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黑豹 HB1027变速器设计 本科论文.pdf

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黑豹 HB1027 变速器 设计 本科 论文
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XXX - I - 黑豹 HB1027 变速器设计论文 摘要 catia 关 键 词 (版权所有,翻版不究 ) XXX - II - Abstract Gearbox is the one main component of the vehicle transmission.The duty of this design is to design a manual transmission used in the Tiny gears, it is the countershaft-type transmission gearbox.This transmission has two prominent merits: firstly, the transmission efficiency of the direct drives keep high ,the attrition and the noise are also slightest;Secondly ,it is allowed to obtain in the bigger gear ratio of the first gear when the center distance in smaller. According to the contour, track, wheel base, the vehicles weight, the all-up weight as well as the highest speed and so on, union the engine model we can obtain the important parameters of the max power, the max torque, the displacement and so on. According to the basic parameters of the certain saloon, choose the suitable final drive ratio. According to the above parameters, combining the knowledge of automobile design, automobile theory , machine design and so on, calculate the correlated parameters of the gearbox and proof the rationality of the design. The design gives a plan of the mechanical gearbox and achieves a kind of mechanical gearbox after rigorous design.The design has passed calibration and Finite element optimization.It has proved to be fit for function and use for reference perfectly. Key Words Automotive engineering,Transmission,Design,Manual XXX - III - 目 录 摘要 ............................................................. I Abstract.......................................................... II 第 1 章 绪论 ...................................................... 1 1.1 ..................................... 1 1.2 ....................................... 1 第 2 章 机 械 式 变 速 器 设 计 .......................................... 3 2.1 ......................................... 3 2.1.1 ................................. 3 2.1.2 ..................................... 3 2.2 ................................................ 7 2.2.1 ..................................... 7 2.2.2 ........................................... 10 2.3 ................................................ 19 2.3.1 ........................................... 19 2.3.2 .......................................... 19 2.4 ...................................... 29 2.4.1 ........................................... 29 2.4.2 ....................................... 31 2.5 .......................................... 32 2.5.1 ........................................ 32 2.5.2 .................................... 34 2.6 ............................................ 34 XXX - IV - 2.6.1 .................................... 34 2.6.2 .............................. 35 2.7 .................................................. 35 第 3 章 有 限 元 优 化 分 析 ........................................... 36 3.1 catia ...................................... 36 3.1.1 catia ........................ 36 3.1.2 catia .......................... 37 3.2 catia .................................. 37 3.2.1 catia ................................ 37 3.2.2 catia ................................ 38 3.3 .................................................. 38 结 论 ........................................................... 39 参 考 文 献 ........................................................ 40 致 谢 ........................................................... 41 - 1 - 第 1 章 绪论 1.1 本 课 题 研 究 的 目 的 和 意 义 , , , , 1. 2. 3. 4. 5. [1] . . , , , , , . , , , 1.2 本 课 题 研 究 现 状 和 发 展 , AT CVT , CVT , , , , , ECT , - 2 - , , AT , AT , , AMT , 2008 , 50 MT AMT , [2] , , , - 3 - 第2 章 机 械 式 变 速 器 设 计 2.1 变 速 器 设 计 基 本 方 案 2.1.1 变 速 器 传 动 机 构 布 置 方 案 1 HB1027 2 3 4 [4] 5 [5] 6 2.1.2 变 速 器 主 要 参 数 选 择 1 - 4 - HB1027 5.0 1.0 2 1 dt du m Gi u A C Gf r i i T a D T g ? ? ? ? ? ? 2 0 emax 15 . 21 2-1 ) sin cos ( max max 0 1 max ? ? ? ? ? f mg r i i T r t g e 2-2 ? ? T tq g i T f Gr i ? ? ? 0 1 sin cos ? ? 1 t e r g i T f mgr i ? ? ? 0 max max max 1 ) sin cos ( ? ? t e r g i T f mgr i ? ? ? 0 max max max 1 ) sin cos ( ? ? =3.488 2-3 2 t e r g i T r G i ? ? 0 max 2 1 ? 2 G 2 G =mg 60% ? =0.7 0.8 =0.75 t e r g i T r G i ? ? 0 max 2 1 ? =5.12 2-4 2-3 2-4 3.488 1 g i 5.12 - 5 - 1 g i =3.5 1 , 4 g i =1.0 [6] q i i i i i i i i i i g g g g g g g g g g ? ? ? ? ? 6 5 5 4 4 3 3 2 2 1 1 1 ? ? n gn g i i q q=1.52 1 g i =3.5 2 g i = 2 q =2.3 3 g i = q =1.5 4 g i =1.0 0.7 0.8 q i g ? 5 =0.78 2-1 2-1 3.5 2.3 1.5 1.0 0.78 3.5 3 A mm 3 max 1 T K A A ? = 3 1 max g e A i T K A ? ? A K , 8.6 9.6 - 6 - max 1 T 1 g g e i T T ? 1 max max 1 ? max e T N ?m 1 g i 1 g ? 0.96 [7] 3 1 max g e A i T K A ? ?=71.22 79.50 mm A=75mm 4 2.2 2.7 A 2.7 3.0 A 3.2 3.5 A A 5 1 4.0mm (2) ? 20 20 20 25 30 30 (3 ? 18 26 24 4 b m k b c ? c k 4.5 8.0 7.0 - 7 - n c m k b ? c k 6.0 8.5 7.0 5 0 f 1.0 2-2 2-2 4 20 24 7 1 2.2 齿 轮 设 计 计 算 2.2.1 各挡齿 轮 齿 数 的 分 配 2-1 2-1 10 1 9 2 1 Z Z Z Z i ? 3-1 11 Z 12 Z h Z - 8 - m A Z h ? cos 2 ? =33.8 34 10 Z =13 9 Z = h Z - 10 Z =21 A 10 9 n 0 cos 2 ? ? ? h Z m A =74.44mm A=75mm 3-1 9 10 1 1 2 Z Z i Z Z ? 3-2 =2.17 ? ? ? cos 2 2 1 Z Z m A n ? ? n m A Z Z ? cos 2 2 1 ? ? 3-3 =34.3 3-2 3-3 1 Z =10.82 2 Z =23.48 1
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