Euler critical load, AISC 360 and NBR 8800 design capacity side by side — with the buckled shape, the effective length K·L and the buckling curve drawn for your column.
End conditions (buckling case)
Pinned – Pinned
Cross-section
Slenderness KL/r
134.7
limit 200 · OK
Euler Pcr (elastic)
310.7 kN
Fe = 108.9 MPa
AISC 360 φcPn
245.2 kN
Fcr = 95.5 MPa · elastic
NBR 8800 Nc,Rd
247.7 kN
χ = 0.382 · λ₀ = 1.52
Code vs code — same column
Nc,Rd / φcPn = 1.010
Both codes share the 0.658 / 0.877 buckling curve — the ~1% gap is purely φc = 0.90 (AISC) vs 1/γa1 = 0.909 (NBR).
Demand check — Nd = 150 kN
Step-by-step derivation — live for YOUR column
IPE 200 · L = 3 m · K = 1 · fy = 250 MPa
Slenderness ratio
λ = K·L/r = 1 × 3000 / 22.28 mm
λ = 134.7 (≤ 200 ✓)
Euler elastic buckling stress and load
Fe = π²E/λ² = π² × 200,000 / 134.7² · Pcr = Fe·A = Fe × 2854 mm²
Fe = 108.9 MPa · Pcr = 310.7 kN
Buckling regime (AISC E3)
4.71·√(E/fy) = 4.71·√(200,000/250) = 133.2 < λ = 134.7
elastic buckling → use E3-3 (0.877·Fe)
Elastic range: capacity no longer depends on fy — only geometry (r, K, L) helps.
AISC 360 critical stress and design capacity
Fcr = 0.877 · Fe = 0.877 × 108.9 = 95.5 MPa · φcPn = 0.9 × Fcr × A
Pn = 272.5 kN · φcPn = 245.2 kN
NBR 8800 reduction factor and design capacity
λ₀ = √(fy/Fe) = 1.515 > 1.5 → χ = 0.877/λ₀² = 0.382 · Nc,Rd = χ·A·fy/1.1
Nc,Rk = 272.5 kN · Nc,Rd = 247.7 kN
Same 0.658/0.877 curve as AISC — the ~1% difference is φc = 0.90 vs 1/γa1 = 0.909.
Sections that work — 3 lightest of 612 catalog profiles carrying Nd = 150 kN at L = 3 m, K = 1
| Section | kg/m | φcPn (kN) | Nc,Rd (kN) | Util. | |
|---|---|---|---|---|---|
| lightestSHS 80x4 | 9.2 | 164 | 166 | 91% | |
| HSS 76x76x4.8 | 9.9 | 165 | 167 | 91% | |
| CHS 88.9x5 | 10.3 | 172 | 174 | 87% |
Pass criterion: φcPn ≥ Nd (AISC 360 LRFD) AND Nc,Rd ≥ Nd (NBR 8800) AND KL/r ≤ 200, using each section's tabulated-mass area and minimum radius of gyration.
Buckling curve — IPE 200, fy = 250 MPa
Capacity of IPE 200 by unbraced length — K = 1, fy = 250 MPa
| L (m) | KL/r | Pcr Euler (kN) | φcPn AISC (kN) | Nc,Rd NBR (kN) | Regime |
|---|---|---|---|---|---|
| 1 | 45 | 2,796 | 577 | 583 | inelastic |
| 2 | 90 | 699 | 419 | 423 | inelastic |
| 3◀ yours | 135 | 311 | 245 | 248 | elastic |
| 4 | 180 | 175 | 138 | 139 | elastic |
| 5 | 224 ⚠ | 112 | 88 | 89 | elastic |
| 6 | 269 ⚠ | 78 | 61 | 62 | elastic |
| 7 | 314 ⚠ | 57 | 45 | 45 | elastic |
| 8 | 359 ⚠ | 44 | 34 | 35 | elastic |
| 9 | 404 ⚠ | 35 | 27 | 28 | elastic |
| 10 | 449 ⚠ | 28 | 22 | 22 | elastic |
A slender column does not fail by squashing — it fails sideways. At a critical value of axial load the straight configuration stops being stable and the column suddenly bows out: that is flexural buckling. Leonhard Euler derived the critical load for an ideal pin-ended elastic column in 1744, and it is still the backbone of every modern column design rule:
Pcr = π² · E · I / (K·L)²
| Symbol | Meaning | Typical value |
|---|---|---|
E | Young's modulus of steel | 200 000 MPa (29 000 ksi) |
I | Moment of inertia about the buckling axis | from the section |
L | Unbraced column length | — |
K | Effective length factor (end conditions) | 0.5 – 2.0 |
Dividing by the area gives the more convenient stress form used by design codes, written with the slenderness ratio λ = KL/r (where r = √(I/A) is the radius of gyration):
Fe = π² · E / (KL/r)²
Three things to notice:
The calculator reports the pure Euler load Pcr = Fe·A (cyan) so you can see how much the design codes shave off it for your specific slenderness.
Euler solved the pin-ended column. Every other support arrangement is mapped back to that solution through the effective length factor K: the buckled shape of any column contains a segment that behaves exactly like a pin-ended column — the segment between inflection points (points of zero bending moment, marked with amber circles in the sketch above). Its length is the effective length K·L.
A fixed–fixed column develops inflection points at the quarter heights, so only the middle half behaves like the Euler column: K = 0.5 and the capacity is four times the pinned case. At the other extreme, a flagpole (fixed–free) contains only a quarter of a buckling half-wave — the full half-wave is completed by a virtual mirror image below the base, so K = 2 and the capacity drops to a quarter.
Fixed – Fixed
design 0.65
Fixed – Pinned
design 0.8
Pinned – Pinned
design 1
Fixed – Guided (sway)
design 1.2
Fixed – Free (flagpole)
design 2.1
Pinned – Guided (sway)
design 2
Theoretical K values assume mathematically perfect fixity. Since a real "fixed" base plate always rotates a little, AISC Commentary Table C-A-7.1 recommends slightly conservative design values — the calculator lets you toggle between the two:
| Case | Sway? | K theoretical | K design | Relative Pcr |
|---|---|---|---|---|
| Fixed – fixed | no | 0.5 | 0.65 | 4.00× |
| Fixed – pinned | no | 0.7 | 0.80 | 2.05× |
| Pinned – pinned | no | 1.0 | 1.0 | 1.00× |
| Fixed – guided | yes | 1.0 | 1.2 | 1.00× |
| Fixed – free | yes | 2.0 | 2.1 | 0.25× |
| Pinned – guided | yes | 2.0 | 2.0 | 0.25× |
In unbraced frames K also depends on the stiffness of the connected beams (alignment charts / Gα–Gβ nomographs, or a direct buckling analysis). For isolated columns and braced frames the six cases above cover practice.
Plot the Euler stress against slenderness and it climbs without limit as columns get shorter — predicting stresses far above yield, which is physically impossible. Real short and intermediate columns fail by an interaction of yielding and buckling, driven by two imperfections the Euler model ignores:
AISC 360-22 §E3 captures both effects with a single empirical curve. With the elastic buckling stress Fe = π²E/(KL/r)²:
KL/r ≤ 4.71·√(E/fy): Fcr = 0.658^(fy/Fe) · fy (inelastic, E3-2)
KL/r > 4.71·√(E/fy): Fcr = 0.877 · Fe (elastic, E3-3)
The 0.877 factor is the elastic-range knockdown for initial crookedness; the 0.658 exponential blends smoothly from yield (short columns, Fcr → fy) into the reduced elastic curve. The transition sits exactly where Fe = 0.44·fy — for fy = 250 MPa that is KL/r ≈ 133, marked on the chart above.
NBR 8800:2008 §5.3 writes the same physics with the reduced slenderness λ₀ = √(Q·fy/Fe):
λ₀ ≤ 1.5: χ = 0.658^(λ₀²)
λ₀ > 1.5: χ = 0.877 / λ₀²
and the design capacity is Nc,Rd = χ·Q·A·fy / γa1 with γa1 = 1.10. Substitute λ₀² = fy/Fe and you find χ·fy is identical to the AISC Fcr — the Brazilian code adopted the AISC/SSRC curve deliberately (replacing the multiple curves of the 1986 edition). Note λ₀ = 1.5 and KL/r = 4.71√(E/fy) are the same point in different clothes.
This calculator assumes Q = 1 (compact/non-slender plate elements — true for all rolled IPE/HEA/HEB/W shapes in compression except a handful of very thin webs). Slender-element sections and cold-formed LSF studs need the local-buckling reduction of AISC §E7 or NBR 14762.
Because the two codes share the buckling curve, comparing them for the same column isolates the safety format — a comparison most calculators never show:
| AISC 360-22 (LRFD) | NBR 8800:2008 | |
|---|---|---|
| Curve | Fcr = 0.658^(fy/Fe)·fy / 0.877·Fe | χ = 0.658^(λ₀²) / 0.877/λ₀² (identical) |
| Slenderness variable | KL/r, transition 4.71·√(E/fy) | λ₀ = √(fy/Fe), transition 1.5 |
| Resistance factor | φc = 0.90 (multiplies) | γa1 = 1.10 (divides) |
| Design capacity | φcPn = 0.90·Fcr·A | Nc,Rd = χ·A·fy/1.10 |
| Net effect | 0.900 × nominal | 0.909 × nominal |
| Slenderness cap | KL/r ≤ 200 (recommended) | KL/r ≤ 200 (mandatory, §5.3.4.1) |
The Brazilian result is therefore always about 1% higher than the American one — visible in the "code vs code" ratio the calculator reports. In the worked example below: φcPn = 245.2 kN vs Nc,Rd = 247.7 kN, ratio 1.010.
Where the codes genuinely diverge is around the column curve: load combinations and factors (ASCE 7 vs NBR 8681), the Q factor details for slender elements, and the treatment of frame stability (AISC's Direct Analysis Method vs NBR's B1/B2 amplification). For an isolated column with given factored load, though, the picture above is the whole story.
Both codes draw a practical line at KL/r = 200 (AISC as a strong recommendation in the §E2 user note, NBR 8800 as a hard requirement). Beyond it, capacity is tiny — at KL/r = 200 the design stress is only about 38 MPa, some 15% of a 250 MPa yield — and the member becomes vulnerable to damage during transport and erection, vibration, and deflection under self-weight. If the calculator flags the limit, the fixes in order of cheapness are: brace the weak axis at mid-height (halves L, quadruples elastic capacity), pick a section with a larger ry (wide-flange H rather than a narrow I), or improve the end restraint (lower K).
Reading the buckling curve above for a typical column:
That last point is the most common (and most expensive) misconception this calculator exists to kill: try it — set a 6 m pinned IPE 200 and switch fy between 250 and 355 MPa. The φPn does not move.
Worked example
Given
1. Slenderness ratio
KL/r = 1.0 × 300 / 2.23
λ = 134.7 (≤ 200 ✓)
2. Euler elastic buckling stress and load
Fe = π²E/λ² = π² × 200 000 / 134.7² · Pcr = Fe·A
Fe = 108.9 MPa · Pcr = 310.7 kN
3. Buckling regime (AISC E3)
4.71·√(E/fy) = 4.71·√800 = 133.2 < 134.7
elastic buckling (just past the transition) → use E3-3
4. AISC 360 critical stress and design capacity
Fcr = 0.877 × Fe = 0.877 × 108.9 = 95.5 MPa · φcPn = 0.9 × 95.5 × 28.54 cm²
Pn = 272.5 kN · φcPn = 245.2 kN
5. NBR 8800 reduction factor and design capacity
λ₀ = √(fy/Fe) = 1.515 > 1.5 → χ = 0.877/λ₀² = 0.382
Nc,Rd = χ·A·fy/1.10 = 247.7 kN
Result
φcPn = 245.2 kN (AISC) · Nc,Rd = 247.7 kN (NBR) · Pcr = 310.7 kN (Euler)
It is the axial load at which an ideal elastic column becomes unstable and bows sideways: Pcr = π²EI/(KL)². It depends only on stiffness (E·I) and effective length — not on the yield strength of the steel. Real design capacities are lower because of residual stresses and initial crookedness, which AISC 360 and NBR 8800 account for with the 0.658/0.877 buckling curve.
Match the real end restraint: pinned–pinned K = 1.0 is the safe default for braced columns; a true fixed base with a fixed top gives K = 0.5 theoretical (0.65 recommended for design); a cantilever/flagpole is K = 2.0. Because perfect fixity does not exist, use the AISC recommended design values — this calculator toggles between theoretical and design K. For columns in unbraced moment frames, K comes from alignment charts or a buckling analysis, not from these six cases.
NBR 8800:2008 adopted the same SSRC column curve as AISC: χ·fy in the Brazilian code is algebraically identical to Fcr in AISC E3. The only difference is the safety format — AISC multiplies by φc = 0.90 while NBR divides by γa1 = 1.10 (equivalent to 0.909) — so NBR results are consistently about 1% higher for the same factored load.
KL/r = 200 is the practical ceiling on column slenderness — recommended by AISC 360 (§E2 user note) and mandatory in NBR 8800 (§5.3.4.1). Beyond it the design stress falls to roughly 15% of yield and the member becomes fragile in handling, erection and service. Fix it by bracing the weak axis, choosing a section with larger minimum r, or improving end fixity.
Use the radius of gyration of the axis that buckles — for equal unbraced lengths, the smaller one (ry, the weak axis, for I/H shapes). If the weak axis is braced at intermediate points, check both axes with their own KL: weak axis with the shorter braced length, strong axis with the full length; the smaller resulting capacity governs. This calculator uses the minimum r of the selected catalog section.
It assumes non-slender (compact) elements, i.e. Q = 1 in NBR 8800 terms — valid for essentially all hot-rolled IPE, HEA, HEB, W and UB/UC shapes in compression. Sections with very slender flanges or webs, and cold-formed profiles (LSF studs, channels), need the local buckling reductions of AISC 360 §E7, NBR 8800 Annex F, or the cold-formed codes NBR 14762 / AISI S100.
Slender columns (KL/r above 4.71·√(E/fy) ≈ 133 for fy = 250 MPa) buckle while fully elastic, so capacity is 0.877·Fe and steel grade is irrelevant. Shorter columns yield partially before buckling — residual stresses make flange tips yield early — and follow the inelastic curve Fcr = 0.658^(fy/Fe)·fy, where both fy and E matter. The transition is marked on the calculator’s buckling curve.
Yes — catalog areas are table-grade. Instead of using the nominal plate rectangles (which miss the flange-to-web fillets and run 2–5% low), the calculator reconciles the geometry against the tabulated mass per meter: A = weight/ρ, which includes the fillets by construction. For an IPE 200 that gives 28.5 cm², matching ArcelorMittal tables, and the radii of gyration are rebuilt from the corrected area (rx = 8.26 cm, ry = 2.23 cm vs tabulated 8.26/2.24). The "table-grade" badge next to the section confirms when this reconciliation is active.
Yes, free and without login. Every input (end conditions, K, L, fy, section, Nd) lives in the URL — copy the permalink and a colleague opens the identical column. You can also download a branded PNG of the column sketch plus buckling curve, and a CSV of the capacity-by-length table for spreadsheets. One more click opens the same column, fully prefilled, in the CalcSteel 3D editor.
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