Let's solve the problem step-by-step. --- ## **Step 1: Gather Given Data** - **Width (\(w\))** = 115 mm = .115 m - **Thickness (\(t\))** = 60 mm = .060 m - **Strength Class = C22** - **Bolt diameter** = 20 mm (but not needed for tensile failure of the member itself) - **Tensile strength of C22 (assumed, from timber tables):** \( f_t = 8. \) MPa (this is a typical value; please confirm with your code or standard if different is required) --- ## **Step 2: Calculate Cross-sectional Area** \[ A = \text{Width} \times \text{Thickness} \] \[ A = 115\,\text{mm} \times 60\,\text{mm} = 690\,\text{mm}^2 \] Convert to \( m^2 \): \[ A = 690 \times 10^{-6} \text{ m}^2 = .0069\,\text{m}^2 \] --- ## **Step 3: Calculate Maximum Axial Load** \[ T_{\text{max}} = f_t \times A \] Where: - \( f_t \) = allowable tensile strength (in MPa or N/mm²) - \( A \) = area (in mm²) \[ T_{\text{max}} = 8 \ \text{N/mm}^2 \times 690\ \text{mm}^2 = 55200\ \text{N} = 55.2\,\text{kN} \] --- ## **Step 4: Final Answer (Check Closest Option)** - The closest answer to 55.2 kN is **53.52 kN**. --- ## **Step 5: Explanation** - The maximum axial load before tensile failure is calculated by multiplying the cross-sectional area by the design tensile strength for the material (C22 timber). - The answer is **53.52 kN** (rounded/closest to the options given). --- ## **Final Answer** \[ \boxed{53.52\ \text{kN}} \] --- ### **Summary Table** | Parameter | Value | |-------------------|-----------------| | Width | 115 mm | | Thickness | 60 mm | | Area | 690 mm² | | Tensile Strength | 8 MPa (N/mm²) | | Max Load | 55.2 kN | | Closest Option | 53.52 kN | --- **Thus, the maximum axial load that can be applied before tensile failure is \(\boxed{53.52\ \text{kN}}\).**