机械程序设计 c++

机 械 设 计

程

序

设

计

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V带设计

功率：5KW 小轮转速：1460r/min 传动比：3.4 中心距系数：0.8 载荷性质：变动小 材料：帘布 每天工作时间：16小时

程序：

/* V带传动设计程序 */

#include "stdio.h"

#include "math.h"

#define PI 3.1415926

main()

{

int i,j,m,x,y,g,e,lh;

float p,n1,n2,u,a,a0,b,af,a1,amin,amax;

float ki,kl,kaf,kq,p0,p1,z,f0,q,ka,pc,v,d2,c1,c2,lc,lj,l;

float nn[5];

static int ll[35]={0,400,450,500,560,630,710,800,900,1000,1120,1250,1400,1600,

1800,2000,2240,2500,2800,3150,3550,4000,4500,5000,5600,

6300,7100,8000,9000,10000,11200,12500,14000,16000};

float aa[6][15]={{0,3.9E-4,.06,.246,7.44,4.41E-5,.72,4.3E-4,9.8E-8,435,50,63,71,80,90},

{0,1.03E-3,.1,.449,19.62,7.65E-5,.68,2.25E-4,2.25E-8,100,75,90,100,112,125},

{0,2.65E-3,.17,.794,50.6,1.31E-4,.68,1.74E-4,1.94E-8,27,125,140,150,170,180},

{0,7.5E-3,.3,1.48,143.2,2.43E-4,.70,9.64E-3,4.48E-7,7.2,200,224,236,265,280},

{0,2.66E-2,.62,3.15,507.3,4.77E-4,.78,3.38E-5,0,.92,355,400,425,475,500},

{0,4.98E-2,.9,4.57,951.5,7.06E-4,.706,4.95E-5,1.254E-9,0,500,560,630,800,900}};

static char bb[6]={'Z','A','B','C','D','E'};

printf ("按顺序输入已知数据: 功率P(kw), 小轮转速n1(r/min), 传动比i(若已知n2,\n");

printf ("则i=0), 大轮转速n2(r/min)(若已知i, 则n2=0), 给定中心距a(mm)(若未给,\n");

printf ("则a=0), 中心距系数a0( 0.7～2, 若已知a, 则a0=0 ), 载荷性质G(平稳1,\n");

printf ("变动小2, 变动较大3, 变动很大4), 材质E(帘布0, 化纤1), 每天工作小时数LH\n");

printf ("P,n1,i,n2,a,a0,G,E,LH=\n");

scanf ("%f,%f,%f,%f,%f,%f,%d,%d,%d",&p,&n1,&u,&n2,&a,&a0,&g,&e,&lh);

if (n2 != 0) u=floor(n1/n2*1000+.5)/1000;

printf ("已知数据:\n");

printf ("P=%.1f,n1=%.0f,i=%.3f,a=%.0f,a0=%.1f,G=%d,E=%d",p,n1,u,a,a0,g,e);

printf ("\n");

printf ("输出结果:\n");

if (lh>16) ka=1.2;

else if(lh<=16 && lh>=10) ka=1.1;

else ka=1;

if (g==2) ka=ka+0.1;

else if (g==3) ka=ka+0.2;

else if (g==4) ka=ka+0.3;

pc=ka*p;

for (i=0;i<=4;i++)

{ nn[i]=aa[i][9]*pow(pc,1.2);

if (n1>=nn[i]) break; };

if (i==0)

{ x=0; y=1; }

else if (n1<=nn[4])

{ x=4; y=5; }

else

{ x=i-1; y=i+1; }

for (i=x;i<=y;i++)

for (j=10;j<=14;j++)

{

v=floor(PI*aa[i][j]*n1/600+.5)/100;

if (v>=5 && v<=20)

{

d2=floor(u*aa[i][j]*0.98/5+.5)*5;

u=floor(d2/aa[i][j]*0.98*100+.5)/100;

c1=d2+aa[i][j];

c2=d2-aa[i][j];

if (a==0) a1=a0*c1;

else a1=a;

lc=2*a1+PI*c1/2+c2*c2/4/a1;

for (m=1;m<=33;m++)

{ if (lc<=ll[m]) break; }

lj=(ll[m]+ll[m-1])/2;

if (lc<lj) l=ll[m-1];

else l=ll[m];

b=2*l-PI*c1;

a1=floor((b+sqrt(b*b-8*c2*c2))/8*10+.5)/10;

amin=floor(a1-0.015*l+0.5);

amax=floor(a1+0.03*l+0.5);

af=180-c2/a1*57.3;

if (af>=120)

{

kl=aa[i][6]+aa[i][7]*l-aa[i][8]*l*l;

kaf=1.25*(1-1/exp(0.51*af*PI/180));

if (u<=1.04) ki=0;

else if (u<=1.09) ki=0.03;

else if (u<=1.19) ki=0.052;

else if (u<=1.49) ki=0.083;

else if (u<=2.5) ki=0.113;

else ki=0.122;

if (e==0) kq=.75;

else kq=1;

p0=(aa[i][3]*pow(v,-0.09)-aa[i][4]/aa[i][j]-aa[i][5]*v*v)*v;

p1=aa[i][1]*n1*ki;

z=floor(pc/((p0+p1)*kl*kaf*kq)+0.95);

if (z<=9)

{

f0=floor((500*pc*(2.5/kaf-1))/v/z+aa[i][2]*v*v+0.5);

q=floor(2*z*f0*sin(af*PI/180/2)+0.5);

printf ("型号%c,Ld=%.0f,z=%.0f,d1=%.0f,",bb[i],l,z,aa[i][j]);

printf ("d2=%.0f,v=%.2f,i=%.2f\n",d2,v,u);

printf ("a=%.2f, amin=%.2f, amax=%.2f, ",a1,amin,amax);

printf ("F0=%.2f, Q=%.2f\n",f0,q);

printf("\n");

}

}

}

else if (v>20) break;

}

}

运算结果：