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最大加工直径320mm的卧式车床的主运动系统设计【Dmax=320mm Nmin=85rmin φ=1.26 Z=10 最大 加工 直径 320 mm 卧式 车床 运动 系统 设计 Dmax Nmin

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最大加工直径320mm的卧式车床的主运动系统设计【Dmax=320mm Nmin=85rmin φ=1.26 Z=10】,最大加工直径320mm的卧式车床的主运动系统设计【Dmax=320mm,Nmin=85rmin,φ=1.26,Z=10,最大,加工,直径,320,mm,卧式,车床,运动,系统,设计,Dmax,Nmin

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课程设计报告320mm的卧式车床的主运动系统设计5设计任务书车床的主参数（规格尺寸）和基本参数如下：完成最大加工直径 320 mm 的卧式车床的主运动系统设计，主要参数如下：转速级数 Z=10公比 1.26 最低转速 nmin=85转/分工件最大回转直径D(mm)正转最低转速nmin( )转速级数公比32085101.26目 录设计任务书2目 录4第1章设计要求及目的6第2章 机床主参数的确定82.1 确定转速范围82.2 确定电动机型号82.3拟定机床传动方案92.3.1 传动系统扩大顺序的安排102.3.2 绘制结构网102.3.3 传动组的变速范围的极限值112.3.4最大扩大组的选择112.4 绘制转速图和传动系统图122.5 确定各变速组此论传动副齿数13第3章 传动件的计算143.1 带传动设计143.2选择带型153.3确定带轮的基准直径并验证带速153.4确定中心距离、带的基准长度并验算小轮包角163.5确定带的根数z173.6确定带轮的结构和尺寸173.7确定带的张紧装置173.8 验算主轴转速误差193.9 计算转速的计算203.10 齿轮模数计算及验算213.11 传动轴最小轴径的初定26第4章 主要零件的设计与验算304.1齿轮强度的校核验算304.2 轴的校核314.3 轴承寿命校核33参考文献35 第1章 设计要求及目的一、设 计 任 务：完成最大加工直径 320 mm 的卧式车床的主运动系统设计，主要参数如下：转速级数 Z=10公比 1.26 最低转速 nmin=85转/分二、设 计 要 求：1.完成传动系统设计：方案合理，运动设计和动力计算正确；2.完成主轴箱草图和展开图设计（计算机打印A0）：布局合理、结构紧凑，内容表达完整、正确，图纸规范整洁。3.完成1张主轴箱的剖面图（A1）：能综合反映各轴的空间位置，操纵机构安排合理、表达清楚，主轴中心高适当；4.完成主轴零件图设计（A1）：结构合理，形位公差和表面粗糙度等技术要求标注合理，尺寸标注完整正确。5.完成计算说明书一份（25页）：包括目录、设计任务书、总论或前言（概述）、参数、运动设计的分析和拟定、动力计算、结构的选择和分析及必要的说明、设计心得体会、参考文献（书目前排列序号，以便正文引用）。要求条理清楚，计算、分析准确。机床技术参数有主参数和基本参数，他们是运动传动和结构设计的依据，影响到机床的加工能力、决定和影响其他基本参数的依据，如车床的最大加工直径，一般在设计题目中给定，基本参数是一些加工件尺寸、机床结构、运动和动力特性有关的参数，可归纳为尺寸参数、运动参数和动力参数。通用车床工艺范围广，所加工的工件形状、尺寸和材料各不相同，有粗加工又有精加工；用硬质合金刀具又用高速钢刀具。因此，必须对所设计的机床工艺范围和使用情况做全面的调研和统计，依据某些典型工艺和加工对象，兼顾其他的可能工艺加工的要求，拟定机床技术参数，拟定参数时，要考虑机床发展趋势和同国内外同类机床的对比，使拟定的参数最大限度地适应各种不同的工艺要求和达到机床加工能力下经济合理。机床主传动系因机床的类型、性能、规格和尺寸等因素的不同，应满足的要求也不一样。设计机床主传动系时最基本的原则就是以最经济、合理的方式满足既定的要求。在设计时应结合具体机床进行具体分析，一般应满足的基本要求有：满足机床使用性能要求。首先应满足机床的运动特性，如机床主轴油足够的转速范围和转速级数；满足机床传递动力的要求。主电动机和传动机构能提供足够的功率和转矩，具有较高的传动效率；满足机床工作性能要求。主传动中所有零部件有足够的刚度、精度和抗震性，热变形特性稳定；满足产品的经济性要求。传动链尽可能简短，零件数目要少，以便节约材料，降低成本。36第2章 机床主参数的确定2.1 确定转速范围根据=1.26因为已知 ，查标准数列表取最大转速因为=1.26=1.064，根据【1】表3-6标准数列。首先找到最小极限转速85，再每跳过3个数取一个转速，即可得到公比为1.26的数列： 85，106，132，170，212，265，335，425，530，6702.2 确定电动机型号合理地确定电机功率N，使用的功率实际情况既能充分的发挥其使用性能，满足生产需要，又不致使电机经常轻载而降低功率因素。目前，确定机床电机功率的常用方法很多，而本次设计中采用的是：估算法，它是一种按典型加工条件（工艺种类、加工材料、刀具、切削用量）进行估算。根据此方法，中型车床典型切削条件下的用量：1）主（垂直）切削力： 2）切削功率： N切 = 3）估算主电动机功率：根据以上条件，选定主电机：a.电机功率N：根据机床切削能力的要求确定电机功率。但电机产品的功率已经标准化，因此，按要求应选取相近的标准值。 N=5.5kwb.电机转速n电机的转速选择的是： n=1440r/min 这个选择是根据电机的转速与主轴最高转速nmax和轴的转速相近或相宜，以免采用过大的升速或过小的降速传动。c.电机的安装和外形 根据电机不同的安装和使用的需要，有四种不同的外形结构，用的最多的有底座式和法兰式两种。本次设计的机床所需选用的是外行安装尺寸之一。具体的安装图可由手册查到。d.常用电机的资料根据常用电机所提供的资料，选用： Y132S-42.3拟定机床传动方案级数为Z的传动系统由若干个顺序的传递组组成，各传动组分别有Z1、Z2、Z3、个传动副.即Z=Z1Z2Z3传动副数为使结构尽量简单以2或3为适合，即变速级数Z应为2和3的因子： 即 Z=2a3b实现12级主轴转速变化的传动系统可以写成多种传动副的组合:1) 12=34 2) 12=433) 12=322 4) 12=2325) 12=223方案1）和方案2)可省掉一根轴。但有一个传动组有四个传动副。若用一个四联滑移齿轮，则将大大增加其轴向尺寸；若用两个双联滑移齿轮，则操纵机构必须互锁以防止两个滑移齿轮同时啮合。将使得结构比较复杂。故在此不予采用。 按照传动副“前）多后少”的原则选择Z=322这一方案，但主轴换向采用双向片式摩擦离合器结构，致使轴的轴向尺寸过大，所以此方案不宜采用，加之主轴对加工精度、表面粗超度的影响最大。因此在主轴的传动副不宜太多，故方案5）亦不采用。而应先择12=232。综上所述： 方案4） 12=232 是比较合理的 2.3.1 传动系统扩大顺序的安排12=232的传动副组合，其传动组的扩大顺序又可以有6种形式：1) 12=213226 2) 12=2134223) 12=233126 4) 12=2631235) 12=223421 6) 12=263221以上各种结构式方案中，由于传动副的极限传动比和传动组的极限变速范围的限制，一般升速时。极限变速范围。检查传动组的变速范围时，只需检查最后一个扩大组。因其他传动组的变速范围都比他小。由式 对于方案2）和 方案5）有：,则对于方案2）和 方案5）不予考虑。对于其余方案有：。然而在可行的结构式方案1）、3）、4）、6）中，为了使中间轴变速范围最小，在各方案同号传动轴的最高转速相同时，变速范围越小，最低转速越高，转矩越小，传动件尺寸也就越小。比较方案1）、3）、4）、6），方案1）的中间传动轴变速范围最小，方案1）最佳。但由于轴装有摩擦离合器，在结构上要求有一齿轮的齿根圆大于离合器的直径因而采用方案3）12=233126 最佳2.3.2 绘制结构网 由上选择的结构式12=233126 画其结构图如下：图2.1结构网2.3.3 传动组的变速范围的极限值齿轮传动最小传动比Umin1/4,最大传动比Umax,决定了一个传动组的最大变速范围rmax=umax/umin。因此，要按照下表，淘汰传动组变速范围超过极限值的所有传动方案。极限传动比及指数X,X,值为：表2.1 公比极限传动比指数1.41X值：Umin=1/44X值：Umax=x, =22(X+ X)值：rmin=x+x=862.3.4最大扩大组的选择正常连续的顺序扩大组的传动的传动结构式为：Z=Z11Z2Z1Z3Z1Z2最后扩大组的变速范围按照r原则，导出系统的最大级数Z和变速范围Rn为：表2.2 Z3 2 3 1.41 Z=12 Rn=44 Z=9 Rn=15.6 最后扩大组的传动副数目Z3=2时的转速范围远比Z3=3时大因此，在机床设计中，因要求的R较大，最后扩大组应取2更为合适。同时，最后传动组与最后扩大组往往是一致的。安装在主轴与主轴前一传动轴的具有极限或接近传动比的齿轮副承受最大扭距，在结构上可获得较为满意的处理，这也就是最后传动组的传动副经常为2的另一原因。2.4 绘制转速图和传动系统图（1）选择电动机：采用Y系列封闭自扇冷式鼠笼型三相异步电动机。（2）绘制转速图：（3）画主传动系统图。根据系统转速图及已知的技术参数，画主传动系统图如图2-3：1-2轴最小中心距：A1_2min1/2(Zmaxm+2m+D)轴最小齿数和:Szmin(Zmax+2+D/m)2.5 确定各变速组此论传动副齿数(1)Sz100-124,中型机床Sz=70-100(2)直齿圆柱齿轮Zmin18-24,m4 图2-3 主传动系统图（7）齿轮齿数的确定。变速组内取模数相等，据设计要求Zmin1824,齿数和Sz100124，由表4.1，根据各变速组公比，可得各传动比和齿轮齿数，各齿轮齿数如表2-2。 表2-2 齿轮齿数传动比基本组第一扩大组第二扩大组1:1.261:21:3.161:1.261:1.581.58:11：1.58代号ZZZZZZZZZ5Z5ZZZ7Z7齿数3342 255018573038 2642 61383861第3章 传动件的计算3.1 带传动设计输出功率P=5.5kW,转速n1=1440r/min,n2=670r/min计算设计功率Pd表4 工作情况系数工作机原动机类类一天工作时间/h10161016载荷平稳液体搅拌机；离心式水泵；通风机和鼓风机（）；离心式压缩机；轻型运输机1.01.11.21.11.21.3载荷变动小带式运输机（运送砂石、谷物），通风机（）；发电机；旋转式水泵；金属切削机床；剪床；压力机；印刷机；振动筛1.11.21.31.21.31.4载荷变动较大螺旋式运输机；斗式上料机；往复式水泵和压缩机；锻锤；磨粉机；锯木机和木工机械；纺织机械1.21.31.41.41.51.6载荷变动很大破碎机（旋转式、颚式等）；球磨机；棒磨机；起重机；挖掘机；橡胶辊压机1.31.41.51.51.61.8根据V带的载荷平稳，两班工作制（16小时），查机械设计P296表4，取KA1.1。即3.2选择带型普通V带的带型根据传动的设计功率Pd和小带轮的转速n1按机械设计P297图1311选取。根据算出的Pd6.05kW及小带轮转速n11440r/min ，查图得：dd=80100可知应选取A型V带。3.3确定带轮的基准直径并验证带速由机械设计P298表137查得，小带轮基准直径为80100mm则取dd1=100mm ddmin.=75 mm（dd1根据P295表13-4查得）表3 V带带轮最小基准直径槽型YZABCDE205075125200355500由机械设计P295表13-4查“V带轮的基准直径”，得=224mm 误差验算传动比： （为弹性滑动率）误差 符合要求 带速 满足5m/sv300mm，所以宜选用E型轮辐式带轮。总之，小带轮选H型孔板式结构，大带轮选择E型轮辐式结构。带轮的材料：选用灰铸铁，HT200。3.7确定带的张紧装置 选用结构简单，调整方便的定期调整中心距的张紧装置。对带轮的主要要求是质量小且分布均匀、工艺性好、与带接触的工作表面加工精度要高，以减少带的磨损。转速高时要进行动平衡，对于铸造和焊接带轮的内应力要小, 带轮由轮缘、腹板（轮辐）和轮毂三部分组成。带轮的外圈环形部分称为轮缘，轮缘是带轮的工作部分，用以安装传动带，制有梯形轮槽。由于普通V带两侧面间的夹角是40，为了适应V带在带轮上弯曲时截面变形而使楔角减小，故规定普通V带轮槽角 为32、34、36、38（按带的型号及带轮直径确定），轮槽尺寸见表7-3。装在轴上的筒形部分称为轮毂，是带轮与轴的联接部分。中间部分称为轮幅（腹板），用来联接轮缘与轮毂成一整体。表 普通V带轮的轮槽尺寸（摘自GB/T13575.1-92） 项目 符号 槽型 Y Z A B C D E 基准宽度 b p 5.3 8.5 11.0 14.0 19.0 27.0 32.0 基准线上槽深 h amin 1.6 2.0 2.75 3.5 4.8 8.1 9.6 基准线下槽深 h fmin 4.7 7.0 8.7 10.8 14.3 19.9 23.4 槽间距 e 8 0.3 120.3 150.3 190.4 25.5 0.5 37 0.6 44.5 0.7 第一槽对称面至端面的距离 f min 6 7 9 11.5 16 23 28 最小轮缘厚 5 5.5 6 7.5 10 12 15 带轮宽 B B =( z -1) e + 2 f z 轮槽数 外径 d a 轮 槽 角 32 对应的基准直径 d d 60 - - - - - - 34 - 80 118 190 315 - - 36 60 - - - - 475 600 38 - 80 118 190 315 475 600 极限偏差 1 0.5 V带轮按腹板（轮辐）结构的不同分为以下几种型式： （1） 实心带轮：用于尺寸较小的带轮(dd(2.53)d时)，如图7 -6a。 （2） 腹板带轮：用于中小尺寸的带轮(dd 300mm 时)，如图7-6b。 （3） 孔板带轮：用于尺寸较大的带轮(ddd) 100 mm 时)，如图7 -6c 。 （4） 椭圆轮辐带轮：用于尺寸大的带轮(dd 500mm 时)，如图7-6d。（a） （b） （c） （d）图7-6 带轮结构类型根据设计结果，可以得出结论：小带轮选择实心带轮，如图（a）,大带轮选择腹板带轮如图（b）3.8 验算主轴转速误差 由于确定的齿轮齿数所得的实际转速与传动设计的理论转速难以完全相符，需要验算主轴各级转速，最大误差不得超过10(-1)%。主轴各级实际转速值用下式计算n实=nd(1-)u1u2u3u4其中： 滑移系数=0.2u1、 u2 、u3 、u4分别为各级的传动比 转速误差用主轴实际转速与标准转速相对误差的绝对值表示n=10(-1)% 实际转速及转速误差如下:表2.5各级传动组的转速误差主轴转速n1n2n3n4n5n6n7n8n9n10理论转速85106132170212265335425530670实际转速85.8107.2134.5173.2214.6266.4336.4418.5535.6674.3转速误差 （%）0.70.4 0.20.320.170.320.320.460.270.3故转速误差满足要求。 3.9 计算转速的计算（1）主轴的计算转速nj，由公式n=n得，主轴的计算转速nj=198.305r/min，取212r/min。(2). 传动轴的计算转速 轴3=132 r/min 轴2=212 r/min，轴1=670r/min。（2）确定各传动轴的计算转速。各计算转速入表3-1。表3-1 各轴计算转速轴 号 轴 轴 轴计算转速 r/min 670212132（3） 确定齿轮副的计算转速。齿轮Z装在主轴上其中只有212r/min传递全功率，故Zj=212r/min。依次可以得出其余齿轮的计算转速，如表3-2。 表3-2 齿轮副计算转速序号ZZZZZn6702122121322123.10 齿轮模数计算及验算（1）模数计算。一般同一变速组内的齿轮取同一模数，选取负荷最重的小齿轮，按简化的接触疲劳强度公式进行计算，即mj=16338可得各组的模数，如表3-3所示。根据和计算齿轮模数，根据其中较大值取相近的标准模数：=16338=16338mm齿轮的最低转速r/min；顶定的齿轮工作期限，中型机床推存：=1524转速变化系数； 功率利用系数；材料强化系数。 （寿命系数）的极值齿轮等转动件在接取和弯曲交边载荷下的疲劳曲线指数m和基准顺环次数C0工作情况系数。中等中级的主运动： 动载荷系数；齿向载荷分布系数；齿形系数； 根据弯曲疲劳计算齿轮模数公式为： 式中：N计算齿轮转动递的额定功率N= 计算齿轮（小齿轮）的计算转速r/min 齿宽系数, Z1计算齿轮的齿数，一般取转动中最小齿轮的齿数： 大齿轮与小齿轮的齿数比，=；（+）用于外啮合，（-）号用于内啮合： 命系数； ：工作期限 ， =； =3.49=1.8=0.84 =0.58 =0.90 =0.55 =0.72 =3.49 0.84 0.58 0.55=0.94=1.80.84 0.90 0.72=0.99 时，取=，当时，取=；=0.85 =1.5； =1.2 =1 =0.378 许用弯曲应力，接触应力，() =354 =1750 按接触疲劳计算齿轮模数m 1轴由公式mj=16338可得mj=2.34mm，取m=3mm2轴由公式mj=16338可得mj=2.31mm，取m=3mm3轴由公式mj=16338可得mj=2.83mm，取m=3mm根据有关文献，也为了便于统一，在这里传动齿轮统一取m=3表3-3 模数组号基本组第一扩大组第二扩大组模数 mm 333（2）基本组齿轮计算。 基本组齿轮几何尺寸见下表齿轮Z1Z1Z2Z2Z3Z3齿数334225501857分度圆直径991267515054171齿顶圆直径1051329115660177齿根圆直径91.5118.567.5142.546.5163.5 齿宽242424242424按基本组最小齿轮计算。小齿轮用40Cr，调质处理，硬度241HB246HB，平均取260HB，大齿轮用45钢，调质处理，硬度229HB246HB，平均取240HB。计算如下： 齿面接触疲劳强度计算： 接触应力验算公式为 弯曲应力验算公式为： 式中 N-传递的额定功率（kW），这里取N为电动机功率，N=5kW; -计算转速（r/min）. =212（r/min）; m-初算的齿轮模数（mm）, m=3（mm）; B-齿宽（mm）;B=24（mm）; z-小齿轮齿数；z=18; u-小齿轮齿数与大齿轮齿数之比,u=3.16; -寿命系数； = -工作期限系数； T-齿轮工作期限，这里取T=15000h.; -齿轮的最低转速（r/min）, =500（r/min） -基准循环次数，接触载荷取=，弯曲载荷取= m-疲劳曲线指数，接触载荷取m=3；弯曲载荷取m=6; -转速变化系数，查【5】2上，取=0.60 -功率利用系数，查【5】2上，取=0.78 -材料强化系数，查【5】2上， =0.60 -工作状况系数，取=1.1 -动载荷系数，查【5】2上，取=1 -齿向载荷分布系数，查【5】2上，=1 Y-齿形系数，查【5】2上，Y=0.386；-许用接触应力（MPa）,查【4】，表4-7，取=650 Mpa；-许用弯曲应力（MPa），查【4】，表4-7，取=275 Mpa；根据上述公式，可求得及查取值可求得：=635 Mpa =78 Mpa（3）第一扩大组齿轮计算。 扩大组齿轮几何尺寸见下表 齿轮Z4Z4Z5Z5齿数30382642分度圆直径9011478126齿顶圆直径9612084132齿根圆直径82.5106.570.5118.5齿宽24242424（4）第二扩大组齿轮计算。 扩大组齿轮几何尺寸见下表 齿轮Z5Z5Z6Z6齿数61383861分度圆直径183114114183齿顶圆直径189120120189齿根圆直径175.5106.5106.5175.5齿宽24242424按扩大组最小齿轮计算。小齿轮用40Cr，调质处理，硬度241HB246HB，平均取260HB，大齿轮用45钢，调质处理，硬度229HB246HB，平均取240HB。 同理根据基本组的计算，查文献【6】，可得 =0.62， =0.77，=0.60，=1.1，=1，=1，m=3.5，=355；可求得：=619 Mpa =135Mpa 3.11 传动轴最小轴径的初定传动轴直径按扭转刚度用下列公式估算传动轴直径： mm其中：N该传动轴的输入功率 KWNd电机额定功率；从电机到该传动轴之间传动件的传动效率的乘积该传动轴的计算转速r/min每米长度上允许的扭转角(deg/m),可根据传动轴的要求选取如表3.2所示：表3.2 刚度要求允许的扭转角 主 轴 一般的传动轴较低的传动轴0.5111.51.52对于一般的传动轴，取=1.5。取估算的传动轴长度为500mm。 对轴有： KW =670r/min 预取mm对轴有：KW=1120 r/min mm 预取 对轴有： KW=140 mm 预取采用花键轴结构，即将估算的传动轴直径d减小7%为花键轴的直径，在选相近的标准花键。=320.93=29.76=380.93=35.34=460.93=42.78查表可以选取花键的型号其尺寸分别为轴取 6-30266轴取 6-383310轴取 6-434012最脆弱轴的计算校核 对于传动轴，除重载轴外，一般无须进行强度校核，只进行刚度验算。轴的抗弯断面惯性矩（）花键轴 =式中 d花键轴的小径（mm）；i花轴的大径（mm）；b、N花键轴键宽，键数；传动轴上弯曲载荷的计算，一般由危险断面上的最大扭矩求得：=式中 N该轴传递的最大功率（kw）; 该轴的计算转速（r/min）。传动轴上的弯矩载荷有输入扭矩齿轮和输出扭矩齿轮的圆周力、径向力，齿轮的圆周力式中 D齿轮节圆直径（mm）,D=mZ。齿轮的径向力：式中 为齿轮的啮合角，20；齿面摩擦角，；齿轮的螺旋角；0故N花键轴键侧挤压应力的验算花键键侧工作表面的挤压应力为：式中 花键传递的最大转矩（）； D、d花键轴的大径和小径（mm）； L花键工作长度； N花键键数； K载荷分布不均匀系数，K=0.70.8； 故此花键轴校核合格轴承疲劳强度校核机床传动轴用滚动轴承，主要是因疲劳破坏而失效，故应进行疲劳验算。其额定寿命的计算公式为： C滚动轴承的额定负载（N）,根据轴承手册或机床设计手册查取，单位用（kgf）应换算成（N）；速度系数， 为滚动轴承的计算转速（r/mm） 寿命系数， 寿命系数，对球轴承=3，对滚子轴承=；工作情况系数，对轻度冲击和振动的机床（车床、铣床、钻床、磨床等多数机床），；功率利用系数，查表33；速度转化系数，查表32；齿轮轮换工作系数，查机床设计手册；P当量动载荷，按机床设计手册。 故轴承校核合格第4章 主要零件的设计与验算4.1齿轮强度的校核验算 齿面接触疲劳强度计算： 接触应力验算公式为 弯曲应力验算公式为： 式中 N-传递的额定功率（kW），这里取N为电动机功率，N=5kW; -计算转速（r/min）. =212（r/min）; m-初算的齿轮模数（mm）, m=3（mm）; B-齿宽（mm）;B=24（mm）; z-小齿轮齿数；z=18; u-小齿轮齿数与大齿轮齿数之比,u=3.16; -寿命系数； = -工作期限系数； T-齿轮工作期限，这里取T=15000h.; -齿轮的最低转速（r/min）, =500（r/min） -基准循环次数，接触载荷取=，弯曲载荷取= m-疲劳曲线指数，接触载荷取m=3；弯曲载荷取m=6; -转速变化系数，查【5】2上，取=0.60 -功率利用系数，查【5】2上，取=0.78 -材料强化系数，查【5】2上， =0.60 -工作状况系数，取=1.1 -动载荷系数，查【5】2上，取=1 -齿向载荷分布系数，查【5】2上，=1 Y-齿形系数，查【5】2上，Y=0.386；-许用接触应力（MPa）,查【4】，表4-7，取=650 Mpa；-许用弯曲应力（MPa），查【4】，表4-7，取=275 Mpa；根据上述公式，可求得及查取值可求得：=635 Mpa =78 Mpa4.2 轴的校核（a） 主轴的前端部挠度（b） 主轴在前轴承处的倾角（c） 在安装齿轮处的倾角E取为，由于小齿轮的传动力大，这里以小齿轮来进行计算将其分解为垂直分力和水平分力由公式可得主轴载荷图如下所示：由上图可知如下数据：a=364mm,b=161mm,l=525mm,c=87mm计算（在垂直平面）,，计算（在水平面）,，合成：4.3 轴承寿命校核由轴最小轴径可取轴承为7008C角接触球轴承,=3；P=XFr+YFaX=1，Y=0。对轴受力分析得：前支承的径向力Fr=2642.32N。 由轴承寿命的计算公式：预期的使用寿命 L10h=15000hL10h=hL10h=15000h 轴承寿命满足要求。参考文献1.段铁群.主轴箱设计，科学出版社；2.于惠力，向敬忠，张春宜.机械设计，科学出版社；3.潘承怡，苏相国. 机械设计课程设计，哈尔滨理工大学；4.戴署.金属切削机床设计，机械工业出版社；5.陈易新，金属切削机床课程设计指导书； 6、机床主轴、变速箱设计简明手册7、机械设计课程设计8、金属切削机床设计9、机械制造装备等An investigation of the effect of counterweight configuration on main bearingload and crankshaft bending stressYasin Yilmaz*, Gunay AnlasDepartment of Mechanical Engineering, Faculty of Engineering, Bogazici University, 34342 Bebek, Istanbul, Turkeya r t i c l ei n f oArticle history:Received 11 February 2008Received in revised form 17 March 2008Accepted 24 March 2008Available online 6 May 2008Keywords:Counterweight configurationCrankshaft modelsBalancing rateBearing loadBending stressa b s t r a c tIn this study, effects of counterweight mass and position on main bearing load and crankshaft bendingstress of an in-line six-cylinder diesel engine is investigated using Multibody System Simulation Program,ADAMS. In the analysis, rigid, beam and 3D solid crankshaft models are used. Main bearing load results ofrigid, beam and 3D solid models are compared and beam model is used in counterweight configurationanalyses. Twelve-counterweight configurations with a zero degree counterweight angle and eight-coun-terweight configurations with 30? counterweight angle, each for 0%, 50% and 100% counterweight balanc-ing rates, are considered. It is found that maximum main bearing load and web bending stress increasewith increasing balancing rate, and average main bearing load decreases with increasing balancing rate.Both configurations show the same trend. The load from gas pressure rather than inertia forces is theparameter with the most important influence on design of the crankshaft. Results of bearing loads andweb bending stresses are tabulated.? 2008 Elsevier Ltd. All rights reserved.1. IntroductionNew internal combustion engines must have high enginepower, good fuel economy, small engine size, and should be asharmless as possible to the environment. Therefore, the effect ofeach component of the engine on its overall performance shouldbe investigated in detail. Crankshaft systems of internal combus-tion engines have important influence on engine performancebeing the main part responsible for power production.Crankshaft system mainly consists of piston, piston pin, con-necting rod, crankshaft, torsional vibration (TV) damper and fly-wheel. Counterweights are placed on the opposite side of eachcrank to balance rotating inertia forces. In general, counterweightsare designed for balancing rates between 50% and 100%. Foracceptable maximum and average main bearing loads, mass ofcounterweights and their positions are important. Maximum andaverage main bearing loads of an engine depend on cylinder pres-sure, counterweight mass, engine speed and other geometricparameters of the crankshaft system.Studies on crankshaft of internal combustion engines mainly fo-cus on vibration and stress analyses 19. Although stress analy-ses of crankshafts are available in literature, there are fewstudies on the effect of counterweight configuration on main bear-ing loads and crankshaft stresses. Sharpe et al. 10 studied balanc-ing of the crankshaft of a V-8 engine using a rigid crankshaft modeland optimized counterweights to minimize main bearing loads.Stanley and Taraza 11 obtained maximum and average mainbearing loads of four and six-cylinder symmetric in-line enginesusing a rigid crankshaft model and estimated ideal counterweightmass that resulted in acceptable maximum bearing load. Rigidcrankshaft models that are used in counterweight analyses donot consider the effect of crankshaft flexibility on main bearingloads and can lead to considerable errors. Therefore, an extensivestudy on effect of counterweight configuration on main bearingloads and crankshaft stresses is still needed.In this study, counterweight positions and masses of an in-linesix-cylinder diesel engine crankshaft system are studied. Maxi-mum and average main bearing forces and crankshaft bendingstresses are calculated for 12-counterweight configurations witha zero degree counterweight angle, and for eight-counterweightconfigurations with 30? counterweight angle for 0%, 50% and100% counterweight balancing rates. Analyses are carried out usingMultibody System Simulation Program, ADAMS/Engine. Simula-tions are carried out at engine speed range of 10002000 rpm.Bending stresses at the centres of each web are also calculated.2. Engine specificationsThe specifications of in-line six-cylinder diesel engine are givenin Table 1. The 9.0 L engine crankshaft has eight counterweights atcrank webs 1, 2, 5, 6, 7, 8, 11 and 12. 3D solid model of the crank-shaft is obtained using Pro/Engineer and is shown in Fig. 1. Sche-matic representation of the crankshaft is given in Fig. 2. Static0965-9978/$ - see front matter ? 2008 Elsevier Ltd. All rights reserved.doi:10.1016/j.advengsoft.2008.03.009* Corresponding author. Tel.: +90 212 359 7534; fax: +90 212 287 2456.E-mail address: yasin.yilmaz.tr (Y. Yilmaz).Advances in Engineering Software 40 (2009) 95104Contents lists available at ScienceDirectAdvances in Engineering Softwarejournal homepage: /locate/advengsoftunbalance of each crank throw (with and w/o counterweights) isdetermined using Pro/Engineer and is given in Table 2. The balanc-ing system data for the crank train are given in Table 3.3. Modeling of crankshaft systemUsing ADAMS/Engine, a crankshaft can be modeled in four dif-ferent ways: rigid crankshaft, torsionalflexible crankshaft, beamcrankshaft and 3D solid crankshaft. Rigid crankshaft model ismainly used to obtain free forces and torques, and for balancingpurposes. Torsionalflexible crankshaft model is used to investi-gate torsional vibrations where each throw is modeled as one rigidpart, and springs are used between each throw to represent tor-sional stiffness. Beam crankshaft model is used to represent thetorsional and bending stiffness of the crankshaft. Using beam mod-el bending stresses at the webs can be calculated 12.Table 1Engine specificationsUnit9.0 L engineBore diametermm115Strokemm144Axial cylinder distancemm134Peak firing pressureMPa19Rated power at speedkW/rpm295/2200Max. torque at speedNm/rpm1600/12001700Main journal/pin diametermm95/81Firing order1-5-3-6-2-4Flywheel masskg47.84Flywheel moment of inertiakg mm21.57E+9Mass of TV damper ringkg4.94Mass of TV damper housingkg6.86Moment of inertia of the ringkg mm21.27E+5Moment of inertia of the housingkg mm20.56E+5Main Bearing #1 Main Bearing #2 MainBearing #3 Main Bearing #4 MainBearing #5 MainBearing #6 Main Bearing #7 CounterweightsFig. 1. 3D solid model of the crankshaft.C3, C4, C5, C6 C1, C2, C7, C8 1, 6 3, 4 2, 5 C1C2C3C4 C5C6C7C8123456Fig. 2. Eight-counterweight arrangement of the 9.0 L engine crankshaft.Table 2Properties of the crank throwsThrow 1Throw 2Throw 3Throw 4Throw 5Throw 6Mass (kg)12.509.2512.5012.509.2812.55CG position from crank rotation axis (mm)12.42331.43511.96711.96631.02711.702Static unbalance (kg mm)155.265290.767149.734149.734287.871146.85696Y. Yilmaz, G. Anlas/Advances in Engineering Software 40 (2009) 95104Elastic 3D solid model of the crankshaft can be obtained usingan additional finite element program. The procedure is lengthyand time consuming and usually one ends up with degrees of free-dom in order of millions. To simplify the finite element model,modal superposition technique is used. The elastic deformationof the structure is approximated by linear combination of suitablemodes which can be shown as follows:u Uq1where q is the vector of modal coordinates and U is the shape func-tion matrix.An elastic body contains two types of nodes, interface nodeswhere forces and boundary conditions interact with the structureduring multibody system simulation (MSS), and interior nodes. InMSS the position of the elastic body is computed by superposingits rigid body motion and elastic deformation. In ADAMS, this isperformed using Component Mode Synthesis” technique basedon CraigBampton method 13,14. The component modes containstatic and dynamic behavior of the structure. These modes are con-straint modes which are static deformation shapes obtained bygiving a unit displacement to each interface degree of freedom(DOF) while keeping all other interface DOFs fixed, and fixedboundary normal modes which are the solution of eigenvalueproblem by fixing the entire interface DOFs. The modal transforma-tion between the physical DOF and the CraigBampton modes andtheir modal coordinates is described by 15u uBuI?I0UCUN?qCqN?2where uBand uIare column vectors and represent boundary DOFand interior DOF, respectively. I, 0 are identity and zero matrices,respectively. UCis the matrix of physical displacements of the inte-rior DOF in the constraint modes. UNis the matrix of physical dis-placements of the interior DOF in the normal modes. qCis thecolumn vector of modal coordinates of the constraint modes. qNisthe column vector of modal coordinates of the fixed boundary nor-mal modes. To obtain decoupled set of modes, constrained modesand normal modes are orthogonalized.Elastic 3D solid crankshaft model of the 9.0 L engine is obtainedin MSC.Nastran using modal superposition technique. First, 3D so-lid model of the crankshaft that is shown in Fig. 1 is exported toMSC.Nastran and finite element model of the crankshaft, which ischaracterized by approximately 300,000 ten-node tetrahedral ele-ments and 500,000 nodes is obtained. The modal model of thecrankshaft is developed with 32 boundary DOFs associated with16 interface nodes. Constrained modes obtained from static analy-sis correspond to these DOFs. Flexible crankshaft model is obtainedthrough modal synthesis considering the first 40 fixed boundarynormal modes. Therefore flexible crankshaft model is character-ized by a total of 72 DOFs. This model is exported to ADAMS/En-gine and crankshaft system model that is shown in Fig. 3 isobtained. 3D finite element model is run with ADAMS.4. Forces acting on crankshaft system and balancingForces in an internal combustion engine may be divided intoinertia forces and pressure forces. Inertia forces are further dividedinto two main categories: rotating inertia forces and reciprocatinginertia forces. The rotating inertia force for each cylinder can bewritten as shown below:FiR;j mR? rR? x2? ?sinhjj coshjk3where mRis the rotating mass that consists of the mass of crank pin,crank webs and mass of rotating portion of the connecting rod; rRisthe distance from the crankshaft centre of rotation to the centre ofgravity of the rotating mass, x is angular velocity of the crankshaft,and hjis the angular position of each crank throw with respect toTop Dead Centre” (TDC). If there are two counterweights per crankthrow, each counterweight force is given by 11FCWi;j ?mCWi;j? rCWi;j? x2? ?sinhj ci;jj coshj ci;jkhi;i 1;2j 1;2;.;64where ci,jis the offset angle of counterweight mass from 180? oppo-site of crank throw j”. There are two counterweights per throw. i”denotes the counterweight number. The counterweight size that isrequired to accomplish an assessed balancing rate isUCWK ? UCrank throw mcr-r? r ? cosc25whereUCWisthestaticunbalanceofeachcounterweight,UCrank_throwis the static unbalance of each crank throw, mcr-risthe mass of connecting rod rotating portion, r is the crank radiusand K is the balancing rate of the internal couple due to rotatingforces. From this formula follows the balancing rate for a givencrankshaft and a given counterweight size:K 2 ? UCWUCrank throw mcr-r? r ? cosc6For a standard in-line six-cylinder engine crankshaft with threepairs of crank throws disposed at angles of 120? that are arrangedsymmetrical to the crankshaft centre, rotating forces, and first andsecond order reciprocating forces are naturally balanced. This canbe explained by the first and second order vector stars shown inFig. 4. The six-cylinder crankshaft generates rotating and firstand second order reciprocating couples in each crankshaft half thatbalance each other but which result in internal bending moment.At high speeds, the two equally directed crank throws, 3 and 4Table 3Crankshaft system dataCrank radius (mm)72Connecting rod length (mm)239Mass of complete piston (kg)3.42Connecting rod reciprocating mass (kg)0.92Reciprocating mass (total per cylinder) (kg)4.32Connecting rod rotating mass (kg)2.01Fig. 3. Model of the crankshaft system.Y. Yilmaz, G. Anlas/Advances in Engineering Software 40 (2009) 9510497yield a high rotating load on centre main bearing. The rotatinginertia force of each cylinder is usually offset at least partially bycounterweights placed on the opposite side of each crank. In gen-eral, the counterweights are designed for balancing rates between50% and 100% of the internal couple.Gas forces in cylinders are acting on piston head, cylinder headand on side walls of the cylinder. These forces are equal toFp;j ?pD24? Pcyl;jh ? Pcc;jh?k;j 1;2;.;671, 6 2, 5 3, 4 3, 4 1, 6 2, 5 Fig. 4. First and second order vector stars.020406080100120140160180200090180270360450540630720Crank Angle (degree) Pressure (bar)1000rpm1200rpm1350rpm1675rpm2000rpmFig. 5. Gas pressure values at different engine speeds for the 9.0 L engine.Bearing #102550751001251500120240360480600720Crank Angle degForce kNRigidBeam3D solidFig. 6. Forces acting on main bearing #1 for rigid, beam and 3D solid crankshaftmodels at 1000 rpm engine speed.Bearing #202550751001251501750120240360480600720Crank Angle degForce kNRigidBeam3D solidFig. 7. Forces acting on main bearing #2 for rigid, beam and 3D solid crankshaftmodels at 1000 rpm engine speed.Bearing #302550751001251500120240360480600720Crank Angle degForce kNRigidBeam3D solidFig. 8. Forces acting on main bearing #3 for rigid, beam and 3D solid crankshaftmodels at 1000 rpm engine speed.Bearing #402550751001251500120240360480600720Crank Angle deg Force kNRigidBeam3D solidFig. 9. Forces acting on main bearing #4 for rigid, beam and 3D solid crankshaftmodels at 1000 rpm engine speed.Bearing #502550751001251500120240360480600720Crank Angle degForce kNRigidBam3D solidFig. 10. Forces acting on main bearing #5 for rigid, beam and 3D solid crankshaftmodels at 1000 rpm engine speed.98Y. Yilmaz, G. Anlas/Advances in Engineering Software 40 (2009) 95104where D is cylinder diameter, Pcylis the gas pressure in the cylinderand Pccis the pressure in the crankcase. The gas forces are transmit-ted to the crankshaft through the piston and connecting rod. Cylin-der pressure curves for the 9.0 L engine studied under full load atdifferent engine speeds are given in Fig. 5. Pressure curves are ob-tained using AVL/Boost engine cycle calculation program whichsimulates thermodynamic processes in the engine taking into ac-count one dimensional gas dynamics in the intake and exhaust sys-tems 16.5. Main bearing loads: comparison of crankshaft modelsMain bearing loads are calculated using ADAMSs rigid, beamand 3D solid crankshaft models and compared. In the rigid model,no vibration effects are considered which can lead to considerableerrors if vibration effects have a major role on the system (like inmultithrow crankshafts). To consider vibration effects beam crank-shaft model is used and main bearing loads and bending stresses atwebs are calculated. Rigid model assumes crankshaft to be stati-cally determinate and reaction force of any given bearing dependson the load exerted on the throws adjacent to that bearing. Beammodel assumes the crankshaft to be statically indeterminate andthe load exerted on a throw affects all bearings. Analyses are car-ried out at an engine speed range of 10002000 rpm. A moresophisticated 3D solid hybrid model that combines FE with ADAMSis used to check the results obtained by beam model.Maximum main bearing load occurs at bearing number two atan engine speed of 1000 rpm, therefore results are plotted in Figs.612 for 1000 rpm only. Rigid crankshaft model overestimates themaximum main bearing load at bearings 1 and 7 with respect tobeam and flexible crankshaft models. However it underestimatesthe maximum main bearing load at other bearings. For exampleat bearing 2, beam model gives a maximum main bearing load thatis 50% more than that of rigid models because the beam model as-sumes the crankshaft to be statically indeterminate and considersBearing #602550751001251500120240360480600720Crank Angle degForce kNRigidBeam3D solidFig. 11. Forces acting on main bearing #6 for rigid, beam and 3D solid crankshaftmodels at 1000 rpm engine speed.Bearing #702550751001251500120240360480600720Crank Angle degForce kNRigidBeam3D solidFig. 12. Forces acting on main bearing #7 for rigid, beam and 3D solid crankshaftmodels at 1000 rpm engine speed.Bearing #1Bearing #14050607080100012001400160018002000100012001400160018002000Crank Angular Velocity (rpm)Crank Angular Velocity (rpm)Maximum Bearing K=0%K=50%K=100%05101520Average Bearing K=0%K=50%K=100%Force (kN)Force (kN)Fig. 13. (a) Maximum and (b) average bearing forces at bearing #1 for 12-counterweight configurations.Bearing #2120130140150160K=0%K=50%K=100%Bearing #22025303540K=0%K=50%K=100%Maximum Bearing Force (kN)100012001400160018002000Crank Angular Velocity (rpm)100012001400160018002000Crank Angular Velocity (rpm)Average Bearing Force (kN)Fig. 14. (a) Maximum and (b) average bearing forces at bearing #2 for 12-counterweight configurations.Y. Yilmaz, G. Anlas/Advances in Engineering Software 40 (2009) 9510499bending vibrations. Maximum main bearing load difference ofbeam and 3D solid models is approximately 5%. Main bearing loadsfor beam and 3D solid crankshaft models are generally in goodagreement. In bearings 3, 5 and 6, 3D solid model gives larger bear-ing loads at firing positions of the cylinders that are not adjacent tobearing. Because obtaining elastic 3D solid models for differentcounterweight configurations is difficult and time consuming,and beam model gives equally valid results, beam model is usedBearing #3100110120130140K=0%K=50%K=100%Bearing #32025303540K=0%K=50%K=100%Maximum Bearing Force (kN)100012001400160018002000Crank Angular Velocity (rpm)100012001400160018002000Crank Angular Velocity (rpm)Average Bearing Force (kN)Fig. 15. (a) Maximum and (b) average bearing forces at bearing #3 for 12-counterweight configurations.Bearing #460708090100110120K=0%K=50%K=100%Bearing #410152025303540K=0%K=50%K=100%Maximum Bearing Force (kN)100012001400160018002000Crank Angular Velocity (rpm)100012001400160018002000Crank Angular Velocity (rpm)Average Bearing Force (kN)Fig. 16. (a) Maximum and (b) average bearing forces at bearing #4 for 12-counterweight configurations.Bearing #6100110120130140K=0%K=50%K=100%Bearing #620253035404550K=0%K=50%K=100%Maximum Bearing Force (kN)100012001400160018002000Crank Angular Velocity (rpm)100012001400160018002000Crank Angular Velocity (rpm)Average Bearing Force (kN)Fig. 18. (a) Maximum and (b) average bearing forces at bearing #6 for 12-counterweight configurations.Bearing #5100110120130140K=0%K=50%K=100%Bearing #52025303540K=0%K=50%K=100%Maximum Bearing Force (kN)100012001400160018002000Crank Angular Velocity (rpm)100012001400160018002000Crank Angular Velocity (rpm)Average Bearing Force (kN)Fig. 17. (a) Maximum and (b) average bearing forces at bearing #5 for 12-counterweight configurations.100Y. Yilmaz, G. Anlas/Advances in Engineering Software 40 (2009) 95104in the rest of the work to study the effect of counterweight config-uration on main bearing loads and crankshaft bending stresses.6. Effect of counterweight configuration on main bearing loadand crankshaft bending stressThe effect of counterweight arrangement on bearing forces andcrankshaft bending stresses is investigated using beam model forthe following cases:? No counterweights, K = 0%.? 12 counterweights with K = 50%, c = 0? and K = 100%, c = 0?.? eight counterweights with K = 50%, c = 30? and K = 100%, c = 30?.6.1. Twelve counterweights, c = 0?, K = 0%, K = 50% and K = 100%In this configuration, counterweights are placed on oppositesides of all webs. Counterweight static unbalance is calculatedusing Eq. (5) for K = 50% and K = 100% balancing rates. Maximumand average main bearing loads are calculated using the beamcrankshaft model considering inertial and gas pressure forces andare plotted in Figs. 1319 as function of crankshaft angular velocityand balancing rate.In Figs. 1319, maximum bearing load increases with increasingbalancing rate. This behavior can be explained as follows: For six-cylinder in-line engine crankshafts, the rotating inertia force andfirst harmonic component of the reciprocating inertia force arein-phase and add in the direction of the cylinder. The pressureforce is almost maximum at TDC position where the reciprocatinginertia force and the component of the rotating inertia force in cyl-inder direction are also at maximum levels. Because the pressureand inertia forces are opposite in-sign, they subtract from eachother which increases the maximum bearing load at high balancingrates. On the other hand, average bearing force increases withdecreasing balancing rate. Maximum main bearing load occurs atbearing number two at engine speed of 1000 rpm and averagemain bearing load of bearing 6 is larger than other bearings aver-age loads because bending vibrations of damper and flywheel oc-cur. At main bearings 3, 4 and 5, where the influence of damperand flywheel bending vibrations is minimal, only torsional vibra-tion occurs and their loads are less. Maximum bending stresses oc-cur at webs 1 and 12 and are given in Table 4 for 0%, 50% and 100%balancing rates.6.2. Eight counterweights, c = 30?, K = 50% and K = 100%Eight-counterweight configuration shows the same trend as the12-counterweight configuration for maximum and average bearingforces: maximum bearing force increases with increasing balanc-ing rate whereas average bearing force increases with decreasingbalancing rate. Maximum and average bearing forces for c = 30?,K = 50% and K = 100% counterweight configurations are calculatedBearing #75060708090K=0%K=50%K=100%Bearing #705101520K=0%K=50%K=100%Maximum Bearing Force (kN)100012001400160018002000Crank Angular Velocity (rpm)100012001400160018002000Crank Angular Velocity (rpm)Average Bearing Force (kN)Fig. 19. (a) Maximum and (b) average bearing forces at bearing #7 for 12-counterweight configurations.Bearing #14050607080K=0%K=50%K=100%Bearing #105101520K=0%K=50%K=100%Maximum Bearing Force (kN)100012001400160018002000Crank Angular Velocity (rpm)100012001400160018002000Crank Angular Velocity (rpm)Average Bearing Force (kN)Fig. 20. (a) Maximum and (b) average bearing forces at bearing #1 for eight-counterweight configurations.Table 4Maximum bending stresses for no counterweight configuration and 12-counter-weight configurations with K = 50% and 100%Maximum bending stress (MPa)1000 rpm1200 rpm1350 rpm1675 rpm2000 rpmK = 0%Web #1135.9133.4136.5135127.8Web #12145.7141.7144.2140.3131.8K = 50%Web #1138.4136.8140.9141.2135.9Web #12147.8144.8148.4146.6140.1K = 100%Web #1140.6140.0145.0147.3143.8Web #12149.9147.8152.3152.7148.4Y. Yilmaz, G. Anlas/Advances in Engineering Software 40 (2009) 95104101Bearing #2120130140150160K=0%K=50%K=100%Bearing #22025303540K=0%K=50%K=100%Maximum Bearing Force (kN)100012001400160018002000Crank Angular Velocity (rpm)100012001400160018002000Crank Angular Velocity (rpm)Average Bearing Force (kN)Fig. 21. (a) Maximum and (b) average bearing forces at bearing #2 for eight-counterweight configurations.Bearing #3100110120130140K=0%K=50%K=100%Bearing #32025303540K=0%K=50%K=100%Maximum Bearing Force (kN)100012001400160018002000Crank Angular Velocity (rpm)100012001400160018002000Crank Angular Velocity (rpm)Average Bearing Force (kN)Fig. 22. (a) Maximum and (b) average bearing forces at bearing #3 for eight-counterweight configurations.Bearing #460708090100110120K=0%K=50%K=100%Bearing #410152025303540K=0%K=50%K=100%Maximum Bearing Force (kN)100012001400160018002000Crank Angular Velocity (rpm)100012001400160018002000Crank Angular Velocity (rpm)Average Bearing Force (kN)Fig. 23. (a) Maximum and (b) average bearing forces at bearing #4 for eight-counterweight configurations.Bearing #5Bearing #5100110120130140K=0%K=50%K=100%2025303540K=0%K=50%K=100%Maximum Bearing Force (kN)100012001400160018002000Crank Angular Velocity (rpm)100012001400160018002000Crank Angular Velocity (rpm)Average Bearing Force (kN)Fig. 24. (a) Maximum and (b) average bearing forces at bearing #5 for eight-counterweight configurations.102Y. Yilmaz, G. Anlas/Advances in Engineering Software 40 (2009) 95104for beam crankshaft model considering inertial and gas pressureforces and given in Figs. 2026.Maximum bending stresses for two crankshaft configurationsare given in Table 5. When compared to 12 counterweights withc = 0? and K = 50% and K = 100% configurations, the maximumbending stresses for eight counterweights with c = 30?, K = 50%and K = 100% configurations are smaller.7. Summary and conclusionsIn this study, the effect of counterweight configuration on bear-ing load and bending stress is investigated for a 9.0 L in-line six-cylinder diesel engine crankshaft system in the presence of inertialand gas pressure forces. Five different counterweight configura-tions are studied in the analyses: no counterweights, 12-counter-weights with K = 50% and 100%, and eight-counterweights with30? counterweight angle, and K = 50% and 100%. Analyses are car-ried out at an engine speed range of 10002000 rpm.First, 3D solid model of the crankshaft is obtained using Pro/Engineer and MSC.Nastran. The crankshaft is also modeled using ri-gid and beam models of ADAMS/Engine. Main bearing loads are ob-tained for the models using ADAMS, and the results are comparedto each other. It is seen that main bearing loads for beam and elas-tic 3D solid crankshaft models are in good agreement. As a result,because obtaining elastic 3D solid models for different counter-weight configurations is difficult and time consuming, beam modelis used in this work to study the effect of counterweight configura-tion on main bearing loads and crankshaft bending stresses.Using beam model of ADAMS/Engine, main bearing reactionloads are obtained for no counterweight configuration, and 12-counterweight configurations with 50% and 100% balancing rates.It is observed that maximum bearing reaction force increases withincreasing balancing rate. Average bearing loads and maximumweb bending stresses are also calculated. Maximum web bendingstress is higher for 100% balancing rate than for 50% and 0% balanc-ing rates. On the other hand, as the balancing rate increases, theaverage bearing load decreases due to lower inertia force.Similarly, maximum and average bearing loads and bendingstresses are calculated for eight-counterweight configuration witha counterweight angle of 30?. In the case of maximum and averagebearing forces, eight-counterweight configurations shows thesame trend as that of 12-counterweight configurations: the maxi-mum bearing load decreases with decreasing bal