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To design the most economical W beam for the given specifications, we need to analyze the moment and shear forces that the beam will encounter under the applied loads, considering both the dead load, live load, and concentrated load. Here’s a step-by-step solution approach:

### Given Information
1. **Span of beam** \( L = 40 \, \text{ft} \)
2. **Uniform dead load** \( w_d = 1.5 \, \text{kip/ft} \) over the entire length
3. **Uniform live load** \( w_l = 2.5 \, \text{kip/ft} \) over the last 32 feet
4. **Concentrated dead load** \( P = 30 \, \text{kips} \) located 8 feet from the left support
5. **Material**: A50 steel, which typically has a yield strength \( F_y = 50 \, \text{ksi} \)

### Step 1: Calculate Maximum Moment due to Uniform Loads
1. **Uniform Dead Load Moment**:
   \[
   M_{d} = \frac{w_d \cdot L^2}{8} = \frac{1.5 \times 40^2}{8} = 300 \, \text{kip-ft}
   \]

2. **Uniform Live Load Moment** (over last 32 feet):
   - Convert the span for this load to an equivalent single span:
     \[
     M_{l} = \frac{w_l \cdot (32)^2}{8} = \frac{2.5 \times 32^2}{8} = 320 \, \text{kip-ft}
     \]

### Step 2: Calculate Moment due to Concentrated Load
For a concentrated load at 8 feet from the left support:
\[
M_{P} = P \times a \times (L - a) / L
\]
Where \( P = 30 \, \text{kips} \), \( a = 8 \, \text{ft} \), \( L = 40 \, \text{ft} \):
\[
M_{P} = 30 \times 8 \times (40 - 8) / 40 = 192 \, \text{kip-ft}
\]

### Step 3: Summing Moments
Total maximum moment \( M_{\text{total}} \) is the sum of all individual moments:
\[
M_{\text{total}} = M_{d} + M_{l} + M_{P} = 300 + 320 + 192 = 812 \, \text{kip-ft}
\]

### Step 4: Select Beam Section
Using the maximum moment, we need to find a W shape with an adequate section modulus \( S \) to resist this moment. The required section modulus is:
\[
S = \frac{M_{\text{total}}}{F_y} = \frac{812}{50} = 16.24 \, \text{in}^3
\]
However, beams are typically selected to provide at least a slight overdesign for safety. You would look up standard W sections with \( S \geq 16.24 \, \text{in}^3 \) in the AISC Steel Manual and select the most economical (lightest) section.

### Step 5: Verification of Shear and Deflection
For completeness, you would also verify that the beam meets shear requirements and any deflection criteria, ensuring adequate lateral support per the problem’s instructions.

Would you like further assistance with choosing a specific W section, calculating deflection, or verifying shear? 

### Relative Questions:
1. How do you calculate the deflection for a beam under uniform load and concentrated load?
2. What is the significance of selecting A50 steel for this design?
3. How does the location of a concentrated load affect the design of a beam?
4. Why is lateral support important in beam design, particularly for flexure?
5. How would you account for combined loading types in more complex beam designs?

#### Tip:
When designing beams, always consider both moment and shear to ensure your design is safe and meets all structural requirements.

Design for flexure a W beam of 40 ft span. The beam has a uniform dead load of 1.5 kip/ft through the entire length and a uniform live load of 2.5 kip/ft over the last 32 ft. There is also a concentrated dead load 30 kips at 8 ft away from the left support. Select the most economical section. The beam has adequate lateral support. Use A50 steel.

This guide details the design process for a W beam under a 40 ft span with specific dead and live loads. It includes calculations for moments due to distributed and concentrated loads, along with section modulus requirements for A50 steel. Ideal for structural engineering students.

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