Certainly! Let’s **derive the moment at both ends** for a couple of cases from the table. I’ll use **Case 1 (Point Load at center)** and **Case 4 (Uniformly Distributed Load, UDL)** as examples. --- ## **Case 1: Point Load at Center** ### **Given:** - Load condition: Point load \( P \) at the center (midspan) of span \( l \). - Both ends fixed. --- ### **Step 1: Draw the Beam and Loading** A beam of length \( l \) with a point load \( P \) at midspan (i.e., at \( l/2 \)), both ends fixed. --- ### **Step 2: Write the Moment Equilibrium Equations** For a fixed-fixed beam with a point load at midspan, the moments at the supports (let’s call them \( M_A \) and \( M_B \)) are **equal in magnitude but opposite in sign** due to symmetry. --- ### **Step 3: Use Known Formulas or Derivation** From structural analysis for a fixed-fixed beam under a central point load, \[ M_A = M_B = -\frac{P l}{8} \] (negative sign indicates a hogging moment at the supports). --- ### **Explanation:** When a point load is applied at midspan on a fixed-fixed beam: - The fixed supports resist both vertical force and moment. - Due to symmetry, the moments at both supports are equal. - The value \(-\frac{P l}{8}\) comes from moment distribution or superposition methods. --- ### **Final Answer:** \[ \boxed{ M_A = M_B = -\frac{P l}{8} } \] --- ## **Case 4: Uniformly Distributed Load (UDL)** ### **Given:** - Load condition: Uniformly distributed load \( q \) over the entire span \( l \). - Both ends fixed. --- ### **Step 1: Draw the Beam and Loading** A beam of length \( l \) with a uniform load \( q \) applied across the entire span, both ends fixed. --- ### **Step 2: Write the Moment Equilibrium Equations** For a fixed-fixed beam with a uniform load: --- ### **Step 3: Use Known Formulas or Derivation** From structural analysis for a fixed-fixed beam under a uniform load, \[ M_A = M_B = -\frac{q l^2}{12} \] (negative sign indicates a hogging moment at the supports). --- ### **Explanation:** For a fixed-fixed beam under uniform load: - Both supports have equal negative moments due to symmetry. - The value \(-\frac{q l^2}{12}\) is a standard result from moment distribution or integration of the moment curve. --- ### **Final Answer:** \[ \boxed{ M_A = M_B = -\frac{q l^2}{12} } \] --- ## **Summary Table (for the two cases):** | Load Case | Moment at A (\(M_A\)) | Moment at B (\(M_B\)) | |--------------------------|-----------------------|-----------------------| | 1. Point load at center | \(-\frac{P l}{8}\) | \(-\frac{P l}{8}\) | | 4. Uniform load over span| \(-\frac{q l^2}{12}\) | \(-\frac{q l^2}{12}\) | --- If you'd like to see the step-by-step derivation for another case from the table, let me know which one!

Certainly! Let's select two more cases from the provided table to derive the moments at both ends, with detailed step-by-step explanations. --- ## **Case 2: Uniformly Varying Load (Triangular Load)** ### **Given:** - Load condition: Triangular load starting from zero at one end to maximum \( w \) at the other end over span \( l \). - Both ends fixed. --- ### **Step 1: Draw the Beam and Loading** The load distribution is triangular, say, starting from 0 at \( x=0 \) to \( w \) at \( x=l \). --- ### **Step 2: Determine Equivalent Load and Its Resultant** - The total load \( W \) is the area under the load diagram: \[ W = \frac{1}{2} \times w \times l \] - The location of the resultant \( W \) acts at a distance \( \frac{l}{3} \) from the smaller end (assuming the load increases linearly from 0 to \( w \)). --- ### **Step 3: Use Superposition and Structural Analysis** Since the load is triangular, the moments at the supports for fixed-fixed conditions can be derived using standard formulas or moment distribution methods. **Standard Result:** \[ M_A = -\frac{w l^2}{30} \] \[ M_B = -\frac{w l^2}{30} \] *(Note: The negative sign indicates hogging moments at the supports.)* --- ### **Explanation:** - The derivation involves integrating the load distribution to find the bending moment diagram. - The maximum moments at the supports for a triangular load on a fixed beam are known to be \(-\frac{w l^2}{30}\). --- ### **Final Answer:** \[ \boxed{ M_A = M_B = -\frac{w l^2}{30} } \] --- ## **Case 3: Point Load at 1/3 Span from Left Support** ### **Given:** - Load condition: Point load \( P \) placed at \( x = \frac{l}{3} \) from support A. - Both ends fixed. --- ### **Step 1: Draw the Beam and Loading** A point load \( P \) located at \( x = l/3 \) from support A, with fixed supports at both ends. --- ### **Step 2: Analyze Support Moments** Since the load is not at the center, the moments at supports are unequal and need to be derived using equilibrium equations. --- ### **Step 3: Apply Equilibrium and Moment Equations** - The vertical equilibrium: \[ \text{Sum of vertical forces} = 0 \Rightarrow R_A + R_B = P \] - Sum of moments about A: \[ \text{Moment at A} = M_A \] - Sum of moments about B: \[ \text{Moment at B} = M_B \] - To find \( M_A \) and \( M_B \), decompose the load's effects: **Support reactions:** \[ R_A = \frac{P (l - a)}{l} \] \[ R_B = \frac{P a}{l} \] where \( a = l/3 \). Calculating: \[ R_A = \frac{P (l - l/3)}{l} = P \left(1 - \frac{1}{3}\right) = \frac{2P}{3} \] \[ R_B = \frac{P \times l/3}{l} = \frac{P}{3} \] --- ### **Step 4: Calculate Moments at Supports** Using the beam's boundary conditions and equilibrium: \[ M_A = \frac{R_A l}{2} - P \times \frac{a}{2} \] Alternatively, for a fixed-fixed beam with a load at \( a \): \[ M_A = -\frac{P a (l - a)^2}{l^2} \] \[ M_B = -\frac{P a^2 (l - a)}{l^2} \] Substitute \( a = l/3 \): \[ M_A = -\frac{P \times (l/3) \times (2l/3)^2}{l^2} = -\frac{P \times \frac{l}{3} \times \frac{4 l^2}{9}}{l^2} = -\frac{P \times l \times 4 l^2}{3 \times 9 l^2} = -\frac{4 P l}{27} \] Similarly, \[ M_B = -\frac{P \times (l/3)^2 \times (2l/3)}{l^2} = -\frac{P \times \frac{l^2}{9} \times \frac{2 l}{3}}{l^2} = -\frac{P \times l^2 \times 2 l}{9 \times 3 l^2} = -\frac{2 P l}{27} \] --- ### **Final Results:** \[ \boxed{ M_A = -\frac{4 P l}{27} } \] \[ \boxed{ M_B = -\frac{2 P l}{27} } \] --- # **Summary of the two additional cases:** | Load Case | Moment at A (\( M_A \)) | Moment at B (\( M_B \)) | |----------------------------------------------|-------------------------------------|-------------------------------------| | 2. Triangular load \( w \) over span \( l \) | \(-\frac{w l^2}{30}\) | \(-\frac{w l^2}{30}\) | | 3. Point load \( P \) at \( l/3 \) from A | \(-\frac{4 P l}{27}\) | \(-\frac{2 P l}{27}\) | --- Let me know if you'd like further clarification or derivations for other cases!