To solve the problem of a steel pipe under axial compression, we'll use the fundamental principles of mechanics of materials, particularly Hooke’s Law for axial stress and strain.

Given:

• Length of the pipe $L = 1 \, \text{m} = 1000 \, \text{mm}$
• Outside diameter $D_o = 0.2 \, \text{m} = 200 \, \text{mm}$
• Wall thickness $t = 10 \, \text{mm}$
• Axial compressive force $P = 200 \, \text{kN} = 200,000 \, \text{N}$
• Modulus of Elasticity $E = 200 \, \text{GPa} = 200 \times 10^3 \, \text{MPa}$
• Poisson’s ratio $\nu = 0.3$

Step 1: Axial stress

The axial stress ($\sigma$) is given by: $\sigma = \frac{P}{A}$ where $A$ is the cross-sectional area of the pipe. For a hollow cylinder (pipe), the cross-sectional area $A$ is: $A = \frac{\pi}{4} \left( D_o^2 - D_i^2 \right)$ The inside diameter $D_i$ is: $D_i = D_o - 2t = 200 \, \text{mm} - 2 \times 10 \, \text{mm} = 180 \, \text{mm}$ Now, calculate the area: $A = \frac{\pi}{4} \left( (200)^2 - (180)^2 \right) = \frac{\pi}{4} (40,000 - 32,400) = \frac{\pi}{4} \times 7,600 \, \text{mm}^2$

$A \approx 5,969 \, \text{mm}^2$

Now calculate the axial stress: $\sigma = \frac{200,000 \, \text{N}}{5,969 \, \text{mm}^2} \approx 33.52 \, \text{MPa}$

Step 2: Axial strain and shortening of the pipe

The axial strain ($\epsilon$) is given by: $\epsilon = \frac{\sigma}{E} = \frac{33.52 \, \text{MPa}}{200,000 \, \text{MPa}} = 1.676 \times 10^{-4}$

Now, the shortening of the pipe ($\Delta L$) is: $\Delta L = \epsilon \cdot L = 1.676 \times 10^{-4} \times 1000 \, \text{mm} = 0.1676 \, \text{mm}$

Thus, the shortening of the pipe is approximately $\Delta L = 0.168 \, \text{mm}$.

Step 3: Increase in the outside diameter

Due to Poisson’s effect, the radial strain ($\epsilon_r$) is related to the axial strain by: $\epsilon_r = - \nu \cdot \epsilon$ Substitute the values: $\epsilon_r = -0.3 \times 1.676 \times 10^{-4} = -5.028 \times 10^{-5}$

The change in the outside diameter ($\Delta D_o$) is: $\Delta D_o = \epsilon_r \cdot D_o = -5.028 \times 10^{-5} \times 200 \, \text{mm} = -0.01006 \, \text{mm}$

Thus, the increase in the outside diameter is approximately $\Delta D_o = 0.0101 \, \text{mm}$.

Step 4: Increase in the wall thickness

Similarly, the change in the wall thickness ($\Delta t$) is related to the radial strain: $\Delta t = \epsilon_r \cdot t = -5.028 \times 10^{-5} \times 10 \, \text{mm} = -0.0005028 \, \text{mm}$

Thus, the increase in wall thickness is approximately $\Delta t = 0.000503 \, \text{mm}$.

Summary of Results:

a. Shortening of the pipe $\Delta L \approx 0.168 \, \text{mm}$
b. Increase in the outside diameter $\Delta D_o \approx 0.0101 \, \text{mm}$
c. Increase in the wall thickness $\Delta t \approx 0.000503 \, \text{mm}$

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Would you like further details or clarification on any part? Here are 5 related questions you might find interesting:

1. How does Poisson's ratio influence the radial expansion of materials under compression?
2. What is the significance of wall thickness in pressure vessels or pipes?
3. How would the results differ if the pipe material had a different modulus of elasticity?
4. How would you calculate the critical buckling load for this pipe under compression?
5. How does temperature change affect the stress and strain in a pipe under compression?

Tip: In structural analysis, understanding the relationship between axial, radial, and circumferential stresses is crucial for ensuring safety in the design of pressure vessels and pipes.