Factor Safety Cable Ac7D24

1. **Problem Statement:** We have a member ABC supported by a pin at C and a cable BD. A load $P = 16$ kN is applied at point A at an angle of $40^\circ$ from the vertical. The cable BD is at $30^\circ$ above the horizontal. The dimensions are: vertical height $0.6$ m, horizontal length $0.8$ m from A to B, and $0.4$ m from B to C. We need to find the factor of safety (FOS) for cable BD given its ultimate load is $100$ kN. 2. **Step 1: Analyze the forces and geometry** - Load $P = 16$ kN at $40^\circ$ from vertical means the components of $P$ are: $$P_x = 16 \sin 40^\circ$$ $$P_y = 16 \cos 40^\circ$$ - Cable BD is at $30^\circ$ above horizontal. 3. **Step 2: Calculate the moment about point C to find tension in cable BD** - Taking moments about C to eliminate reaction at C. - Coordinates: - A is $0.8 + 0.4 = 1.2$ m horizontally from C. - Vertical height from C to A is $0.6$ m. - Moment arm of load $P$ about C: Horizontal distance from C to A: $1.2$ m Vertical distance from C to A: $0.6$ m - Moment of $P$ about C: $$M_P = P_x \times 0.6 - P_y \times 1.2$$ Substitute: $$P_x = 16 \sin 40^\circ = 16 \times 0.6428 = 10.285\, \text{kN}$$ $$P_y = 16 \cos 40^\circ = 16 \times 0.7660 = 12.256\, \text{kN}$$ So, $$M_P = 10.285 \times 0.6 - 12.256 \times 1.2 = 6.171 - 14.707 = -8.536\, \text{kN}\cdot\text{m}$$ 4. **Step 3: Calculate moment arm of cable BD about C** - Cable BD is attached at B, which is $0.4$ m horizontally from C and $0$ m vertically (same horizontal line). - The cable force $T$ acts at $30^\circ$ above horizontal. - The vertical component of tension creates moment about C: $$M_T = T \times 0.4 \times \sin 30^\circ = T \times 0.4 \times 0.5 = 0.2 T$$ 5. **Step 4: Set moment equilibrium about C** - Sum of moments = 0: $$M_P + M_T = 0$$ $$-8.536 + 0.2 T = 0$$ $$0.2 T = 8.536$$ $$T = \frac{8.536}{0.2} = 42.68\, \text{kN}$$ 6. **Step 5: Calculate factor of safety (FOS)** - Ultimate load of cable BD is $100$ kN. - Factor of safety: $$\text{FOS} = \frac{\text{Ultimate Load}}{\text{Actual Load}} = \frac{100}{42.68} = 2.344$$ 7. **Final answer:** The factor of safety with respect to cable failure is **2.344** (rounded to three decimal places).