Let's solve **Problem 7.36** step by step: --- # Problem 7.36 **Given:** - Simply supported beam \( AB \), length \( 1.7 \) m (\( 1.5 + .2 \)), supports at \( A \) and \( C \). - Three point loads: - \( 40 \) N at \( B \) (at \( .2 \) m from right support \( C \)) - \( 32 \) N at \( D \) (\( 1. \) m from \( A \)) - \( 16 \) N at \( E \) (\( 1.5 \) m from \( A \)) We are to: 1. Draw the **shear force** and **bending moment diagrams**. 2. Determine the maximum absolute values of the shear force and bending moment. --- ## **Step 1: Find Reactions at Supports** Let: - \( A \): Left support (at \( x = \)) - \( C \): Right support (at \( x = 1.7\,\text{m} \)) - \( B \): .2 m left of \( C \) (\( x = 1.5\,\text{m} \)) - \( D \): \( x = 1. \,\text{m} \) - \( E \): \( x = 1.5 \,\text{m} \) Let \( R_A \) and \( R_C \) be the vertical reactions at \( A \) and \( C \), respectively. **Sum of Forces:** \[ R_A + R_C = 40 + 32 + 16 = 88\,\text{N} \] **Take Moment about \( A \):** \[ \sum M_A = \] \[ -40 \cdot 1.5 - 32 \cdot 1. - 16 \cdot 1.5 + R_C \cdot 1.7 = \] \[ -60 - 32 - 24 + 1.7 R_C = \] \[ -116 + 1.7 R_C = \] \[ 1.7 R_C = 116 \] \[ R_C = \frac{116}{1.7} = 68.24\,\text{N} \] \[ R_A = 88 - 68.24 = 19.76\,\text{N} \] --- ## **Step 2: Shear Force and Bending Moment Diagrams** ### **A. Shear Force Diagram (SFD)** **Key Points:** - At \( A \) (\( x = \)): \( +19.76\,\text{N} \) (upward reaction) - Just left of \( D \) (\( x = 1. \)): \( +19.76\,\text{N} \) - Just right of \( D \): \( 19.76 - 32 = -12.24\,\text{N} \) - Just right of \( E \) (\( x = 1.5 \)): \( -12.24 - 16 = -28.24\,\text{N} \) - Just right of \( B \) (\( x = 1.5 \)): \( -28.24 - 40 = -68.24\,\text{N} \) - At \( C \): \( -68.24 + 68.24 = \) **Shear force values at critical points:** | Point | Position (m) | Shear Just Right (N) | |-------|--------------|----------------------| | A | | +19.76 | | D | 1. | -12.24 | | E | 1.5 | -28.24 | | B | 1.5 | -68.24 | | C | 1.7 | | #### **Graph (SFD):** ```plaintext Shear (N) | 20| ______ | | | | | | |______|_______________ | | | -10| | | | | | -30| |______|___________ | | | -70| |______|____ 1. 1.5 1.7 (m) ``` --- ### **B. Bending Moment Diagram (BMD)** Calculate at critical points: #### At \( A \) (\( x = \)): \( M_A = \) #### At \( D \) (\( x = 1. \)): \[ M_D = R_A \cdot 1. = 19.76 \cdot 1. = 19.76\,\text{Nm} \] #### At \( E \) (\( x = 1.5 \)): \[ M_E = R_A \cdot 1.5 - 32 \cdot (1.5-1.) = 19.76 \cdot 1.5 - 32 \cdot .5 \] \[ = 29.64 - 16 = 13.64\,\text{Nm} \] #### At \( B \) (\( x = 1.5 \)): (Same calculation as for E, as both are at \( x = 1.5 \)) #### At \( C \) (\( x = 1.7 \)): \[ M_C = \,\text{Nm} \text{ (Simply supported)} \] --- ## **Final Answers** ### **1. Maximum Absolute Shear Force:** \[ \boxed{68.24\,\text{N}} \] (Occurs just left of support \( C \)) ### **2. Maximum Absolute Bending Moment:** \[ \boxed{19.76\,\text{Nm}} \] (Occurs at \( D \), \( x = 1.\,\text{m} \)) --- ## **Summary Table** | Location | Shear (N) | Moment (Nm) | |----------|-----------|-------------| | A () | +19.76 | | | D (1.) | -12.24 | 19.76 | | E/B (1.5)| -68.24 | 13.64 | | C (1.7) | | | --- ## **Shear and Bending Moment Diagrams** ### Shear Force Diagram ``` SFD: | | _______ 20| | | | | |_____________ | | | | -12| | | _______________| -28| | | | -68|__|______________ 1. 1.5 1.7 (m) ``` ### Bending Moment Diagram ``` BMD: | 20| * | / | / 10| * | / | / |___*______ 1. 1.5 1.7 (m) ``` --- Let me know if you need the solution for **Problem 7.35** or further clarifications!