calcnote ACI 318-19
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How to construct a column interaction diagram per ACI 318-19

A P-M interaction diagram is built one point at a time: pick a neutral-axis depth, work out the strains, turn them into forces, and sum. This page walks that construction end-to-end on a real section — a 16×16 in tied column with 8-#9 bars — deriving three of the seven control points in full and tabulating the rest. The same section's results are cross-checked against four independent authorities on the verification page; here the focus is the mechanics.

The example section

Section16 × 16 in (Ag = 256 in²)
Reinforcement8-#9, two-face layout → 4 bars at d = 2.5 in, 4 bars at d = 13.5 in (Ast = 8.00 in²)
Materialsf′c = 5,000 psi, Grade 60 (Es = 29,000 ksi)
Derived constantsβ1 = 0.80 · εty = 60/29,000 = 0.00207 · tension-controlled limit εty + 0.003 = 0.00507

The recipe

Every point on the curve comes from the same five steps, anchored in two assumptions: plane sections remain plane (strain varies linearly through the depth, with the extreme compression fiber at the crushing strain εcu = 0.003), and concrete stress is represented by the Whitney stress block.

  1. Pick a neutral-axis depth c. Each c is one point on the curve.
  2. Strains by similar triangles: each bar layer at depth d gets ε = 0.003·(c − d)/c.
  3. Stresses: fs = Es·ε, capped at ±fy. Bars inside the stress block get a net contribution of As·(fs − 0.85·f′c) — the concrete they displace is already counted in the block.
  4. Forces and moments: concrete block Cc = 0.85·f′c·a·b with a = β1·c; sum axial forces for Pn, sum moments about the section centroid for Mn.
  5. Apply φ from the net tensile strain at the extreme tension layer per Table 21.2.2 (0.65 compression-controlled → 0.90 tension-controlled).

Worked point 1: the axial cap (φPn,max)

The top of the usable diagram isn't a strain-compatibility point at all — it's the code's accidental-eccentricity ceiling. Pure-compression capacity first (§22.4.2.2):

$$P_o = 0.85\,f'_c\,(A_g - A_{st}) + f_y\,A_{st} = 0.85(5)(248) + 60(8) = 1{,}534\ \text{kip}$$

Then the tied-column cap (§22.4.2.1, with φ = 0.65):

$$\phi P_{n,\max} = 0.80\,\phi\,P_o = 0.80(0.65)(1{,}534) = 797.7\ \text{kip}$$

On the diagram this is the horizontal line truncating the top of the curve.

Worked point 2: the balanced point

At the balanced point the concrete crushes exactly as the extreme tension steel reaches yield. Similar triangles fix c directly:

$$c_b = \frac{\varepsilon_{cu}}{\varepsilon_{cu} + \varepsilon_{ty}}\,d = \frac{0.003}{0.003 + 0.00207}(13.5) = 7.99\ \text{in}, \qquad a = \beta_1 c_b = 6.39\ \text{in}$$

Forces, layer by layer:

$$C_c = 0.85(5)(6.39)(16) = 434.6\ \text{kip}$$
$$\text{top bars } (d=2.5)\!: \varepsilon = 0.003\tfrac{7.99-2.5}{7.99} = 0.00206 \Rightarrow f_s = 59.8\ \text{ksi (elastic)}, \quad C_s = 4(1.00)(59.8 - 4.25) = 222.2\ \text{kip}$$
$$\text{bottom bars } (d=13.5)\!: \varepsilon = \varepsilon_{ty} \Rightarrow T_s = 4(1.00)(60) = 240\ \text{kip}$$
$$P_n = 434.6 + 222.2 - 240 = 416.8\ \text{kip}$$

Moments about the section centroid (block arm 8 − 6.39/2 = 4.805 in; both bar layers at 5.5 in):

$$M_n = 434.6(4.805) + 222.2(5.5) + 240(5.5) = 4{,}630\ \text{kip-in} = 385.8\ \text{kip-ft}$$

εt = εty means compression-controlled, so φ = 0.65: φPn = 270.9 kip, φMn = 250.8 kip-ft.

Worked point 3: the tension-controlled limit

The strain state where φ first reaches 0.90 — the endpoint of the transition zone. Same similar triangles, new target strain:

$$c = \frac{0.003}{0.003 + 0.00507}(13.5) = 5.02\ \text{in}, \qquad a = 4.02\ \text{in}$$
$$C_c = 0.85(5)(4.02)(16) = 273.0\ \text{kip}; \quad \text{top: } \varepsilon = 0.00151 \Rightarrow f_s = 43.7\ \text{ksi}, \ C_s = 4(1.00)(43.7-4.25) = 157.7\ \text{kip}$$
$$P_n = 273.0 + 157.7 - 240 = 190.7\ \text{kip}, \qquad M_n = 318.6\ \text{kip-ft}$$

With εt = 0.00507 the section is tension-controlled, φ = 0.90: φPn = 171.6 kip, φMn = 286.8 kip-ft. Notice the reward for ductility — φ jumped from 0.65 to 0.90 between the last two points.

All seven control points

Point c (in) φ φPn (kip) φMn (kip-ft)
Cap φPn,max (worked above)0.65797.7
fs = 0 at extreme tension bars13.500.65622.3169.9
fs = 0.5·fy10.040.65421.9220.1
Balanced point (worked above)7.990.65270.9250.8
Tension-controlled limit (worked above)5.020.90171.6286.8
Pure bending (Pn = 0) — c found iteratively3.250.900.0214.0
Maximum tension Pnt0.90−432.00.0

These values are computed by calcnote's engine and cross-checked against StructurePoint's spColumn-validated table, Wight's Example 11-1, and an independent first-principles baseline — agreement ≤ 0.03% on the directly-tabulated points.

The finished diagram

ACI 318-19 P-M interaction diagram for a 16x16 in tied column with 8-#9 bars: phi-reduced capacity polygon with the axial cap, balanced point, tension-controlled limit, and a factored demand point plotted
The φ-reduced capacity polygon traced through the seven control points, with a sample factored demand (Pu = 200 kip, M2 = 60 kip-ft) plotted inside it. See this exact calc as a live result page.

Sources