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+discussion/matlab-basic-functions-reference.pdf filter=lfs diff=lfs merge=lfs -text
+discussion/RC-SinglyReinforced_no-image_cleaned.pdf filter=lfs diff=lfs merge=lfs -text
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.gitignore

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+pdf-to-image-main/

LICENSE

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+MIT License
+
+Copyright (c) 2026 Rembrant Oyangoren Albeos
+
+Permission is hereby granted, free of his/her choice to any person obtaining a copy
+of this software and associated documentation files (the "Software"), to deal
+in the Software without restriction, including without limitation the rights
+to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
+copies of the Software, and to permit persons to whom the Software is
+furnished to do so, subject to the following conditions:
+
+The above copyright notice and this permission notice shall be included in all
+copies or substantial portions of the Software.
+
+THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
+IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
+FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
+AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
+LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
+OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN THE
+SOFTWARE.

RC_SRP_Calculator.m

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+classdef RC_SRP_Calculator < handle
+    % This application provides a modern, peach-themed interface
+    % to calculate singly reinforced sections step-by-step.
+    % We avoid heavy plots and focus purely on real-time equation updates.
+    
+    properties
+        % Window and main grid
+        UIFigure
+        GridLayout
+        
+        % The three main sections
+        LeftPanel       % Sidebar for configurable settings
+        CenterPanel     % Main panel for step-by-step equations
+        RightPanel      % Right sidebar for our activity log
+        
+        % Left Panel Inputs (using f(c)', f(y), A(s) etc. for UI labels)
+        ModeDropdown
+        TitleFc, FieldFc
+        TitleFy, FieldFy
+        TitleB, FieldB
+        TitleH, FieldH
+        TitleBars, FieldBars
+        TitleArea, FieldArea
+        TitleCover, FieldCover
+        
+        % Extra Inputs for Double Layer
+        TitleBars2, FieldBars2
+        TitleArea2, FieldArea2
+        TitleSpacing, FieldSpacing
+        
+        % Center Panel Output
+        EquationHTML
+        
+        % Right Panel Output
+        LogTextArea
+        
+        % Keep track of our log messages
+        LogHistory = {}
+    end
+    
+    methods (Access = public)
+        % Constructor. This runs when we launch the calculator.
+        function app = RC_SRP_Calculator()
+            app.createApp();
+            app.addLog('Calculator app initialized! Welcome.');
+            app.updateEquations(); % trigger first calculation
+        end
+        
+        % Clean up when the app is nicely closed
+        function delete(app)
+            if isvalid(app.UIFigure)
+                delete(app.UIFigure);
+            end
+        end
+    end
+    
+    methods (Access = private)
+        % Let's build our user interface step-by-step
+        function createApp(app)
+            % Define our peach color palette (very warm, minimalistic aesthetic)
+            peachBase  = [1.00, 0.89, 0.80];  % Main background
+            peachDark  = [0.95, 0.76, 0.65];  % Panels
+            peachLight = [1.00, 0.95, 0.90];  % Highlights
+            peachText  = [0.20, 0.20, 0.20];  % Text color for readability
+            
+            % Setup the main application window
+            app.UIFigure = uifigure('Name', 'RC-SRP Calculator', 'Position', [100, 100, 1200, 750]);
+            app.UIFigure.Color = peachBase;
+            
+            % Setup a 3-column layout
+            app.GridLayout = uigridlayout(app.UIFigure);
+            app.GridLayout.ColumnWidth = {300, '1x', 280}; 
+            app.GridLayout.RowHeight = {'1x'};
+            app.GridLayout.BackgroundColor = peachBase;
+            
+            % --- 1. LEFT SIDEBAR: Configurable Settings ---
+            app.LeftPanel = uipanel(app.GridLayout);
+            app.LeftPanel.Layout.Row = 1;
+            app.LeftPanel.Layout.Column = 1;
+            app.LeftPanel.BackgroundColor = peachDark;
+            app.LeftPanel.Title = 'Configurable Settings';
+            app.LeftPanel.ForegroundColor = peachText;
+            app.LeftPanel.FontWeight = 'bold';
+            
+            % Grid inside the left panel to align our inputs beautifully
+            leftGrid = uigridlayout(app.LeftPanel);
+            leftGrid.ColumnWidth = {'1x', '1x'};
+            leftGrid.RowHeight = repmat({32}, 1, 15); % 15 rows of 32px height
+            leftGrid.BackgroundColor = peachDark;
+            
+            % Calculation mode dropdown
+            uilabel(leftGrid, 'Text', 'Mode:', 'FontWeight', 'bold', 'FontColor', peachText);
+            app.ModeDropdown = uidropdown(leftGrid, ...
+                'Items', {'Singly Reinforced', 'Single Layer', 'Double Layer'}, ...
+                'ValueChangedFcn', @(src, event) app.onInputChanged(), ...
+                'FontColor', peachText, 'BackgroundColor', peachLight);
+                
+            % f(c)' input (concrete compressive strength)
+            app.TitleFc = uilabel(leftGrid, 'Text', 'f(c)'' (psi):', 'FontColor', peachText, 'FontWeight', 'bold');
+            app.FieldFc = uieditfield(leftGrid, 'numeric', 'Value', 4000, 'ValueChangedFcn', @(src, event) app.onInputChanged());
+            
+            % f(y) input (steel yield strength)
+            app.TitleFy = uilabel(leftGrid, 'Text', 'f(y) (psi):', 'FontColor', peachText, 'FontWeight', 'bold');
+            app.FieldFy = uieditfield(leftGrid, 'numeric', 'Value', 60000, 'ValueChangedFcn', @(src, event) app.onInputChanged());
+            
+            % Beam width
+            app.TitleB = uilabel(leftGrid, 'Text', 'b (in):', 'FontColor', peachText, 'FontWeight', 'bold');
+            app.FieldB = uieditfield(leftGrid, 'numeric', 'Value', 12, 'ValueChangedFcn', @(src, event) app.onInputChanged());
+            
+            % Beam height
+            app.TitleH = uilabel(leftGrid, 'Text', 'h (in):', 'FontColor', peachText, 'FontWeight', 'bold');
+            app.FieldH = uieditfield(leftGrid, 'numeric', 'Value', 20, 'ValueChangedFcn', @(src, event) app.onInputChanged());
+            
+            % Bars configuration
+            app.TitleBars = uilabel(leftGrid, 'Text', 'Number of Bars:', 'FontColor', peachText, 'FontWeight', 'bold');
+            app.FieldBars = uieditfield(leftGrid, 'numeric', 'Value', 4, 'ValueChangedFcn', @(src, event) app.onInputChanged());
+            
+            app.TitleArea = uilabel(leftGrid, 'Text', 'Area per A(s):', 'FontColor', peachText, 'FontWeight', 'bold');
+            app.FieldArea = uieditfield(leftGrid, 'numeric', 'Value', 0.79, 'ValueChangedFcn', @(src, event) app.onInputChanged());
+            
+            app.TitleCover = uilabel(leftGrid, 'Text', 'Cover (in):', 'FontColor', peachText, 'FontWeight', 'bold');
+            app.FieldCover = uieditfield(leftGrid, 'numeric', 'Value', 2.5, 'ValueChangedFcn', @(src, event) app.onInputChanged());
+            
+            % Double layer specific fields (hidden by default)
+            app.TitleBars2 = uilabel(leftGrid, 'Text', 'L2 Bars:', 'FontColor', peachText, 'Visible', 'off');
+            app.FieldBars2 = uieditfield(leftGrid, 'numeric', 'Value', 2, 'Visible', 'off', 'ValueChangedFcn', @(src, event) app.onInputChanged());
+            
+            app.TitleArea2 = uilabel(leftGrid, 'Text', 'L2 Area A(s):', 'FontColor', peachText, 'Visible', 'off');
+            app.FieldArea2 = uieditfield(leftGrid, 'numeric', 'Value', 1.00, 'Visible', 'off', 'ValueChangedFcn', @(src, event) app.onInputChanged());
+            
+            app.TitleSpacing = uilabel(leftGrid, 'Text', 'Spacing (in):', 'FontColor', peachText, 'Visible', 'off');
+            app.FieldSpacing = uieditfield(leftGrid, 'numeric', 'Value', 2.13, 'Visible', 'off', 'ValueChangedFcn', @(src, event) app.onInputChanged());
+            
+            % --- 2. CENTER PANEL: Step-by-Step Equations ---
+            app.CenterPanel = uipanel(app.GridLayout);
+            app.CenterPanel.Layout.Row = 1;
+            app.CenterPanel.Layout.Column = 2;
+            app.CenterPanel.BackgroundColor = peachLight;
+            app.CenterPanel.Title = 'Formula & Equations';
+            app.CenterPanel.ForegroundColor = peachText;
+            app.CenterPanel.FontWeight = 'bold';
+            
+            % We use an HTML view to render nice Math/LaTeX-like output with CSS animations
+            centerGrid = uigridlayout(app.CenterPanel);
+            centerGrid.ColumnWidth = {'1x'};
+            centerGrid.RowHeight = {'1x'};
+            
+            app.EquationHTML = uihtml(centerGrid);
+            
+            % --- 3. RIGHT SIDEBAR: Activity Log ---
+            app.RightPanel = uipanel(app.GridLayout);
+            app.RightPanel.Layout.Row = 1;
+            app.RightPanel.Layout.Column = 3;
+            app.RightPanel.BackgroundColor = peachDark;
+            app.RightPanel.Title = 'Log & Summary';
+            app.RightPanel.ForegroundColor = peachText;
+            app.RightPanel.FontWeight = 'bold';
+            
+            rightGrid = uigridlayout(app.RightPanel);
+            rightGrid.ColumnWidth = {'1x'};
+            rightGrid.RowHeight = {'1x'};
+            
+            app.LogTextArea = uitextarea(rightGrid);
+            app.LogTextArea.Editable = 'off';
+            app.LogTextArea.BackgroundColor = peachLight;
+            app.LogTextArea.FontColor = peachText;
+            app.LogTextArea.FontName = 'Courier New';
+        end
+        
+        % Whenever the user touches a variable, we update instantly
+        function onInputChanged(app)
+            % Gently check if we need to show Double Layer inputs
+            if strcmp(app.ModeDropdown.Value, 'Double Layer')
+                app.TitleBars2.Visible = 'on'; app.FieldBars2.Visible = 'on';
+                app.TitleArea2.Visible = 'on'; app.FieldArea2.Visible = 'on';
+                app.TitleSpacing.Visible = 'on'; app.FieldSpacing.Visible = 'on';
+            else
+                app.TitleBars2.Visible = 'off'; app.FieldBars2.Visible = 'off';
+                app.TitleArea2.Visible = 'off'; app.FieldArea2.Visible = 'off';
+                app.TitleSpacing.Visible = 'off'; app.FieldSpacing.Visible = 'off';
+            end
+            
+            app.addLog('Settings modified, recalculating formulas...');
+            app.updateEquations();
+        end
+        
+        % This is the core logic where all the math happens, gracefully formatting results
+        function updateEquations(app)
+            % 1. Grab variables from the visual fields
+            fc = app.FieldFc.Value;
+            fy = app.FieldFy.Value;
+            b  = app.FieldB.Value;
+            h  = app.FieldH.Value;
+            n_bars = app.FieldBars.Value;
+            a_bar  = app.FieldArea.Value;
+            cover  = app.FieldCover.Value;
+            mode   = app.ModeDropdown.Value;
+            
+            % Convert stress to ksi for smaller, readable numbers
+            fc_ksi = fc / 1000;
+            fy_ksi = fy / 1000;
+            Es_ksi = 29000;
+            E_cu   = 0.003; % standard concrete strain parameter
+            
+            % 2. Setup our HTML framework with a lovely peach CSS style and transition animations
+            html = [ ...
+                '<html><head><style>' ...
+                'body { font-family: "Segoe UI", Tahoma, Geneva, Verdana, sans-serif; ' ...
+                'color: #4A3B32; background-color: #FFF2E6; padding: 20px; line-height: 1.6; } ' ...
+                'h2 { color: #D35400; border-bottom: 2px solid #FAD7A1; padding-bottom: 5px; } ' ...
+                'h3 { color: #E67E22; margin-top: 25px; } ' ...
+                '@keyframes popup { from { opacity: 0; transform: translateY(15px); } to { opacity: 1; transform: translateY(0); } } ' ...
+                '.equation { background-color: #FFE6CC; padding: 15px; border-radius: 12px; ' ...
+                'margin: 12px 0; font-family: "Courier New", Courier, monospace; font-weight: bold; ' ...
+                'box-shadow: 0 4px 6px rgba(0,0,0,0.06); animation: popup 0.4s ease-out; } ' ...
+                '.highlight { color: #C0392B; font-size: 1.1em; } ' ...
+                '.success { color: #27AE60; font-weight: bold; } ' ...
+                '.danger { color: #E74C3C; font-weight: bold; } ' ...
+                '</style></head><body>' ...
+            ];
+            
+            html = [html, '<h2>Analysis Mode: ', mode, '</h2>'];
+            html = [html, '<h3>Given Variables</h3>'];
+            html = [html, '<div class="equation">', ...
+                                'f_c'' = ', num2str(fc), ' psi &nbsp;&nbsp;&nbsp; f_y = ', num2str(fy), ' psi<br>', ...
+                                'b = ', num2str(b), ' in &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; h = ', num2str(h), ' in<br>', ...
+                                'Bars = ', num2str(n_bars), ' (A_b = ', num2str(a_bar), ' in&sup2;)<br>', ...
+                                'Cover = ', num2str(cover), ' in</div>'];
+            
+            % 3. Common mathematical terms
+            % Beta_1 decreases gently for higher strength concrete
+            beta1 = 0.85;
+            if fc > 4000
+                beta1 = 0.85 - 0.05 * ((fc - 4000) / 1000);
+            end
+            beta1 = max(beta1, 0.65); % Never drops below this rule
+            
+            % Steel minimum base requirement
+            limit_val = 3 * sqrt(fc);
+            As_min_factor = max(limit_val, 200); % chooses the dominating rule automatically
+            
+            % 4. Logic splits neatly based on chosen calculation mode
+            if strcmp(mode, 'Singly Reinforced') || strcmp(mode, 'Single Layer')
+                % Simple straightforward geometry
+                d = h - cover;
+                html = [html, '<h3>1. Solve Effective Depth (d)</h3>'];
+                html = [html, '<div class="equation">d = h - cover<br>', ...
+                              '= ', num2str(h), ' - ', num2str(cover), '<br>', ...
+                              '= <span class="highlight">', num2str(d), ' in</span></div>'];
+                
+                As = n_bars * a_bar;
+                html = [html, '<h3>2. Solve Steel Area (A_s)</h3>'];
+                html = [html, '<div class="equation">A_s = bars &times; a_bar<br>', ...
+                              '= ', num2str(n_bars), ' &times; ', num2str(a_bar), '<br>', ...
+                              '= <span class="highlight">', num2str(As), ' in&sup2;</span></div>'];
+                
+                % Standard assumption: the steel is already yielding
+                T = As * fy_ksi;
+                html = [html, '<h3>3. Assume steel yields (T = A_s &middot; f_y)</h3>'];
+                html = [html, '<div class="equation">T = ', num2str(As), ' &times; ', num2str(fy_ksi), ' ksi<br>', ...
+                              '= <span class="highlight">', num2str(T), ' kips</span></div>'];
+                
+                % Concrete compression block depth
+                a = T / (0.85 * fc_ksi * b);
+                html = [html, '<h3>4. Solve Compression Block (a &amp; c)</h3>'];
+                html = [html, 'By matching concrete volume force to tension:<br>'];
+                html = [html, '<div class="equation">a = T / (0.85 &middot; f_c'' &middot; b)<br>', ...
+                              '= <span class="highlight">', num2str(a, '%.2f'), ' in</span></div>'];
+                                    
+                c = a / beta1;
+                html = [html, '<div class="equation">c = a / &beta;&sub1; = ', num2str(a, '%.2f'), ' / ', num2str(beta1, '%.2f'), '<br>', ...
+                              '= <span class="highlight">', num2str(c, '%.2f'), ' in</span></div>'];
+                                    
+                % Strain compatibility check
+                eps_y = fy_ksi / Es_ksi;
+                eps_s = ((d - c) / c) * E_cu;
+                html = [html, '<h3>5. Strain Compatibility Check (&epsilon;_s &ge; &epsilon;_y)</h3>'];
+                html = [html, '<div class="equation">&epsilon;_y = f_y / E_s = ', num2str(eps_y, '%.5f'), '<br>', ...
+                              '&epsilon;_s = ((d - c) / c) &middot; &epsilon;_cu<br>', ...
+                              '= <span class="highlight">', num2str(eps_s, '%.5f'), '</span></div>'];
+                                    
+                if eps_s >= eps_y
+                    html = [html, '<p class="success">Tension steel yielded. Assumption confirmed gracefully.</p>'];
+                else
+                    html = [html, '<p class="danger">Assumption failed! Steel did not yield.</p>'];
+                end
+                
+                % Calculate the raw moment capacity
+                Mn = T * (d - a/2) / 12;
+                html = [html, '<h3>6. Nominal Moment (M_n)</h3>'];
+                html = [html, '<div class="equation">M_n = T &middot; (d - a/2) / 12<br>', ...
+                              '= <span class="highlight">', num2str(Mn, '%.2f'), ' kip-ft</span></div>'];
+                                    
+                % Checking code requirements for minimum reinforcements
+                As_min = (As_min_factor / fy) * b * d;
+                html = [html, '<h3>7. Check Minimum Code Rules</h3>'];
+                html = [html, '<div class="equation">A_s,min = (', num2str(As_min_factor), ' / f_y) &middot; b &middot; d<br>', ...
+                              '= <span class="highlight">', num2str(As_min, '%.2f'), ' in&sup2;</span></div>'];
+                if As >= As_min
+                    html = [html, '<p class="success">A_s exceeds minimum rules. Wonderful.</p>'];
+                else
+                    html = [html, '<p class="danger">A_s falls short of required minimum!</p>'];
+                end
+                
+                % Extra steps for single layer reduction factor calculation
+                if strcmp(mode, 'Single Layer')
+                    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;
+                    
+                    html = [html, '<h3>8. Strength Reduction Factor (&phi;)</h3>'];
+                    html = [html, '<div class="equation">&epsilon;_t = ', num2str(eps_t, '%.5f'), '<br>', ...
+                                  '&phi; = <span class="highlight">', num2str(phi, '%.2f'), '</span></div>'];
+                                        
+                    html = [html, '<h3>9. Designed Maximum Moment (&phi;M_n)</h3>'];
+                    html = [html, '<div class="equation">&phi;M_n = ', num2str(phi, '%.2f'), ' &times; ', num2str(Mn, '%.2f'), '<br>', ...
+                                  '= <span class="highlight">', num2str(phi_Mn, '%.2f'), ' kip-ft</span></div>'];
+                end
+                
+            elseif strcmp(mode, 'Double Layer')
+                % --- Advanced Multiple Layers processing ---
+                n_bars2 = app.FieldBars2.Value;
+                a_bar2  = app.FieldArea2.Value;
+                spacing = app.FieldSpacing.Value;
+                
+                % Centroid gymnastics for double layers
+                As1 = n_bars * a_bar;
+                As2 = n_bars2 * a_bar2;
+                As = As1 + As2;
+                
+                dist1 = cover; 
+                dist2 = cover + spacing;
+                
+                g = (As1 * dist1 + As2 * dist2) / As;
+                d = h - g;
+                d_t = h - cover; 
+                
+                html = [html, '<h3>1. Resolve the Steel Centroids</h3>'];
+                html = [html, '<div class="equation">g = (A_s1 &middot; y1 + A_s2 &middot; y2) / A_s<br>', ...
+                              '= (', num2str(As1), ' &times; ', num2str(dist1), ' + ', num2str(As2), ' &times; ', num2str(dist2), ') / ', num2str(As), '<br>', ...
+                              '= ', num2str(g, '%.2f'), ' in</div>'];
+                                    
+                html = [html, '<div class="equation">d = h - g = ', num2str(h), ' - ', num2str(g, '%.2f'), '<br>', ...
+                              '= <span class="highlight">', num2str(d, '%.2f'), ' in</span></div>'];
+                
+                % Assume Yield again
+                T = As * fy_ksi;
+                a = T / (0.85 * fc_ksi * b);
+                c = a / beta1;
+                
+                html = [html, '<h3>2. Solve Basic Block (Assuming Yield)</h3>'];
+                html = [html, '<div class="equation">T = ', num2str(T, '%.2f'), ' kips<br>', ...
+                              'a = ', num2str(a, '%.2f'), ' in<br>', ...
+                              'c = <span class="highlight">', num2str(c, '%.2f'), ' in</span></div>'];
+                                    
+                eps_y = fy_ksi / Es_ksi;
+                eps_s = ((d - c) / c) * E_cu;
+                
+                html = [html, '<h3>3. Verify the Yield Strain</h3>'];
+                html = [html, '<div class="equation">&epsilon;_s = ', num2str(eps_s, '%.5f'), ' | &epsilon;_y = ', num2str(eps_y, '%.5f'), '</div>'];
+                                    
+                % More complex quadratic solving requested for over-reinforced double layers
+                if eps_s < eps_y
+                    html = [html, '<p class="danger">Assumption Failed! Solving via Quadratic Exact Match...</p>'];
+                    
+                    A_quad = 0.85 * fc_ksi * b * beta1;
+                    B_quad = As * Es_ksi * E_cu;
+                    C_quad = -1 * As * Es_ksi * E_cu * d;
+                    
+                    % Find actual real root
+                    roots_c = roots([A_quad, B_quad, C_quad]);
+                    c_new = max(roots_c(roots_c > 0));
+                    a_new = beta1 * c_new;
+                    
+                    html = [html, '<div class="equation">Quadratic Form Evaluated for c:<br>', ...
+                                  'c = <span class="highlight">', num2str(c_new, '%.2f'), ' in</span><br>', ...
+                                  'a = <span class="highlight">', num2str(a_new, '%.2f'), ' in</span></div>'];
+                                        
+                    c = c_new;
+                    a = a_new;
+                    
+                    % Recalculate accurately
+                    T = As * Es_ksi * ((d - c)/c) * E_cu;
+                    html = [html, '<div class="equation">Accurate Balance Tension T = C_c = <span class="highlight">', num2str(T, '%.2f'), ' kips</span></div>'];
+                else
+                    html = [html, '<p class="success">Assumption smoothly verified.</p>'];
+                end
+                
+                % Compute derived raw moment
+                Mn = T * (d - a/2) / 12;
+                html = [html, '<h3>4. Base Moment Geometry (M_n)</h3>'];
+                html = [html, '<div class="equation">M_n = T &middot; (d - a/2) / 12<br>', ...
+                              '= <span class="highlight">', num2str(Mn, '%.2f'), ' kip-ft</span></div>'];
+                                    
+                % For double layers, compute reduction on the most extreme layer strain
+                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
+                
+                html = [html, '<h3>5. Reduction Factor for Design Result</h3>'];
+                html = [html, '<div class="equation">&epsilon;_t (extreme) = ', num2str(eps_t, '%.5f'), '<br>', ...
+                              '&phi; = ', num2str(phi, '%.2f'), '<br>', ...
+                              '&phi;M_n = <span class="highlight">', num2str(phi*Mn, '%.2f'), ' kip-ft</span></div>'];
+            end
+            
+            % Close and assign to the beautiful HTML widget
+            html = [html, '</body></html>'];
+            app.EquationHTML.HTMLSource = html;
+            
+            % Add to our active log so we have a trace
+            app.addLog(sprintf('Computed [%s] cleanly. End M_n: %.2f k-ft.', mode, Mn));
+        end
+        
+        % This simple helper lets us comfortably add logs with timestamps
+        function addLog(app, msg)
+            timestamp = datestr(now, 'HH:MM:SS');
+            logEntry = sprintf('[%s] %s', timestamp, msg);
+            app.LogHistory = [app.LogHistory; {logEntry}];
+            
+            % Keep logs fairly recent to keep memory efficient
+            if length(app.LogHistory) > 40
+                app.LogHistory(1) = [];
+            end
+            
+            app.LogTextArea.Value = app.LogHistory;
+        end
+        
+    end
+end

README.md

@@ -0,0 +1,57 @@
+# RC-SRP Calculator
+
+[![License: MIT](https://img.shields.io/badge/License-MIT-yellow.svg)](https://opensource.org/licenses/MIT)
+![Repo Size](https://img.shields.io/github/repo-size/unban-algorembrant/RC-SRP)
+![Last Commit](https://img.shields.io/github/last-commit/unban-algorembrant/RC-SRP)
+![MATLAB](https://img.shields.io/badge/Language-MATLAB-orange.svg)
+
+A comprehensive MATLAB-based calculator for Reinforcement Concrete Singly Reinforced Rectangular Beams (RC-SRP). This project provides both a lightweight command-line script and a professional Graphical User Interface (GUI) featuring a modern peach-themed aesthetic.
+
+## Features
+
+- Singly reinforced rectangular beam analysis.
+- Support for single and double layers of tension reinforcement.
+- Automated strain compatibility checks (tension-controlled, transition, or compression-controlled).
+- Accurate compression block calculations using Beta 1 modifiers for different concrete strengths.
+- Nominal and design moment strength calculations.
+- Minimum reinforcement area validation according to code standards.
+
+## Usage
+
+### Command Window Version (`singly_reinforced.m`)
+For users who prefer a direct, text-based interaction, use the script version:
+
+1. Open MATLAB and navigate to the project directory.
+2. Run the script by typing:
+   ```matlab
+   singly_reinforced
+   ```
+3. Follow the prompts in the Command Window to input material properties (fc, fy), beam dimensions (b, d, dt), and reinforcement details.
+
+### GUI Application Version (`RC_SRP_Calculator.m`)
+For a professional, interactive experience with real-time formula updates and a peach-themed interface:
+
+1. Open MATLAB and navigate to the project directory.
+2. Launch the calculator by typing:
+   ```matlab
+   RC_SRP_Calculator
+   ```
+3. Select the analysis mode (Singly Reinforced, Single Layer, or Double Layer).
+4. Adjust the configurable settings in the left sidebar. The center panel will update instantly with step-by-step equations and results.
+
+## Technical Details
+
+The calculator follows standard reinforced concrete design principles, calculating the depth of the equivalent rectangular stress block (a) and the neutral axis (c). It accounts for the strength reduction factor (phi) based on the net tensile strain in the extreme tension steel.
+
+## Citation
+
+If you use this software in your research or project, please cite it using the following BibTeX entry:
+
+```bibtex
+@software{Albeos_RC_SRP_Calculator_2026,
+  author = {Rembrant Oyangoren Albeos},
+  title = {RC-SRP Calculator: Singly Reinforced Rectangular Beam Analysis for MATLAB},
+  year = {2026},
+  url = {https://github.com/unban-algorembrant/RC-SRP}
+}
+```

STRUCTURE.md

@@ -0,0 +1,97 @@
+## Project Structure
+
+```text
+RC-SRP/
+├── discussion/
+│   ├── combined_document.md
+│   ├── main.md
+│   ├── main_corrected.md
+│   ├── matlab-basic-functions-reference.pdf
+│   ├── RC-SinglyReinforced_no-image_cleaned.pdf
+│   └── revision_discussion.md
+├── matlab/
+│   ├── inline_RC_SRP.m
+│   ├── simple_RC_SRP.m
+│   ├── singly_reinforced copy.m
+│   └── tae.m
+├── output_images/
+│   ├── page_01.png
+│   ├── page_02.png
+│   ├── page_03.png
+│   ├── page_04.png
+│   ├── page_05.png
+│   ├── page_06.png
+│   ├── page_07.png
+│   ├── page_08.png
+│   ├── page_09.png
+│   ├── page_10.png
+│   ├── page_11.png
+│   ├── page_12.png
+│   ├── page_13.png
+│   ├── page_14.png
+│   ├── page_15.png
+│   ├── page_16.png
+│   ├── page_17.png
+│   ├── page_18.png
+│   ├── page_19.png
+│   ├── page_20.png
+│   ├── page_21.png
+│   ├── page_22.png
+│   ├── page_23.png
+│   ├── page_24.png
+│   ├── page_25.png
+│   ├── page_26.png
+│   ├── page_27.png
+│   ├── page_28.png
+│   ├── page_29.png
+│   ├── page_30.png
+│   ├── page_31.png
+│   ├── page_32.png
+│   ├── page_33.png
+│   ├── page_34.png
+│   ├── page_35.png
+│   └── page_36.png
+├── output_mds/
+│   ├── page_01.md
+│   ├── page_02.md
+│   ├── page_03.md
+│   ├── page_04.md
+│   ├── page_05.md
+│   ├── page_06.md
+│   ├── page_07.md
+│   ├── page_08.md
+│   ├── page_09.md
+│   ├── page_10.md
+│   ├── page_11.md
+│   ├── page_12.md
+│   ├── page_13.md
+│   ├── page_14.md
+│   ├── page_15.md
+│   ├── page_16.md
+│   ├── page_17.md
+│   ├── page_18.md
+│   ├── page_19.md
+│   ├── page_20.md
+│   ├── page_21.md
+│   ├── page_22.md
+│   ├── page_23.md
+│   ├── page_24.md
+│   ├── page_25.md
+│   ├── page_26.md
+│   ├── page_27.md
+│   ├── page_28.md
+│   ├── page_29.md
+│   ├── page_30.md
+│   ├── page_31.md
+│   ├── page_32.md
+│   ├── page_33.md
+│   ├── page_34.md
+│   ├── page_35.md
+│   └── page_36.md
+├── .gitignore
+├── LICENSE
+├── RC_SRP_Calculator.m
+├── README.md
+├── singly_reinforced.m
+└── TECHSTACK.md
+```

TECHSTACK.md

@@ -0,0 +1,12 @@
+## Techstack
+
+Audit of **RC-SRP** project files (excluding environment and cache):
+
+| File Type | Count | Size (KB) |
+| :--- | :--- | :--- |
+| Markdown (.md) | 41 | 206.8 |
+| PNG Image (.png) | 36 | 9,112.9 |
+| Objective-C / MATLAB (.m) | 6 | 45.4 |
+| (no extension) | 2 | 1.1 |
+| PDF (.pdf) | 2 | 5,476.7 |
+| **Total** | **87** | **14,843.0** |

discussion/RC-SinglyReinforced_no-image_cleaned.pdf

@@ -0,0 +1,3 @@
+version https://git-lfs.github.com/spec/v1
+oid sha256:d84b641cdc01285429a6ab23c3604b207012aa4aa2f9348c000bbd641b32efd8
+size 4918009

discussion/combined_document.md

@@ -0,0 +1,875 @@
+# Flexure: Behavior and Nominal Strength of Beam Sections
+
+## 4-1 INTRODUCTION
+In this chapter, the stress-strain relationships for concrete and reinforcement from Chapter 3 are used to develop an understanding of the flexural behavior of rectangular beam sections. The effect of changes in material and section properties on the flexure behavior (moment versus curvature relationship) of beam sections will be presented. A good understanding of how changes in these primary design variables affect section behavior will be important for making good design decisions concerning material and section properties, as will be covered in the next chapter.
+
+After gaining a good understanding of the entire range of flexural behavior, a general procedure will be developed to evaluate the nominal flexural strength, $M_n$, for a variety of beam sections. Simplifications for modeling material properties, which correspond to the ACI Code definitions for nominal strength, will be presented. Emphasis will be placed on developing a fundamental approach that can be applied to any beam or slab section.
+
+In Chapter 11, the section analysis procedures developed in this chapter will be extended to sections subjected to combined bending and axial load to permit the analysis and design of column sections.
+
+Most reinforced concrete structures can be subdivided into beams and slabs, which are subjected primarily to flexure (bending), and columns, which are subjected to axial compression and bending. Typical examples of flexural members are the slab and beams shown in Fig. 4-1. The load, $P$, applied at Point $A$ is carried by the strip of slab shown shaded. The end reactions due to the load $P$ and the weight of the slab strip load the beams at $B$ and $C$. The beams, in turn, carry the slab reactions and their own weight to the columns at $D$, $E$, $F$, and $G$. The beam reactions normally cause axial load and bending in the columns. The slab in Fig. 4-1 is assumed to transfer loads in one direction and hence is called a one-way slab. The design of such slabs will be discussed in the next chapter. If there were no beams in the floor system shown in Fig. 4-1, the slab would carry the load in two directions. Such a slab is referred to as a two-way slab. The design of such slabs will be discussed in Chapter 13.
+
+Fig. 4-1
+One-way flexure.
+
+### Analysis versus Design
+
+Two different types of problems arise in the study of reinforced concrete:
+
+1. **Analysis.** Given a cross section, concrete strength, reinforcement size and location, and yield strength, compute the resistance or strength. In analysis there should be one unique answer.
+2. **Design.** Given a factored design moment, normally designated as $M_u$, select a suitable cross section, including dimensions, concrete strength, reinforcement, and so on. In design there are many possible solutions.
+
+Although both types of problem are based on the same principles, the procedure is different in each case. Analysis is easier, because all of the decisions concerning reinforcement, beam size, and so on have been made, and it is only necessary to apply the strength-calculation principles to determine the capacity. Design, on the other hand, involves the choice of section dimensions, material strengths, and reinforcement placement to produce a cross section that can resist the moments due to factored loads. Because the analysis problem is easier, this chapter deals with section analysis to develop the fundamental concepts before considering design in the next chapter.
+
+### Required Strength and Design Strength
+
+The basic safety equation for flexure is:
+
+$$ \text{Reduced nominal strength} \ge \text{factored load effects} \tag{4-1a} $$
+
+or for flexure,
+
+$$ \phi M_n \ge M_u \tag{4-1b} $$
+
+where $M_u$ is the moment due to the factored loads, which commonly is referred to as the factored design moment. This is a load effect computed by structural analysis from the governing combination of factored loads given in ACI Code Section 9.2. The term $M_n$ refers to the nominal moment strength of a cross section, computed from the nominal dimensions and specified material strengths. The factor $\phi$ in Eq. (4-1b) is a strength-reduction factor (ACI Code Section 9.3) to account for possible variations in dimensions and material strengths and possible inaccuracies in the strength equations. Since the mid 1990s, the ACI Code has referred to the load factors and load combinations developed by ASCE Committee 7, which is responsible for the ASCE Standard for Minimum Design Loads for Buildings and Other Structures [4-1]. The load factors and load combinations given in ACI Code Section 9.2 are essentially the same as those
+
+developed by ASCE Committee 7. The strength-reduction factors given in ACI Code Section 9.3 are based on statistical studies of material properties [4-2] and were selected to give approximately the same level of safety as obtained with the load and strength-reduction factors used in earlier editions of the code. Those former load and strength-reduction factors are still presented as an alternative design procedure in Appendix C of the latest edition of the ACI Code, ACI 318-08 [4-3]. However, they will not be discussed in this book.
+
+For flexure without axial load, ACI Code Section 9.3.2.1 gives $\phi = 0.90$ for what are called tension-controlled sections. Most practical beams will be tension-controlled sections, and $\phi$ will be equal to $0.90$. The concept of tension-controlled sections will be discussed later in this chapter. The product, $\phi M_n$, commonly is referred to as the reduced nominal moment strength.
+
+### Positive and Negative Moments
+
+A moment that causes compression on the top surface of a beam and tension on the bottom surface will be called a positive moment. The compression zones for positive and negative moments are shown shaded in Fig. 4-2. In this textbook, bending-moment diagrams will be plotted on the compression side of the member.
+
+### Symbols and Notation
+
+Although symbols are defined as they are first used and are summarized in Appendix B, several symbols should essentially be memorized because they are commonly used in discussions of reinforced concrete members. These include the terms $M_u$ and $M_n$ (defined earlier) and the cross-sectional dimensions illustrated in Fig. 4-2. The following is a list of common symbols used in this book:
+
+- $A_s$ is the area of reinforcement near the tension face of the beam, tension reinforcement, $\text{in.}^2$.
+- $A_s'$ is the area of reinforcement on the compression side of the beam, compression reinforcement, $\text{in.}^2$.
+- $b$ is a general symbol for the width of the compression zone in a beam, in. This is illustrated in Fig. 4-2 for positive and negative moment regions. For flanged sections this symbol will normally be replaced with $b_e$ or $b_w$.
+
+(a) Positive moment (compression on top).
+(b) Negative moment (compression on bottom).
+
+Fig. 4-2
+Cross-sectional dimensions.
+
+- $b_e$ is the effective width of a compression zone for a flanged section with compression in the flange, in.
+- $b_w$ is the width of the web of the beam (and may or may not be the same as $b$), in.
+- $d$ is the distance from the extreme fiber in compression to the centroid of the longitudinal reinforcement on the tension side of the member, in. In the positive-moment region (Fig. 4-2a), the tension steel is near the bottom of the beam, while in the negative-moment region (Fig. 4-2b) it is near the top.
+- $d'$ is the distance from the extreme compression fiber to the centroid of the longitudinal compression steel, in.
+- $d_t$ is the distance from the extreme compression fiber to the farthest layer of tension steel, in. For a single layer of tension reinforcement, $d_t = d$, as shown in Fig. 4-2b.
+- $f_c'$ is the specified compressive strength of the concrete, psi.
+- $f_c$ is the stress in the concrete, psi.
+- $f_s$ is the stress in the tension reinforcement, psi.
+- $f_y$ is the specified yield strength of the reinforcement, psi.
+- $h$ is the overall height of a beam cross section.
+- $jd$ is the *lever arm*, the distance between the resultant compressive force and the resultant tensile force, in.
+- $j$ is a dimensionless ratio used to define the lever arm, $jd$. It varies depending on the moment acting on the beam section.
+- $\epsilon_{cu}$ is the assumed maximum useable compression strain in the concrete.
+- $\epsilon_s$ is the strain in the tension reinforcement.
+- $\epsilon_t$ is the strain in the extreme layer of tension reinforcement.
+- $\rho$ is the longitudinal tension reinforcement ratio, $\rho = A_s/bd$.
+
+## 4-2 FLEXURE THEORY
+
+### Statics of Beam Action
+
+A *beam* is a structural member that supports applied loads and its own weight primarily by internal moments and shears. Figure 4-3a shows a simple beam that supports its own dead weight, $w$ per unit length, plus a concentrated load, $P$. If the axial applied load, $N$, is equal to zero, as shown, the member is referred to as a beam. If $N$ is a compressive force, the member is called a *beam-column*. This chapter will be restricted to the very common case where $N = 0$.
+
+The loads $w$ and $P$ cause *bending moments*, distributed as shown in Fig. 4-3b. The bending moment is a *load effect* calculated from the loads by using the laws of statics. For a simply supported beam of a given span and for a given set of loads $w$ and $P$, the moments are independent of the composition and size of the beam.
+
+At any section within the beam, the *internal resisting moment*, $M$, shown in Fig. 4-3c is necessary to equilibrate the bending moment. An internal resisting shear, $V$, also is required, as shown.
+
+The internal resisting moment, $M$, results from an internal compressive force, $C$, and an internal tensile force, $T$, separated by a lever arm, $jd$, as shown in Fig. 4-3d. Because there are no external axial loads, summation of the horizontal forces gives
+
+$$ C - T = 0 \quad \text{or} \quad C = T \tag{4-2} $$
+
+(a) Beam.
+
+(b) Bending moment diagram.
+
+(c) Free body diagrams showing internal moment and shear force.
+
+Fig. 4-3
+Internal forces in a beam.
+
+(d) Free body diagrams showing internal moment as a compression-tension force couple.
+
+If moments are summed about an axis through the point of application of the compressive force, $C$, the moment equilibrium of the free body gives
+
+$$ M = T \times jd \tag{4-3a} $$
+
+Similarly, if moments are summed about the point of application of the tensile force, $T$,
+
+$$ M = C \times jd \tag{4-3b} $$
+
+Because $C = T$, these two equations are identical. Eqs. (4-2) and (4-3) come directly from statics and are equally applicable to beams made of steel, wood, or reinforced concrete.
+
+The conventional *elastic* beam theory results in the equation $\sigma = My/I$, which, for an uncracked, homogeneous rectangular beam without reinforcement, gives the distribution of stresses shown in Fig. 4-4. The stress diagram shown in Fig. 4-4c and d may be visualized as having a "volume"; hence, one frequently refers to the *compressive stress*
+
+(a)
+(b)
+(c)
+(d)
+
+Fig. 4-4
+Elastic beam stresses and stress blocks.
+
+*block*. The resultant compressive force $C$, which is equal to the volume of the compressive stress block in Fig. 4-4d, is given by
+
+$$ C = \frac{\sigma_{c(\text{max})}}{2} \left( b\frac{h}{2} \right) \tag{4-4} $$
+
+In a similar manner, one could compute the force $T$ from the tensile stress block. The forces $C$ and $T$ act through the centroids of the volumes of the respective stress blocks. In the *elastic* case, these forces act at $h/3$ above or below the neutral axis, so that $jd = 2h/3$. From Eqs. (4-3b) and (4-4) and Fig. 4-4, we can write
+
+$$ M = \sigma_{c(\text{max})} \frac{bh}{4} \left( \frac{2h}{3} \right) \tag{4-5a} $$
+
+$$ M = \sigma_{c(\text{max})} \frac{bh^2/12}{h/2} \tag{4-5b} $$
+
+or, because
+
+$$ I = \frac{bh^3}{12} $$
+
+and
+
+$$ y_{\text{max}} = h/2 $$
+
+it follows that
+
+$$ M = \frac{\sigma_{c(\text{max})} I}{y_{\text{max}}} \tag{4-5c} $$
+
+Thus, for the elastic case, identical answers are obtained from the traditional beam stress equation, Eq. (4-5c), and when the stress block concept is used in Eq. (4-5a).
+
+The elastic beam theory in Eq. (4-5c) is not used in the design of reinforced concrete beams, because the compressive stress-strain relationship for concrete becomes nonlinear at higher strain values, as shown in Fig. 3-18. What is even more important is
+
+that concrete cracks at low tensile stresses, making it necessary to provide steel reinforcement to carry the tensile force, $T$. These two factors are easily handled by the stress-block concept, combined with Eqs. (4-4) and (4-5).
+
+### Flexure Theory for Reinforced Concrete
+
+The theory of flexure for reinforced concrete is based on three basic assumptions, which are sufficient to allow one to calculate the moment resistance of a beam. These are presented first and used to illustrate flexural behavior, i.e., the *moment–curvature relationship* for a beam cross section under increasing moment. After gaining an understanding of the general development of a moment–curvature relationship for a typical beam section, a series of moment–curvature relationships will be developed to illustrate how changes in section properties and material strengths affect flexural behavior.
+
+#### Basic Assumptions in Flexure Theory
+
+Three basic assumptions are made:
+
+1. Sections perpendicular to the axis of bending that are plane before bending remain plane after bending.
+2. The strain in the reinforcement is equal to the strain in the concrete at the same level.
+3. The stresses in the concrete and reinforcement can be computed from the strains by using stress–strain curves for concrete and steel.
+
+The first of these is the traditional “plane sections remain plane” assumption made in the development of flexural theory for beams constructed with any material. The second assumption is necessary, because the concrete and the reinforcement must act together to carry load. This assumption implies a perfect bond between the concrete and the steel. The third assumption will be demonstrated in the following development of moment–curvature relationships for beam sections.
+
+#### Flexural Behavior
+
+General moment–curvature relationships will be used to describe and discuss the flexural behavior of a variety of beam sections. The initial discussion will be for *singly reinforced* sections, i.e., sections that have reinforcement only in their tension zone, as shown in Fig. 4-5. After singly reinforced sections have been discussed, a short discussion will be given on
+
+Fig. 4-5
+Typical singly reinforced sections in positive bending (tension in bottom).
+
+Fig. 4-6
+Typical doubly reinforced sections in positive bending.
+
+how adding steel in the compression zone to create a *doubly reinforced* section, as shown in Fig. 4-6, will affect flexural behavior. All of the sections considered here will be *under-reinforced*. Although this may sound like a bad design, this is exactly the type of cross section we will want to design to obtain the preferred type of flexural behavior. The meaning of an under-reinforced beam section is that, when the section is loaded in bending beyond its elastic range, the tension zone steel will *yield* before the concrete in the compression zone reaches its maximum useable strain, $\epsilon_{cu}$.
+
+To analytically create a moment–curvature relationship for any beam section, assumptions must be made for material stress–strain relationships. A simple elastic-perfectly plastic model will be assumed for the reinforcing steel in tension or compression, as shown in Fig. 4-7. The steel elastic modulus, $E_s$, is assumed to be 29,000 ksi.
+
+The stress–strain relationship assumed for concrete in compression is shown in Fig. 4-8. This model consists of a parabola from zero stress to the compressive strength of the concrete, $f_c'$. The strain that corresponds to the peak compressive stress, $\epsilon_o$, is often assumed to be 0.002 for normal strength concrete. The equation for this parabola, which was originally introduced by Hognestad [4-4], is
+
+$$ f_c = f_c' \left[ 2 \left( \frac{\epsilon_c}{\epsilon_o} \right) - \left( \frac{\epsilon_c}{\epsilon_o} \right)^2 \right] \tag{4-6} $$
+
+Fig. 4-7
+Assumed stress–strain relationship for reinforcing steel.
+
+Fig. 4-8
+Assumed stress–strain relationship for concrete.
+
+Beyond the strain, $\epsilon_o$, the stress is assumed to decrease linearly as the strain increases. An equation for this portion of the relationship can be expressed as
+
+$$ f_c = f_c' \left[ 1 - \frac{Z}{1000} \left( \frac{\epsilon_c - \epsilon_o}{\epsilon_o} \right) \right] \tag{4-7} $$
+
+where $Z$ is a constant to control the slope of the line. For this discussion, $Z$ will be set equal to a commonly used value of 150. Lower values for $Z$ (i.e., a shallower unloading slope) can be used if longitudinal and transverse reinforcement are added to confine the concrete in the compression zone.
+
+In tension the concrete is assumed to have a linear stress–strain relationship (Fig. 4-8) up to the concrete modulus of rupture, $f_r$, defined in Chapter 3.
+
+Consider a singly reinforced rectangular section subjected to positive bending, as shown in Fig. 4-9a. In this figure, $A_s$ represents the total *area of tension reinforcement*, and $d$ represents the *effective flexural depth* of the section, i.e., the distance from the extreme compression fiber to the centroid of the tension reinforcement. A complete moment–curvature relationship, as shown in Fig. 4-10, can be generated for this section by continuously increasing the section curvature (slope of the strain diagram) and using the assumed material stress–strain relationships to determine the resulting section stresses and forces, as will be discussed in the following paragraphs.
+
+(a) Basic section.
+(b) Strain distribution.
+(c) Stress distribution.
+(d) Internal forces.
+
+Fig. 4-9
+Steps in analysis of moment and curvature for a singly reinforced section.
+
+Fig. 4-10
+Moment–curvature relationship for the section in Fig. 4.9(a) using $f_c' = 4000$ psi and $f_y = 60$ ksi.
+
+The calculation of specific points on the moment–curvature curve follows the process represented in Fig. 4-9b through 4-9d. Each point is usually determined by selecting a specific value for the maximum compression strain at the extreme compression fiber of the section, $\epsilon_c(\text{max})$. From the assumption that plane sections before bending remain plane, the strain distribution through the depth of the section is linear. From the strain diagram and the assumed material stress–strain relationships, the distribution of stresses is determined. Finally, by integration, the volume under the stress distributions (i.e., the section forces) and their points of action can be determined.
+
+After the section forces are determined, the following steps are required to complete the calculation. First, the distance from the extreme compression fiber to the section neutral axis (shown as $x$ in Fig. 4-9b) must be adjusted up or down until section equilibrium is established, as given by Eq. (4-2). When Eq. (4-2) is satisfied, the curvature, $\Phi$, for this point is calculated as the slope of the strain diagram,
+
+$$ \Phi = \frac{\epsilon_c(\text{max})}{x} \tag{4-8} $$
+
+The corresponding moment is determined by summing the moments of the internal forces about a convenient point—often selected to be the centroid of the tension reinforcement. This process can be repeated for several values of maximum compression strain. A few maximum compression strain values are indicated at selected points in Fig. 4-10. Exceptions to this general procedure will be discussed for the cracking and yield points.
+
+#### Cracking Point
+
+Flexural tension cracking will occur in the section when the stress in the extreme tension fiber equals the modulus of rupture, $f_r$. Up to this point, the moment–curvature relationship is linear and is referred to as the *uncracked-elastic* range of behavior (from $O$ to $C$ in Fig. 4-10). The moment and curvature at cracking can be calculated directly from elastic
+
+flexural theory, as expressed in Eq. 4-5c. In most cases, the contribution of the reinforcement can be ignored in this range of behavior, and the cracking moment can be calculated using only the concrete section, normally referred to as the *gross section*. If the moment of inertia for the gross section is defined as $I_g$, and the distance from the section centroid to the extreme tension fiber is defined as $y_t$, then the stress at the extreme tension fiber in a modified version of Eq. (4-5c) is
+
+$$ f = \frac{M y_t}{I_g} \tag{4-5d} $$
+
+The *cracking moment* is defined as the moment that causes the stress in the extreme tension fiber to reach the modulus of rupture, i.e.,
+
+$$ M_{cr} = \frac{f_r I_g}{y_t} \tag{4-9} $$
+
+This expression is the same as used in ACI Code Section 9.5. When calculating $M_{cr}$, it is recommended to take the modulus of rupture, $f_r$, equal to $7.5\sqrt{f_c'}$. The reinforcement could be included in this calculation by using a transformed section method to define the section properties, but for typical sections, this would result in a relatively small change in the value for $M_{cr}$.
+
+The section curvature at cracking, $\Phi_{cr}$, can be calculated for this point using the elastic bending theory,
+
+$$ \Phi_{cr} = \frac{M_{cr}}{E_c I_g} \tag{4-10} $$
+
+where the elastic concrete modulus, $E_c$, can be taken as $57,000\sqrt{f_c'}$, in psi units.
+
+When a beam section cracks in tension, the crack usually propagates to a point near the centroid of the section and there is a sudden transfer of tension force from the concrete to the reinforcing steel in the tension zone. Unless a minimum amount of reinforcement is present in the tension zone, the beam would fail suddenly. To prevent such a brittle failure, the minimum moment strength for a *reinforced* concrete beam section should be equal to or greater than the cracking moment strength for the *plain* concrete section. Such a specific recommendation is not given in the ACI Code for reinforced concrete beam sections. However, based on a calculation that sets the moment capacity of a reinforced section equal to approximately twice that of a plain section, ACI Code Section 10.5 specifies the following minimum area of longitudinal reinforcement for beam sections in positive bending as
+
+$$ A_{s,\text{min}} = \frac{3\sqrt{f_c'}}{f_y} b_w d \tag{4-11} $$
+
+where the quantity $3\sqrt{f_c'}$ is not to be taken less than 200 psi. The notation for *web width*, $b_w$, is used here to make the equation applicable for both rectangular and flanged sections. Additional discussion of this minimum area requirement is given in Section 4-8 for flanged sections with the flange in tension.
+
+From the metric version of ACI Code Section 10.5, using MPa units for $f_c'$ and $f_y$ the expression for $A_{s,\text{min}}$ is:
+
+$$ A_{s,\text{min}} = \frac{0.25\sqrt{f_c'}}{f_y} b_w d \ge \frac{1.4 b_w d}{f_y} \tag{4-11M} $$
+
+After cracking but before yielding of the tension reinforcement, the relationship between moment and curvature is again approximately linear, but with a different slope than before cracking. This is referred to as the *cracked-elastic* range of behavior (from C to Y in Fig. 4-10). This linear relationship is important for the calculation of deflections, as will be discussed in Chapter 9.
+
+#### Yield Point
+
+The yield point represents the end of the elastic range of behavior. As the moment applied to the section continues to increase after the cracking point, the tension stress in the reinforcement and the compression stress in the concrete compression zone will steadily increase. Eventually, either the steel or the concrete will reach its respective capacity and start to yield (steel) or crush (concrete). Because the section under consideration here is assumed to be *under-reinforced*, the steel will yield before the concrete reaches its maximum useable strain.
+
+To calculate moment and curvature values for the yield point, the strain at the level of the tension steel is set equal to the yield strain ($\epsilon_y = f_y/E_s$). As discussed previously for the general procedure, the neutral axis needs to be adjusted up or down until section equilibrium is established. At this stage of flexural behavior the contribution of the concrete in tension is not significant for section equilibrium and moment calculations, so the vector, $T_c$, shown in Fig. 4-9d can be ignored. After section equilibrium is established, the section *yield moment*, $M_y$, is then calculated as the sum of the moments of the internal forces about a convenient point. Referring to Fig. 4-9b, the *yield curvature* is calculated as the slope of the strain diagram, which can be calculated by setting the strain at the level of the tension reinforcement equal to the yield strain,
+
+$$ \Phi_y = \frac{\epsilon_y}{d - x} \tag{4-12} $$
+
+#### Points beyond the Yield Point
+
+Additional points on the moment–curvature relationship can be determined by steadily increasing the maximum strain in the extreme compression fiber, following the general procedure described previously. Usually, points are generated until some predefined maximum useable compression strain is reached or until the section moment capacity drops significantly below the maximum calculated value. Points representing maximum compression strains of 0.003, 0.004, 0.005, and 0.006 are noted in Fig. 4-10. For each successive point beyond the compression strain of 0.003, the section moment capacity is decreasing at an increasing rate. If a more realistic model was used for the stress–strain properties of the reinforcing steel, i.e. a model that includes strain hardening (Fig. 3-29), the moment capacity would increase beyond the yield point and would hold steady or at least show a less significant decrease in moment capacity for maximum compressive strain values greater than 0.003.
+
+Most concrete design codes specify a maximum useable compression strain at which the *nominal moment strength* of the section is to be calculated. For the ACI Code, this maximum useable strain value is specified as 0.003. For the Canadian Concrete Code [4-5], a maximum useable compressive strain value of 0.0035 is specified. It should be clear from Fig. 4-10 that the calculation of a nominal moment capacity for this section would not be affected significantly by selection of either one of these values. Also, the beam section shown here has considerable deformation capacity beyond the limit corresponding to either of the maximum compression strains discussed here.
+
+Any discussion of flexural behavior of a reinforced concrete beam section usually involves a discussion of *ductility*, i.e., the ability of a section to deform beyond its yield point without a significant strength loss. Ductility can be expressed in terms of displacement, rotation, or curvature ratios. For this discussion, section ductility will be expressed in terms of the ratio of the curvature at maximum useable compression strain to the curvature at yield. The maximum useable strain can be expressed as a specific value, as done by most codes, or it could be defined as the strain at which the moment capacity of the section has dropped below some specified percentage of the maximum moment capacity of the section. By either measure, the moment–curvature relationship given in Fig. 4-10 represents good ductile behavior.
+
+#### Effect of Major Section Variables on Strength and Ductility
+
+In this subsection, a series of systematic changes are made to section parameters for the beam given in Fig. 4-9a to demonstrate the effect of such parametric changes on the moment–curvature response of the beam section. Values of material strengths and section parameters are given for seven different beams in Table 4-1. The first column (Basic Section) represents the original values that correspond to the $M-\Phi$ curve given in Fig. 4-10. Each successive beam section (represented by a column in Table 4-1) represents a modification of either the material properties or section dimensions from those for the *basic section*. Note that for each new beam section (column in table) only one of the parameters has been changed from those used for the basic section.
+
+$M-\Phi$ plots that correspond to the first three sections given in Table 4-1 are shown in Fig. 4-11. The only change for these sections is an increase in the area of tension reinforcement, $A_s$. It is clear that increasing the tension steel area causes a proportional increase in the strength of the section. However, the higher tension steel areas also causes a less ductile behavior for the section. Because of this loss of ductility as the tension steel area is increased, the ACI Code places an upper limit on the permissible area of tension reinforcement, as will be discussed in detail in Section 4-6.
+
+Figure 4-12 shows $M-\Phi$ plots for the basic section and for the sections defined in the last four columns of Table 4-1. It is interesting from a design perspective to observe how changes in the different section variables affect the flexural strength, stiffness, and ductility of the beam sections. An increase in the steel yield strength has essentially the same effect as increasing the tension steel area—that is the section moment strength increases and the section ductility decreases. Increases in either the steel yield strength or the tension steel area have very little effect on the stiffness of the section before yield (as represented by the elastic slope of the $M-\Phi$ relationship).
+
+Increasing the effective flexural depth of the section, $d$, increases the section moment strength without decreasing the section ductility. Increasing the effective flexural depth
+
+TABLE 4-1 Material and Section Properties for Various Beam Sections
+| Primary Variables | Basic Section | Moderate* $A_s$ | High* $A_s$ | High* $f_y$ | Large* $d$ | High* $f_c'$ | Large* $b$ |
+| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
+| $A_s$ (sq.in.) | 2.5 | 4.5 | 6.5 | 2.5 | 2.5 | 2.5 | 2.5 |
+| $f_y$ (ksi) | 60 | 60 | 60 | 80 | 60 | 60 | 60 |
+| $d$ (in.) | 21.5 | 21.5 | 21.5 | 21.5 | 32.5 | 21.5 | 21.5 |
+| $f_c'$ (psi) | 4000 | 4000 | 4000 | 4000 | 4000 | 6000 | 4000 |
+| $b$ (in.) | 12 | 12 | 12 | 12 | 12 | 12 | 18 |
+
+*Relative to values in the basic section.
+
+Fig. 4-11
+Effect of increasing tension steel area, $A_s$.
+
+also increases the elastic stiffness of the section, because the section moment of inertia is significantly affected by the depth of the section. These results clearly indicate the importance of the effective flexural depth of a member, so proper placement of reinforcement during construction should be a priority item for field inspectors.
+
+Changes in concrete strength and section width have a smaller effect on moment strength than might be initially expected. These two variables will have only a small
+
+Fig. 4-12
+Effect of increasing $f_y$, $d$, $f_c'$, $b$, and $A_s'$.
+
+affect on the moment arm between the tension and compression forces shown in Fig. 4-9d, but they do not affect the value of the tension (and thus the compression) force. Thus, these variables have a significantly smaller effect on moment strength of the section than the tension steel area, steel yield strength, and effective flexural depth. Because final failure in bending for these sections is governed by reaching the maximum useable compression strain in the extreme concrete compression fiber, increases in either the concrete strength or section width do cause a significant increase in curvature at failure, as calculated in Eq. (4-8), by decreasing the neutral axis depth required to balance the tension force from the reinforcing steel.
+
+The last variable discussed here is the addition of compression zone reinforcement, $A_s'$, equal to one-half of the area of tension reinforcement, $A_s$. This variable is not listed in Table 4-1, because all of the other cross-section values are set equal to those listed for the basic section. As shown in Fig. 4-12, the addition of compression reinforcement has very little effect on the moment strength of the beam section. However, because the compression reinforcement carries part of the compression force that would be carried by the concrete in a singly reinforced beam, the required depth of the neutral axis is decreased and the section reaches a much higher curvature (higher ductility) before the concrete reaches its maximum useable strain. Thus, one of the primary reasons for using compression reinforcement will be to increase the ductility of a given beam section.
+
+## 4-3 SIMPLIFICATIONS IN FLEXURE THEORY FOR DESIGN
+
+The three assumptions already made are sufficient to allow calculation of the strength and behavior of reinforced concrete elements. For design purposes, however, the following additional assumptions are introduced to simplify the problem with little loss of accuracy.
+
+1. The tensile strength of concrete is neglected in flexural-strength calculations (ACI Code Section 10.2.5).
+
+The strength of concrete in tension is roughly one-tenth of the compressive strength, and the tensile force in the concrete below the zero strain axis, shown as $T_c$ in Fig. 4-9d, is small compared with the tensile force in the steel. Hence, the contribution of the tensile stresses in the concrete to the flexural capacity of the beam is small and can be neglected. It should be noted that this assumption is made primarily to simplify flexural calculations. In some instances, particularly shear, bond, deflection, and service-load calculations for prestressed concrete, the tensile resistance of concrete is not neglected.
+
+2. The section is assumed to have reached its nominal flexural strength when the strain in the extreme concrete compression fiber reaches the maximum useable compression strain, $\epsilon_{cu}$.
+
+Strictly speaking, this is an artificial limit developed by code committees to define at what point on the general moment–curvature relationship the nominal strength of the section is to be calculated. As shown in Fig. 4-10, the moment–curvature relationship for a typical beam section is relatively flat after passing the yield point, so the selection of a specific value for $\epsilon_{cu}$ will not significantly affect the calculated value for the nominal flexural strength of the section. Thus, design calculations are simplified when a limiting strain is assumed.
+
+The maximum compressive strains, $\epsilon_{cu}$, from tests of beams and eccentrically loaded columns of normal-strength, normal-density concrete are plotted in Fig. 4-13a [4-6], [4-7]. Similar data from tests of normal-density and lightweight concrete are compared in Fig. 4-13b. ACI Code Section 10.2.3 specifies a limiting compressive strain, $\epsilon_{cu}$, equal to 0.003, which approximates the smallest measured values plotted in Fig. 4-13a and b. In
+
+Fig. 4-13
+Limiting compressive strain.
+(From [4-6] and [4-7].)
+
+(a) Ultimate strain from tests of reinforced members.
+
+(b) Ultimate strain from tests of plain concrete specimens.
+
+Canada, the CSA Standard [4-5] uses $\epsilon_{cu} = 0.0035$ for beams and eccentrically loaded columns. Higher limiting strains have been measured in members with a significant moment gradient and in members in which the concrete is confined by spirals or closely spaced hoops [4-8], [4-9]. Throughout this book, however, a constant maximum useable compressive strain equal to 0.003 will be used.
+
+3. The compressive stress–strain relationship for concrete may be based on measured stress–strain curves or may be assumed to be rectangular, trapezoidal, parabolic, or any other shape that results in prediction of flexural strength in substantial agreement with the results of comprehensive tests (ACI Code Section 10.2.6).
+
+Thus, rather than using a closely representative stress–strain curve (such as that given in Fig. 4-8), other diagrams that are easier to use in computations are acceptable, provided they adequately predict test results. As is illustrated in Fig. 4-14, the shape of the
+
+Fig. 4-14
+Mathematical description of compression stress block.
+
+stress block in a beam at the ultimate moment can be expressed mathematically in terms of three constants:
+
+$k_3 =$ ratio of the maximum stress, $f_c''$, in the compression zone of a beam to the cylinder strength, $f_c'$
+$k_1 =$ ratio of the average compressive stress to the maximum stress (this is equal to the ratio of the shaded area in Fig. 4-15 to the area of the rectangle, $c \times k_3 f_c'$)
+$k_2 =$ ratio of the distance between the extreme compression fiber and the resultant of the compressive force to the depth of the neutral axis, $c$, as shown in Figs. 4-14 and 4-15.
+
+For a rectangular compression zone of width $b$ and depth to the neutral axis $c$, the resultant compressive force is
+
+$$ C = k_1 k_3 f_c' b c \tag{4-13a} $$
+
+Values of $k_1$ and $k_2$ are given in Fig. 4-15 for various assumed compressive stress–strain diagrams or *stress blocks*. The use of the constant $k_3$ essentially has disappeared from the flexural theory of the ACI Code. As shown in Fig. 4-12, a large change in the concrete compressive strength did not cause a significant change in the beam section moment capacity. Thus, the use of either $f_c'$ or $f_c'' = k_3 f_c'$, with $k_3$ typically taken equal to 0.85, is not significant for the flexural analysis of beams. The use of $f_c''$ is more significant for column sections subjected to high axial load and bending. Early papers by Hognestad [4-10] and
+
+Fig. 4-15
+Values of $k_1$ and $k_2$ for various stress distributions.
+
+(a) Concrete.
+(b) Triangle.
+(c) Parabola.
+
+Phrang, Siess, and Sozen [4-11] recommended the use of $f_c'' = k_3 f_c'$ when analyzing the combined axial and bending strength for column sections. However, the ACI Code does not refer to the use of $f_c''$ except for column sections subjected to pure axial load (no bending), as will be discussed in Chapter 11.
+
+#### Whitney Stress Block
+
+As a further simplification, ACI Code Section 10.2.7 permits the use of an equivalent rectangular concrete stress distribution shown in Fig. 4-16 for nominal flexural strength calculations. This rectangular stress block, originally proposed by Whitney [4-12], is defined by the following:
+
+1. A uniform compressive stress of $0.85 f_c'$ shall be assumed distributed over an equivalent compression zone bounded by the edges of the cross section and a straight line located parallel to the neutral axis at a distance $a = \beta_1 c$ from the concrete fiber with the maximum compressive strain. Thus, $k_2 = \beta_1 / 2$, as shown in Fig. 4-16.
+2. The distance $c$ from the fiber of maximum compressive strain to the neutral axis is measured perpendicular to that axis.
+3. The factor $\beta_1$ shall be taken as follows [4-7]:
+
+(a) For concrete strengths, $f_c'$, up to and including 4000 psi,
+
+$$ \beta_1 = 0.85 \tag{4-14a} $$
+
+(b) For 4000 psi $< f_c' \le$ 8000 psi,
+
+$$ \beta_1 = 0.85 - 0.05 \frac{f_c' - 4000 \text{ psi}}{1000 \text{ psi}} \tag{4-14b} $$
+
+(c) For $f_c'$ greater than 8000 psi,
+
+$$ \beta_1 = 0.65 \tag{4-14c} $$
+
+For a rectangular compression zone of constant width $b$ and depth to the neutral axis $c$, the resultant compressive force is
+
+$$ C = 0.85 f_c' b \beta_1 c = 0.85 \beta_1 f_c' b c \tag{4-13b} $$
+
+Comparing Eqs. 4-13a and 4-13b, and setting $k_3 = 1.0$, results in $k_1 = 0.85 \beta_1$.
+
+Fig. 4-16
+Equivalent rectangular stress block.
+
+(a) Stresses
+(b) Forces
+
+In metric units (MPa), the factor $\beta_1$ shall be taken as follows:
+
+(a) For concrete strengths, $f_c'$, up to and including 28 MPa,
+
+$$ \beta_1 = 0.85 \tag{4-14Ma} $$
+
+(b) For 28 MPa $< f_c' \le$ 56 MPa,
+
+$$ \beta_1 = 0.85 - 0.05 \frac{f_c' - 28 \text{ MPa}}{7 \text{ MPa}} \tag{4-14Mb} $$
+
+(c) For $f_c'$ greater than 58 MPa,
+
+$$ \beta_1 = 0.65 \tag{4-14Mc} $$
+
+The dashed line in Fig. 4-17 is a lower-bound line corresponding to a rectangular stress block with a height of $0.85 f_c'$ and by using $\beta_1$ as given by Eq. (4-14). This equivalent rectangular stress block has been shown [4-6], [4-7] to give very good agreement with test data for calculation of the nominal flexural strength of beams. For columns, the agreement is good up to a concrete strength of about 6000 psi. For columns loaded with small eccentricities and having strengths greater than 6000 psi, the moment capacity tends to be overestimated by the ACI Code stress block. This is because Eq. (4-14) for $\beta_1$ was chosen as a lower bound on the test data, as indicated by the dashed line in Fig. 4-17. The internal moment arm of the compression force in the concrete about the *centroidal axis* of a rectangular column is $(h/2 - \beta_1 c / 2)$, where $c$ is the depth to the neutral axis (axis of zero strain). If $\beta_1$ is too small, the moment arm will be too large, and the moment capacity will be overestimated.
+
+To correct this potential error, which can lead to unconservative designs of columns constructed with high-strength concrete, an ACI Task Group [4-13] has recommended the
+
+Fig. 4-17
+Values of $\beta_1$ from tests of concrete prisms. (From [4-7].)
+
+use of a coefficient, $\alpha_1$, to replace the constant $0.85$ as the definition for the height of the stress block shown in Fig. 4-16a. This new coefficient is defined as follows:
+
+(a) For concrete strengths, $f_c'$, up to and including $8000$ psi,
+
+$$ \alpha_1 = 0.85 \tag{4-15a} $$
+
+(b) For $8000\text{ psi} < f_c' \le \text{between } 8000 \text{ and } 18,000 \text{ psi}$,
+
+$$ \alpha_1 = 0.97 - 0.015 \frac{f_c'}{1000} \tag{4-15b} $$
+
+(c) For $f_c'$ greater than $18,000$ psi,
+
+$$ \alpha_1 = 0.70 \tag{4-15c} $$
+
+Until the ACI Code adopts a modification of the stress block shown in Fig. 4-16a, the authors recommend the use of this coefficient, $\alpha_1$, when analyzing the flexural strength of columns constructed with concrete strengths exceeding $8000$ psi.
+
+## 4-4 ANALYSIS OF NOMINAL MOMENT STRENGTH FOR SINGLY REINFORCED BEAM SECTIONS
+
+### Stress and Strain Compatibility and Section Equilibrium
+
+Two requirements are satisfied throughout the flexural analysis and design of reinforced concrete beams and columns:
+
+1. **Stress and strain compatibility.** The stress at any point in a member must correspond to the strain at that point. Except for short, deep beams, the distribution of strains over the depth of the member is assumed to be linear.
+2. **Equilibrium.** Internal forces must balance the external load effects, as illustrated in Fig. 4-3 and Eq. (4-2).
+
+### Analysis of Nominal Moment Strength, $M_n$
+
+Consider the singly reinforced beam section shown in Fig. 4-18a subjected to positive bending (tension at the bottom). As was done in the previous section, it will be assumed that this is an *under-reinforced* section, i.e., the tension steel will yield before the extreme concrete compression fiber reaches the maximum useable compression strain. In Section 4-5, a definition will be given for a “balanced” steel area, which results in a beam section where the tension steel will just be reaching the yield strain when the extreme concrete compression fiber is reaching the maximum useable compression strain. Because the ACI Code requires that beam sections be under-reinforced, this initial discussion will concentrate on the nominal moment strength evaluation for under-reinforced sections.
+
+As was done in the general analysis, a linear strain distribution is assumed for the section in Fig. 4-18b. For the evaluation of the nominal moment capacity of the section, the strain in the extreme compression fiber is set equal to the maximum useable strain, $\epsilon_{cu}$. The depth to the neutral axis, $c$, is unknown at this stage of the analysis. The strain at the level of the tension reinforcement is also unknown, but it is assumed to be greater than the yield strain. This assumption *must be confirmed* later in the calculation.
+
+(a) Singly reinforced section.
+(b) Strain distribution.
+(c) Stress distribution.
+(d) Internal forces.
+
+Fig. 4-18
+Steps in analysis of $M_n$ for singly reinforced rectangular sections.
+
+The assumed stress distribution is given in Fig. 4-18c. Above the neutral axis, the stress-block model from Fig. 4-16 is used to replace the actual concrete stress distribution. The coefficient $\beta_1$ is multiplied by the depth to the neutral axis, $c$, to get the depth of the stress block, $a$. The concrete is assumed to carry no tension, so there is no concrete stress distribution below the neutral axis. At the level of the steel, the stress, $f_s$, is assumed to be equal to the steel yield stress, $f_y$. This corresponds to the assumptions that the steel strain exceeds the yield strain and that the steel stress remains constant after yielding occurs (Fig. 4-7).
+
+The final step is to go from the stress distributions to the equivalent section forces shown in Fig. 4-18d. The concrete compression force, $C_c$, is equal to the volume under the stress block. For the rectangular section used here,
+
+$$ C_c = 0.85 f_c' b \beta_1 c = 0.85 f_c' b a \tag{4-13b} $$
+
+The compression force in the concrete cannot be evaluated at this stage, because the depth to the neutral axis is still unknown. The tension force shown in Fig. 4-18d is equal to the tension steel area, $A_s$, multiplied by the yield stress, $f_y$. Based on the assumption that the steel has yielded, this force is known.
+
+A key step in section analysis is to enforce section equilibrium. For this section, which is assumed to be subject to only bending (no axial force), the sum of the compression forces must be equal to the sum of the tension forces. So,
+
+$$ C_c = T \tag{4-2} $$
+
+or
+
+$$ 0.85 f_c' b \beta_1 c = 0.85 f_c' b a = A_s f_y $$
+
+The only unknown in this equilibrium equation is the depth of the stress block. So, solving for the unknown value of $a$,
+
+$$ a = \beta_1 c = \frac{A_s f_y}{0.85 f_c' b} \tag{4-16} $$
+
+and
+
+$$ c = \frac{a}{\beta_1} \tag{4-17} $$
+
+With the depth to the neutral axis known, the assumption of yielding of the tension steel can be checked. From similar triangles in the linear strain distribution in Fig. 4-18b, the following expression can be derived:
+
+$$ \frac{\epsilon_s}{d - c} = \frac{\epsilon_{cu}}{c} $$
+
+$$ \epsilon_s = \left( \frac{d - c}{c} \right) \epsilon_{cu} \tag{4-18} $$
+
+To confirm the assumption that the section is under-reinforced and the steel is yielding, show
+
+$$ \epsilon_s \ge \epsilon_y = \frac{f_y}{E_s} = \frac{f_y \text{ (ksi)}}{29,000 \text{ ksi}} \tag{4-19} $$
+
+Once this assumption is confirmed, the nominal-section moment capacity can be calculated by referring back to the section forces in Fig. 4-18d. The compression force is acting at the middepth of the stress block, and the tension force is acting at a distance $d$ from the extreme compression fiber. Thus, the nominal moment strength can be expressed as either the tension force or the compression force multiplied by the moment arm, $d - a/2$:
+
+$$ M_n = T \left( d - \frac{a}{2} \right) = C_c \left( d - \frac{a}{2} \right) \tag{4-20} $$
+
+For singly reinforced sections, it is more common to express the nominal moment strength using the definition of the tension force as
+
+$$ M_n = A_s f_y \left( d - \frac{a}{2} \right) \tag{4-21} $$
+
+This simple expression can be used for all singly reinforced sections with a rectangular (constant width) compression zone after it has been confirmed that the tension steel is yielding. The same fundamental process as used here to determine $M_n$ for singly reinforced rectangular sections will be applied to other types of beam sections in the following parts of this chapter. However, the reader is urged to concentrate on the process rather than the resulting equations. If the process is understood, it can be applied to any beam section that may be encountered.
+
+### Example 4-1 Calculation of $M_n$ for a Singly Reinforced Rectangular Section
+
+For the beam shown in Fig. 4-19a, calculate $M_n$ and confirm that the area of tension steel exceeds the required minimum steel area given by Eq. (4-11). The beam section is made of concrete with a compressive strength, $f_c' = 4000$ psi, and has four No. 8 bars with a yield strength of $f_y = 60$ ksi.
+
+For this beam with a single layer of tension reinforcement, it is reasonable to assume that the effective flexural depth, $d$, is approximately equal to the total beam depth minus 2.5 in. This accounts for a typical concrete clear cover of 1.5 in., the diameter of the stirrup (typically a No. 3 or No. 4 bar) and half the diameter of the beam longitudinal reinforcement. Depending on the sizes of the stirrup and longitudinal bar, the dimension to the center of the steel layer will vary slightly, but the use of 2.5 in. will be accurate enough for most design work unless adjustments in reinforcement location are required to avoid rebar interference at connections with other members. Small bars are often used in the compression zone to hold the stirrups in position, but these bars normally are ignored unless they were specifically designed to serve as compression-zone reinforcement.
+
+Fig. 4-19
+(a) Beam sections for Examples 4-1 and 4-1M.
+
+1. **Following the procedure summarized in Fig. 4-18, assume that the steel strain exceeds the yield strain, and thus, the stress $f_s$ in the tension reinforcement equals the yield strength, $f_y$. Compute the steel tension force:**
+
+$$ A_s = 4 \text{ No. 8 bars} = 4 \times 0.79 \text{ in.}^2 = 3.16 \text{ in.}^2 $$
+
+$$ T = A_s f_y = 3.16 \text{ in.}^2 \times 60 \text{ ksi} = 190 \text{ kips} $$
+
+The assumption that $\epsilon_s > \epsilon_y$ will be checked in step 3. This assumption generally should be true, because the ACI Code requires that the steel area be small enough in beam sections such that the steel will yield before the concrete reaches the maximum useable compression strain.
+
+2. **Compute the area of the compression stress block so that $C_c = T$.** This is done for the equivalent rectangular stress block shown in Fig. 4-16a. The stress block consists of a uniform stress of $0.85 f_c'$ distributed over a depth $a = \beta_1 c$ which is measured from the extreme compression fiber. For $f_c' = 4000$ psi, Eq. (4-14a) gives $\beta_1 = 0.85$. Using Eq. (4-16), which was developed from section equilibrium,
+
+$$ a = \beta_1 c = \frac{A_s f_y}{0.85 f_c' b} = \frac{190 \text{ kips}}{0.85 \times 4 \text{ ksi} \times 12 \text{ in.}} = 4.66 \text{ in.} $$
+
+3. **Check that the tension steel is yielding.** The yield strain is
+
+$$ \epsilon_y = \frac{f_y}{E_s} = \frac{60 \text{ ksi}}{29,000 \text{ ksi}} = 0.00207 $$
+
+From above, $c = a / \beta_1 = 5.48$ in. Now, use strain compatibility, as expressed in Eq. (4-18), to find
+
+$$ \epsilon_s = \left( \frac{d - c}{c} \right) \epsilon_{cu} = \left( \frac{17.5 - 5.48}{5.48} \right) 0.003 = 0.00658 $$
+
+Clearly, $\epsilon_s$ exceeds $\epsilon_y$, so the assumption used above to establish section equilibrium is confirmed. Remember that you *must make this check* before proceeding to calculate the section nominal moment strength.
+
+4. **Compute $M_n$.** Using Eq. (4-21), which was derived for sections with constant width compression zones,
+
+$$ M_n = A_s f_y \left( d - \frac{a}{2} \right) = 190 \text{ kips} \left( 17.5 \text{ in.} - \frac{4.66 \text{ in.}}{2} \right) $$
+
+$$ M_n = 2880 \text{ k-in.} = 240 \text{ k-ft} $$
+
+5. **Confirm that tension steel area exceeds $A_{s,\min}$.** For Eq. (4-11), there is a requirement to use the larger of $3\sqrt{f_c'}$ or 200 psi in the numerator. In this case, $3\sqrt{4000 \text{ psi}} = 190$ psi, so use 200 psi. Thus,
+
+$$ A_{s,\min} = \frac{200 \text{ psi}}{f_y} b_w d = \frac{200 \text{ psi}}{60,000 \text{ psi}} \times 12 \text{ in.} \times 17.5 \text{ in.} = 0.70 \text{ in.}^2 $$
+
+$A_s$ exceeds $A_{s,\min}$, so this section satisfies the ACI Code requirement for minimum tension reinforcement.
+
+### Example 4-1M Analysis of Singly Reinforced Beams: Tension Steel Yielding—SI Units
+
+Compute the nominal moment strength, $M_n$, of a beam (Fig. 4-19b) with $f_c' = 20$ MPa ($\beta_1 = 0.85$), $f_y = 420$ MPa, $b = 250$ mm, $d = 500$ mm, and three No. 25 bars (Table A-1M) giving $A_s = 3 \times 510 = 1530 \text{ mm}^2$. Note that the difference between the total section depth, $h$, and the effect depth, $d$, is 65 mm, which is a typical value for beam sections designed with metric dimensions.
+
+1. **Compute $a$ (assuming the tension steel is yielding).**
+
+$$ a = \frac{A_s f_y}{0.85 f_c' b} $$
+
+$$ = \frac{1530 \text{ mm}^2 \times 420 \text{ MPa}}{0.85 \times 20 \text{ MPa} \times 250 \text{ mm}} = 151 \text{ mm} $$
+
+Therefore, $c = a / \beta_1 = 151 / 0.85 = 178$ mm.
+
+2. **Check whether the tension steel is yielding.** The yield strain for the reinforcing steel is
+
+$$ \epsilon_y = \frac{f_y}{E_s} = \frac{420 \text{ MPa}}{200,000 \text{ MPa}} = 0.0021 $$
+
+From Eq. (4-18),
+
+$$ \epsilon_s = \left( \frac{500 \text{ mm} - 178 \text{ mm}}{178 \text{ mm}} \right) \times 0.003 = 0.00543 $$
+
+Thus, the steel is yielding as assumed in step 1.
+
+3. **Compute the nominal moment strength, $M_n$.** From Eq. (4-21), $M_n$ is (where 1 MPa $= 1 \text{ N/mm}^2$)
+
+$$ M_n = A_s f_y \left( d - \frac{a}{2} \right) = 1530 \text{ mm}^2 \times 420 \text{ N/mm}^2 \left( 500 - \frac{151}{2} \right) \text{mm} $$
+
+$$ = 273 \times 10^6 \text{ N-mm} = 273 \text{ kN-m} $$
+
+Therefore, the design or factored moment strength, $\phi M_n$, of this beam is $0.9 \times 273 = 246$ kN-m.
+
+4. **Confirm that the tension steel area exceeds $A_{s,\min}$.** For the given concrete strength of 20 MPa, the quantity $0.25\sqrt{f_c'} = 1.12$ MPa, which is less than 1.4 MPa. Therefore, the second part of Eq. (4-11M) governs for $A_{s,\min}$ as
+
+$$ A_{s,\min} = \frac{1.4 b_w d}{f_y} = \frac{1.4 \text{ MPa} \times 250 \text{ mm} \times 500 \text{ mm}}{420 \text{ MPa}} = 417 \text{ mm}^2 $$
+
+$A_s$ exceeds $A_{s,\min}$, so this section satisfies the ACI Code requirement for minimum tension reinforcement.
+
+### Example 4-2 Calculation of the Nominal Moment Strength for an Irregular Cross Section
+
+The beam shown in Fig. 4-20 is made of concrete with a compressive strength, $f_c' = 3000$ psi and has three No. 8 bars with a yield strength, $f_y = 60$ ksi. This example is presented to demonstrate the general use of strain compatibility and section equilibrium equations for any type of beam section.
+
+1. **Initially, assume that the stress $f_s$ in the tension reinforcement equals the yield strength $f_y$, and compute the tension force $T = A_s f_y$:**
+
+$$ A_s = 3 \text{ No. 8 bars} = 3 \times 0.79 \text{ in.}^2 = 2.37 \text{ in.}^2 $$
+
+$$ T = A_s f_y = 2.37 \text{ in.}^2 \times 60 \text{ ksi} = 142 \text{ kips} $$
+
+The assumption that the tension steel is yielding will be checked in step 3.
+
+2. **Compute the area of the compression stress block so that $C_c = T$.** As in the prior problem, this is done using the equivalent rectangular stress block shown in Fig. 4-16a. The stress block consists of a uniform stress of $0.85 f_c'$ distributed over a depth $a = \beta_1 c$, which is measured from the extreme compression fiber. For $f_c' = 3000$ psi, Eq. (4-14a) gives $\beta_1 = 0.85$. The magnitude of the compression force is obtained from equilibrium as
+
+$$ C_c = T = 142 \text{ kips} = 142,000 \text{ lbs} $$
+
+By the geometry of this particular triangular beam, shown in Fig. 4-20, if the depth of the compression zone is $a$, the width at the bottom of the compression zone is also $a$, and the area is $a^2 / 2$. This is, of course, true only for a beam of this particular triangular shape.
+
+Therefore, $C_c = (0.85 f_c')(a^2/2)$ and
+
+$$ a = \sqrt{\frac{142,000 \text{ lb} \times 2}{0.85 \times 3000 \text{ psi}}} = 10.6 \text{ in.} $$
+
+3. **Check whether $f_s = f_y$.** This is done by using strain compatibility. The strain distribution at ultimate is shown in Fig. 4-20c. As before,
+
+$$ c = \frac{a}{\beta_1} = \frac{10.6 \text{ in.}}{0.85} = 12.4 \text{ in.} $$
+
+(a) Cross section.
+(b) Elevation.
+(c) Strain distribution.
+
+Fig. 4-20
+Analysis of arbitrary cross section—Example 4-2.
+
+Using strain compatibility, as expressed in Eq. (4-18), calculate
+
+$$ \epsilon_s = \left( \frac{21.5 - 12.4}{12.4} \right) 0.003 = 0.00220 $$
+
+Although this is close to the yield strain, it does exceed the yield strain as calculated in step 3 of Example 4-1 for Grade-60 reinforcement. Thus, the assumption made in step 1 is satisfied.
+
+4. **Compute $M_n$.**
+
+$$ M_n = C_c jd = T jd $$
+
+where $jd$ is a general expression for the lever arm, i.e., the distance from the resultant tensile force (at the centroid of the reinforcement) to the resultant compressive force $C_c$. Because the area on which the compression stress block acts is triangular in this example, $C_c$ acts at $2a/3$ from top of the beam. Therefore,
+
+$$ jd = d - \frac{2a}{3} $$
+
+$$ M_n = A_s f_y \left( d - \frac{2a}{3} \right) $$
+
+$$ = 2.37 \text{ in.}^2 \times 60 \text{ ksi} \left( 21.5 - \frac{2 \times 10.6}{3} \right) \text{ in.} $$
+
+$$ = 2060 \text{ lb-in.} = 171 \text{ k-ft} $$
+
+**Note:** If we wanted to calculate $A_{s,\min}$ for this section, we should base the calculation on the average width of the portion of the section that would be cracking in tension. It is not easy to determine this value, because the distance that the flexural crack penetrates into the section is difficult to evaluate. However, it would be conservative to use the width of the section at the extreme tension fiber, 24 in., in Eq. (4-11). As in the Example 4-1, 200 psi will govern for the numerator in this equation. So,
+
+$$ A_{s,\min} = \frac{200 \text{ psi}}{f_y} b_w d = \frac{200 \text{ psi}}{60,000 \text{ psi}} \times 24 \text{ in.} \times 21.5 \text{ in.} = 1.72 \text{ in.}^2 $$
+
+Therefore, even with a conservative assumption for the effective width of the cracked tension zone, this section has a tension steel area that exceeds the required minimum area of tension reinforcement.
+
+## 4-5 DEFINITION OF BALANCED CONDITIONS
+
+The prior sections dealt with under-reinforced beam sections. To confirm that a particular section was under-reinforced, the section was put into equilibrium, and the steel strain evaluated using Eq. (4-18) was shown to be greater than the yield strain. This discussion will concentrate on the condition when the steel strain corresponding to section equilibrium is equal to the yield strain, $\epsilon_y$, and the strain in the extreme concrete fiber is equal to the maximum useable compression strain, $\epsilon_{cu}$. The area of tension steel required to cause this strain condition in a beam section will be defined as the *balanced area* of tension reinforcement. The balanced area is an important parameter for design of beam and slab sections and will be referred to in later chapters of this book. The analysis procedure to find the balanced area of tension reinforcement is similar to the analysis for $M_n$.
+
+The key starting point for the analysis of the balanced area of tension reinforcement is the *balanced strain diagram*, as shown in Fig. 4-21b. This strain diagram corresponds to a *balanced failure*, i.e., the tension reinforcement is just reaching its yield strain, $\epsilon_y$, at the same time the extreme concrete compression fiber is reaching the maximum useable compression strain, $\epsilon_{cu}$. Understanding and using the balanced strain diagram is important for
+
+(a) Beam section.
+(b) Balanced strain distribution.
+(c) Stress distribution.
+(d) Internal forces.
+
+Fig. 4-21
+Steps in analysis of $A_s (\text{bal})$, singly reinforced rectangular section.
+
+the analysis of both beam sections subjected to only bending and column sections subjected to bending plus axial load (Chapter 11).
+
+The balanced strain diagram is being applied to the singly reinforced beam section shown in Fig. 4-21a. There are some important differences between this analysis and the analysis for $M_n$ discussed in the previous sections. First, the major unknown now is the balanced area of steel, $A_s (\text{bal})$. Second, everything is known in the strain diagram, including the depth to the neutral axis, $c(\text{bal})$. This is calculated with the use of similar triangles from the strain diagram:
+
+$$ \frac{c(\text{bal})}{\epsilon_{cu}} = \frac{d}{\epsilon_{cu} + \epsilon_y} $$
+
+$$ c(\text{bal}) = \left( \frac{\epsilon_{cu}}{\epsilon_{cu} + \epsilon_y} \right) d \tag{4-22} $$
+
+The next steps through the stress distribution (Fig. 4-21c) and the force diagram (Fig. 4-21d) are similar to what was done for the analysis of $M_n$ in the previous sections. The only difference is that the forces have been labeled as $C_c (\text{bal})$ and $T(\text{bal})$ to distinguish them from the forces in the procedure for the analysis of $M_n$.
+Enforcing section equilibrium,
+
+$$ T(\text{bal}) = C_c (\text{bal}) $$
+
+$$ A_s (\text{bal}) f_y = 0.85 f_c' b \beta_1 c(\text{bal}) $$
+
+Solving for the only unknown $A_s (\text{bal})$,
+
+$$ A_s (\text{bal}) = \frac{1}{f_y} [0.85 f_c' b \beta_1 c(\text{bal})] \tag{4-23} $$
+
+This general expression applies *only* to singly reinforced rectangular sections and will not be used frequently. However, the reinforcement ratio at balanced conditions, $\rho_b$, is a parameter that often is used in design. Recalling that the reinforcement ratio is the tension
+
+steel area divided by the effective area of concrete, $bd$, and using the definition of $c(\text{bal})$ from Eq. (4-22), we get
+
+$$ \rho_b = \frac{A_s (\text{bal})}{bd} = \frac{0.85 \beta_1 f_c'}{f_y} \times \frac{b}{bd} \times \left( \frac{\epsilon_{cu}}{\epsilon_{cu} + \epsilon_y} \right) d $$
+
+$$ \rho_b = \frac{0.85 \beta_1 f_c'}{f_y} \left( \frac{\epsilon_{cu}}{\epsilon_{cu} + \epsilon_y} \right) \tag{4-24} $$
+
+Although this form is acceptable, the more common form is obtained by substituting in $\epsilon_{cu}$ equal to 0.003 and then multiplying both the numerator and denominator by $E_s = 29,000,000$ psi to obtain
+
+$$ \rho_b = \frac{0.85 \beta_1 f_c'}{f_y} \left( \frac{87,000}{87,000 + f_y} \right) \tag{4-25} $$
+
+where $f_y$ and $f_c'$ are used in psi units. Equations (4-24) and (4-25) represent classic definitions for the *balanced reinforcement ratio*. Some references to this reinforcement ratio will be made in later chapters of this book.
+
+## 4-6 CODE DEFINITIONS OF TENSION-CONTROLLED AND COMPRESSION-CONTROLLED SECTIONS
+
+Recall, the general design strength equation for flexure is
+
+$$ \phi M_n \ge M_u \tag{4-1b} $$
+
+where $\phi$ is the strength reduction factor. For beams, the factor $\phi$ is defined in ACI Code Section 9.3.2 and is based on the expected behavior of the beam section, as represented by the moment–curvature curves in Figs. 4-11 and 4-12. Because of the monolithic nature of reinforced concrete construction, most beams are part of a continuous floor system, as shown in Fig. 1-6. If a beam section with good ductile behavior was overloaded accidentally, it would soften and experience some plastic rotations that would permit loads to be redistributed to other portions of the continuous floor system. This type of behavior essentially creates an increased level of safety in the structural system, so a higher $\phi$-value is permitted for beams designed to exhibit ductile behavior. For beams that exhibit less ductile behavior, as indicated for the sections with larger tension steel areas in Fig. 4-11, the ability to redistribute loads away from an overloaded section is reduced. Thus, to maintain an acceptable level of safety in design, a lower $\phi$-value is required for such sections.
+
+Until 2002, the ACI Code defined only a single $\phi$-value for the design of reinforced concrete beam sections, but the behavior was controlled by limiting the permitted area of tension reinforcement. The design procedure was to keep the reinforcement ratio, $\rho$, less than or equal to 0.75 times the *balanced reinforcement ratio* defined in Eq. (4-25). This procedure, which is still permitted by Appendix C of the ACI Code, is easy to apply to singly reinforced rectangular sections, but becomes more complicated for flanged sections and sections that use compression reinforcement. When the same criteria is applied to beam sections that contain both normal reinforcement and prestressing tendons, the definition for the permitted area of tension reinforcement becomes quite complex.
+
+Another method for controlling the ductility of a section is to control the value of tension strain reached at the level of the tension reinforcement when the extreme concrete compression fiber reaches the maximum useable compression strain, i.e., at *nominal strength*
+
+*conditions* (Fig. 4-18b). Requiring higher tension strains at the level of tension steel is a universal method for controlling the ductility of all sections, as initially presented by Robert Mast [4-13]. Starting with the 2002 edition, this is the procedure used in Chapters 9 and 10 of the ACI Code to control section ductility, and thus, specify the corresponding values for the strength-reduction factor, $\phi$.
+
+### Definitions of Effective Depth and Distance to Extreme Layer of Tension Reinforcement
+
+The *effective depth*, $d$, is measured from the extreme compressive fiber to the *centroid* of the longitudinal reinforcement. This is the distance used in calculations of the nominal moment strength, as demonstrated in prior examples. To have consistency in controlling tension strains for a variety of beam and column sections, the ACI Code defines a distance, $d_t$, which is measured from the extreme compression fiber to the *extreme layer of tension reinforcement*, as shown in Fig. 4-22a. The strain at this level of reinforcement, $\epsilon_t$, is defined as the net strain at the extreme layer of tension reinforcement at nominal-strength conditions, excluding strains due to effective prestress, creep, shrinkage, and temperature. For beam sections with more than one layer of reinforcement, $\epsilon_t$ will be slightly larger than the strain at the centroid of the tension reinforcement, $\epsilon_s$, as shown in Fig. 4-22b. The ACI Code uses the strain $\epsilon_t$ to define the behavior of the section at nominal conditions, and thus, to define the value of $\phi$.
+
+### Definitions of Tension-Controlled and Compression-Controlled Sections
+
+A *tension-controlled section* has a tension-reinforcement area such that when the beam reaches its nominal flexural strength, the net tensile strain in the extreme layer of tensile steel, $\epsilon_t$, is greater than or equal to 0.005. For Grade-60 reinforcement with a yield strength $f_y = 60$ ksi, the tensile yield strain is $\epsilon_y = 60/29,000 = 0.00207$. The tension-controlled limit strain of 0.005 was chosen to be approximately 2.5 times the yield strain of the reinforcement, giving a moment–curvature diagram similar to that shown in Fig. 4-11 for the section with an area of tension reinforcement equal to 4.50 in.$^2$ The strain diagram corresponding to the tension-controlled limit (TCL) is demonstrated in Fig. 4-23b, with the depth from the extreme compression fiber to the neutral axis defined as $c(\text{TCL})$. From the strain diagram it can be shown that
+
+$$ c(\text{TCL}) = \frac{3}{8} d_t = 0.375 d_t \tag{4-26} $$
+
+(a) Beam section.
+(b) Strain distribution.
+
+Fig. 4-22
+Definitions for $d_t$ and $\epsilon_t$.
+
+Fig. 4-23
+Strain distributions at tension-controlled and compression-controlled limits.
+
+(a) Beam section.
+(b) Strain distribution at tension-controlled limit.
+(c) Strain distribution at compression-controlled limit.
+
+Clearly, if a calculated value of $c$ is *less* that $3/8 d_t$, the strain, $\epsilon_t$, will *exceed* 0.005. Thus, when analyzing the nominal flexural strength of a beam section, demonstrating that the depth to the neutral axis obtained from section equilibrium is less than $3/8 d_t$, as given in Eq. (4-26), will be one method to verify that the section is tension-controlled.
+
+A *compression-controlled section* has a tension-reinforcement area such that when the beam reaches its nominal flexural strength, the net tensile strain in the extreme layer of tensile steel, $\epsilon_t$, is less than or equal to the yield strain. For beams with Grade-60 reinforcement ($\epsilon_y = 0.00207$) and beams with prestressed reinforcement, ACI Code Section 10.3.3 permits the use of 0.002 in place of the yield strain. A beam section with this amount of tension reinforcement would exhibit a moment–curvature relationship similar to that shown in Fig. 4-11 for the section with the largest steel area. The strain diagram corresponding to the compression-controlled limit (CCL) is demonstrated in Fig. 4-23c, with the depth from the extreme compression fiber to the neutral axis defined as $c(\text{CCL})$. From the strain diagram it can be shown that
+
+$$ c(\text{CCL}) = \frac{3}{5} d_t = 0.60 d_t \tag{4-27} $$
+
+Clearly, if a calculated value of $c$ is *greater* that $3/5 d_t$, the strain $\epsilon_t$ will be *less* than 0.002.
+
+A *transition-zone section* has a tension-reinforcement area such that when the beam reaches its nominal flexural strength, the net tensile strain in the extreme layer of tensile steel, $\epsilon_t$, is between 0.002 and 0.005. A beam section with this amount of tension reinforcement would exhibit a moment–curvature relationship in between those shown in Fig. 4-11 for sections with tension steel areas of 4.50 and 6.50 in.$^2$
+
+Because tension-controlled sections demonstrate good ductile behavior if overloaded, they are analyzed and designed using a strength-reduction factor, $\phi$, of 0.9. Because of their brittle behavior if overloaded, compression-controlled sections are analyzed and designed with $\phi$ equal to 0.65. (*Note:* This is the value for beams with standard stirrup-tie reinforcement similar to that shown in Fig. 4-19. As will be discussed in Chapter 11, for column sections with *spiral reinforcement*, the value of $\phi$ is 0.75 if the section is compression-controlled). When analyzing a transition-zone section, the value of $\phi$ varies linearly between 0.65 and 0.90 as the value of $\epsilon_t$ varies from 0.002 to 0.005. A simple expression for this variation is given in Eq. (4-28).
+
+$$ \phi = 0.65 + (\epsilon_t - 0.002)\frac{250}{3} \tag{4-28} $$
+
+In the prior examples, the values of $\phi$ now can be calculated. In all three examples, there was only one layer of tension steel, so $\epsilon_t$ is equal to $\epsilon_s$. For the rectangular beams in Examples 4-1 and 4-1M, the value of $\epsilon_s$ exceeded 0.005, so the $\phi$-value would be 0.9. For the triangular beam section in Example 4-2, the value of $\epsilon_s$ was 0.00220. Using that as the value for $\epsilon_t$ in Eq. (4-28) results in a $\phi$-value of 0.67 (the authors recommend using only two significant figures for $\phi$).
+
+### Upper Limit on Beam Reinforcement
+
+Prior to the 2002 edition of the ACI Code, the maximum-tension steel area in beams was limited to 0.75 times the steel area corresponding to balanced conditions (Fig. 4-21). In the latest edition of the ACI Code (ACI 318-08), Section 10.3.5 requires that for reinforced concrete (nonprestressed) beam sections (stated as members with axial compressive load less than $0.10 f_c' A_g$) the value of $\epsilon_t$ at nominal flexural strength conditions shall be greater than or equal to 0.004. This strain value was selected to approximately correspond to the former ACI Code requirement of limiting the tension steel area to 0.75 times the balanced-tension steel area. A beam section with a tension steel area resulting in $\epsilon_t = 0.004$ at nominal conditions would have a higher $M_n$ value than a beam section with a lower-tension steel area that resulted in $\epsilon_t = 0.005$ (the tension-controlled limit) at nominal strength conditions. However, because there are different $\phi$-values for these two beam sections, the resulting values of $\phi M_n$ for the two sections will be approximately equal.
+
+The rectangular beam section in Fig. 4-24 will be used to demonstrate the change in the reduced nominal moment strength, $\phi M_n$, as the amount of tension-reinforcing steel is increased. Table 4-2 gives the results from a series of moment strength calculations for constantly increasing values for the reinforcement ratio, $\rho$. The corresponding steel areas are given in the second column of Table 4-2, and the depth to the neutral axis, $c$, obtained from Eqs. (4-16) and (4-17) are given in the third column. The beam section represented by the last row in Table 4-2 is over-reinforced and a strain-compatibility procedure is required to establish equilibrium and find the corresponding depth to the neutral axis, $c$. The details of this analysis procedure will be discussed at the end of this subsection.
+
+Values for $\epsilon_t$, which are equal to $\epsilon_s$ for a single layer of reinforcement, are obtained from Eq. (4-18) and then used to determine the corresponding values of the strength reduction factor, $\phi$. If $\epsilon_t$ is greater than or equal to 0.005 (signifying a tension-controlled section), $\phi$ is set equal to 0.9. For $\epsilon_t$ values between 0.005 and 0.002, Eq. (4-28) is used to calculate the corresponding $\phi$-value to three significant figures for this comparison. For the largest $\rho$-value in Table 4-2 (last row), the calculated value of $\epsilon_t$ is equal to the
+
+Fig. 4-24
+Typical beam section with $A_s$ as a variable.
+
+**TABLE 4-2** Relationship Between Reinforcement Ratio and Nominal Moment Strength
+
+| $\rho$ | $A_s$ (in.$^2$) | $c$ (in.) | $\epsilon_t$ | $\phi$ | $M_n$ (k-ft) | $\phi M_n$ (k-ft) |
+| :--- | :--- | :--- | :--- | :--- | :--- | :--- |
+| 0.005 | 1.37 | 2.03 | 0.0258 | 0.900 | 128 | 115 |
+| 0.010 | 2.73 | 4.05 | 0.0115 | 0.900 | 243 | 218 |
+| 0.015 | 4.10 | 6.08 | 0.00662 | 0.900 | 347 | 312 |
+| 0.0181 | 4.93 | 7.31 | 0.0050 | 0.900 | 404 | 364 |
+| 0.0207 | 5.64 | 8.36 | 0.0040 | 0.817 | 449 | 367 |
+| 0.025 | 6.83 | 10.1 | 0.00278 | 0.715 | 519 | 371 |
+| 0.0285 | 7.78 | 11.5 | 0.00207 | 0.656 | 568 | 372 |
+| 0.030 | 8.19 | 11.7 | 0.00200 | 0.650 | 575 | 374 |
+
+compression-controlled limit of 0.002, so $\phi$ was set equal to 0.65. Finally, Eq. (4-20) was used to calculate the nominal moment strength, $M_n$, which was multiplied by $\phi$ to get the values of $\phi M_n$ given in the last column of the Table 4-2.
+
+Some interesting results can be observed in the plots of $\rho$ versus $M_n$ and versus $\phi M_n$ in Fig. 4-25. There is an almost linear increase in $M_n$ and $\phi M_n$ for increasing values of $\rho$ up to the point where the tension strain, $\epsilon_t$, reaches the tension-controlled limit of 0.005. Beyond this point, $M_n$ continues to increase almost linearly for increasing values of $\rho$, but the value of $\phi M_n$ tends to stay almost constant due the decrease in the value of $\phi$ obtained from Eq. (4-28). This is a very important result that diminishes the significance of the limit set on $\epsilon_t$ in ACI Code Section 10.3.5 ($\epsilon_t \ge 0.004$). The authors believe that the **important limit for the amount of tension steel** to use in the design of beam sections will be to keep
+
+Fig. 4-25
+Relationship between $\rho$ and values for $M_n$ and $\phi M_n$.
+
+$\epsilon_t$ at or above the tension-controlled limit of 0.005, because Fig. 4-25 clearly shows that beyond this point it is not economical to add more tension steel to the section. Thus, for the flexural design procedures discussed in Chapter 5, the authors will always check that the final section design is classified as a tension-controlled section ($\epsilon_t \ge 0.005$), and thus, the $\phi$-value always will be 0.90.
+
+One final point of interest in Fig. 4-25 occurs in the plot of $\rho$ versus $M_n$ for a steel area larger that the balanced steel area given in Eq. (4-25). This section (last row of values in Table 4-2) is referred to as being *over-reinforced*, but the value for $M_n$ does not increase for this larger area of tension steel because the concrete compression zone will start to fail before the steel reaches its yield stress. Thus, the values for the compression force, $C_c$, and therefore the tension force, $T$, tend to stay relatively constant. Exact values for the steel stress and strain can be determined using the fundamental procedure of satisfying section equilibrium (Eq. 4-2) and strain compatibility (Eq. 4-18). Then the section nominal moment strength, $M_n$, can be calculated by the more general expression in Eq. (4-20). For over-reinforced sections, the nominal moment strength will tend to decrease as more tension steel is added to the section, because the moment arm ($d - a/2$) will decrease as $A_s$ is increased. An analysis of an over-reinforced beam section is presented as Beam 3 in the following example.
+
+**Example 4-3 Analysis of Singly Reinforced Rectangular Beams**
+
+Compute the nominal moment strengths, $M_n$, and the strength reduction factor, $\phi$, for three singly reinforced rectangular beams, each with a width $b = 12$ in. and a total height $h = 20$ in. As shown in Fig. 4-26 for the first beam section to be analyzed, a beam normally will have small longitudinal bars in the compression zone to hold the stirrups (shear reinforcement) in place. These bars typically are ignored in the calculation of the section nominal moment strength. Assuming that the beam has $1 \frac{1}{2}$ in. of clear cover and uses No. 3 stirrups, we will assume the distance from the tension edge to the centroid of the lowest layer of tension reinforcement is $2.5$ in.
+
+**Beam 1: $f_c' = 4000$ psi and $f_y = 60$ ksi. The tension steel area, $A_s = 4 \; (1.00 \; \text{in.}^2) = 4.00 \; \text{in.}^2$**
+
+**1. Compute $a$, $c$, and $\epsilon_s$ (same as $\epsilon_t$ for single layer of reinforcement).** As before, assume that the tension steel is yielding, so $f_s = f_y$. Using Eq. (4-16), which was developed from section equilibrium for a rectangular compression zone,
+
+$$ a = \beta_1 c = \frac{A_s f_y}{0.85 f_c' b} $$
+$$ = \frac{4.00 \text{ in.}^2 \times 60 \text{ ksi}}{0.85 \times 4 \text{ ksi} \times 12 \text{ in.}} = 5.88 \text{ in.} $$
+
+For $f_c' = 4000$ psi, $\beta_1$ is equal to 0.85. Thus, $c = a/\beta_1 = 6.92$ in., and using strain compatibility as expressed in Eq. (4-18), find
+
+$$ \epsilon_s = \left(\frac{d - c}{c}\right) \epsilon_{cu} = \left(\frac{17.5 - 6.92}{6.92}\right) 0.003 = 0.00459 $$
+
+Fig. 4-26
+Section used for Beams 1 and 2 of Example 4-3.
+
+This exceeds the yield strain for Grade-60 steel ($\epsilon_y = 0.00207$, previously calculated), so the assumption that the tension steel is yielding is confirmed.
+
+**2. Compute the nominal moment strength, $M_n$.** As in Example 4-1, use Eq. (4-21), which applies to sections with rectangular compression zones for
+
+$$ M_n = A_s f_y \left(d - \frac{a}{2}\right) = 4.0 \text{ in.}^2 \times 60 \text{ ksi} \left(17.5 \text{ in.} - \frac{5.88 \text{ in.}}{2}\right) $$
+$$ M_n = 3490 \text{ k-in.} = 291 \text{ k-ft} $$
+
+**3. Confirm that tension steel area exceeds $A_{s,\min}$.** Although this is seldom a problem with most beam sections, it is good practice to make this check. The expression for $A_{s,\min}$ is given in Eq. (4-11) and includes a numerator that is to be taken equal to $3\sqrt{f_c'}$, but not less than 200 psi. As was shown in Example 4-1, the value of 200 psi governs for beams constructed with 4000 psi concrete. Thus,
+
+$$ A_{s,\min} = \frac{200 \text{ psi}}{f_y} b_w d = \frac{200}{60,000} \times 12 \text{ in.} \times 17.5 \text{ in.} = 0.70 \text{ in.}^2 $$
+
+Clearly, $A_s$ for this section satisfies the ACI Code requirement for minimum tension reinforcement.
+
+**4. Compute the strength reduction factor, $\phi$, and the resulting value of $\phi M_n$.** As stated previously, for a single layer of tension reinforcement, $\epsilon_t$ is equal to $\epsilon_s$, which was calculated in step 1. Because $\epsilon_t$ is between 0.002 and 0.005, this is a transition-zone section. Thus, Eq. (4-28) is used to calculate $\phi$:
+
+$$ \phi = 0.65 = 0.65 + (0.00459 - 0.002)\frac{250}{3} = 0.87 $$
+
+Then,
+
+$$ \phi M_n = 0.87 \times 291 \text{ k-ft} = 253 \text{ k-ft} $$
+
+**Beam 2: Same as Beam 1, except that $f_c' = 6000$ psi.** As shown in Fig. 4-12, changing the concrete compressive strength will not produce a large change in the nominal moment strength, but it does increase the ductility of the section. Thus, increasing the concrete compressive strength might change the beam section in Fig. 4-26 from a *transition-zone section* to a *tension-controlled section*.
+
+**1. Compute $a$, $c$, and $\epsilon_s$.** Again, assume that the tension steel is yielding, so $f_s = f_y$. For this compressive strength, Eq. (4-14b) is used to determine that $\beta_1 = 0.75$. Then, using Eq. (4-16),
+
+$$ a = \beta_1 c = \frac{A_s f_y}{0.85 f_c' b} $$
+$$ = \frac{4.00 \text{ in.}^2 \times 60 \text{ ksi}}{0.85 \times 6 \text{ ksi} \times 12 \text{ in.}} = 3.92 \text{ in.} $$
+
+Thus, $c = a/\beta_1 = 5.23$ in., and using strain compatibility as expressed in Eq. (4-18), find
+
+$$ \epsilon_s = \left(\frac{d - c}{c}\right) \epsilon_{cu} = \left(\frac{17.5 - 5.23}{5.23}\right) 0.003 = 0.00704 $$
+
+This exceeds the yield strain for Grade-60 steel ($\epsilon_y = 0.00207$), confirming the assumption that the tension steel is yielding.
+
+**2. Compute the nominal moment strength, $M_n$.** As in Example 4-1, use Eq. (4-21), which applies to sections with rectangular compression zones:
+
+$$ M_n = A_s f_y \left(d - \frac{a}{2}\right) = 4.0 \text{ in.}^2 \times 60 \text{ ksi} \left(17.5 \text{ in.} - \frac{3.92 \text{ in.}}{2}\right) $$
+$$ M_n = 3730 \text{ k-in.} = 311 \text{ k-ft} \text{ (7\% increase from Beam 1)} $$
+
+**3. Confirm that tension steel area exceeds $A_{s,\min}$.** For this beam section with 6000 psi concrete, the value of $3\sqrt{f_c'}$ exceeds 200 psi, and will govern in Eq. (4-11). Thus,
+
+$$ A_{s,\min} = \frac{3\sqrt{f_c'}}{f_y} b_w d = \frac{3\sqrt{6000}}{60,000} \times 12 \text{ in.} \times 17.5 \text{ in.} = 0.81 \text{ in.}^2 $$
+
+Again, $A_s$ for this section easily satisfies the ACI Code requirement for minimum tension reinforcement.
+
+**4. Compute the strength reduction factor, $\phi$, and the resulting value of $\phi M_n$.** As before, $\epsilon_t$ is equal to $\epsilon_s$, which was calculated in step 1. This beam section is clearly a tension-controlled section, so $\phi = 0.9$. Then,
+
+$$ \phi M_n = 0.9 \times 311 \text{ k-ft} = 280 \text{ k-ft} \text{ (11\% increase from beam 1)} $$
+
+**Beam 3: Same as Beam 1, except increase tension steel to six No. 9 bars in two layers (Fig. 4-27).** For this section, $\epsilon_t$ will be larger than $\epsilon_s$ and will be calculated using the distance to the extreme layer of tension reinforcement, $d_t$, with the same cover and size of stirrup ($d_t = 17.5$ in.) as used for $d$ in Beams 1 and 2. The value of $d$ for this section involves a centroid calculation for the six No. 9 bars. ACI Code Section 7.6.2 requires a clear spacing between layers of reinforcement greater than or equal to 1 in. Thus, we can assume that the second layer of steel (two bars) is one bar diameter plus 1 in. above the lowest layer,—or a total of 2.5 in. + 1.128 in. + 1 in. $\approx$ 4.63 in. from the extreme tension fiber. A simple calculation is used to find the distance from the bottom of the beam to the centroid of the tension reinforcement, $g$, and then find the value of $d = h - g$.
+
+$$ g = \frac{4.0 \text{ in.}^2 \times 2.5 \text{ in.} + 2.0 \text{ in.}^2 \times 4.63 \text{ in.}}{6.0 \text{ in.}^2} = 3.21 \text{ in.} $$
+$$ d = h - g = 20 \text{ in.} - 3.21 \text{ in.} \approx 16.8 \text{ in.} $$
+
+**1. Compute $a$, $c$, and $\epsilon_s$.** Again, assume that the tension steel is yielding, so $f_s = f_y$. Then, using Eq. (4-16):
+
+$$ a = \beta_1 c = \frac{A_s f_y}{0.85 f_c' b} $$
+$$ = \frac{6.00 \text{ in.}^2 \times 60 \text{ ksi}}{0.85 \times 4 \text{ ksi} \times 12 \text{ in.}} = 8.82 \text{ in.} $$
+
+As for Beam 1, $\beta_1 = 0.85$. Thus, $c = a/\beta_1 = 10.4$ in. and using strain compatibility, as expressed in Eq. (4-18), find
+
+$$ \epsilon_s = \left(\frac{d - c}{c}\right) \epsilon_{cu} = \left(\frac{16.8 - 10.4}{10.4}\right) 0.003 = 0.00186 $$
+
+Fig. 4-27
+Section used for Beam 3 of Example 4-3.
+
+This is less than the yield strain for Grade-60 steel ($\epsilon_y = 0.00207$), so the assumption that the tension steel is yielding is *not confirmed*. Because the tension steel is not yielding, this is referred to as an *over-reinforced* section, and the previously developed procedure for calculating the nominal moment strength *does not apply*. A procedure that enforces strain compatibility and section equilibrium will be demonstrated in the following.
+
+**2. Compute the nominal moment strength, $M_n$, by enforcing strain compatibility and section equilibrium.** Referring to Fig. 4-18, we must now assume that the steel stress, $f_s$, is an unknown but is equal to the steel strain, $\epsilon_s$, multiplied by the steel modulus, $E_s$. Strain compatibility as expressed in Eq. (4-18) still applies, so the steel stress and thus the tension force can be expressed as a function of the unknown neutral axis depth, $c$.
+
+$$ T = A_s f_s = A_s E_s \epsilon_s = A_s E_s \left(\frac{d - c}{c}\right) \epsilon_{cu} $$
+
+Similarly, the concrete compression force can be expressed as a function of the neutral axis depth, $c$.
+
+$$ C_c = 0.85 f_c' b \beta_1 c $$
+
+Enforcing equilibrium by setting $T = C_c$, we can solve a second degree equation for the unknown value of $c$. The solution normally results in one positive and one negative value for $c$; the positive value will be selected. Using all of the given section and material properties and recalling that $E_s = 29,000$ ksi and $\epsilon_{cu} = 0.003$, the resulting value for $c$ is 10.1 in. Using this value, the authors obtained
+
+$$ T = 346 \text{ kips} \cong C_c = 350 \text{ kips} $$
+
+An average value of $T = C_c = 348$ kips will be used to calculate $M_n$. Then, using $a = \beta_1 c = 0.85 \times 10.1 \text{ in.} = 8.59 \text{ in.}$, calculate $M_n$ using the more general expression in Eq. (4-20).
+
+$$ M_n = T \left(d - \frac{a}{2}\right) = 348 \text{ kips} \left(16.8 \text{ in.} - \frac{8.59 \text{ in.}}{2}\right) $$
+$$ M_n = 4350 \text{ k-in.} = 363 \text{ k-ft} $$
+
+**3. Confirm that tension steel area exceeds $A_{s,\min}$.** For this beam section, the concrete compressive strength is 4000 psi, as was the case for Beam 1. However, the effective flexural depth $d$ has been reduced to 16.8 in. Using this new value of $d$, the value for $A_{s,\min}$ is 0.67 in., which is well below the provided tension steel area $A_s$.
+
+**4. Compute the strength reduction factor, $\phi$, and the resulting value of $\phi M_n$.** For this section, the value of $\epsilon_t$ will be slightly larger than $\epsilon_s$ and should be used to determine the value of $\phi$. The strain compatibility of Eq. (4-18) can be modified to calculate $\epsilon_t$ by using $d_t$ in place of $d$. Then,
+
+$$ \epsilon_t = \left(\frac{d_t - c}{c}\right) \epsilon_{cu} = \left(\frac{17.5 - 10.1}{10.1}\right) 0.003 = 0.00220 $$
+
+This is an interesting result, because we have previously considered this to be an over-reinforced section based on the strain, $\epsilon_s$, calculated at the centroid of the tension reinforcement. However, because of the difference between $d$ and $d_t$, we now have found the value of $\epsilon_t$ to be between 0.002 and 0.005. Thus, this is a transition-zone section, and we must use Eq. (4-28) to calculate $\phi$.
+
+$$ \phi = 0.65 + (0.00220 - 0.002)\frac{250}{3} = 0.67 $$
+
+Then,
+
+$$ \phi M_n = 0.67 \times 363 = 243 \text{ k-ft} $$
+

discussion/main.md

@@ -0,0 +1,496 @@
+# Note:
+
+- **"ε"** is epsilon
+- **β₁** is beta₁
+
+- **A_s** is the area of reinforcement near the tension face of the beam, tension reinforcement, in.².
+
+- **A'_s** is the area of reinforcement on the compression side of the beam, compression reinforcement, in.².
+
+- **b** is a general symbol for the width of the compression zone in a beam, in. This is illustrated in Fig. 4-2 for positive and negative moment regions. For flanged sections this symbol will normally be replace with b_e or b_w.
+
+- **b_e** is the effective width of a compression zone for a flanged section with compression in the flange, in.
+
+- **b_w** is the width of the web of the beam (and may or may not be the same as b), in.
+
+- **d** is the distance from the extreme fiber in compression to the centroid of the longitudinal reinforcement on the tension side of the member, in. In the positive-moment region (Fig. 4-2a), the tension steel is near the bottom of the beam, while in the negative-moment region (Fig. 4-2b) it is near the top.
+
+- **d'** is the distance from the extreme compression fiber to the centroid of the longitudinal compression steel, in.
+
+- **d_t** is the distance from the extreme compression fiber to the farthest layer of tension steel, in. For a single layer of tension reinforcement, d_t = d, as shown in Fig. 4-2b.
+
+- **f'_c** is the specified compressive strength of the concrete, psi.
+
+- **f_c** is the stress in the concrete, psi.
+
+- **f_s** is the stress in the tension reinforcement, psi.
+
+- **f_y** is the specified yield strength of the reinforcement, psi.
+
+- **h** is the overall height of a beam cross section.
+
+- **jd** is the *lever arm*, the distance between the resultant compressive force and the resultant tensile force, in.
+
+- **j** is a dimensionless ratio used to define the lever arm, *jd*. It varies depending on the moment acting on the beam section.
+
+- **ε_cu** is the assumed maximum useable compression strain in the concrete.
+
+- **ε_s** is the strain in the tension reinforcement.
+
+- **ε_t** is the strain in the extreme layer of tension reinforcement.
+
+- **ρ** is the longitudinal tension reinforcement ratio, ρ = A_s / bd.
+
+---
+
+# Calculation of Mₙ for a Singly Reinforced Rectangular Section
+
+For the beam shown in Fig. 4-19a, calculate Mₙ and confirm that the area of tension steel
+exceeds the required minimum steel area given by Eq. (4-11). The beam section is made of
+concrete with a compressive strength, f'_c = 4000 psi, and has four No. 8 bars with a yield
+strength of f_y = 60 ksi.
+
+For this beam with a single layer of tension reinforcement, it is reasonable to assume that
+the effective flexural depth, d, is approximately equal to the total beam depth minus 2.5 in.
+This accounts for a typical concrete clear cover of 1.5 in., the diameter of the stirrup
+(typically a No. 3 or No. 4 bar) and half the diameter of the beam longitudinal reinforcement.
+Depending on the sizes of the stirrup and longitudinal bar, the dimension to the center of the
+steel layer will vary slightly, but the use of 2.5 in. will be accurate enough for most design
+work unless adjustments in reinforcement location are required to avoid rebar interference at
+connections with other members. Small bars are often used in the compression zone to hold the
+stirrups in position, but these bars normally are ignored unless they were specifically designed
+to serve as compression-zone reinforcement.
+
+## Given:
+
+- f'_c = 4000 psi
+- f_y = 60000 psi
+- b = 12 in
+- h = 20 in
+- 4 No. 8 bars
+- concrete cover: 2.5 in (standard if not stated, 1 bar diameter)
+
+## Goal is to solve for:
+
+a. d
+b. A_s
+c. T
+d. a
+e. C
+f. ε_y
+g. ε_s
+h. Mₙ
+i. A_s,min
+
+> solve d
+
+d = h − concrete cover
+= 20 − 2.5
+**d = 17.5 in**
+
+## STEPS
+
+### 1. Assume tension bars are yielded
+
+f_s = f_y, ε_s ≥ or equal to ε_y
+= 60000 psi
+
+> Solve A_s:
+
+A_s = [(πd²) / 4] (no. of pcs of bars)
+= [(π(8/8)²) / 4] (4)
+**A_s = 3.16 in²**
+
+> Solve T:
+
+T = A_s · f_y
+= 3.16 in² (60 ksi)
+**T = 189.9 kips**
+
+### 2. C = T
+
+C = volume of compression zone = 0.85 · f'_c · b · a
+
+> Solve a:
+
+0.85 (4 ksi)(12 in)(a)(4 bars) = 189.9 kips
+**a = 4.66 in**
+
+> Solve C
+
+(Note: Whitney Stress Block)
+- a. f'_c up to and including 4000 psi, β₁ = 0.85
+- b. For 4000 psi < f'_c
+- c. For f'_c greater than 8000 psi, β₁ = 0.65
+
+a = β₁ · C
+4.66 in = 0.85 (C)
+**C = 5.48 in**
+
+
+### 3. Check Assumption
+
+> Solve ε_y
+
+ε_y = f_y / E_s
+ε_y = 60 ksi / 29000 ksi
+
+ε_y = 0.00207
+
+> Solve ε_s
+
+(ε_s / d − C) = (ε_cu / C)
+(ε_s / 17.5 − 5.48) = (0.003 / 5.48)
+**ε_s = 0.0067**
+
+Therefore, ε_s ≥ ε_y,
+**ASSUMPTION OK**
+
+### 4. Compute Nominal Moment (Mₙ)
+
+Mₙ = T · [d − (a/2)]
+= 189.6 kips [17.5 in − (4.66 in / 2)]
+Mₙ = 2876.23 kips-in
+divide by 12
+**Mₙ = 239.69 kip-ft**
+
+### 5. Check if it exceeds the required minimum of the code (A_s,min)
+
+> A. A_s,min = (200 / f_y)(b)(d) , for 200 < 3√(f'_c)
+> B. A_s,min = [(3√(f'_c)) / f_y] (b)(d)
+
+> check 3√(f'_c)
+= 3√4000 psi
+= 189.74 < 200 , therefore A. governs, use A.
+
+> A. A_s,min = (200 / f_y)(b)(d) , for 200 < 3√(f'_c)
+= (200 / 60000 psi)(12 in)(17.5 in)
+A_s,min = 0.7
+Therefore, **A_s,min = 0.7**
+
+---
+
+# SINGLE LAYER REINFORCEMENT
+
+Compute the nominal moment strengths, Mₙ, and the strength reduction factor, φ, for three
+singly reinforced rectangular beams, each with a width b = 12 in. and a total height h = 20 in.
+As shown in Fig. 4-26 for the first beam section to be analyzed, a beam normally will have small
+longitudinal bars in the compression zone to hold the stirrups (shear reinforcement) in place.
+These bars typically are ignored in the calculation of the section nominal moment strength.
+Assuming that the beam has 1½ in. of clear cover and uses No. 3 stirrups, we will assume the
+distance from the tension edge to the centroid of the lowest layer of tension reinforcement is 2.5 in.
+
+**Beam 1:** f'_c = 4000 psi and f_y = 60 ksi. The tension steel area, A_s = 4 (1.00 in.²) = 4.00 in.²
+
+## Given:
+
+- f'_c = 4000 psi
+- f_y = 60 ksi
+- b = 12 in
+- h = 20 in
+- 4 No. 9 bars
+
+- concrete cover: 2.5 in (standard if not stated, 1 bar diameter)
+
+## Goal is to solve for:
+
+a. d
+b. A_s
+c. T
+d. a
+e. C
+f. ε_y
+g. ε_s
+h. Mₙ
+i. A_s,min
+
+> solve d
+
+d = h − concrete cover
+= 20 − 2.5
+**d = 17.5 in**
+
+## STEPS
+
+### 1. Assume tension bars are yielded
+
+f_s = f_y, ε_s ≥ ε_y
+
+> Solve A_s:
+
+A_s = [(πd²) / 4] (no. of pcs of bars)
+= [(π(9/8)²) / 4] (4)
+**A_s = 3.98 in²**
+
+> Solve T:
+
+T = A_s · f_y
+= 3.98 in² (60 ksi)
+**T = 238.56 kips**
+
+### 2. C = T
+
+C = volume of compression zone = 0.85 · f'_c · b · a
+
+> Solve a:
+
+0.85 (4 ksi)(12 in)(a)(4 bars) = 189.9 kips
+**a = 5.85 in**
+
+> Solve C
+
+(Note: Whitney Stress Block)
+- a. f'_c up to and including 4000 psi, β₁ = 0.85
+- b. For 4000 psi < f'_c
+- c. For f'_c greater than 8000 psi, β₁ = 0.65
+
+a = β₁ · C
+5.85 in = 0.85 (C)
+**C = 6.88 in**
+
+### 3. Check Assumption
+
+> Solve ε_y
+
+ε_y = f_y / E_s
+ε_y = 60 ksi / 29000 ksi
+**ε_y = 0.00207**
+
+> Solve ε_s
+
+(ε_s / d − C) = (ε_cu / C)
+(ε_s / 17.5 − 6.88) = (0.003 / 6.88)
+**ε_s = 0.0046**
+
+Therefore, ε_s ≥ ε_y,
+**ASSUMPTION OK**
+
+### 4. Compute Nominal Moment (Mₙ)
+
+Mₙ = T · [d − (a/2)]
+= 238.56 kips [17.5 in − (5.85 in / 2)]
+Mₙ = 3477.017 kips-in
+divide by 12
+**Mₙ = 289.751 kip-ft**
+
+### 5. Check if it exceeds the required minimum of the code (A_s,min)
+
+> A. A_s,min = (200 / f_y)(b)(d) , for 200 < 3√(f'_c)
+> B. A_s,min = [(3√(f'_c)) / f_y] (b)(d)
+
+> check 3√(f'_c)
+= 3√4000 psi
+= 189.74 < 200 , therefore A. governs, use A.
+
+> A. A_s,min = (200 / f_y)(b)(d) , for 200 < 3√(f'_c)
+= (200 / 60000 psi)(12 in)(17.5 in)
+A_s,min = 0.7
+Therefore, **A_s,min = 0.7**
+
+### 6. Solve Strength Reduction Factor φ
+
+(Note: for a single layer of tension reinforcement, ε_t = ε_s. Because ε_t is between
+0.002 and 0.005, this is a transition zone section)
+
+Thus, φ = 0.65 = 0.65 + (ε_t − 0.002)(250/3)
+= 0.65 + (0.0046 − 0.002)(250/3)
+**φ = 0.87**
+
+### 7. Solve φ(Mₙ)
+
+φ(Mₙ) = 0.87 **(289.751 kip-ft)**
+**φ(Mₙ) = 252.08 k-ft**
+
+---
+
+# DOUBLE / MULTIPLE LAYER Reinforcement
+
+Compute the nominal moment strengths, Mₙ, and the strength reduction factor, φ, for three
+singly reinforced rectangular beams, each with a width b = 12 in. and a total height h = 20 in.
+As shown in Fig. 4-26 for the first beam section to be analyzed, a beam normally will have small
+longitudinal bars in the compression zone to hold the stirrups (shear reinforcement) in place.
+These bars typically are ignored in the calculation of the section nominal moment strength.
+Assuming that the beam has 1½ in. of clear cover and uses No. 3 stirrups, we will assume the
+distance from the tension edge to the centroid of the lowest layer of tension reinforcement is 2.5 in.
+
+**Beam 1:** f'_c = 4000 psi and f_y = 60 ksi. The tension steel area, A_s = 4 (1.00 in.²) = 4.00 in.²
+
+**Beam 3:** Same as Beam 1, except increase tension steel to six No. 9 bars in two layers
+
+## Given:
+
+- f'_c = 4000 psi
+- f_y = 60 ksi
+- b = 12 in
+- h = 20 in
+- 6 No. 9 bars
+- concrete cover: 2.5 in (standard if not stated, 1 bar diameter)
+
+## Goal is to solve for:
+
+a. d
+b. g
+c. A_s
+d. T
+e. a
+f. C
+g. ε_y
+h. ε_s
+i. Mₙ
+j. A_s,min
+
+> solve d
+
+d = h − g
+
+**Beam 3:** Same as Beam 1, except increase tension steel to six No. 9 bars in two layers
+(Fig. 4-27). For this section, ε_t will be larger than ε_s and will be calculated using the distance
+to the extreme layer of tension reinforcement, d_t, with the same cover and size of stirrup
+(d_t = 17.5 in.) as used for d in Beams 1 and 2. The value of d for this section involves a
+centroid calculation for the six No. 9 bars. ACI Code Section 7.6.2 requires a clear spacing between
+layers of reinforcement greater than or equal to 1 in. Thus, we can assume that the second
+layer of steel (two bars) is one bar diameter plus 1 in. above the lowest layer — or a total of
+2.5 in. + 1.128 in. + 1 in. ≈ 4.63 in. from the extreme tension fiber. A simple calculation is
+used to find the distance from the bottom of the beam to the centroid of the tension reinforcement,
+g, and then find the value of d = h − g.
+
+g = (4.0 in.² × 2.5 in. + 2.0 in.² × 4.63 in.) / 6.0 in.² = 3.21 in.
+
+d = h − g = 20 in. − 3.21 in. ≈ **16.8 in.**
+
+### 1. Assume tension bars are yielded
+
+f_s = f_y, ε_s ≥ ε_y
+
+> Solve A_s:
+
+A_s = [(πd²) / 4] (no. of pcs of bars)
+= [(π(9/8)²) / 4] (6 bars)
+**A_s = 5.96 in²**
+
+> Solve T:
+
+T = A_s · f_y
+= 5.96 in² (60 ksi)
+
+**T = 357.85 kips**
+
+### 2. C = T
+
+C = volume of compression zone = 0.85 · f'_c · b · a
+
+> Solve a:
+
+0.85 (4 ksi)(12 in)(a)(4 bars) = 357.85 kips
+**a = 8.77 in**
+
+> Solve C
+
+(Note: Whitney Stress Block)
+- a. f'_c up to and including 4000 psi, β₁ = 0.85
+- b. For 4000 psi < f'_c
+- c. For f'_c greater than 8000 psi, β₁ = 0.65
+
+a = β₁ · C
+8.77 in = 0.85 (C)
+**C = 10.32 in**
+
+### 3. Check Assumption
+
+> Solve ε_y
+
+ε_y = f_y / E_s
+ε_y = 60 ksi / 29000 ksi
+**ε_y = 0.00207**
+
+> Solve ε_s
+
+(ε_s / d − C) = (ε_cu / C)
+(ε_s / 16.8 − 8.77) = (0.003 / 8.77)
+**ε_s = 0.00188**
+
+Therefore, ε_s < ε_y,
+**ASSUMPTION NOT CONFIRMED, OVER REINFORCED STEEL**
+
+### 4. Compute Nominal Moment (Mₙ) by enforcing strain compatibility and section equilibrium
+
+> Solve new value C
+
+T = A_s · f_s
+T = A_s · E_s · ε_s
+T = A_s · E_s · [(d − C) / C] · ε_cu
+
+357.85 kips = 5.96 in² (29000 ksi) [(16.8 − C) / C] (0.003)
+
+**C = 9.94**
+
+cc = 0.85 · f'_c · (b)(β₁ · C)
+T = cc
+
+**section equilibrium.** Referring to Fig. 4-18, we must now assume that the steel stress, f_s, is an
+unknown but is equal to the steel strain, ε_s, multiplied by the steel modulus, E_s. Strain
+compatibility as expressed in Eq. (4-18) still applies, so the steel stress and thus the tension
+force can be expressed as a function of the unknown neutral axis depth, c.
+
+T = A_s · f_s = A_s · E_s · ε_s = A_s · E_s · ((d − c) / c) · ε_cu
+
+Similarly, the concrete compression force can be expressed as a function of the neutral axis
+depth, c.
+
+C_c = 0.85 · f'_c · b · β₁ · c
+
+Enforcing equilibrium by setting T = C_c, we can solve a second degree equation for the unknown
+value of c. The solution normally results in one positive and one negative value for c; the positive
+value will be selected. Using all of the given section and material properties and recalling that
+E_s = 29,000 ksi and ε_cu = 0.003, the resulting value for c is 10.1 in. Using this value, the
+authors obtained:
+
+T = 346 kips ≅ C_c = 350 kips
+
+An average value of T = C_c = 348 kips will be used to calculate Mₙ. Then, using
+a = β₁ · c = 0.85 × 10.1 in. = 8.59 in., calculate Mₙ using the more general expression in
+Eq. (4-20).
+
+Mₙ = T · (d − a/2) = 348 kips (16.8 in. − 8.59 in. / 2)
+Mₙ = 4350 k-in. = **363 k-ft**
+
+### 5. Check if it exceeds the required minimum of the code (A_s,min)
+
+> A. A_s,min = (200 / f_y)(b)(d) , for 200 < 3√(f'_c)
+> B. A_s,min = [(3√(f'_c)) / f_y] (b)(d)
+
+### 6. Compute ε_t Again using new value of C
+
+### 7. Solve Strength Reduction Factor φ
+
+(Because ε_t is between 0.002 and 0.005, this is a transition zone section)
+
+Thus, φ = 0.65 = 0.65 + (ε_t − 0.002)(250/3)
+
+### 8. Solve φ(Mₙ)
+
+
+**3. Confirm that tension steel area exceeds A_s,min.** For this beam section, the concrete
+compressive strength is 4000 psi, as was the case for Beam 1. However, the effective flexural
+depth d has been reduced to 16.8 in. Using this new value of d, the value for A_s,min is 0.67 in.,
+which is well below the provided tension steel area A_s.
+
+**4. Compute the strength reduction factor, φ, and the resulting value of φMₙ.** For this
+section, the value of ε_t will be slightly larger than ε_s and should be used to determine the value
+of φ. The strain compatibility of Eq. (4-18) can be modified to calculate ε_t by using d_t in place
+of d. Then,
+
+ε_t = ((d_t − c) / c) · ε_cu = ((17.5 − 10.1) / 10.1) · 0.003 = 0.00220
+
+This is an interesting result, because we have previously considered this to be an over-reinforced
+section based on the strain, ε_s, calculated at the centroid of the tension reinforcement. However,
+because of the difference between d and d_t, we now have found the value of ε_t to be between
+0.002 and 0.005. Thus, this is a transition-zone section, and we must use Eq. (4-28) to calculate φ.
+
+φ = 0.65 + (0.00220 − 0.002) · (250 / 3) = 0.67
+
+Then,
+
+φMₙ = 0.67 × 363 = **243 k-ft**
+
+---

discussion/main_corrected.md

@@ -0,0 +1,486 @@
+# Note:
+
+- **"ε"** is epsilon
+- **β₁** is beta₁
+
+- **A_s** is the area of reinforcement near the tension face of the beam, tension reinforcement, in.².
+
+- **A'_s** is the area of reinforcement on the compression side of the beam, compression reinforcement, in.².
+
+- **b** is a general symbol for the width of the compression zone in a beam, in. This is illustrated in Fig. 4-2 for positive and negative moment regions. For flanged sections this symbol will normally be replace with b_e or b_w.
+
+- **b_e** is the effective width of a compression zone for a flanged section with compression in the flange, in.
+
+- **b_w** is the width of the web of the beam (and may or may not be the same as b), in.
+
+- **d** is the distance from the extreme fiber in compression to the centroid of the longitudinal reinforcement on the tension side of the member, in. In the positive-moment region (Fig. 4-2a), the tension steel is near the bottom of the beam, while in the negative-moment region (Fig. 4-2b) it is near the top.
+
+- **d'** is the distance from the extreme compression fiber to the centroid of the longitudinal compression steel, in.
+
+- **d_t** is the distance from the extreme compression fiber to the farthest layer of tension steel, in. For a single layer of tension reinforcement, d_t = d, as shown in Fig. 4-2b.
+
+- **f'_c** is the specified compressive strength of the concrete, psi.
+
+- **f_c** is the stress in the concrete, psi.
+
+- **f_s** is the stress in the tension reinforcement, psi.
+
+- **f_y** is the specified yield strength of the reinforcement, psi.
+
+- **h** is the overall height of a beam cross section.
+
+- **jd** is the *lever arm*, the distance between the resultant compressive force and the resultant tensile force, in.
+
+- **j** is a dimensionless ratio used to define the lever arm, *jd*. It varies depending on the moment acting on the beam section.
+
+- **ε_cu** is the assumed maximum useable compression strain in the concrete.
+
+- **ε_s** is the strain in the tension reinforcement.
+
+- **ε_t** is the strain in the extreme layer of tension reinforcement.
+
+- **ρ** is the longitudinal tension reinforcement ratio, ρ = A_s / bd.
+
+---
+
+# Calculation of Mₙ for a Singly Reinforced Rectangular Section
+
+For the beam shown in Fig. 4-19a, calculate Mₙ and confirm that the area of tension steel
+exceeds the required minimum steel area given by Eq. (4-11). The beam section is made of
+concrete with a compressive strength, f'_c = 4000 psi, and has four No. 8 bars with a yield
+strength of f_y = 60 ksi.
+
+For this beam with a single layer of tension reinforcement, it is reasonable to assume that
+the effective flexural depth, d, is approximately equal to the total beam depth minus 2.5 in.
+This accounts for a typical concrete clear cover of 1.5 in., the diameter of the stirrup
+(typically a No. 3 or No. 4 bar) and half the diameter of the beam longitudinal reinforcement.
+Depending on the sizes of the stirrup and longitudinal bar, the dimension to the center of the
+steel layer will vary slightly, but the use of 2.5 in. will be accurate enough for most design
+work unless adjustments in reinforcement location are required to avoid rebar interference at
+connections with other members. Small bars are often used in the compression zone to hold the
+stirrups in position, but these bars normally are ignored unless they were specifically designed
+to serve as compression-zone reinforcement.
+
+## Given:
+
+- f'_c = 4000 psi
+- f_y = 60000 psi
+- b = 12 in
+- h = 20 in
+- 4 No. 8 bars
+- concrete cover: 2.5 in (standard if not stated, distance to centroid)
+
+## Goal is to solve for:
+
+a. d
+b. A_s
+c. T
+d. a
+e. c
+f. ε_y
+g. ε_s
+h. Mₙ
+i. A_s,min
+
+> solve d
+
+d = h − concrete cover
+= 20 − 2.5
+**d = 17.5 in**
+
+## STEPS
+
+### 1. Assume tension bars are yielded
+
+f_s = f_y, ε_s ≥ or equal to ε_y
+= 60000 psi
+
+> Solve A_s:
+
+A_s = (4 bars) × (0.79 in²/bar from standard nominal area)
+**A_s = 3.16 in²**
+
+> Solve T:
+
+T = A_s · f_y
+= 3.16 in² (60 ksi)
+**T = 189.6 kips**
+
+### 2. C_c = T
+
+C_c = volume of compression zone = 0.85 · f'_c · b · a
+
+> Solve a:
+
+0.85 (4 ksi)(12 in)(a) = 189.6 kips
+40.8 (a) = 189.6
+**a = 4.65 in**
+
+> Solve c
+
+(Note: Whitney Stress Block)
+- a. f'_c up to and including 4000 psi, β₁ = 0.85
+- b. For 4000 psi < f'_c ≤ 8000 psi
+- c. For f'_c greater than 8000 psi, β₁ = 0.65
+
+a = β₁ · c
+4.65 in = 0.85 (c)
+**c = 5.47 in**
+
+
+### 3. Check Assumption
+
+> Solve ε_y
+
+ε_y = f_y / E_s
+ε_y = 60 ksi / 29000 ksi
+**ε_y = 0.00207**
+
+> Solve ε_s
+
+(ε_s / (d − c)) = (ε_cu / c)
+(ε_s / (17.5 − 5.47)) = (0.003 / 5.47)
+**ε_s = 0.0066**
+
+Therefore, ε_s ≥ ε_y,
+**ASSUMPTION OK**
+
+### 4. Compute Nominal Moment (Mₙ)
+
+Mₙ = T · [d − (a/2)]
+= 189.6 kips [17.5 in − (4.65 in / 2)]
+Mₙ = 2877 kips-in
+divide by 12
+**Mₙ = 239.75 kip-ft**
+
+### 5. Check if it exceeds the required minimum of the code (A_s,min)
+
+> A. A_s,min = (200 / f_y)(b)(d) , for 200 > 3√(f'_c)
+> B. A_s,min = [(3√(f'_c)) / f_y] (b)(d)
+
+> check 3√(f'_c)
+= 3√4000 psi
+= 189.74 < 200 , therefore A. governs, use 200.
+
+> A. A_s,min = (200 / f_y)(b)(d)
+= (200 / 60000 psi)(12 in)(17.5 in)
+A_s,min = 0.70 in²
+Therefore, **A_s,min = 0.70 in²**
+
+Since A_s (3.16 in²) > A_s,min (0.70 in²), **OK**.
+
+---
+
+# SINGLE LAYER REINFORCEMENT
+
+Compute the nominal moment strengths, Mₙ, and the strength reduction factor, φ, for three
+singly reinforced rectangular beams, each with a width b = 12 in. and a total height h = 20 in.
+As shown in Fig. 4-26 for the first beam section to be analyzed, a beam normally will have small
+longitudinal bars in the compression zone to hold the stirrups (shear reinforcement) in place.
+These bars typically are ignored in the calculation of the section nominal moment strength.
+Assuming that the beam has 1½ in. of clear cover and uses No. 3 stirrups, we will assume the
+distance from the tension edge to the centroid of the lowest layer of tension reinforcement is 2.5 in.
+
+**Beam 1:** f'_c = 4000 psi and f_y = 60 ksi. The tension steel area, A_s = 4 (1.00 in.²) = 4.00 in.²
+
+## Given:
+
+- f'_c = 4000 psi
+- f_y = 60 ksi
+- b = 12 in
+- h = 20 in
+- 4 No. 9 bars
+
+- concrete cover: 2.5 in (distance to centroid)
+
+## Goal is to solve for:
+
+a. d
+b. A_s
+c. T
+d. a
+e. c
+f. ε_y
+g. ε_s
+h. Mₙ
+i. A_s,min
+
+> solve d
+
+d = h − concrete cover
+= 20 − 2.5
+**d = 17.5 in**
+
+## STEPS
+
+### 1. Assume tension bars are yielded
+
+f_s = f_y, ε_s ≥ ε_y
+
+> Solve A_s:
+
+A_s = (4 bars) × (1.00 in²/bar from nominal area)
+**A_s = 4.00 in²**
+
+> Solve T:
+
+T = A_s · f_y
+= 4.00 in² (60 ksi)
+**T = 240 kips**
+
+### 2. C_c = T
+
+C_c = volume of compression zone = 0.85 · f'_c · b · a
+
+> Solve a:
+
+0.85 (4 ksi)(12 in)(a) = 240 kips
+40.8 (a) = 240
+**a = 5.88 in**
+
+> Solve c
+
+(Note: Whitney Stress Block)
+- a. f'_c up to and including 4000 psi, β₁ = 0.85
+
+a = β₁ · c
+5.88 in = 0.85 (c)
+**c = 6.92 in**
+
+### 3. Check Assumption
+
+> Solve ε_y
+
+ε_y = f_y / E_s
+ε_y = 60 ksi / 29000 ksi
+**ε_y = 0.00207**
+
+> Solve ε_s
+
+(ε_s / (d − c)) = (ε_cu / c)
+(ε_s / (17.5 − 6.92)) = (0.003 / 6.92)
+**ε_s = 0.00459**
+
+Therefore, ε_s ≥ ε_y,
+**ASSUMPTION OK**
+
+### 4. Compute Nominal Moment (Mₙ)
+
+Mₙ = T · [d − (a/2)]
+= 240 kips [17.5 in − (5.88 in / 2)]
+Mₙ = 3494 kips-in
+divide by 12
+**Mₙ = 291.17 kip-ft**
+
+### 5. Check if it exceeds the required minimum of the code (A_s,min)
+
+> A. A_s,min = (200 / f_y)(b)(d) , for 200 > 3√(f'_c)
+> B. A_s,min = [(3√(f'_c)) / f_y] (b)(d)
+
+> check 3√(f'_c)
+= 3√4000 psi
+= 189.74 < 200 , therefore A. governs, use 200.
+
+> A. A_s,min = (200 / f_y)(b)(d)
+= (200 / 60000 psi)(12 in)(17.5 in)
+**A_s,min = 0.70 in²**
+
+Since A_s (4.00 in²) > A_s,min (0.70 in²), **OK**.
+
+### 6. Solve Strength Reduction Factor φ
+
+(Note: for a single layer of tension reinforcement, ε_t = ε_s. Because ε_t is between
+0.002 and 0.005, this is a transition zone section)
+
+Thus, φ = 0.65 + (ε_t − 0.002)(250/3)
+= 0.65 + (0.00459 − 0.002)(250/3)
+**φ = 0.87**
+
+### 7. Solve φ(Mₙ)
+
+φ(Mₙ) = 0.87 **(291.17 kip-ft)**
+**φ(Mₙ) = 253.3 k-ft**
+
+---
+
+# DOUBLE / MULTIPLE LAYER Reinforcement
+
+Compute the nominal moment strengths, Mₙ, and the strength reduction factor, φ, for three
+singly reinforced rectangular beams, each with a width b = 12 in. and a total height h = 20 in.
+As shown in Fig. 4-26 for the first beam section to be analyzed, a beam normally will have small
+longitudinal bars in the compression zone to hold the stirrups (shear reinforcement) in place.
+These bars typically are ignored in the calculation of the section nominal moment strength.
+Assuming that the beam has 1½ in. of clear cover and uses No. 3 stirrups, we will assume the
+distance from the tension edge to the centroid of the lowest layer of tension reinforcement is 2.5 in.
+
+**Beam 1:** f'_c = 4000 psi and f_y = 60 ksi. The tension steel area, A_s = 4 (1.00 in.²) = 4.00 in.²
+
+**Beam 3:** Same as Beam 1, except increase tension steel to six No. 9 bars in two layers
+
+## Given:
+
+- f'_c = 4000 psi
+- f_y = 60 ksi
+- b = 12 in
+- h = 20 in
+- 6 No. 9 bars
+- concrete cover: 2.5 in (distance to bottom layer centroid)
+
+## Goal is to solve for:
+
+a. d
+b. g
+c. A_s
+d. T
+e. a
+f. c
+g. ε_y
+h. ε_s
+i. Mₙ
+j. A_s,min
+k. ε_t
+l. φ
+m. φ(Mₙ)
+
+> solve d
+
+d = h − g
+
+**Beam 3:** Same as Beam 1, except increase tension steel to six No. 9 bars in two layers
+(Fig. 4-27). For this section, ε_t will be larger than ε_s and will be calculated using the distance
+to the extreme layer of tension reinforcement, d_t, with the same cover and size of stirrup
+(d_t = 17.5 in.) as used for d in Beams 1 and 2. The value of d for this section involves a
+centroid calculation for the six No. 9 bars. ACI Code Section 7.6.2 requires a clear spacing between
+layers of reinforcement greater than or equal to 1 in. Thus, we can assume that the second
+layer of steel (two bars) is one bar diameter plus 1 in. above the lowest layer — or a total of
+2.5 in. + 1.128 in. + 1 in. ≈ 4.63 in. from the extreme tension fiber. A simple calculation is
+used to find the distance from the bottom of the beam to the centroid of the tension reinforcement,
+g, and then find the value of d = h − g.
+
+g = (4.0 in.² × 2.5 in. + 2.0 in.² × 4.63 in.) / 6.0 in.² = 3.21 in.
+
+d = h − g = 20 in. − 3.21 in.
+**d = 16.8 in.**
+
+## STEPS
+
+### 1. Assume tension bars are yielded
+
+f_s = f_y, ε_s ≥ ε_y
+
+> Solve A_s:
+
+A_s = (6 bars) × (1.00 in²/bar from nominal area)
+**A_s = 6.00 in²**
+
+> Solve T:
+
+T = A_s · f_y
+= 6.00 in² (60 ksi)
+**T = 360 kips**
+
+### 2. C_c = T
+
+C_c = volume of compression zone = 0.85 · f'_c · b · a
+
+> Solve a:
+
+0.85 (4 ksi)(12 in)(a) = 360 kips
+40.8 (a) = 360
+**a = 8.82 in**
+
+> Solve c
+
+(Note: Whitney Stress Block)
+- a. f'_c up to and including 4000 psi, β₁ = 0.85
+
+a = β₁ · c
+8.82 in = 0.85 (c)
+**c = 10.38 in (Use 10.4 in)**
+
+### 3. Check Assumption
+
+> Solve ε_y
+
+ε_y = f_y / E_s
+ε_y = 60 ksi / 29000 ksi
+**ε_y = 0.00207**
+
+> Solve ε_s
+
+(ε_s / (d − c)) = (ε_cu / c)
+(ε_s / (16.8 − 10.4)) = (0.003 / 10.4)
+**ε_s = 0.00185**
+
+Therefore, ε_s < ε_y,
+**ASSUMPTION NOT CONFIRMED, OVER REINFORCED SECTION**
+
+### 4. Compute Nominal Moment (Mₙ) by enforcing strain compatibility and section equilibrium
+
+> Solve new value for c
+
+T = A_s · f_s
+T = A_s · E_s · ε_s
+T = A_s · E_s · [(d − c) / c] · ε_cu
+
+Similarly, the concrete compression force is:
+C_c = 0.85 · f'_c · b · β₁ · c
+
+Enforcing section equilibrium, T = C_c:
+6.00 in² (29000 ksi) [(16.8 − c) / c] (0.003) = 0.85 (4 ksi)(12 in)(0.85 · c)
+522 [(16.8 − c) / c] = 34.68 c
+
+Enforcing equilibrium by setting T = C_c, we can solve a second-degree equation for the unknown
+value of c. The result is the following quadratic equation:
+34.68 c² + 522 c - 8769.6 = 0
+
+Solving for the positive root yields:
+**c = 10.1 in**
+
+Using this value:
+T = 6.00 in² (29000 ksi) [(16.8 - 10.1) / 10.1] (0.003) = 346 kips
+C_c = 34.68 (10.1) = 350 kips
+
+An average value of T = C_c = 348 kips will be used to calculate Mₙ. Then, using
+a = β₁ · c = 0.85 × 10.1 in. = 8.59 in., calculate Mₙ using the more general expression in Eq. (4-20).
+
+Mₙ = T · (d − a/2) = 348 kips (16.8 in. − 8.59 in. / 2)
+Mₙ = 4351 k-in.
+divide by 12
+**Mₙ = 363 k-ft**
+
+### 5. Check if it exceeds the required minimum of the code (A_s,min)
+
+> A. A_s,min = (200 / f_y)(b)(d) , for 200 > 3√(f'_c)
+> B. A_s,min = [(3√(f'_c)) / f_y] (b)(d)
+
+> check 3√(f'_c)
+= 3√4000 psi
+= 189.74 < 200 , therefore A. governs, use 200.
+
+> A. A_s,min = (200 / f_y)(b)(d)
+= (200 / 60000 psi)(12 in)(16.8 in)
+**A_s,min = 0.67 in²**
+
+Since A_s (6.00 in²) > A_s,min (0.67 in²). **OK.**
+
+### 6. Compute ε_t using extreme tension layer
+The value of ε_t is calculated using the distance to the extreme layer of tension reinforcement,
+d_t = 17.5 in.
+
+ε_t = [(d_t − c) / c] · ε_cu = [(17.5 − 10.1) / 10.1] · 0.003
+**ε_t = 0.00220**
+
+### 7. Solve Strength Reduction Factor φ
+
+Because of the difference between d and d_t, we now have found the value of ε_t to be between
+0.002 and 0.005. Thus, this is a transition-zone section, and we must use Eq. (4-28) to calculate φ.
+
+Thus, φ = 0.65 + (ε_t − 0.002)(250/3)
+φ = 0.65 + (0.00220 − 0.002)(250/3)
+**φ = 0.67**
+
+### 8. Solve φ(Mₙ)
+
+φ(Mₙ) = 0.67 × 363 k-ft
+**φ(Mₙ) = 243.2 k-ft**

discussion/matlab-basic-functions-reference.pdf

@@ -0,0 +1,3 @@
+version https://git-lfs.github.com/spec/v1
+oid sha256:f767754646523838acbb06bb5bb7bd3ff5aaa3f0666c6153aa3de63383ca6035
+size 690126

discussion/revision_discussion.md

@@ -0,0 +1,31 @@
+# Discussion of Revisions
+
+The initial document (`main.md`) contained several missing segments and minor calculation or notation errors that required addressing to ensure complete accuracy based on standard reinforced concrete design principles.
+
+The following revisions were made in `main_corrected.md`:
+
+### 1. Revision of Reinforcement Areas (A_s)
+In the initial document, the area of reinforcement was calculated manually using a precise circular area equation (`A_s = [(pi * d^2) / 4] * number of bars`). While theoretically close, standard concrete design dictates using ACI nominal areas.
+- For **No. 8 bars**, the nominal area is exactly `0.79 in²`. The updated evaluation utilizes `4 * 0.79 in² = 3.16 in²`.
+- For **No. 9 bars**, the nominal area is exactly `1.00 in²`. Noticeable deviations in calculations resulted from computing this as `3.98 in²` instead of `4.00 in²` (for 4 bars) or `5.96 in²` instead of `6.00 in²` (for 6 bars). These were corrected, resulting in fully precise values for tension force `T` and compression block depth `a`.
+
+### 2. Standardization of Notation
+The initial document interchangeably used uppercase `C` to denote both the generic compressive force and the neutral axis depth. This was rectified.
+- Compressive force is clearly designated as `C_c` to distinguish it from tension force `T`.
+- The depth to the neutral axis is accurately labeled with lowercase `c`, according strictly to ACI Code nomenclature.
+
+### 3. Completion of Over-Reinforced Section Evaluation (Beam 3)
+The major incompletion occurred in the "Double / Multiple Layer Reinforcement" evaluation for Beam 3.
+- When `ε_s` was found to be lower than `ε_y` (`ASSUMPTION NOT CONFIRMED`), the section was properly classified as over-reinforced.
+- The missing quadratic equation enforcing `T = C_c` through strain compatibility was populated: `34.68 c² + 522 c - 8769.6 = 0`.
+- Solving for positive `c` provided the correct value of `10.1 in`.
+- Nominal moment (`Mₙ`) was finalized using the average tension/compression force of `348 kips`.
+
+### 4. Completion of Steps 5 to 8 for Beam 3
+Steps 5 to 8 corresponding to verification limits and capacity factors were missing their calculations entirely.
+- **A_s,min Validation**: Completed utilizing actualized terms (`A_s,min = 0.67 in²`), validating it successfully handles the required reinforcement constraints.
+- **Computation of extreme layer strain (`ε_t`)**: Calculated properly using the difference between `d` and `d_t` (`17.5 in`). This transitioned the assessment classification from an over-reinforced generic to a "transition zone section" due to `ε_t = 0.00220`.
+- **Strength Reduction Factor (φ)**: Executed the accurate transformation standard evaluating `φ = 0.65 + (0.00220 - 0.002) * (250/3)`, yielding `φ = 0.67`.
+- **Reduced Nominal Moment (φMₙ)**: Generated the final limit capacity (`φMₙ = 243.2 k-ft`).
+
+With these corrections, `main_corrected.md` aligns precisely with the source reference material step-by-step.

matlab/inline_RC_SRP.m

@@ -0,0 +1,200 @@
+% 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

matlab/simple_RC_SRP.m

@@ -0,0 +1,193 @@
+% Simple RC-SRP Script Calculator
+
+clear; clc;
+
+disp('--- Simple RC-SRP Calculator ---');
+disp('Follow the steps to enter the given variables.');
+
+% Get inputs from the user through the command window
+fc = input('Enter concrete compressive strength f_c'' (psi): ');
+fy = input('Enter steel yield strength f_y (psi): ');
+b = input('Enter beam width b (in): ');
+h = input('Enter beam height h (in): ');
+cover = input('Enter cover distance to extreme tension layer (in): ');
+
+n_bars = input('Enter total number of bars: ');
+a_bar = input('Enter area per bar (in^2): ');
+
+% Ask if there's a second layer
+double_layer_ans = input('Is this a double layer reinforcement? (yes/no): ', 's');
+
+if strcmpi(double_layer_ans, 'yes')
+    layer_type = 'Double Layer Singly Reinforced';
+    n_bars1 = input('Enter number of bars in layer 1 (bottom): ');
+    n_bars2 = input('Enter number of bars in layer 2 (top): ');
+    spacing = input('Enter spacing between layers (in): ');
+else
+    layer_type = 'Single Layer Singly Reinforced';
+end
+
+% Ask user if they want to calculate and show output
+show_output = input('Ready to calculate and show the output? (yes/no): ', 's');
+
+if strcmpi(show_output, 'yes')
+    disp(' ');
+    disp('--- CALCULATION OUTPUT ---');
+    
+    % Identify the type of calculated section
+    disp(['Type of section: ', layer_type]);
+    disp(' ');
+    
+    % 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);
+    
+    if strcmpi(double_layer_ans, 'no')
+        % --- SINGLE LAYER CALCULATION ---
+        d = h - cover;
+        As = n_bars * a_bar;
+        
+        disp(['1. Effective depth (d): ', num2str(d), ' in']);
+        disp(['2. Area of steel (A_s): ', num2str(As), ' in^2']);
+        
+        % Force and depth of compression block
+        T = As * fy_ksi;
+        disp(['3. Tension force (T, assuming yield): ', num2str(T), ' kips']);
+        
+        a = T / (0.85 * fc_ksi * b);
+        c = a / beta1;
+        disp(['4. 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(['5. 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(['6. Nominal Moment (M_n): ', num2str(Mn), ' kip-ft']);
+        
+        % Check minimum code requirement for steel
+        As_min = (As_min_factor / fy) * b * d;
+        disp(['7. Minimum steel (A_s,min): ', num2str(As_min), ' in^2']);
+        if As >= As_min
+            disp('   Status: A_s >= A_s,min. OK.');
+        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(['8. Reduction factor (phi): ', num2str(phi)]);
+        disp(['9. Design Moment (phi*M_n): ', num2str(phi_Mn), ' kip-ft']);
+        
+    else
+        % --- 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(['2. 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(['3. Tension force (T, assuming yield): ', num2str(T), ' kips']);
+        disp(['4. 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(['5. 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(['6. Nominal Moment (M_n): ', num2str(Mn), ' kip-ft']);
+        
+        % Check minimum code requirement for steel
+        As_min = (As_min_factor / fy) * b * d;
+        disp(['7. Minimum steel (A_s,min): ', num2str(As_min), ' in^2']);
+        if As >= As_min
+            disp('   Status: A_s >= A_s,min. OK.');
+        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(['8. Extreme tension strain (eps_t): ', num2str(eps_t)]);
+        disp(['   Reduction factor (phi): ', num2str(phi)]);
+        disp(['9. Design Moment (phi*M_n): ', num2str(phi_Mn), ' kip-ft']);
+    end
+else
+    disp('Calculation stopped. Have a great day!');
+end

matlab/singly_reinforced copy.m

@@ -0,0 +1,97 @@
+% Name: Rembrant Oyangoren Albeos
+% Year: 2026
+% Singly Reinforced Rectangular Beam Calculator
+
+clc; clear;
+disp('Singly Reinforced Beam Calculator');
+
+% Inputs
+fc = input('Concrete strength f''c (ksi): ');
+fy = input('Steel yield strength fy (ksi): ');
+b = input('Beam width b (in): ');
+h = input('Beam depth h (in): ');
+cover = input('Clear cover to centroid of bottom bars (in): ');
+n_bars = input('Total number of bars: ');
+a_bar = input('Area per bar (in^2): ');
+
+% As calculation
+As = n_bars * a_bar;
+fprintf('Total As: %.3f in^2\n', As);
+
+% Handle layers
+layers = input('Double layer? (y/n): ', 's');
+if lower(layers) == 'y'
+    n_bot = input('Number of bars in bottom layer: ');
+    n_top = input('Number of bars in top layer: ');
+    spacing = input('Spacing between layers (in): ');
+    
+    % Centroid and depths
+    g = (n_bot * a_bar * cover + n_top * a_bar * (cover + spacing)) / As;
+    d = h - g;
+    dt = h - cover;
+    fprintf('Centroid g: %.3f in\n', g);
+    fprintf('Effective d: %.3f in\n', d);
+else
+    d = h - cover;
+    dt = d;
+end
+
+% Constants
+Es = 29000;
+Ecu = 0.003;
+
+% Beta 1
+if fc <= 4
+    beta1 = 0.85;
+elseif fc > 4 && fc <= 8
+    beta1 = 0.85 - 0.05 * (fc - 4);
+else
+    beta1 = 0.65;
+end
+
+% Initial assumption: steel yields
+T = As * fy;
+a = T / (0.85 * fc * b);
+c = a / beta1;
+
+% Verify yield
+eps_y = fy / Es;
+eps_s = Ecu * (d - c) / c;
+
+if eps_s < eps_y
+    fprintf('Steel did not yield. Solving for exact c...\n');
+    A_q = 0.85 * fc * b * beta1;
+    B_q = As * Es * Ecu;
+    C_q = -As * Es * Ecu * d;
+    c = (-B_q + sqrt(B_q^2 - 4 * A_q * C_q)) / (2 * A_q);
+    a = beta1 * c;
+    eps_s = Ecu * (d - c) / c;
+    T = As * (Es * eps_s);
+end
+
+% Nominal Moment
+Mn = T * (d - a/2) / 12; % in kip-ft
+fprintf('a: %.3f in\n', a);
+fprintf('c: %.3f in\n', c);
+fprintf('Mn: %.2f kip-ft\n', Mn);
+
+% Minimum Area Check
+As_min = max(3 * sqrt(fc * 1000), 200) / (fy * 1000) * b * d;
+fprintf('As_min: %.3f in^2\n', As_min);
+if As >= As_min
+    disp('As >= As_min: OK');
+else
+    disp('As < As_min: FAILED');
+end
+
+% Reduction Factor
+eps_t = Ecu * (dt - c) / c;
+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
+fprintf('phi: %.3f\n', phi);
+fprintf('Design Strength (phi*Mn): %.2f kip-ft\n', phi * Mn);

matlab/tae.m

@@ -0,0 +1,112 @@
+%Name: reference
+%Course & Year: BSCE 3-D
+
+%SINGLY REINFORCED BEAM
+
+%inputs
+b= input( 'Width of the beam in inches: '); %value of b
+d= input('Effective depth of the beam in inches: '); %value of d
+fc= input('Compressive strength of concrete in ksi: '); %value of fc
+fy= input('Yield strength of steel in ksi:  '); %value of fy
+Asq= input('The number of steel in the beam: '); %number of steel
+Asd= input('The diameter of the steel:  '); %diameter of steel
+B= input('The Length of Beam in ksf: '); %beamspan
+DL= input("Dead Load in ksf: "); %value of dead load
+LL= input("Live Load in ksf: "); %value of live load
+ME= input ("Modulus of Elasticity, in ksi: "); %value of ME
+
+%Calculation of As (cross sectional area of steel)
+As= Asq*((pi*(Asd/8)^2)/4); %cross sectional area of steel
+    fprintf('Value of As: %0.2f in^2\n',As)
+
+%Equilibrium Equation T=Cc
+T=As*fy; %Tensile Force in Steel
+
+% Assigning the value of B based on fc
+if fc<=4
+    B= 0.85;
+elseif (4<fc)&&(fc<=8)
+    B= 0.85-0.05*(fc-4);
+else
+    B= 0.65;
+end
+
+%Compute depth of neutral axis 'c'
+c=T/(0.85*fc*B*b); % Neutral axis depth
+    fprintf('Value of c: %0.2f in\n',c) 
+
+%Compute compression block depth 'a'
+a=B*c; % Compression Block Depth
+    fprintf('Value of a: %0.2f in^2\n',a)
+
+%Compute compression force in concrete
+Cc=0.85*fc*(b*a); % Compression Force in Concrete
+
+%Check Assumption through Strain
+    fprintf('Assume the steel bar is yeilding, Strain_s>=Strain_y\n')
+
+Ecu=0.003;
+
+Strain_y=fy/ME; %Vlaue of Ey
+    fprintf ('Value of Strain_y is: %0.4f\n', Strain_y);
+
+Strain_s=(Ecu*(d-c))/(c); %Value of Es
+    fprintf ('Value of Strain_s is: %0.4f\n', Strain_s);
+
+if Strain_s>Strain_y
+    c=(As*Strain_s*ME)/(0.85*fc*B*b); 
+end
+     fprintf ('Value of Strain_s is: %0.4f\n', Strain_s);
+
+if Strain_s>=Strain_y
+    fprintf('Considering that Strain_s>=Strain_y. Therefore, the Steel Bars are Yielding. The assumption is correct.\n')
+else
+    fprintf('Considering that Strain_s<Strain_y. Therefore, the Steel Bars are not Yielding. The assumption is wrong.\n')
+end
+
+%Compute for Nominal Moment, Mn
+Mn=(T)*(d-(a/2));
+    fprintf('Value of Mn is: %0.2f kips-in\n', Mn);
+
+%Compute for Reduction Factor, phi
+if Strain_s<0.002
+    phi=0.65;
+elseif (Strain_s>=0.002)&&(Strain_s<0.005)
+    phi=0.65+(Strain_s-0.002)*(250/3);
+elseif Strain_s>=0.005
+    phi=0.9;
+end
+    fprintf('Value of phi is: %0.2f kips-in\n', phi)
+    
+%Compute for Reduced Nominal Monent, M
+M=phi*Mn; 
+    fprintf('Value of M is: %0.2f kips-in\n', M)
+
+%Check for Minimum Area of Steel Bars, As_min
+As_m=3*(sqrt(fc));
+
+if As_m>=200
+    As_min=(3*(sqrt(fc)*b*d))/(fy);
+elseif As_m<=200
+    As_min=(200*b*d)/(fy*1000);
+end
+
+fprintf ('Value of As_min is: %0.2f in^2\n', As_min)
+fprintf ('Value of As is: %0.2f in^2\n', As)
+
+if As_min>As
+    fprintf ('Considering that As_min>As, therefore the area of steel bars does not achieve the required area.\n');
+elseif As_min <= As
+    fprintf ('Considering that As_min<=As, therefore area of steel bars does achieve the required area.\n');
+end
+
+%Calculation for Required Maximum Nominal Moment, Mn_max
+Mn_max=(((1.4*DL)+(1.6*LL))*(As)*(As)*12)/(8); 
+    fprintf ('Value of Mn_max is: %.2f kips*in\n', Mn_max);
+
+%General Design Strength Equation for Flexure
+if M>=Mn_max
+    fprintf('Since M>=Mn_max. Therefore, the flexural strength of the beam is achieved.\n');
+else
+    fprintf('Since M<Mn_max. Therefore, the flexural strength of the beam is not achieved.\n');
+end