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Diffstat (limited to '3761/CH4/EX4.5')
-rw-r--r-- | 3761/CH4/EX4.5/Ex4_5.sce | 67 |
1 files changed, 67 insertions, 0 deletions
diff --git a/3761/CH4/EX4.5/Ex4_5.sce b/3761/CH4/EX4.5/Ex4_5.sce new file mode 100644 index 000000000..27366a9e7 --- /dev/null +++ b/3761/CH4/EX4.5/Ex4_5.sce @@ -0,0 +1,67 @@ +disp("Example 4.5")
+disp("Calculating allowable moment capacity for beam of section as in 4.4")
+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("The neutral axis depth factor, k")
+k=kd1/d
+disp(k,"k=")
+disp("For a balanced section as per Eq 4.23, kb")
+sigmast=130
+kb=280/(280+3*sigmast)
+disp(kb,"kb=")
+disp("As k<kb the section is under-reinforced section(WSM)")
+disp("fst=sgimgast=130MPa (For Fe 250 Steel, dia>20mm")
+fst=sigmast
+fc=(kd1/(d-kd1))*(fst/m)
+fc1=0.526*fc //As derived in previous example 4.4
+disp("MPa",fc,"fc=")
+disp("MPa",fc1,"fc1=")
+disp("Substituting in Eq 4.34")
+fact=0.526
+M=(fc*((0.5*bf*(kd1)*(d-kd1/3)-(0.5*fact*(bf-bw)*(kd1-Df))*(d-Df-((kd1-Df)/3)))))/10^6
+disp("Therefore, Moment carrying capacity Mall=")
+disp("kNm", M , "Mall=")
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