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+disp("Example 4.15")
+disp("Asc=2-25dia bars","Ast=3-36dia bars","M20 Grade of concrete and Fe250steel","dd = 50mm","d=550mm","b=300mm","Given:")
+b=300
+d=550
+dd=50
+Ast= %pi*36*36*3/4
+Asc=2*%pi*25*25/4
+m=13.33 //(280/(3*sigmacbc))
+Es=2*10^5
+fck=20
+fy=415
+xumaxd=0.0035/(0.0055+(0.87*fy)/Es)
+disp(xumaxd,"xumax/d for Fe250=")
+xumax=xumaxd*d
+disp("mm",xumax,"xu,max")
+disp("Assuming fsc=fst=0.87*fy,, and considering force equilibrium")
+disp("Cus+Cuc = Tu")
+Cuc=0.362*fck*b
+Cus=(0.87*fy-0.447*fck)*Asc
+Tu=0.87*fy*Ast
+xu=(Tu-Cus)/Cuc
+disp("mm",xu,"xu=")
+disp("xu>xu,max, hence the section is over-reinforced")
+disp("Exact Solution considering strain compatibility")
+disp("Applying Eq. 4.81: xu = fst*Ast - (fsc-0.447*fck)*Asc/(0.362*fck*b)")
+disp("Therefore,xu=(3054*fst - 982*fsc+8779)/2172")
+
+disp("First Cycle")
+disp("1. xu lies within the two limits above; 263.5 mm < xu < 348.5mm")
+disp("2. xu = (xu,max+xu)/2")
+xu1=(xumax+xu)/2
+disp("mm",xu1,"xu=")
+disp("3.Esc = 00035*(1-dd/xu1)")
+Esc = 0.0035*(1-dd/xu1)
+disp(Esc,"Esc=")
+disp("4.Est = 0.0035*(d/xu1-1)")
+Est = 0.0035*(d/xu1-1)
+disp(Est,"Est=")
+disp("for Esc= 0.00380 fsc = 360.9 and for Esc = 0.00276 fsc=351.8")
+fst1=351.8
+fst2=360.9
+fsc=fst1+((fst2-fst1)*((Esc*10^5-276)/(380-276)))
+disp("MPa",fsc,"fsc=")
+fst=fst1+((fst2-fst1)*((Est*10^5-276)/(380-276)))
+disp("MPa",fst,"fst=")
+disp("Therefore, xu = ")
+xu2=(3054*fst - 982*fsc+8779)/2172
+disp("mm",xu2,"xu=")
+
+disp("Second Cycle")
+disp("Assume xu= ")
+xu3=(xu2+xu1)/2
+disp("mm",xu3,"xu=")
+Esc = 0.0035*(1-dd/xu3)
+disp(Esc,"Esc=")
+Est1=0.0035*(d/xu3-1)
+disp(Est1,"Est=")
+disp("for Esc= 0.00380 fsc = 360.9 and for Esc = 0.00276 fsc=351.8")
+fst1=351.8
+fst2=360.9
+fsc=fst1+((fst2-fst1)*((Esc*10^5-276)/(380-276)))
+disp("MPa",fsc,"fsc=")
+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=")
+xu4=(3054*fst11- 982*fsc+8779)/2172
+disp("mm",xu4,"xu=")
+
+disp("Third Cycle")
+disp("1.Assume xu=")
+xu5=(xu3+xu4)/2
+disp("mm",xu5,"xu=")
+Esc = 0.0035*(1-dd/xu5)
+disp(Esc,"Esc=")
+Est2=0.0035*(d/xu5-1)
+disp(Est2, "Est=")
+disp("for Esc= 0.00380 fsc = 360.9 and for Esc = 0.00276 fsc=351.8")
+fst1=351.8
+fst2=360.9
+fsc=fst1+((fst2-fst1)*((Esc*10^5-276)/(380-276)))
+disp("For strain, 0.00276 fst = 351.8 and for strain 0.00241 fst=342.8  From table 3.2")
+fst12=342.8
+disp("MPa",fst12,"fst2=")
+xu6=(3054*fst12- 982*fsc+8779)/2172
+disp("mm",xu6,"xu,final=")
+Cuc=0.362*fck*b
+Cus=(fsc-0.447*fck)*Asc
+MuR=(Cuc*xu6*(d-0.416*xu6)+Cus*(d-dd))/10^6
+disp("kNm",MuR,"MuR,final=")
+
+disp("Approximate Solution")
+disp("As an approximate and conservative estimate limiting xu to xu,max=263.5mm,")
+Esc=0.0035*(1-dd/xumax)
+disp(Esc,"Esc=")
+fsc=352.5
+disp("MPa",fsc,"fsc=")
+disp("This value is alternatively obtainable from Table 4.5 for dd/d=0.09 and Fe415")
+disp("Accordingly, limiting the ultimate moment of resistance MuR to the limiting moment Mu,lim")
+Mulim=(0.362*fck*b*xumax*(d-0.416*xumax)+(fsc-0.447*fck)*Asc*(d-dd))/10^6
+disp("kNm",Mulim,"Mu,lim=")