Let's solve the problem step-by-step: --- ## **Step 1: Free Body DiagramFBD) & Reactions** Given:** - \( F_ = 15\, \text{kN} \) (downward at 75 mm from C) - \( F_2 = 25\, \text{kN} \) at an angle (3-4-5 triangle) at 125 mm from C - Pins at **A** and **C** - Factor of Safety (FS) = **1.5** - Failure Shear Stress \( \tau_{fail} = 82.5\, \text{MPa} \) ### **Determine Pin Forces (Reactions at A and C)** #### **Sum Moments about C:** Let: - \( A_x, A_y \): Reaction at A (horizontal/vertical) - \( C_x, C_y \): Reaction at C \[ \sum M_C = \] Distances: - \( \overrightarrow{CA} = 200\, \text{mm} \) up - \( \overrightarrow{CF} = 75\, \text{mm} \) right (for 15 kN) - \( \overrightarrow{CF_2} = 75 + 50 = 125\, \text{mm} \) right (for 25 kN) Resolve \( F_2 \) using the 3-4-5 triangle: - Horizontal: \( 25 \times \frac{4}{5} = 20\, \text{kN} \) - Vertical: \( 25 \times \frac{3}{5} = 15\, \text{kN} \) **down** \[ \sum M_C = = A_y \cdot 200 - 15 \cdot 75 - 25 \cdot 125 \cdot \sin\theta \] But let's get the moment arms: - For 15 kN: \( 75\, \text{mm} \) - For 25 kN: \( 125\, \text{mm} \) - 25 kN is at an angle: vertical component causes moment about C. \[ \sum M_C = = A_y \cdot 200 - 15 \cdot 75 - (25 \cdot \sin\theta) \cdot 125 \] But \( \sin\theta = \frac{3}{5} \): \[ \sum M_C = = A_y \cdot 200 - 15 \cdot 75 - (25 \cdot .6) \cdot 125 \] \[ A_y \cdot 200 = 1125 + 1875 \] \[ A_y \cdot 200 = 300 \] \[ A_y = \frac{300}{200} = 15\, \text{kN} \] #### **Sum Forces in Y:** \[ \sum F_y = : A_y + C_y - 15 - 25 \cdot .6 = \] \[ A_y + C_y = 15 + 15 = 30\, \text{kN} \] \[ C_y = 30 - A_y = 30 - 15 = 15\, \text{kN} \] #### **Sum Forces in X:** \[ \sum F_x = : A_x + C_x - 25 \cdot .8 = \] \[ A_x + C_x = 20\, \text{kN} \] We need another equation—let's consider moments about A for \( C_x \): #### **Sum Moments about A:** - 15 kN: 200 mm below A - 25 kN: Find moment arm from A For 15 kN: \[ M_A = 15 \cdot (200) = 300\, \text{kN-mm} \text{ (CW)} \] For 25 kN: the perpendicular distance from A is: \[ \text{Vertical drop from A to F2} = 200\, \text{mm} \] \[ M_{A,25} = 25 \cdot .8 \cdot 200 = 400\, \text{kN-mm} \text{ (CW)} \] Pin C resists with \( C_x \) and \( C_y \). The perpendicular distance for \( C_x \) from A is 200 mm. \[ \sum M_A = = C_y \cdot - C_x \cdot 200 + 300 + 400 \] \[ -C_x \cdot 200 + 700 = \] \[ C_x = \frac{700}{200} = 35\, \text{kN} \] Now, from above: \[ A_x + C_x = 20 \] \[ A_x = 20 - 35 = -15\, \text{kN} \] --- ## **Step 2: Resultant Shear at Pins** - At **A**: \( A_x = -15\, \text{kN} \), \( A_y = 15\, \text{kN} \) - At **C**: \( C_x = 35\, \text{kN} \), \( C_y = 15\, \text{kN} \) \[ R_A = \sqrt{A_x^2 + A_y^2} = \sqrt{(-15)^2 + 15^2} = \sqrt{225 + 225} = \sqrt{450} \approx 21.2\, \text{kN} \] \[ R_C = \sqrt{C_x^2 + C_y^2} = \sqrt{35^2 + 15^2} = \sqrt{1225 + 225} = \sqrt{145} \approx 38.1\, \text{kN} \] --- ## **Step 3: Pin Shear (Single or Double Shear?)** - **A**: Pin at A is in **double shear** (since AB is a two-force member). - **C**: Pin at C is in **single shear**. --- ## **Step 4: Allowable Shear Stress** \[ \tau_{allow} = \frac{\tau_{fail}}{FS} = \frac{82.5}{1.5} = 55\, \text{MPa} \] --- ## **Step 5: Pin Diameter Calculation** Shear force per section: - **Double Shear:** \( V_{section} = \frac{R}{2} \) - **Single Shear:** \( V_{section} = R \) Shear area \( A = \frac{\pi}{4} d^2 \) ### **Pin A (Double Shear):** \[ \tau_{allow} = \frac{R_A/2}{A} \] \[ 55 = \frac{21.2 \times 10^3 / 2}{\frac{\pi}{4} d^2} \] \[ 55 = \frac{10,600}{.785 d^2} \] \[ 55 \times .785 d^2 = 10,600 \] \[ 43.175 d^2 = 10,600 \] \[ d^2 = \frac{10,600}{43.175} \approx 245.5 \] \[ d = \sqrt{245.5} \approx 15.7\, \text{mm} \] Round up to nearest 5 mm: **20 mm** --- ### **Pin C (Single Shear):** \[ \tau_{allow} = \frac{R_C}{A} \] \[ 55 = \frac{38.1 \times 10^3}{.785 d^2} \] \[ 55 \times .785 d^2 = 38,100 \] \[ 43.175 d^2 = 38,100 \] \[ d^2 = \frac{38,100}{43.175} \approx 882.7 \] \[ d = \sqrt{882.7} \approx 29.73\, \text{mm} \] Round up to nearest 5 mm: **30 mm** --- ## **Final Answers** ### **To the Nearest 5 mm:** - **Diameter of pin at A:** \( \boxed{20\, \text{mm}} \) - **Diameter of pin at C:** \( \boxed{30\, \text{mm}} \) --- ### **Summary Table** | Pin | Shear Type | Resultant Force (kN) | Diameter (mm) | |-----|--------------|----------------------|---------------| | A | Double Shear | 21.2 | **20** | | C | Single Shear | 38.1 | **30** | --- ### **Shear Stress Calculation Recap** - Allowable shear stress: \( 55\, \text{MPa} \) (with FS = 1.5) - Used resultant forces and checked shear type at each pin. --- ### **Boxed Final Answers** \[ \boxed{ \begin{aligned} \text{Diameter of pin at A} & :\, 20\, \text{mm} \\ \text{Diameter of pin at C} & :\, 30\, \text{mm} \\ \end{aligned} } \]