Let's break down the problem step by step. --- ## **Step 1: Understanding the Problem** - **Material**: Titanium (Ti 6Al-4V) - **Dimensions**: - Width (\( w \)) = 25 cm - Thickness (\( t \)) = 1 cm - Hole diameter (\( d \)) = 6 cm (supposed to be 4 cm) - **Load**: Find the maximum tensile load before yielding. --- ## **Step 2: Find the Yield Strength of Ti 6Al-4V** Yield strength (\( \sigma_{y} \)) ≈ **880 MPa** (can vary slightly, but this is a typical value). --- ## **Step 3: Calculate the Critical Cross-sectional Area** When there's a hole, the weakest section is across the hole. \[ A = (w - d) \times t \] Where: - \( w \) = width = 25 cm - \( d \) = hole diameter = 6 cm - \( t \) = thickness = 1 cm So, \[ A = (25\, \text{cm} - 6\, \text{cm}) \times 1\, \text{cm} = 19\, \text{cm}^2 \] Convert to \( \text{mm}^2 \) (since 1 cm\(^2\) = 100 mm\(^2\), and yield strength is in MPa): \[ A = 19\, \text{cm}^2 \times 100 = 190\, \text{mm}^2 \] --- ## **Step 4: Find the Maximum Load** The maximum load at yield (\( F_{max} \)) is: \[ F_{max} = \sigma_{y} \times A \] Where: - \( \sigma_{y} \) = 880 MPa = 880 N/mm\(^2\) - \( A \) = 190 mm\(^2\) \[ F_{max} = 880\, \text{N/mm}^2 \times 190\, \text{mm}^2 = 1,672,000\, \text{N} \] --- ## **Step 5: Final Answer** ### **The maximum load in tension that can be applied to the piece before it yields is:** \[ \boxed{1,672,000~\text{N}} \] --- ### **Summary Table** | Quantity | Value | |------------------|----------| | Width (\(w\)) | 25 cm | | Thickness (\(t\))| 1 cm | | Hole (\(d\)) | 6 cm | | Yield Strength | 880 MPa | | Area (\(A\)) | 190 mm² | | Max Load (\(F\)) | 1,672,000 N | --- **Let me know if you have any questions or want to see what would happen with the original 4 cm hole!**