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+b=225//width in mm

+D=300//depth in mm

+fck=15//in MPa

+fy=415//in MPa

+l=4.2//span, in m

+W1=(b/10^3)*(D/10^3)*25//self-weight, in kN/m

+W2=6//live load, in kN/m

+W=W1+W2//in kN/m

+Wu=1.5*W//in kN/m

+Mu=Wu*l^2/8//in kN-m

+d=270//assume, in mm

+Mulim=0.138*fck*b*d^2/10^6//in kN-m

+//as Mulim > Mu, it will be a singly reinforced beam

+//Xu=0.87*fy*Ast/0.36/fck/b = a*Ast

+a=0.87*fy/0.36/fck/b

+//using Mu=0.87 fy Ast (d-0.416 Xu), we get a quadratic equation

+p=0.87*fy*0.416*a

+q=-0.87*fy*d

+r=Mu*10^6

+Ast=(-q-sqrt(q^2-4*p*r))/2/p//in sq mm

+//provide 12 mm dia bars

+n=Ast/0.785/12^2

+n=3//assume

+Ast=n*0.785*12^2//in sq mm

+Vu=Wu*l/2//in kN

+Tv=Vu*10^3/b/d//in MPa

+pt=Ast/b/d*100//pt=0.56

+//for M15 and pt=0.56

+Tc=0.46//in MPa

+//as Tc>Tv, no shear reinforcement required

+//provide nominal stirrups and provide 6 mm stirrups

+Asv=2*0.785*6^2//in sq mm

+Sv=Asv*fy/0.4/b//in mm

+Sv=260//assume, in mm

+Svmax=0.75*d//in mm

+Svmax=200//round-off, in mm

+Sv=min(Sv,Svmax)//in mm

+mprintf("Summary of design:\nBeam size - %d x %d mm\nCover - 25 mm\nSteel - %d-12 mm dia bars\nStirrups - 6 mm dia @ %d mm c/c",b,D,n,Sv)

+//deflection check

+Ec=5700*sqrt(fck)//in MPa

+Es=2*10^5//in MPa

+m=Es/Ec

+fcr=0.7*sqrt(fck)//in MPa

+//using b x x/2 = m Ast (d-x), we get a quadratic equation

+//solving the quadratic equation

+p=b/2

+q=m*Ast

+r=-m*Ast*d

+x=(-q+sqrt(q^2-4*p*r))/2/p//in mm

+z=d-x/3//in mm

+Ir=b*x^3/12+b*x*(x/2)^2+m*Ast*(d-x)^2//in mm^4

+Igr=b*D^3/12//in mm^4

+yt=D/2//in mm

+Mr=fcr*Igr/yt//in N-mm

+M=W*l^2/8*10^6//in N-mm

+Ieff=Ir/(1.2-Mr/M*z/d*(1-x/d)*b/b)//in mm^4

+//Ir<Ieff<Igr, hence OK

+W1=W*l//in kN

+u1=5/384*(W1*10^3)*(l*10^3)^3/Ec/Ieff//short-term deflection, in mm

+//long-term deflection

+//(i) deflection due to shrinkage

+k3=0.125//for simply supported beam

+pt=0.56//in %

+pc=0//in %

+k4=0.72*(pt-pc)/sqrt(pt)

+phi=k4*0.0003/D

+u2=k3*phi*(l*10^3)^2//in mm

+//(ii) deflection due to creep

+Ecc=Ec/(1+1.6)//in MPa

+//assuming a permanent load of 60%

+W2=0.6*W*l//in kN

+u3=5/384*(W2*10^3)*(l*10^3)^3/Ecc/Ieff//in mm

+u4=5/384*(W2*10^3)*(l*10^3)^3/Ec/Ieff//in mm

+u5=u3-u4//in mm

+u=u1+u2+u5//total deflection, in mm

+v1=l*10^3/250//permissible deflection, in mm

+v2=l*10^3/350//in mm

+//assuming half the shrinkage strain occurs within the first 28 days, the deflection occurring after this time

+v3=u2/2+u5//< permissible value, hence OK