1. **Problem Statement:** Determine the forces in members BD, BE, and CE of the given truss using the method of sections. 2. **Method of Sections Overview:** The method of sections involves cutting through the truss to expose the internal forces in specific members and then applying equilibrium equations to solve for those forces. 3. **Step 1: Identify the section to cut** Cut through members BD, BE, and CE to isolate a section of the truss. We will consider the right section containing points E, F, and G. 4. **Step 2: Calculate support reactions** Sum of vertical forces and moments about G to find reactions at F and G. - Let reaction at F be $F_y$ (vertical) and at G be $G_y$ (vertical). - Taking moments about G: $$\sum M_G = 0 = F_y \times 12 - 12 \times 9 - 12 \times 6 - 12 \times 3$$ $$F_y \times 12 = 12 \times (9 + 6 + 3) = 12 \times 18 = 216$$ $$F_y = \frac{216}{12} = 18\text{ kN}$$ - Sum of vertical forces: $$F_y + G_y = 12 + 12 + 12 = 36$$ $$18 + G_y = 36 \Rightarrow G_y = 18\text{ kN}$$ 5. **Step 3: Analyze the right section** Cut through members BD, BE, and CE. The cut exposes forces $F_{BD}$, $F_{BE}$, and $F_{CE}$. 6. **Step 4: Geometry of members** - Horizontal distance between verticals = 3 m - Vertical height = 4 m Calculate lengths and direction cosines: - Member BD: diagonal from B to D, length $= \sqrt{3^2 + 4^2} = 5$ m - Member BE: vertical member, length = 4 m - Member CE: horizontal member, length = 3 m 7. **Step 5: Apply equilibrium equations to the right section** Sum of forces in x and y directions and moments about a point (choose point E to eliminate $F_{BE}$ and $F_{CE}$): - Sum of moments about E: $$\sum M_E = 0 = -F_{BD} \times 4 + 18 \times 3$$ $$-4 F_{BD} + 54 = 0$$ $$F_{BD} = \frac{54}{4} = 13.5\text{ kN}$$ - Sum of vertical forces: $$\sum F_y = 0 = 18 - F_{BE} - F_{BD} \times \frac{4}{5}$$ $$18 - F_{BE} - 13.5 \times 0.8 = 0$$ $$18 - F_{BE} - 10.8 = 0$$ $$F_{BE} = 7.2\text{ kN}$$ - Sum of horizontal forces: $$\sum F_x = 0 = F_{CE} - F_{BD} \times \frac{3}{5}$$ $$F_{CE} = 13.5 \times 0.6 = 8.1\text{ kN}$$ 8. **Step 6: Interpret results** - $F_{BD} = 13.5$ kN (tension if assumed direction is away from joint) - $F_{BE} = 7.2$ kN (tension or compression depending on assumed direction) - $F_{CE} = 8.1$ kN (tension or compression depending on assumed direction) **Final answer:** $$F_{BD} = 13.5\text{ kN}, \quad F_{BE} = 7.2\text{ kN}, \quad F_{CE} = 8.1\text{ kN}$$