水电站镇墩抗滑稳定计算.c

对于水轮机阀门开启、关闭的一瞬间

/* Note:Your choice is C IDE */
#include <iostream.h>
#define PI 3.1415927
#define g  9.81
main()
{float D1,D2,D3,L1,L2,L3,L,H0,Hm,c,c1,c2,c3,V,V1,V2,V3;
 float Q,GDD,Ts,T0,ns,N,Qmax;
 float p,a,sum,b,tr;
 float t[100],smax,s[100],time[100];
 int i,I;
 printf("请输入已知条件及数据:");
 printf("\n Hm="); scanf("%f",&Hm);
 printf("\n N="); scanf("%f",&N); 
 printf("\n Qmax="); scanf("%f",&Qmax); 
 printf("\n ns="); scanf("%f",&ns); 
 printf("\n GDD="); scanf("%f",&GDD); 
 printf("\n Ts="); scanf("%f",&Ts);
 H0=Hm;
 D1=1;D2=D3=0.6;
 L1=200;L2=24;L3=18;
 c1=800;c2=900;c3=1000;
 V1=2*Qmax/(PI*D1*D1/4);
 V2=Qmax/(PI*D2*D2/4);
 V3=Qmax/(PI*D3*D3/4);
 L=L1+L2+L3;
 c=L/(L1/c1+L2/c2+L3/c3);
 V=(L1*V1+L2*V2+L3*V3)/L;
 p=c*V/(2*g*H0);
 b=L*V/(g*H0*Ts);
 tr=2*L/c;
 printf("\n V=%.4f\n c=%.4f\n p=%.4f\n b=%.4f\n Tr=%.4f\n",V,c,p,b,tr); 
 t[0]=1;sum=0;time[0]=0;I=0;smax=0;
 i=1;
 for(;tr*i<=Ts;i++)
  {time[i]=tr*i;	
   t[i]=t[i-1]-b/p;
   s[i]=2*(t[0]-t[i]-sum/p)/(t[i]+1/p); 
   sum+=s[i];
   if(smax<=s[i]){smax=s[i];I=i;}
   printf("第%2d次,时间为t=%.2fs:   s%-2d=%2.5f\n",i,time[i],i,s[i]);
  }
 printf("最后算得：smax=%.5f\n",smax);

}