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LecturesModule 25

RC — Compression Members (Columns)

Short columns under axial load and uniaxial bending; tied vs spiral detailing per ACI Chapters 10 and 22.

AISC Reference Box
  • ACI 318-19 Ch. 10 — Columns — General
  • ACI 318-19 §22.4 — Axial Strength
  • ACI 318-19 §25.7 — Transverse Reinforcement (Ties & Spirals)
  • ACI Table 21.2.2 — φ — tied = 0.65, spiral = 0.75
  • ACI §6.6 — Slenderness — Moment Magnification

Lecture Notes

Reinforced concrete columns carry combinations of axial load and bending moment. ACI 318-19 Chapter 10 distinguishes SHORT columns (slenderness effects negligible — kℓu/r ≤ 22 for unbraced or ≤ 34 − 12·M1/M2 for braced) from SLENDER columns (require moment magnification per ACI §6.6 or P-Δ analysis).

Two column types: TIED columns use a rectangular or square cross-section confined by individual lateral ties (#3 or #4) — φ = 0.65 (ACI Table 21.2.2); SPIRAL columns use a circular core wrapped by a continuous spiral (#3 minimum) — φ = 0.75, reflecting the more ductile post-peak behavior of spirally confined concrete.

Maximum nominal axial load (ACI §22.4.2): Pn,max = α·[0.85·f'c·(Ag − Ast) + fy·Ast], with α = 0.80 for tied and α = 0.85 for spiral. The α factor accounts for unintended eccentricity. Design strength φPn,max = φ·α·[...].

Longitudinal steel ratio ρg = Ast/Ag must satisfy 0.01 ≤ ρg ≤ 0.08 (ACI §10.6.1.1); for practical purposes ρg ≤ 0.04 simplifies splicing. Minimum bar count: 4 in tied rectangular columns; 6 in spiral round columns (ACI §10.7.3.1).

Tie spacing (ACI §25.7.2): smax = min(16·db,long, 48·db,tie, least column dimension). Spirals (ACI §25.7.3) must satisfy ρs ≥ 0.45·(Ag/Ach − 1)·f'c/fyt with clear spacing 1 in. to 3 in.

Combined axial + bending uses an interaction diagram (P-M diagram) constructed from strain compatibility. Key points: pure compression (Po), balanced point (εt = εty), pure bending (Pn = 0). For preliminary design, ACI Design Handbook SP-17 columns charts give Pn/(f'c·Ag) vs Mn/(f'c·Ag·h) curves for ρg families.

Project case study — Cardinal Square RC — 4-story cast-in-place concrete 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 @ 30 ft × 30 ft. Stories: 4 @ 13 ft (52 ft roof). Floor: 7 in one-way slab on RC beams (b = 14 in, h = 24 in) framing to RC columns (16 in × 16 in tied) on grid. Roof: same beam grid + 6 in slab. Materials: Normal-weight concrete f'c = 4,000 psi (λ = 1.0). Reinforcement Grade 60 (fy = 60,000 psi); ties/stirrups Grade 60 #3 / #4. Clear cover: 1.5 in interior beams/slabs, 2 in columns, 3 in footings (ACI §20.5.1).

Chapter 25 — Tied column under axial + bending
Typical first-story interior tied column, 16 × 16 in, 8 #8 longitudinal
Demand carried forward
Trib. area = 30 × 30 = 900 ft²/floor Pu = (1.2·110 + 1.6·50)·0.9 ksf·4 ≈ (132+80)·3.6 ≈ 763 k Edge column adds Mu ≈ 65 k-ft from frame action.
This chapter contributes
Computes φPn,max = 0.80·φ·[0.85·f'c·(Ag − Ast) + fy·Ast] = 0.80·0.65·[0.85·4·(256 − 6.32) + 60·6.32] ≈ 638 k → upsize to 18 × 18 with 8 #9 (Ast = 8.0 in²): φPn,max ≈ 838 k > 763 k. Checks the P–M interaction (φPn, φMn) against the demand point.
Feeds into next chapter
Pu and column dimensions feed the spread-footing / mat sizing in the capstone, and the joint detail in Chapter 24.
Tied (φ = 0.65, α = 0.80)Spiral (φ = 0.75, α = 0.85)
Tied (φ = 0.65, α = 0.80) vs spiral (φ = 0.75, α = 0.85) — same Pn,max formula, different ductility.

Formula Sheet

NameEquationAISC Ref
Pn,max (tied)Pn,max = 0.80·[0.85·f'c·(Ag − Ast) + fy·Ast]ACI §22.4.2.1
Pn,max (spiral)Pn,max = 0.85·[0.85·f'c·(Ag − Ast) + fy·Ast]ACI §22.4.2.1
Design strengthφPn,max (φ = 0.65 tied, 0.75 spiral)ACI Table 21.2.2
Steel ratio0.01 ≤ ρg = Ast/Ag ≤ 0.08ACI §10.6.1.1
Tie spacingsmax = min(16·db,long, 48·db,tie, least dim.)ACI §25.7.2.1
Spiral ratioρs ≥ 0.45·(Ag/Ach − 1)·f'c/fytACI §25.7.3.3
Slenderness (braced)kℓu/r ≤ 34 − 12·(M1/M2) (else slender)ACI §6.2.5

Worked Example

Axial capacity of a tied square column

Given
Square tied column 18 × 18 in. (Ag = 324 in²) reinforced with 8 #9 bars (Ast = 8·1.00 = 8.00 in²). Materials: f'c = 4000 psi, fy = 60,000 psi. Ties #3 at spacing s.
Load combination
Pu = 850 kips (factored axial, no significant moment).
Required strength
Verify φPn,max ≥ Pu = 850 kips; design tie spacing.
Limit states
  • Axial strength φPn,max
  • Steel ratio limits
  • Tie spacing
AISC reference
ACI 318-19 §22.4.2 and §25.7.2
Solution steps
  1. 1. Given
    b × h = 18 × 18 in. Ag = 324 in². 8 #9 → Ast = 8.00 in². f'c = 4000 psi. fy = 60,000 psi. Tied column → α = 0.80, φ = 0.65.
  2. 2. Find
    φPn,max and tie spacing s.
  3. 3. ACI reference
    ACI §22.4.2.1 (Pn,max) and §25.7.2 (tie spacing).
  4. 4. Assumptions
    Short column (kℓu/r ≤ 22 assumed). Concentric load (no significant moment). Grade 60 longitudinal and tie steel.
  5. 5. Steel ratio
    ρg = Ast/Ag. ρg = 8.00/324. ρg = 0.0247. 0.01 ≤ 0.0247 ≤ 0.08 ✓
  6. 6. Concrete contribution
    Pc = 0.85·f'c·(Ag − Ast). Pc = 0.85·4·(324 − 8.00). Pc = 0.85·4·316. Pc = 1074.4 kips.
  7. 7. Steel contribution
    Ps = fy·Ast. Ps = 60·8.00. Ps = 480.0 kips.
  8. 8. Nominal Pn,max
    Pn,max = α·(Pc + Ps). Pn,max = 0.80·(1074.4 + 480.0). Pn,max = 0.80·1554.4. Pn,max = 1243.5 kips.
  9. 9. Design φPn,max
    φPn,max = 0.65·1243.5. φPn,max = 808.3 kips. NG: 808.3 < 850 → need more steel or larger column.
  10. 10. Re-try 8 #10 bars
    Ast = 8·1.27 = 10.16 in². Pc = 0.85·4·(324 − 10.16) = 1067.0 k. Ps = 60·10.16 = 609.6 k. Pn,max = 0.80·(1067.0 + 609.6) = 1341.3 k. φPn,max = 0.65·1341.3 = 871.8 k ≥ 850 ✓
  11. 11. Tie spacing
    db,long = 1.27 in. (#10). db,tie = 0.375 in. (#3). smax = min(16·1.27, 48·0.375, 18). smax = min(20.3, 18.0, 18). smax = 18 in. Use #3 ties at 18 in. o.c.
Final design decision
Use 18 × 18 in. tied column with 8 #10 longitudinal bars and #3 ties at 18 in. o.c. φPn,max = 872 k ≥ Pu = 850 k.
Common mistakes in this example
  • Using α = 0.85 for a tied column (only spirals get 0.85).
  • Forgetting to subtract Ast when computing the concrete area (Ag − Ast).
  • Using φ = 0.90 (flexure value) instead of 0.65 (tied compression).
  • Missing the least column dimension as a tie-spacing limit.

FE-Style Worked Examples (5)

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
16 in. square tied column; 8·#8 (Ast = 6.32 in²); f'c = 4 ksi; fy = 60 ksi.
AISC Reference
ACI 318-19 §22.4.2.2
Step-by-step solution
  1. 1. Problem Statement
    Compute φPn,max for a short tied column at e ≈ 0.
  2. 2. Variables & Values
    Ag = 16·16 = 256 in² Ast = 6.32 f'c = 4,000 fy = 60,000 α = 0.80 (tied) φ = 0.65.
  3. 3. Approach
    Use Pn,max = α·[0.85·f'c·(Ag−Ast) + fy·Ast] apply φ.
  4. 4. Formula Setup
    φPn,max = φ·α·[0.85·f'c·(Ag − Ast) + fy·Ast]
  5. 5. Substitution
    φPn,max = 0.65·0.80·[0.85·4·(256−6.32) + 60·6.32] = 0.52·[0.85·4·249.68 + 379.2] = 0.52·[848.9 + 379.2] = 0.52·1,228.1 = 638.6 k.
  6. 6. Solution Steps
    φPn,max ≈ 639 k.
Answer φPn,max ≈ 639 k.
PuLcAxially loaded column

Practice Problems

  1. [E] State Pn,max = α[0.85·f'c·(Ag − Ast) + fy·Ast]; α = 0.80 tied, 0.85 spiral.
  2. [E] State φ = 0.65 tied, 0.75 spiral.
  3. [E] 0.01 ≤ ρg ≤ 0.08.
  4. [E] Min bar count: 4 tied rectangular, 6 spiral round.
  5. [E] Tie spacing = min(16·db,long, 48·db,tie, least dim.).
  6. [M] φPn,max for 16 x 16 tied with 6 #8, f'c = 4, fy = 60.
  7. [M] φPn,max for 18 in. dia. spiral with 6 #9, f'c = 5, fy = 60.
  8. [M] Max Pu for 14 x 14 tied with 6 #8, f'c = 4.
  9. [M] Tie spacing for 18 x 18 column with 8 #9 longitudinal and #3 ties.
  10. [M] Slenderness for braced 16 x 16, kℓu = 14 ft, M1/M2 = 0.3.
  11. [H] Design circular spiral column for Pu = 1200 k, f'c = 5, fy = 60.
  12. [H] P-M interaction: three key points (Po, balanced, pure bending) for 16 x 16 with 8 #9 (f'c = 4).
  13. [H] Slender column moment magnification: Cm = 0.85, Pu = 220 k, Pc = 1100 k. Compute δns (§6.6.4).
  14. [H] Biaxial 18 x 18 tied, Pu = 600 k, Mux = 120, Muy = 80 k-ft — reciprocal-load (Bresler).
  15. [H] Spiral confinement ρs required for 18 in. dia., f'c = 5, fyt = 60, cover 1.5 in.
Structured Clues
  • α = 0.80 for tied, 0.85 for spiral (NOT the φ-factor).
  • Subtract Ast from Ag before multiplying by 0.85·f'c.
  • Tie spacing smax = min(16·db,long, 48·db,tie, least column dimension).
Code References
  • ACI 318-19 §22.4, §10.6, §25.7, Table 21.2.2

Quiz

1. The φ factor for a tied column controlled by compression is:
2. For a spiral column the maximum axial reduction factor α is:
3. ACI longitudinal steel ratio limits for columns are:
4. Minimum number of longitudinal bars in a rectangular tied column is:
5. Tie vertical spacing must not exceed the SMALLEST of 16db (long), 48db (tie), and:
6. Pn,max (kips) of a 16×16 tied column, 6 #8 (As = 4.74 in²), f'c = 4 ksi, fy = 60 ksi:
7. φPn,max for the column above is:
8. ρg for the column above:
9. α for a spiral column is:
10. φ for a spiral compression-controlled column:
11. Maximum tie spacing for an 18×18 column with #9 longitudinal bars and #3 ties:
12. Minimum number of longitudinal bars in a circular spiral column is:
13. Required spiral ratio ρs = 0.45·(Ag/Ach − 1)·f'c/fyt for Ag/Ach = 1.30, f'c = 4 ksi, fyt = 60 ksi:
14. Slenderness limit for a BRACED column with M1/M2 = +0.4 is kℓu/r ≤
15. Radius of gyration r for a 16-in. square column (r = 0.30·h):

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 rc — compression members (columns). 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.