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To determine the axial compressive stress (\(\sigma\)) at the base of the wall, we will follow these steps:

### Step 1: Calculate the total load on the wall due to the canopy.
The canopy is 8 ft by 8 ft, and the given load is 100 psf (pounds per square foot). The total load from the canopy is:

\[
\text{Total load from canopy} = \text{Area of canopy} \times \text{Load intensity} = (8 \, \text{ft} \times 8 \, \text{ft}) \times 100 \, \text{psf}
\]
\[
= 64 \, \text{ft}^2 \times 100 \, \text{psf} = 6400 \, \text{lb}
\]

### Step 2: Calculate the weight of the wall.
The wall has dimensions \(6 \, \text{in} \times 8 \, \text{ft} \times 15 \, \text{ft}\). We need to convert 6 inches to feet for consistent units:

\[
6 \, \text{in} = 0.5 \, \text{ft}
\]

Now, find the volume of the wall:

\[
\text{Volume of wall} = 0.5 \, \text{ft} \times 8 \, \text{ft} \times 15 \, \text{ft} = 60 \, \text{ft}^3
\]

Given the unit weight of the wall is \(250 \, \text{lb/ft}^3\):

\[
\text{Weight of wall} = 60 \, \text{ft}^3 \times 250 \, \text{lb/ft}^3 = 15000 \, \text{lb}
\]

### Step 3: Calculate the total axial load at the base of the wall.
The total load (\(P\)) at the base is the sum of the canopy load and the wall's own weight:

\[
P = 6400 \, \text{lb} + 15000 \, \text{lb} = 21400 \, \text{lb}
\]

### Step 4: Calculate the cross-sectional area of the wall.
The wall has a cross-sectional area given by:

\[
\text{Area} = 0.5 \, \text{ft} \times 8 \, \text{ft} = 4 \, \text{ft}^2
\]

Convert the area to square inches (\(\text{in}^2\)):

\[
4 \, \text{ft}^2 = 4 \times 144 \, \text{in}^2 = 576 \, \text{in}^2
\]

### Step 5: Calculate the axial compressive stress.
The formula for axial compressive stress is:

\[
\sigma = \frac{P}{A} = \frac{21400 \, \text{lb}}{576 \, \text{in}^2}
\]

Calculating this gives:

\[
\sigma \approx 37.15 \, \text{psi}
\]

### Final Answer:
The axial compressive stress at the base of the wall is approximately **37.15 psi**.

Would you like a more detailed breakdown or have any questions about the process?

### Related Questions:
1. How would the stress change if the canopy load increased to 150 psf?
2. What would be the impact on stress if the wall height increased to 20 ft?
3. How would you calculate the stress if the cross-section of the wall was circular?
4. What if the unit weight of the wall was 200 lb/ft³ instead—how would the result change?
5. How do different materials' unit weights affect the design of structures like walls?

### Tip:
Always double-check unit consistency (like converting between feet and inches) in structural calculations to ensure accurate results.

The 8 ft x 8 ft canopy shown in the figure is supported by a 6 in x 8 ft x 15 ft wall. The load on the canopy (including its self-weight) is 100 psf. Given the unit weight of the wall is 250 lb/ft³, determine the axial compressive stress at the base of the wall.

Learn how to calculate the axial compressive stress at the base of a wall supporting a canopy. This problem involves calculating total load, wall weight, and cross-sectional area, followed by applying the formula σ = P / A. Ideal for college-level engineering students.

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