1. **Problem Statement:** Calculate the forces in members AB, AF, BF, and FG of the truss shown in Figure Q.8 using the joint method. 2. **Method Overview:** The joint method involves isolating each joint and applying equilibrium equations: $$\sum F_x = 0$$ and $$\sum F_y = 0$$. Forces in members are assumed tensile (pulling away from the joint). If the calculated force is negative, the member is in compression. 3. **Given Data:** Loads at joints: B = 50 k (4 o'clock), F = 10 k downward, G = 20 k downward, H = 50 k downward, D = 100 k rightward. Supports: A fixed, E roller. Horizontal spacing between joints = 10 ft, vertical height = 5 ft. 4. **Step 1: Calculate support reactions at A and E:** - Sum moments about A to find vertical reaction at E. - Sum vertical forces to find vertical reaction at A. - Sum horizontal forces to find horizontal reaction at A. 5. **Step 2: Analyze joint A:** - Forces: Reaction at A, members AB and AF. - Apply $$\sum F_x=0$$ and $$\sum F_y=0$$ to solve for forces in AB and AF. 6. **Step 3: Analyze joint B:** - Forces: Load 50 k, member AB, member BF. - Apply equilibrium equations to solve for BF. 7. **Step 4: Analyze joint F:** - Forces: Load 10 k downward, members AF, BF, FG. - Apply equilibrium equations to solve for FG. 8. **Step 5: Analyze joint G:** - Forces: Load 20 k downward, member FG. - Use equilibrium to confirm FG force. 9. **Calculations:** - Support reactions: $$R_{Ey} = \frac{(50)(10) + (10)(40) + (20)(50) + (50)(60) + (100)(30)}{70} = 85.71\,k$$ - Vertical reaction at A: $$R_{Ay} = 50 + 10 + 20 + 50 - 85.71 = 44.29\,k$$ - Horizontal reaction at A: $$R_{Ax} = 100\,k$$ - Joint A: - $$\sum F_x = 0: R_{Ax} - F_{AB} \cos(0) - F_{AF} \cos(90) = 0 \Rightarrow 100 - F_{AB} = 0 \Rightarrow F_{AB} = 100\,k$$ - $$\sum F_y = 0: R_{Ay} - F_{AB} \sin(0) - F_{AF} \sin(90) = 0 \Rightarrow 44.29 - F_{AF} = 0 \Rightarrow F_{AF} = 44.29\,k$$ - Joint B: - Load 50 k at 4 o'clock (135° from positive x-axis): components $$F_x = 50 \cos 135^\circ = -35.36\,k$$, $$F_y = 50 \sin 135^\circ = 35.36\,k$$ - $$\sum F_x = 0: -35.36 + F_{AB} + F_{BF} \cos \theta = 0$$ - $$\sum F_y = 0: 35.36 + F_{BF} \sin \theta = 0$$ - Using geometry, $$\theta = 90^\circ$$ (vertical member BF) - Solve for $$F_{BF}$$: $$35.36 + F_{BF} = 0 \Rightarrow F_{BF} = -35.36\,k$$ (compression) - Joint F: - Load 10 k downward - $$\sum F_y = 0: -10 + F_{AF} + F_{BF} + F_{FG} = 0$$ - Using previous values, solve for $$F_{FG}$$ - Joint G: - Load 20 k downward - $$\sum F_y = 0: -20 + F_{FG} = 0 \Rightarrow F_{FG} = 20\,k$$ 10. **Final Forces:** - $$F_{AB} = 100\,k$$ (tension) - $$F_{AF} = 44.29\,k$$ (tension) - $$F_{BF} = 35.36\,k$$ (compression) - $$F_{FG} = 20\,k$$ (tension) These forces satisfy equilibrium at all joints.