Worked examples

Circular Spiral Column — ACI 318-19 Interaction Diagram Worked Example

By Michael Flynn, PE · water resources & dam safety engineer ·

A 20-in round column in a braced frame carries Pu = 450 kip and Mu = 140 kip-ft. We check slenderness, build the interaction diagram, read the moment capacity at the factored axial load, size the spiral pitch, and price what the spiral buys over ordinary ties: 23% more axial capacity and 17% more moment at this load.

The column

SectionD = 20 in round, 8-#8 bars equally spaced, #3 spiral, 1.5 in clear cover
Materialsf′c = 5,000 psi (β1 = 0.80), Grade 60 bars and spiral
FrameBraced (nonsway), unsupported length 14 ft, k = 1.0, double curvature with M1/M2 = +0.3
LoadsPu = 450 kip, Mu = 140 kip-ft

Column tool: Circular, Spiral, D 20, 8 bars #8, tie #3, f′c 5000, Pu 450, Mu 140, ℓu 14, M1/M2 0.3.

Open the column calculator →
Step 1 · Slenderness

Can second-order effects be ignored?

r = 0.25D = 5.0 in  ·  kℓu/r = 1.0 × 168/5.0 = 33.6
Nonsway limit = 34 + 12(M1/M2) = 34 + 12(0.3) = 37.6 (≤ 40) → short column, use Mu directly
Mind the sign. ACI 318-19 takes M1/M2 negative for single curvature. Had this column been bent in single curvature (−0.3), the limit would drop to 30.4 and the column would be slender — moments would need magnifying.
Step 2 · Axial capacity

P0 and the spiral cap

Ag = π(20)²/4 = 314.2 in²  ·  Ast = 8 × 0.79 = 6.32 in²  ·  ρg = 2.01% (1–8% ✓; ≥ 6 bars with a spiral ✓)
P0 = 0.85 × 5(314.2 − 6.32) + 60 × 6.32 = 1,308.3 + 379.2 = 1,688 kip
φPn,max = 0.75 × 0.85 × 1,688 = 1,076 kip  (ties: 0.65 × 0.80 × 1,688 = 878 kip)
Step 3 · Interaction diagram

Key points and the capacity at Pu

Bars sit on a 15.25-in circle (D − 2 × 1.5 − 2 × 0.375 − 1.0), one at the compression face. For each neutral-axis depth the concrete block is a circular segment of depth a = 0.80c; its area and centroid come from the segment formulas, and each bar's stress from its own strain. Selected points of the design curve:

Pu (kip)φMn (kip-ft)Note
−3410pure tension, 0.9fyAst
0199pure bending
200257transition zone
300266near the peak
401259balanced: Pb = 535, Mb = 345 nominal, × 0.75
450257this column
700232compression-controlled
900181
1,076—φPn,max plateau
At Pu = 450 kip: φMn = 257 kip-ft → Mu/φMn = 140/257 = 0.54 ✓

The same bars with ties instead of a spiral give φMn = 219 kip-ft at 450 kip (0.64). The spiral's higher φ is worth 17% more moment here and 23% more axial cap.

Step 4 · Spiral

Pitch from the volumetric ratio

Dc = 20 − 2(1.5) = 17.0 in  ·  Ach = π(17)²/4 = 227.0 in²
ρs,min = 0.45(314.2/227.0 − 1)(5/60) = 0.0144
s = 4Asp/(Dcρs) = 4(0.11)/(17.0 × 0.0144) = 1.80 in → use #3 spiral at 1¾ in pitch
Clear pitch = 1.75 − 0.375 = 1.375 in, within 1–3 in ✓
Higher f′c means tighter spirals. ρs scales with f′c/fyt. At 6,000 psi, ρs,min = 0.0173 and a #3 spiral needs a 1.50-in pitch (1.125 in clear, close to the 1-in minimum, which makes concrete placement hard). A #4 spiral (0.20 in²) relaxes that to 2.72 in.
Step 5 · Bar spacing

Clear distance between longitudinal bars

Chord = 2(7.625) sin(22.5°) − 1.0 = 4.84 in ≥ max(1.5 in, 1.5db, 4/3 dagg) ✓

Tools used in this example

RC column interaction diagram. Guide: How to build a column interaction diagram.

Frequently asked questions

Why do spiral columns get a higher phi?

Spirals confine the core so the column keeps carrying load after the shell spalls, giving a more ductile failure. ACI 318-19 rewards that with φ = 0.75 for compression-controlled sections (0.65 for tied) and Pn,max = 0.85P0 (0.80P0 for tied).

How do you calculate spiral pitch?

From the minimum volumetric ratio ρs = 0.45(Ag/Ach − 1)f′c/fyt (ACI 318-19 §25.7.3.3) and the relation ρs = 4Asp/(Dc·s), where Dc is the core diameter to the outside of the spiral. Solve s = 4Asp/(Dc·ρs) and keep the clear pitch between 1 and 3 in.

Why does moment capacity rise and then fall as axial load increases?

Below the balanced point, axial compression offsets tension in the steel, so the section can take more moment before the steel yields. Above it, the concrete is already highly stressed by axial load and has less capacity left for bending. In this column φMn peaks at about 266 kip-ft near Pu = 300 kip.

What is the radius of gyration of a round column?

For slenderness, ACI 318-19 §6.2.5.2 allows r = 0.25D for circular columns (0.30h for rectangular). A 20-in column has r = 5 in.

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