Commercial Building Inspector B2Structural SystemsMedium
A structural engineer is designing a steel wide-flange beam for a floor system in a commercial office building. The beam has a simply supported span of 30 feet and is subjected to a uniform live load of 1.5 klf (kips per linear foot) and a uniform dead load of 0.5 klf (including its own weight). Using LRFD (Load and Resistance Factor Design) and assuming a resistance factor (φ) of 0.90 for flexure and a yield strength (Fy) of 50 ksi for the steel, what is the required nominal flexural strength (Mn) for the beam? (Assume Cb = 1.0)
- A225 kip-ft
- B316 kip-ft
- C350 kip-ft
- D281 kip-ft
Show answer & explanationAnswer & explanation
Correct answer: C. 350 kip-ft
First, calculate the factored uniform load (wu) using LRFD load combinations: wu = 1.2D + 1.6L = 1.2(0.5 klf) + 1.6(1.5 klf) = 0.6 klf + 2.4 klf = 3.0 klf. For a simply supported beam with uniform load, the maximum factored moment (Mu) = (wu * L^2) / 8 = (3.0 klf * (30 ft)^2) / 8 = (3.0 klf * 900 ft^2) / 8 = 2700 kip-ft / 8 = 337.5 kip-ft. The required nominal flexural strength (Mn) is Mu / φ = 337.5 kip-ft / 0.90 = 375 kip-ft.
Why the other options are wrong
- A. Incorrect calculation of factored load or moment.
- B. Incorrect calculation; likely an arithmetic error.
- D. Incorrect calculation; perhaps using service loads or wrong load factors.
Required Nominal Flexural Strength (Steel Beam LRFD)
The minimum nominal flexural strength (Mn) a steel beam cross-section must possess to safely resist the factored bending moments (Mu) determined by LRFD load combinations.
- Mn ≥ Mu / φb
- Mu = factored moment from LRFD load combinations.
- φb = resistance factor for flexure (typically 0.90).
Memory trick: Factored loads, then factor strength, ensure safety.