## Problem Breakdown Let's break down the problem step by step: **Given:** - Hollow square tube - Height = 42 mm - Width = 42 mm - Wall thickness = 5 mm - Axial load = 110 kN (compression) - Find: Axial stress (to one decimal place) in MPa --- ## Step 1: Find the Cross-Sectional Area The **axial stress** is calculated as: \[ \text{Stress} = \frac{\text{Force}}{\text{Area}} \] For a hollow square tube: - **Outer dimension:** \( a = 42 \) mm - **Wall thickness:** \( t = 5 \) mm - **Inner dimension:** \( a_{in} = a - 2t = 42 - 2 \times 5 = 32 \) mm ### **Area Calculation** \[ \text{Area} = (\text{outer area}) - (\text{inner area}) \] \[ = a^2 - (a_{in})^2 \] \[ = 42^2 - 32^2 \] \[ = 1764 - 1024 = 740 \text{ mm}^2 \] --- ## Step 2: Convert Load to Newtons Given: \( 110 \) kN \( = 110,000 \) N --- ## Step 3: Calculate Axial Stress \[ \text{Stress} = \frac{F}{A} \] \[ = \frac{110,000 \text{ N}}{740 \text{ mm}^2} \] - 1 MPa = 1 N/mm², so the units are already correct. \[ \text{Stress} = \frac{110,000}{740} \approx 148.6 \text{ MPa} \] --- ## Step 4: **Final Answer** \[ \boxed{148.6 \text{ MPa}} \] --- ### **Summary Table** | Quantity | Value | |----------------|----------------| | Outer side | 42 mm | | Inner side | 32 mm | | Area | 740 mm² | | Load | 110,000 N | | Axial Stress | 148.6 MPa | --- ### **Explanation** The stress in a hollow square tube under axial load is found by dividing the load by the area of material that resists the load (the "walls" of the tube). We subtract the area of the hollow part from the total square to get this effective area. If you have more questions about this calculation, let me know!