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diff --git a/3683/CH12/EX12.2/Ex12_2.sce b/3683/CH12/EX12.2/Ex12_2.sce
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+sigma_cbc=5//in MPa
+sigma_st=140//in MPa
+phi=35//angle of repose, in degrees
+H=6//height of wall, in m
+B=0.4*H//assume, in m
+T=B/4//assume toe to base ratio as 1:4
+W=18//density of retained earth, in kN/cu m
+P=W*H^2/2*(1-sind(phi))/(1+sind(phi))//in kN
+P=88//assume, in kN
+M1=P*H/3//in kN-m
+//bending moment at 3 m below the top
+h=3//in m
+M2=W*h^2/2*(1-sind(phi))/(1+sind(phi))*h/3//in kN-m
+M2=22//in kN-m
+//thickness of stem (at the base)
+d=sqrt(M1*10^6/0.87/1000)//in mm
+d=450//in mm
+dia=20//assume 20 mm dia bars
+D1=d+dia/2+25//in mm
+D2=200//thickness at top, in mm
+D3=D2+(D1-D2)*h/H//in mm
+d3=sqrt(M2*10^6/0.87/1000)//in mm
+D3=d3+dia/2+25//< 342.5 mm (provided), hence OK
+D3=342.5//in mm
+d3=D3-dia/2-25//in mm
+//main steel
+//(a) 6 m below the top
+Ast=M1*10^6/sigma_st/0.87/d//in sq mm
+//provide 20 mm dia bars
+s1=1000*0.785*20^2/Ast//in mm
+s1=95//assume, in mm
+//(b) 3 m below the top
+Ast=M2*10^6/sigma_st/0.87/d3//in sq mm
+//provide 10 mm dia bars
+s2=1000*0.785*10^2/Ast//in mm
+s2=130//assume, in mm
+//distribution steel
+Ads=0.15/100*10^3*D3//in sq mm
+//provide 10 mm dia bars
+s3=1000*0.785*10^2/Ads//in mm
+s3=150//assume, in mm
+//check for shear
+V=P//in kN
+Tv=V*10^3/10^3/d//in MPa
+//for M15 grade concrete and pt=0.71
+Tc=0.34//in MPa
+//as Tc > Tv, no shear reinforcement required
+//development length
+//(a) At the base of stem
+dia=20//in mm
+Tbd=0.6//in MPa
+Ld=dia*sigma_st/4/Tbd//in mm
+Ld=1170//assume, in mm
+//(b) At 3 m below the top
+dia=10//in mm
+Ld=dia*sigma_st/4/Tbd//in mm
+Ld=590//assume, in mm
+//check for stability
+D4=500//thickness of base, in mm (assume)
+V1=1/2*(D1-D2)/10^3*H*25//in kN
+V2=(D2/10^3)*H*25//in kN
+V3=(D4/10^3)*B*25//weight of base, in kN
+V4=(B-T-D1/10^3)*H*W//in kN
+V=V1+V2+V3+V4//in kN
+M=V1*(T+2/3*(D1-D2)/10^3)+V2*(T+(D1-D2)/10^3+D2/10^3/2)+V3*B/2+V4*(B-(B-T-D1/10^3)/2)//in kN-m
+x=M/V//in m
+//factor of safety
+//for overturning
+F1=V*x/P/(H/3)//> 1.5, hence OK
+mu=0.5
+//for sliding
+F2=mu*V/P//< 1.5, hence it is not safe against sliding
+mprintf("Summary of design:\nThickness of stem (at base) = %d mm\nThickness of stem at top = %d mm\nRefer Fig. 12.7 of textbook for reinforcement details",D1,D2)
+//answers in textbook for factor of safety against overturning and sliding are incorrect