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(bfDf, the assumption kdDf, 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 k20mm") 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=")