1. **Problem Statement:** Determine the force in each member of the truss and state whether each member is in tension (T) or compression (C). 2. **Key Concepts and Formulas:** - Use equilibrium equations at each joint: $$\sum F_x = 0$$ and $$\sum F_y = 0$$. - Forces in members can be resolved into components using trigonometric functions based on given angles. - Positive force indicates tension (pulling), negative indicates compression (pushing). 3. **Joint E Analysis:** - Vertical equilibrium: $$\sum F_y = 0 \Rightarrow F_{EA} = F_{EC}$$ (since vertical components balance). - Horizontal equilibrium: $$\sum F_x = 0 \Rightarrow 2.31 - 2 F_{EA} \sin 30^\circ = 0$$ Calculate $$F_{EA}$$: $$2.31 = 2 F_{EA} \times 0.5$$ $$F_{EA} = \frac{2.31}{1} = 2.31\, \text{kN}$$ Since $$F_{EA}$$ balances the force at E, and the force is pushing towards the joint, it is compression (C). Therefore: $$F_{EA} = 2.31\, \text{kN (C)}$$ $$F_{EC} = 2.31\, \text{kN (T)}$$ (equal magnitude, opposite nature). 4. **Joint A Analysis:** - Horizontal equilibrium: $$2.31 - 2.31 \sin 30^\circ - F_{AB} \cos 30^\circ + F_{AD} \cos 45^\circ = 0$$ Calculate known values: $$2.31 - 2.31 \times 0.5 - F_{AB} \times 0.866 + F_{AD} \times 0.707 = 0$$ $$2.31 - 1.155 - 0.866 F_{AB} + 0.707 F_{AD} = 0$$ - Vertical equilibrium: $$2 - 2.31 \cos 30^\circ + F_{AD} \sin 45^\circ - F_{AB} \sin 30^\circ = 0$$ Calculate known values: $$2 - 2.31 \times 0.866 + F_{AD} \times 0.707 - F_{AB} \times 0.5 = 0$$ $$2 - 2.000 + 0.707 F_{AD} - 0.5 F_{AB} = 0$$ $$0 + 0.707 F_{AD} - 0.5 F_{AB} = 0$$ 5. **Solve the system of equations:** From vertical: $$0.707 F_{AD} = 0.5 F_{AB} \Rightarrow F_{AD} = \frac{0.5}{0.707} F_{AB} = 0.707 F_{AB}$$ Substitute into horizontal: $$2.31 - 1.155 - 0.866 F_{AB} + 0.707 (0.707 F_{AB}) = 0$$ $$1.155 - 0.866 F_{AB} + 0.5 F_{AB} = 0$$ $$1.155 - 0.366 F_{AB} = 0$$ Solve for $$F_{AB}$$: $$0.366 F_{AB} = 1.155$$ $$F_{AB} = \frac{1.155}{0.366} = 3.16\, \text{kN}$$ Calculate $$F_{AD}$$: $$F_{AD} = 0.707 \times 3.16 = 2.24\, \text{kN}$$ Interpretation: - $$F_{AB}$$ is compression (C) because it resists the horizontal force pushing inward. - $$F_{AD}$$ is tension (T) as it pulls away from the joint. 6. **Joint B Analysis:** - Horizontal equilibrium: $$\sum F_x = 0 \Rightarrow F_{BC} = 3.16\, \text{kN (C)}$$ (given directly from equilibrium with member AB). - Vertical equilibrium: $$2(3.16) \sin 30^\circ - F_{BD} = 0$$ Calculate: $$2 \times 3.16 \times 0.5 - F_{BD} = 0$$ $$3.16 - F_{BD} = 0$$ $$F_{BD} = 3.16\, \text{kN}$$ Since it balances upward force, $$F_{BD}$$ is compression (C). 7. **Joint D Analysis:** - Given: $$F_{DC} = 2.24\, \text{kN (T)}$$ 8. **Summary of Forces:** - $$F_{EA} = 2.31\, \text{kN (C)}$$ - $$F_{EC} = 2.31\, \text{kN (T)}$$ - $$F_{AB} = 3.16\, \text{kN (C)}$$ - $$F_{AD} = 2.24\, \text{kN (T)}$$ - $$F_{BC} = 3.16\, \text{kN (C)}$$ - $$F_{BD} = 3.16\, \text{kN (C)}$$ - $$F_{DC} = 2.24\, \text{kN (T)}$$ Each step uses equilibrium equations and trigonometric resolution of forces to find magnitudes and nature (tension or compression) of each member force.