牛頭刨床（機(jī)械原理課程設(shè)計(jì)）完整版

機(jī)械原理課程設(shè)計(jì)題目：牛頭刨床作者：張晴2/30機(jī)械原理 2 4 52、導(dǎo)桿機(jī)構(gòu)的運(yùn)動(dòng)分析2.1位置2的速度分析 62.4位置2的加速度分析 72.3位置4的速度分析 2.4位置4的加速度分析 3、導(dǎo)桿機(jī)構(gòu)的動(dòng)態(tài)靜力分析3.1位置2的慣性力計(jì)算 (1)進(jìn)一步加深學(xué)生所學(xué)的理論知識(shí)，培養(yǎng)學(xué)生獨(dú)立解決有關(guān)本課程(2)使學(xué)生對(duì)于機(jī)械運(yùn)動(dòng)學(xué)和動(dòng)力學(xué)的分析設(shè)計(jì)有一較完整的概念。(3)使學(xué)生得到擬定運(yùn)動(dòng)方案的訓(xùn)練，并具有初步設(shè)計(jì)選型與組合以(4)通過(guò)課程設(shè)計(jì)，進(jìn)一步提高學(xué)生運(yùn)算、繪圖、表達(dá)、運(yùn)用計(jì)算機(jī)機(jī)械原理課程設(shè)計(jì)的任務(wù)是對(duì)機(jī)械的主體機(jī)構(gòu)(連桿機(jī)構(gòu)、凸輪機(jī)構(gòu)、齒輪機(jī)構(gòu)以及其他機(jī)構(gòu))進(jìn)行設(shè)計(jì)和運(yùn)動(dòng)分析、動(dòng)態(tài)靜力分析，并根據(jù)給定機(jī)器的工作要求，在此基礎(chǔ)上設(shè)計(jì)凸輪、齒輪；或?qū)Ω鳈C(jī)構(gòu)進(jìn)行運(yùn)動(dòng)分析。牛頭刨床的簡(jiǎn)介工作臺(tái)輪機(jī)構(gòu)等組成，如圖所示。電動(dòng)機(jī)經(jīng)過(guò)減速裝置(圖中只畫(huà)出了齒輪z?,z?)使曲柄2轉(zhuǎn)動(dòng)，再通過(guò)導(dǎo)桿機(jī)構(gòu)2-3-4-5-6帶動(dòng)刨頭6和刨刀作往復(fù)切削運(yùn)動(dòng)。工作行程時(shí)，刨刀速度要平穩(wěn)，空回行程時(shí)，刨刀切削階段刨刀應(yīng)近似勻速運(yùn)動(dòng)，以提高刨刀的使有關(guān)機(jī)構(gòu)(圖中未畫(huà)出來(lái))完成的。為減小機(jī)器的速度波動(dòng)，在曲柄軸0?上安裝原始數(shù)據(jù)導(dǎo)桿機(jī)構(gòu)的運(yùn)動(dòng)分析導(dǎo)桿機(jī)構(gòu)的動(dòng)態(tài)靜力分析符號(hào)Xs?Ys?G?單位N方案內(nèi)容凸輪機(jī)構(gòu)設(shè)計(jì)齒輪機(jī)構(gòu)設(shè)計(jì)符號(hào)0?Dδ0δδ0運(yùn)動(dòng)規(guī)律Z?mα單位mmmm方案1)66/301.1位置2的速度分析(左極限位置的曲柄對(duì)應(yīng)位置順時(shí)針轉(zhuǎn)30°)數(shù)據(jù)：n?=60r/min,lo?o?=380mm,lo?A=110mm,lo?g=540mm,lgc對(duì)位置2:選取尺寸比例尺μ?=1,作機(jī)構(gòu)運(yùn)動(dòng)簡(jiǎn)圖。CXCX6X6A2U?由CAD測(cè)量得：lo?A=0.41899m速度分析：取構(gòu)件3和4的重合點(diǎn)(A2,A3,A4)進(jìn)行速度分析。對(duì)構(gòu)件2:VA?=O2lo?A=(60×2π)160×0.11=0.69m/s對(duì)構(gòu)件3:構(gòu)件3和構(gòu)件2在A處構(gòu)成轉(zhuǎn)動(dòng)副，VA3=VA2=0.69m/s對(duì)構(gòu)件4:VA4=VA3+VA4A3?//AO??//AO?方向：⊥AO?⊥AO?取速度極點(diǎn)P,速度比例尺μv=0.001(m/s)/mm。作速度多邊形如下圖所示。計(jì)算結(jié)果Pb4得：VA?=0.3237m/s,w?=VA?/lo?A=0.323710.41899=0.7726rad/sVA4A3=0.60932m/s(A指向O?)VB?=W?*lo?B=0.7727*0.54=0.4172m/sVB5=VB4=0.4172m/s對(duì)構(gòu)件4:Vcs=VB5+VC5B5大?。?0.4172?得：Vcs=0.39785m/s對(duì)構(gòu)件2:aAz=w?2*lo?A=6.282*0.11=4.338224m/s2對(duì)構(gòu)件4:a?a=a?4+aA4=aA3+a64A3+aA?A3m/s2aa?=0.2502Tm/s2m/sPa/4A?=2o?VA?A?=2*0.7727*0.60932=0.94164aA?=042*lo?A=0.7727*0.7727*0.41899=0.2502m/s2作加速度多邊形如上圖所示。得：aA?=2.8894m/s2,a?4=2.90m/s2aB?=lo?B*βBO?=6.896*0.54=3.72384m/s2aB?=lBO?*042=0.7727*0.7727*0.54=0.33244.5在B處構(gòu)成轉(zhuǎn)動(dòng)副，所以aB?=aB:對(duì)構(gòu)件5::acs=aBs+ass+aCsB?+acsgs方向：//XXB→O?取加速度極點(diǎn)p’,加速度比例尺μa=0.01(m/s2)/mm。作加速度多邊形如上圖所示。得：acs=3.57m/s2(方向水平向右)ac6=acs=3.57m/s2數(shù)據(jù)：n?=60r/min,lo?O?=380mm,lo?A=110mm,lo?B=540mm,lgc=0.25lo?B,lo?s?=0.5lo?B11A4l??A=0.48633m對(duì)構(gòu)件2:VA?=W?*lo?A=(60*2π)/60*0.11=0.69m/s示。VA2=0.69m/sVA4=0.67212m/sO?=1.382rad/sVA4A?=0.15607m/s(A指向O?)VB?=04*l??B=1.382*0.54=0.74628m/sVBs=VB?=0.74628m/s對(duì)構(gòu)件4:Vcs=Vps+Vc?B?對(duì)構(gòu)件3:構(gòu)件3和構(gòu)件2在AVB4=0.74628Vcs=0.74845m/sm/sa?z=4.338a?3=4.338m/s2m/s2af?A?=20o?0A4A?=2*1.382*0.15607=0.4314m/s2aA?=042*lo?A=1.382*1.382*0.48633=0.92889m/s2作加速度多邊形如上圖所示。得：aA?=0.5498m/s2,aA4=1.1079m/s2aR4=lBO?*βBO?=1.1305*0.54=0.61407m/s2aB?=lgO?*042=1.382*1.382*0.54=1.03136m/s//xxB//xxB→O?大?。喝〖铀俣葮O點(diǎn)p',加速度比例尺μa=0.01(m/s2)/mm作加速度多邊形如上圖所示。得acs=0.5851m/s2ac6=acs=0.5851m/siacs=0.5851m/s2n2=60r/min,LO2O4=380mm,LO2A=110mm,LO4B=540mm,LBC=0.25LO4B,LO4S4=0B位置2:對(duì)各構(gòu)件進(jìn)行受力分析，按靜定條件將機(jī)構(gòu)分解為兩個(gè)基本桿組及作用有未知平衡力的構(gòu)件2,并對(duì)桿組進(jìn)行分析。根據(jù)影像原理a=1.86192m/s2對(duì)構(gòu)件4:慣性力FI4=M4·as?=(G4/g)·as?=200/9.8*1.86192=37.998367N對(duì)構(gòu)件6:慣性力FI6=M6·ac?=(Gb/g)·ac?=700/9.8*3.57=255N2.1桿組.5,6的動(dòng)態(tài)靜力分析示力體(注：FR45是桿5給6的力)66CCX力多邊形力多也又ZF=P+Gs+Fi?+FR?s+FR?6=0,作為多邊行如圖所示，μv=100N/mm。力多也2.2桿組3,4的動(dòng)態(tài)靜力分析FR4s=7256.955NFR16=1080.842NFR54=7256.955NFR23=9067.6577N示力體如下圖Ms?=Js4·Qs4=7.5856N·m(逆時(shí)針)對(duì)O4點(diǎn)取矩得：ZMo4=FR54*ho4B+Ms4+G4ls?o?FR23=9067.657703N力多邊形力多邊形由ZF=0,作力的多邊形如上圖所示，μy=10N/mm.得：FR?=2597.369081NFR|?=FR??=FR2?=9067.657703N2.3平衡力矩的計(jì)算在下圖中，對(duì)o2點(diǎn)取矩得：2Mo2=9021.6814*0.05162=464.43N.m2Mo2=FR?×lh-Mp=0Mh=9067.657703×FR14=2597.369081NMb=498.7211737N.MB導(dǎo)桿機(jī)構(gòu)的動(dòng)態(tài)靜力分析(位置4號(hào))Bn2=60r/min,LO2O4=380mm,LO2A=110mm,LO4B=540mm,LBC=0.25LO4B,LO4S4=0.5LO4B位置4:對(duì)各構(gòu)件進(jìn)行受力分析，按靜定條件將機(jī)構(gòu)分解為兩個(gè)基本桿組及作用有未知平衡力的構(gòu)件2,并有桿組進(jìn)行分析。2.4位置4的慣性力計(jì)算根據(jù)影像原理as?=LO4S4·αs=0.27*1.1305=0.305235m/s對(duì)構(gòu)件4:慣性力FI4=M4·as?=(G?/g)·a?=200/9.8*0.305235=6.229285714N:對(duì)構(gòu)件6:慣性力FI6=m6·ac?=(Go/g)·ac?=700/9.8*0.6176=44.11428571N,示力體FI4=6.2292857FI6==44.11428as?=0.305235m/s2力多邊形力多邊形又ZF=P+G?+F?+FR4s+FR?6=0,作為多邊行如圖所示，μv=100N/mm。力多邊形由力多邊形可得：FR16=FR16μN(yùn)=100*2.0742745=207.427545NFR45=7061.315Fg?6=207.4275FR54-7061.3152FR2?=7854.343FR14=14910.3N2.6桿組3,4的動(dòng)態(tài)靜力分析示力體如下圖Ms?=Js4·as4=1.24355N·m(逆時(shí)針)對(duì)O4點(diǎn)取矩得：R23lo4A^=0FR3=7854.343996N力多邊形由ZF=0,作力的多邊形如上圖所示，μy=10N/mm得：Fpl=14910.3NFRI=FR=FR=7854.343996N在右圖中，對(duì)o2點(diǎn)取矩得：2Mo2=FR?2×1h-Mb=)1h=lo?A×cos4?=0.109732mMb=7854.343996×0.109732=861.873233N.m3.解析法校核機(jī)構(gòu)運(yùn)動(dòng)分析和動(dòng)態(tài)靜力分析結(jié)果3.1運(yùn)動(dòng)分析(程序見(jiàn)附錄)程序截圖如下：Q1C3C4SeW310493.129184.75?-0.1641.392-6.3050.754-6.934-7.5500.464140100.590181.190-0.2341.164-0.8560.620-3.780-2.5342.035144101.317180.637-G-0.541-0.884一一位移速度速度店品加速度加速度1050-109g位置基準(zhǔn)參見(jiàn)上頁(yè)圖片3.2動(dòng)態(tài)靜力分析(程序見(jiàn)附錄)S:S:NDebug\B.ereR?2f?2R?42.0881226.329592372.5819527.922612333.0768528.216252233.5?24127.411552014.0685825.410061684.5654922.311731415.0634218.214541285.562?313.217341226.063827.519961216.567061.522691238.51198-41.62441639.01506-37.23761669.51680-34.547617610.01389-35.942?19312.5108240.4266197?281-2.71043-4.12?10?3.2303-14.1870664.4157-27.4070244.9101-42.6369784.6141-30.4969293.5271-15.866820-0.1717-6.006759-2.2963-4.46438-4.9?370.23545-4.57420.23622-2.6?280.236130.46960.244453.46730.25764.96930.243623.96750.25650-2.07220.23554-4.27400.239218824980548222-1111-1202-930-160475.1632.6 486.5-106.9-70.5G.Gf74R45f45R?6h?6R23MB已知2位置時(shí)M=9067.657703×0.055=498.7211737N4位置時(shí)M=7854.343996×0.109732=861.873233N.m因此2位置相對(duì)誤差為4.9%,4位置相對(duì)誤差為0.02%23/30由此可算出其平均驅(qū)動(dòng)矩Ma=312.04NM?！鱓?=278.487NM△W?=-537.815NM△W?=259.328NM24/305.凸輪機(jī)構(gòu)的設(shè)計(jì)1)基圓半徑的確定已知凸輪運(yùn)動(dòng)規(guī)律為正弦加速度，最大壓力角amax=40,推程運(yùn)動(dòng)角δmax=15,律的標(biāo)尺于0.8處，于是，根據(jù)h/r?=0.8,即可求得r?≥40.79mm,所以可以取凸輪為42mm牛頭刨床運(yùn)動(dòng)循環(huán)圖213.6°刨頭工作行程空回行程工作臺(tái)停止進(jìn)給5.1圖解法看附錄手繪圖5.2解析法幾何畫(huà)板-[凸輪]f(θ)=15(sing(θ)=15.X(θ)=15·sin(θ)-12.5·sin(θ+f(θ)+21)Y(θ)=15·cos(θ)-12.5·cos(θ+f(θ)+21)r(θ)=15·sin(θ)-12.5·sin(θ+g(θ)+21)s(θ)=15·cos(θ)-12.5·cos(θ+g(θ)+21)HG55由幾何畫(huà)板編寫函數(shù)可得上圖25/30牛頭刨床動(dòng)力傳輸裝置，電動(dòng)機(jī)和減速器(齒輪)的選?。哼x用交流異步電動(dòng)機(jī)轉(zhuǎn)速1440r/min;,選用傳動(dòng)比為3的帶傳動(dòng)，主輪D1=50mm,副輪D2=150mm;再選用一個(gè)兩級(jí)齒輪傳動(dòng)即可?？梢匀?=18,z?=45,z?=15,z?=48;設(shè)計(jì)的齒輪為標(biāo)準(zhǔn)直齒圓柱齒輪，模數(shù)為2,ha*=1分度圓直徑d=mz齒頂高h(yuǎn)a=ha*m齒全高h(yuǎn)=ha*+hr=(2ha*+c*)mdr=d-2hr=(z-2ha*-2c*)m基圓直徑d,=dcosα=mzcosa根據(jù)上述公式可得：26/30da?=40,dri=31,dp?=33.8da?=94,drz=85,do?=84.6da?=34,dr?=25,dp?=28.2da?=100,df?=91,dp?=90.2通過(guò)查表，我組選用Y132M-4型電動(dòng)機(jī)，功率7.5Kw,轉(zhuǎn)速附錄3.1運(yùn)動(dòng)分析(程序)#include<stdio.h>voidmain(){/*a=AB,b=CD,c=DE,d=AC,e=H,f=w1*/doubleB,C,D,E,F,G,I,L,M,0;/*B=q3,C=q4,E=Se,F=w3,G=w4,I=Ve,L=a3,M=a4,o=ae*/printf("@1\t@3\t@4\tSe\tW3\tW4\tVe\tA3\tA4\tAe\t\n");while(x<6.3)B=atan((d+a*sin(x))/(a*cos(x)));/*求q3*/if(B<0)B=PI+B;C=PI-asin((e-b*sin(B))/c);/*求q4*/if(C<0)C=PI+C;E=b*cos(B)+c*cos(C);/*求Se*/F=(a*f*(a+d*sin(x)))/(d*d+a*a+2*d*a*sin(x));/*求w3*/G=-(F*b*cos(B))/(c*cos(C));/*求w4*/I=(F*b*sin(B-C))/cos(C);/*求Ve*/L=((d*d-a*a)*d*a*f*f*cos(x))/((d*d+a*a+2*d*a*sin(x))*(d*d+a*a+2*dM=(F*F*b*sin(B)+G*G*c*sin(C)-L*b*cos(B))/(c*cos(C));O=-(L*b*sin(B-C)+F*F*b*cos(B-C)-G*G*c)/cos(C);/*求ae*/printf("%3.0f\t%3.3f\t%3.3f\t%3.3f\t%3.3f\t%3.3f\t%3.3f\t%3.3f\t%3.3f\t%3.3f\n",x*180/PI,(B*180)/PI,(C*180)/PI,E,F,G,I,L,M,O);x=x+PI*4/180;}3.2動(dòng)態(tài)靜力分析(程序)#include<stdio.h>pi,p,K,H,Ll,BFO3B,QMAX,L4,L2,L5,fq,Y;f2,f40,g,f50,n,f4,i,fz,w2,s,w4,v,ak,ar,e4;floatL,f5,w5,e5,xf,vf,afloatFC,G6,PI6,PI6X,PI6Y,R45,R45X,R45Y,R76;floatR54,R54X,R54Y,G4,R23,R23X,R23Y,f23,PI4,PI4X,PI4Y;floatR74,R74X,R74Y,f74,R32,R32X,R32Y,PB,PBX,PBY,G2;floatR72,R72X,R72Y,f72,MI4,MB,M4,M6,JS4,YP;gan()p=pi/180;K=1.459;H=0.32;M4=20.4;M6=71.4;JS4=1.1;YP=0.8;QMAX=180*(K-1)/(K+1)*p;L4=H/(2*sin(QMAX/2));L2=L1*sin(QMAX/2);Y=L4-L4*(1-cos(QMAX/2))/2;fq=180+QMAX/2/p;fz=360-QMAX/2/p;YP=Y-L1-YP;}main()f2=fq*p;f40=180*p+f40;g=Y/L5-L4/L5*sin(f40);f50=atan(-g/sqrt(1-g*g));f50=180*p+f50;printf("nR72f72R74”);printf("h76R23MB\n");for(n=1;n<=13;n=++n-0.5)elseif(n==10.5){f2=270*p;f4=90*p;}elseif(n==8.5){f2=fz*p;f4=atan((Ll+L2*sin(f2))/L2/cos(f2));}f2=(fq-i)*p;{if(f2<0)f2=360*p+f2;f4=atan((Ll+L2*sin(f2))/L2/cos(f2));}{if(f2/p<270&&f2/p>90)f4=180*p+f4;}}w2=-2*pi*60/60;s=sqrt(L2*L2+L1*L1+2*L1*L2*sin(f2));v=-L2*w2*sin(f2-f4);ak=2*v*w4;e4=-(ak+L2*w2*w2*sin(f2-f4))/s;L=L4/L5;g=Y/L5-L*sin(f4);f5=atan(-g/sqrt(1-g*g));f5=180*p+f5;w5=-L*w4*cos(f4)/cos(f5);e5=w5*w5*sin(f5)/c