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diff --git a/3761/CH4/EX4.9/Ex4_9.sce b/3761/CH4/EX4.9/Ex4_9.sce
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+disp("Example 4.9")
+disp("Grade of Steel,fy = Fe415","Grade of Concrete,fck = M20","D=600mm","d=550mm","b=300mm","Bars used = 4 - 25 dia")
+b=300
+d=550
+D=600
+fck=20
+Ast=%pi*4*25*25/4
+disp("mm^2",Ast,"Ast=")
+disp("For Fe415 Steel,")
+Es=2*10^5
+fy=415
+Est=0.87*fy/Es
+xumaxd=(0.0035/(0.0055+Est))
+disp(xumaxd,"xumax/d")
+xumax=xumaxd*d
+disp("mm",xumax,"xu,max=")
+disp("Assuming, xu</xu,max and applying the force equilibrium condition Cu=Tu")
+xu= (0.87*fy*Ast)/(0.362*fck*b)
+disp("mm",xu,"xu")
+disp("xu>xu,max, 326.3mm>263.5mm")
+disp("As xu>xu,max steel would not have yielded accordingly the strain compatibility method is adopted to obtain the correct value of xu")
+disp("FIRST CYCLE")
+disp("1. Assume xu = (xu+xu,max)/2 ")
+xu1=(xu+xumax)/2
+disp("mm",xu1,"xu,1=")
+disp("2. Strain Compatibility = Est = 0.0035*(d/xu1-1)")
+//Est=strainst, ephselon st
+Est =0.0035*(d/xu1-1)
+disp(Est,"Est=")
+disp(Est,"Interpoating for value of fst, corresponding to strain ,Fe415 and Est = ")
+disp("For strain, 0.00276 fst = 351.8 and for strain >/0.00380 fst=360.9  From table 3.2")
+fst1=351.8
+fst2=360.9
+disp("fst= ")
+fst=fst1+((fst2-fst1)*((Est*10^5-276)/(380-276)))
+disp("MPa",fst,"fst=")
+disp("Cu=Tu")
+xu2=fst*(Ast/(0.362*fck*b))
+disp("mm",xu2,"xu,2=")
+
+disp("SECOND CYCLE")
+disp("Assume xu= ")
+xu3=(xu2+xu1)/2
+disp("mm",xu3,"xu,3=")
+Est1=0.0035*(d/xu3-1)
+disp(Est1,"Est=")
+disp(Est,"Interpoating for value of fst, corresponding to strain ,Fe415 and Est = ")
+disp("For strain, 0.00276 fst = 351.8 and for strain 0.00241 fst=342.8  From table 3.2")
+fst4=351.8
+fst3=342.8
+fst11=(fst3+(fst4-fst3)*((Est1*10^5-241)/(276-241)))
+disp("MPa",fst11,"fst1=")
+
+disp("Cu=Tu")
+fact=Ast/(0.362*fck*b)
+//disp(fact)
+xu4=fst11*(fact)
+disp("mm",xu4,"xu,4=")
+
+
+disp("THIRD CYCLE")
+disp("1.Assume xu=")
+xu5=(xu4+xu3)/2
+disp("mm",xu5,"xu,5=")
+Est2=0.0035*(d/xu5-1)
+disp(Est2, "Est=")
+disp(Est,"Interpoating for value of fst, corresponding to strain ,Fe415 and Est = ")
+disp("For strain, 0.00276 fst = 351.8 and for strain 0.00241 fst=342.8  From table 3.2")
+fst4=351.8
+fst3=342.8
+fst12=(fst3+(fst4-fst3)*((Est2*10^5-241)/(276-241)))
+disp("MPa",fst12,"fst2=")
+
+disp("Cu=Tu")
+fact=Ast/(0.362*fck*b)
+//disp(fact)
+xu6=fst12*(fact)
+disp("mm",xu6,"xu,6=")
+disp("Therefore, the final value of xu may be takaen as xu=315mm")
+