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+b=0.2//column width, in m
+D=0.3//column depth, in m
+fck=15//in MPa
+fy=415//in MPa
+P1=600//load on column, in kN
+P2=0.05*P1//weight of footing, in kN
+P=P1+P2//in kN
+Pu=1.5*P//in kN
+q=150//bearing capacity of soil, in kN/sq m
+qu=2*q//ultimate bearing capacity of soil, in kN/sq m
+A=Pu/qu//in sq m
+L=sqrt(A)//assuming footing to be square, in m
+L=1.8//round-off, in m
+p=P1*1.5/L^2//soil pressure, in kN/sq m
+p=277.8//round-off, in kN/sq m
+bc=b/D
+ks=0.5+bc//>1
+ks=1
+Tc=0.25*sqrt(fck)*10^3//in kN/sq m
+Tv=Tc
+//let d be the depth of footing in metres
+//case I: consider greater width of shaded portion in Fig. 19.6 of textbook
+d1=L*(L-b)/2*p/(Tc*L+L*p)//in m
+//case II: refer Fig. 19.7 of textbook; we get a quadratic equation of the form e d^2 + f d + g = 0
+e=p+4*Tc
+f=b*p+D*p+2*(b+D)*Tc
+g=-(L^2-b*D)*p
+d2=(-f+sqrt(f^2-4*e*g))/2/e//in m
+d2=0.35//round-off, in m
+//bending moment consideration, refer Fig. 19.8 of textbook
+Mx=1*((L-b)/2)^2/2*p//in kN-m
+My=1*((L-D)/2)^2/2*p//in kN-m
+d3=sqrt(Mx*10^6/0.138/fck/10^3)//<350 mm, hence OK
+//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*d2*10^3
+r=Mx*10^6
+Ast=(-q-sqrt(q^2-4*p*r))/2/p//in sq mm
+Ast=L*Ast//steel required for full width of 1.8 m
+//provide 12 mm dia bars
+dia=12//in mm
+n=Ast/0.785/dia^2//no. of 12 mm dia bars
+n=12//round-off
+Tbd=1.6//in MPa
+Ld=dia*0.87*fy/4/Tbd//in mm
+Ld=677//assume, in mm
+//this length is available from the face of the column in both directions
+D=d2*10^3+dia/2+100//in mm
+mprintf("Summary of design:\nOverall depth of footing=%d mm\nCover=100 mm\nSteel-%d bars of 12 mm dia both ways",D,n)