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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
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+disp("Example 4.10")
+disp("Grade of Steel,fy = Fe250","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=250
+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("mm",xu,"xu<xu,max, therefore xu=")