级东北大学机械原理课程设计.docx

1、级东北大学机械原理课程设计机械原理课程设计说明书题目： 牛头刨床机构方案分析 班 级 ：车辆11XX姓 名 ：XX学 号 ：2011XXXX指导教师 ： 成 绩 ：2013 年 10 月 31 日1机构简图和已知条件如图所示为一种牛头刨床主机构的运动简图，已知，l1=0.1m，l0=0.4m，l3=0.75m，l4=0.15m，ly=0.738m，l3=0.375m，a=0.05m，b=0.15，c=0.4m，d=0.1m。只计构件3、5的质量，其余略去不计，m3=30kg，JS3=0.7kgm2，m5=95kg。工艺阻力Q如图所示，Q=9000N。主轴1的转速为60r/min(顺时针方向)，

2、许用运转不均匀系数=0.03。2滑枕初始位置及行程H 的确定方法因为题目中L1L0,所以滑枕处于初始位置的时候，构件和构件相互垂直，示意图如下： 3杆组的拆分方法及所调用的杆组子程序中虚参与实参对照表3-1 杆组拆分方法关键点：1点,2点,3点,4点,5点,6点（rrpk函数中选取参考点）,7,8,9,10（rrpf函数选取移动副反力作用点，未标出）。构件号：，。3-2 杆组子程序中虚参与实参对照表3-2-1机构运动分析程序虚参与实参对照表1)调用bark函数，求2点的运动参数。形式参数n1n2n3kr1r2gamtwepvpap实 值1201r120.00.0twepvpap 2)调用rpr

3、k函数，求构件3的运动参数。形式参数mn1n2k1k2r1r2vr2ar2twepvpap实 值132320.0&r2&vr2&ar2twepvpap3)调用bark函数，求4点的运动参数。形式参数n1n2n3kr1r2gamtwepvpap实 值3403r340.00.0twepvpap4)调用rrpk函数，求5点的位置和5构件的运动参数。形式参数mn1n2n3k1k2k3r1r2vr2ar2twepvpap实 值-1465456r45&r2&vr2&ar2twepvpap3-2-2机构力分析程序虚参与实参对照表力分析是在运动分析基础上来的，力分析中对照表省略运动分析对照表1)调用bark函

4、数，求构件3的质心7点的运动参数。形式参数n1n2n3kr1r2gamtwepvpap实 值3703r370.00.0twepvpap2)调用bark函数，求构件5的质心8点的运动参数。形式参数n1n2n3kr1r2gamtwepvpap实 值50850.0r58-161.57*drtwepvpap3)调用bark函数，求构件5的工艺阻力作用点9点的运动参数。形式参数n1n2n3kr1r2gamtwepvpap实 值50950.0r59165.96*drtwepvpap4) 调用rrpf函数，求4、5点的反作用力。形式参数n1n2n3ns1ns2nn1nn2nexfk1k2pvpaptwefr

5、实 值41050809945pvpaptwefr5) 调用rprf函数求2点的反作用力。形式参数n1n2ns1ns2nn1nn2nexfk1k2pvpaptwefrfkpk实 值327040032pvpaptwefrfkpk6）调用barf函数，求1点运动副反力及平衡力矩。形式参数n1ns1nn1k1papefrtb实 值1021papefr&tb 4飞轮转动惯量的计算方法飞轮转动惯量的计算方法 1、一周期内驱动力矩功等于阻力功，所以：t*Td=1/2(Tr0+Tr1)+1/2(Tr1+Tr2)+.+1/2(Trn-1+Trn)因为Trn=Tr0，所以由上式可得：Td=/tTri=1/nTri

6、2、间隔i-1、i内的盈亏功变化量E:E=Td-1/2*（Tri+Tri-1）3、计算各点的盈亏功Ei：Ei=Ei-1+Ei4、找出最大和最小盈亏功：Emax,Emin5、计算飞轮的转动惯量Jf: Jf=(Emax-Emin)/W。5自编程序及计算结果清单（包括线图）5-1运动分析5-1-1运动分析主程序#include graphics.h#include subk.c#include draw.cmain() static double p202,vp202,ap202,del; static double t10,w10,e10,pdraw370,vpdraw370,apdraw370;

7、 static int ic; double r12,r13,r34,r45; double pi,dr; double r2,vr2,ar2; int i; FILE *fp; char *m=p,vp,ap; r12=0.1, r13=0.4, r34=0.75, r45=0.15; pi=4.0*atan(1.0); dr=pi/180.0; t6=0.0*dr; w6=0.0; e6=0.0; w1=-2.0*pi; e1=0.0; del=10.0; p11=0.0; p12=0.0; p31=0.0; p32=-0.4; p61=-100.0; p62=0.338; printf(

8、 n The Kinematic Parameters of Point 6n); printf(No THETA1 S6 V6 A6n); printf( deg m m/s m/s/sn); if(fp=fopen(file1,w)=NULL) printf( Cant open this file.n); exit(0); fprintf(fp, n The Kinematic Parameters of Point 6n);fprintf(fp,No THETA1 S6 V6 A6n);fprintf(fp, deg m m/s m/s/s); ic=(int)(360.0/del);

9、 for(i=0;i=(-0.31825)&p510) fenexf1=-9000.0; else fenexf1=0.0;main() static double p202,vp202,ap202,del; static double t10,w10,e10,tbdraw370,tb1draw370; static double sita1370,fr1draw370,sita2370,fr2draw370,sita3370,fr3draw370; static double fr202,fe202,fk202,pk202, tb,tb1,fr1,bt1,fr2,bt2,we1,we2,we

10、3,we4,we5; static int ic; double r12,r13,r34,r45,r37,r58,r59,gam1,gam2; double pi,dr; double r2,vr2,ar2; int i; FILE *fp; char *m=tb,tb1,fr1,fr2; sm1=0.0;sm2=0.0;sm3=30.0;sm4=0.0;sm5=95.0; sj1=0.0;sj2=0.0;sj3=0.7;sj4=0.0; r12=0.1;r34=0.75;r45=0.15;r37=0.375; r59=sqrt(0.1*0.1+0.25*0.25); r58=sqrt(0.1

11、5*0.15+0.05*0.05);r13=0.4; pi=4.0*atan(1.0); dr=pi/180.0; gam1=(atan(1.0/3.0)-pi)*dr; gam2=(pi-atan(1.0/4.0)*dr; t6=0.0;w6=0.0;e6=0.0; w1=-2*pi;e1=0.0;del=10.0; t6=0.0*dr; p31=0.0; p32=-0.4; p11=0.0; p12=0.0; p61=-100; p62=0.338; printf(n The Kineto-static Analysis of a Six-bar Linkasen);printf( NO

12、THETA1 FR1 BT1 FR2 BT2 TB TB1n);printf( (deg.) (N) (deg.) (N) (deg.) (N.m) (N.m)n); if(fp=fopen(file,w)=NULL) printf(Cant open this file.n); exit(0); fprintf(fp,n The Kineto-static Analysis of a Six-bar Linkasen);fprintf(fp, NO THETA1 FR1 BT1 FR2 BT2 TB TB1n);fprintf(fp, (deg.) (N) (deg.) (N) (deg.)

13、 (N.m) (N.m)n); ic=(int)(360.0/del); for(i=0;i=ic;i+) t1=(-(pi-asin(r12/r13)+(-i)*del*dr; bark(1,2,0,1,r12,0.0,0.0,t,w,e,p,vp,ap); rprk(1,3,2,3,2,0.0,&r2,&vr2,&ar2,t,w,e,p,vp,ap); bark(3,4,0,3,r34,0.0,0.0,t,w,e,p,vp,ap); rrpk(-1,4,6,5,4,5,6,r45,&r2,&vr2,&ar2,t,w,e,p,vp,ap); bark(3,7,0,3,r37,0.0,0.0,

14、t,w,e,p,vp,ap); bark(5,0,8,5,0.0,r58,gam1,t,w,e,p,vp,ap); bark(5,0,9,5,0.0,r59,gam2,t,w,e,p,vp,ap); rrpf(4,6,5,0,8,0,9,9,4,5,p,vp,ap,t,w,e,fr); rprf(3,2,7,0,4,0,0,3,2,p,vp,ap,t,w,e,fr,fk,pk); barf(1,0,2,1,p,ap,e,fr,&tb); fr1=sqrt(fr11*fr11+fr12*fr12); bt1=atan2(fr12,fr11); fr2=sqrt(fr31*fr31+fr32*fr

15、32); bt2=atan2(fr32,fr31); /*求反力的大小和方向*/ we1=0; we2=0; we3=-(ap71*vp71+(ap72+9.81)*vp72)*sm3-e3*w3*sj3; we4=0; extf(p,vp,ap,t,w,e,9,fe); we5=-(ap91*vp91+(ap92+9.81)*vp92)*sm5+fe91*vp91; tb1=-(we1+we2+we3+we4+we5)/w1; printf(n%2d%10.3f%10.3f%10.3f%10.3f%10.3f%10.3f%10.3f,i+1,t1/dr,fr1,bt1/dr,fr2,bt2/

16、dr,tb,tb1); fprintf(fp,n%2d%10.3f%10.3f%10.3f%10.3f%10.3f%10.3f%10.3f,i+1,t1/dr,fr1,bt1/dr,fr2,bt2/dr,tb,tb1); tbdrawi=tb; tb1drawi=tb1; fr1drawi=fr1; sita1i=bt1; fr2drawi=fr2; sita2i=bt2; fr3drawi=fr2; sita3i=bt2; if(i%16)=0)getch(); fclose(fp); getch(); draw2(del,tbdraw,tb1draw,ic,m); / draw3(del,

17、sita1,fr1draw,sita2,fr2draw,sita3,fr3draw,ic,m); 5-2-2力分析程序计算结果（包括线图） The Kineto-static Analysis of a Six-bar Linkase NO THETA1 FR1 BT1 FR2 BT2 TB TB1 (deg.) (N) (deg.) (N) (deg.) (N.m) (N.m) 1 -165.522 1465.961 14.478 627.211 -171.015 0.000 0.000 2 -175.522 1231.137 14.262 498.156 -176.556 -20.923

18、-20.923 3 -185.522 1032.934 13.658 403.975 174.931 -33.936 -33.936 4 -195.522 15767.371 12.723 6867.477 -148.658 -746.184 -746.184 5 -205.522 15279.446 11.512 6271.810 -152.002 -920.269 -920.269 6 -215.522 14868.908 10.074 5753.760 -156.636 -1062.285 -1062.285 7 -225.522 14516.560 8.455 5324.070 -16

19、2.398 -1174.074 -1174.074 8 -235.522 14211.026 6.692 4992.651 -169.019 -1257.253 -1257.253 9 -245.522 13947.394 4.823 4764.261 -176.106 -1313.482 -1313.48210 -255.522 13725.674 2.881 4635.881 176.812 -1344.548 -1344.54811 -265.522 13549.308 0.895 4597.093 170.195 -1352.283 -1352.28312 -275.522 13423.811 -1.104 4633.369 164.397 -1338.391 -1338.39113 -285.522 13355.544 -3.086 4730.512 159.624 -1304.217 -1304.21714 -295.522 13350.657 -5.023 4878.175 155.939 -1250.521 -1250.52115 -305.522 13414.175 -6.882 5071.434 153.300 -1177.289 -1177.28916 -315.522 13