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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)