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+disp("Example 4.12")

+disp("fck=20MPa","fy=Fe250","Ast=3695mm^2","d=520mm","bw=250mm","Df=100mm","bf=850mm","Given:")

+bf=850

+Df=100

+bw=250

+d=520

+Ast=3695

+fy=250

+fck=20

+Es=2*10^5

+xumaxd=0.0035/(0.0055+0.87*(fy/Es))

+xumax=xumaxd*d

+disp("First assuming xu</Df and xu</xu,max and considering force equilibrium")

+disp("Cu=Tu=0.362*fck*bf*xu=0.87*fy*Ast")

+disp("Therefore,xu=")

+xu=(0.87*fy*Ast)/(0.362*bf*xu*fck)

+disp(xu,"mm")

+disp("Since, xu>Df, the value is incorrect ")

+

+disp("Asxu>Df, the compression in web is given by: Cuw=0.362*fck*bw*xu")

+Cuw=0.362*fck*bw

+disp("*xu N",Cuw)

+disp("Assuming xu>/7/3*Df = 233.3, the commpression in the flange is given by: Cuf=0.447*fck*(bf-bw)*Df")

+Cuf=0.447*fck*(bf-bw)*Df

+disp("Cuf=")

+disp("N",Cuf)

+disp("Also assuming xu</xu,max = 276.1mm fst= 0.87*fy, and Tu=0.87*fy*Ast")

+Tu=0.87*fy*Ast

+disp("Tu=")

+disp("N",Tu)

+disp("Aplying the equilibrium conditions (Cuw+Cuf=Tu")

+xu=(Tu-Cuf)/Cuw

+disp("Since, xu<7/3*Df, hence this value of xu is also not correct")

+disp("As Df<xu<7/3*Df, the depth yf</Df of the equivalent concrete stress block is obtained as: yf=0.15*xu+0.65*Df = (0.15*xu+65)mm")

+disp("Cuf=Cuf*(yf/Df)= (804.6*xu+348660)N")

+disp("Cuw+Cuf=Tu")

+xu1=(Tu-(Cuf*(65/100)))/(Cuw+(Cuf*0.15/100))

+disp("xu=")

+disp("mm",xu1)

+disp("As, xu<xu,max; Hence the assumption fst=0.87*fy is OK")

+yf=(0.15*xu1)+65

+disp("mm",yf,"yf=")

+disp("Taking moments of Cuw and Cuf about the centroid of tension steel, MuR= Cuw*(d-0.416*xu)+Cuf*(d-yf/2)")

+Cuf=804.6*xu1+348660

+MuR=(Cuw*xu1*(d-0.416*xu1)+(Cuf*(d-yf/2)))/10^6

+disp("kNm",MuR,"MuR=")

+

+

+

+