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+Bf=1600//width, in mm
+Df=100//thickness of slab, in mm
+d=400//effective depth, in mm
+Bw=225//breadth of web, in mm
+b=Bw
+W=30//UDL including self-weight, in kN/m
+Ast=5*0.785*22^2//five 22 mm dia bars, in sq mm
+l=9.2//span, in m
+sigma_cbc=5//in MPa
+sigma_sv=230//in MPa
+Fy=415//in MPa
+V=W*l/2//in kN
+Tv=(V*10^3)/(b*d)//in MPa
+Tcmax=1.6//in MPa
+//Tv<Tcmax; OK
+Pt=Ast/(b*d)*100//percentage tensile steel
+//for given Pt and for M15 grade concrete
+Tc=0.44//in MPa
+Vs=V-Tc*b*d/10^3//in kN
+//providing bent-up bars
+Asv=0.785*22^2//in sq mm
+Vs1=Asv*sigma_sv*sind(45)/10^3//in kN
+//but shear taken up by bent-up bar is limited to Vs/2
+Vs1=Vs/2//in kN
+//providing 6 mm dia stirrups, which will take up remaining shear force
+Vs2=Vs-Vs1//in kN
+dia=6//in mm
+Asv=2*0.785*dia^2//in sq mm
+Sv=Asv*sigma_sv*d/(Vs2*10^3)//in mm
+Sv1=105//round-off, in mm
+//Sv<0.75d or 450 mm, whichever is less; hence OK
+//calculating minimum spacing of shear reinforcement
+Sv2=Asv*Fy/(b*0.4)//in mm
+Sv2=260//round-off, in mm
+//to calculate distance 'x' from support where shear stress in concrete is equal to Tc
+x=Tc/Tv*l/2//in m
+mprintf("Provide 6 mm dia stirrups at %d mm c/c upto %f m from both ends\nFor the remaining portion, provide 6 mm dia stirrups at %d mm", Sv1,(l/2-x),Sv2)