For the circular aluminium tube referred to in Question 2, if the compressive stress in the tube is intended to be 40 MPa, then determine what the load should be.

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Question restatement: For the circular aluminium tube referred to in Question 2, if the compressive stress in the tube is intended to be 40 MPa, then determine what the load should be.

Option 1: 34.6 kN. This option, if converted through stress = load / area, or equivalently load = stress × cross-sectional area, would align with a specific cross-sectional area value for the circular tube. If the tube’s area is A such that 40 MPa × A = 34.6 kN, then A must be 34.6 kN / 40 MPa = 0.000865 m^2, i.e., 865 mm^2. This implies a particular diameter for the circular section that yields that area, which could be consistent with typical tube dimensions. In other words, it reflects a plausible calculation outcome given a 40 MPa target stress.

Option 2: 31.4 kN. If the required stress is 40 MPa, this load would correspond to A = 31.4 kN / 40 MPa = 0.000785 m^2, or 785 mm^2. This is a smaller cross-sectional area than in option 1, and thus implies a slightly smaller diameter for the circular tube. The discrepancy suggests either a different tube geometry assumption or a misalignment with the supposed area that yields exact 40 MPa for the given tube.

Option 3: 28.1 kN. This would require A = 28.1 kN / 40 MPa = 0.0007025 m^2, or about 703 mm^2. This is a noticeably smaller area, leading to a noticeably smaller diameter. If the tube’s actual cross-sectional area is larger than this, then applying 28.1 kN would produce a higher stress than 40 MPa, so this value would be inconsistent with the target stress for the given geometry.

Option 4: 125.7 kN. This load corresponds to A = 125.7 kN / 40 MPa = 0.0031425 m^2, or about 3142 mm^2. That is a substantially larger cross-sectional area, implying a much larger diameter than typical small circular tubes. If the tube’s diameter were this large, the resulting stress would match 40 MPa, but the area suggests an unusually thick or wide tube, making this option less plausible without additional context.

Summary of reasoning trends: The core relationship is stress = load / area, so for a fixed target stress of 40 MPa, the load scales with the cross-sectional area of the tube. Each option corresponds to a different assumed cross-sectional area derived from the same 40 MPa target. The most plausible option is the one whose implied cross-sectional area matches the actual tube geometry referenced in Question 2. The other values would require geometries (areas) that deviate from the tube’s known dimensions, leading to mismatches with the 40 MPa target. The examination relies on ensuring unit consistency (N, MPa, m^2) and converting kN to N when computing the implied area.

INU1127 (24/25) Knowledge Check: Stress-strain basics

For the circular aluminium tube referred to in Question 2, if the compressive stress in the tube is intended to be 40 MPa, then determine what the load should be.

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