1. **State the problem:** We need to design member AB of a truss subjected to various loads and reactions. Given loads are vertical forces at points A, B, C, D, F, and G, with angles of 30° at A and G. The span is divided into three segments of 3 m each. 2. **Calculate support reactions R1 and R2:** Sum of vertical forces must be zero for equilibrium: $$R1 + R2 = 10 + 50 + 30 + 150 + 50 + 10 = 300\text{ kN}$$ Taking moments about point A (counterclockwise positive): $$\sum M_A = 0 = -50 \times 3 - 30 \times 6 - 150 \times 9 - 50 \times 12 - 10 \times 15 + R2 \times 15$$ Calculate moments: $$-150 - 180 - 1350 - 600 - 150 + 15 R2 = 0$$ $$-2430 + 15 R2 = 0$$ $$15 R2 = 2430$$ $$R2 = \frac{2430}{15} = 162\text{ kN}$$ Then, $$R1 = 300 - 162 = 138\text{ kN}$$ 3. **Determine forces in member AB:** Member AB is inclined at 30°. The vertical component of the reaction at A is $R1 = 138$ kN. The force in member AB, $F_{AB}$, can be found by resolving vertical forces: $$F_{AB} \sin 30^\circ = R1$$ $$F_{AB} = \frac{R1}{\sin 30^\circ} = \frac{138}{0.5} = 276\text{ kN}$$ 4. **Apply safety factors and design parameters:** Given specific gravity $s.g. = 60\% = 0.6$ and $k_c = 1.2$. The design force considering safety factor: $$F_{design} = F_{AB} \times k_c = 276 \times 1.2 = 331.2\text{ kN}$$ 5. **Final answer:** The design force for member AB is approximately **331.2 kN** considering the given safety factor and load conditions.