機械原理課程設計(臺式電風扇的搖頭裝置)

1、江西農業(yè)大學機械原理課程設計目 錄1臺式電風扇搖頭裝置的功能與設計要求·································31.1工作原理及工藝過程·········

2、·······································31.2功能分解·········

3、3;················································31.3原始數據及設

4、計要求················································31.3.1 原始

5、數據·················································

6、3;······31.3.2 設計要求··········································

7、;··············31.4設計任務··································

8、83;·······················32執(zhí)行機構的設計·························

9、····························42.1（方案）····················

10、3;·····································42.2（方案）···········&#

11、183;··············································42.3（方案）··

12、··················································

13、······52.4（方案）··········································

14、3;···············63執(zhí)行機構的輔助構件設計 ································&#

15、183;···········63.1滑銷控制機構（方案） ····································

16、;········63.2齒輪控制機構（方案）·······································

17、83;······74減速機構的設計··········································

18、···········74.1蝸桿減速機構（方案）····································

19、3;·········74.2錐齒輪減速機構（方案）······································&

20、#183;·····74.3行星輪系減速機構（方案）·········································

21、83;75方案的確定················································&#

22、183;·······85.1原動機的選擇········································

23、3;············85.2傳動方案確定····································

24、·················85.3有關參數及相關計算 ······························

25、3;················85.3.1相關計算································

26、························85.3.2傳動構件的尺寸確定·······················

27、83;·························86尺寸與運動綜合·······················

28、·····························96.1執(zhí)行機構尺寸設計···················

29、·······························96.2驗算曲柄存在條件即最小傳動角················

30、······················106.2.1曲柄存在條件··························

31、;···························106.2.2最小傳動角驗算····················

32、83;·······························117.系統總圖·················&

33、#183;·········································118.總體評價·······

34、;··················································

35、;··118.1課題總結··············································&#

36、183;··········118.2存在問題······································

37、;···················12參考文獻······························

38、;·······························121臺式電風扇搖頭裝置的功能與設計要求1.1工作原理及工藝過程1.2功能分解電風扇的工作原理是將電風扇的送風區(qū)域進行周期性變換，達到增大送風區(qū)域的目的。顯然，為了完成電風扇的擺頭動作，需實現下列運動功能要求：（1）風扇需要按

39、運動規(guī)律做左右擺動，因此需要設計相應的擺動機構。（2）風扇需要轉換傳動軸線和改變轉速，因此需要設計相應的齒輪系機構。對這兩個機構的運動功能作進一步分析，可知它們分別應該實現下列基本運動：（3）左右擺動有三個基本運動：運動軸線變換、傳動比降低和周期性擺動。（4）轉換運動軸線和改變傳動比有一個基本動作：運動軸線變換。此外，還要滿足傳動性能要求：改變電風扇的送風區(qū)域時，在急回系數K1.015、擺動角度=85°的要求下，盡量保持運動的平穩(wěn)轉換和減小機構間的摩擦。運動功能圖1.3原始數據及設計要求1.3.1 原始數據風扇直徑為300mm，電扇電動機轉速n=1450r/min，電扇搖頭周期T=1

40、0s。電扇擺動角度=85°與急回系數k=1.015。1.3.2 設計要求設計臺式電風扇的搖頭裝置要求能按給定的急回系數和擺角左右擺動，以實現一個動力下扇葉旋轉和搖頭動作的聯合運動。1.4設計任務1按給定主要參數，擬定機械傳動系統總體方案。2畫機構運動簡圖。3分配蝸輪蝸桿、齒輪傳動比，確定他們的基本參數，設計計算幾何尺寸。4解析法確定平面連桿機構的運動學尺寸，它應滿足擺角及形成速比系數k。并對平面連桿機構進行運動分析，繪制運動線圖。并驗算曲柄存在條件，驗算最小傳動角（最大壓力角）。5提出調節(jié)擺角的結構方案，并進行分析計算。6編寫設計說明書。7進一步完成臺式電風扇搖頭機構的計算機動態(tài)演示

41、驗證。2執(zhí)行機構的設計2.1（方案）通過構件2對構件3做相對的圓周轉動使構件3擺動，從而實現電風扇的搖頭動作，如圖1。圖1圖1 方案立體圖2.2（方案）當圓柱凸輪回轉時，凹槽側面迫使構件2擺動，從而實現電風扇的搖頭功能，如圖2。圖22.3（方案）當圓盤回轉時，凹槽帶動槽輪使導桿擺動，從而實現電風扇的搖頭功能，如圖3。圖32.4（方案）在曲柄2回轉的過程中，桿1實現擺動，從而產生使電風扇搖頭的功能，如圖4。圖4由以上四個方案可以看出：方案二、三、四的機構都很難難制造且精度要求高，制造成本也相對要高。傳動機構也相對比較復雜，且齒輪數目太多，制造起來麻煩。方案一結構更簡單，成本相對較低。方案三和方案

42、四相似，結構復雜制造成本太高較二還高。因此不難看出，方案一相對較好。所以搖頭機構選擇方案一。3執(zhí)行機構的輔助構件設計 3.1滑銷控制機構（方案） 通過滑銷和錐齒輪卡和以實現是否搖頭的運動。當滑銷下滑實現搖頭，上提則停止搖頭，如圖5。圖53.2齒輪控制機構（方案）通過兩個直齒輪的嚙合與否來控制是否搖頭，兩個齒輪嚙合時電風扇搖頭、不嚙合時就不搖頭，從而實現了是否搖頭的控制，如圖6。圖64減速機構的設計4.1蝸桿減速機構（方案）如圖7圖74.2錐齒輪減速機構（方案）如圖8圖84.3行星輪系減速機構（方案）如圖9圖95方案的確定5.1原動機的選擇籠式三相異步電動機，使用三相交流電、轉速與旋轉磁場轉速不

43、同，可以進行幾檔變速。而且籠式電動機的具有簡單、體積小、易維護、價格低、壽命長連續(xù)運動特性好、轉速受負載轉矩波動的影響小和硬機械特性等優(yōu)點。這些特性能夠滿足臺式電風扇搖頭裝置的工作特性，所以選擇籠式三相異步電動機作為原動機。5.2傳動方案確定綜合考慮后得出以下傳動方案：在電動機主軸尾部連接蝸輪蝸桿速機構以實現減速，蝸輪在與小齒輪連成一體，小齒輪帶動大齒輪，大齒輪與鉸鏈四桿機構的連桿做成一體，并以 鉸鏈四桿機構的連桿為原動件，則機架、兩個連架桿都做擺動，其中一個連架桿相對機架的擺動即是搖頭動作。扇葉直接接到原動機上，既可以實現電風扇的功能。5.3有關參數及相關計算 5.3.1相關計算電動機轉速r

44、=1450r/min 搖頭周期T=10s 總的傳動比n=，蝸桿頭數Z1=1，蝸輪的齒數Z2=62，小齒輪齒數Z3與大齒輪齒數Z4 的確定由n 12 =62；n 34 =；n=n12 xn34；則 n34 =。則取Z3 =18；Z4 =70。5.3.2傳動構件的尺寸確定查閱相關資料后取：漸開線圓柱齒輪基本參數取齒輪的齒數分別為z3=18，z4=70。模數m=1.25，ha*=1， c*=0.25（ha*為齒頂高系數，c*為頂隙系數），壓力角=200。分度圓直徑d3=m z3=1.25x18=22.5（mm）；d4=m z4=1.25x70=87.5（mm）。齒頂高ha= ha*m=m=1.25m

45、m。齒根高hf=（ ha*+ c*）m=（1+0.25）x1.25=2.8125（mm）。齒全高h=ha+hf=1.25+2.8125=4.0625（mm）。齒頂圓直徑da3= d3+2ha=22.5+2x1.25=25（mm）；da4= d4+2ha=87.5+2x1.25=90（mm）。齒根圓直徑df3= d3-2hf=22.5-2x2.8125=16.875（mm）；df4= d4-2 hf=87.5-2x2.8125=81.875（mm）。基圓直徑db3= d3cos=22.5xcos200=21.14（mm）；db4= d4cos=87.5x cos200=82.22（mm）。齒距p

46、=m=3.14x1.25=3.93（mm）。齒厚s=m/2=1.96（mm）。齒槽寬e=m/2=1.96（mm）。中心距a=（d3+ d4）/2=55mm。頂隙c= c*m=0.25x1.25=0.3125（mm）。普通圓柱蝸桿傳動幾何尺寸計算查表取得：中心距a取40mm，i=62，m=1，d1=18，z1=1，z2=62，x2=0.0000，=3010/47/，ha*=1， c*=0.25。齒形角=200（ZA型）。蝸輪變?yōu)橄禂祒2=0.0000。蝸桿軸向齒距px=m=3.14mm。蝸桿齒頂高ha1= ha*m=1mm。頂隙c= c*m=0.25mm。蝸桿齒根高hf1=（ ha*+ c*）m

47、=1.25mm。蝸桿齒高h1=ha1+hf1=2.25mm。蝸桿齒寬b1取19.5mm。蝸輪分度圓直徑d2=m z2=62mm。蝸輪齒根圓直徑df2=d2-2hf2=59.5mm。蝸輪齒頂高ha2=m（ha*+ x2）=1mm。蝸輪齒根高hf2=m（ha*- x2+ c*）=1.25mm。蝸輪齒高h2= ha2+hf2=2.25mm。蝸輪齒頂圓弧半徑Ra2= d1/2-m=8mm。蝸桿軸向齒厚sx1= px/2=1.57mm。蝸桿法向齒厚sn1= sx1cos=1.568mm。蝸桿節(jié)圓直徑d1/=d1+2x2m=18mm。蝸輪節(jié)圓直徑d2/=d2=62mm。6尺寸與運動綜合6.1執(zhí)行機構尺寸設

48、計設計與計算 （1）由急回系數K計算極位夾角。由式=180，其中K=1.015；得=1.340（2）選定比例尺按已知條件畫出搖桿BC的兩個極限位置C1D和C2D，連C1,C2兩點，以C1為直角頂做C2C1M=85º,再做C1C2N=85º,C1M與C2N交于P點，以PC2為直徑做C1C2P的外接圓。在此圓上任取一點A作為曲柄的固定鉸鏈中心，連接AC1和AC2因同一圓弧的圓周角相等，故C1AC2=C1PC2=，令AB=a,BC=b,CD=c,DA=d.則AC1 =b-a,AC2=b+a。圖一（總體示意圖）圖二（局部圖）（3）上圖四桿機構若取AD為機架則為曲柄搖桿機構，而電風扇的搖頭四桿機構是要是雙搖桿機構，所以取以CD為機架的倒置四桿機構。（4）計算各桿的實際長度。分別量取圖中AB、B2C2、C1D、AD的長度，計算得a=µAB1=0.6µ，b=µB2C2=1.94µ，c=µC2D=µ，d=µAD=2.02µ。（C2D=1mm）因為本設計使用曲柄連桿機構的倒置機構（即雙搖桿機構），所以取桿