in detail. Failure occurredat the keyway of the shaft. Microstructural, mechanical and chemical properties of theshaft are determined. After visual investigation of the fracture surface it is concluded thatfracture occurred due to torsional-bending fatigue. Fatigue crack has initiated at the key-way edge. Considering elevator and driving systems, forces and torques acting on the shaftare determined； stresses occurring at the failure surface are calculated. Stress analysis isalso carried out by using finite element method (FEM) and the results are compared withthe calculated values. Endurance limit and fatigue safety factor is calculated, fatigue cycleanalysis of the shaft is estimated. Reason for failure is investigated and concluded that frac-Failure analysis of an elevator drive shaftA. Gksenli*, I.B. Eryrek1012 A. Gksenli, I.B. Eryrek/Engineering Failure Analysis 16 (2009) 101110193. Visual investigation of fracture surfaceAfter primary visual investigation it is revealed that fracture occurredat the keyway where the pulley is fasten at the shaft(Fig. 3).After analyzing fracture surface (Fig. 4), a typical torsional-bending fatigue fracture surface is detected 1. Fatigue cracksinitiated at the corners of the keyway and moved almost along the whole surface. The small area of brittle fracture surfaceindicates a low applied stress. Fatigue lines were detected only near to the brittle fracture region. This might be due to thefriction of two separated surfaces with each other, causing the formed fatigue lines to disappear.4. Properties of materialAs no information with respect to the chemical composition of the shaft material was available, the first task in the failureanalysis was the material identification. To determine shaft material, chemical, mechanical properties and microstructuralanalysis was carried out.Fig. 1. Elevator system inside the building.Fig. 2. Elevator drive system.A. Gksenli, I.B. Eryrek/Engineering Failure Analysis 16 (2009) 10111019 1013Fig. 3. Failed shaft.4.1. Chemical analysisThe chemical analysis of the shaft was carried out by atomic absorption spectroscopy and is reported in Table 1.4.2. MicrostructureThe microstructure of the shaft material was developed by etching, after diamond polishing, with 2% Nitral solution andwas observed under the microscope (Fig. 5). A ferriticpearlitic and fine grain structure can be clearly seen.4.3. Mechanical propertiesTensile and hardness tests were performed to determine the mechanical properties of the shaft which can be seen inTable 2.Considering mechanical, chemical and microstructural analyze results, shaft material is estimated as St52.0. Tensile, yield,elongation and hardness values are suitable for the catalogue values of St52.0 according DIN 1629 2.Fig. 4. Fracture surface.Table 1Chemical composition of the shaftC 0.22 Al 0.012 P 0.031Si 0.40 Mn 0.13 S 0.0291014 A. Gksenli, I.B. Eryrek/Engineering Failure Analysis 16 (2009) 101110195. Stress analysisBy stress analysis, minimum and maximum normal and shear stress values occurring at the fracture surface during oper-ation is investigated. At first, forces and torques acting on the shaft are determined. By analyzing minimum stress value, onlythe weight of the empty cabin (420 kg) and balance weight (580 kg) is considered. In this case, reaction forces causes a bend-ing moment of 437.4 N m at the fracture surface resulting as normal stress value of 20.6 MPa. Shear forces, caused due to theloads of empty cabin, ropes and balance weights, forms a shear stress of 3.5 MPa. By analyzing maximum stress value, bal-ance weight, cabin weight with four persons inside (each person is 80 kg and the total weight of the cabin is 740 kg), torsionmoment and impact ratio is considered. In this case bending moment of 571 N m occurs at the fracture surface causing anormal stress value of 27 MPa. Shear stress value, due to shear force, is 4.7 MPa. The 6.5 HP electric engine rotates at1500 rot/min, conversion ratio of the worm gear is 28.6 and drive system efficiency (due to worm gear mechanism) isTable 2Mechanical properties of the shaftYield strength (MPa) Tensile strength (MPa) Rapture elongation (%) Hardness (BHN)339 569 18 1650.7. ConsidericircumstancByapproximatTheoretical,is takenevanttigueFig. 5. Microstructure of the shaft material (C2200).ng these parameters, torsion moment is calculated as 887.7 N m. Total shear stress is calculated under thesees as 25.7 MPa.visual examination it was determined that the transition from keyway ground to keyway side surface (corner) wasely perpendicular and almost no radius of curvature (RC) was observed (Fig. 6).RC cannot be zero, it can only reach a value of 0.4 mm by fine milling cutter 3. Therefore by calculations RCas 0.4 mm, causing an enormous high notch effect. The theoretical notch effect is analyzed in two states (using rel-tables 4)； shear notch effect (aS) which is determined as 2.93 and bending notch effect aBis determined as 2.72. Fa-notch factor (b) considering geometry and material of the shaft can be calculated as 4:b 1aC01C1g 1Fig. 6. Failed shaft and fracture surface.where g is notch sensitivity factor and its value is 0.85 4. Using Eq. (1) shear fatigue notch effect (bS) is calculated as 2.64and normal fatigue notch effect (bN) as 2.46. Impact ratio coefficient is taken as 1.2. Considering shear and normal stresses,impact ratio (q) and fatigue notch effects, equivalent stress (rEQ) is calculated using Shape Deformation Energy Hypothesis”5 asrEQrC1qC1 bN23C1 sC1qC1 bS2q2The summarized results can be seen in Table 3.5.1. Fatigue strength analysisity is zero) and with no acceleration, maximum stress value by acceleration of the elevator and four persons (each person isassumedEachestimated by following assumptions: The total elevator displacement inside the building is by eight floors approximately21 m.ForTable 3Stress, forceA. Gksenli, I.B. Eryrek/Engineering Failure Analysis 16 (2009) 10111019 1015Normal stress (MPa) Shear stress (MPa) Equivalent stress (MPa)Shear force Torque TotalMinimum 50.6 9.2 9.2 54Maximum 79.7 15.1 67.3 82.4 162and torque values occurring at the fracture surface during operationing to Juvinall and Shigley 7,8, stress value (rF) where fatigue failure cycle at 103cycles occur, can be calculated asrF m C1UTS 7further fatigue analysis, stress-cycle (SN) curve of the shaft is estimated. To draw the SN curve of the shaft, accord-living. Each person is using the elevator twice a day. The elevator has been used 340 days in a year and shaft is in operationfor 30 years. The diameter of the pulley is 400 mm and by each rotation the elevator is moving 1.257 m. Considering theseexplanations, total life of the shaft was approximately 6.8 C2 106cycles.The building consists of 16 apartments. In half of the apartments there are two and other half there are three personsempty, each maximum peak demonstrates the stress value occurring at the time of cruise of the elevator with person(s) in-side the cabin and considering acceleration.To calculate average stress value, we have to transform Variable-amplitude stress” into Constant-amplitude stress” byassuming that always maximum stress occurs at the fracture surface (always four persons are transported). According to thisassumption, change of stress value versus time can be seen in Fig. 8. In this case average stress value is 108 MPa.Considering average stress value, before calculated endurance limit is modified. By the analysis, criteria of Goodman 6 isregarded (Fig. 9) and modified endurance limit rIIEC0C1is calculated asrIIE rIEC1 1C0rM=UTS208C11C0108=569170 5Considering modified endurance limit rIIEC0C1and equivalent stress (rEQ), fatigue safety factor (t) can be calculated ast rIE=rEQ 170=162 1:05 65.2. Fatigue life analysisFrom Eq. (6) can be seen that fatigue safety factor is quite very low (t = 1.05). For detailed fatigue life analysis, life cycle isminimum peak demonstrates the stress value at the fracture surface at the time the elevator is not moving and ismum transportation number of person (four). Therefore a Variable-amplitude stress” occurs at the fracture surface depend-ing on the number of persons inside the cabin (Fig. 7).to be 80 kg) inside the cabin. But in real, number of persons transported inside the cabin is not always the maxi-Fatigue strength (endurance limit rE) of the shaft material was calculated as 6rE 0:5C1UTS 0:5C1568 284 MPa 3Considering size factor (kD) for 60 mm diameter as kD= 0.77 and surface factor (fine polished) kS= 0.95 6； the newendurance limit rIEC0C1is calculated asrIE rEC1kSC1 kD 284C10:77C10:95 208 MPa 4rIE 208 MPa. From our calculations done before we know that average stress value (rM) is not zero. To determine theeffect of rMon endurance limit, we have to know or at least to estimate rM. To calculate rM, we have to consider the changeof stress value occurring at the fracture surface in time. But this is quite difficult. By our calculations done before we calcu-lated maximum and minimum stress values. Minimum stress value occurred at empty cabin and stationary position (veloc-1016 A. Gksenli, I.B. Eryrek/Engineering Failure Analysis 16 (2009) 10111019where m = 0.9 for bending. The stress value at Nf=103cycles according Eq. (7) is 512 MPa. The second point at the graphis the endurance limit value, which is 106cycles for steel. According to these explanations, estimated SN diagram can beseen in Fig. 10.By low stress values and high cycle fatigue, failures can occur between 106and 107cycles 9. Therefore estimating lifecycle value of 6.8 C2 106cycles for fatigue failure supports our thesis.Fig. 7. Stresstime graphic acting on the fracture surface depending on the number of persons inside the cabin (Variable-amplitude load).Fig. 8. Stresstime graphic acting on the fracture surface (Constant-amplitude load).Fig. 9. Determining modified endurance limit considering average stress value of 108 MPa (according to Goodman).Fig. 10. Estimated stress-cycle diagram.A. Gksenli, I.B. Eryrek/Engineering Failure Analysis 16 (2009) 10111019 10176. Finite element analysisTo examine stress distribution at the keyway and fracture surface, finite element method (FEM) was applied. By the anal-ysis ANSYS program was used. A precise geometrical model of the shaft was built up. Since the shaft is too long to be ana-lyzed completely as a three-dimensional object, only the keyway and fracture region was modeled in detail (Fig. 11).High stress regions, especially at the corner of the keyway can be clearly seen. The dominant effect of the raise in stressvalues is the low radius of curvature causing a high notch effect. The aim of stress analysis using FEM was to verify our stresscalculations done before. From Fig. 11 can be seen that stress values at the fracture surface were close to the calculated ones.7. DiscussionBy increasing radius of curvature (RC) value, stresses occurring at the keyway corner could be decreased effectively. Todetermine the effect of RC on stress distribution, finite element analysis is carried out. By this examination, RC-value wasincreased stepwise for visual analysis of decrease in stress values, which can be seen in Fig. 12.Dramatic decrease of stress values at keyway corner can be clearly seen. For further investigation, the effect of change inRC on stress and fatigue safety factor is analyzed in detail which can be seen in Figs. 13 and 14.By increasing radius of curvature even from 0.5 mm to 2 mm would decrease stress value from 163 to 104 MPa and anincrease in fatigue safety factor from 1.05 to 1.78. Figs. 13 and 14 demonstrate that an increase of radius of curvature wouldFig. 11. Finite element mesh of the shaft.1018 A. Gksenli, I.B. Eryrek/Engineering Failure Analysis 16 (2009) 10111019probably prevent the failure of the elevator drive shaft. In conclusion it is determined that fracture of the shaft occurred dueto faulty design or manufacturing of the keyway (low radius of curvature), causing a high notch effect.Fig. 12. Effect of radius of curvature (RC) on stress distribution using FEM.Fig. 13. Effect of RC on stress distribution.Fig. 14. Effect of RC on fatigue safety factor.8. SummaryFailure analysis of an elevator drive shaft is investigated in detail. Microstructural, mechanical