Buckling per Euler and Tetmajer
Calculate the buckling safety of straight compression members: from the Euler case, member length and cross-section follow the slenderness ratio, buckling stress, critical buckling load and the available safety factor. The calculator automatically distinguishes elastic buckling (Euler), inelastic buckling (Tetmajer) and the crushing range, plots the working point in the σ-λ chart and additionally checks the compressive stress, live with every input.
Calculation
Model: ideal straight member with constant cross-section under static compression, centric or with a given eccentricity e; imperfections are covered globally by the buckling safety factor. No torsional or lateral-torsional buckling of open thin-walled sections, no transverse load. Sizing tool for machine and fixture design - structural members must be verified per EN 1993-1-1 (buckling curves); do not mix the two concepts. Use the spring calculator for compression springs and the beam calculator for members under transverse load.
Results
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Formulas and fundamentals
Euler's formula and effective length
A centrically loaded ideal column becomes unstable once the compressive force reaches the critical buckling load. In the elastic range Euler's formula applies:
The smallest second moment of area I_min always governs, because the member buckles about its weak axis. The end conditions enter through the effective (buckling) length l_k: the four Euler cases give β = 2 (fixed at the base, free at the top - the classic piston rod), β = 1 (pinned at both ends, the basic case), β = 0.699 (fixed / pinned) and β = 0.5 (fixed at both ends).
Slenderness ratio and limit slenderness
Whether Euler applies at all is decided by the slenderness ratio λ with the radius of gyration i:
Above the material-dependent limit slenderness λ_g (S235: 105, E295/S355: 89, grey cast iron: 80) the buckling stress stays below the proportional limit - the member buckles elastically:
Tetmajer range and crushing limit
Below it the material partially yields before buckling; here the empirical Tetmajer equations apply:
Below the crushing limit λ_0 (S235: 65.8) the straight line reaches the yield strength: buckling is no longer the issue and the check becomes a plain compressive stress check with σ_k = R_e.
Where the formula actually changes
Both curves are meant to meet at λ_g: λ_g = π·√(E/σ_P) arises precisely from equating the Euler stress with the proportional limit, and that is where the Tetmajer line starts. In the tables, however, λ_g and the coefficients a and b come from two different sources - λ_g from that formula, a and b from Tetmajer's tests - so they do not always meet exactly: for the line 335 − 0.62·λ the intersection lies at λ = 85.8 while the tables state λ_g = 89 (some sources 85, others 88). Between 85.8 and 89 the line would sit up to 6.9 % above the Euler hyperbola.
That cannot be a buckling stress: in the inelastic range σ_k = π²·T_K/λ² applies, with the buckling modulus T_K after Engesser and v. Kármán, and T_K is always smaller than E - the Euler hyperbola is the upper bound of any buckling stress. The calculator therefore switches at the intersection of the two curves rather than at the tabulated value (E295/E335/S355: 85.8; S235: 104.1; grey cast iron: 77.5). The transition is thus continuous, the result independent of which source supplies λ_g, and within the affected band up to 6.9 % more conservative than a hard switch at λ_g.
Buckling safety and compressive stress check
The critical buckling load and the available safety factor follow from:
In mechanical engineering, buckling safety factors of 3 to 6 are customary because of the high sensitivity to imperfections (initial curvature, load eccentricity, flexible end fixity), tending towards the upper end in the slender Euler range - the calculator defaults to 4. In addition the compressive stress σ_d = F/A is checked against the yield strength. The method applies to straight, prismatic members under static load; structural members governed by building codes must instead be verified per Eurocode 3 with buckling curves.
Buckling lengths per principal axis
A member is rarely supported the same way in both planes: intermediate bracing often acts in one direction only. Calculating with I_min alone is then insufficient, because what governs is not the weak axis but the axis with the larger slenderness ratio. Brace the weak plane closely enough and the strong axis takes over, even though its second moment of area is several times larger. The calculator therefore accepts separate end conditions and lengths per principal axis on request and reports both slenderness ratios.
Eccentric compressive load
If the compressive force does not act along the member axis, a bending moment F·e is present from the outset and amplifies itself with the deflection. The extreme fibre stress therefore does not follow simple superposition but grows with the reciprocal of a cosine:
That cosine reaches zero as soon as the compressive force attains the Euler buckling load - the relation has its pole there and an extreme fibre stress is no longer defined; the calculator then reports no figure but flags buckling instead. The eccentricity is applied at both ends in the same sense: entering it for one end only leaves you roughly a factor of two on the safe side. The relation is derived for the pin-ended member and is offered here for buckling cases 1 and 2 only. With a fixed base and free top the largest moment migrates to the base, and with both ends fixed no bending share remains at all, because the moment F·e passes entirely into the fixities.
Worked example
A round bar of S235 with 30 mm diameter and 1500 mm free length is pinned at both ends (Euler case 2, β = 1) and carries a centric compressive force of 7 kN. Section properties: A = 706.9 mm², I_min = 39,761 mm⁴, radius of gyration i = √(I_min/A) = 7.5 mm.
With the effective length l_k = 1·1500 mm = 1500 mm the slenderness ratio is λ = 1500/7.5 = 200. Since λ = 200 ≥ λ_g = 105 the member buckles elastically - the Euler range governs. The buckling stress is σ_k = π²·210,000/200² = 51.8 N/mm², the critical buckling load F_k = σ_k·A = 36.6 kN.
The available safety factor is S_avail = F_k/F = 36,626/7000 = 5.23 and exceeds the required buckling safety S_req = 4 - the check passes. The compressive stress check is uncritical with σ_d = F/A = 9.9 N/mm² against R_e = 235 N/mm², as is typical for slender members: stability, not strength, limits the load capacity.
Frequently asked questions
When does Euler apply, when Tetmajer?
Above the limit slenderness λ_g the member buckles elastically per Euler (σ_k = π²·E/λ²), below it inelastically per the empirical Tetmajer equations. Below the crushing limit λ_0 there is no stability problem at all, just yielding under compression. The calculator assigns the range automatically and shows it as a badge. The switch happens at the intersection of the two curves rather than at the tabulated λ_g, because the tabulated values do not match everywhere - see the section "Where the formula actually changes".
Which Euler case should I choose?
Always by the actual end conditions, not by the case number - the numbering is not consistent across the literature. An extended piston rod corresponds to case 1 (β = 2), a pin-ended strut to case 2 (β = 1). Real fixed ends are rarely ideally rigid: when in doubt, conservatively use the larger β value, since the buckling load drops with 1/β².
Why does the smallest second moment of area I_min govern?
The member buckles about the axis with the least resistance. For a rectangular section that is the axis parallel to the long side (I_min = h·b³/12 with b ≤ h). Mixing up I_min and I_max is the most common user error in buckling checks - the calculator derives I_min automatically from the dimensions. This holds only while the member has the same length and end conditions in both planes: as soon as intermediate bracing acts in one direction only, the larger slenderness ratio governs, and that can be the strong axis. Buckling lengths can be entered separately per principal axis for exactly this case.
How is an eccentric compressive load accounted for?
Through the eccentricity field e. The calculator then checks not the compressive stress F/A against the yield strength but the extreme fibre stress from compression and bending, which grows disproportionately with load and is undefined once the buckling load is reached. The denominator requires the section modulus; when area and second moment of area are entered directly it cannot be derived from them and must be supplied as well. The field is available for buckling cases 1 and 2 only, because the relation is derived for the pin-ended member: with a fixed base and free top the largest moment sits at the base, and with both ends fixed there is no bending share at all. For those end conditions a Eurocode 3 verification is the way to go, where imperfections enter through the buckling curve and the end conditions through the buckling length.
Why is the required buckling safety of 3 to 6 so high?
Because the buckling load is extremely sensitive to imperfections: initial curvature, eccentric load introduction and flexible end fixity reduce the real capacity well below the ideal value. These effects are covered globally by the safety factor. The more slender the member, the higher it should be chosen; cylinder manufacturers often specify their own values for piston rods.
What does the badge "crushing range - compression check governs" mean?
For very stocky members (λ < λ_0) the Tetmajer line would yield stresses above the yield strength - physically meaningless, the material yields first. The calculator therefore caps the buckling stress at R_e; the verification becomes a plain compressive stress check. For grey cast iron the compressive strength σ_dB takes the place of the yield strength.
Does the calculator replace a Eurocode 3 verification?
No. Euler/Tetmajer with a global safety factor is the classic sizing practice in machine and fixture design. Structures governed by building regulations must be verified per EN 1993-1-1 (flexural buckling with buckling curves and partial safety factors) - a different concept that must not be mixed with the buckling safety S = 3…6. The historical ω-method per DIN 4114, still found in old calculations, was superseded in the 1970s.
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