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Buckling of compression members

A member in compression can fail long before its material reaches yield: it suddenly bends sideways. That is buckling, a stability failure rather than a strength failure, and it depends mostly on the length, the shape of the section and how the ends are held. Here you will see the Euler critical load, the effective length, slenderness and the Johnson formula for short columns, with examples that open in the Buckling tab of the beam simulator.

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What buckling is

Think of a plastic ruler pushed at both ends. With a small force it stays straight and, if you nudge it sideways, it comes back: the equilibrium is stable. At a certain load, the critical load $P_{cr}$ , the ruler no longer comes back: any disturbance bends it more and more. From then on the straight position is unstable.

For an ideal column (straight, elastic material and the load on the centroid) pinned at both ends, the slightly bent column is in equilibrium when the moment $P y$ produced by the load is balanced by its bending stiffness:

E I \frac{d ^{2} y}{d x ^{2}} + P y = 0 ⟹ y = A sin \frac{π x}{L} con P_{cr} = \frac{π ^{2} E I}{L ^{2}}

The equation only has a bent solution ( $A \neq = 0$ ) for certain values of $P$ ; the smallest is the Euler load. It does not depend on the strength of the material, only on its stiffness $E$ : a high-strength steel buckles at the same load as an ordinary one if they have the same section and length.

The column buckles about its weakest axis: if both planes are held the same way, the smallest second moment of area governs. The simulator checks both planes (x–y and x–z) and marks the governing one.

Effective length

With other supports the buckled shape changes. Every case reduces to the pinned column by using the effective length $L_{e} = K L$ : the distance between two points of zero moment (inflection points) of the buckled shape, which behaves like a pinned column.

P_{cr} = \frac{π ^{2} E I}{( K L ) ^{2}}

Exact buckled shapes of the four classic columns. In the fixed–free column there is no second inflection point inside the column: the shape is half a half-wave, like half of a pinned column of length

2 L

Ends	Theoretical $K$	Recommended design $K$ (AISC)
Pinned–pinned	1	1.0
Fixed–free	2	2.1
Fixed–pinned	0.7	0.8
Fixed–fixed	0.5	0.65

The recommended values are larger because a real fixed end always rotates a little. The beam tool's Buckling tab computes $K$ with a stability analysis of your model (supports, hinges and the actual axial-force distribution), so it gives the theoretical value; the Columns tool lets you use the recommended one.

Slenderness and the Johnson formula

Dividing the Euler load by the area brings in the radius of gyration $r = I / A$ and, with it, the slenderness $L_{e} / r$ , the number that decides how a column behaves:

σ_{cr} = \frac{P _{cr}}{A} = \frac{π ^{2} E}{( L _{e} / r ) ^{2}}

For short columns this formula would give stresses above yield, which is impossible: before buckling elastically the material already yields in part (and real imperfections make things worse). For that range the Johnson parabola is used: it starts at $S_{y}$ for zero slenderness and joins the Euler curve, tangent to it, at $σ = S_{y} /2$ :

σ_{cr} = S_{y} - \frac{1}{E} (\frac{S _{y} L _{e} / r}{2 π})^{2} si \frac{L _{e}}{r} < C_{c} = π \frac{2 E}{S _{y}}

Design curve for A36 steel: Johnson for short and intermediate columns, Euler for long ones. The dots are the three examples below.

Slenderness	Behavior	Formula
$L_{e} / r \geq C_{c}$	long column: elastic buckling	Euler
$L_{e} / r < C_{c}$	short or intermediate column: inelastic buckling	Johnson
very small $L_{e} / r$	practically crushing: $σ_{cr} \to S_{y}$	Johnson

For A36 steel ( $E = 200$ GPa, $S_{y} = 250$ MPa), $C_{c} = 125.7$ . With a stronger steel $C_{c}$ drops: the benefit of its higher $S_{y}$ only shows in short columns.

Buckling safety factor

F S_{pandeo} = \frac{P _{cr}}{P}

A higher safety factor than for strength is usually required, because buckling is sudden and very sensitive to imperfections (a column is never perfectly straight nor the load perfectly centered). The simulator marks Buckles if $FS < 1$ , Low margin if $FS < 2$ and Stable otherwise, and in the Design tab you can require your own buckling FS.

If there are lateral loads besides the compression (a beam-column), the deflection makes the axial load add moment, and that moment increases the deflection. Approximately, the moments are multiplied by

\frac{1}{1 - P / P _{cr}}

: with

P = P_{cr} /2

they already double. The simulator does not include this amplification and warns about it: with a low buckling FS the real moments will be larger than the computed ones.

Common mistakes

Using the strong-axis second moment of area: the column buckles about the weak axis (unless it is braced in that plane).
Applying Euler to a short column: it gives impossible loads, above yield. Always check $L_{e} / r$ against $C_{c}$ (example 2).
Confusing $L$ with $L_{e}$ , especially for cantilevers: a fixed–free column buckles like a pinned one twice as long.
Mixing units of $E$ : in GPa it must be converted to MPa ( $\times 1000$ ) to use it with mm.
Checking buckling of a member in tension: only compression can buckle.

Worked examples

All three examples use an HSS 100×100×5 square tube of A36 steel: $A = 1836$ mm², $I = 2.711 \times 1 0^{6}$ mm⁴ about both axes (including the rounded corners), $r = I / A = 38.43$ mm, $E = 200$ GPa and $S_{y} = 250$ MPa, so $C_{c} = π 2 \cdot 200 000/250 = 125.7$ .

Example 1Long pinned column (Euler)

The column is 6 m long, pinned at both ends and carries an axial load of 60 kN. Find the critical load and the buckling safety factor.

Solution

Pinned at both ends: $K = 1$ and $L_{e} = 6$ m.
Slenderness: $L_{e} / r = 6000/38.43 = 156.1 > C_{c} = 125.7$ , a long column: Euler applies.
$σ_{cr} = \frac{π ^{2} \cdot 200 000}{156. 1 ^{2}} = 80.98$ MPa and $P_{cr} = 80.98 \cdot 1836 = 148.6$ kN (the same as $π^{2} E I / L_{e}^{2}$ ).
$FS = \frac{148.6}{60} = 2.48$ . The compressive stress is only $60 000/1836 = 32.7$ MPa: strength would be ample, buckling governs.

P_{cr} = 148.6

kN (Euler) and

F S_{buckling} = 2.48

Solve it in Simuveliq

In Engineering → Beams, length 6 m, Pin at 0 and Roller at 6 (the roller lets the column shorten).
In Section and material: A36 steel, section type Rectangular tube (HSS) and standard shape HSS 100×100×5.
In Loads, a 60 kN + Point load at $x = 6$ with direction ← −x: it pushes toward support A and compresses the whole member.
Open the Buckling tab: Buckling FS 2.48, Critical load 148.6 kN (Euler) and Effective length 6 m; the table of the explanation gives $L_{e} / r = 156.1$ and «Euler». The point lies on the Euler curve, and the «Buckled shape» is half a sine wave.

Open in the simulator →

Example 2Intermediate column (Johnson)

The same pinned column, now 3 m long, carries 150 kN. Find the critical load and the safety factor, and compare with what Euler would give.

Solution

$L_{e} = 3$ m and $L_{e} / r = 3000/38.43 = 78.06 < C_{c}$ : an intermediate column, so Johnson is used.
$σ_{cr} = 250 - \frac{1}{200 000} (\frac{250 \cdot 78.06}{2 π})^{2} = 250 - 48.2 = 201.8$ MPa.
$P_{cr} = 201.8 \cdot 1836 = 370.4$ kN and $FS = \frac{370.4}{150} = 2.47$ .
Euler would give $P = π^{2} E I / L^{2} = 594.6$ kN, that is $σ = 323.9$ MPa, more than yield: that is why it does not apply here. Using it would have overestimated the load by 60 %.

P_{cr} = 370.4

kN (Johnson) and

F S_{buckling} = 2.47

Solve it in Simuveliq

The same model as example 1 with length 3 m, the roller at 3 and the 150 kN load at $x = 3$ .
In the Buckling tab, Critical load 370.4 kN with «Johnson» and Buckling FS 2.47. The header of the buckled shape gives the Euler load, 594.6 kN, for comparison, and the table of the explanation shows $σ_{cr} = 201.8$ MPa.

Open in the simulator →

Example 3Fixed–free column ( $K = 2$ )

A 2 m column is fixed at the base and free at the top, where it takes 100 kN. Find the effective length, the critical load and the safety factor.

Solution

Fixed–free: $K = 2$ , so $L_{e} = 4$ m even though the column is 2 m long.
$L_{e} / r = 4000/38.43 = 104.1 < C_{c}$ : Johnson.
$σ_{cr} = 250 - \frac{1}{200 000} (\frac{250 \cdot 104.1}{2 π})^{2} = 164.2$ MPa and $P_{cr} = 164.2 \cdot 1836 = 301.5$ kN.
$FS = \frac{301.5}{100} = 3.01$ . Forgetting the factor $K$ , with $L_{e} = 2$ m, would give $FS = 4.2$ : an error on the unsafe side.

L_{e} = 4

P_{cr} = 301.5

kN (Johnson) and

F S_{buckling} = 3.01

Solve it in Simuveliq

Length 2 m and a single Fixed support at 0; the same section and material; a 100 kN load at $x = 2$ with direction ← −x.
In the Buckling tab: Effective length 4 m, Critical load 301.5 kN and Buckling FS 3.01. The header of the buckled shape says $K = 2$ .

Open in the simulator →

Buckling of compression members

What buckling is

Effective length

Slenderness and the Johnson formula

Buckling safety factor

Common mistakes

Worked examples

Example 1Long pinned column (Euler)

Solution

Solve it in Simuveliq

Example 2Intermediate column (Johnson)

Solution

Solve it in Simuveliq

Example 3Fixed–free column (K=2)

Solution

Solve it in Simuveliq

Related topics

Example 3Fixed–free column ( $K = 2$ )