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Bf=1200//width of flange, in mm
Bw=300//breadth of web, in mm
Df=120//thickness of flange, in mm
d=500//effective depth, in mm
sigma_cbc=7//in MPa
sigma_st=190//in MPa
m=13.33//modular ratio
Ast=5*0.785*20^2//five 20 mm dia bars, in sq mm
l=6//span, in m
//assume depth of neutral axis is less than or equal to thickness of flange; find x using Bf(x^2)/2=mAst(d-x), which becomes of the form px^2+qx+r=0
p=Bf/2
q=m*Ast
r=-m*Ast*d
//solving quadratic equation
x=(-q+sqrt(q^2-4*p*r))/(2*p)//in mm
//x < Df; hence our assumption is correct
//to find critical depth of neutral axis
Xc=d/(1+sigma_st/(m*sigma_cbc))//in mm
//as x<Xc, beam is under-reinforced
sigma_cbc=sigma_st/m*x/(d-x)//in MPa
//taking moments about tensile steel
Mr=Bf*x*sigma_cbc*(d-x/3)/2//in N-mm
W=(Mr/10^6)*8/l^2//in kN/m
mprintf("Moment of resistance of the beam=%f kN-m\nCapacity to take uniformly distributed load(including self-weight)=%f kN/m", Mr/10^6,W)
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