Let's solve the problem step by step, with diagrams and explanations. --- ## **Given Data** - **Pin diameter:** \( d = 20\,\text{mm} \) - **Clevis thickness:** \( t_c = 15\,\text{mm} \) - **Blade thickness:** \( t_b = 30\,\text{mm} \) - **Lever arm length:** \( L = 300\,\text{mm} \) - **Clamping length:** \( e = 100\,\text{mm} \) - **Pin material:** Unhardened pin (ISO 234) - **Breaking strength:** \( R_m = 400\,\text{N/mm}^2 \) --- ## **Step 1: Find Stress Concentration Factor \( K_t \)** From the graph for "nitriding steel, mild steel (breaking strength)": \[ K_t = 1 - .23 \cdot \log_{10}(d/100) \] where \( d = 20\,\text{mm} \): \[ K_t = 1 - .23 \cdot \log_{10}(20/100) = 1 - .23 \cdot \log_{10}(.2) \] \[ \log_{10}(.2) = -.69897 \] \[ K_t = 1 - .23 \cdot (-.69897) = 1 + .16076 = 1.16076 \] But from the graph, for \( d = 20\,\text{mm} \), the value is approximately \( K_t \approx .82 \) (let's use the graph value). --- ## **Step 2: Maximum Permissible Shear Force in the Pin** The pin is subjected to **double shear** (see diagram below): ``` [Lever(1)] ||--- Pin ---|| [Base Plate(2)] ``` For double shear: \[ F_{\text{max}} = 2 \cdot \tau_{\text{max}} \cdot A \] Where: - \( \tau_{\text{max}} = K_t \cdot R_m \) - Shear area for each plane: \( A = \frac{\pi}{4} d^2 \) But in many standards, the breaking load is calculated as: \[ F_{\text{max}} = K_t \cdot R_m \cdot A \] where \( A = \frac{\pi}{4} d^2 \) (for single shear; double shear, area is doubled). But let's use the static shear strength, which for mild steel is typically \( \tau_{\text{max}} \approx .6 R_m \). But here, since we are told to use the breaking strength directly (since the device is checked for breaking, not yield): \[ F_{\text{max}} = K_t \cdot R_m \cdot A_{\text{shear}} \] For double shear: \[ A_{\text{shear}} = 2 \cdot \left(\frac{\pi}{4} d^2 \right) \] \[ A_{\text{shear}} = 2 \cdot \left(\frac{\pi}{4} (20)^2 \right) = 2 \cdot \left(\frac{\pi}{4} \cdot 400 \right) = 2 \cdot 314.16 = 628.32\,\text{mm}^2 \] So, \[ F_{\text{max}} = .82 \times 400\,\text{N/mm}^2 \times 628.32\,\text{mm}^2 \] \[ F_{\text{max}} = .82 \times 400 \times 628.32 = .82 \times 251,328 = 206,089\,\text{N} \] --- ## **Step 3: Relationship Between \( F \) and \( F_{cl} \)** From the lever diagram: \[ F \cdot L = F_{cl} \cdot e \] \[ F_{cl} = \frac{F \cdot L}{e} \] --- ## **Step 4: Maximum Load \( F \) on the Lever** The maximum force on the pin is \( F_{cl,\text{max}} = F_{\text{max}} \). So, \[ F = \frac{F_{\text{max}} \cdot e}{L} \] --- ## **Step 5: Calculate Final Answers** - \( F_{\text{max}} = 206,089\,\text{N} \) - \( e = 100\,\text{mm} \) - \( L = 300\,\text{mm} \) \[ F = \frac{206,089 \times 100}{300} = \frac{20,608,900}{300} = 68,696.3\,\text{N} \] - Maximum clamping force: \( F_{cl,\text{max}} = 206,089\,\text{N} \) --- ## **Diagrams** **Pin Joint in Double Shear** ``` [Lever Arm] | V ————————————|——————— | |3| | | | | | <--- Pin (Double Shear) ————————————|——————— [Base Plate] ``` **Lever Arm Action** ``` F (applied) | V [Lever Arm]---L---(Pin Joint)---e---[Clamped Part] ``` --- ## **Final Answers** ### **Maximum Load \( F \) to be applied:** \[ \boxed{F_{\text{max}} = 68,696\,\text{N}} \] ### **Resulting Clamping Load \( F_{cl} \):** \[ \boxed{F_{cl} = 206,089\,\text{N}} \] --- **If you need the solution with a free-body diagram or want to check other failure modes (bearing, etc.), let me know!**