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The image contains a calculation related to the cross-sectional area of a tube under tensile force, showing formulas and steps for determining the inner and outer diameters of the tube based on the applied force and material strength. Here's a breakdown:

1. **Formulas**:
   - The diameter of the outer circle \( D \) is given as \( D = \lambda d \), where \( \lambda > 1 \).
   - \( \sigma_x = \frac{N}{A} = \frac{N}{(D^2 - d^2)\pi} \), where \( N \) is the applied force, \( A \) is the area, \( D \) is the outer diameter, and \( d \) is the inner diameter. 
   - Given:
     - \( N = 80000 \, \text{N} \)
     - \( \sigma_{\text{max}} = 100 \, \text{MPa} \)
     - \( D = 1.4d \)

2. **Steps**:
   - The goal is to determine both the outer diameter \( D \) and inner diameter \( d \).
   - The calculation for \( d \) starts by rearranging the stress formula \( \sigma_x \leq \sigma_{\text{max}} \) to solve for the inner diameter.
   - Once \( d \) is found (rounded up to 32.6 mm), the outer diameter is determined by multiplying \( d \) by 1.4, yielding \( D \approx 45.7 \, \text{mm} \).

3. **Final verification**:
   - The calculation verifies that the resulting stress \( \sigma_x \) (99.304 MPa) is below the maximum allowable stress (100 MPa).

Let me know if you need any more details or further explanation on any part of this problem.

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Here are five related questions you might consider:
1. What would happen if the stress \( \sigma_x \) exceeded the maximum allowable stress?
2. How would the result change if \( \lambda \) had a different value?
3. What is the significance of rounding up or down in mechanical design, especially for diameters?
4. How does the applied force \( N \) influence the choice of diameters for the tube?
5. How could you calculate the wall thickness of the tube from the given diameters?

**Tip**: Always double-check unit conversions, especially in stress and force calculations, to avoid significant errors in mechanical design.

Determine the outer (D) and inner (d) diameters of a tube under tensile stress, given an applied force N=80000 N and maximum allowable stress σ_max=100 MPa.

Learn how to calculate the inner (d) and outer (D) diameters of a tube under an applied tensile force of 80000 N and a maximum allowable stress of 100 MPa. This problem involves key formulas in stress analysis and mechanical design, perfect for undergraduate engineering students.

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