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diff --git a/3683/CH19/EX19.6/Ex19_6.sce b/3683/CH19/EX19.6/Ex19_6.sce
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+b=225//width in mm
+D=300//depth in mm
+fck=15//in MPa
+fy=415//in MPa
+l=4.2//span, in m
+W1=(b/10^3)*(D/10^3)*25//self-weight, in kN/m
+W2=6//live load, in kN/m
+W=W1+W2//in kN/m
+Wu=1.5*W//in kN/m
+Mu=Wu*l^2/8//in kN-m
+d=270//assume, in mm
+Mulim=0.138*fck*b*d^2/10^6//in kN-m
+//as Mulim > Mu, it will be a singly reinforced beam
+//Xu=0.87*fy*Ast/0.36/fck/b = a*Ast
+a=0.87*fy/0.36/fck/b
+//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 12 mm dia bars
+n=Ast/0.785/12^2
+n=3//assume
+Ast=n*0.785*12^2//in sq mm
+Vu=Wu*l/2//in kN
+Tv=Vu*10^3/b/d//in MPa
+pt=Ast/b/d*100//pt=0.56
+//for M15 and pt=0.56
+Tc=0.46//in MPa
+//as Tc>Tv, no shear reinforcement required
+//provide nominal stirrups and provide 6 mm stirrups
+Asv=2*0.785*6^2//in sq mm
+Sv=Asv*fy/0.4/b//in mm
+Sv=260//assume, in mm
+Svmax=0.75*d//in mm
+Svmax=200//round-off, in mm
+Sv=min(Sv,Svmax)//in mm
+mprintf("Summary of design:\nBeam size - %d x %d mm\nCover - 25 mm\nSteel - %d-12 mm dia bars\nStirrups - 6 mm dia @ %d mm c/c",b,D,n,Sv)
+//deflection check
+Ec=5700*sqrt(fck)//in MPa
+Es=2*10^5//in MPa
+m=Es/Ec
+fcr=0.7*sqrt(fck)//in MPa
+//using b x x/2 = m Ast (d-x), we get a quadratic equation
+//solving the quadratic equation
+p=b/2
+q=m*Ast
+r=-m*Ast*d
+x=(-q+sqrt(q^2-4*p*r))/2/p//in mm
+z=d-x/3//in mm
+Ir=b*x^3/12+b*x*(x/2)^2+m*Ast*(d-x)^2//in mm^4
+Igr=b*D^3/12//in mm^4
+yt=D/2//in mm
+Mr=fcr*Igr/yt//in N-mm
+M=W*l^2/8*10^6//in N-mm
+Ieff=Ir/(1.2-Mr/M*z/d*(1-x/d)*b/b)//in mm^4
+//Ir<Ieff<Igr, hence OK
+W1=W*l//in kN
+u1=5/384*(W1*10^3)*(l*10^3)^3/Ec/Ieff//short-term deflection, in mm
+//long-term deflection
+//(i) deflection due to shrinkage
+k3=0.125//for simply supported beam
+pt=0.56//in %
+pc=0//in %
+k4=0.72*(pt-pc)/sqrt(pt)
+phi=k4*0.0003/D
+u2=k3*phi*(l*10^3)^2//in mm
+//(ii) deflection due to creep
+Ecc=Ec/(1+1.6)//in MPa
+//assuming a permanent load of 60%
+W2=0.6*W*l//in kN
+u3=5/384*(W2*10^3)*(l*10^3)^3/Ecc/Ieff//in mm
+u4=5/384*(W2*10^3)*(l*10^3)^3/Ec/Ieff//in mm
+u5=u3-u4//in mm
+u=u1+u2+u5//total deflection, in mm
+v1=l*10^3/250//permissible deflection, in mm
+v2=l*10^3/350//in mm
+//assuming half the shrinkage strain occurs within the first 28 days, the deflection occurring after this time
+v3=u2/2+u5//< permissible value, hence OK