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To solve this problem, we need to calculate the critical load (\( P_{cr} \)) for both short and long column conditions based on the formulas provided. Here's a step-by-step approach:

### Given Data
1. **Effective Length, \( L \) of Column**: 15 ft (which we need to convert to inches for some calculations since typical steel measurements are in inches).
2. **Modulus of Elasticity, \( E \)**: 29,600 ksi.
3. **Assumed Yield Stress \( F_y \)** for A36 steel: 36 ksi (this is a standard assumption for A36 steel unless otherwise specified).
4. **Formulas**:
   - **Short Column**: \( P_{cr} = F_y \times A \)
   - **Long Column**: \( P_{cr} = \frac{\pi^2 \times E \times I}{(kL)^2} \)
5. **Determine the Lightest Square Shape**: We are looking for the minimum cross-sectional area \( A \) (for short column) and minimum moment of inertia \( I \) (for long column) that can safely support the load.

### Approach
1. **Short Column Calculation**:
   - We need to find \( A \), the cross-sectional area.
   - For a square shape, if each side of the square is \( b \), then \( A = b^2 \).
   - We use the formula \( P_{cr} = F_y \times A \).

2. **Long Column Calculation**:
   - We need to find \( I \), the moment of inertia for a square cross-section.
   - For a square cross-section with side \( b \), \( I = \frac{b^4}{12} \).
   - Substitute into the formula \( P_{cr} = \frac{\pi^2 \times E \times I}{(kL)^2} \) to solve for \( b \).

#### Steps to Solve

1. **Convert L to inches**: \( L = 15 \, \text{ft} = 15 \times 12 = 180 \, \text{inches} \).
2. **Determine k value**: \( k \) depends on the boundary conditions. Assuming pinned ends, \( k = 1.0 \).

Let's start by performing these calculations in detail. I'll proceed with this computation.I have displayed the calculations for various side lengths, showing the cross-sectional area and the corresponding critical load (\( P_{cr} \)) for both short and long column cases.

If you'd like to delve into any specific results or need further explanation on how these calculations help determine the lightest square shape, please let me know.

Here are some questions that could further deepen understanding:

1. How does the choice of boundary conditions (such as pinned or fixed) affect the effective length factor \( k \) and subsequently the critical load?
2. Why is the modulus of elasticity important in calculating the critical load for a long column?
3. How do different cross-sectional shapes (e.g., circular vs. square) affect the critical load calculations for columns?
4. Why is the yield strength \( F_y \) used in short column calculations but not in long column calculations?
5. What are the implications of using a larger side length \( b \) beyond just increasing the area?

**Tip:** Remember that slenderness ratio (\( kL/r \)) is a key factor in deciding whether a column is classified as "short" or "long," influencing which failure mode is more likely.

Refer to Figure 1 again. The trusses are supported by columns at either end at points E and K. The support reactions that you get from the trusses become the Pcr on the columns. Determine the lightest square shape that would be safe for this load for both short AND long column options. L = 15 ft and E = 29,600 ksi.

This solution details the steps to determine the lightest square cross-sectional shape that can safely support a truss load for both short and long columns. Given an effective length L = 15 ft and modulus of elasticity E = 29,600 ksi, the solution applies formulas for critical load and buckling analysis, including the use of Fy for short columns and Euler’s formula for long columns.

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