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| author | prashantsinalkar | 2017-10-10 12:27:19 +0530 |
|---|---|---|
| committer | prashantsinalkar | 2017-10-10 12:27:19 +0530 |
| commit | 7f60ea012dd2524dae921a2a35adbf7ef21f2bb6 (patch) | |
| tree | dbb9e3ddb5fc829e7c5c7e6be99b2c4ba356132c /3761/CH4/EX4.11/Ex4_11.sce | |
| parent | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (diff) | |
| download | Scilab-TBC-Uploads-7f60ea012dd2524dae921a2a35adbf7ef21f2bb6.tar.gz Scilab-TBC-Uploads-7f60ea012dd2524dae921a2a35adbf7ef21f2bb6.tar.bz2 Scilab-TBC-Uploads-7f60ea012dd2524dae921a2a35adbf7ef21f2bb6.zip | |
initial commit / add all books
Diffstat (limited to '3761/CH4/EX4.11/Ex4_11.sce')
| -rw-r--r-- | 3761/CH4/EX4.11/Ex4_11.sce | 71 |
1 files changed, 71 insertions, 0 deletions
diff --git a/3761/CH4/EX4.11/Ex4_11.sce b/3761/CH4/EX4.11/Ex4_11.sce new file mode 100644 index 000000000..4bf023034 --- /dev/null +++ b/3761/CH4/EX4.11/Ex4_11.sce @@ -0,0 +1,71 @@ +disp("Example 4.11")
+disp("4.11.a (Referring to Example 4.9)")
+disp("Grade of Steel,fy = Fe415","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=415
+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)
+xu=315
+disp("mm",xu,"xu")
+disp("Taking moments about the tension steel centroid")
+MuR=0.362*fck*b*xu*(d-0.416*xu)/10^6
+disp("kNm",MuR,"MuR=")
+disp("The value of MuR can also be calculated in terms of the steel tensile stress fst, whish is less than 0.87fy, as xu>xu,max. The value of fst obtained from last itteration is obtained as 349MPa, therefore, MuR=fst*Ast*(d-0.416*xu)")
+MuR1=fst*Ast*(d-0.416*xu)/10^6
+disp("kNm",MuR1,"Therefore, MuR=")
+disp("Alternative (using analysis aid)")
+pt=(100*Ast)/(b*d)
+disp(pt,"Referring Table A.2(a)-for M20 concrete and Fe415 steel for pt=")
+disp("MuR/bd^2 for pt,1.18 = 3.145 and for pt, 1.20 = 3.170, therefore, for M20 concrete and Fe415 steel and pt=1.19 MuR/bd^2=")
+MuR1bd2=(3.145+3.170)/2
+MuR1=MuR1bd2*b*d^2/10^6
+disp("kNm",MuR1,"MuR=")
+
+disp("Example 4.11.b, (Refering 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=")
+disp("Taking moments about the tension steel centroid")
+MuR2=0.362*fck*b*xu*(d-0.416*xu)/10^6
+disp("kNm",MuR2,"MuR=")
+
+disp("Alternatively, as xu<xu,max, it follows that fst=0.87fy, and")
+MuR3=0.87*Ast*fy*(d-0.416*xu)/10^6
+disp("kNm",MuR3,"MuR=")
+disp("pt=1.190 as calculated above")
+disp("Referring to table A.2(a) for M20 concrete and Fe 250 Steel, for pt=1.180 Mu/bd^2= 2.188 and for pt= 1.20 it is = 2.219, therefore for pt=1.19, Mu/bd^2=")
+MuRbd2=(2.188+2.219)/2
+MuR4=MuRbd2*b*d^2/10^6
+disp("kNm",MuR4)
+
+
+
+
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