Let's solve the problem step by step: --- ## **a) Required Diameter of the Pins at A and B** ### **Step 1: Calculate the Total Load Applied** - The uniformly distributed load on member AB is \( w = 2 \) kN/m. - The length of AB is the radius, \( L = 4 \) m. \[ \text{Total Load} = w \times L = 2 \times 4 = 8 \text{ kN} \] --- ### **Step 2: Find the Reaction Forces at A and B** Since AB is a 4-m horizontal beam with a pin at A and B and curved member CB (which acts as a tie), let's draw the free-body diagram: - The distributed load acts downward. - Both A and B have vertical reactions \( R_A \) and \( R_B \). - The sum of vertical forces must be zero: \[ R_A + R_B = 8 \text{ kN} \] Take moments about A to solve for \( R_B \): - The total load acts at the center of AB, which is 2 m from A. - Distance from A to B is 4 m. \[ \sum M_A = \implies R_B \times 4 - 8 \times 2 = \implies R_B = \frac{16}{4} = 4 \text{ kN} \] \[ R_A = 8 - 4 = 4 \text{ kN} \] --- ### **Step 3: Maximum Shear Force on Pins at A and B** - The force transferred to the pin at A or B is 4 kN. - Each pin is in double shear. --- ### **Step 4: Shear Force in Each Shear Plane** Double shear means the force is divided equally between two shear planes: \[ \text{Shear force per plane} = \frac{4 \text{ kN}}{2} = 2 \text{ kN} = 200 \text{ N} \] --- ### **Step 5: Shear Stress Equation** \[ \tau = \frac{P}{A} \] Where: - \( \tau \) = Shear stress - \( P \) = Shear force per plane (200 N) - \( A \) = Area of pin in shear (\( = \frac{\pi}{4} d^2 \), where \( d \) is the pin diameter) Given allowable shear stress \( \tau_{allow} = 100 \) MPa \( = 100 \times 10^6 \) N/m². \[ \tau_{allow} = \frac{P}{A} \implies 100 \times 10^6 = \frac{200}{\frac{\pi}{4} d^2} \] \[ \frac{\pi}{4} d^2 = \frac{200}{100 \times 10^6} \] \[ d^2 = \frac{4 \times 200}{\pi \times 100 \times 10^6} \] \[ d^2 = \frac{800}{314 \times 10^6} \] \[ d^2 = 2.548 \times 10^{-5} \] \[ d = \sqrt{2.548 \times 10^{-5}} \] \[ d = .00505 \text{ m} = 5.05 \text{ mm} \] --- ### **Final Answer (a):** **The required diameter of the pins at A and B is \(\boxed{5.05 \text{ mm}}\).** --- ## **b) Overall Factor of Safety of the Structure** Given: - Failure stress of pins \( = 160 \) MPa. - Cross-sectional area of AB \( = 55 \) mm². - Failure stress of AB \( = 110 \) MPa. ### **Step 1: Factor of Safety for Pins** \[ \text{FOS}_{\text{pins}} = \frac{\text{Failure Stress}}{\text{Actual Stress}} \] Actual stress on pins: \[ \tau_{\text{actual}} = \frac{P}{A} = \frac{200}{\frac{\pi}{4} \times (5.05)^2} \] \[ A = \frac{\pi}{4} \times (5.05)^2 = 20.05 \text{ mm}^2 \] \[ \tau_{\text{actual}} = \frac{200}{20.05} = 99.8 \text{ MPa} \] \[ \text{FOS}_{\text{pins}} = \frac{160}{99.8} = 1.60 \] --- ### **Step 2: Factor of Safety for Member AB** Axial force in AB \( = R_A = 4 \) kN \( = 400 \) N. \[ \sigma_{\text{AB}} = \frac{F}{A} = \frac{400}{55} = 72.73 \text{ MPa} \] \[ \text{FOS}_{\text{AB}} = \frac{110}{72.73} = 1.51 \] --- ### **Step 3: Overall Factor of Safety** The **overall factor of safety** is the lowest among all critical components: \[ \boxed{1.51} \] --- ## **Summary** ### **a) Required diameter of pins at A and B:** \[ \boxed{5.05 \text{ mm}} \] ### **b) Overall factor of safety of the structure:** \[ \boxed{1.51} \]