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Columns: two axes, AISC and eccentric loads

The Buckling topic explains the Euler load, the effective length and the Johnson formula for a single axis. A real column has two axes with different stiffness and may be braced about only one, it is designed with a code (here AISC 360) and it almost never takes the load exactly on its centroid. This topic covers those three things, which are what the Columns tool solves.

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Two axes, two effective lengths

A section has a second moment of area about each principal axis: in an I-section, $I_{x}$ (the strong axis, the one of the depth) is much larger than $I_{y}$ (the weak axis). The column can buckle about either one, each with its own supports and bracing, so you compute the slenderness of both and the larger one governs:

\frac{K _{x} L _{x}}{r _{x}} y \frac{K _{y} L _{y}}{r _{y}} r = \frac{I}{A}

A brace (a beam, a wall or a tie that prevents lateral movement at some height) splits the buckling length in that plane. Since the weak axis usually governs, buildings brace exactly that axis: with a brace at mid-height its effective length is halved and its Euler load is multiplied by four.

I-section pinned at both ends, with a brace at mid-height on the weak axis only (example 1).

A brace only works if it is stiff in that direction and tied to something that does not move. A brace that restrains bending about x–x does nothing for buckling about y–y: each axis is checked with what restrains it.

Steel columns per AISC 360

The Euler and Johnson formulas assume a perfect column. Real shapes have small initial crookedness and residual stresses from rolling and welding, which make them yield earlier. The AISC 360 specification (chapter E) includes them through a curve calibrated with tests:

F_{e} = \frac{π ^{2} E}{( K L / r ) ^{2}} F_{cr} = ⎩ ⎨ ⎧ 0.65 8^{F_{y} / F_{e}} F_{y} 0.877 F_{e} \frac{K L}{r} \leq 4.71 \frac{E}{F _{y}} \frac{K L}{r} > 4.71 \frac{E}{F _{y}}

The nominal strength is $P_{n} = F_{cr} A$ . Design does not use a single safety factor but one of these two formats:

Format	Requirement	Load used
LRFD	$ϕ P_{n} \geq P_{u}$ , $ϕ = 0.90$	factored $P_{u}$ (for example 1.2D + 1.6L)
ASD	$P_{n} /Ω \geq P_{a}$ , $Ω = 1.67$	service $P_{a}$

Column curves for A36 steel. AISC stays below Johnson over the whole range: that is the price of real imperfections.

In the long-column range AISC takes 87.7 % of Euler ( $0.877 F_{e}$ ) to cover the initial crookedness. In the short-column range the exponential drops somewhat faster than Johnson's parabola. Both curves are used in class, but for structural steel a code asks for the AISC one.

Eccentric load: the secant formula

If the load is offset a distance $e$ from the centroid, the column is no longer «straight until it buckles»: the moment $P e$ bends it from the very start, and that deflection increases the lever arm of the load. Solving the elastic-curve equation with that moment, the maximum stress (at mid-height of a pinned column) is

σ_{m \overset{a}{ˊ} x} = \frac{P}{A} [1 + \frac{e c}{r ^{2}} sec (\frac{L _{e}}{2 r} \frac{P}{E A})]

$\frac{ec}{r ^{2}}$ is the eccentricity ratio: with $e = 0$ it gives $σ = P / A$ .
The secant grows without bound as $P$ approaches the Euler load: buckling is still the ceiling.
The relation between $P$ and $σ_{max}$ is not linear. That is why the safety factor is applied to the load: you find the load $P_{y}$ that takes $σ_{max}$ to $F_{y}$ and compare it with the actual one, $n = P_{y} / P$ .

Bar of example 3: it bends from the start, and its maximum stress grows faster and faster with the load, up to the Euler asymptote (269 kN, dotted line).

Higher modes and where to put a brace

Besides the first mode, a column has others: the second is a full wave (two half-waves) with a load four times larger, the third three half-waves with nine times. A brace at the point where the second mode does not move (the middle, in a pinned column) rules out the first mode and forces the column to buckle in the second. Putting it elsewhere helps less: the longest half-wave governs.

The tool draws the first three modes of each axis («Mode» buttons), with the inflection points and the length of each half-wave.

Common mistakes

Checking only the weak axis when that axis is braced: then the strong one may govern.
Comparing $P_{u}$ (factored) with $P_{n} /Ω$ , or $P_{a}$ (service) with $ϕ P_{n}$ : each format goes with its own load.
Applying the safety factor to the stress in the secant formula: with an eccentric load the response is not linear.
Measuring the eccentricity from the edge instead of the centroid.
Using the radius of gyration of one axis with the effective length of the other.

Worked examples

Example 1IPE 300 braced on the weak axis (AISC)

An IPE 300 column of S275 steel (

E = 210

GPa,

F_{y} = 275

MPa) is 6 m long and pinned at both ends. About the weak axis it has a brace at mid-height. With

A = 5381

mm²,

I_{x} = 8356 \times 1 0^{4}

mm⁴ and

I_{y} = 603.8 \times 1 0^{4}

mm⁴, find the LRFD design strength and the utilization for

P_{u} = 250

kN.

Solution

Radii of gyration: $r_{x} = 8356 \times 1 0^{4} /5381 = 124.6$ mm and $r_{y} = 603.8 \times 1 0^{4} /5381 = 33.50$ mm.
Slenderness: strong axis $6000/124.6 = 48.15$ ; weak axis, with the brace, $3000/33.50 = 89.56$ . The weak axis governs.
$F_{e} = \frac{π ^{2} \cdot 210 000}{89.5 6 ^{2}} = 258.4$ MPa. Since $89.56 \leq 4.71 210 000/275 = 130.2$ : $F_{cr} = 0.65 8^{275/258.4} \cdot 275 = 176.1$ MPa.
$P_{n} = 176.1 \cdot 5381 = 947.9$ kN and $ϕ P_{n} = 0.9 \cdot 947.9 = 853.1$ kN. Utilization: $250/853.1 = 0.293$ .
Without the brace, $K L / r_{y} = 6000/33.50 = 179$ $F_{cr}$ would become $0.877 F_{e} = 56.7$ MPa, almost a third: the brace is what makes the shape useful.

The y–y axis governs:

ϕ P_{n} = 853.1

kN and utilization

0.293

(OK).

Solve it in Simuveliq

In Engineering → Columns, in Examples, «Load column» → IPE shape braced on the weak axis (AISC). Check what it sets: IPE 300, S275, 6 m, pinned at the bottom and top about both axes and a Brace at 3 m about the y axis only.
With the AISC 360 (LRFD / ASD) method, the cards give Design strength (LRFD) 853.1 kN (y–y axis) and Utilization 0.293. The «Results per axis» table shows both slenderness ratios, and the note below it, $P_{n} /Ω = 567.6$ kN for ASD.
Delete the brace and watch how the y–y buckled shape and the strength change.

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Example 2Steel box: Johnson versus AISC

A 100 × 150 × 6 mm box of A36 steel (

E = 200

GPa,

F_{y} = 250

MPa), 4 m long and pinned at both ends, carries 200 kN. Find the critical load with Johnson, the safety factor and the AISC strength.

Solution

$A = 100 \cdot 150 - 88 \cdot 138 = 2856$ mm². Weak axis (the 100 width): $I_{y} = \frac{150 \cdot 10 0 ^{3} - 138 \cdot 8 8 ^{3}}{12} = 4.663 \times 1 0^{6}$ mm⁴ and $r_{y} = 40.41$ mm.
$K L / r_{y} = 4000/40.41 = 98.99 < C_{c} = 125.7$ : Johnson. $σ_{cr} = 250 - \frac{1}{200 000} (\frac{250 \cdot 98.99}{2 π})^{2} = 172.4$ MPa.
$P_{cr} = 172.4 \cdot 2856 = 492.5$ kN and $n = 492.5/200 = 2.46$ . (The strong axis, with $K L / r = 71.85$ , would give 597.3 kN.)
AISC: $F_{e} = \frac{π ^{2} \cdot 200 000}{98.9 9 ^{2}} = 201.4$ MPa, $F_{cr} = 0.65 8^{250/201.4} \cdot 250 = 148.7$ MPa, $P_{n} = 424.7$ kN (14 % less than Johnson) and $ϕ P_{n} = 382.2$ kN.

Johnson:

P_{cr} = 492.5

kN and

n = 2.46

(y–y axis). AISC:

P_{n} = 424.7

kN,

ϕ P_{n} = 382.2

kN.

Solve it in Simuveliq

In Examples, «Load column» → Box pinned at both ends.
With the «Euler + Johnson» method, the cards give Critical load 492.5 kN (y–y axis) and Factor of safety 2.46.
The «Results per axis» table has Euler, Johnson and AISC for each axis at once: about y–y, 575.3, 492.5 and 424.7 kN (φ 382.2). Switch the method to AISC 360 to see its step-by-step solution.

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Example 3Bar with an eccentric load (secant)

A 1.5 m solid bar, 50 mm in diameter, pinned at both ends and made of A36 steel, takes 60 kN with an eccentricity of 10 mm. Find the maximum stress, the load that causes yielding and the safety factor.

Solution

$A = π \cdot 2 5^{2} = 1963.5$ mm², $r = d /4 = 12.5$ mm and $c = 25$ mm: eccentricity ratio $\frac{ec}{r ^{2}} = \frac{10 \cdot 25}{12. 5 ^{2}} = 1.6$ .
Secant argument: $\frac{1500}{2 \cdot 12.5} \frac{60 000}{200 000 \cdot 1963.5} = 0.7417$ rad, $sec 0.7417 = 1.356$ .
$σ_{max} = \frac{60 000}{1963.5} (1 + 1.6 \cdot 1.356) = 30.56 \cdot 3.170 = 96.87$ MPa.
The load that takes $σ_{max}$ to 250 MPa has no closed form: it is found by trial (or bisection) and comes out as $P_{y} = 118.0$ kN. Safety factor: $n = 118.0/60 = 1.97$ .
Dividing $250/96.87 = 2.58$ would give a safety factor 30 % larger than the real one: that is why it is applied to the load.

σ_{max} = 96.87

MPa,

P_{y} = 118.0

kN and

n = 1.97

Solve it in Simuveliq

In Examples, «Load column» → Bar with an eccentric load (secant): method «Secant (eccentric load)» and Eccentricity e = 10 mm about the x axis.
The cards give Load that causes yielding 118 kN and Factor of safety 1.97; the step-by-step solution shows the σ_max of 96.87 MPa under the actual load.

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Columns: two axes, AISC and eccentric loads

Two axes, two effective lengths

Steel columns per AISC 360

Eccentric load: the secant formula

Higher modes and where to put a brace

Common mistakes

Worked examples

Example 1IPE 300 braced on the weak axis (AISC)

Solution

Solve it in Simuveliq

Example 2Steel box: Johnson versus AISC

Solution

Solve it in Simuveliq

Example 3Bar with an eccentric load (secant)

Solution

Solve it in Simuveliq

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