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Reinforced Concrete Column Calculator — P–M Interaction Diagram (ACI 318-19)

Lay out the longitudinal bars in a rectangular or circular column, choose ties or a spiral, and get the full axial load–moment interaction diagram from strain compatibility. Your factored load is plotted on the design curve, with the moment capacity at that axial load, the balanced point, the axial cap φP_n,max, the reinforcement ratio, tie spacing or spiral pitch, bar spacing and a slenderness screen.

Section (US units: in, psi, kip)

Shape

Transverse reinforcement

Width, b (⊥ to bending)in

Depth, h (in bending direction)in

Diameter, Din

Bars on each b-face (top & bottom)incl. corners

Bars on each h-face (sides)incl. corners

Number of barsequally spaced, one bar at the compression face

Bar size

Tie / spiral bar

Clear cover to tiein

Max aggregate sizein

f′_cpsi

f_ypsi

f_yt (spiral)psi

Factored loads

Axial load, P_ukip (compression +)

Moment, M_ukip·ft (magnified, if slender)

Slenderness screen (ACI 318-19 §6.2.5)

Frame

Effective length factor, k—

Unsupported length, ℓ_uft

M₁/M₂negative = single curvature

Results

Gross area, A_g—in²

Steel area, A_st (bars)—in² · — bars

ρ_g (§10.6.1.1)—

P₀ = 0.85f′_c(A_g − A_st) + f_yA_st—kip

φP_n,max—kip

Balanced point P_b / M_b (nominal)—kip / — kip·ft

Pure bending, φM_n (P = 0)—kip·ft

Pure tension, φP_nt—kip

φM_n at P_u—kip·ft

Demand / capacity, M_u/φM_n——

Strength check—

Transverse steel—max spacing / pitch — in

Clear spacing, long. bars (§25.2.3)—

Slenderness—

Solid = design strength (φP_n, φM_n) with the φP_n,max cap; dashed = nominal strength. The orange dot is the factored load. Points inside the solid curve pass.

For each neutral-axis depth c, strain compatibility (ε_cu = 0.003) gives every bar's stress, and:

$$ P_n = 0.85 f'_c A_{c}(a) + \sum A_{s,i}\, f_{s,i}, \qquad M_n = 0.85 f'_c A_{c}\left(\tfrac{h}{2} - \bar{y}_c\right) + \sum A_{s,i}\, f_{s,i}\left(\tfrac{h}{2} - d_i\right) $$

$$ f_{s,i} = E_s\,\varepsilon_{s,i} \le f_y, \qquad \varepsilon_{s,i} = 0.003\,\frac{c - d_i}{c}, \qquad a = \beta_1 c $$

Upper limit on axial strength (ACI 318-19 Table 22.4.2.1):

$$ P_{n,max} = 0.80 P_0 \;(\text{ties}), \quad 0.85 P_0 \;(\text{spirals}), \qquad P_0 = 0.85 f'_c (A_g - A_{st}) + f_y A_{st} $$

A_c(a) concrete area within depth a of the compression face (a circular segment for round columns) · ȳ_c its centroid · d_i depth of bar i · bars inside the stress block have 0.85f′_c deducted for displaced concrete · φ from the strain ε_t at the extreme tension bar.

How the interaction diagram is built

A column carries axial load and bending together, and the two interact: a little axial compression actually raises the moment capacity (it pre-compresses the tension side), while a lot of axial load lowers it. ACI 318 handles this with the same assumptions used for beams — plane sections, ε_cu = 0.003, the Whitney block and elastic–plastic steel. The calculator picks a neutral-axis depth c, finds the strain and stress in every bar, adds the concrete block, and sums forces (P_n) and moments about the section centroid (M_n). Repeating this for about 400 values of c, from deep enough that the whole section is compressed to shallow enough that every bar yields in tension, traces the nominal curve.

The design curve multiplies each point by φ. φ depends on the net tensile strain ε_t in the bar farthest from the compression face: 0.65 for tied columns (0.75 for spirals) when ε_t ≤ ε_ty, rising linearly to 0.90 at ε_t = ε_ty + 0.003. The top of the design curve is cut off flat at φP_n,max. For a given P_u, the moment capacity is read from the design curve by solving for the c at which φP_n = P_u exactly.

Key points on the interaction diagram
Point	Neutral axis	Strain condition	φ (tied / spiral)
Pure compression P₀	c → ∞	uniform 0.003	0.65 / 0.75, capped at 0.80P₀ / 0.85P₀
Zero tension (ε_t = 0)	c = d_t	extreme bar unstressed	0.65 / 0.75
Balanced	c_b = 0.003d_t/(0.003 + ε_ty)	ε_t = ε_ty	0.65 / 0.75
Tension-controlled limit	c = 0.003d_t/(0.006 + ε_ty)	ε_t = ε_ty + 0.003	0.90
Pure bending	c where P_n = 0	—	usually 0.90
Pure tension	c → 0	all bars yield	0.90, φP_nt = 0.9f_yA_st

Detailing checks included

ACI 318-19 column detailing (nonseismic)
Item	Requirement	Section
Longitudinal steel ratio	0.01 ≤ ρ_g ≤ 0.08	10.6.1.1
Minimum number of bars	4 (rectangular/circular ties), 3 (triangular ties), 6 (spirals)	10.7.3.1
Clear spacing between bars	≥ max(1.5 in, 1.5d_b, 4/3 d_agg)	25.2.3
Tie size	#3 for #10 and smaller; #4 for #11, #14, #18, bundled	25.7.2.2
Tie spacing	≤ min(16d_b, 48d_tie, least dimension)	25.7.2.1
Spiral ratio	ρ_s ≥ 0.45(A_g/A_ch − 1)f′_c/f_yt	25.7.3.3
Spiral clear pitch	1 in to 3 in	25.7.3.1

For a circular spiral, the maximum pitch comes from ρ_s = 4A_sp/(D_cs), where D_c is the core diameter to the outside of the spiral. The calculator reports the pitch that satisfies ρ_s,min, limited to 3 in clear.

Worked example

16 × 20 in tied column, 8-#9 bars, bending about the 20-in direction

Given: f′_c = 5,000 psi (β₁ = 0.80), Grade 60, #3 ties, 1.5 in cover. Three bars on each 16-in face plus one bar at mid-depth on each side: layers at d = 2.44, 10.0 and 17.56 in with 3, 2 and 3 bars. P_u = 500 kip, M_u = 200 kip·ft.

A_g = 320 in², A_st = 8.00 in², ρ_g = 2.5%. P₀ = 0.85(5)(320 − 8) + 60(8) = 1,326 + 480 = 1,806 kip; φP_n,max = 0.65 × 0.80 × 1,806 = 939 kip.

Balanced: c_b = 0.003(17.56)/(0.003 + 0.00207) = 10.39 in, a = 8.31 in. C_c = 0.85(5)(16)(8.31) = 565.4 kip. Top bars: ε = 0.00230 > ε_y → 3(1.0)(60 − 4.25) = 167.3 kip. Middle bars: ε = 0.00011 → 2(1.0)(3.3) = 6.6 kip. Bottom bars: −3(60) = −180 kip.

P_b = 565.4 + 167.3 + 6.6 − 180 = 559 kip; M_b = [565.4(10 − 4.16) + 167.3(10 − 2.44) + 180(17.56 − 10)] / 12 = 494 kip·ft.

φP_b = 0.65 × 559 = 363 kip, so P_u = 500 kip lies above the balanced point: the section is compression-controlled and φ = 0.65. Solving 0.65P_n(c) = 500 gives c = 12.4 in (ε_t = 0.0012 < ε_ty) and φM_n = 297 kip·ft.

Slenderness (nonsway, k = 1.0, ℓ_u = 12 ft, M₁/M₂ = −0.5): kℓ_u/r = 144/(0.3 × 20) = 24 ≤ 34 + 12(−0.5) = 28 → short column. Ties: min(16 × 1.128, 48 × 0.375, 16) = 16 in.

M_u/φM_n = 200/297 = 0.67 — the column is adequate for P_u = 500 kip, M_u = 200 kip·ft.

Limits of this tool

Uniaxial bending only. For biaxial bending, check each axis and combine with the Bresler reciprocal load method or a full biaxial analysis.
Slenderness is screened, not designed: if the column is slender, magnify M_u per §6.6.4 (nonsway) or §6.6.4.6 (sway), or run a second-order analysis, and enter the magnified moment.
Circular columns place one bar at the extreme compression face. Rotating the bar pattern changes the capacity slightly; the worst orientation can be a few percent lower.
Nonseismic detailing. Special moment frame columns need the confinement and spacing rules of ACI 318-19 Ch. 18.
Only bar area is counted for A_st; ties and spirals are not included in the axial strength.

Frequently asked questions

How do you calculate the axial capacity of a concrete column?

The concentric strength is P0 = 0.85f′c(Ag − Ast) + fy·Ast. ACI 318-19 Table 22.4.2.1 caps the usable nominal strength at 0.80P0 for tied columns and 0.85P0 for spirally reinforced columns, to cover accidental eccentricity, and the design strength is φ times that with φ = 0.65 (tied) or 0.75 (spiral). For a 16 × 20 in tied column with 8-#9 bars and f′c = 5,000 psi, P0 = 1,806 kip and φPn,max = 939 kip.

What is a column interaction diagram?

A plot of every combination of axial load P and moment M that just brings the column to its strength limit (concrete strain 0.003). It is generated by sweeping the neutral-axis depth c from very large (pure compression) to very small (pure tension), computing Pn and Mn at each step. Multiplying by φ, which depends on the strain in the extreme tension bar, and capping at φPn,max gives the design diagram. A load point inside the design curve is adequate.

What is the balanced point?

The point where the extreme tension steel just reaches yield (εt = εty = fy/Es) as the concrete reaches 0.003. Above it, failure is compression-controlled and φ is 0.65 (tied) or 0.75 (spiral). Below it, φ rises linearly to 0.90 at εt = εty + 0.003. It is also roughly where a column has its largest moment capacity.

What are the minimum and maximum reinforcement ratios for columns?

ACI 318-19 §10.6.1.1 requires 0.01Ag ≤ Ast ≤ 0.08Ag. In practice designers rarely exceed about 4% because lap splices double the steel at the splice. At least four bars are required with rectangular or circular ties, three with triangular ties and six with spirals (§10.7.3.1).

How far apart can column ties be?

Vertical tie spacing may not exceed the smallest of 16 longitudinal bar diameters, 48 tie bar diameters and the least column dimension (§25.7.2.1). Use #3 ties for #10 and smaller bars and #4 ties for #11, #14, #18 and bundled bars. Every corner bar and every alternate bar must be laterally supported, and no unsupported bar may be more than 6 in clear from a supported one.

When is a concrete column slender?

Per ACI 318-19 §6.2.5, slenderness may be neglected in a nonsway frame when kℓu/r ≤ 34 + 12(M1/M2) and ≤ 40 (M1/M2 negative for single curvature), and in a sway frame when kℓu/r ≤ 22. r may be taken as 0.30h for rectangles and 0.25D for circles. Otherwise the moments must be magnified (§6.6.4) before using the interaction diagram.

References: ACI 318-19, Building Code Requirements for Structural Concrete — §6.2.5, §6.6.4, §10.6.1, §10.7.3, Table 21.2.2, §22.2, Table 22.4.2.1, §25.2.3, §25.7.2–25.7.3. Wight, J.K., Reinforced Concrete: Mechanics and Design, 8th ed., Pearson (2022), ch. 11–12. Bar data per ASTM A615.

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How to build a column interaction diagram — step-by-step guide
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