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+fck=15//in MPa
+fy=415//in MPa
+l=4.5//span, in m
+MF=1.4
+a=MF*20
+D=l*10^3/a//in mm
+D=160//in mm
+W1=(D/10^3)*25//self-weight, in kN/m
+W2=1//floor finish, in kN/m
+W3=1//partitions, in kN/m
+W4=4//live load, in kN/m
+W=W1+W2+W3+W4//in kN/m
+Wu=1.5*W//in kN/m
+lef=l+0.16//in m
+Mu=Wu*lef^2/8//in kN-m
+d=sqrt(Mu*10^6/0.138/fck/10^3)//in mm
+dia=12//assume 12 mm dia bars
+D=d+dia/2+15//=160 mm(assumed value), approximately
+D=160//in mm
+d=140//in mm
+//steel
+//Xu=0.87*fy*Ast/0.36/fck/b = a*Ast
+a=0.87*fy/0.36/fck/10^3
+//using Mu=0.87 fy Ast (d-0.416 Xu), we get a quadratic equation
+p=0.87*fy*0.416*a
+q=-0.87*fy*d
+r=Mu*10^6
+Ast=(-q-sqrt(q^2-4*p*r))/2/p//in sq mm
+s1=1000*0.785*dia^2/Ast//in mm
+s1=112//in mm
+pt=Ast/10^3/d*100//in %
+Ads=0.12/100*10^3*D//in sq mm
+//provide 8 mm dia bars
+s2=1000*0.785*8^2/Ads//in mm
+s2=260//in mm
+Vu=Wu*lef/2//in kN
+Tv=Vu*10^3/10^3/d//in MPa
+//for M15 and pt=0.718
+Tc=0.53//in MPa
+//for solid slabs
+Tc=1.25*Tc//in MPa
+//as Tc>Tv, no shear reinforcement required
+mprintf("Summary of design:\nSlab thickness= %d mm\nCover = 15 mm\nMain steel = 12 mm dia @ %d mm c/c\nDistribution steel = 8 mm dia @ %d mm c/c",D,s1,s2)