Flexural Design of Beam with Compression Rebar : ACI 318-19

% Beam parameters
fc = 30;  % MPa
fy = 500; % MPa
Es = 200000; % MPa
b = 300; % mm
d = 550; % mm
dp = 50; % mm
Mu = 400e6; % N.mm

phi = 0.9;
if fc <= 28
    beta1 = 0.85;
else
    beta1 = max(0.65, 0.85 - 0.05 * (fc - 28)/7);
end

% Step 1: Assume yield strain reached
a = 0.375 * d; % assume tension steel yields
As1 = 0.85 * fc * b * a / fy;

% Step 2: Nominal moment from As1
Mn = As1 * fy * (d - a/2);

% Step 3: Check Mu vs phi*Mn
phi_Mn = phi * Mn;
M1u = max(0, Mu - phi_Mn);

% Step 4: Compute fsc from strain
c = a / beta1;
eps_sc = 0.003 * (c - dp) / c;
fsc = min(Es * eps_sc, fy);

% Step 5: Asc if needed
if M1u > 0
    Asc = M1u / (phi * fsc * (d - dp));
else
    Asc = 0;
end

% Step 6: Total steel
As_total = As1 + Asc;

% Display results
fprintf('As1 = %.2f mm^2\n', As1);
fprintf('phi*Mn = %.2f kNm\n', phi_Mn/1e6);
fprintf('Mu = %.2f kNm\n', Mu/1e6);
fprintf('fsc = %.2f MPa\n', fsc);
fprintf('Asc = %.2f mm^2\n', Asc);
fprintf('As_total = %.2f mm^2\n', As_total);

Step	Description
1	Set cmax = 0.375 d & a = beta1*c
2	Compute As1 = 0.85 fc.*a* b / fy
3	Compute Mn1 = As1 fy (d - a / 2)
4	Compute M2u = Mu - φ * Mn1
5	Decision: Is M2u > 0?
6A	Yes → Compute fs = 0.003 * (c - d') / c
7A	Compute Asc = M2u / (0.9 fs (d - d'))
8A	Compute Astotal = As1 + Asc
6B	No → Compute a = β1 * cmax
7B	Compute Astotal = Mu / (0.9 fy (d - a / 2))
9	End