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.

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.