calcnote ACI 318-19
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Maximum axial strength Pn,max in ACI 318-19: 0.80Po tied, 0.85Po spiral

A concrete column is never allowed its full squash load. ACI 318-19 §22.4.2.1 caps the nominal axial strength Pn at Pn,max, a fraction of the pure compression strength Po set by Table 22.4.2.1. On an interaction diagram the cap is the flat top that cuts off the pointed peak of the curve.

Pure compression strength Po (§22.4.2.2)

$$P_o = 0.85\,f'_c\,(A_g - A_{st}) + f_y\,A_{st}$$

The concrete term uses the net concrete area Ag − Ast, so the steel is not counted twice. The 0.85 here is the concrete stress factor, not a cap.

The cap (Table 22.4.2.1)

Transverse reinforcement Pn,max φ (Table 21.2.2) φPn,max
Tied0.80Po0.650.52Po
Spiral0.85Po0.750.6375Po

The φ values are those for compression-controlled sections, which is where the cap sits in practice. Written out for a tied column, the design cap is

$$\phi P_{n,max} = 0.65 \times 0.80\,\big[0.85\,f'_c\,(A_g - A_{st}) + f_y\,A_{st}\big]$$

Why the cap exists

No real column is loaded exactly through its centroid. Out-of-plumb forms, bar placement tolerances and unbalanced framing always add some eccentricity. The cap accounts for that accidental eccentricity: it keeps the design from relying on the concentric peak that a real column never reaches.

Spiral columns get the higher factor, and a higher φ, because a conforming spiral confines the core: after the cover spalls the column keeps carrying load and deforms a good deal before failing, where a tied column fails more abruptly. That behavior is only assured when the spiral meets §25.7.3, which is why both rewards are tied to conformance.

Worked example: 20 in column, spiral vs tied

D = 20 in, 8-#10 bars (Ast = 10.16 in²), f′c = 5,000 psi, Grade 60, the StructurePoint column calcnote is verified against. Ag = π(20)²/4 = 314.16 in².

$$P_o = 0.85(5)(314.16 - 10.16) + 60(10.16) = 1292.0 + 609.6 = 1901.6\ \text{kip}$$
$$\text{Spiral:}\quad \phi P_{n,max} = 0.75 \times 0.85 \times 1901.6 = 1212.3\ \text{kip}$$
$$\text{Tied:}\quad \phi P_{n,max} = 0.65 \times 0.80 \times 1901.6 = 988.8\ \text{kip}$$

StructurePoint's published solution prints Po = 1902 kip and φPn,max = 1212.3 kip for the spiral column; calcnote's engine returns the same. The same section with ties instead of a spiral is capped at 988.8 kip, so the conforming spiral is worth 22.6% more usable axial capacity. In calcnote a demand of Pu = 1,200 kip with a nominal 1 kip-ft moment passes this spiral column at an axial D/C of 0.99 and fails the tied version at 1.21.

For a rectangular tied section, calcnote's live demo column (16×16 in, 8-#9, 5,000 psi, Grade 60) has Po = 1534.0 kip, Pn,max = 1227.2 kip and φPn,max = 797.7 kip.

What calcnote does

calcnote computes Po from the entered section and applies 0.80 with φ = 0.65 for tied columns, or 0.85 with φ = 0.75 for circular columns with a spiral that passes its §25.7.3 checks. The capped value is the top of the interaction diagram the demand point is checked against, and it appears as its own line in the result's equation list.

Sources