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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/EX4.4/Ex4_4.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/EX4.4/Ex4_4.sce')
| -rw-r--r-- | 3761/CH4/EX4.4/Ex4_4.sce | 62 |
1 files changed, 62 insertions, 0 deletions
diff --git a/3761/CH4/EX4.4/Ex4_4.sce b/3761/CH4/EX4.4/Ex4_4.sce new file mode 100644 index 000000000..025fbc195 --- /dev/null +++ b/3761/CH4/EX4.4/Ex4_4.sce @@ -0,0 +1,62 @@ +L=6
+Df=100
+bw=250
+b=1000
+D=600
+d=520
+M=200
+disp("Example 4.4")
+disp("Service Load Moment = 200kNm","Ast=6-28mmdia bar","Effective depth d=520 mm","Depth D=600mm","b=1000mm","bw=250mm","Depth of Flange Df=100mm","Span of Beam L=6m","Given Data:")
+disp("Grade of Steel =Fe250","Grade of Concrete=M20")
+disp("Verifying for the effective flange width b")
+disp("Refering IS456:2000, Clause 23.1.2 c or Eq 4.30(b) from TB")
+disp("For T-beams, bf = (lo/((lo/b)+4)+bw")
+disp("bw=250mm","lo=6000mm")
+lo=6000
+bf=((lo)/((lo/b)+4)+bw)
+disp("mm",bf,"bf=")
+
+if(bf<b)
+ disp(bf,"Since, bf<b,b=")
+
+sigmacbc=7
+m=280/(3*sigmacbc)
+disp(m,"m=")
+dia=28
+Ast= %pi*dia*dia*6/4
+mAst=m*Ast
+disp("Assuming kd</Df, and equating moments of compression and transformed tension areas about the neutral axis,")
+disp("bf*(kd)^2/2 = m*Ast*(d-kd)")
+b1=(2*mAst/bf)
+c=(-mAst*2*d/bf)
+a=1
+kd1=((-b1+sqrt((b1*b1)-(4*a*c)))/(2*a))
+kd2=((-b1-sqrt((b1*b1)-(4*a*c)))/(2*a))
+disp("mm",kd1,"therefore, kd=")
+disp("kd1>Df, the assumption kd</Df is incorrect")
+disp(" For kd>Df, neutral axis located in the web")
+disp("Using Equation 4.31 of TB")
+disp("(bf-bw)*Df*(kd-Df/2)+ bw*(kd)^2/2 = mAst*(d-kd)")
+a=bw/2
+B=(bf*Df-bw*Df+mAst)
+c=(bw*Df*Df/2 - bf*Df*Df/2-mAst*d)
+
+Dis=(B*B)-(4*a*c)
+kd1=((-B+sqrt(Dis))/(2*a))
+disp("mm",kd1,"kd=")
+disp("Relating the compressive stress fc1 at the flange bottom to fc,")
+disp("fc1=fc*((kd-Df)/kd)")
+fact=((kd1-Df)/kd1)
+disp("*fc",fact,"fc1=")
+disp("Compressive Force C= 0.5*fc*bf*(kd)-0.5*fc1*(bf-bw)*(kd-Df)")
+disp("Taking moments of forces about the tension steel centriod, using equation 4.34")
+fc=((M*10^6)/((0.5*bf*(kd1)*(d-kd1/3)-(0.5*fact*(bf-bw)*(kd1-Df))*(d-Df-((kd1-Df)/3)))))
+disp("MPa",fc,"Therefore, on solving we get, fc=")
+disp("Applying C= T")
+fst= (0.5*fc*bf*(kd1)-0.5*fact*fc*(bf-bw)*(kd1-Df))/Ast
+disp("MPa",fst,"Therefore, fst = ")
+disp("From the stress distribution diagram: fst = m*fc*((d-kd)/kd)")
+fst1=m*fc*((d-kd1)/kd1)
+disp("MPa",fst1, "As before")
+
+
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