Design for bending for flanged sections (beams: ACI 318)
IF hf < 0.5*bw THEN treat the beam as rectangular1
where
bw = web width
Depth of the equivalent stress block is given 2;
a = ρ*d*fy/ (0.85*f’c) (= 1.18*ω*d)
where
ρ = 0.85*f’c/fy * [1-(1- 2*Rn/√(0.85*f’c)]
Rn = (Mu/φ) / (beff*d2) assumption φ = 0.9
IF a ≤ hf THEN the rectangular compression block is wholly in the depth of the flange and the section can be designed as a rectangular section with tension reinforcement only by setting b = beff and checking the φ-factor as followed;
IF (a/β1)/d < 0.375 THEN φ=0.9 (section tension controlled)
IF 0.375 > (a/β1)/d < 0.600 THEN φ=0.7 + (εt -0.002 *(200/3)
IF (a/β1)/d > 0.6 THEN φ=0.65 (section comp. controlled)
where
εt = [(d*β1)/a-1]*0.003
IF a > hf THEN the rectangular compression block extends into the rib of the flanged section and the following design method is to be used;
Required reinforcement is given;
Asf = 0.85* f`c*(beff-b)*hf / fy
Nominal moment strength of flange;
Mnf = [Asf*fy(d-hf /2)]
Required nominal moment strength to be carried by the beam web is given;
Mnw = Mu - Mnf
Can be written as;
Mnw = Mu - [(0.85* f`c*(beff-b)*hf / fy)*fy(d-hf /2)] = Mu - [(0.85* f`c*(beff-b)*hf)*(d-hf /2)]
Reinforcement Asw required to develop the moment strength to be carried by the web;
Asw = ωw* f`c*b*d / fy
where
ωw = ρ*fy / f`c = 0.85*f`c/fy * [1-(1- 2*(Mnw / (b*d2))/√(0.85*f`c)] * fy / fc`
Can be written as;
Asw = b*d*0.85*f`c/fy*[1-(1- 2*(Mnw / (b*d2))/√(0.85*f`c)]
Total required reinforcement is given;
As = Asf + Asw
Check to see if the section is tension-controlled;
IF
ρw ≤ ρt section is tension- controlled (φ=0.9)
ELSE add compression reinforcement where
ρw = ωw f`c/ fy ρt = 0.319* β1* f`c/ fy
Can be simplified as;
ωw ≤ 0.319* β1 section is tension-controlled (φ=0.9)
ELSE add compression reinforcement
1 ACI 318-08 and ACI 318-11 Section 8.12.4
2 Notes on ACI 318-08 Section 7 (1)