How to Build a Column Interaction Diagram — ACI 318-19 Step by Step
An interaction diagram is a set of strain-compatibility calculations, one per neutral-axis depth. This guide walks through the method by hand for the balanced point of a 16 × 20 in tied column with 8-#9 bars and f′c = 5,000 psi, then shows how the full curve, the φ factor and the axial cap are assembled. The free column calculator repeats it for 400 depths and plots the result.
Example column
| Section | 16 in wide × 20 in deep (bending about the 20-in direction), tied, #3 ties, 1.5 in cover |
| Bars | 8-#9: 3 at each 16-in face, 1 at mid-depth on each side |
| Materials | f′c = 5,000 psi (β1 = 0.80), Grade 60, Es = 29,000 ksi |
Screen slenderness first
The diagram gives cross-section strength. If the column is slender, the moment you check against it must be magnified first. ACI 318-19 §6.2.5: neglect slenderness when kℓu/r ≤ 34 + 12(M1/M2) ≤ 40 in a braced frame (M1/M2 negative for single curvature) or ≤ 22 in a sway frame, with r = 0.30h or 0.25D.
Compute P0 and the axial cap
φPn,max = 0.65 × 0.80 × 1,806 = 939 kip (tied) · spiral would be 0.75 × 0.85P0
Locate the bar layers
Layer 1: 3 bars at d1 = 2.44 in · Layer 2: 2 bars at d2 = 10.0 in · Layer 3: 3 bars at dt = 17.56 in
Pick a neutral-axis depth c
Each c gives one point on the diagram. The balanced point is the most useful one to do by hand: the extreme tension bar is just at yield when the concrete reaches 0.003.
Find each bar's strain and stress
Layer 1: ε = 0.00230 → 60 ksi, inside the block so deduct 0.85f′c: F1 = 3(1.0)(60 − 4.25) = 167.3 kip
Layer 2: ε = 0.000113 → 3.3 ksi (below the block): F2 = 2(1.0)(3.3) = 6.6 kip
Layer 3: ε = −0.00207 → −60 ksi: F3 = 3(1.0)(−60) = −180 kip
Add the concrete block
For a circular column the block is a circular segment of depth a; use the segment area R²(θ − sinθcosθ) and its centroid, with cosθ = (R − a)/R.
Sum forces and moments about the centroid
Mn = 565.4(10 − 4.16) + 167.3(10 − 2.44) + 6.6(10 − 10) + 180(17.56 − 10) = 5,930 kip-in = 494 kip-ft
Apply φ and repeat for other c values
| Point | c | εt | φ (tied) |
|---|---|---|---|
| Pure compression | ∞ | compression | 0.65, capped at 0.80P0 |
| Zero tension | dt = 17.56 in | 0 | 0.65 |
| Balanced | 10.39 in | 0.00207 | 0.65 |
| Tension-controlled limit | 0.003dt/0.00807 = 6.53 in | 0.00507 | 0.90 |
| Pure bending | solve Pn = 0 | > 0.00507 | 0.90 |
| Pure tension | → 0 | all yield | 0.90 |
Multiply each (Pn, Mn) by its φ, then cut the top of the curve off at φPn,max. For this column the balanced design point is φPb = 0.65 × 559 = 363 kip with φMb = 321 kip-ft.
Check the factored load
Find the point on the design curve at φPn = Pu and compare its moment with Mu. For Pu = 500 kip (above the balanced point, so φ = 0.65), solving 0.65Pn(c) = 500 gives c = 12.4 in and φMn = 297 kip-ft. Mu = 200 kip-ft passes at 0.67.
Rectangular or circular, tied or spiral: full diagram, your load point, φMn at Pu, tie spacing, spiral pitch and slenderness.
Open the column calculator →Worked examples
Circular spiral column, 8-#8 · P-Δ column (second-order moments)
Frequently asked questions
How many points do I need for an interaction diagram?
By hand, five or six: pure compression, the zero-tension point (c = dt), the balanced point, the tension-controlled limit, pure bending and pure tension, joined with smooth curves. A program can sweep hundreds of neutral-axis depths and capture the shape exactly, including the kink where φ starts to rise.
About which axis are moments taken?
The plastic centroid, which for symmetric reinforcement is the geometric center of the section. Taking moments about another point changes Mn for every point except pure bending.
Why is there a flat top on the design curve?
ACI 318-19 Table 22.4.2.1 caps axial strength at 0.80P0 for tied and 0.85P0 for spiral columns to allow for accidental eccentricity. The cap is applied to φPn, producing a horizontal line across the top of the design diagram.
Why does the design curve have a bend below the balanced point?
φ changes from 0.65 (tied) to 0.90 as the extreme tension strain goes from εty to εty + 0.003. Multiplying the nominal curve by a varying φ produces the visible bend, sometimes a small bump, in the design curve.