# Step-by-Step Solution Let's analyze the problem step by step: - **Given Data:** - Width of wood, \( b = 40 \) mm - Distance from top to hole center, \( c = 55 \) mm - Pin diameter, \( d = 12 \) mm - Ultimate strength of wood tension, \( f_{tu,wood} = 60 \) MPa - Ultimate strength of wood in shear, \( f_{su,wood} = 7.5 \) MPa - Ultimate strength of steel in shear, \( f_{su,steel} = 145 \) MPa - Factor of safety, \( FS = 3.2 \) ## 1. **Modes of Failure** The load \( P \) can cause: 1. **Tension failure** in the wooden member at the hole. 2. **Shear failure** in the wood at the hole. 3. **Shear failure** in the steel pin. We need to check all these and use the smallest value for \( P \) to ensure safety. --- ## 2. **Allowable Stresses** Divide the ultimate strengths by the factor of safety: - **Wood, Tension:** \( f_{t,allow,wood} = \frac{60}{3.2} = 18.75 \) MPa - **Wood, Shear:** \( f_{s,allow,wood} = \frac{7.5}{3.2} = 2.344 \) MPa - **Steel, Shear:** \( f_{s,allow,steel} = \frac{145}{3.2} = 45.31 \) MPa --- ## 3. **Tension Failure in Wood at Pin Hole** The net area in tension is the area left after subtracting the pin hole: \[ A_{t,net} = b \times (c - d) \] But usually, \( c \) is the *distance from the top edge to the pin center*, so the net width is \( b - d \): \[ A_{t,net} = (b - d) \times t \] But the problem gives no thickness. For a general solution, let the thickness of the member be \( t \). \[ A_{t,net} = (b - d) \cdot t \] \[ P_{t,allow} = f_{t,allow,wood} \cdot (b - d) \cdot t \] --- ## 4. **Shear Failure in Wood at Pin Hole** The area subject to shear is the area across the width at the pin (single shear): \[ A_{s,wood} = d \cdot t \] \[ P_{s,allow,wood} = f_{s,allow,wood} \cdot d \cdot t \] --- ## 5. **Shear Failure in Steel Pin** The pin is in *single shear*: \[ A_{s,steel} = \frac{\pi}{4} d^2 \] \[ P_{s,allow,steel} = f_{s,allow,steel} \cdot \frac{\pi}{4} d^2 \] --- ## 6. **Calculate All Allowable Loads (in terms of \( t \))** Plug in values: - \( b = 40 \) mm - \( d = 12 \) mm **Tension in wood:** \[ P_{t,allow} = 18.75 \times (40 - 12) \times t = 18.75 \times 28 \times t = 525t \ \text{N} \] **Shear in wood:** \[ P_{s,allow,wood} = 2.344 \times 12 \times t = 28.13t \ \text{N} \] **Shear in steel pin:** \[ A_{s,steel} = \frac{\pi}{4} \times 12^2 = \frac{\pi}{4} \times 144 = 113.097 \ \text{mm}^2 \] \[ P_{s,allow,steel} = 45.31 \times 113.097 = 5124 \ \text{N} \] --- ## 7. **Find Governing Load** - \( P_{t,allow} = 525t \) N - \( P_{s,allow,wood} = 28.13t \) N - \( P_{s,allow,steel} = 5124 \) N The *minimum* of these will determine the maximum safe load. Compare \( 28.13t \) and \( 5124 \). For \( t \) (thickness) around 200 mm, both would be equal; but in practice, \( t \) is likely much less (say, 40-60 mm). Let's keep it general. - If \( 28.13t < 5124 \), then shear in wood governs. - \( t < \frac{5124}{28.13} = 182 \) mm, which is almost always true. --- ## 8. **Final Answer (General Form)** ### **The maximum safe load is:** \[ \boxed{P_{max} = 28.13 \; t \ \text{N}} \] where \( t \) is the thickness of the wooden member in mm. ### **For a specific thickness, substitute \( t \) into the above formula.** --- ## 9. **Summary Table** | Failure Mode | Allowable Load (\(P\)) | |---------------------------|----------------------------------| | Wood: Tension | \( 525 t \) N | | Wood: Shear | \( 28.13 t \) N | | Steel Pin: Shear | \( 5124 \) N | - **Shear in wood is the most critical.** --- ## 10. **If Thickness is Known** If, for example, \( t = 40 \) mm: \[ P_{max} = 28.13 \times 40 = 1125 \ \text{N} \] --- ## **Conclusion** - **The limiting factor is shear in the wood at the pin.** - **Allowable load:** \(\boxed{P_{max} = 28.13\, t \ \text{N}}\), where \(t\) is in mm. --- **Note:** If the thickness \( t \) is provided, substitute to get a numerical answer.