Civil Engineering PE Steel Structure Design Methods 3 — Questions and Answers
Question 1: A steel column with KL/r = 80 and Fy = 50 ksi falls into which AISC buckling category?
- Inelastic buckling (KL/r < 4.71√(E/Fy) ≈ 113) (Correct answer)
- Elastic buckling (KL/r > 113)
- No buckling concern (KL/r < 25)
- Post-buckling reserve applies
Correct answer: Inelastic buckling (KL/r < 4.71√(E/Fy) ≈ 113)
For Fy = 50 ksi, the transition slenderness is 4.71√(E/Fy) = 4.71√(29000/50) ≈ 113; KL/r = 80 is in the inelastic buckling range.
Question 2: Which effective length factor K applies to a column with both ends pinned (no rotational fixity)?
- K = 1.0 (Correct answer)
- K = 0.5
- K = 0.7
- K = 2.0
Correct answer: K = 1.0
A pin-pin column has no end restraint against rotation, giving K = 1.0 and an effective length equal to the actual length.
Question 3: What is the purpose of the Cb factor in AISC beam design?
- It accounts for non-uniform moment along the unbraced length, allowing higher capacity (Correct answer)
- It reduces nominal strength for compact sections
- It adjusts for weak-axis bending
- It accounts for residual stresses in rolled shapes
Correct answer: It accounts for non-uniform moment along the unbraced length, allowing higher capacity
Cb is the lateral-torsional buckling modification factor; moments varying along the unbraced length are less severe than uniform moment, so Cb ≥ 1.0 increases the allowable strength.
Question 4: In AISC Chapter F, which condition defines a compact section regarding the compression flange?
- λ ≤ λp where λ = bf/2tf ≤ 0.38√(E/Fy) (Correct answer)
- λ ≤ λr where λ = bf/2tf ≤ 1.0√(E/Fy)
- λ ≤ λp where λ = h/tw ≤ 2.24√(E/Fy)
- No flange slenderness limit applies to rolled W-shapes
Correct answer: λ ≤ λp where λ = bf/2tf ≤ 0.38√(E/Fy)
A compact flange requires bf/(2tf) ≤ 0.38√(E/Fy), allowing the section to reach the plastic moment before local buckling.
Question 5: For a steel beam experiencing lateral-torsional buckling in the elastic range, nominal moment Mn is proportional to:
- 1/Lb² (inversely proportional to unbraced length squared) (Correct answer)
- Lb (proportional to unbraced length)
- Lb⁰·⁵ (square root of unbraced length)
- Constant (independent of unbraced length)
Correct answer: 1/Lb² (inversely proportional to unbraced length squared)
In the elastic LTB range, Mn decreases with 1/Lb via the elastic buckling moment Mcr ∝ (1/Lb)√(EIyGJ + (πE/Lb)²IyCw).
Question 6: Which AISC design method uses Ω (omega) as the safety factor applied to nominal strength?
- ASD (Allowable Strength Design) (Correct answer)
- LRFD (Load and Resistance Factor Design)
- Both methods equally
- Neither; both use φ factors
Correct answer: ASD (Allowable Strength Design)
In modern AISC ASD, allowable strength = Rn/Ω, where Ω is the safety factor (e.g., Ω = 1.67 for yielding).
Question 7: A W14×82 steel column (A = 24.0 in²) with Fy = 50 ksi and φcFcr = 38.2 ksi has a design axial strength of:
- 917 kips (Correct answer)
- 1,200 kips
- 760 kips
- 1,000 kips
Correct answer: 917 kips
φcPn = φcFcr × A = 38.2 ksi × 24.0 in² = 916.8 ≈ 917 kips.
Civil Engineering PE Steel Structure Design Methods 3
A steel column with KL/r = 80 and Fy = 50 ksi falls into which AISC buckling category?