RC column design

Worked example: check a reinforced concrete column under axial load and biaxial bending against the N-M interaction surface, with hand verification of the squash load, the balanced point and the Bresler reciprocal.

Problem statement

An interior column in a multi-storey office building carries combined axial compression and biaxial bending from frame action. The column is 400 $\times$ 400 mm and 3.5 m tall between floor levels. The exposure classification is A1 (interior, non-aggressive).

Check the column section capacity under the critical ULS load combination to AS 3600:2018.

Given data

On the cover. AS 3600:2018 Table 4.10.3.2 gives 20 mm for exposure classification A1 at every strength grade, so 35 mm is not that table’s number. It is a deliberate choice above the minimum: Cl 4.10.2 requires cover no less than the bar diameter, which is 24 mm here, and 35 mm leaves margin for the fitment cage and placing tolerance. The distinction matters because the cover sets the bar inset, and the bar inset moves every capacity on this page.

Design actions (critical ULS combination)

$N^*$ is positive in compression, $M^*_x$ positive puts the top face in compression and $M^*_y$ positive puts the left face in compression, so the two together put the peak compression in the top-left quadrant. See Platform — Sign and axis conventions.

These actions include second-order effects (moment magnification per AS 3600 Cl. 10.4 has been applied externally). The eccentricities they imply are worth writing down now, because one of the three biaxial methods is built on them:

Reinforcement details

• 8-N24 bars: $A_{st} = 8 \times 452.4 = 3619$ mm$^2$
• Reinforcement ratio: $\rho = 3619 / (400 \times 400) = 2.26\%$ — inside the Cl 10.7.1 range of 1%—4%
• Bar inset from each face: $35 + 10 + 24/2 = 57$ mm, so the bars sit at $\pm 143$ mm and $0$ about the section centroid
• Depth to the outermost bar layer: $d = 400 - 57 = 343$ mm

Step-by-step solution

Step 1: Define section geometry

Apply the Rectangular template with $b = 400$ mm, $D = 400$ mm.

Step 2: Set materials and member type

On the General tab:

• Design Standard: AS 3600
• Member Type: Column
• Under Materials — Concrete grade: N50 ($f'_c = 50$ MPa), Rebar grade: D500N ($f_{sy} = 500$ MPa)

On the Rebar/PT tab, under Cover:

• Cover: 35 mm all sides

On the Settings tab (the gear icon), under Analysis models:

• ULS model: Rectangular — the AS 3600 Cl 8.1.3 block, which is what the interaction sweep integrates and what the hand checks at the end of this page derive

Member Type is not cosmetic. It selects which checks run and which AS 3600 Section 5 axis distance applies, and it is what makes the Cl 10.1.2 minimum-moment row appear at all. ACS refuses to analyse a section that has none saved rather than assuming “beam” on your behalf. At $N^* = 2500$ kN against the $0.1 f'_c A_g = 800$ kN threshold, the load regime agrees with the declaration, so no mismatch note renders.

Step 3: Place reinforcement

Use the Perimeter Pattern tool:

• Bar diameter: 24 mm (N24)
• Number of bars: 8 (3 per face with corner sharing)

This distributes 8 bars evenly around the perimeter at the cover + fitment inset — three at $d = 57$ mm, two at $d = 200$ mm, three at $d = 343$ mm.

Then the fitments. This section is a convex rectangle, so the whole outline is a single cell and the Cell Tie tool opens its dialog with no cell to pick first. In the dialog:

• Stirrup Diameter: 10 mm (N10)
• Grade: D500N
• Spacing (mm): 300
• Hook ends: 135° both ends

There is no “legs” input. $A_{sv}$ is derived from the tie geometry you drew — a closed perimeter tie on a rectangle crosses any horizontal cut twice. Nothing on this page depends on it (the combination carries no shear), but the cage has to exist for the bars to be placed against.

Step 4: Enter the design combination

Loads live on the Actions tab, under the ULS combinations heading. Add one row:

The $k_\phi$ column is the one to read twice. AS 3600 Amd 2:2021 Table 2.2.2 has two separate entries for a column, and they do not carry the same $\phi$:

• item (a)(ii), pure axial compression — a fixed $\phi = 0.65$, with no $k_\phi$ in it
• item (d), bending with axial compression — a floor of $\phi_o = 0.65 \, k_\phi$

So $k_\phi$ sets $\phi$ everywhere on the interaction curve except its squash endpoint. The grid offers 12/13 and 1.0, and 12/13 is both the default it renders and the value it saves into the design. 1.0 is available only where Cl 10.3 makes the column short and $Q/G \geq 0.25$; taking it otherwise is unconservative. This example designs at 12/13, which gives $\phi_o = 0.65 \times 12/13 = 0.60$.

API and MCP callers get no default at all: an AS 3600 ULS combination with no $k_\phi$ saved is refused with k_phi_not_set rather than analysed at 0.60 silently.

Because the member type is Column, AS 3600 Cl. 10.1.2 imposes a minimum design moment about each principal axis:

Both entered moments exceed it, so the floor does not govern and the results below are for the moments as entered. Where a floor does govern, the panel says so explicitly (”$M^*_y$ taken as … — minimum design moment per AS 3600 Cl. 10.1.2”) rather than substituting silently.

Step 5: Review the interaction results

Switch to the ULS results tab and find the Interaction Diagram section.

That section is a set of readouts, not a plot. The plotted envelope lives on the canvas — the Interaction Surface tab, whose 2D / 3D toggle sits at the bottom of the canvas. Step 6 covers both views.

What the panel reports

The panel shows two axial end-caps and the biaxial utilisation checks. It does not report a uniaxial $N$-$M_x$ utilisation: that row was removed deliberately, because $M^*_y$ was never one of its arguments and it therefore read lower than the true demand whenever there was out-of-plane bending. Biaxial (rigorous) is the utilisation for this envelope.

All three are below 1.0, so the section is adequate. The Bresler reciprocal governs at 0.905, and the order the three come in is the subject of the Discussion below — it is not the order you might expect.

The panel’s contour label rounds the exponent to one decimal, so for this section it renders “Bresler contour (α=1.5)”. That 1.5 is a coincidence of rounding, not the method’s constant — see the α_n note under Hand calculation verification below.

Features of the curve

These are not on screen in the editor. They come from the design report, section N-M Interaction Diagram, whose Key Points table lists each labelled point with the $\phi$ it was factored by, above a chart plotting the nominal and design $N$-$M_x$ curves together. The canvas plots the design envelope only — there is no nominal surface in the engine, and the 2D curve marks no point but $\phi N_{u0}$.

The nominal column is the unfactored capacity; the design column carries $\phi$ from Table 2.2.2 at each point.

Two things to notice in that table, both of them $\phi$ rather than mechanics:

• The squash load is the same under either $k_\phi$ class, because Table 2.2.2(a)(ii) pins $\phi = 0.65$ at the pure-axial endpoint and carries no $k_\phi$.
• Pure bending takes $\phi = 0.85$, not 0.60 — Table 2.2.2(b), because at $N = 0$ there is no compression-controlled floor to apply. Its ductility parameter is $k_{uo} =$ 0.218, comfortably under the 0.36 limit.

$N^* = 2500$ kN sits above the nominal balanced axial load of 2241 kN, so the section is compression-controlled: the concrete crushes before the tension steel yields, and adding reinforcement buys less than it would on a beam.

Step 6: View the interaction envelope on the canvas

Click the Interaction Surface tab on the canvas. The toggle at the bottom of the canvas switches between two views of the same envelope.

3D ($N$-$M_x$-$M_y$ surface). Left-drag to rotate the surface and see the capacity envelope from different angles; scroll to move in and out, and middle-drag (or Shift / Ctrl / ⌘ + left-drag) to pan. The $M_x$-$M_y$ contour at the applied axial load level shows the remaining moment capacity in all directions — and it is where the difference between the three methods below becomes visible rather than numerical.

2D (one curve, one combination). The curve is swept in the resultant-moment plane at this combination’s own bending angle $\theta$, so the x-axis is $M_R = \sqrt{M_x^2 + M_y^2}$ and not $M_x$. The design point’s radial position against the curve is the printed utilisation, by construction. One combination at a time is deliberate: each has its own $\theta$ and therefore its own curve, and a point overlaid on a curve from another plane could read comfortably inside an envelope that is not its own.

The design-point marker is coloured by verdict, not by identity: green when the utilisation is at or below 1.0, red when it is above, and a neutral colour when no ratio was returned at all. This combination passes at 0.747, so the marker renders green — a red marker here would mean the section had failed, not that you had found the design point.

Results summary

The section is adequate as detailed: 400 $\times$ 400 N50, 8-N24, N10-300 fitments. The Bresler reciprocal governs at 0.905, which leaves about 10% reserve — materially tighter than the rigorous surface’s 0.747 suggests.

Discussion

The three methods do not bracket each other

The ordering is contour < rigorous < reciprocal, and it holds under both $k_\phi$ classes. The Bresler methods are not “slightly conservative” relative to the rigorous surface: one of them is markedly less conservative and the other markedly more. They are different approximations to the same surface, built from different read-offs, and for a square section with symmetric reinforcement at this eccentricity ratio they land 70% apart.

That matters because a designer who checks the contour row, sees 0.533, and stops has read the most generous of the three. Design to the governing row — here the reciprocal at 0.905 — and use the rigorous surface as the physical answer the two approximations are trying to estimate.

What $k_\phi$ is worth

Every utilisation in the table above falls on moving from 12/13 to 1.0 — the two biaxial rows by about 11.5%, the Bresler contour row by 15.2% — because $\phi_o$ rises from 0.60 to 0.65 across the whole compression-controlled region of the curve. That is not a rounding matter, and 1.0 is not available by default: Cl 10.3 has to make the column short and $Q/G$ has to reach 0.25. If neither has been established, 12/13 is the classification, and the app saves it rather than leaving the question open.

Other observations

• The reinforcement ratio of 2.26% is within the Cl 10.7.1 range for columns (1%—4%).
• For a symmetric section with symmetric reinforcement the Bresler methods are cheap sanity checks; for asymmetric sections the rigorous 3D surface is the only one of the three that reflects the actual geometry.

If the utilisation were too high, in order of effectiveness for a compression-dominated section:

• Increase the column size — $N_{u0}$ scales with $A_g$ and the lever arms grow with $D$
• Increase the concrete grade — $N_{u0}$ scales close to directly with $f'_c$
• Add reinforcement (up to the 4% Cl 10.7.1 maximum) — the weakest of the three here, because the section is above the balanced point and the extra steel is not yielding

Hand calculation verification

Squash load ($\phi N_{u0}$)

AS 3600:2018 Cl 10.6.2.2. Note the uniform-compression intensity factor $\alpha_1$, which is a different quantity from the bending block’s $\alpha_2$:

Matches the reported 8455.7 / 5496.2 kN. The $\phi$ here is Table 2.2.2(a)(ii)‘s fixed 0.65 — $k_\phi$ does not enter.

Balanced point

The balanced state has the extreme tension bar reaching $\varepsilon_{sy} = 0.0025$ as the extreme compression fibre reaches $\varepsilon_{cu} = 0.003$:

With the Cl 8.1.3 block at $f'_c = 50$: $\alpha_2 = 0.85 - 0.0015 \times 50 = 0.775$, $\gamma = 0.97 - 0.0025 \times 50 = 0.845$, so $a = \gamma c = 158$ mm. Taking moments about the gross centroid, with the three bar rows at $y = +143$, $0$ and $-143$ mm:

Cc   = 0.775 x 50 x 400 x 158.1 = 2451 kN   at arm 121 mm
row 1 (3 bars, y = +143, compression, net of displaced concrete)  = +514 kN at arm +143 mm
row 2 (2 bars, y = 0, elastic tension)                            =   -38 kN at arm 0
row 3 (3 bars, y = -143, yielded tension)                         =  -679 kN at arm -143 mm

N = 2451 + 514 - 38 - 679 = 2248 kN
M = 2451(0.121) + 514(0.143) + 679(0.143)  = 467 kN.m

Against the reported nominal balanced point of (2241.0, 466.6) that is 0.3% on both ordinates — two independent constructions of the same state.

Pure bending ($\phi M_{uo}$)

At $N = 0$ the solve returns $c = 74.8$ mm ($k_{uo} = 74.8/343 = 0.218$), so $a = 0.845 \times 74.8 = 63.2$ mm. The neutral axis sits at $y = 200 - 74.8 = 125$ mm, which puts the $y = +143$ row in compression at a strain of $0.003 \times (143-125)/74.8 = 0.00071$ — elastic, at $\sigma = 143$ MPa:

Cc                  = 0.775 x 50 x 400 x 63.2       = 980 kN  at arm +168 mm
row 1 (compression, net of displaced concrete)      = +141 kN  at arm +143 mm
row 2 (2 bars, y = 0, yielded tension)              = -452 kN  at arm 0
row 3 (3 bars, y = -143, yielded tension)           = -679 kN  at arm -143 mm

N = 980 + 141 - 452 - 679 = -10 kN     (~0 as required)
M = 980(0.168) + 141(0.143) + 679(0.143) = 282 kN.m

Against the reported $M_{uo} = 283.6$ kN.m that is 0.5%. $\phi M_{uo} = 0.85 \times 283.6 = 241.0$ kN.m — and note again that the 0.85 is Table 2.2.2(b), because at $N = 0$ the item-(d) floor does not apply.

Bresler reciprocal

The reciprocal is a constant-eccentricity construction, which is what makes it a hand check you can actually reproduce. $N_{ux}$ and $N_{uy}$ are the uniaxial capacities on the radial line through the design point in each plane — at $e_x = 48$ mm and $e_y = 32$ mm respectively — not the capacities read off the diagram at constant moment. Reading them at constant moment gives numbers 15%—20% high and an unconservative utilisation.

From the envelope at those two eccentricities, with the squash load from above:

That confirms the reciprocal row — 0.905 — and nothing else. Comparing a reciprocal hand check against the rigorous surface’s 0.747 and calling the agreement confirmation would be a category error: they are different constructions and they are supposed to differ.

Bresler contour exponent

The contour exponent is computed per AS 3600:2018 Cl 10.6.4, not taken as a constant. The clause pins $\phi = 0.65$ inside its own definition, so $\alpha_n$ does not move with $k_\phi$:

with the uniaxial moment capacities at $N^*$ both 247.6 kN.m (equal, because the section is square and symmetrically reinforced). Matches the reported 0.533.

A fixed $\alpha = 1.5$ is the ACI 318 value. Under AS 3600 — and under EN 1992 — the exponent is a function of the axial utilisation, and it happens to round to 1.5 for this particular column at this particular $N^*$. It will not for the next one.

Related pages

• M-N interaction theory — interaction diagram background
• Section analysis — overview of all analysis types
• RC beam example — beam design with flexure and shear
• Glossary — definitions for M-N interaction, ductility parameter, and ULS