Steel Structure Design Methods Flashcards
7 cards from real Civil Engineering PE practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.
Read the first 7 Steel Structure Design Methods flashcards as text
In LRFD design, which load combination governs for a steel beam subjected to dead load D = 20 kips and live load L = 40 kips?
Answer: 1.2D + 1.6L = 88 kips
AISC LRFD load combination 1.2D + 1.6L = 1.2(20) + 1.6(40) = 88 kips governs over 1.4D = 28 kips.
What does the resistance factor φ = 0.90 represent in LRFD steel design for tension yielding?
Answer: A safety margin accounting for variability in material strength and fabrication
φ = 0.90 for tension yielding represents the probability-based reduction accounting for material and fabrication variability to ensure adequate reliability.
A W-shape beam has a plastic section modulus Zx = 120 in³ and Fy = 50 ksi. What is the plastic moment Mp?
Answer: 6,000 kip-in
Mp = Fy × Zx = 50 ksi × 120 in³ = 6,000 kip-in.
In ASD for steel, the allowable bending stress Fb for a compact section in a braced frame with Fy = 36 ksi is typically:
Answer: 0.66Fy = 23.76 ksi
ASD allowable bending stress for compact laterally braced beams is Fb = 0.66Fy per AISC 9th edition.
Which steel design method explicitly accounts for the probability of failure through reliability index β?
Answer: LRFD
LRFD was calibrated using reliability theory to achieve a target reliability index β ≈ 3.0 for typical members.
For a tension member, AISC LRFD requires checking two limit states. Which pair is correct?
Answer: Yielding on gross area and fracture on net area
AISC requires φPn for both gross-section yielding (φ=0.90, Pn=FyAg) and net-section fracture (φ=0.75, Pn=FuAeU).
In steel design, the shear factor of safety in ASD is typically set at approximately:
Answer: 1.67 (Fv = 0.40Fy)
ASD allowable shear stress is Fv = 0.40Fy, corresponding to a safety factor of about 1.67 against shear yielding.