油冷式电动滚筒设计

油冷式电动滚筒设计 目 录 摘要???????????????????????????????1 Abstract?????????????????????????????1 绪论???????????????????????????????2 第一章 查阅相关资料???????????????????????3 第二章 装置整体设计?????????????????????4 第三章 传动方案设计???????????????????????5 第四章 分配传动比????????????????????????5 一、电动机选择??????????????????????5 二、确定传动装置的总传动比????????????????5 第五章 传动机构结构设计?????????????????????6 一、高速级齿轮传动的几何计算???????????????6 二、低速级齿轮传动的几何计算???????????????15 三、齿轮轴的设计计算???????????????????26 四、齿轮轴上的轴承设计计算????????????????30 五、法兰轴的设计计算???????????????????32 六、法兰轴上的轴承设计计算????????????????36 七、弹性挡圈的选用????????????????????38 八、吊环的选用及几何参数?????????????????39 九、键的选用及几何参数??????????????????40 十、螺栓的计算??????????????????????40 第六章 滚筒其它结构设计?????????????????????41 一、滚筒设计???????????????????????41 二、端盖设计???????????????????????42 总结???????????????????????????????45 致谢???????????????????????????????46 参考文献?????????????????????????????47 附录???????????????????????????????48 油冷式电动滚筒设计 摘 要 带式输送机是最重要的现代散状物料输送设备，它广泛的应用电力、粮食、冶金、 化工、煤炭、矿山、港口、建材等领域。驱动滚筒是带式输送机的心脏，其性能直接影 响到带式输送机工作的可靠性与整体性能。电动滚筒是驱动滚筒其中的一种形式，它具 有结构紧凑、传动效率高、噪声低、使用寿命长、运转平稳、工作可靠、密封性好、占 据空间位置少以及安装维修方便等优点，因此研究电动滚筒具有非常重要的意义。在该 篇论文主要包括电动滚筒的整体方案设计、传动方案设计，以及主要零部件轴、齿轮、 轴承等的设计及计算。 关键词 带式输送机;电动滚筒; 设计 Oil-cool design electric drum Abstract Belt conveyor is the most important modern bulk material handling equipment.Its wider use of electricity,food,metallurgical,chemical,coal,mining,ports,building materials and other fields.Drum-driven belt conveyors is the heart of its direct impact on the performance of the work of the belt conveyor to the overall reliability and performance.Electric drum-driven drum which is a form,it has a compact,high transmission efficiency,low noise,long life,smooth operation,reliable,tightness,and occupy less space location of maintenance and installation of the advantages of convenience,Research electric drum has very important significance.In the papers,including electric drum major of the overall programme design,transmission design,and the main shaft parts,gears,bearings,such as the design and calculation. Key words: belt conveyor;electric drum;design 1 绪 论 经过大学四年的时间，我们先后学习了公共文化、设计基础和零件设计等课程，比较系统地学习了所需的专业知识，已初步掌握本专业的各类专门技能，根据教学目标和教学计划要求，我们进行了这次毕业设计课程。 毕业设计时教学计划中最后一个综合性实践环节，时学生在教师的指导下，独立从事设计工作的又一次尝试，其基本目的时培养学生综合运用所学的基础理论、专业知识、基本技能研究和处理问的能力。时我们对所学知识进行系统化、综合化运用、总结和深化的过程。 电动滚筒设计，因为在当今社会中电动滚筒有着非常重要的意设计题目选油冷式 义。 带式输送机是最重要的现代散状物料输送设备，它广泛的应用电力、粮食、冶金、化工、煤炭、矿山、港口、建材等领域。近年来，带式输送机因为它所拥有的输送料类广泛、 输送能力范围宽、输送路线的适应性强以及灵活的装卸料和可靠性强费用低的特点，已经在某些领域逐渐开始取代汽车、机车运输。成为散料运输的主要装备，在社会经济结构中扮演越来越重要的角色。特别是电动滚筒驱动的带式输送机在粮库的散料输送过程中更加有无可比拟的优势和发展潜力因此我们开拓思维、努力创新并结合自己原有的知识和现有的资料对其进行创新完善。在此过程中检验自己的创新能力使其应用的范围更加广泛,在国民经济的各个领域起到更加重要的作用。 以电动滚筒作为驱动装置的带式输送机有着极其重要的意义。因其拥有结构紧凑、传动效率高、噪声低、使用寿命长、运转稳定、工作可靠性和密封性好、占据空间小等特点，并能适应在各种恶劣工作环境下工作包括潮湿、泥泞、粉尘多等。因此国内外将带式输送机(电动滚筒驱动)广泛应用于采矿、粮食、冶金等各个生产领域，思维的不断开阔、制造技术的不断提高和制造材料的不断改进，带式输送机将以前所未有的速度发展。保障散料输送工作高效、安全、可靠的运转，并将在社会和经济发展领域继续起到更加重要的意义。 这次毕业设计的任务是设计一个滚筒驱动器，包括减速器、零件及滚筒外观的设计，给定该滚筒驱动器的大致尺寸从而计算滚筒内部具体的尺寸，即电动滚筒受到空间的限制，这是这次设计主要的考查之处。 由于我的经验有限，所以在设计中难免出现错误，恳请老师批评改正。 2 油冷式电动滚筒设计 第一章 查阅相关资料 经过查阅相关资料，滚筒的大致尺寸如下: 图1 滚筒外观尺寸 其它原始数据: B,650适用胶带速度 mm D, 滚筒直径 500 mm 胶带运行速度 1.6 v,mm 电机功率 N,5.5KW 980r/min 电机转速 n, 电机型号 YGMB1326,2 3 第二章 装置整体方案设计 图(2)为油冷式电动滚筒装置图。图中1——接线盒;2——支座;3——端盖;4——滚筒;5、11——法兰轴;6——电机;7、8——齿轮;9——齿轮轴;10——内齿轮。 4 第三章 传动方案设计 图(3)为传动装置减速器的简图 图3 传动装置 传动顺序为外啮合小齿轮——外啮合大齿轮——内啮合小齿轮——内齿轮——滚筒 第四章 分配传动比 一 电动机选择 Y按工作条件和要求，选用三相型异步电动机，封闭式结构，电压380V，型，电机型号。 YGMB1326,2 二 确定传动装置的总传动比 滚筒轴工作转速为 601000，v n,,D 5 式中——胶带运行速度， ; ms/v D——滚筒直径，。 mm 6010001.6，， n,,61.153.14500， 由选定电动机满载转速和工作机主动轴转速，可得传动装置总传动比为 nnm nm i,an iii,a12 式中、分别为高速级和低速级的传动比。 ii12 n980m故 i,,,16a61.15n 由于滚筒内部结构受空间的限制，经过查阅资料，直径500mm的滚筒内减速 器齿轮中心距为，因此综合考虑取高速级和低速级的传动比分别为，130i,3.3mm1 。 i,4.82 第五章 传动机构结构设计 一 高速级齿轮传动的几何计算 小齿轮用，调质处理，硬度241286HBHB，平均取为260HB;大齿轮用4540Cr 钢，调质处理，硬度229286HBHB，平均取为240HB。计算步骤如下: 1 高速级齿轮的校核计算 计算项目 计算内容 计算结果 齿面接触疲劳强度计算 1.初步计算 P转矩 T61T,，，9.5510 1n1 5.56 TNmm,53587 ,，，9.55101980 P式中——电机功率，KW r/min ——高速轴转速， n1 6 齿宽系数 由《机械设计》表，取 ,,1.012.13,,,1.0ddd 由《机械设计》图12.17c ,,710MP接触疲劳极限 ,Halim1Hlim ,,580MPHalim2初步计算的许用接 ] [,,0.9,H1Hlim1 触应力 [] ,H ,，0.9710] [,,639MPH1a [] ,0.9,,H2Hlim2 [] ,,522MPH2a ,，0.9580 A,85值 由《机械设计》表，取 12.16AA,85ddd初步计算小齿轮直 1Tu，113 dA,() d12径直径 d,,[]u1dH 1535973.31，3 ,，85() 21.05223.3， 取dmm,601 ,53.99 b,60 mm初步齿宽 b=160， bd,,d1 2.校核计算 dn,圆周速度 v11v , 601000， 3.1460980，，vms,3.1/ ,601000， 精度等级 由《机械设计》表12.6 选8级精度 齿数和模数 zm,24初取齿数 z1 ziz,,，,3.3248021 d601m,,,2.5 ,2.5 mz241 z,241 2.5,80取 ，则， z,24zm,12 ,80 z2 12.9 由《机械设计》表 K,1.5使用系数K AA 7 由《机械设计》图12.9 K,1.16动载系数 Kvv 齿间载荷分配系数，先求 由《机械设计》表 12.10 2253597T， k1H, F,,,1786.6N td601 KF1.51786.6， At Nmm/,,44.7 60b ,100/Nmm 11,,,,， [1.883.2()]cos ,zz12 11 ,1.71, ,,，，，,1.883.2()11.71,2480 4,,41.71,,Z,,,0.87 , ,0.87Z33, 11K,,,由此得1.32 H,22 ,1.32KZ0.87H,, 齿向载荷分布系数12.11 由《机械设计》表 bK ,3H,,，，()10KABCb H,d1 23,K,1.37 ,，，，，，1.170.1610.611060H, K载荷系数 KKKKK, AVHH,, ,，，，1.51.161.321.37 K,3.15 12.12 由《机械设计》表弹性系数 ZZMP,189.8 EEa 12.16 由《机械设计》图 Z,2.5节点区域系数 ZHH 接触最小安全系数12.14 由《机械设计》表 S,1.05Hmin SHmin ,24000ht,，，103008 t 总工作时间t hhh Nrnt,60 应力循环次数 NLh1L19N,，1.410 L1,，，，60198024000 8 99 NNi,,，/1.410/3.3N,，0.42410LLL212 由《机械设计》图12.18 Z,0.97接触寿命系数 ZNN1 Z,1.1N2 ,Z7100.97，许用接触应力 [],HNlim11H [],,, H1S1.05Hmin [],,688.7MPHa1 ,Z5801.1，HNlim22 [],,,H2, [],607MPS1.05HaHmin2验算 2KTu，11 ,,ZZZg HEH,2bdu1 23.15535973.31，，， ,,589MP,，，189.82.50.87gHa6060603.3，， ,[],H2 计算结果表明，接触疲劳强度较为合适，齿轮尺寸无需调整。 3.确定传动主要尺寸 实际分度圆直径 d dmzmm,,，,2.52460dmm,60111 , 200mm dmzmm,,，,2.580200d222 dd，，60200中心距 a12 a,,,130mm a22 齿宽 b bdmm,,,，,,16060d1 为了便于装配和调整，根据和求,d1d 取 bmm,651出齿宽后，将小齿轮宽度再加大b 510:mm 齿根弯曲疲劳强度验算 0.750.75重合度系数 Y,,，,，Y0.250.25 ,,0.67 Y,1.71,, 12.10齿间载荷分布系数 由《机械设计》表 11 KF,K,, F,,1.49 KY0.67F,, 9 齿向载荷分布系数 bh/65/(2.252.5)10.7,，, KK,1.45F,F,由《机械设计》图 12.14 载荷系数K KKKKK,AVFF,, ,，，，1.51.161.491.45K ,3.76 由《机械设计》图12.21 Y,2.65齿形系数 YFa1Fa Y,2.23Fa2 由《机械设计》图12.22 Y,1.57应力修正系数 YSa1Sa Y,1.77Sa2 12.23c 由《机械设计》图 ,,600MP弯曲疲劳极限 ,Falim1Flim ,,450MPFalim2弯曲最小安全系数 由《机械设计》表12.14 S,1.25Fmin SFmin Nrnt,60 应力循环次数 NLh1L19N,，1.410 L1,，，，60198024000 99NNi,,，/1.410/3.3 N ,，0.42410LLL212 12.24由《机械设计》图 Y,0.87弯曲寿命系数 Y N1N Y,1.07N2 由《机械设计》图12.25 Y,1.0尺寸系数 Y XX ,YY许用弯曲应力 [],FNXlim11F[], , F1SFmin 6000.871，， , 1.25 ,YYFNXlim22[], ,F2 [],,417.6MPSF1aFmin 4501.071，，[],,385.2MP ,F2a 1.25 10 验算 2KT1 ,,YYY FFaSa,111bdm1 ,,125.2MPF1a23.7653597，， ,60602.5，， ,[],F1 YYFaSa22,,, FF21YYFaSa11 2.231.77， ,118.4MP, ,，125.2F2a2.651.57， ,[],F2传动无严重过载，故不作静强度校核 齿轮图形及其几何参数如下: 图4 外啮合小齿轮 表2 外啮合小齿轮的基本尺寸 齿轮项目名称 几何参数 ,,:20 齿形角 , ,,齿顶高系数h h,1 aa ,,cc,0.25顶隙系数 ,,p,0.38p齿根圆半径系数 ff ,,0 ,分度圆螺旋角 bmm,65 b齿宽 11 ,齿顶高 h ,，,12.52.5mmhhm,aaa dzmmm,，,，，,(2)(242)2.565齿顶圆直径 daa ,,齿根高 hhhcm,，,，，,()(10.25)2.53.125 mmffa 齿高 h hhh,，,，,2.53.1255.625mmaf o基圆直径 d ddd,,,coscos2056.382,mmbb 988,,HK(1009588)GB,精度等级 ()齿圈径向跳动公差 查《互换性与技术测量》表得 118,Fmm,0.071Frr ()公法线长度变动公差 查《互换性与技术测量》表得 117,FFmm,0.056ww ()1111,齿形公差 查《互换性与技术测量》表得 ffmm,0.014ff ()基节极限偏差 查《互换性与技术测量》表得 ffmm,,0.0181113,pbpb ()1114,齿向公差F Fmm,0.025查《互换性与技术测量》表得 ,, 公法线长度W ,0.205，其计算过程如下: W,19.29,0.283 Wmkz,,，，[1.476(21)0.014] ,，，，2.5[1.476(23)0.01424] ,19.29 1112,查《互换性与技术测量》表得fmm,,0.02 pt Efm,,,,，,,101020200, sspt Efm,,,,，,,161620320, sipt EEF,,cos0.72sin,,wsssr oo,,,，200cos200.7271sin20 ,,205,m EEF,，cos0.72sin,,wisir 12 ,,205,m 式中——齿距极限偏差 fpt ——齿厚上偏差 Ess ——齿厚下偏差 Esi ——公法线平均长度上偏差 Ews ——公法线平均长度下偏差 Ewi kz,，/90.5,，,24/90.53跨齿数 k 图5 外啮合大齿轮 表3 外啮合大齿轮基本尺寸 齿轮项目名称 几何参数 ,,:20 齿形角 , ,,齿顶高系数h h,1 aa ,,cc,0.25顶隙系数 ,,p,0.38p齿根圆半径系数 ff ,,0 ,分度圆螺旋角 bmm,60 b齿宽 13 ,齿顶高 h ,，,12.52.5mmhhm,aaa dzm,，,，，,(2)(802)2.5205齿顶圆直径 mmdaa ,,齿根高 hhhcm,，,，，,()(10.25)2.53.125 mmffa 齿高 h hhh,，,，,2.53.1255.625mmaf o基圆直径 d ddmm,,,cos200cos20187.94,bb 988,,HK(1009588)GB,精度等级 ()齿圈径向跳动公差 查《互换性与技术测量》表得 118,Fmm,0.08Frr ()公法线长度变动公差 查《互换性与技术测量》表得 117,FFmm,0.071ww ()1111,齿形公差 查《互换性与技术测量》表得 ffmm,0.018ff ()基节极限偏差 查《互换性与技术测量》表得 ffmm,,0.0201113,pbpb ()1114,齿向公差F Fmm,0.025查《互换性与技术测量》表得 ,, 公法线长度W ,0.227，其计算过程如下: W,59.93,0.311 Wmkz,,，，[1.476(21)0.014] ,2.5[1.476(291)0.01480]，,，， ,59.93 1112,查《互换性与技术测量》表得fmm,,0.022 pt Efm,,,,，,,101022220, sspt Efm,,,,，,,161622352, sipt EEF,,cos0.72sin,,wsssr oo,,,，220cos200.7280sin20 ,,227,m EEF,，cos0.72sin,,wisir 14 oo,,，，352cos200.7280sin20 ,,311,m 式中——齿距极限偏差 fpt ——齿厚上偏差 Ess ——齿厚下偏差 Esi ——公法线平均长度上偏差 Ews ——公法线平均长度下偏差 Ewi kz,，,，,/90.580/90.59跨齿数 k 二 低速级齿轮传动的几何计算 小齿轮用，调质处理，硬度，平均取为;大齿轮用241286HBHB260HB4540Cr 钢，调质处理，硬度229286HBHB，平均取为240HB。计算步骤如下: 表4低速级齿轮传动的校核计算 计算项目 计算内容 计算结果 齿面接触疲劳强度计算 1.初步计算 n980转速 n12n,, 2 nr,297/mini3.321 TTi,gg,转矩 T2112 ,,171564Nmm T2,，，535973.30.97 式中——传动效率 , ,,1.1齿宽系数 由《机械设计》表12.13，取 ,,,1.1ddd 12.17c 由《机械设计》图,,710MP 接触疲劳极限 ,Halim1Hlim ,,580MPHalim2初步计算的许用接 [] ,,0.9,H1Hlim1 15 ,，0.9710触应力 [] ] [,,,639MPHH1a [] ,0.9,,H2Hlim2 [] ,,522MPH2a ,，0.9580 A,85值 由《机械设计》表，取 12.16AA,85ddd 1初步计算小齿轮直 3,,Tu,12 dA,g,,d12径直径 d1,,[]udH,, 1 3,,1715644.81, ,85g,,2855224.8，,, 取,68mmd1 ,65.3 b,75mm初步齿宽 b= 1.16874.8，,mmbd,,d1 2.校核计算 dn,圆周速度 v12v , 601000， 3.1468297，，,1.1/ms v ,601000， 精度等级 由《机械设计》表 选8级精度 12.6 齿数和模数 zm初取齿数 ,17z1 ziz,,，,4.81782221 ,4 md681m,,,4 17 z,z1711 82 z,4取，则，17;82 z,z,m,212 12.9 由《机械设计》表 K,1.5使用系数 KAA 12.9 由《机械设计》图 K,1.15动载系数 Kvv 齿间载荷分配系12.10，先求 由《机械设计》表 22171564T，数 k2H,FN,,,5046 td681 KF1.55046， At ,,100.92/ms b75 ,100/ms 16 11 ,,,,,[1.883.2()]cos ,zz12 11,,,[1.883.2]cos0,, ,,1782,, ,1.03 ,,1.03, 4,,41.03,,Z,,,0.995 , Z,0.99533, 由《机械设计》表 12.10 K,1.2H,齿向载荷分布系 由《机械设计》表 12.11 b数 K,3H,,，，()10KABCb H,d1 23,K,1.36 ,，，，，，1.170.161.10.611075H, K载荷系数 KKKKK,AVHH,, ,，，，1.51.151.21.36K,2.82 由《机械设计》表12.12弹性系数 ZZMP,189.8 EEa 12.16 由《机械设计》图 Z,2.5节点区域系数 ZHH 接触最小安全系12.14 由《机械设计》表 S,1.05Hmin数 SHmin ,24000h t,，，103008t总工作时间 thhh Nrnt,60 应力循环次数 NLh1L18N,，4.310 L1,，，，60129724000 97NNi,,，/1.410/3.3 N,，0.910 LLL221 12.18 由《机械设计》图Z,1.1 接触寿命系数 ZNN2 Z,1.31N2许用接触应力 ,Z7101.1，HNlim11[],, , H1[], S1.05HHmin [], ,743.8MPHa1 17 Z,5801.31， ,723.6MP[],HNlim22H2 [],,,H2S1.05Hmin 验算 2KTu,12 ,,ZZZg HEH,2bdu1 ,,699MP22.821715644.81，，,Ha,，，189.92.50.995g275684.8， ,[],H2 计算结果表明，接触疲劳强度较为合适，齿轮尺寸无需调整。 3.确定传动主要尺寸 实际分度圆直径 d ,68mmdmzmm,,，,41768d111 , 328mmdmzmm,,，,482328d222 dd,,32868中心距 a21 a,,,130mm a22 齿宽 b,75 bd,,，,1.168mmd1 为了便于装配和调整，根据和求,d1d 取80mm b,1出齿宽后，将小齿轮宽度再加大b 510:mm 齿根弯曲疲劳强度验算 0.750.75重合度系数 Y,,，,，Y0.250.25 ,,0.98 Y,1.03,, 齿间载荷分布系数12.10 由《机械设计》表 ,1.2 KKF,F,齿向载荷分布系数 bh/75/(2.254)8.3,，, K K,1.28 F,F,由《机械设计》图12.14 K载荷系数 KKKKK, AVFF,, ,，，，1.51.151.21.28 K,2.65 12.21 由《机械设计》图Y,2.95 齿形系数 YFa1Fa Y,2.26 Fa2 18 由《机械设计》图12.22 Y,1.52应力修正系数 YSa1Sa Y,1.78Sa2 由《机械设计》图12.23c ,,600MP弯曲疲劳极限 ,Falim1Flim ,,450MPFalim2 弯曲最小安全系数 由《机械设计》表12.14 S,1.25Fmin SFmin Nrnt,60应力循环次数 NLh1L19N,，1.410 L1,，，，60198024000 99NNi,,，/1.410/3.3 N ,，0.42410LLL212 12.24由《机械设计》图 Y,0.95弯曲寿命系数 Y N1N Y,0.99N2 由《机械设计》图12.25 Y,1.0尺寸系数 Y XX YY,许用弯曲应力 [],FNXlim11F[], , F1SFmin 6000.951，， , 1.25 YY,FNXlim22[], ,F2 ,456MP[],SF1aFmin 4500.991，， [],,356.4MP ,F2a 1.25 验算 2KT2,,YYY FFaSa111,bdm1 ,,196MPF1a22.65171564，， ,，，，2.951.520.9875684，，,[], F1 YYFaSa22,,, FF21YYFaSa11 2.231.77， ,,104MP ,，196F2a2.651.57， 19 ,[],F2传动无严重过载，故不作静强度校核 齿轮图形及其几何参数如下: 锻造小齿轮结构适用于内啮合小齿轮，适用于的齿轮。当时，da,200mmXm,2.5t齿轮与轴做成一体;当时，齿轮与轴分开制造。 Xm,2.5t 若内啮合小齿轮与轴分开制造，则齿轮轴直径选为，键槽尺寸，40mmtmm1,3.8齿轮分度圆直径为，齿根圆直径为dmzmm1,,，,41758 dd,,f,,Xt,,，1dzhcmmm,,,,,，,，，,(22)(172120.25)2.558。如图(6)所示，,,fa22,, 5840,,Xmm,,，,85.3.2式中为轴的直径，故，所以齿轮与轴做成一体。 d,,5102.m,,22,, 图6 20 图7 齿轮轴 表5 外齿轮几何参数 齿轮项目名称 几何参数 ,,:20齿形角 , ,,齿顶高系数 hh,1aa ,,顶隙系数 cc,0.25 ,,齿根圆半径系数 p,0.38 pff ,,0 ,分度圆螺旋角 bmm,80齿宽 b ,齿顶高 h ,，,144mm hhm,aaa dzm,，,，，,(2)(172)476齿顶圆直径 mmdaa ,,齿根高h hhcm,，,，，,()(10.25)45 mmffa 齿高 h hhh,，,，,459mmaf o基圆直径 d ddmm,,,cos68cos2063.899,bb 988,,HK (1009588)GB,精度等级 ()齿圈径向跳动公差 查《互换性与技术测量》表118,得 Fmm,0.08Frr ()公法线长度变动公差 查《互换性与技术测量》表117,得 FFmm,0.056ww ()1111,ffmm,0.020齿形公差 查《互换性与技术测量》表得 ff (ffmm,,0.0221113,基节极限偏差 查《互换性与技术测量》表得pbpb ) ()1114,FFmm,0.025齿向公差 查《互换性与技术测量》表得 ,, 公法线长度W ,0.227W,59.93，其计算过程如下: ,0.311 Wmkz,,，，[1.476(21)0.014] 21 ,4[1.476(21.41)0.01417]，,，， ,11.58mm 查《互换性与技术测量》表1112,得 fmm,,0.025pt Efm,,,,，,,101025250,sspt Efm,,,,，,,161625400,sipt EEF,,cos0.72sin,,wsssr oo,,,，250cos200.7280sin20 ,,255,m EEF,，cos0.72sin,,wisir oo,,，，352cos200.7280sin20 ,,378,m 式中——齿距极限偏差 fpt ——齿厚上偏差 Ess ——齿厚下偏差 Esi ——公法线平均长度上偏差 Ews ——公法线平均长度下偏差 Ewi kz,，,，,/90.517/90.52 跨齿数 k 表6 内齿轮几何参数 齿轮项目名称 几何参数 ,,:20 齿形角 , ,,齿顶高系数h h,1 aa ,,cc,0.25顶隙系数 ,,p,0.38p齿根圆半径系数 ff 22 ,,0 ,分度圆螺旋角 bmm,75齿宽 b ,齿顶高 h ,，,144mmhhm,aaa dzm,，,，，,(2)(822)4336齿顶圆直径 mmdaa ,,齿根高 hhhcm,，,，，,()(10.25)45 mmffa 齿高 h hhh,，,，,459mmaf o基圆直径 d ddmm,,,cos68cos20308.219,bb 988,,HK(1009588)GB,精度等级 ()齿圈径向跳动公差 查《互换性与技术测量》表得 118,FFmm,0.100rr ()公法线长度变动公差 查《互换性与技术测量》表得 117,FFmm,0.071ww ()1111,齿形公差f fmm,0.022查《互换性与技术测量》表得 ff (基节极限偏差f fmm,,0.025查《互换性与技术测量》表1113,得pbpb ) ()1114,齿向公差F Fmm,0.025查《互换性与技术测量》表得 ,, 公法线长度W ，0.396W,116.77，其计算过程如下: ，0.289Wmkz,,，，[1.476(21)0.014] ,4[1.476(21.41)0.01417]，,，， ,11.58mm 1112,fmm,,0.028查《互换性与技术测量》表得 pt Efm,,,,，,,101028280, sspt Efm,,,,，,,161628448, sipt EEF,,,cos0.72sin,,wssir 23 oo,,，448cos200.72100sin20 ,，396,m EEF,,，cos0.72sin,,wissr oo,，，280cos200.72100sin20 ,，288,m 式中——齿距极限偏差 fpt ——齿厚上偏差 Ess ——齿厚下偏差 Esi ——公法线平均长度上偏差 Ews ——公法线平均长度下偏差 Ewi kz,，,，,/90.582/90.510 跨齿数 k 24 图8 内齿轮 25 三 齿轮轴的设计计算 图9 齿轮轴 轴的材料为，轴速为。设计过程如下: nr,297/min40Cr2 表7齿轮轴的计算 计算项目 计算内容 计算结果 计算齿轮受力 估算轴径 由《机械设计》表16.2得C,102，故 P5.533d, 26.7mm dmm,,,10210226.7 n297 外圆周力 22171564T，,1715.64N F2t2F,, 啮 t2d200合 大 径向力 ,624.5N FFF,,tan1715.64tan20, r2rt22齿 轮 F1715.64法向力 t2,1826N F ,,Fn2n2,coscos20 内圆周力 22171564T，,5046N F2t3F,, 啮t3d68合 小径向力 ,1837N FFF,,tan5046tan20, r3rt33齿 轮 F5046法向力 t3,5370N F ,,Fn3n3,coscos20 26 画齿轮轴受 力图 计算支撑反力 FF，,，49142.5水平面反力 'rr32 F,R198.5',10.4N FR1183749624.5142.5，,， ,98.5' ,,1202NFR2'' FFFF,，,,，,10.4624.51837RRrr2123 FF，,，142.549垂直面反力 "tt23 F,R198.5",,28.2N FR11715.64142.5504649，,， , 98.5 """FFFF,,,,,，1715.64504628.2 ,,3302.2N FRttRR22231 水平面受力 图 垂直面受力 图 画轴弯矩图 27 水平面弯矩 图 垂直面弯矩 图 合成弯矩图 22合成弯矩 MMM,，XyXz 画轴转矩图 T,,171564Nmm 轴受转矩 TT,2 28 转矩图 许用应力 许用应力值 用插入法由《机械设计》表查得: 16.3 ,,102.5MP;,,60MP ,,,,0ba,1ba 应力校正系 ,,,60,1b ,,, ,0.59,数 102.5,,,0b 画当量弯矩图 ,T,,101223Nmm 当量转矩 ，见转矩图 ,T,，0.59171564 当量弯矩 在轴齿轮中间截面处 2'2MMT,，, ，， 22 ,，172194101223M ,,199742Nmm 当量弯矩图 校核轴径 齿根圆直径 ,,ddhcm,,，2 ，，fa11 d,58mm ,,，，68210.254 ，，f1轴径 199742M3,,d 3d,32.2mm 0.160，0.1,,,,b1,58mm 29 四 齿轮轴上轴承的设计计算 和的深沟球轴承。其尺寸和主要参数如下: 根据轴径，分别选用内径4540mmmm 表8 轴承参数 轴承代号 基本尺寸 基本额定载荷 极限转速 /mm/KN db 脂//minrD CCr0r 620945851931.520.57000 620840801829.518.08000 图10 轴承示意图 表9 轴承计算 寿命计算 YX、值 F XX,,1a120,,e由《机械设计》表18.7得 Fr YY,,012冲击载荷系数18.8得 考虑中等冲击，由《机械设计》表 f,1.5d fd 当量动载荷 PfXFYF,， ，，11111dra ,，，1.5110.4 ,15.6N P1 PfXFYF,， ，，22222dra ,1803N P2,，，1.511202 轴承寿命 2因，只计算轴承的寿命 PP,21 30 , ,,16670C L,,,10h ,245843hLnP10h22,, ,24000h31667029500,, ,,,2971803,, 静载荷计算 由《机械设计》表18.12 X,0.6、 XY000 Y,0.50 '当量静载荷 ,PXF,,0101rR'取大者则 PFN,,10.4,rR011' ,10.4NPPF,01r,011rR, ' ,PXF,,0202rR取大者则,' ,1202NPPF,02r,022rR, ' PFN,,1202rR022 18.14 正常使用球轴承，由《机械设计》表 S,1.3安全系数 S00 计算额定静载'' ,1562.6N CSP,,，1.31202Crr0r02002荷 '2(,只计算轴承) PP,轴承 CC,0201rr002rr 许用转速验算 载荷系数 P15.61,,0.0005 C31500r1 f,111由《机械设计》图18.19 P18032,,0.06 C29500r2 f,11218.19由《机械设计》图 载荷分布系数 Fa10,，由《机械设计》图18.20 ' f,1.5F21R1 Fa20,18.20，由《机械设计》图 ' f,1.5F22R2 许用转速 ,10500/minrNffN,,，，11.57000 N 1112101 31 ,12000/minrNffN,,，，11.58000N2122202 均大于工作转速 297/minr结论:所选轴承满足寿命、静载荷与许用转速的要求，且各项指标潜力都很大。 32 五 法兰轴的设计计算 图11右法兰轴 为了便于计算分析，可以把整个滚筒的法兰周电机轴简化为如图(12) 33 图12 表10 法兰轴的计算 计算项目 计算内容 计算结果 计算滚筒受力 NKW5.5滚筒所受 牵引力 FN,,,3437.5皮带的拉 vms1.6/ 力计算 式中N——电机功率; ——胶带运行速度 v ,,Se,1，，L 摩擦条件 F,n 式中——输送带在分离点张力 SL ——驱动滚筒与输送带间的摩, 擦系数 ——驱动滚筒的围包角 , ——摩擦力备用系数 n,2635.8N SL 其数值为均为已知 Fn3437.51.2，S,6073.3N 则 ,S,yL,,0.3,1,ee,1 SSF,，,，2635.83437.5 yL T,12146N S式中——输送带在相遇点张力 y TS,,，226073.3滚筒所受拉力 y 2S为了便于计算，中包括滚筒的重力等其y 它力 34 轴受力的平移 简化图 计算支撑反力 水平面反力 TT''856.5277.5165143.5，,，,，，，FFRR12 22 F,y11000 6073.3856.510.2277.512021656073.3143.5，,，,，，， F,5872.2N,y11000 TT''143.5722.5835856.5.5，,，,，，，FFRR12 22F,y21000 F,5062.2Ny26073.3143.510.2722.512028356073.3856.5，,，,，，， ,1000 垂直面反力 ""FF，，，277.5165RR12 F, z11000 28.2277.53302.2165，，，,552.7N F ,z11000 ""FF，，，722.5835RR12 F, z21000 28.2722.53302.2835，，，,2777.7N F ,z21000 水平面受力图 垂直面受力图 35 画弯矩图 水平面弯矩图 垂直面弯矩图 合成弯矩图 22合成弯矩 MMM,，XyXz 画轴转矩图 T,,53597Nmm 轴受转矩 TT,1 转矩图 许用应力 16.3许用应力用插入法由《机械设计》表查得: 值 36 ; ,,102.5MP,,60MP,,,,0ba,1ba 应力校正 ,,,60,1b ,,, ,0.59,系数 102.5,,,0b 画当量弯矩图 ,T,,31622Nmm当量转矩 ，见转矩图 ,T,，0.5953597 当量弯矩 在轴两端轴承中间截面处 2'2MMT,，, ，， 22 ,，84635631622M ,,846946Nmm当量弯矩 图 校核轴径 轴径 '846946M33,d, d,52mm0.160，0.1,,,,1b ,70mm 六 法兰轴上的轴承设计 53597 轴承结构尺寸见图(10) 53597 表11 轴承参数 /mm/KN轴承代号 基本尺寸 基本额定载荷 极限转速 d b //minr 脂D CCr0r6315 75 160 37 76.8 4000 112 表12轴承计算 寿命计算 YX、值 FXX,,1 a12018.7,,e由《机械设计》表得 Fr YY,,0 12 37 冲击载荷系数得 考虑中等冲击，由《机械设计》表18.8 f,1.5d fd 当量动载荷 PfXFYF,，，，11111dra ,8808.3NP1 ,，，1.515872.2 由于两个轴承对称布置，为了便于分析近 似认为 ,8808.3NPP,P212 ,轴承寿命 ,,16670C L, ,,10hnP22,, 3 ,34969hL16670112000,,10h ,,,9808808.3,, ,24000h静载荷计算 由《机械设计》表18.12 X,0.6、 XY000 Y,0.50 '当量静载荷 ,PXF,,0101rR取大者则,' ,5872.2NPPF,01r,011rR, ' PFN,,5872.2rR011 18.14 正常使用球轴承，由《机械设计》表 S,1.3安全系数 S00 计算额定静载'' ,7634N CSP,,，1.35872.2C0rrr02002荷 '轴承CC, 002rr许用转速验算 载荷系数 P8808.31,,0.079 C112000r1 f,0.99118.19由《机械设计》图 载荷分布系数 Fa10,18.20，由《机械设计》图 ' f,1.5F21R1 许用转速 ,5940/minr NffN,,，，0.991.54000N11201 大于工作转速 38 980/minr结论:所选轴承满足寿命、静载荷与许用转速的要求，且各项指标潜力都很大。 七 弹性挡圈的选用 挡圈几何参数如下: 图14 孔用弹性挡圈A型 表13 孔用挡圈参数 孔径/mm D/mm S/mm b/mm d1/mm 85 90.5 2.5 6.8 3 80 85.5 2.5 6.5 3 图15 轴用弹性挡圈A型 39 表14 轴用挡圈参数 轴径/mm d/mm S/mm b/mm d1/mm 70 65.5 2.5 6.32 3 八 吊环的选用及几何参数 图16 吊环螺钉A型 表15 吊环参数 /mm D1/mm d2/mm d4/mm l/mm b/mm d 20 40 41.4 72 35 19 九 键的选用及几何参数 图17 键 (一) 外啮合大齿轮所选用的键 40 111'，，，， ThldhlbdNmm,,,,,,，,，，,,8401240120268800171564，，，，pp，，，，444 (二) 外啮合小齿轮所选用的键 111'，，，， ThldhlbdNmm,,,,，,，，,,,,73082812012936053597，，，，pp，，，，444 (三) 法兰轴在支座处所选用的键 111'，，，， ThldhlbdNmm,,,,,,，,，，,,1260207012010800053597，，，，pp，，，，444 表16 键的参数 d/mm b/mm h/mm t/mm t1/mm L/mm ,384412 8 5.0 3.3 选用L=40 ,22110 ,22308 7 4.0 3.3 选用L=30 ,1890 ,657520 12 7.5 4.9 选用L=60 ,56220 十 螺母的计算 当用受拉螺栓联接时，需要的螺栓预紧力: kTf' ,F,，，，rrr，，s126 式中k——考虑摩擦传力的可靠系数，k,1.11.5 ff ——接合面摩擦系数，当接合面干燥时，,0.100.16;当接合面有油时，,,ss ,0.060.10 ,s 、——各螺栓中心至底板旋转中心的距离 rrr126 1.353597，'F,故,385N 0.132326，，， 4F,ss ,,,,,2dm，，,sp，，式中——接合面数目 m ,——屈服极限 s ，，ss,1.2 ——安全系数， ,,ps，， 5.6螺栓性能等级选用级 41 ,Bmin故， ,5,,，,5100500MPBamin100 式中——材料最小拉伸强度极限 ,Bmin ,,,smin故， ,,0.6,,，,0.6500300MP106,,,sminBmina,Bmin,, ,300s ,,,MP250,,,1.2，，sp，， 4F6s ,，25010Pa2,dm 4385，6 ,，2501023.14d 4385，,3d,，10= 1.4mm3.14250， 端盖与滚筒联接选用M8螺栓，6个均布;内齿轮选用M8螺栓，6个均布;电机与法兰 轴联接选用M8，共10个。 第六章 滚筒其它结构设计 一 滚筒设计 滚筒材料选用Q235钢板卷起焊接而成。两根钢条分别焊在滚筒两端，在滚筒内部均 布六个刮油板，与滚筒焊接联接。如图: 图18焊接滚筒 二 端盖设计 42 端盖材料为HT200。 43 图19 大端盖 图20 左小端盖 图21 右小端盖 44 总 结 随着毕业日子的到来，毕业设计也接近了尾声。经过努力我的毕业设计终于完成了。在没有做毕业设计以前觉得毕业设计只是对这几年来所学知识的单纯总结，但是通过这次设计发现自己的看法有点太片面。毕业设计不仅是对前面所学知识的一种检验，而且也是对自己能力的一种提高。通过这次毕业设计使我明白了自己原来只是还比较欠缺。自己要学习的东西还太多，通过这次毕业设计，我才明白学习时一个长期积累的过程，在以后的工作、生活中都应该不断的学习，努力提高自己只是和综合素质。 在这半学期里，我们不断学习、不断积累并且不断的提高，在马素平老师的悉心指导下，我们从最初的开题报告开始做起，进行设计方案的确定;之后尺寸拟定、结构设计计算、工程制图、论文整理。这次的毕业设计，是对我这四年来所学的专业知识是否踏实的检验，让我对这四年中所学知识进行了综合，也让我温习了一些已经快要淡忘的专业知识，并且还学到了一些实际工程经验。 这次的油冷式电动滚筒设计中重点在于滚筒要受到空间的限制，但在空间的范围内还要满足滚筒的性能要求，比如减速器的传动比、轴承的寿命等，这就给设计制定了一定的难度，但是在设计中不能盲目，应该以资料为基础在加以自己的创新想法。 毕业设计收获很多，比如学会了查找相关资料相关，分析数据，提高了自己的绘图能力，懂得了许多经验的获得是前人不懈努力的结果。同时，仍有很多课题需要后辈去努力去完善。在这次毕业设计也使我们的同学关系更进一步了，同学之间互相 帮助，有什么不懂的大家在一起商量，听听ti不同的看法对我们更好的理解知识。 在设计过程中，我通过查阅大量有关资料，与同学交流经验和自学，并向老师请教等方式，使自己学到了不少知识，也经历了不少艰辛，但收获同样巨大。在整个设计中我懂得了许多东西，也培养了我独立工作的能力，树立了对自己工作能力的信心，相信会对今后的学习工作有非常重要的影响。而且大大提高了动手的能力，使我充分体会到了在创造过程中探索的艰难和成功时的影响。 但是毕业设计也暴露处自己专业基础的很多不足之处，比如缺乏综合应用专业知识的能力，对材料的不了解，等等。这次实践是对自己大学四年所学的一次大检阅，使我明白自己知识还很浅薄，虽然马上要毕业了，但是自己的求学之路还很长，以后更应该在工作中学习，努力使自己成为一个对社会有所贡献的人。 虽然设计中难免会出现错误，但是在设计过程中所学到的东西是这次毕业设计的最大收获和财富，使我终身受益。 45 谢 致 作者在设计(论文)期间都是在马素平老师全面、具体指导下完成进行的。马老师渊博的学识、 敏锐的思维、民主而严谨的作风使学生受益非浅，并终生难忘。 感谢刘英林教授等在毕业设计工作中给予的帮助。 感谢我的学友和朋友对我的关心和帮助。 46 参考文献 1 廖念钊主编.互换性与技术测量(第四版).北京:中国计量出版社，2006 2 东南大学机械学学科组 郑文纬、吴克坚主编.机械原理(第七版).北京:高等教育出版社 3 陈昭怡、吴桂英主编.材料力学.北京:中国建设工业出版社，2005 4 邱宣怀主编.机械设计(第四版).北京:高等教育出版社 5 王明珠主编.工程制图学及计算机绘图.北京:国防工业出版社 6 谢锡纯、李晓豁主编.矿山机械与设备.中国矿业大学出版社 7 成大先主编.机械设计手册(第三版).化学工业出版社 8 东北工学院《机械零件设计手册》编写组编.机械零件设计手册.冶金工业出版社 47 附 录 Kinetostatic analysis of a roller drive Kuen-Bao Sheu, , a, Chih-Wei Chienb, Shen-Tarng Chioub and Ta-Shi Laia a Department of Vehicle Engineering, National Huwei Institute of Technology, Huwei, Yuenlin 63208, Taiwan b Department of Mechanical Engineering, National Cheng Kung University, Tainan 70101, Taiwan Received 15 April 2003; Revised 8 March 2004; accepted 23 March 2004. Available online 13 May 2004. Abstract A roller drive has a topological structure similar to an epicyclic gearing, where both the planet gear and the ring gear have rollers as their teeth. The roller drive thus has the advantages of easy manufacture and low cost. In this paper, the kinematic model based on the vector loop approach is developed, and models for analyzing the kinetostatics and mechanical efficiency of roller drives considering friction are proposed. By using these models, the velocity, acceleration, normal forces acting on the conjugated pairs and efficiency of roller drives can be obtained. An example is given, and experimental testing is applied to verify the analysis results. The results show that the roller drive has a speed fluctuation that corresponds with the number of ring gear rollers, and the magnitudes of the forces vary periodically with the crank angular displacement; hence, the roller drive has an input torque ripple. It is also found that the roller drive has high efficiency under the operation of high speed. Moreover, the influence of the geometric parameters on efficiency and speed fluctuation of the roller drives is discussed. Author Keywords: Roller drives; Kinematics; Kinetostatics; Efficiency Article Outline 1. Introduction 2. Kinematic analysis 2.1. Angular velocity and acceleration analysis 2.2. Relative speed of the conjugated pairs 3. Kinetostatic analysis 3.1. Force acting on the crank 3.2. Force acting on the planet gear 3.3. Force acting on the disc plate 3.4. Nonlinear equations 3.5. Frictional powers and efficiency analysis 4. Analysis examples and test 4.1. Angular velocity and acceleration 4.2. Forces and efficiency 48 4.3. Influence of the geometric parameters 4.4. Test 5. Conclusions Acknowledgements References 1. Introduction Epicyclic gearings are widely used in industry for the purposes of speed and torque conversion such as planetary gear speed reducers, harmonic drives, and cycloid drives [1, 2, 3 and 4]. Compactness, large speed reduction ratio, and high efficiency are the advantages of this type of reducers. Such mechanisms must be manufactured to exact standards to offer high performance levels. A roller drive with the advantages of easy manufacture and low cost has been created [5 and 6]. The topological structure of a typical single-stage roller drive is shown in Fig. 1. A crank, which is eccentrically mounted on the planet gear set serves as the input. A ring gear that has cylindrical rollers as its teeth is mounted on the housing; and a planet gear that has cylindrical rollers as its teeth engages with the ring gear. A constant velocity joint consisting of a set of pins attached to a disc plate and mated with an equal number of holes in the planet gear is used for converting the rotary motion to the output shaft. Display Full Size version of this image (11K) Fig. 1. Topological structure of the roller drive. In the past, the kinematic analysis of the planetary gear trains has been the subject of a number of studies [7, 8, 9, 10, 11, 12 and 13]. Blanche and Yang [14] developed an analytical model of the cycloid drives with machining tolerances and investigated the effects of machining tolerances on backlash and torque ripple; and they [15] also presented a computer-aided analysis procedure to verify the performance of cycloid drives. In addition, Romiti et al. [16] performed the kinematic analysis of a novel gearless articulated speed reducer with machining tolerances. Moreover, kinetostatic analysis of the planetary gear trains has been performed by many investigators [17, 18, 19 and 20]. Malhotra and Parameswaran [21] presented a procedure to calculate the forces on various elements of the cycloid drives. However, there have been only a few studies on the roller drives. Yan and Lai [22] described the characteristics of roller drives, and by using the basic efficiency of the mesh pair to derive the formulas for the mechanical efficiency of roller drives. In this work, firstly, the mathematical model for analyzing the kinematics is introduced; secondly, models for kinetostatics and mechanical efficiency of roller drives considering friction are developed. The forces on various elements of the roller drive as well as the mechanical efficiency are then presented. Finally, an experimental testing is used to verify the analysis results. 2. Kinematic analysis 49 The following is a discussion of the angular velocity and acceleration of the roller drives and the relative speed at the contact point of the conjugated pairs. 2.1. Angular velocity and acceleration analysis As depicted in Fig. 2, for an individual contact roller pair of the planet gear and ring gear, if we connect the four centers of the crank O4 (which is also coincident with the ring gear center, O2), a roller of the ring gear O2i, the planet gear O3, and a roller of the planet gear O3i, the quadrilateral O4O3O3iO3i forms a 4-bar equivalent linkage. By using the vector loop approach [23], the position vector loop equation can be expressed as (1) where , , and represent the vectors from O4 to O3, O3 to O3i, O2i to O3i, and O4 to O2i, respectively. Based on Eq. (1), its scalar component equations are R4cosθ4+R3icosθ3i?Racosθa?R2icosθ2i=0 (2) R4sinθ4+R3isinθ3i?Rasinθa?R2isinθ2i=0 (3) where θ2i, θ4, θ3i and θa denote the angular positions (measured counterclockwise from positive X axis) of the ring gear, crank, planet gear, and the segment O2iO3i, respectively. Display Full Size version of this image (5K) Fig. 2. Mathematical modeling for kinematic analysis. For the analysis of planet gear position, one problem is to find out which pair of the planet gear roller and ring gear roller makes the first contact. One way to determine this is to calculate the distance from the roller center of the ring gear to the roller center of the planet gear Ra based on Eqs. ((2) and (3)). For a given contact pair of the planet gear roller and ring gear roller, the distance from the roller center of the ring gear to the roller center of the planet gear Ra is equal to the sum of the roller radii of the ring gear r2 and the planet gear r3, i.e., Ra=r2+r3. Additionally, Ra>r2+r3 indicate that the roller of the planet gear and that of the ring gear are not in contact. The results show that the roller drive has only a single tooth meshed while it is gearing. By differentiating Eqs. ((2) and (3)) with respect to the angle of the crank θ4, the kinematic coefficients [23] can be determined as (4) (5) 50 where . Based on the chain rule of differentiation, we have , and the angular velocity of the planet gear and segment O2iO3i can be expressed as follows. ωi=hiω4 (i=3,a) (6) By differentiating Eqs. ((2) and (3)) with respect to the angle of the crank θ4 twice, gives (7) (8) where , and the angular acceleration of the planet gear and segment O2iO3i can be written as αi=hiα4+h′iω42 (i=3,a) (9) As shown in Fig. 3, the relative speed at contact point P5i3 of the disc pin and disc pin hole can be considered to be belonging to the planet gear and to the disc pin , respectively, which can be written as (10) (11) Assuming both the planet gear and the disc pin are rigid bodies, the normal speed component between the planet gear and disc pin at the contact point is 0. Hence, the normal speed component with respect to the disc pin hole and to the disc pin are the same; that is, VP35in=ω3R3ccosβc, VP35in=ω5R5i3cosβP and ω3R3ccosβc=ω5R5i3cosβP. As shown in Fig. 3, since R3ccosβc=R5i3cosβP, therefore ω3=ω5 (12) Display Full Size version of this image (7K) Fig. 3. Relative speeds at conjugated joints. 2.2. Relative speed of the conjugated pairs 51 From Eqs. ((10) and (11)), the relative tangential speed component between the planet gear and disc pin at the contact point P5i3 can be obtained as (13) Once the location of the instantaneous center is found, the position vector from O4 to the instantaneous center can be expressed as : , and the angular speed ratio can be written as (14) Substituting Eqs. ((12) and (14)) into Eq. (13), yields (15) As shown in Fig. 3, the relative speed at contact point P2i3 of the ring gear roller and planet gear roller can be expressed as (16) where is the vector from the planet gear center O3 to the contact point P2i3, and it can be expressed as: . Based on Eq. (16), the tangential speed component at the planet gear roller and ring gear roller at the contact point P2i3 can be determined as (17) 3. Kinetostatic analysis The model of kinetostatic analysis considering friction of the roller drives is developed in this section. It is assumed that the gravity forces are relative small and can be neglected, and the centers of mass are located at the geometrics center of the components. 3.1. Force acting on the crank The free-body diagram of the crank is as shown in Fig. 4, and a known input torque is applied to the crank. Vectors and point from O3, the planet gear center, to points P43 and P′43 (they are the contact points of the crank and planet gear), respectively. is the vector of the force exerted by frame 1 on crank 4, additionally, there is a frictional torque . Similarly, and are, respectively, the force and frictional torque exerted by the planet gear on the crank at the contact point P34 (P′34). Frictional torques , t′34 and t14 can be expressed as t34=μ34F34Rb3D34 (18) t′34=μ34F′34R′b3D′34 (19) 52 t14=μ14F14Rb4D14 (20) where Rb3 (R′b3) and Rb4 are the radii of the roller bearings placed between the crank and planet gear and the frame, respectively, and μ34 and μ14 are the friction coefficients of the roller bearings. Furthermore, D34, D′34 and D14 represent the direction indicators that can be expressed as: , and , respectively [23]. Display Full Size version of this image (5K) Fig. 4. Free-body diagram of the crank. As shown in Fig. 4, the three scalar equations of motion that are applied to the crank body can be written as F34x+F′34x+F14x=m4a4x (21) F34y+F′34y+F14y=m4a4y (22) (23) where m4 is the mass of the crank, I4 is the moment of inertia of the crank about the center of mass, and a4x and a4y represent, respectively, the magnitudes of the X and Y components of the center of mass acceleration of the crank. Furthermore, R34x and R34y can be expressed as: R34x=R4cosθ4?Rb3cosθ34 and R34y=R4sinθ4?Rb3sinθ34, and θ34 (θ34=tan?1F34y/F34x), is measured counterclockwise from the X axis to the common normal at the contact point P34. 3.2. Force acting on the planet gear In Fig. 5, since the constant velocity joint that behaves kinematically like an articulated parallelogram, the common normal at the contact point P5i3 of the planet gear and ith disc pin is parallel to . Assuming that an external load acts on the planet gear, and the elastic deflection at the contact point causes the planet gear to rotate a small angle Δ. Since the elastic deflection at the contact point is very small, it is also assume that the force exerted by the ith disc pin on the planet gear is proportional to the deflection Δ along the common normal at the contact point P5i3; i.e., Δ=R5 Δ. Thus (24) Eq. (24) then can be rewritten as (25) 53 where F53 is the resultant force exerted by the disc pins on the planet gear, and n5 is the contact number of the disc pins and planet gear. When the centers of the crank, O4, the planet gear, O3, and the disc pin, O3d, are aligned, n5=(Z5/2)+1 for an even number Z5 or n5=(Z5?1)/2 for an odd number Z5. Where Z5 denotes the number of the disc pin. Moreover, as the centers O4, O3 and O3d are not aligned, n5=Z5/2 for an even number Z5 or n5=(Z5?1)/2 for an odd number Z5 [15]. In addition, the line of action of the force is in the direction of the common normal at the contact point P5i3 and it is parallel to . Thus (26) Based on Eqs. ((25) and (26)), the resultant force exerted by the disc pins on the planet gear can be determined as (27) where Display Full Size version of this image (11K) Fig. 5. Free-body diagram of the planet gear. Moreover, the frictional force exerted by the ith disc pin on the planet gear can be expressed as (28) where μ53 is the friction coefficient between the planet gear and disc pin. The direction indicator D5i3 can be written as: , where , which represents the unit vector in the direction of common tangent at the contact point P5i3 can be expressed as: . Therefore, the resultant frictional force exerted by the disc pins on the planet gear can be expressed as (29) where 54 The total torque acting on the planet gear caused by the two forces can be determined as (30) where which represents the vector from O3 to the contact point P5i3 can be expressed as: . Here R3a and r3a are the radius of the circle containing rollers in the planet gear and the radius of the disc pin, respectively. Substituting Eqs. ((25) and (28)) into Eq. (30), gives (31) where In addition to the force exerted by the disc pins, the ring gear rollers also exert force on the planet gear. As sown in Fig. 5, when the crank rotates counterclockwise, force and frictional force exerted by the ring gear roller on the planet gear, and these forces cause torque and frictional torque with respect to the center of the planet gear. Since the roller drive has only a single tooth meshed while it is gearing, the forces acting on the planet gear exerted by the ring gear roller can be written as (32) (33) where D23=ω3/|ω3|. Also, denotes the unit vector of common tangent of the contact point P2i3 and μ23 is the frictional coefficient between the planet gear roller and ring gear roller. The total torque acting on the planet gear caused by the forces can be determined as (34) where K23t=R3sin(θa?θ3)+μ23D23[r3?R3cos(θa?θ3)] and is the vector from O3 to P2i3. With reference to Fig. 5, three scalar equations of motion which apply to the planet gear can be written as F53x+f53x+F23x+f23x?F34x=(m3+m3a)a3x (35) F53y+f53y+F23y+f23y?F34y=(m3+m3a)a3y (36) T53+t53+T23+t23?t34=I3α3 (37) 55 where a3x and a3y, which represent the magnitudes of the X and Y components of the center of mass acceleration of the planet gear, respectively,can be expressed as: a3x=?α4R4sinθ4D41?ω42R4cosθ4 and a3y=α4R4cosθ4D41?ω42R4sinθ4, where D41=?D14=ω4/|ω4|. In addition, m3, m3a and I3 denote the mass of the planet gear, the planet gear roller and the moment of inertia of the planet gear about the mass center. 3.3. Force acting on the disc plate The free-body diagram of the disc plate is as shown in Fig. 6, torque is the external load on the disc plate. Force and frictional force are exerted by the planet gear on the disc plate, as well as the force is exerted by the frame on the disc plate. In addition, T , and represent the torque caused by the force , and , respectively. According to Fig. 6, the force and moment equations can be written as ?F53x?f53x?F′53x?f′53x+F15x=m5a5x (38) ?F53y?f53y?f53y?F′53y?f′53y+F15y=m5a5y (39) (40) where m5 is the mass of the disc plate, I5 is the moment of inertia of the disc plate about the center of mass; a5x and a5y represent, respectively, the magnitudes of the X and Y components of the center of mass acceleration of the disc plate. The friction torque t15 that is dependent on the force F15 and angular speed ω14 can be expressed as: t15=μ15F15Rb5D15, where D15=?ω5/|ω5|. Also, Rb5 is the roller bearing radius of the output shaft and μ15 is the friction coefficient between the frame and the disc plate. Display Full Size version of this image (3K) Fig. 6. Free-body diagram of the output shaft. 3.4. Nonlinear equations From the free-body diagrams of the crank, planet gear, and disc plate, the equations of motion for each moving link can be written as shown in Eqs. ((21), (22) and (23)), ((35), (36) and (37)), and ((38), (39) and (40)), respectively. These equations form a set of nine nonlinear equations with the unknown F34x, F34y, F′34x, F′34y, F14x, F14y, Tin, F53, F′53, T23, F15x and F15y. Since the gravity forces are neglected in this paper, the forces exerted by the two planet gears on the crank and disc plate are equal in magnitude but opposite in sense; that is, F′34x=?F34x, F′34y=?F34y, F′53x=?F53x and F′53y=?F53y. In addition, the torques exerted by the two planet gears on the crank and disc plate are equal in magnitude and in direction; that is, t34=t′34, t53=t′53 and T53=T′53. Therefore, the nine equations with nine unknowns can be rewritten as follows. F14x=m4a4x (41) 56 F14y=m4a4y (42) 2(R34xF34y?R34yF34x)+Tin=I4α4?2t34?t14 (43) (K53x+L53x)F53+(cosθa+μ23D23sinθa)F23?F34x=(m3+m3a)a3x (44) (K53y+L53y)F53+(sinθa?μ23D23cosθa)F23?F34y=(m3+m3a)a3y (45) K53tT53+K23tT23=I3α3+t34 (46) F15x=m5a5x (47) F15y=m5a5y (48) Tl?2K53tF53=I5α5?t15 (49) 3.5. Frictional powers and efficiency analysis Based on the power balance for a machine, the power equation for the roller drive can be written as where dU/dt and dT/dt represent the rates of changes of stored potential and kinetic energy, respectively, and Pl is the rate of change of the dissipation energy which can be expressed as Pl=|t14ω4|+|t15ω5|+|t34ω34|+2|f53V53|+2|f23V23| If the rates of changes of stored potential and kinetic energy are neglected here, the mechanical efficiency of the roller drive can be expressed as (50) 4. Analysis examples and test 4.1. Angular velocity and acceleration A roller drive, whose dimensions of the geometric parameters are listed in Table 1, is adopted here. The external load Tl=60 N m, input angular velocity ω4=1800 rpm, and input angular acceleration α4=0 rad/s2 are used in the example. In addition, the values of the friction coefficients of the roller bearings and the conjugated joints of the roller drive are μb=0.0015, μ23=0.02 and μ35=0.02, respectively. The angular velocity and acceleration of the roller drive can be obtained from Eqs. ((6) and (9)). 57 Table 1. Geometric parameters of the roller drive Based on the analysis results, the curves of the angular velocity and acceleration of the planet gear versus the crank angular displacement are shown in Fig. 7 and Fig. 8, respectively. It can be noted in Fig. 7 that the angular velocity of the roller drive is not a constant, and has a speed fluctuation corresponding to the number of the ring gear rollers. The speed fluctuation factor is defined as: (51) where ω3max and ω3min denote the maximum and minimum angular speed of the planet gear and ω3c is the ideal angular speed of the planet gear of the roller drive, which can be expressed as ω3c=[Z3/(Z3?Z2)]ω4. Here Z2 and Z3 are the number of the ring gear roller and planet gear roller, respectively. Based on Eq. (51), the speed fluctuation factor as shown in Fig. 7 is about 6%. Display Full Size version of this image (5K) 58 Fig. 7. Angular velocity for one crank revolution. Display Full Size version of this image (5K) Fig. 8. Angular acceleration for one crank revolution. 4.2. Forces and efficiency Since the set of force and moment equations is nonlinear, an iterative technique is used here, whose algorithm is as follows. 1. Set all friction terms t14, t34 and t15 equal to 0; then use Eqs. ((41), (42), (43), (44), (45), (46), (47), (48) and (49)) to solve the forces F14x, F14y, F34x, F34y, F15x and F15y. 2. Use the results of step 1 to calculate the friction terms t14, t34 and t15, as well as θ34. 3. Substitute the results of the frictional terms into the equations and solve for the forces. 4. Check for the convergence. If the specified accuracy is satisfied, then stop; otherwise, go to step 2. Once the forces and friction torques are obtained, the mechanical efficiency of the roller drive can be determined from Eq. (50). The curves of the analysis results of F34 (force exerted by the planet gear on the crank), F53 (force exerted by the disc plate on the planet gear), and F23 (force exerted by the ring gear rollers on the planet gear) are shown in Fig. 9, Fig. 10 and Fig. 11, respectively. The results show that the magnitudes of the forces vary periodically with crank angular displacement; and of all the conjugated joints of the roller drive, the force exerted by the planet gear on the crank is the largest. The input torque with an applied constant external load is shown in Fig. 12, indicating that the roller drive has an input torque ripple caused by the forces acting on the conjugated joints with periodic variation and fluctuation in the output speed. In Fig. 13, the mechanical efficiency of the roller drive is reported; and the mean value of efficiency is about 93% considering friction. Display Full Size version of this image (4K) Fig. 9. Force exerted by the planet gear on crank. Display Full Size version of this image (5K) Fig. 10. Force exerted by the disc plate on planet gear. 59 Display Full Size version of this image (5K) Fig. 11. Force exerted by the ring gear roller on planet gear roller. Display Full Size version of this image (5K) Fig. 12. Input torque with a constant external load. Display Full Size version of this image (4K) Fig. 13. Efficiency of the roller drive. 4.3. Influence of the geometric parameters In order to investigate the influences of the geometric parameters on the speed fluctuation and mechanical efficiency of the roller drives, curves of further results are given in Fig. 14. It is shown in Fig. 14(a)–(e) that reducing R2 (radius of the ring gear) and R4 (crank eccentricity), and increasing R3 (radius of the planet gear), r2 (radius of the ring gear roller) and r3 (radius of the planet gear roller), respectively, can reduce the speed fluctuation of the roller drives; unfortunately, in gaining lower speed fluctuation we suffer a reduction in the mechanical efficiency. Furthermore as shown in Fig. 2, the model of the 4-bar equivalent linkage is adopted to analyze the kinematics of the roller drive. It can be observed that link R4 is always located on one side of the link R2i while the roller drive is running. Therefore, the 4-bar linkage is not a Grashof mechanism and the relationship of the 4 link lengths can be expressed as: R2i+R4>R3i+(r2+r3) (52) Based on Eq. (52) and analysis results, it is found that as the geometric parameters of the roller drive, R2i, R4, R3i, r2 and r3, approach the relationship R2i+R4=R3i+(r2+r3), then there is less speed fluctuation as shown in Fig. 15. The value of abscissa in Fig. 15 is the value of dg=R2i+R4?R3i?(r2+r3). Each curve in the plot corresponds to change the value of one of the parameters from the set value in the equal condition while the other parameters are kept fixed. Display Full Size version of this image (16K) 60 Fig. 14. Influence of efficiency and speed fluctuation caused by geometric parameters. Display Full Size version of this image (4K) Fig. 15. Influence of speed fluctuation caused by geometric parameters. 4.4. Test A roller drive, whose dimensions are as shown in Table 1, was built for the test. As shown in Fig. 16(a), a rig was constructed to test the roller drive consisting of a power source, an external load and associated instruments. A detailed arrangement of the test rig is illustrated in the schematic diagram as shown in Fig. 16(b). An AC motor having two magnetic poles (FUKUTA, 7.5 kW) with an inverter (ADLEE AP2G3-75) is used as the power source. The external load of the roller drive is provided by a powder brake (MITSUBISHI ZKB-20XN, 200 N m). The instruments of the test rig include two torque transducers (KYOWA TP-2KMCB, 20 N m and TP-20KMCB, 200 N m) and two optical encoders (NIDEC NEMICON CORP. OEK-50-2, 50 P/R and OEK2-05-2, 500P/R) for measuring the input and output torques and angular displacements of the roller drive, respectively. One of the torque transducers is located between the AC motor and the roller drive input shaft to measure the input torque. The other one is placed between the roller drive and the powder brake to monitor the output torque. Each torque transducer was connected to the shaft though flexible coupling in order to avoid any bending moments that would invalidate the measurement. Two optical encoders are employed for the measurement of the input and output angular displacements of the roller drive. Each encoder was connected to the roller drive shaft though a toothed belt. The data coming from the torque transducers were sent by a dynamic strain amplifier (KYOWA DPM-712B-M33) and while the signals also from the encoders both via an acquisition card (NI PCI-6070-E, 1.25 MHz) and PC for data acquisition and storage. Display Full Size version of this image (30K) Fig. 16. Roller drive testing rig. The curve of the output angular velocity for one revolution of the input shaft is shown in Fig. 17. Since the speed fluctuation is the same for every period in the angular velocity analysis, we take only one value for the maximum and minimum angular speed of the planet gear from for every revolution. The speed fluctuation factor in the example as shown in Fig. 7 is about 6%. However, the roller drive consists of two planet gears for the physical test. The tolerances between two planet gears are different due to manufacturing limitation. In addition, when the roller drive is exerted by an external load, the corresponding factors such as elastic 61 deformation, manufactured error and lubrication oil etc. may influence the results. Hence, for the physical test the speed fluctuations for each cycle are quite different. It is more appropriate to take different values of the maximum and minimum angular speed of the planet gear for each period. The results show the values of the speed fluctuation factor are distributed between 7% and 21%. Display Full Size version of this image (6K) Fig. 17. Test data of angular velocity for one crank revolution. The input torque and mechanical efficiency of the roller drive are demonstrated in Fig. 18 and Fig. 19, respectively. The mean value of efficiency is about 85%. The values of the mechanical efficiency obtained from the theoretical simulations are different those from the physical test. In this paper, we assume the elements of the roller drive are rigid body. Actually, when the roller drive is exerted by an external load, the corresponding factors mentioned above may influence the results and make the friction coefficient remain indetermination. Moreover, the changing rates of stored potential and kinetic energy are neglected in the theoretical predictions. However, we measure torques and angular velocities to calculate the mechanical efficiency for the physical test. The equation of mechanical efficiency is This factor is also a reason that cause the efficiency physical test lower than that from theoretical predictions. Display Full Size version of this image (5K) Fig. 18. Test data of input torque of the roller drive. Display Full Size version of this image (6K) Fig. 19. Test data of efficiency of the roller drive. 5. Conclusions We present here the kinetostatics analysis of a novel mechanical transmission, the roller drive, for speed reduction. The roller drive is an epicyclic gearing in which both the planet gear and the ring gear have rollers as their teeth. The roller drive thus has the advantages of easy manufacture and low cost, and also has attractive attributes including high speed reduction ratio, compact size, and light weight as an epicyclical gearing. 62 In this work, the kinematic modeling on the basis of vector loop approach is developed, and models for analyzing the kinetostatics and mechanical efficiency of roller drives considering friction are proposed. For a given a roller drive, the loads acting on the conjugated joints, such as the planet gear and crank, the planet gear and ring gear roller, and the planet gear and disc plate, can be investigated by using the models. The results show that the roller drive has a speed fluctuation that corresponds to the number of ring gear rollers, and the magnitudes of the forces vary periodically with the crank angular displacement. It is also found that the roller drive has an input torque ripple caused by the forces acting on the conjugated joints with periodic variation and fluctuation in the output speed. In addition, the influence of the geometric parameters on efficiency and speed fluctuation of the roller drives is discussed. It is also shown that the speed fluctuation of the roller drives can be decreased by reducing the radius of the ring gear and the crank eccentricity, and by increasing the radius of the planet gear, the radius of the ring gear roller and planet gear roller. Acknowledgements The authors are grateful to the National Science Council of the Republic of China for its support of this research through Grant NSC 90-2218-E-006-138 to National Cheng Kung University and Grant NSC 90-2212-E-150-014 to the National Huwei Institute of Technology. References 1. C. Musser, Strain wave gearing, US Patent 2,906,143, 1995 2. C. Musser, Breakthrough in mechanical drive design: the harmonic drive. Machine Design (1960), pp. 160–172. 3. D.W. Botsiber and L. Kingston, Design and performance of the cycloid speed reducer. Machine Design (1956), pp. 65–69. 4. E.P. Pollitt, Some applications of the cycloid machine design. ASME Journal of Engineering for Industry (1960), pp. 407–414. 5. C.-C. Ko, Roladrive, ROC (Taiwan) Patent 51906, 1991 6. C.-C. Ko, Speed reducer which employs rolling means, US Patent 5,431,605, 1995 7. F. Freudenstein, An application of Boolean algebra to the motion of planetary drives. 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Abstract | View Record in Scopus | Cited By in Scopus (3) 64 滚筒驱动的动态静力学分析 陈坤保莲，陈志暐 ，沉堂和拉美一体化协会 a车辆工程系，国家护胃技术学院，护胃， 台湾 b机械工程系，国立成功大学，台南70101 ，台湾 收到2003年4月15日;修订2004年3月8日;接受， 2004年3月23日。网上提供2004年5月13日。 摘要 电动滚筒式驱动器类似于一个行星齿轮带有拓扑结构，而这个行星齿轮和齿圈都用辊作为它们的齿。因此该辊筒传动具有制造容易和成本低的优点。在这个论文中，运动学模型是建立在矢量回路发展的基础上，并且建议滚筒驱动的运动静力学分析和机械效率模型考虑摩擦力。通过利用这些模型，可以得到滚筒的速度、加速度、作用在变化齿对的正常受力和效率。给出了一个例子与实验测试用来验证分析结果。研究结果表明，滚筒驱动带有与滚齿数相对应的转速波动并且力大小的周期性变化与曲柄角度转移相对应，因此，这个滚筒驱动拥有一个输入扭矩脉动。并且我们也发现在高速运作下这个滚筒驱动具有高的效率。并且论文也讨论了滚筒的几何参数对效率和速度变化的影响。 作者关键词:滚筒驱动;运动学; 运动静力学 ;效率 文章概要 1 导言 2 运动学分析 2.1 角速度和加速度分析 2.2 相对共轭运动速度 3 运动静力学分析 3.1 作用于曲柄上的力 3.2 作用于行星齿轮上的力 3.3 作用于圆盘上的力 3.4 非线性方程 3.5 摩擦力及效率分析 4 分析实例与试验 4.1 角速度和加速度 4.2 受力和效率 4.3 几何参数的影响 4.4 测试 5 结论 鸣谢 参考文献 1 导言 行星齿轮广泛应用于以速度和扭矩转换为主的工业，如行星齿轮减速机，谐波传动，摆线驱动器[ 1 ， 2 ， 3 ， 4 〕 ，这种类型的减速机具有简洁、减速比率大以及高效率的 65 优点。这种机械装置必须制造出精密的规格从而提供高性能标准，现在已经创造出了制造简单成本低的滚筒驱动器[5和6, 。典型的单级滚筒驱动器的拓扑结构的如图1。安装于行星齿轮组的偏心曲柄作为输入，环形齿轮以圆柱滚子作为它的齿安装在机架上。在行星轮上速度恒定点是由一系列粘在圆盘上的插脚与成对的等量的孔相配套共同构成了输出轴的旋转运动。 显示全图( 1.1K) 图1 滚筒驱动的拓扑结构 在过去，运动学分析行星齿轮火车一直是研究的主题[ 7 ， 8 ， 9 ，10 ，11 ，12和13 ] 。 布兰奇和杨[ 14 ]研制出一种摆线驱动器的解析模型与加工公差和调查的影响，加工公差就反弹和转矩脉动，他们[ 15 ]为核实摆线驱动器的性能还提出了一项电脑辅助分析程序。此外，罗米蒂等人 [ 16 ]在加工工差范围内演示了一种无齿轮铰接式减速器的运动学分析。此外许多调查员也展示了行星齿轮火车的运动静力学分析[ 17 ，第18 ，第19和20 ] 。马罗特拉和帕拉 [ 21 ]介绍了一套程序来计算作用在摆线驱动器上各元素上的各种力，但也有只有少数研究是针对滚筒驱动器。闫丽[ 22 ]描述了滚筒驱动器的特性，并利用基本效率为滚筒驱动器的机械效率推导出公式。 在这项工作中，第一，介绍分析运动学的数学模型;其次，滚筒驱动器考虑存在摩擦力时的运动静力学模型和机械效率;然后对滚筒驱动器的机械效率和作用在各元素上的力进行介绍;最后，通过实验测试来验证分析结果。 2 运动学分析 以下是一个有关滚筒驱动器的角速度和加速度以及在接触点上一对共轭相对速度的讨论。 2.1 角速度和加速度分析 如图2所示，一对单独接触滚筒的行星齿轮和齿圈，如果我们连接曲柄的四个中心O4的(这也是齿圈的中心 O2 ) ，齿圈的一个辊子o2i ，行星齿轮O3，行星齿轮的一个辊子o3i ，四边形o4o3o3io3i 形成了一个四杆联动。用矢量回路方法[ 23 ] ，位置矢量回路方程可以表示为 ( 1 ) 分别画出向量从O4到O3，O3到o3i ， o2i到o3i ， O4到o2i代表、、和。因此其分量方程为 66 R4cosθ4+R3icosθ3i?Racosθa?R2icosθ2i=0 (2) R4sinθ4+R3isinθ3i?Rasinθa?R2isinθ2i=0 (3) θ2i,θ4 , θ3i和θa分别指齿圈、曲柄、行星齿轮和O2iO3i段的角位置(从X轴反时针方向测量) 显示全图( 5 K ) 图2 运动学分析的数学模型 对于行星齿轮位置的分析，其中一个问题是要找出第一个接触的那对行星齿轮辊子和齿圈辊子。解决这个问题的方法是利用公式(2)(3)计算齿圈辊子中心到行星齿轮辊子中心的距离Ra。由于给定行星齿轮辊子和齿圈辊子的接触点因此星星棍子中心和齿圈棍子中心的距地Ra等于齿圈棍子半径r2和齿盘辊子半径r3之和。即Ra = r2+ r3。如果Ra> r2+ r3则表示齿盘辊子与齿圈辊子没有接触。研究结果表明滚筒驱动器在传动时只有一对齿是啮合的。 通过区分公式 ( ( 2 )及( 3 ) )与考虑曲柄角度θ4 ，运动学系数[ 23 ]可写为 ( 4 ) ( 5 ) 式中。由微分法则我们可知，并且行星齿轮的角速度和扇形齿轮O2iO3i可以表达如下。 ωi=hiω4 (i=3,a) (6) 再次利用公式( ( 2 )及( 3 ) )并考虑曲柄角度θ4，给出 ( 7 ) ( 8 ) 67 式中，行星齿轮角速度和扇形齿轮O2iO3i可以写为 αi=hiα4+h′iω42 (i=3,a) (9) 如图3所示，针和孔接触点P5i3的相对速度也可以看作是行星齿轮的速度和针的速度 ，它们分别可以写为 ( 10 ) ( 11 ) 假设行星齿轮和圆盘针是刚体，行星齿轮和圆盘针接触点之间正常速度分量是0，因此，对于针孔速度和针的速度是相同的，也就是说VP35in=ω3R3ccosβc, VP35in=ω5R5i3cosβP and ω3R3ccosβc=ω5R5i3cosβP。如图3所示，since R3ccosβc=R5i3cosβP, 因此 ω3 = ω5 ( 12 ) 全图显示( 7K) 图3 共轭点的相对速度 2.2 共轭偶的相对运动速度 由公式 ( ( 10 )及( 11 )) ，行星齿轮和圆盘针在接触点p5i3的相对切向速度分量可写为 ( 13 ) 一旦找到瞬心的位置，O4到瞬心的方向矢量可以表述为，角速比就可以写为 ( 14 ) 把式子( ( 12 )及( 14 ) )代入式子 ( 13 ) ，得到 68 ( 15 ) 如图 3 所示，齿圈辊子和行星齿轮辊子在接触点P2i3的相对速度可以表示为 ( 16 ) 式中表示行星齿轮中心O3到接触点P2i3的向量。它可以表示为: 由式子( 16 ) ，行星齿轮辊子和齿圈辊子在接触点P2i3切向速度分量可确定为 ( 17 ) 3 运动静力学分析 考虑摩擦力的滚筒驱动器运动静力学模型在这节中讲解。假设重力相对较小并可以忽略不计，并且质心为物体的几何图案中心。 3.1 作用在曲柄上的力 曲柄的自由体示意图如图 4 ，一个已知的输入扭矩作用在曲柄上。向量和分别是从点O3、齿轮中心到点P43 and P′43，(它们曲柄和行星齿轮的接触点) 。相量车架1作用在曲柄4上的外力，此外还有一个摩擦扭矩。同样地，和分别是行星齿轮作用在曲柄上接触点P34 (P′34)的力和摩擦力矩，摩擦力矩t14、t′34和t34可表示为 t34=μ34F34Rb3D34 (18) t′34=μ34F′34R′b3D′34 (19) t14=μ14F14Rb4D14 (20) 式中Rb3 (R′b3) 和Rb4分别是放在曲柄、行星齿轮和机架之间滚子轴承的半径。此外，μ34 和μ14代表滚子轴承的摩擦系数。此外，D34, D′34 和 D14 是方向标，可分别表示为 , 和 显示全图( 5 K ) 69 图 4 曲柄自由体示意图曲柄。 如图四所示，适用于曲柄机构的三标运动方程可以写为 F34x+F′34x+F14x=m4a4x (21) F34y+F′34y+F14y=m4a4y (22) ( 23 ) 式中m4的是曲柄的质量， i4是曲柄相对于质心的转动惯量，a4x 和 a4y分别代表，曲柄质心加速度大小。此外，, R34x和 R34y可表示为:R34x=R4cosθ4?Rb3cosθ34 和 R34y=R4sinθ4?Rb3sinθ34，θ34 (θ34=tan?1F34y/F34x)是表示在接触点P34 从x轴向公法线反时针方向转的角度。 3.2 作用于行星齿轮上的力 在图5中，等速度点连接起来在运动学上像铰接式平行四边形，行星齿轮接触点P5i3上的公法线和盘针是平行的。假设一个外部负载作用在行星齿轮上，由于弹性变形，在接触点会导致行星齿轮旋转一个小角度Δ。由于弹性变形在接触点非常小，这也是假定了在接触点沿着公法线方向光盘针对行星齿轮的力与变形Δ是成正比，即Δ=R5 Δ。因此， ( 24 ) 然后式(24)改写为 ( 25 ) 式中F53圆盘针作用在行星齿轮上的合力，n5是圆盘针与行星齿轮接触的数量。当曲柄的中心O4、行星齿轮O3和光盘针O3d 在一条直线上，当Z5为偶数时n5=(Z5/2)+1;当Z5为奇数时n5=(Z5?1)/2。式中Z5表示圆盘针的数量。另外，因为中心O4, O3 和 O3d不在一条直线上，当Z5为偶数时n5=Z5/2;当Z5为奇数时n5=(Z5?1)/2。另外，作用线方向是在接触点P5i3的公法线方向。因此 ( 26 ) 由公式( ( 25 )和( 26 ) ) ，圆盘针作用在行星齿轮上的合力可确定为 70 ( 27 ) 式中 显示全图(11K) 图5 行星齿轮自由体图 此外，光盘针作用在行星齿轮上的摩擦力可表示为 ( 28 ) 式中μ53是香型齿轮和光盘真的摩擦系数。方向指标D5i3可以写为，式中是在接触点沿公切线方向的相量，它可表示为 。因此，由此产生的摩擦力可以表示为 ( 29 ) 式中 两个力作用在行星齿轮上的总力矩加以确定 ( 30 ) 式中是从O3到接触点P5i3的方向相量，可表示为 71 ，这里R3a和r3a分别是行星齿轮牵引滚筒的半径和圆盘针的半径。 把式子( ( 25 )和( 28 ) )代入 ( 30 ) ，得 ( 31 ) 式中 除作用在圆盘针上的力外，齿轮滚柱也在行星齿轮上施加力。如图5，当曲柄反时针方向旋转时 ，齿轮滚柱作用在行星齿轮上的力和摩擦力以及这些力对于行星齿轮中心所产生的力矩和摩擦力矩，由于滚筒驱动器在传动时只有一个齿是啮合的因此齿轮滚柱作用在行星齿轮上的力为 ( 32 ) ( 33 ) 式中D23=ω3/|ω3| ，是接触点P2i3的公切线的单位相量，μ23是行星齿轮辊子和齿圈辊子和行星齿轮辊子的摩擦系数。 外力作用在行星齿轮总的力矩可确定为 ( 34 ) 式中K23t=R3sin(θa?θ3)+μ23D23[r3?R3cos(θa?θ3)]，它是从O3到P2i3的相量。 参考图5 ，作用于行星齿轮的三标运动方程可以写为 F53x+f53x+F23x+f23x?F34x=(m3+m3a)a3x( 35 ) F53y+f53y+F23y+f23y?F34y=(m3+m3a)a3y( 36 ) T53+t53+T23+t23?t34=I3α3( 37 ) 式中a3x和a3y 代表行星齿轮质心X和Y组成部分的加速度，它们分别可以表示为:a3x=?α4R4sinθ4D41?ω42R4cosθ4和a3y=α4R4cosθ4D41?ω42R4sinθ4，式中D41=?D14=ω4/|ω4| 。此外，m3, m3a and I3是指行星齿轮的质量，行星齿轮辊子和该行星齿轮相对于质心的转动惯量。 72 3.3 作用于圆盘上的力 圆盘自由体示意图如图 6 ，扭矩是外部载荷相对于圆盘的扭矩。力和摩擦力是行星齿轮作用在圆盘上的，此外，力矩、和分别是力、和所产生的力矩。根据图6，力和力矩方程可以写为 ?F53x?f53x?F′53x?f′53x+F15x=m5a5x (38) ?F53y?f53y?f53y?F′53y?f′53y+F15y=m5a5y (39) ( 40 ) 式中m5是圆盘的质量，I5是圆盘相对于质心的转动惯量，a5x and a5y分别代表光盘质心X和Y组成部分的加速度大小。摩擦力矩t15取决于力F15和角速度ω14，可表示为:t15=μ15F15Rb5D15，式中D15=?ω5/|ω5| 。此外，Rb5是输出轴的滚子轴承半径，μ15是机架与圆盘间的摩擦系数。 显示全图( 3K ) 图 6 输出轴自由体示意图。 3.4 非线性方程 从自由体的图可以看出曲柄、行星齿轮、圆盘的移动并可相应地写出式子 ( ( 21 ) ， ( 22 )及( 23 ) ) ， ( ( 35 ) ， ( 36 )和( 37 ) ) ，和( ( 38 ) ， ( 39 )和( 40 ) ) ，这九个非线性方程含有未知数F34x, F34y, F′34x, F′34y, F14x, F14y, Tin, F53, F′53, T23, F15x 和 F15y 。由于在这个论文中重力是忽略的，因此两个行星齿轮作用在曲柄和圆盘上的力都是大小相等但方向相反的，即F′34x=?F34x, F′34y=?F34y, F′53x=?F53x 和F′53y=?F53y 。此外，两个行星齿轮作用在曲柄和圆盘上的力矩在大小和方向上都是相同的，即t34=t′34, t53=t′53 和 T53=T′53。因此，含有这九个未知数的九个可改写如下 F14x=m4a4x (41) F14y=m4a4y (42) 2(R34xF34y?R34yF34x)+Tin=I4α4?2t34?t14 (43) (K53x+L53x)F53+(cosθa+μ23D23sinθa)F23?F34x=(m3+m3a)a3x (44) 73 (K53y+L53y)F53+(sinθa?μ23D23cosθa)F23?F34y=(m3+m3a)a3y (45) K53tT53+K23tT23=I3α3+t34 (46) F15x=m5a5x (47) F15y=m5a5y (48) Tl?2K53tF53=I5α5?t15 (49) 3.5 摩擦功率及效率分析 由于机器受力平衡，滚筒驱动的力平衡方程式可写为 式中dU/dt和dT/dt分别指动能和势能的变化率，Pl是散逸能并可以表示为 Pl=|t14ω4|+|t15ω5|+|t34ω34|+2|f53V53|+2|f23V23| 如果势能和动能的变化率在这里可以忽略，则滚筒驱动的机械效率可表示为 ( 50 ) 4 实例与试验分析 4.1角速度和加速度 这里采用一个电动滚筒式驱动器，其尺寸的几何参数列于表1 。在例子中，外载荷Tl=60 N m，输入角速度ω4=1800 rpm时，输入角加速度α4=0 rad/s2，另外，滚筒传动的摩擦系数和滚筒驱动器的共轭点分别为μb=0.0015, μ23=0.02 和 μ35=0.02。滚筒驱动器的角速度和加速度可以从公式( ( 6 )和( 9 ) )中求出 。 74 表1滚筒传动的几何参数 根据分析结果，该行星齿轮饿角速度和加速度以及曲柄角位移曲线如如图7和图8所示。从图7可以注意到滚筒驱动器的角速度不是一个常量，而是对应齿圈辊子数有一个速度的波动，转速波动系数定义为: ( 51 ) 式中ω3max 和 ω3min是指行星齿轮的最大和最小角速度，ω3c是滚筒传动中行星齿轮的理想角速度，它可以表示为ω3c=[Z3/(Z3?Z2)]ω4。这里的Z2 和 Z3分别是齿圈辊子和行星齿轮辊子的数量。由公式( 51 )速度波动的因素在图7上约为6%。 显示全图( 5 K ) 75 图7 一曲柄旋转的角速度 显示全图( 5 K ) 图 8一曲柄旋转的角加速度 4.2 力和效率 因为力矩方程组是非线性的，这里用迭代方法，其算法如下 1 令所有摩擦力t14, t34 和 t15 等于 0 ，然后利用式子( ( 41 ) ， ( 42 ) ， ( 43 ) ， ( 44 ) ， ( 45 ) ， ( 46 ) ， ( 47 ) ， ( 48 )和( 49 ) ) 求出力F14x, F14y, F34x, F34y, F15x 和 F15y 2 使用第一步的结果，计算摩擦力t14, t34 和 t15, ，以及θ34 。 3 将摩擦力的数代入方程求出未知力。 4 检查，如果满足指定的准确性，则停止，否则，转到第2步。 求出了摩擦力矩则根据式子(50)就可以计算出滚筒驱动的机械效率。 F34(行星齿轮作用在曲柄上的外力)、F53(圆盘作用在行星齿轮上的力)和F23(齿圈辊子作用在行星齿轮上的力)的分析结果曲线分别如图9、图10和图11。结果表明力的大小随着曲柄的叫位移周期性变化;滚筒驱动的所有共轭点中，行星齿轮作用在曲柄上的力是最大的。恒定负载荷输入扭矩显示在图12，该图表明该滚筒驱动器有输入转矩脉动，该脉动是由共轭点的周期性变化所产生。在图13显示联了滚筒传动的机械效率，考虑摩擦力后效率的均值为93%。 显示全图(4K) 图 9 行星齿轮作用在曲柄上的力 显示全图( 5 K ) 图10 圆盘作用在行星齿轮上的力 76 显示全图( 5 K ) 图11齿圈辊子作用在行星齿轮辊子上的力 显示全图( 5 K ) 图12 恒定外载荷输入扭矩 显示全图(4K) 图13 滚筒传动的效率 4.3 几何参数的影响 为了研究滚筒驱动器几何参数对速度波动和机械效率的影响，进一步的曲线结果在图14中给出 。在图14中(a)-(e)分别可以通过减小R2(齿圈半径)和R4(曲柄偏心值)、增大R3(行星齿轮半径)、r2(齿圈辊子的半径)和r3(行星齿轮辊子的半径)来减小滚筒传动的速度波动。但是在获得较低的转速波动，我们将很大程度减小机械效率。此外，如图2显示 ，通过利用四杆连接模型来对滚筒驱动器进行运动学分析。我们可以观察到，在滚筒驱动器运转时R4总是在R2的一旁，然而，四杆联动不是一个安全的结构，四杆长度的关系可以表示为: R2i+R4>R3i+(r2+r3) (52) 由式子( 52 )并分析结果，可以发现，滚筒驱动器的几何参数R2i, R4, R3i, r2 和 r34近似的满足关系R2i+R4=R3i+(r2+r3)，那么就如图15所示速度波动较小，图15中横坐标值就是dg=R2i+R4?R3i?(r2+r3) 。图15每一条曲线都是在同等条件下一个参数在相应的变化而其它参数则保持不变。 显示全图( 16k ) 77 图14 几何参数对效率和速度波动的影响 显示全图(4K) 图15 几何参数对转速波动的影响 4.4 。测试 这个实验采用尺寸如表1的电动滚筒式驱动器。如图16(a)所示，钻机是为了测试滚筒传动由一个动力源，外部载荷和关联器械组成。如图16(b)是对试验的装备一份详细的安排。交流电动机动力源有两个磁极(FUKUTA, 7.5 kW)与一个逆变器(ADLEE AP2G3-75 ) ，滚筒驱动器的外部负荷是由磁粉制动器(MITSUBISHI ZKB-20XN, 200 N m)提供。该仪器的测试装置包括两个扭矩传感器(KYOWA TP-2KMCB, 20 N m and TP-20KMCB, 200 N m)和两个搭载光学编码器((NIDEC NEMICON CORP. OEK-50-2, 50 P/R and OEK2-05-2, 500P/R)，它们分别用来测量滚筒驱动器输入和输出力矩和角位移。其中一个扭矩传感器是位于交流电机和滚筒驱动器之间来衡量输入扭矩;另一个是放在滚筒驱动器和磁粉制动器之间用来监测输出扭矩。每个扭矩传感器都是用弹性联轴器与轴相连以避免任何弯矩将测量数据无效。两个光学编码器是用于测量滚筒驱动器的输入和输出角位移。每个编码器用有齿的带子连接到辊传动轴。转矩传感器的数据通过一个动态应变放大器(KYOWA DPM-712B-M33)发送，而编码器的信号都通过一个采集卡(NI PCI-6070-E, 1.25 MHz)和PC来采集和储存数据。 显示全图(30K) 图16 滚筒传动试验台。 输入轴的一个输出角位移曲线如图17所示。由于每一个时期角速度分析的速度波动是一样的，因此在行星齿轮的每一个旋转我们只需要角速度的一个最大值和最小值，例子中转速波动因素图17中可以看出为76,左右。然而在这个物理实验中滚筒驱动器由两个行星齿轮组成，由于制造业的局限性两个行星齿轮的公差是不同的，而且在滚筒驱动器在施加外部压力时还存在一些因素如弹性变形、制造误差和润滑油等都可能影响结果。因此，在物理实验中对于每一个周期的速度波动都有很大的不同，行星齿轮每个周期角速度的最大值和最小值不同是合乎常规的。结果表明，该转速波动因素值分布在7,与21,之间。 78 显示全图( 6k ) 图17 曲柄旋转的角速度数据试验 滚筒传动的输入扭矩和机械效率分别如图18和图19 。效率的均值大约是85 ,，机械效率的值在物理试验中与理论仿真是不同的。在这篇文章中，我们假设滚筒驱动器是刚体。其实，当滚筒驱动器施加外部载荷时，上述因素都相应的可能会影响结果，使得摩擦系数无法确定。此外，在理论中势能和动能都是忽略的，而我们在物理试验中测量力矩和角速度来计算机械效率。机械效率计算等式为 这个因素也是一个导致物理实验中效率比理论低的原因。 显示全图( 5 K ) 图18 驱动传动器输入扭矩的数据实验 展出全尺寸版的形象( 6k ) 图19 滚筒驱动器效率数据实验 5 结论 我们目前在这里用运动静力学分析了一种新型的机械传动，即滚筒驱动器，一个减速装置。滚筒传动是一个行星齿轮系传动，其中两个行星齿轮和齿圈都用辊作为自己的齿。滚筒传动具有容易制造、成本低的优点，作为一个齿轮传动它也有其它吸引人的属性，其中包括高减速比、体积小、重量轻。 在这项工作中，运动静力学建模在矢量回路的基础上，而且运动静力学模型和机械效率考虑摩擦的建议。一个给定的滚筒驱动器作用在共轭点上的荷载如行星齿轮和曲轴，行星齿轮和齿圈滚子，以及行星齿轮和圆盘，可以利用模型进行研究。研究结果表明，该滚筒驱动器对应的齿圈辊子数带有转速波动，并且力的大小随着曲柄的角位移作周期性变化。同时也发现滚筒驱动器随着作用在共轭点的周期性变化外力会导致输入扭矩和输出扭矩的 79 波动。此外，文章还讨论了滚筒驱动器几何参数对效率和速度波动的影响。也可以看出滚筒驱动器速度波动可以通过减小齿圈的半径和曲柄偏心率而减小，也可以通过增加行星齿轮半径、齿圈辊子半径和行星齿轮辊子半径来减小。 80