Let's break down and solve the problem step by step, including **explanations for each step** and the **final answer** as requested. --- ## **Step 1: Understanding and Visualizing the Problem** ### **Explanation:** - The beam is fixed at the left end (point 1), simply supported at points 2 and 3. - The span between supports is 3 meters (1 m from 1 to 2, 1 m from 2 to 3, and 1 m from 3 to the end). - A **partial uniform load** of \( 12 \) kN/m is applied from point 2 to point 3 (over 1 m). - Material and geometric properties: \( E = 200 \) GPa, \( I = 4 \times 10^6 \) mm\(^4\). --- ## **Step 2: Converting Units** ### **Explanation:** Let's convert everything to consistent SI units (N, m): - \( E = 200 \) GPa = \( 200 \times 10^9 \) N/m\(^2\) - \( I = 4 \times 10^6 \) mm\(^4\) = \( 4 \times 10^6 \times 10^{-12} \) m\(^4\) = \( 4 \times 10^{-6} \) m\(^4\) - Load \( w = 12 \) kN/m = \( 12,000 \) N/m --- ## **Step 3: Calculating Support Reactions** ### **Explanation:** Since this is a statically indeterminate beam (fixed + two simple supports), we use the **Three-Moment Equation** or superposition, or use finite element/MATLAB for slopes. But, for manual calculation, let's proceed with the **moment-area method** or equations for indeterminate beams. --- ## **Step 4: Writing the MATLAB Script Structure** ### **Explanation:** Since the question asks for a **MATLAB script** for slopes at points 2 and 3 and maximum stress, here's a general outline: 1. **Input Parameters** 2. **Stiffness Matrix Assembly** 3. **Load Vector** 4. **Solving for Deflections and Rotations** 5. **Calculating Maximum Stress** --- ## **Step 5: MATLAB Script for Slope Calculation** ### **Explanation:** We will set up the **beam finite element method** for the three spans with the given loading. The goal is to find the rotations (slopes) at points 2 and 3. #### **MATLAB Script:** ```matlab % Beam and Material Properties E = 200e9; % Young's modulus in Pa I = 4e-6; % Moment of inertia in m^4 w = 12000; % Load in N/m % Node and element setup L1 = 1; L2 = 1; L3 = 1; % Lengths in m % Stiffness matrices for each span k1 = E*I/L1^3 * [12 6*L1 -12 6*L1; 6*L1 4*L1^2 -6*L1 2*L1^2; ... -12 -6*L1 12 -6*L1; 6*L1 2*L1^2 -6*L1 4*L1^2]; k2 = E*I/L2^3 * [12 6*L2 -12 6*L2; 6*L2 4*L2^2 -6*L2 2*L2^2; ... -12 -6*L2 12 -6*L2; 6*L2 2*L2^2 -6*L2 4*L2^2]; k3 = E*I/L3^3 * [12 6*L3 -12 6*L3; 6*L3 4*L3^2 -6*L3 2*L3^2; ... -12 -6*L3 12 -6*L3; 6*L3 2*L3^2 -6*L3 4*L3^2]; % Assemble global stiffness matrix (6 DOF: 3 nodes x 2 DOF each) K = zeros(6); K(1:4,1:4) = K(1:4,1:4) + k1; K(3:6,3:6) = K(3:6,3:6) + k2; K(5:6,5:6) = K(5:6,5:6) + k3(3:4,3:4); % Global force vector F = zeros(6,1); % For uniform distributed load on span 2-3 only (element 2) % Equivalent nodal loads for uniform load on beam element of length L: % [w*L/2; w*L^2/12; w*L/2; -w*L^2/12] fe2 = [w*L2/2; w*L2^2/12; w*L2/2; -w*L2^2/12]; F(3:6) = F(3:6) + fe2; % Apply boundary conditions: % Node 1 fixed (both displacement and rotation = 0) K(1,:) = 0; K(:,1) = 0; K(1,1) = 1; F(1) = 0; K(2,:) = 0; K(:,2) = 0; K(2,2) = 1; F(2) = 0; % Solve for displacements and rotations d = K\F; % Slopes at points 2 and 3 (rotation DOFs at nodes 2 and 3) theta2 = d(4); % at node 2 (radians) theta3 = d(6); % at node 3 (radians) fprintf('Slope at point 2: %.6e radians\n', theta2); fprintf('Slope at point 3: %.6e radians\n', theta3); % Maximum moment for max stress % For this loading, max moment occurs at point 2 or 3 or under the load Mmax = (w*L2^2)/8; % For simply supported beam, approx. y_max = 0.05; % Assume half depth of beam in meters (example) sigma_max = Mmax * y_max / I; fprintf('Maximum stress in the beam: %.2f MPa\n', sigma_max/1e6); ``` --- ## **Step 6: Maximum Stress Calculation** ### **Explanation:** Maximum bending stress: \( \sigma_{max} = \frac{M_{max} \cdot y_{max}}{I} \) - \( M_{max} \) for a simply supported beam with uniform load is at midspan: \( M_{max} = \frac{wL^2}{8} \) - \( y_{max} \) is the distance from neutral axis to outer fiber (need beam height; assume for example). --- ## **Step 7: Final Answers** ### **Explanation:** 1. **MATLAB script** provided above. 2. **Slopes at points 2 and 3** will be printed by the script. 3. **Maximum stress** will be printed by the script. **Note:** For exact values, run the script in MATLAB. You should substitute the actual beam height for `y_max` for more accurate stress. --- ## **Summary Table** | Quantity | Expression / Value (to be computed by MATLAB) | |-------------------------|-----------------------------------------------| | Slope at point 2 | `theta2` (radians) | | Slope at point 3 | `theta3` (radians) | | Maximum bending stress | `sigma_max` (MPa) | ### **Final Answer:** - Use the MATLAB script to obtain the precise numerical values for the slopes at points 2 and 3 and the maximum bending stress in the beam, using the above method. - The **maximum stress** and **slopes** will be displayed in the MATLAB output. --- If you need **manual hand-calculation** for slopes and stress, let me know!

Certainly! Let's proceed step-by-step to manually determine the slopes at points 2 and 3, as well as the maximum stress in the beam based on the given data. --- ## **Step 1: Understanding the problem** - The beam is supported at points 1 (fixed support), 2, and 3 (simply supported). - Lengths: - 1 m from point 1 to 2 - 1 m from point 2 to 3 - 1 m from point 3 to the free end - A **partial uniform load** of 12 kN/m acts over the span from point 2 to point 3 (length = 1 m). - Material properties: - \( E = 200\, \text{GPa} = 200 \times 10^9\, \text{Pa} \) - Moment of inertia: - \( I = 4 \times 10^6\, \text{mm}^4 = 4 \times 10^{-6}\, \text{m}^4 \) --- ## **Step 2: Convert units and understand loading** - Load \(w = 12\, \text{kN/m} = 12,000\, \text{N/m}\) - Length of loaded span: \(L = 1\, \text{m}\) --- ## **Step 3: Analyze the loading** Since the load acts only on the span from point 2 to point 3, the reactions at supports will be: - \( R_1 \): Reaction at support 1 - \( R_2 \): Reaction at support 2 - \( R_3 \): Reaction at support 3 **Note:** Support at 1 is fixed, so it provides both vertical reaction and moment restraint. --- ## **Step 4: Find reactions** ### **Assumption:** - Support 1 is fixed (cantilever-like), so it can carry moment and vertical force. - Supports 2 and 3 are simple supports. ### **Equilibrium equations:** Sum of vertical forces: \[ R_1 + R_2 + R_3 = \text{total load} = 12 \times 1 = 12\, \text{kN} \] Sum of moments about support 1: - Moment contributions: - \( R_2 \) at 1 m from support 1 - \( R_3 \) at 2 m from support 1 - Load \(w\) on span 2-3 (from 2 m to 3 m): \[ \text{Moment at support 1}: \] \[ 0 = R_2 \times 1 + R_3 \times 2 - \left( w \times \text{length of load} \times \text{center of load} \right) \] Since load acts from point 2 to point 3 (length 1 m): - The load acts over 1 m starting at 2 m from support 1, so its centroid is at 2.5 m: \[ \text{Moment due to load} = w \times 1\, \text{m} \times 2.5\, \text{m} = 12,000 \times 1 \times 2.5 = 30,000\, \text{N·m} \] **Balance of moments:** \[ R_2 \times 1 + R_3 \times 2 = 30,000\, \text{N·m} \] --- ## **Step 5: Simplify reactions** Additionally, the sum of vertical reactions: \[ R_1 + R_2 + R_3 = 12,000\, \text{N} \] But since support 1 is fixed, it can carry a moment and reaction, but for simplicity, in statics, the reactions at supports 2 and 3 can be obtained by assuming support 1 is fixed. Alternatively, for a more straightforward approach, assume **support 1 is fixed**, and support 3 is simply supported, with support 2 acting as a roller. This leads to: - Reactions at 2 and 3: \[ \text{Sum of vertical forces}: \] \[ R_2 + R_3 = 12,000\, \text{N} \] - Moments about 2: \[ R_3 \times 1 - w \times 1 \times 0.5 = 0 \] (since load centroid at 2.5 m from support 1, but relative to support 2, load acts at 0.5 m from 2) Calculations: \[ R_3 \times 1 = 12,000 \times 0.5 = 6,000\, \text{N} \] \[ R_3 = 6,000\, \text{N} \] Then, \[ R_2 = 12,000 - R_3 = 6,000\, \text{N} \] --- ## **Step 6: Determine bending moments** Now, calculate maximum bending moments in each span. ### **Span 2-3 (from point 2 to point 3):** - The load \(w=12\, \text{kN/m}\) acts over 1 m. - Reaction at support 3: \( R_3 = 6\, \text{kN} \). Moment at support 2: \[ M_2 = R_3 \times 1 - w \times 1 \times 0.5 = 6,000 \times 1 - 12,000 \times 0.5 = 6,000 - 6,000 = 0\, \text{kN·m} \] This indicates the maximum bending moment occurs somewhere within the span, possibly at the load point. --- ## **Step 7: Find maximum bending moment** For the loaded span: Maximum moment occurs under the uniform load: \[ M_{max} = \frac{w L^2}{8} = \frac{12,000 \times 1^2}{8} = 1,500\, \text{kN·mm} \] Convert to N·m: \[ M_{max} = 1,500\, \text{kN·mm} = 1,500,000\, \text{N·mm} = 1,500\, \text{N·m} \] --- ## **Step 8: Calculate maximum stress** Assuming the beam has a **height \(h\)**, and the **outer fiber distance \(y_{max}\)** is: - For a typical beam, assume \(h = 300\, \text{mm} = 0.3\, \text{m}\). Then, \[ y_{max} = \frac{h}{2} = 0.15\, \text{m} \] Calculate maximum bending stress: \[ \sigma_{max} = \frac{M_{max} \times y_{max}}{I} \] Where: - \( M_{max} = 1,500\, \text{N·m} \) - \( y_{max} = 0.15\, \text{m} \) - \( I = 4 \times 10^{-6}\, \text{m}^4 \) Plug in: \[ \sigma_{max} = \frac{1,500 \times 0.15}{4 \times 10^{-6}} = \frac{225}{4 \times 10^{-6}} = 56,250,000\, \text{Pa} = 56.25\, \text{MPa} \] --- ## **## **Final Results:** ### **1. Slopes at points 2 and 3:** - To calculate slopes precisely, you'd typically use the bending moment equations and beam deflection formulas, but for a rough estimate, the slopes are small. - Approximate maximum slope at the load point: \[ \theta_{max} \approx \frac{w L^3}{24 E I} = \frac{12,000 \times 1^3}{24 \times 200 \times 10^9 \times 4 \times 10^{-6}} \] Calculate numerator: \[ 12,000 \times 1 = 12,000 \] Denominator: \[ 24 \times 200 \times 10^9 \times 4 \times 10^{-6} = 24 \times 200 \times 4 \times 10^{9 - 6} = 24 \times 200 \times 4 \times 10^{3} \] \[ = 24 \times 200 \times 4 \times 1000 = 24 \times 200 \times 4000 \] \[ = 24 \times 800,000 = 19,200,000 \] Hence, \[ \theta_{max} \approx \frac{12,000}{19,200,000} \approx 6.25 \times 10^{-4}\, \text{radians} \] which is approximately **0.000625 radians** (~0.036°). - Slopes at points 2 and 3 will be small and similar, with the maximum at the center of span 2-3. ### **2. Maximum stress:** \[ \sigma_{max} \approx 56.25\, \text{MPa} \] --- ## **## Final Summary:** | Quantity | Approximate Value | |----------------------------------------------|-------------------------------------------------| | **Slope at points 2 and 3** | **~0.000625 radians (~0.036°)** | | **Maximum bending stress in the beam** | **~56.25 MPa** | --- **Note:** These are approximate manual calculations; for precise results, detailed finite element analysis or more advanced structural analysis methods are recommended.