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LecturesModule 5

Compression Members and Column Buckling

Euler buckling, AISC E3/E4, slender elements, and effective length.

AISC Reference Box
  • AISC 360-22 Chapter E — Design of Members for Compression
  • AISC §E3 — Flexural Buckling of Members Without Slender Elements
  • AISC §E4 — Torsional & Flexural-Torsional Buckling
  • AISC §E7 — Members with Slender Elements

Lecture Notes

This module introduces compression members and column buckling. Lecture content here covers the governing physics, LRFD philosophy, and how the relevant AISC 360-22 chapter organizes the limit states.

Instructors can replace this text in Admin Mode. Each section is structured around: (1) behavior, (2) failure modes, (3) AISC limit-state equations, (4) design workflow, (5) detailing requirements.

A short comparison to ASD is included only where the resistance factor / safety factor relationship clarifies the LRFD design check.

Project case study — Cardinal Square — 4-story braced-frame office

Every chapter's worked example is one step in the design of the same building: Plan: 4 bays N–S × 3 bays E–W, each 30 ft × 30 ft. Stories: 4 @ 13 ft (52 ft roof). Composite floor: 4.5 in NW concrete on 3 VLI20 deck. Roof: 1.5 in B-deck + insulation + membrane. Materials: Wide-flange members A992 (Fy = 50 ksi, Fu = 65 ksi). Plates A572 Gr. 50. HSS bracing A500 Gr. C. Bolts A325-N 7/8 in dia. Welds E70XX. Concrete f'c = 4 ksi. Anchor rods F1554 Gr. 36.

Chapter 5 — Compression (interior gravity column)
Typical first-story interior column, KL = 13 ft, A992 W-shape
Demand carried forward
Tributary area = 30 × 30 = 900 ft²/floor Pu = (1.2·75 + 1.6·50)·0.9 ksf·4 = (90+80)·3.6 = ~612 kips at the base (live-load reduction not yet applied).
This chapter contributes
Picks a trial W12/W14 from AISC Manual Table 4-1a using KL = 13 ft. Checks Fcr per §E3 and confirms φc Pn ≥ Pu.
Feeds into next chapter
Column Pu becomes the bearing demand on the base plate in Chapter 16 and the splice demand in Chapter 12.
Slenderness Lc/rFcr (ksi)Lc/r = 4.71√(E/Fy)Inelastic (0.658^(Fy/Fe))·FyElastic 0.877 FeFy
AISC §E3 column curve: inelastic (0.658^(Fy/Fe))·Fy below 4.71√(E/Fy), elastic 0.877·Fe above.

Formula Sheet

NameEquationAISC Ref
Elastic buckling stressFe = π² E / (Lc/r)²AISC §E3
InelasticFcr = (0.658^(Fy/Fe)) Fy if Lc/r ≤ 4.71√(E/Fy)AISC §E3
ElasticFcr = 0.877 Fe if Lc/r > 4.71√(E/Fy)AISC §E3
Design strengthφc Pn = 0.90 · Fcr · AgAISC §E1

Worked Example

Compression Members and Column Buckling

Given
Replace with project-specific given data (loads, geometry, material).
Load combination
Controlling LRFD load combination from ASCE 7.
Required strength
Compute required strength Ru from the controlling combination.
Limit states
  • Limit state 1
  • Limit state 2
AISC reference
AISC 360-22 — applicable chapter
Solution steps
  1. 1. Required strength
    Compute Ru.
  2. 2. Trial section
    Pick a trial from AISC shape tables Instructor should verify with official AISC Manual.
  3. 3. Check each limit state
    Apply φ Rn ≥ Ru for every governing limit state.
  4. 4. Iterate
    Resize until the most economical section satisfies all checks.
Final design decision
Select the lightest section that satisfies all LRFD limit states.
Common mistakes in this example
  • Skipping a limit state
  • Using the wrong φ factor
  • Forgetting serviceability checks

FE-Style Worked Examples (7)

Each example mirrors the NCEES FE Civil Reference Handbook style: brief givens, a labeled figure, AISC section reference, step-by-step numeric solution, and a single boxed answer.

Given
W10×49 column, L = 14 ft, K = 1.0, ry = 2.54 in.
AISC Reference
AISC Commentary Table C-A-7.1
Step-by-step solution
  1. Lc/ry
    1.0(14×12)/2.54 = 66.1
Answer Slenderness Lc/ry = 66.1.
PuLcAxially loaded column

Textbook — Aghayere & Vigil (2009)(5 worked examples with figures + numerical answers)

Worked examples scanned directly from the CEGR 436 course textbook. Each card shows the original page (figure + full step-by-step solution) and adds an FE-style numerical multiple-choice prompt with answer key.

Chapter summary

Chapter 5 covers concentrically loaded columns. The design strength φcPn = 0.90·Fcr·Ag uses AISC §E3 (no slender elements) — the elastic/inelastic split at Lc/r = 4.71·√(E/Fy). Effective length Lc = K·L from §C2 / Appendix 7; rules of thumb: K = 1.0 (pinned), 0.65 (fixed–fixed), 2.0 (cantilever).

  • Fe = π²E / (Lc/r)² (Euler).
  • Inelastic: Fcr = 0.658^(Fy/Fe) · Fy for Lc/r ≤ 4.71√(E/Fy) (≈ 113 for A992).
  • Elastic: Fcr = 0.877·Fe for Lc/r > 4.71√(E/Fy).
  • φc = 0.90; check both axes (rx, ry) — weak axis usually controls.
  • Slender elements (Chapter B4) reduce φPn — use AISC §E7.
  • L/r ≤ 200 recommended for compression members.
Setup
W10×33 column, KL = 14 ft, A992 steel (Fy = 50 ksi). ry = 1.94 in. Compute Fe and classify (elastic vs inelastic).
AISC Reference
AISC §E3
Numerical practice
KL/ry and regime?
Textbook page 191 — Critical buckling load by Euler
Aghayere & Vigil (2009), p. 191 — full worked solution & sketch.

FE Practice Bank (10)

Multiple-choice problems pulled from the instructor's CEGR 492 FE Review packet (EasyFEExam © 2025, Steve Efe, PhD). Pick an answer, then click Reveal solution.

P16D. RC Column φPnEasyFEExam Structural Design – Problem 16
Ag = 400, Ast = 8, f′c = 4, fy = 60. φPn?
P17D. Required AstEasyFEExam Structural Design – Problem 17
Pu = 900 k, Ag = 500, f′c = 5, fy = 60. Required Ast?
P18D. Required AgEasyFEExam Structural Design – Problem 18
Ast = 10, f′c = 4, fy = 60, Pu = 1000. Min Ag?
P19D. Ag at ρmaxEasyFEExam Structural Design – Problem 19
Pu = 1200 k, f′c = 4, fy = 60, ρ = 0.08. Min Ag?
P20D. Reinforcement Ratio ρEasyFEExam Structural Design – Problem 20
20 × 30 in column with 4 #9 (As = 4.0 in²). ρ?
P21D. Euler BucklingEasyFEExam Structural Design – Problem 21
Steel column 6 × 10 in, L = 12 ft, pinned-fixed (K = 0.7), E = 29,000 ksi. Pcr about weak axis?
P22D. Slenderness RatiosEasyFEExam Structural Design – Problem 22
W14×74, L = 20 ft, K = 1, rx = 6.04, ry = 2.48. KL and KL/r?
P23D. K from Alignment ChartEasyFEExam Structural Design – Problem 23
Interior column L = 20 ft, sway permitted, Ga ≈ Gb ≈ 1. Kx and KxL?
P24D. φPn W12×50EasyFEExam Structural Design – Problem 24
W12×50, L = 12 ft, fixed-pinned (K = 0.8), Fy = 50, A = 14.7, rx = 5.47, ry = 1.96 in. φPn?
P25D. W14×132 φPnEasyFEExam Structural Design – Problem 25
W14×132, L = 30 ft pinned, Fy = 50, ry = 3.76. Governing KL/r and φPn?

Interactive Calculator

Compression Member Design

AISC §E3
Lc/r96.6
Fe = π²E/(Lc/r)²30.64 ksi
Regime (limit 113.4)Inelastic
Fcr25.26 ksi
φc Pn = 0.90 Fcr Ag201.2 kipsOK

Practice Problems

  1. [E] State Euler stress Fe = π²E/(Lc/r)².
  2. [E] Define K for the four idealized end conditions (1.0, 0.7, 0.5, 2.0).
  3. [E] State the AISC slenderness boundary Lc/r = 4.71√(E/Fy).
  4. [E] State φc = 0.90 per §E1.
  5. [E] State the recommended slenderness L/r ≤ 200 for compression members.
  6. [M] W12x72 (A992) column, L = 14 ft, K = 1.0. Compute Pn per §E3.
  7. [M] HSS 8x8x3/8 (A500 Gr C), L = 12 ft, K = 1.0. Compute φPn.
  8. [M] W14x90 column carries Pu = 950 k, KxLx = KyLy = 14 ft. Check using Table 4-1.
  9. [M] W12x96: Lx = 18 ft Kx = 1.0; Ly = 6 ft Ky = 1.0. Compute KL/r both axes; identify governing axis.
  10. [M] 2L4x4x3/8 SLBB double-angle column, L = 10 ft, K = 1.0. Compute φPn per §E6.
  11. [H] Design lightest W14 for Pu = 720 k, KxLx = KyLy = 16 ft, A992.
  12. [H] Slender-element column WT9x27.5 (b/t > λr). Compute Qs (§E7) and effective Pn.
  13. [H] Built-up box column (4) PL 1/2 x 12 in. Compute Ix, ry; check stitch spacing (§E6.2).
  14. [H] Pipe column 6 in. Sch 80 (A53 Gr B), KL = 22 ft. Determine φPn and check D/t.
  15. [H] Multi-story column splice: lower W14x120, upper W14x82. Justify splice elevation by axial demand reduction.
Structured Clues
  • Compute KL/r for BOTH axes, then use the larger.
  • Slenderness boundary: 4.71√(E/Fy) = 113 for Fy = 50.
  • AISC Manual Table 4-1a gives φcPn directly vs KL.
Code References
  • AISC 360-22 §E3, E7
  • AISC Manual Part 4 — Tables 4-1 to 4-22

Quiz (10 FE-bank + 2 concept)

1. [P16] Ag = 400, Ast = 8, f′c = 4, fy = 60. φPn?
2. [P17] Pu = 900 k, Ag = 500, f′c = 5, fy = 60. Required Ast?
3. [P18] Ast = 10, f′c = 4, fy = 60, Pu = 1000. Min Ag?
4. [P19] Pu = 1200 k, f′c = 4, fy = 60, ρ = 0.08. Min Ag?
5. [P20] 20 × 30 in column with 4 #9 (As = 4.0 in²). ρ?
6. [P21] Steel column 6 × 10 in, L = 12 ft, pinned-fixed (K = 0.7), E = 29,000 ksi. Pcr about weak axis?
7. [P22] W14×74, L = 20 ft, K = 1, rx = 6.04, ry = 2.48. KL and KL/r?
8. [P23] Interior column L = 20 ft, sway permitted, Ga ≈ Gb ≈ 1. Kx and KxL?
9. [P24] W12×50, L = 12 ft, fixed-pinned (K = 0.8), Fy = 50, A = 14.7, rx = 5.47, ry = 1.96 in. φPn?
10. [P25] W14×132, L = 30 ft pinned, Fy = 50, ry = 3.76. Governing KL/r and φPn?
11. Which AISC 360-22 chapter primarily governs compression members and column buckling?
12. In LRFD, the basic design inequality is:

Common Student Mistakes

  • Mixing ASD and LRFD load combinations in the same problem.
  • Using nominal strength Rn instead of design strength φRn.
  • Forgetting to check every limit state listed in the AISC chapter.

"Professor Explains" Script

Today we're talking about compression members and column buckling. Think of this topic as one step in the LRFD workflow: identify the demand, identify the limit states from the relevant AISC chapter, then check that φ·Rn is at least equal to Ru. We'll walk through the failure modes, the equations, and a worked example. Pay close attention to where the resistance factor changes — that's where students lose points on exams.