matlab/inline_RC_SRP.m

% Simple RC-SRP Script Calculator (Inline Configuration)
% Author: Rembrant Oyangoren Albeos
% Year: 2026

clear; clc;

%% --- CONFIGURABLE SETTINGS ---
% Modify these variables directly before running the script.

% Concrete and Steel Properties
fc = 4000;      % Concrete compressive strength f_c' (psi)
fy = 60000;     % Steel yield strength f_y (psi)

% Beam Dimensions
b = 12;         % Beam width b (in)
h = 20;         % Beam height h (in)
cover = 2.5;    % Cover distance to extreme tension layer (in)

% Layer 1 (Bottom) Reinforcement
n_bars1 = 4;    % Number of bars in layer 1
a_bar = 0.79;   % Area per bar (in^2)

% Layer 2 (Top) Reinforcement
% Set n_bars2 to 0 for a Single Layer computation.
n_bars2 = 0;    % Number of bars in layer 2
spacing = 0;    % Spacing between layers (in)

% Output Option
show_output = true; % Set to true to display calculation output in Command Window

%% --- CALCULATION LOGIC ---

if show_output
    disp(' ');
    disp('--- CALCULATION OUTPUT ---');
    
    % Basic constants
    fc_ksi = fc / 1000;
    fy_ksi = fy / 1000;
    Es_ksi = 29000;
    E_cu = 0.003;
    
    % Calculate beta1 factor
    beta1 = 0.85;
    if fc > 4000
        beta1 = 0.85 - 0.05 * ((fc - 4000) / 1000);
    end
    beta1 = max(beta1, 0.65);
    
    % Minimum steel factor
    limit_val = 3 * sqrt(fc);
    As_min_factor = max(limit_val, 200);
    
    % Determine if it's Single or Double Layer automatically
    if n_bars2 == 0
        layer_type = 'Single Layer Singly Reinforced';
        
        % --- SINGLE LAYER CALCULATION ---
        d = h - cover;
        As = n_bars1 * a_bar;
        
        disp(['1. Effective depth (d): ', num2str(d), ' in']);
        disp(['   Area of steel (A_s): ', num2str(As), ' in^2']);
        
        % Force and depth of compression block
        T = As * fy_ksi;
        disp(['2. Tension force (T, assuming yield): ', num2str(T), ' kips']);
        
        a = T / (0.85 * fc_ksi * b);
        c = a / beta1;
        disp(['3. Compression block depth (a): ', num2str(a), ' in']);
        disp(['   Neutral axis depth (c): ', num2str(c), ' in']);
        
        % Check if steel actually yields
        eps_y = fy_ksi / Es_ksi;
        eps_s = ((d - c) / c) * E_cu;
        disp(['4. Yield strain (eps_y): ', num2str(eps_y)]);
        disp(['   Steel strain (eps_s): ', num2str(eps_s)]);
        
        if eps_s >= eps_y
            disp('   Status: Steel yielded. Assumption OK.');
        else
            disp('   Status: Steel did NOT yield. Assumption failed.');
        end
        
        % Calculate Nominal Moment
        Mn = T * (d - a/2) / 12;
        disp(['5. Nominal Moment (M_n): ', num2str(Mn), ' kip-ft']);
        
        % Check minimum code requirement for steel
        As_min = (As_min_factor / fy) * b * d;
        disp(['6. Minimum steel (A_s,min): ', num2str(As_min), ' in^2']);
        if As >= As_min
            disp('   Status: A_s >= A_s,min. OK.');
        else
            disp('   Status: A_s < A_s,min. Does not meet minimum requirement.');
        end
        
        % Compute reduction factor
        eps_t = eps_s;
        if eps_t >= 0.005
            phi = 0.90;
        elseif eps_t <= 0.002
            phi = 0.65;
        else
            phi = 0.65 + (eps_t - 0.002) * (250/3);
        end
        
        phi_Mn = phi * Mn;
        disp(['7. Reduction factor (phi): ', num2str(phi)]);
        disp(['8. Design Moment (phi*M_n): ', num2str(phi_Mn), ' kip-ft']);
        
    else
        layer_type = 'Double Layer Singly Reinforced';
        
        % --- DOUBLE LAYER CALCULATION ---
        As1 = n_bars1 * a_bar;
        As2 = n_bars2 * a_bar;
        As = As1 + As2;
        
        dist1 = cover;
        dist2 = cover + spacing;
        
        % Calculate centroid of steel
        g = (As1 * dist1 + As2 * dist2) / As;
        d = h - g;
        d_t = h - cover;
        
        disp(['1. Centroid distance (g): ', num2str(g), ' in']);
        disp(['   Effective depth (d): ', num2str(d), ' in']);
        disp(['   Area of steel (A_s): ', num2str(As), ' in^2']);
        
        % Force and depth of compression block
        T = As * fy_ksi;
        a = T / (0.85 * fc_ksi * b);
        c = a / beta1;
        disp(['2. Tension force (T, assuming yield): ', num2str(T), ' kips']);
        disp(['3. Compression block depth (a): ', num2str(a), ' in']);
        disp(['   Neutral axis depth (c): ', num2str(c), ' in']);
        
        % Check if steel yields
        eps_y = fy_ksi / Es_ksi;
        eps_s = ((d - c) / c) * E_cu;
        disp(['4. Yield strain (eps_y): ', num2str(eps_y)]);
        disp(['   Steel strain (eps_s): ', num2str(eps_s)]);
        
        if eps_s < eps_y
            disp('   Status: Over-reinforced section. Assumption Failed.');
            disp('   Solving exact c using quadratic method...');
            
            % Use quadratic for exact answer if not yielded
            A_quad = 0.85 * fc_ksi * b * beta1;
            B_quad = As * Es_ksi * E_cu;
            C_quad = -1 * As * Es_ksi * E_cu * d;
            
            roots_c = roots([A_quad, B_quad, C_quad]);
            c = max(roots_c(roots_c > 0));
            a = beta1 * c;
            
            disp(['   Exact Neutral axis (c): ', num2str(c), ' in']);
            disp(['   Exact Compression block (a): ', num2str(a), ' in']);
            
            T = As * Es_ksi * ((d - c)/c) * E_cu;
            disp(['   Exact Tension force (T): ', num2str(T), ' kips']);
        else
            disp('   Status: Steel yielded. Assumption OK.');
        end
        
        % Calculate Nominal Moment
        Mn = T * (d - a/2) / 12;
        disp(['5. Nominal Moment (M_n): ', num2str(Mn), ' kip-ft']);
        
        % Check minimum code requirement for steel
        As_min = (As_min_factor / fy) * b * d;
        disp(['6. Minimum steel (A_s,min): ', num2str(As_min), ' in^2']);
        if As >= As_min
            disp('   Status: A_s >= A_s,min. OK.');
        else
            disp('   Status: A_s < A_s,min. Does not meet minimum requirement.');
        end
        
        % Compute reduction factor using extreme tension layer
        eps_t = ((d_t - c) / c) * E_cu;
        if eps_t >= 0.005
            phi = 0.90;
        elseif eps_t <= 0.002
            phi = 0.65;
        else
            phi = 0.65 + (eps_t - 0.002) * (250/3);
        end
        
        phi_Mn = phi * Mn;
        disp(['7. Extreme tension strain (eps_t): ', num2str(eps_t)]);
        disp(['   Reduction factor (phi): ', num2str(phi)]);
        disp(['8. Design Moment (phi*M_n): ', num2str(phi_Mn), ' kip-ft']);
    end
    
    disp(' ');
    disp(['CONCLUSION: The calculated section is [', layer_type, '].']);
end