Truss Member Forces A3B954

1. **Problem Statement:** Determine the forces in members $CD$ and $GD$ of the truss shown in Figure 3 using the method of joints, and identify if these forces are tension or compression. 2. **Given Data:** - Loads: $12\text{kN}$ downward at joint $E$, $18\text{kN}$ downward at joint $G$ - Supports: Roller at $A$, pin at $B$ - Geometry: Angles $30^\circ$ and $60^\circ$, spans $6\text{m}$ between $A$-$B$ and $C$-$D$ 3. **Step 1: Calculate support reactions** Sum of vertical forces and moments to find reactions at $A$ and $B$. Taking moments about $A$: $$\sum M_A = 0 = -12 \times 6 - 18 \times 12 + R_B \times 12$$ $$R_B = \frac{12 \times 6 + 18 \times 12}{12} = \frac{72 + 216}{12} = \frac{288}{12} = 24\text{kN}$$ Sum of vertical forces: $$R_A + R_B = 12 + 18 = 30\text{kN}$$ $$R_A = 30 - 24 = 6\text{kN}$$ 4. **Step 2: Analyze joint $D$ to find forces in $CD$ and $GD$** At joint $D$, members $CD$ and $GD$ meet. Assume tension positive (pulling away from joint). Sum of vertical forces at $D$: $$\sum F_y = 0 = F_{CD} \sin 30^\circ + F_{GD} \sin 60^\circ$$ Sum of horizontal forces at $D$: $$\sum F_x = 0 = F_{GD} \cos 60^\circ - F_{CD} \cos 30^\circ$$ 5. **Step 3: Solve the system of equations** From horizontal forces: $$F_{GD} \cos 60^\circ = F_{CD} \cos 30^\circ$$ $$F_{GD} = F_{CD} \frac{\cos 30^\circ}{\cos 60^\circ} = F_{CD} \frac{\sqrt{3}/2}{1/2} = F_{CD} \sqrt{3}$$ Substitute into vertical forces: $$F_{CD} \sin 30^\circ + (F_{CD} \sqrt{3}) \sin 60^\circ = 0$$ $$F_{CD} \left( \frac{1}{2} + \sqrt{3} \times \frac{\sqrt{3}}{2} \right) = 0$$ $$F_{CD} \left( \frac{1}{2} + \frac{3}{2} \right) = 0$$ $$F_{CD} \times 2 = 0 \Rightarrow F_{CD} = 0$$ Then: $$F_{GD} = 0$$ 6. **Step 4: Interpretation** Forces in members $CD$ and $GD$ are zero, meaning these members are **zero-force members** under the given loading. **Final answer:** - $F_{CD} = 0$ (zero force member) - $F_{GD} = 0$ (zero force member) Both members carry no tension or compression under the given loads.