To determine the wall thickness of a copper column with an annular cross section, we can apply the formula for compressive stress:

$\sigma = \frac{P}{A}$

where:

• $\sigma$ is the compressive stress (in $\text{lb/ft}^2$),
• $P$ is the applied force (in pounds),
• $A$ is the cross-sectional area (in $\text{ft}^2$).

Step-by-Step Solution

1. Convert kips to pounds: A kip is equal to 1000 pounds. Therefore, the applied force is:

   $45 \, \text{kips} = 45 \times 1000 = 45000 \, \text{lb}$

2. Set up the formula: The allowable compressive stress is 300 lb/ft². Using the stress formula:

   $300 = \frac{45000}{A}$

   Solve for $A$:

   $A = \frac{45000}{300} = 150 \, \text{ft}^2$

3. Determine the area of the annular cross-section: The cross-sectional area of an annulus (a ring-shaped region) is given by:

   $A = \pi (R_{\text{outer}}^2 - R_{\text{inner}}^2)$

   where:

   • $R_{\text{outer}}$ is the outer radius of the column,
   • $R_{\text{inner}}$ is the inner radius of the column.

   The outer diameter is 15 ft, so:

   $R_{\text{outer}} = \frac{15}{2} = 7.5 \, \text{ft}$

   Let the inner radius be $R_{\text{inner}} = R_{\text{outer}} - t$, where $t$ is the wall thickness we need to find.

   Substituting into the area equation:

   $A = \pi (7.5^2 - (7.5 - t)^2)$

   And we know from step 2 that $A = 150 \, \text{ft}^2$:

   $150 = \pi (7.5^2 - (7.5 - t)^2)$

4. Solve for $t$: Expand and solve the equation for the thickness $t$.

   $150 = \pi \left( 56.25 - (7.5 - t)^2 \right)$

   Dividing by $\pi$:

   $\frac{150}{\pi} = 56.25 - (7.5 - t)^2$

   $\frac{150}{\pi} \approx 47.75$

   Therefore:

   $47.75 = 56.25 - (7.5 - t)^2$

   $(7.5 - t)^2 = 56.25 - 47.75 = 8.5$

   Taking the square root of both sides:

   $7.5 - t = \sqrt{8.5} \approx 2.915$

   Finally, solving for $t$:

   $t = 7.5 - 2.915 = 4.585 \, \text{ft}$

Conclusion:

The wall thickness of the copper column is approximately 4.59 ft.

Would you like further details or clarifications?

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Related Questions:

1. How would the wall thickness change if the compressive stress limit increased?
2. What if the column were made from a different material with a higher stress tolerance?
3. How does the outer diameter affect the wall thickness for the same load and stress?
4. How is the allowable compressive stress determined for copper in different applications?
5. What are typical units for measuring compressive stress in engineering?

Tip: Always double-check unit conversions, especially between kips, pounds, and stress values, to avoid mistakes in force and stress calculations.