To design a W beam for flexure with the given specifications, let's go through the analysis step-by-step. The main steps involved are calculating the maximum moment and then selecting a W section that satisfies the strength and economy requirements.

Given Data:

• Span, $L$: 40 ft
• Uniform dead load, $w_D$: 1.5 kip/ft (applies over the entire length of the beam)
• Uniform live load, $w_L$: 2.5 kip/ft (applies over 32 ft from the right end)
• Concentrated dead load, $P_D$: 30 kips (located 8 ft from the left support)
• Material: A50 steel (yield strength $F_y = 50$ ksi)
• Adequate lateral support: No lateral-torsional buckling concerns.

Step 1: Calculate Maximum Moment Due to Dead and Live Loads

1. Dead Load Moment

1. Uniform Dead Load $w_D = 1.5$ kip/ft over 40 ft: $M_{D, \text{uniform}} = \frac{w_D \cdot L^2}{8} = \frac{1.5 \cdot (40)^2}{8} = 300 \text{ kip-ft}$

2. Concentrated Dead Load of 30 kips at 8 ft: $M_{D, \text{concentrated}} = P_D \cdot a \cdot \left(1 - \frac{a}{L}\right) = 30 \cdot 8 \cdot \left(1 - \frac{8}{40}\right)$ $= 30 \cdot 8 \cdot 0.8 = 192 \text{ kip-ft}$

3. Total Dead Load Moment: $M_{D, \text{total}} = M_{D, \text{uniform}} + M_{D, \text{concentrated}} = 300 + 192 = 492 \text{ kip-ft}$

2. Live Load Moment

1. Uniform Live Load $w_L = 2.5$ kip/ft over the last 32 ft: $M_L = \frac{w_L \cdot L_{\text{live}}^2}{8} = \frac{2.5 \cdot (32)^2}{8} = 320 \text{ kip-ft}$

3. Total Factored Moment

Using the load factors (for AISC LRFD), typically: $M_u = 1.2 M_D + 1.6 M_L$ $M_u = 1.2 \cdot 492 + 1.6 \cdot 320 = 590.4 + 512 = 1102.4 \text{ kip-ft}$

Step 2: Select a W-Shape Based on Required Moment Capacity

To resist the factored moment $M_u$, the required section modulus $S_{req}$ can be calculated as: $S_{req} = \frac{M_u}{\phi F_y}$ Where:

• $\phi = 0.9$ (resistance factor for bending in AISC)
• $F_y = 50$ ksi

$S_{req} = \frac{1102.4 \times 12}{0.9 \times 50} = \frac{13228.8}{45} = 293.97 \text{ in}^3$

Step 3: Choose the Most Economical W-Shape

Using the AISC Steel Manual, we would look for a W shape with $S_x \geq 293.97 \, \text{in}^3$ and select the lightest section.

Example Selection:
The W27x94 has:

• $S_x = 302 \, \text{in}^3$ (adequate for our design moment)
• Weight: 94 lb/ft (an economical choice based on strength and weight)

Conclusion

The most economical section that satisfies the bending requirements is W27x94.

Would you like further details on this design or have any questions?

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1. How would you determine deflection limits for this beam?
2. What other checks would be necessary if lateral-torsional buckling was a concern?
3. How can you verify the shear capacity of this W-beam under the given loads?
4. What would change if A36 steel was used instead of A50?
5. How does the load combination factor affect the selection of beam size?

Tip: Always verify shear and deflection limits after selecting a section to ensure comprehensive safety and performance.