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C   this is the calculation process of the number of the aggregates 
      program main 
      print*,'请输入截面高度、宽度' 
      read*,height,broad 
	print*,'请输入最大骨料粒径' 
      read*,dmax 
10    print*,'请输入骨料尺寸' 
      read*,a1,a2,a3 
	print*,a1,a2,a3 
      bi=1/((a3/2)*(a3/2)*3.14159265/broad/height) 
      dd=a1/dmax 
      pc1=0.75*(1.065*dd**0.5-0.053*dd**4-0.012*dd**6-0.0045*dd**8+ 
     & 0.0025*dd**10) 
      print*,pc1 
      dd=a2/dmax 
      pc2=0.75*(1.065*dd**0.5-0.053*dd**4-0.012*dd**6-0.0045*dd**8+ 
     & 0.0025*dd**10) 
      print*,pc2 
      xn=(pc1-pc2)*bi 
      if (xn.LT.(INT(xn)+0.5))then 
      xn=INT(xn) 
      else 
      xn=INT(xn)+1 
      end if 
      print*,'骨料颗粒数' 
	print*,xn 
	print*,'请输入骨料尺寸' 
      read*,a1,a2,a3 
	print*,'骨料的砂浆含量' 
	read*,cm1,cm2,cm3 
	w1=(2600*cm1)/((100-cm1)*2000) 
	w2=(2600*cm2)/((100-cm2)*2000) 
	w3=(2600*cm3)/((100-cm3)*2000) 
	print*,w1,w2,w3 
      h1=(a1*(w1+1)-a1*sqrt(1+w1))/2/(w1+1) 
	h2=(a2*(w2+1)-a2*sqrt(1+w2))/2/(w2+1) 
	h3=(a3*(w3+1)-a3*sqrt(1+w3))/2/(w3+1) 
      h1=INT(h1)+1 
      h2=INT(h2)+1 
      h3=INT(h3)+1 
	write(*,*)'砂浆层厚度' 
	print*,h1,h2,h3 
      goto 10 
      end program