Let's break down the solution step by step for the truss problem using the **Method of Sections**. --- ## **Given:** - Truss with pin at **A** and roller at **E** - Vertical load of **100 kN** at **J** - All panels are 1 m high and 2 m wide (assuming square/rectangular panels for simplicity) We need to find the force in members **CF**, **BC**, and **FG**. --- ## **Step 1: Find Support Reactions** ### **Sum of Moments at A:** Let’s denote the distance between each vertical member as 2 m (since there are 4 panels and 8 m total length). Distance from A to J = 8 m. \[ \sum M_A = 0 \implies E_y \times 8 - 100 \times 8 = 0 \] \[ E_y = 100 \text{ kN} \] ### **Sum of Vertical Forces:** \[ A_y + E_y = 100 \text{ kN} \] \[ A_y + 100 = 100 \] \[ A_y = 0 \] --- ## **Step 2: Method of Sections** Let's pass a section through members **CF**, **BC**, and **FG**. You'll want to cut through the truss between **B, C, F, G**, and then analyze either the left or right portion. Let's take the **left section** (A, B, C, E, F, G). --- ### **Free-Body Diagram (Left Section: A-B-C-E-F-G)** - Known: Support at A (A_y = 0), E_y = 100 kN (upward) - Unknowns: Forces in CF, BC, FG --- ### **Write Equilibrium Equations** Let's denote: - \( F_{BC} \): Force in member BC - \( F_{CF} \): Force in member CF - \( F_{FG} \): Force in member FG #### **Take Moments about Point F (eliminates \( F_{FG} \) and \( F_{CF} \)):** Let’s assign coordinates: - Each panel is 2 m wide. - F is at (2, 0), G at (4, 0), C at (4, 1), B at (2, 1). Take moments about F: \[ \sum M_F = 0 \] - **E_y** acts at E, which is 0 m from F (doesn't create a moment). - **F_{BC}**: Line of action is 2 m above F (vertical distance). - **F_{CF}** and **F_{FG}**: Both pass through F (no moment). \[ F_{BC} \times 2 = 0 \implies F_{BC} = 0 \] --- ### **Sum of Forces in Y (Vertical):** \[ \sum F_y = 0 \] \[ E_y + F_{CF} \cdot \sin\theta = 0 \] Find angle θ for CF: - CF: from F(2,0) to C(4,1): dx = 2, dy = 1 - \(\sin\theta = \frac{1}{\sqrt{2^2+1^2}} = \frac{1}{\sqrt{5}}\) So, \[ 100 + F_{CF} \cdot \frac{1}{\sqrt{5}} = 0 \] \[ F_{CF} = -100 \sqrt{5} \approx -223.6 \text{ kN} \] (Negative means **compression**) --- ### **Sum of Forces in X (Horizontal):** \[ \sum F_x = 0 \] \[ F_{FG} + F_{CF} \cdot \cos\theta = 0 \] \(\cos\theta = \frac{2}{\sqrt{5}}\) \[ F_{FG} + (-223.6) \cdot \frac{2}{\sqrt{5}} = 0 \] \[ F_{FG} = 223.6 \cdot \frac{2}{\sqrt{5}} \] \[ F_{FG} = 223.6 \cdot 0.8944 \approx 200 \text{ kN} \] (**Tension**) --- ## **Summary of Forces** - **CF**: \(-223.6 \text{ kN}\) (**Compression**) - **BC**: \(0 \text{ kN}\) (**Zero-force member**) - **FG**: \(200 \text{ kN}\) (**Tension**) --- ## **Boxed Answers** \[ \boxed{ \begin{aligned} \text{CF:} & \ -223.6 \ \text{kN} \ (\text{Compression}) \\ \text{BC:} & \ 0 \ \text{kN} \\ \text{FG:} & \ 200 \ \text{kN} \ (\text{Tension}) \end{aligned} } \] If you need a more detailed diagram or clarification, let me know!

**Step 1: Conceptual Introduction** **Explanation:** The **Method of Sections** is a powerful technique used in structural analysis to determine the internal forces in specific members of a truss without analyzing the entire structure in detail. By making an imaginary cut through the truss that passes through the members of interest, we can isolate a section of the truss and apply equilibrium equations to solve for the unknown forces in those members. This method is especially useful when only a few member forces are required, saving time and effort compared to analyzing the entire structure. --- **Step 2: Relevant Formulae and Principles** **Explanation:** The primary equilibrium equations used are: 1. **Sum of Vertical Forces:** \[ \sum F_y = 0 \] 2. **Sum of Horizontal Forces:** \[ \sum F_x = 0 \] 3. **Sum of Moments about a Point:** \[ \sum M = 0 \] When analyzing a section of a truss, these equations help determine the unknown member forces. The forces in members are assumed to be either tension (pulling away from the joint) or compression (pushing towards the joint). The geometry of the members (lengths and angles) allows us to resolve these forces into their components using sine and cosine functions. --- **Step 3: Step-by-Step Calculation** **Detailed Explanation:** 1. **Support Reactions:** - The truss has a pin support at **A** and a roller support at **E**. - A vertical load of **100 kN** is applied at point **J**. Using moments about **A** to find the reaction at **E**: \[ \sum M_A = 0 \] - The load at **J** causes a moment about **A**: \[ E_y \times 8\, \text{m} - 100\, \text{kN} \times 8\, \text{m} = 0 \] \[ E_y = \frac{100 \times 8}{8} = 100\, \text{kN} \] - Vertical equilibrium: \[ A_y + E_y = 100\, \text{kN} \] \[ A_y + 100 = 100 \] \[ A_y = 0 \] 2. **Cutting the Truss:** - Cut through members **CF**, **BC**, and **FG**. - The cut divides the truss into two parts; analyze the **left section**. 3. **Analyzing the Left Section:** - Known: - Support reactions: \(A_y = 0\), \(E_y = 100\, \text{kN}\) - Member forces to find: **CF**, **BC**, **FG** - Assign coordinates: - F at (2,0) - C at (4,1) - B at (2,1) - G at (4,0) 4. **Calculating Member Forces:** - **Member CF:** - From F to C: dx = 2, dy = 1 - Length: \(\sqrt{2^2 + 1^2} = \sqrt{5}\) - \(\sin \theta = \frac{1}{\sqrt{5}}\), \(\cos \theta = \frac{2}{\sqrt{5}}\) - **Vertical equilibrium:** Summing vertical forces: \[ E_y + F_{CF} \times \sin \theta = 0 \] \[ 100 + F_{CF} \times \frac{1}{\sqrt{5}} = 0 \] \[ F_{CF} = -100 \times \sqrt{5} \approx -223.6\, \text{kN} \] Negative sign indicates **compression**. - **Horizontal equilibrium:** Summing horizontal forces: \[ F_{FG} + F_{CF} \times \cos \theta = 0 \] \[ F_{FG} + (-223.6) \times \frac{2}{\sqrt{5}} = 0 \] \[ F_{FG} = 223.6 \times \frac{2}{\sqrt{5}} \approx 200\, \text{kN} \] Positive sign indicates **tension**. - **Member BC:** - No moment created (since it is vertical), and the calculations show \(F_{BC} = 0\), indicating it is a **zero-force member**. --- **Final Results:** | Member | Force (kN) | Nature | |----------|--------------|--------------| | **CF** | **−223.6** | Compression | | **BC** | **0** | Zero-force | | **FG** | **200** | Tension | --- **Summary:** - Member **CF** is in compression (~224 kN). - Member **BC** carries no force. - Member **FG** is in tension (~200 kN). --- **Diagram (Sketch):** *(Please visualize or draw based on the description)* ``` J (100 kN load) | | A ----- B ----- C | | | | E (roller) F ----- G ``` - Members: AB, BC, CF, FG, and others. - The cut passes through CF, BC, and FG. - Supports at A (pin) and E (roller). --- If you'd like, I can prepare a detailed diagram or clarify further steps!

Certainly! Here's a simplified virtual diagram of the truss and the cut you would make for the Method of Sections. Please note that this is a schematic representation to help visualize the problem: ``` J (100 kN load) | | * (J) | | G (4,0) | / \ | / \ | / \ *--*-------*--* (C at 4,1) | | | | | | *--*-------*--* (F at 2,0) | | | | | | *--*-------*--* (B at 2,1) | | | | | | *--*-------*--* (A at 0,0) (Support at A) ``` ### Key points: - **Supports**: Pin at A (bottom-left), roller at E (not shown in diagram but assumed somewhere along the base). - **Members**: - Horizontal members: AB, FG - Vertical members: BC, AE (not explicitly shown) - Diagonal members: CF, others as per the actual truss. ### Cut for Method of Sections: - Imagine slicing through **members CF, BC, and FG** (shown as a dashed line in the actual diagram). - You analyze the **left section** (A, B, C, E, F, G) to find forces in these members. --- **Note:** For precise analysis, it's best to draw the actual truss with exact measurements, but this schematic helps visualize the concept. --- If you'd like, I can generate a more detailed diagram or help draw this in a drawing tool!