机械原理课程设计(台式电风扇的摇头装置)

目录1．台式电风扇摇头装置的功能与设计要求·································31.1工作原理及工艺过程················································31.2功能分解··························································31.3原始数据及设计要求················································31.3.1原始数据·····················································1.3.2设计要求·····················································1.4设计任务··························································32．执行机构的设计·····················································42.1〔方案Ⅰ〕··························································42.2〔方案Ⅱ〕··························································42.3〔方案Ⅲ〕··························································52.4〔方案Ⅳ〕··························································63．执行机构的辅助构件设计············································63.1滑销控制机构〔方案Ⅰ〕············································63.2齿轮控制机构〔方案Ⅱ〕··············································74．减速机构的设计·····················································74.1蜗杆减速机构〔方案Ⅰ〕··············································74.2锥齿轮减速机构〔方案Ⅱ〕············································74.3行星轮系减速机构〔方案Ⅲ〕··········································75．方案确实定························································85.1原动机的选择·····················································85.2传动方案确定·····················································85.3有关参数及相关计算···············································8相关计算························································8传动构件的尺寸确定·················································86．尺寸与运动综合····················································96.1执行机构尺寸设计··················································96.2验算曲柄存在条件即最小传动角······································10曲柄存在条件·····················································106.2.2最小传动角验算····················································7.系统总图···························································118.总体评价···························································118.1课题总结·························································118.2存在问题·························································12参考文献·····························································121．台式电风扇摇头装置的功能与设计要求1.1工作原理及工艺过程1.2功能分解电风扇的工作原理是将电风扇的送风区域进行周期性变换，到达增大送风区域的目的。显然，为了完成电风扇的摆头动作，需实现以下运动功能要求：〔1〕风扇需要按运动规律做左右摆动，因此需要设计相应的摆动机构。

〔2〕风扇需要转换传动轴线和改变转速，因此需要设计相应的齿轮系机构。对这两个机构的运动功能作进一步分析，可知它们分别应该实现以下根本运动：

〔3〕左右摆动有三个根本运动：运动轴线变换、传动比降低和周期性摆动。〔4〕转换运动轴线和改变传动比有一个根本动作：运动轴线变换。此外，还要满足传动性能要求：改变电风扇的送风区域时，在急回系数K＝1.015、摆动角度φ=85°的要求下，尽量保持运动的平稳转换和减小机构间的摩擦。运动功能图1.3原始数据及设计要求1.3风扇直径为φ300mm，电扇电动机转速n=1450r/min，电扇摇头周期T=10s。电扇摆动角度Ψ=85°与急回系数k=1.015。1.3设计台式电风扇的摇头装置要求能按给定的急回系数和摆角左右摆动，以实现一个动力下扇叶旋转和摇头动作的联合运动。1.4设计任务1．按给定主要参数，拟定机械传动系统总体方案。2．画机构运动简图。3．分配蜗轮蜗杆、齿轮传动比，确定他们的根本参数，设计计算几何尺寸。4．解析法确定平面连杆机构的运动学尺寸，它应满足摆角Ψ及形成速比系数k。并对平面连杆机构进行运动分析，绘制运动线图。并验算曲柄存在条件，验算最小传动角〔最大压力角〕。5．提出调节摆角的结构方案，并进行分析计算。6．编写设计说明书。7．进一步完成台式电风扇摇头机构的计算机动态演示验证。2．执行机构的设计2.1〔方案Ⅰ〕通过构件2对构件3做相对的圆周转动使构件3摆动，从而实现电风扇的摇头动作，如图1。图1图1方案立体图2.2〔方案Ⅱ〕当圆柱凸轮回转时，凹槽侧面迫使构件2摆动，从而实现电风扇的摇头功能，如图2。图22.3〔方案Ⅲ〕当圆盘回转时，凹槽带动槽轮使导杆摆动，从而实现电风扇的摇头功能，如图3。图32.4〔方案Ⅳ〕在曲柄2回转的过程中，杆1实现摆动，从而产生使电风扇摇头的功能，如图4。图4由以上四个方案可以看出：方案二、三、四的机构都很难难制造且精度要求高，制造本钱也相对要高。传动机构也相比照拟复杂，且齿轮数目太多，制造起来麻烦。方案一结构更简单，本钱相对较低。方案三和方案四相似，结构复杂制造本钱太高较二还高。因此不难看出，方案一相对较好。所以摇头机构选择方案一。3．执行机构的辅助构件设计3.1滑销控制机构〔方案Ⅰ〕通过滑销和锥齿轮卡和以实现是否摇头的运动。当滑销下滑实现摇头，上提那么停止摇头，如图5。图53.2齿轮控制机构〔方案Ⅱ〕通过两个直齿轮的啮合与否来控制是否摇头，两个齿轮啮合时电风扇摇头、不啮合时就不摇头，从而实现了是否摇头的控制，如图6。图64．减速机构的设计4.1蜗杆减速机构〔方案Ⅰ〕如图7图74.2锥齿轮减速机构〔方案Ⅱ〕如图8图84.3行星轮系减速机构〔方案Ⅲ〕如图9图95．方案确实定5.1原动机的选择笼式三相异步电动机，使用三相交流电、转速与旋转磁场转速不同，可以进行几档变速。而且笼式电动机的具有简单、体积小、易维护、价格低、寿命长连续运动特性好、转速受负载转矩波动的影响小和硬机械特性等优点。这些特性能够满足台式电风扇摇头装置的工作特性，所以选择笼式三相异步电动机作为原动机。5.2传动方案确定综合考虑后得出以下传动方案：在电动机主轴尾部连接蜗轮蜗杆速机构以实现减速，蜗轮在与小齿轮连成一体，小齿轮带动大齿轮，大齿轮与铰链四杆机构的连杆做成一体，并以铰链四杆机构的连杆为原动件，那么机架、两个连架杆都做摆动，其中一个连架杆相对机架的摆动即是摇头动作。扇叶直接接到原动机上，既可以实现电风扇的功能。5.3有关参数及相关计算相关计算电动机转速r=1450r/min摇头周期T=10s总的传动比n==，蜗杆头数Z1=1，蜗轮的齿数Z2=62，小齿轮齿数Z3与大齿轮齿数Z4确实定由n12===62；n34=；n=n12xn34；那么n34=EQ。那么取Z3=18；Z4=70。传动构件的尺寸确定查阅相关资料后取：渐开线圆柱齿轮根本参数取齿轮的齿数分别为z3=18，z4=70。模数m=1.25，ha*=1，c*=0.25〔ha*为齿顶高系数，c*为顶隙系数〕，压力角α=200。分度圆直径d3=mz3=1.25x18=22.5〔mm〕；d4=mz4=1.25x70=87.5〔mm〕。齿顶高ha=ha*m=m=1.25mm。齿根高hf=〔ha*+c*〕m=〔1+0.25〕x1.25=2.8125〔mm〕。齿全高h=ha+hf=1.25+2.8125=4.0625〔mm〕。齿顶圆直径da3=d3+2hda4=d4+2h齿根圆直径df3=d3-2hf=22.5-2x2.8125=16.875〔mm〕；df4=d4-2hf=87.5-2x2.8125=81.875〔mm〕。基圆直径db3=d3cosα=22.5xcos200=21.14〔mm〕；db4=d4cosα=87.5xcos200=82.22〔mm〕。齿距p=πm=3.14x1.25=3.93〔mm〕。齿厚s=πm/2=1.96〔mm〕。齿槽宽e=πm/2=1.96〔mm〕。中心距a=〔d3+d4〕/2=55mm。顶隙c=c*m=0.25x1.25=0.3125〔mm〕。普通圆柱蜗杆传动几何尺寸计算查表取得：中心距a取40mm，i=62，m=1，d1=18，z1=1，z2=62，x2=0.0000，γ=3010/47//，ha*=1，c*=0.25。齿形角α=200〔ZA型〕。蜗轮变为系数x2=0.0000。蜗杆轴向齿距px=πm=3.14mm。蜗杆齿顶高ha1=ha*m=1mm。顶隙c=c*m=0.25mm。蜗杆齿根高hf1=〔ha*+c*〕m=1.25mm。蜗杆齿高h1=ha1+hf1=2.25mm。蜗杆齿宽b1取19.5mm。蜗轮分度圆直径d2=mz2=62mm。蜗轮齿根圆直径df2=d2-2hf2=59.5mm。蜗轮齿顶高ha2=m〔ha*+x2〕=1mm。蜗轮齿根高hf2=m〔ha*-x2+c*〕=1.25mm。蜗轮齿高h2=ha2+hf2=2.25mm。蜗轮齿顶圆弧半径Ra2=d1/2-m=8mm。蜗杆轴向齿厚sx1=px/2=1.57mm。蜗杆法向齿厚sn1=sx1cosγ=1.568mm。蜗杆节圆直径d1/=d1+2x2m=18mm蜗轮节圆直径d2/=d2=62mm。6．尺寸与运动综合6.1执行机构尺寸设计设计与计算〔1〕由急回系数K计算极位夹角。由式=180，其中K=1.015；得=1.340〔2〕选定比例尺μ按条件画出摇杆BC的两个极限位置C1D和C2D，连C1,C2两点，以C1为直角顶做∠C2C1M=85º,再做∠C1C2N=85º,C1M与C2N交于P点，以PC2为直径做ΔC1C2P的外接圆。在此圆上任取一点A作为曲柄的固定铰链中心，连接AC1和AC2因同一圆弧的圆周角相等，故∠C1AC2=∠C1PC2=，令AB=a,BC=b,CD=c,DA=d.那么AC1=b-a,AC2=b+a。图一〔总体示意图〕图二〔局部图〕〔3〕上图四杆机构假设取AD为机架那么为曲柄摇杆机构，而电风扇的摇头四杆机构是要是双摇杆机构，所以取以CD为机架的倒置四杆机构。〔4〕计算各杆的实际长度。分别量取图中AB、B2C2、C1D、AD的长度，计算得a=µAB1=0.6µ，b=µB2C2=1.94µ，c=µC2D=µ，d=µAD=2.02µ。〔C2D=1mm〕因为本设计使用曲柄连杆机构的倒置机构〔即双摇杆机构〕，所以取杆CD为机架。那么取d=88mm，即µ=88。所以：a=74.77mm，b=81.62mm，c=88mm6.2验算曲柄存在条件即最小传动角曲柄存在条件〔1〕由上图可知A为整转副，那么以A为圆心a为半径的圆上各点都能与B构成整转副，即a+d≤b+c①,│d-a│≥│b-c│②假设a≤d,由式②可得：a+b≤c+d或a+c≤b+d，又由式①可得：a≤b,a≤c,a≤d；假设d≤a,由式②可得：d+b≤a+c,d+c≤a+d，又由式①可得d≤a,d≤b,d≤c。〔2〕由以上各式可得曲柄存在条件：组成该转动副的两个构件中必有一个构件为最短构件，且四杆机构的长度满足杆长之和条件。6.2.2cosδmin=[b²+c²-(d-a)²]/2bc=0.906那么δmin=82.46º；cosδmax=[b²+c²-(d+a)²]/2bc=-1.428那么δmax=126.34º；那么传动角EQ的范围为〔53.66º，82.46º〕，符合EQmin≥40º的条件。7.系统总图台式电风扇摇头系统结构简图8.总体评价8.1课题总结通过这次课程设计，让我对机械原理这门课程有了更深入的了解，对以前不熟悉的环节理解。虽然在设计的过程中遇到了好多麻烦，但是经过自己认真的思考和查阅资料，以及和同学一起讨论最终把问题都解决了。这次设计给我一个感受，学习的过程中要懂得把所学的东西联系起来并运用到实践中来，而不是把每个章节分开来理解。通过这个实践我学得了好多，同时认识到理论联系实际的重要性，不仅加深了我对课程的理解程度而且也激起了我学习的兴趣。机械原理课程设计是使我们较全面系统的掌握和深化机械原理课程的根本原理和方法的重要环节，是培养我们机械运动方案设计创新设计和应用计算机对工程实际中各种机构进行分析和设计能力的一门课程。经过这几天的设计，让我初步了解了机械设计的全过程，可以初步的进行机构选型组合和确定运动方案；使我将机械原理课程各章的理论和方法融会贯穿起来，进一步稳固和加深了所学的理论知识；并对动力分析与设计有了一个较完整的概念；提高了运算绘图遗迹运