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+disp("Example 4.7")
+disp("M20 Grade of concrete and Fe250steel","sigmast=130MPa","Sigmacbc=7 MPa","dd = 50mm","d=550mm","b=300mm","Given:")
+b=300
+d=550
+dd=50 //Mentioning the top cover d' as dd throughout the example
+sigmacbc=7
+sigmast=130
+disp("Transformed Section Properties")
+m=13.33 //(280/(3*sigmacbc))
+disp("Neutral Axis Depth, here in this case k=kb, corresponding to balanced section")
+disp("For a balanced (WSM) section, kb= 280/(280+3*sigmast)")
+kb=280/(280+(3*sigmast))
+disp(kb,"kb=")
+kbd=kb*d
+disp("mm",kbd,"kb d=")
+disp("Considering moments of areas about the neutral axis,")
+disp("b*kb d^2/2 + CSA*(kb d-dd) = mAst*(d-kb d)")
+disp("On replacing values stated above we get equation as : Ast = (0.8Asc + 1856)mm^2")
+disp("Asc is to be determined using equation 4.39 as stated")
+disp("M= Cc(d-kb d/3)+Cs(d-dd)")
+disp("Cc=0.5*fc*b*(kb.d)")
+disp("Cs=(1.5*m-1)*Asc*fc*((kd-dd)/kd)")
+M=175*10^6
+fc=7
+Cc=0.5*fc*b*(kbd)
+Cs1=(1.5*m-1)*fc*((kbd-dd)/kbd)
+Asc=(M-(Cc*(d-kbd/3)))/(Cs1*(d-dd))
+disp("mm^2",Asc,"Therefore, Asc=")
+Ast=(0.8*Asc+1856)
+disp("mm^2", Ast,"Therefore, Ast=")
+
+disp("Alternate Solution to Example 4.7")
+disp("Calculating Mwb=0.5*kb*(1-kb/3)*sigmacbc*b*d*d")
+Mwb=0.5*kb*(1-kb/3)*sigmacbc*b*d*d/10^6
+disp("kNm",Mwb,"Mwb=")
+disp("Calculating Ast1=Mwb/(sigmast*d*(1-kb/3)")
+Ast1=(Mwb*10^6)/(sigmast*d*(1-kb/3))
+disp("mm^2",Ast1,"Ast1=")
+disp("Calculating Ast2, Ast2= (M-Mwb)/(sigmast*(d-dd))")
+Ast2= (M-Mwb*10^6)/(sigmast*(d-dd))
+disp("mm^2",Ast2,"Ast2=")
+disp("Therefore, Ast= Ast1+Ast2")
+Astf=Ast1+Ast2
+disp("mm^2",Astf,"Therefore, final Ast=")
+disp("Calculating fcsc=sigmacbc*(1-dd/kbd)")
+fcsc=sigmacbc*(1-dd/kbd)
+disp(fcsc)
+disp("Calculating Asc=(M-Mwb)/(1.5*m-1)*fcsc*(d-dd)")
+Asc=(M-Mwb*10^6)/((1.5*m-1)*fcsc*(d-dd))
+disp("mm^2",Asc,"Asc=")