What is the critical buckling load of a column?

The critical buckling load is the compressive force at which a straight column will suddenly lose its stability and bend sideways. Below this load, the column remains straight, but when this load is reached it can take on a bent shape, with the lateral displacement increasing even without additional loading. For slender columns, this critical load does not depend on material strength, but rather on stiffness and geometry, and equals the value resulting from pi squared times E * I divided by the square of the effective length.

Why do the boundary conditions have such a big impact on the result?

The boundary conditions determine the effective length and this is squared in the formula. For a column with fixed ends, the effective length is half of its actual length. If the same column were made into a cantilever, the effective length would double. The manner in which the ends are restrained can change the value of K by as much as four times, resulting in a sixteen-fold change in load. Therefore, a column that allows lateral movement is much weaker than one that prevents such movement.

When should you use the Johnson formula instead of the Euler formula?

If the length to width ratio is below a certain threshold then Johnson's formula will be used. This threshold is the square root of the product of two times pi squared and E divided by the yield stress. If this value is exceeded, elastic buckling occurs so that Euler's formula gives the correct result. If this value is below, the stress predicted by the Euler model exceeds the stress achievable by the material resulting in ductile failure. In this case Johnson's parabolic formula will give the correct result. This calculator automatically switches between the formulas and also outputs the pure Euler load so that the magnitude of error can be checked.

What must be the length-width ratio for it to be too high?

There is no fixed value. The threshold varies depending on the material. However, there are practical rules of thumb for structural steel according to which a length-width ratio of about 125 indicates elastic buckling. In addition, many standards stipulate an upper limit for the length-width ratio of compression members at around 200, regardless of the calculation method. If the length-width ratio falls below about 50, then the column is thick and short enough that it should not be considered for buckling but rather for crushing.

How can you prevent a column buckling?

First the unsupported length should be reduced. By adding an additional support in the middle of the column the strength can be quadrupled with only one extra support. Next the cross section shape should be considered. For a given weight a hollow profile has significantly more strength than a solid material because the area is distributed further from the center of gravity. It is also effective to avoid lateral vibration and make end connections very stiff. Choosing a higher grade of the same metal is usually not very effective since the yield point is not taken into account in the Euler formula.

Buckling Calculator

End conditions and length

End support conditions

How the two ends of the column are held. The more rotation and sway the ends prevent, the shorter the effective buckling length and the higher the load the column carries.

Use recommended design K values instead of theoretical

The classical eigenvalue solution gives K = 0.5 for fixed-fixed and 0.7 for fixed-pinned. Real connections are never perfectly rigid, so AISC 360 Table C-A-7.1 recommends higher values (0.65 and 0.80). Turn this on for design work, leave it off to match textbook answers.

Unsupported column length (L)

Cross-section

Rectangle or square

Solid round bar

Round tube or pipe

Enter I and A directly

Width (mm)

Depth (mm)

Material

Sets Young's modulus. Buckling depends on stiffness, so this is the single most important material property here.

Yield or crushing strength (MPa)

Load and checks

Applied compressive load

Check an off-centre (eccentric) load

A load that misses the centroid bends the column from the very first newton. The secant formula returns the extra sideways deflection and the real peak compressive stress.

Size the column for a target factor of safety

Works the problem backwards: the longest length you can use, and the moment of inertia you would need, to keep your target margin against buckling.

Critical buckling load (kN)

Critical stress (MPa)

Slenderness ratio (KL/r)

Factor of safety against buckling

Critical buckling load

Critical stress (MPa)
Slenderness ratio (KL/r)
Factor of safety against buckling

This column is slender. Its slenderness ratio of 207.8 is above the transition value of 125.7, so it buckles elastically and Euler's formula governs. Stiffness decides the answer here, not strength.

The applied load uses 35 percent of the critical load, a factor of safety of 2.86. That is the sort of margin buckling checks usually ask for, because the failure gives no warning.

Column properties and reference values

Effective length factor (K)
Effective length (KL): m
Radius of gyration (mm)

Moment of inertia, weak axis: mm⁴
Cross-section area: mm²
Young's modulus in use (GPa)

Transition slenderness ratio
Transition length: m
Elastic Euler load: kN

Crushing load at yield: kN
Applied compressive stress (MPa)
Load used, percent of critical

The radius of gyration is the section's own measure of how far its area sits from the centroid, r = sqrt(I / A). Divide the effective length by it and you get the slenderness ratio, the single number that decides which buckling curve applies.

The transition length is where the two curves meet for this material: shorter than that and the column crushes, longer and it buckles.

Capacity curves

The buckling stability calculator answers the question of how much compressive load a straight structural member can withstand before it loses its stability. By entering length, cross-section, material and boundary conditions at both ends, the tool provides the critical buckling load, corresponding stress, deflection ratio and safety factor with respect to the actual applied load.

Buckling is a failure mechanism that is difficult to assess because it depends little on the strength of the material. Steel columns for example can fail due to their length and small cross-section even though the applied load is only a fraction of the yield stress.

This calculator selects the appropriate formula. For slender columns, the elastic Euler's formula is used while for thick short columns, Johnson parabolic formula is applied. The choice of which formula to use depends on the slenderness ratio.

What is buckling?

If a compressive force is applied to a short thick block it will break when the stress reaches the material strength. For a long thin member however a different phenomenon occurs. When a certain load is reached the member suddenly bends sideways and continues to deform in the horizontal direction even if the load is not increased.

This load is the critical buckling load. Below this load, the strut remains stable and straight. Above this load, a straight shape is still mathematically possible but unstable, similar to a ball on top of a hill.

This leads to two results that both go against our intuition. The deflection depends on the stiffness so the modulus of elasticity and cross-section shape are important while the grade of steel is not really relevant. Also, since the deflection happens suddenly there is no warning from sagging or cracking. This is why the safety factor in checking for deflection is higher than when checking for bending moment.

The critical load according to Euler

Leonhard Euler solved this problem in 1757 by examining when a curved structure can regain equilibrium. For a column with pinned ends the solution is a half-wavelength sine wave and only one load allows it to remain in equilibrium.

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

E is the modulus of elasticity, I is the moment of inertia of the cross-section with respect to the axis about which the column will bend, L is the unbraced length and K is the effective length factor that accounts for end conditions. The product KL is the effective length and is denoted as Le.

Also note what the formula does not include. There is no mention of tensile strength, fracture toughness or other properties of the material. The formula only deals with the hardness of the material and how that material is arranged within the cross-section.

In addition, the effective length is squared. If the length doubles, then the load capacity reduces to a quarter, which means that length is the most controlling factor in this problem while adding bracing is the most economical repair method.

Effective length and boundary conditions

The effective length is the distance that a strut will travel before it reaches its first deflection. Ends which resist rotation reduce this distance and increase the load. Ends which allow lateral displacement of the strut increase this distance and significantly decrease the load.

There are currently two types of K-values and this calculation tool provides both. The theoretical column model is based on classical eigenvalue solutions and assumes that the ends are either fully fixed or fully pinned. The recommended design column model is based on AISC 360 Table C-A-7.1 and takes into account the fact that real connections can be neither fully fixed nor fully pinned.

End conditions	Theoretical K	Recommended design K	Buckled shape
Pinned at both ends	1.0	1.0	One half sine wave
Fixed at both ends, no sway	0.5	0.65	Full sine wave, two inflection points
Fixed at one end, pinned at the other	0.7	0.80	About seven tenths of a half wave
Fixed at one end, free at the other	2.0	2.10	Quarter sine wave
Fixed at both ends, top free to sway	1.0	1.2	S shape with sway
Pinned at one end, guided at the other	2.0	2.0	Sway dominated

It is important to note this difference. The load that a fixed-fixed beam can withstand is 16 times the load of an equal length cantilevered beam. This is because the K-value changes from 0.5 to 2.0 and the K-value appears in a squared term. The member itself remains unchanged, only the way it is supported changes.

The ratio of length to cross-sectional area and when the Euler formula fails

The Euler formula has a serious flaw in the area of short columns. When length approaches zero, the predicted critical stress approaches infinity. There is no material that can withstand this. Before the theoretical load is reached, the material will fail by yielding or crushing.

The basis for the evaluation is the length-width ratio, i.e. the quotient of effective length and moment of inertia radius of cross-section.

\lambda = \frac{K L}{r} \qquad r = \sqrt{\frac{I}{A}}

The moment of inertia radius indicates how far the center of gravity of a cross section is from its geometric axis. Tubes have a large moment of inertia in relation to their surface, which is why tubes are used rather than solid bars for scaffolding and bicycle frames.

The length-width ratio is then compared to a transition value that depends only on the material. This ratio corresponds to the point at which the stress predicted by Euler falls below half of the yield strength.

\lambda_{c} = \sqrt{\frac{2 \pi^2 E}{\sigma_{y}}}

If this value is exceeded the column is long and slender, and the Euler formula applies. If it is below this value, the column is in an intermediate or short state, and the calculation takes over the parabolic formula of J.B. Johnson. This formula starts at the yield point for thick, short columns and gradually decreases until it connects with the Euler curve at the transition point.

\sigma_{cr} = \sigma_{y} \left( 1 - \frac{\sigma_{y} \, \lambda^2}{4 \pi^2 E} \right) \qquad P_{cr} = \sigma_{cr} \, A

Material	E (GPa)	Yield (MPa)	Transition slenderness
Structural steel	200	250	About 126
S355 steel	200	355	About 105
Stainless steel 304	193	215	About 133
Aluminium 6061-T6	69	276	About 70
Timber, softwood	11	40	About 74

Two points can be drawn from this table: High strength steel reduces the transition length-width ratio, so high strength grades are useful for thick short columns but have no effect on long slender ones. Aluminium has only one third of the stiffness of steel and therefore reaches the transition state earlier.

Cross-section properties of typical profiles.

Bending always occurs in the weaker axis, i.e., in the direction with the lower moment of inertia. This is because this is the direction of least resistance. For rectangular cross-sections, this means that the smaller dimension is taken to the power of three.

Section	Moment of inertia (weak axis)	Area	Radius of gyration
Rectangle, b wide and h deep, h smaller	$I = \dfrac{b \, h^3}{12}$	$A = b \, h$	$r = \dfrac{h}{\sqrt{12}}$
Solid round bar, diameter d	$I = \dfrac{\pi \, d^4}{64}$	$A = \dfrac{\pi \, d^2}{4}$	$r = \dfrac{d}{4}$
Tube, outside D, inside d	$I = \dfrac{\pi \, (D^4 - d^4)}{64}$	$A = \dfrac{\pi \, (D^2 - d^2)}{4}$	$r = \dfrac{\sqrt{D^2 + d^2}}{4}$

For rolled steel sections you can take the values for I (Moment of Inertia) and A (Cross-sectional Area) from the table and enter them directly. Unless there is a support in the direction of the minor axis, you must use the moment of inertia of the two axes that is smaller. If there is a support in the direction of the minor axis, then the major axis may be the dominant direction.

Calculation example (slender steel column)

Consider a steel column with a square cross-section of side length 50 mm, an unsupported length of 3 m, both ends pinned, E (elasticity modulus) is 200 GPa and the yield strength is 250 MPa. First we calculate the properties of the cross-section.

I = \frac{50 \times 50^3}{12} = 520\,833 \ \text{mm}^4 \qquad A = 50 \times 50 = 2\,500 \ \text{mm}^2

r = \sqrt{\frac{520\,833}{2\,500}} = 14.43 \ \text{mm}

If both ends are supported, k is equal to 1 so the effective length is the full length of 3 meters and the slenderness ratio becomes:

\lambda = \frac{1 \times 3000}{14.43} = 207.8

The critical slenderness ratio for this material is approximately 125.7 but 207.8 is well above that so the column is slender and determined by the Euler formula.

P_{cr} = \frac{\pi^2 \times 200\,000 \times 520\,833}{3000^2} = 114\,200 \ \text{N} \approx 114.2 \ \text{kN}

\sigma_{cr} = \frac{114\,200}{2\,500} = 45.7 \ \text{MPa}

So this column becomes unstable by buckling at a load of about 114 kN and an average stress of 45.7 MPa. The material begins to flow at 250 MPa, which corresponds to a load of 625 kN. This means that more than four-fifths of the strength of the material is lost because of buckling. That difference is exactly what makes carrying out a stability check so important.

If an additional support is added in the middle of the column, the unsupported length is halved to 1.5 m. A calculation using only the Euler formula would give a four times higher critical load of 457 kN, but since the ratio of height to cross-section for a supported column is 104 and below the transition value, the column is already in the plastic deformation range. The actual result, as shown by Johnson's parabolic diagram, is about 411 kN. By adding just one support without increasing the amount of material, an almost fourfold improvement can be achieved, which illustrates why both curves are important.

Example calculation (if the Euler formula is inaccurate for thick, short columns)

Now consider a steel strut with a square cross section of 100 mm edge length and a short length of only 1.5 m, both ends pinned. Since this cross section is four times as deep in each direction, the stiffness is much higher.

I = \frac{100 \times 100^3}{12} = 8\,333\,333 \ \text{mm}^4 \qquad A = 10\,000 \ \text{mm}^2 \qquad r = 28.87 \ \text{mm}

\lambda = \frac{1 \times 1500}{28.87} = 52.0

Since this is well below the transition value of 125.7 it is a thick short column. The Euler formula gives a critical stress of just over 731 MPa which is nearly three times the yield strength so obviously not correct. The actual result as shown by Johnson's parabolic diagram is:

\sigma_{cr} = 250 \left( 1 - \frac{250 \times 52^2}{4 \pi^2 \times 200\,000} \right) = 228.6 \ \text{MPa}

P_{cr} = 228.6 \times 10\,000 = 2\,286 \ \text{kN}

For such thick short columns the load is close to the crushing value of 2,500 kN. This is the expected result and this small difference represents the "penalty" effect of buckling. The use of the Euler formula would give a three times overestimate of the capacity.

Security factors and factors to consider when breaking up:

When the actual load acting on a column is inputted, this load is divided by the critical load. As buckling can occur without warning and real columns are not perfectly straight, stability checking usually requires a larger safety factor than bending checking, with the critical load typically being increased two to three times.

When the safety factor is too small, various adjustment measures have different effects. This table evaluates them based on the ratio of input and effect.

Change	Effect on critical load	Comment
Brace at mid-height	Up to four times higher	Halves the unsupported length. Almost always the cheapest fix, though the gain is capped once the shorter column turns stocky.
Fix the ends instead of pinning them	Up to four times higher	K falls from 1.0 to 0.5, but only if the connection is genuinely rigid and cannot sway.
Swap a solid bar for a tube of the same area	Two to six times higher	Moves material away from the centroid, which raises I and r. A 60 mm tube with a 5 mm wall has five and a half times the inertia of the solid bar that weighs the same.
Increase the section size	Rises with the fourth power of the depth	Very effective, and the heaviest and most expensive option.
Use a stronger grade of the same metal	No change when slender	Yield strength is absent from Euler's formula. It only helps once the column is stocky.

Off-axis loading and actual defects:

Previous statements assume that the column is perfectly straight and the load acts exactly through its center of gravity. These conditions do not exist in reality. Steel structures already have slight deformations when they leave the factory, connection points can deviate by a few millimeters from the center, and even loads are rarely perfectly axially aligned.

If an eccentric load is applied the problem changes fundamentally. In this case the column will begin to bend at the first Newtonian load so there is no longer a distinct critical load. This situation can be described by a secant formula.

\delta_{max} = e \left[ \sec\left( \frac{L_e}{2} \sqrt{\frac{P}{E I}} \right) - 1 \right]

\sigma_{max} = \frac{P}{A} \left[ 1 + \frac{e \, c}{r^2} \sec\left( \frac{L_e}{2} \sqrt{\frac{P}{E I}} \right) \right]

Here 'e' is the eccentricity and 'c' is the distance from the center of gravity to the extreme fiber. As the load approaches the critical value, the secant term goes to infinity, causing both deflection and stress to go to infinity. This represents a simulation of actual failure as determined by calculations.

When the eccentricity check is activated, the calculation tool provides the deflection in the middle of the span as well as the maximum compressive stress. The latter figure should be compared with the yield strength. Curved columns also show similar behavior; for usual design assumptions an initial deflection of one thousandth of the length is assumed.

How to use this calculator:

First start with the boundary conditions and unbraced length as these two factors have the most effect on results. Then select a cross section and enter its dimensions in millimeters. If it is one of the cross sections listed in the chart, you can also directly enter values for 'I' (moment of inertia) and 'A' (cross sectional area).

Select the material and set the modulus of elasticity. Also check the yield strength. The yield strength only determines when to replace the Euler formula with the Johnson formula. By adding a load, the safety factor is shown together with the critical value.

In the "Column properties" section you can check intermediate values such as effective length, radius of gyration, transition strain and transition length. In addition two reference loads are shown: the pure Euler load and the crushing load. The two graphs show how the bearing capacity changes with length, respectively the position of the column on the classical stress-strain curve.

This calculation tool is for educational purposes and initial sizing only. It deals with flexural buckling of prismatic members subjected to an axial load at the center or off-center, but does not account for torsional buckling or bending-strengthened torsional buckling, local plate buckling, member assemblies or members with variable cross-sections, residual stresses, combined tension and bending loads, nor the reduction factors for strength specified in the respective design codes. Structural members must be designed and checked by a qualified engineer.

Frequently asked questions

What is the critical buckling load of a column?: The critical buckling load is the compressive force at which a straight column will suddenly lose its stability and bend sideways. Below this load, the column remains straight, but when this load is reached it can take on a bent shape, with the lateral displacement increasing even without additional loading. For slender columns, this critical load does not depend on material strength, but rather on stiffness and geometry, and equals the value resulting from pi squared times E * I divided by the square of the effective length.
Why do the boundary conditions have such a big impact on the result?: The boundary conditions determine the effective length and this is squared in the formula. For a column with fixed ends, the effective length is half of its actual length. If the same column were made into a cantilever, the effective length would double. The manner in which the ends are restrained can change the value of K by as much as four times, resulting in a sixteen-fold change in load. Therefore, a column that allows lateral movement is much weaker than one that prevents such movement.
When should you use the Johnson formula instead of the Euler formula?: If the length to width ratio is below a certain threshold then Johnson's formula will be used. This threshold is the square root of the product of two times pi squared and E divided by the yield stress. If this value is exceeded, elastic buckling occurs so that Euler's formula gives the correct result. If this value is below, the stress predicted by the Euler model exceeds the stress achievable by the material resulting in ductile failure. In this case Johnson's parabolic formula will give the correct result. This calculator automatically switches between the formulas and also outputs the pure Euler load so that the magnitude of error can be checked.
What must be the length-width ratio for it to be too high?: There is no fixed value. The threshold varies depending on the material. However, there are practical rules of thumb for structural steel according to which a length-width ratio of about 125 indicates elastic buckling. In addition, many standards stipulate an upper limit for the length-width ratio of compression members at around 200, regardless of the calculation method. If the length-width ratio falls below about 50, then the column is thick and short enough that it should not be considered for buckling but rather for crushing.
How can you prevent a column buckling?: First the unsupported length should be reduced. By adding an additional support in the middle of the column the strength can be quadrupled with only one extra support. Next the cross section shape should be considered. For a given weight a hollow profile has significantly more strength than a solid material because the area is distributed further from the center of gravity. It is also effective to avoid lateral vibration and make end connections very stiff. Choosing a higher grade of the same metal is usually not very effective since the yield point is not taken into account in the Euler formula.

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Disclaimer: This calculator is provided for general informational and educational purposes only. Our calculators are under active development, and results may be inaccurate, incomplete, or unsuitable for your situation. Always verify the figures independently and seek advice from a qualified professional before relying on them. We make no warranties and accept no liability for any loss or decision arising from use of this tool.

References

Wikipedia: Buckling
Overview of buckling modes, the critical load, and where elastic theory stops applying.
Wikipedia: Euler's critical load
Derivation of the critical load and the effective length factors for the standard end conditions.
Wikipedia: Johnson's parabolic formula
The inelastic branch used below the transition slenderness ratio, and how it meets Euler's curve.
Wikipedia: Radius of gyration
Definition of the radius of gyration that turns a section's moment of inertia into a slenderness ratio.
Wikipedia: Slenderness ratio
The ratio that decides which buckling curve governs a compression member.