diff options
Diffstat (limited to '3761/CH4/EX4.13')
-rw-r--r-- | 3761/CH4/EX4.13/Ex4_13.sce | 50 |
1 files changed, 50 insertions, 0 deletions
diff --git a/3761/CH4/EX4.13/Ex4_13.sce b/3761/CH4/EX4.13/Ex4_13.sce new file mode 100644 index 000000000..e9b3561e5 --- /dev/null +++ b/3761/CH4/EX4.13/Ex4_13.sce @@ -0,0 +1,50 @@ +disp("Example 4.13")
+disp("fck=20MPa","fy=Fe250","Ast=4926mm^2","d=520mm","bw=250mm","Df=100mm","bf=850mm","Given:")
+bf=850
+Df=100
+bw=250
+d=520
+Ast=4926
+fy=250
+fck=20
+Es=2*10^5
+xumaxd=0.0035/(0.0055+0.87*(fy/Es))
+xumax=xumaxd*d
+disp("mm",xumax,"xumax=")
+disp("First assuming xu</Df and xu</xu,max")
+disp("xu=(0.87*fy*Ast)/(0.362*fck*bf)")
+xu=(0.87*fy*Ast)/(0.362*fck*bf)
+disp("mm",xu,"xu=")
+disp("xu >Df, Hence this value of xu is not correct")
+disp("As xu>Df, Cuw = 0.362*fck*fy*bw*xu")
+Cuw=0.362*fck*fy
+disp("xu N", Cuw,"Cuw=")
+disp("ASssuming xu>/7/3*Df = 233.33mm, yf=Df=100mm and Cuf=0.447*fck*(bf-bw)*Df")
+Cuf=0.447*fck*(bf-bw)*Df
+disp("N",Cuf,"Cuf=")
+disp("Further assuming xu</xu,max = 276.1 mm, fst=0.87*fy, and")
+Tu=0.87*fy*Ast
+disp("N",Tu,"Tu=")
+disp("Applying the force equilibrium condition Cuw+Cuf=Tu")
+xu=(Tu-Cuf)/Cuw
+disp("mm",xu,"xu=")
+disp("which implies xu>7/3Df =233.3mm, but not xu</xu,max=276.1mm. The section is over-reinforcedas per the Code provisions")
+
+disp("Exact Solution (considering strain compatibility)")
+disp("mm",xu,"Corresponding to xu=")
+disp("Est=0.0035*(d/xu-1)")
+Est=0.0035*(d/xu-1)
+disp(Est,"Est=")
+strainyield=0.87*fy/Es
+disp("Est is greater than strain at yield for Fe250")
+disp(strainyield)
+disp("Hence the design steel stress isindeed fst=0.87*fy and the so calculated xu above, is the correct depth of the neutral axis")
+disp("Accordingly,MuR= Cuw*(d-0.416*xu)+Cuf*(d-Df/2)")
+MuR=(Cuw*xu*(d-0.416*xu)+Cuf*(d-Df/2))/10^6
+disp("kNm",MuR,"MuR=")
+
+disp("APPROXIMATE SOLUTION")
+disp("Limiting xu to xu,max=276.1 mm and taking moments of Cuw and Cuf about the centroid of the tension steel.(Note that, following the Code procedure, Df/d=100/520=0.192<0.2, yf=Df=100mm")
+xumax
+MuRl=((Cuw*xumax*(d-0.416*xumax))+(Cuf*(d-Df/2)))/10^6
+disp("kNm",MuRl,"MuR,lim=")
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