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diff --git a/3683/CH19/EX19.4/Ex19_4.sce b/3683/CH19/EX19.4/Ex19_4.sce
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+fck=15//in MPa
+fy=415//in MPa
+MF=1.4//modification factor
+//let a be span to depth ratio
+l=1//span, in m
+a=MF*7
+D=l*1000/a//in mm
+D=105//assume, in mm
+//to calculate loading
+W1=25*(D/10^3)*1.5//self-weight, in kN/m
+W2=0.5*1.5//finish, in kN/m
+W3=0.75*1.5//live load, in kN/m
+W=W1+W2+W3//in kN/m
+Wu=1.5*W//in kN/m
+lef=l+0.23/2//effective span, in m
+Mu=Wu*lef/2//in kN-m
+//check for depth
+d=sqrt(Mu*10^6/(0.138*fck*1500))//in mm
+dia=12//assume 12 mm dia bars
+D=d+12/2+15//<105, hence OK
+D=100//assume, in mm
+d=D-dia/2-15//in mm
+//steel
+//Xu=0.87*fy*Ast/0.36/fck/b = a*Ast
+a=0.87*fy/0.36/fck/1.5/10^3
+//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
+//provide 8 mm dia bars
+dia=8//in mm
+s1=1500*0.785*dia^2/Ast//>3d=3x79=237 mm
+s1=235//in mm
+Ads=0.12/100*1000*D//distribution steel, in sq mm
+//assume 6 mm dia bars
+s2=1000*0.785*6^2/Ads//in mm
+s2=235//round-off, in mm
+Tbd=1.6//in MPa
+Ld=dia*0.87*fy/4/Tbd//in mm
+Ld=452//in mm
+Tv=Wu*10^3/1500/d//in MPa
+Ast=1500*0.785*8^2/235//in sq mm
+pt=Ast/1500/d*100//in %
+//for M15 and pt=0.26
+Tc=0.35//in MPa
+//as Tc>Tv, no shear reinforcement required
+mprintf("Summary of design\nThickness of slab = %d mm\nCover = 15 mm\nMain steel = 8 mm dia @ %d mm c/c\nDevelopment length = %d mm\nDistribution steel = 6 mm dia @ %d mm c/c",D,s1,Ld,s2)