Civil Engineering

Column Buckling Calculator

Check the compressive capacity of a column against buckling. Enter the section shape and dimensions, the unbraced length and the end condition factor, and this calculator returns the least radius of gyration, the effective slenderness ratio, the critical buckling stress from either the Euler formula or the Johnson parabola, the IS 800 buckling curve reduction factor and the design compressive resistance. Rectangular, solid circular, I-section and circular hollow sections are supported.

Euler and Johnson parabola · IS 800:2007 Cl.7.1.2 buckling curves · KL/r limit 180

🏯 Column, Section and Load

Inputs convert when you switch. Results are always reported in SI — see the note under the results.

Unbraced Length L (m)

Actual length between points of lateral restraint, before the K factor.

Effective Length Factor K

Section Shape

Flange Width b (mm)

Overall Depth d (mm)

Web Thickness t_w (mm)

Flange Thickness t_f (mm)

Elastic Modulus E (GPa)

Yield Stress f_y (MPa)

Applied Load N (kN)

📊 Buckling Checks

Press Check Column to run the checks.

ƒ Governing Formulae

Slenderness

r_min = √(I_min / A) I_min = min(I_x, I_y)
λ = KL / r_min limit 180

Critical stress — Euler above the transition, Johnson below it

λ_c = π√(2E / f_y)

λ ≥ λ_c: σ_cr = π²E / λ² (Euler)
λ < λ_c: σ_cr = f_y[1 − f_yλ² / (4π²E)] (Johnson)

P_cr = σ_cr · A

IS 800 design resistance

σ_e = π²E / λ² λ̄ = √(f_y / σ_e)
φ = 0.5[1 + α(λ̄ − 0.2) + λ̄²]
χ = 1 / [φ + √(φ² − λ̄²)] ≤ 1.0
f_cd = χ f_y / γ_M0, γ_M0 = 1.10
P_d = f_cd · A

L — unbraced length between points of lateral restraint

K — effective length factor for the end conditions

λ — effective slenderness ratio KL/r

λ_c — Euler–Johnson transition slenderness, about 126 for mild steel

λ̄ — non-dimensional slenderness used by the IS 800 buckling curves

α — imperfection factor selecting the buckling curve

χ — stress reduction factor, between 0 and 1

γ_M0 — partial safety factor on material, 1.10

P_cr — theoretical critical load, no safety factor, for comparison only

P_d — design compressive resistance, the value to compare N against

Compare your load against P_d, not P_cr. The critical load is the theoretical elastic buckling load of a perfectly straight, perfectly centred column and carries no allowance for initial bow, residual stress, or accidental eccentricity. It is shown so you can see how far the design equation reduces it — typically to between 60 and 85 percent. Sizing a real column against P_cr would be a serious error.

How the buckling curve is chosen here. The imperfection factor is assigned by section shape alone: 0.21 for I-sections and solid circular, 0.34 for circular hollow sections, and 0.49 for rectangular. IS 800 Table 10 actually selects the curve from the section shape, the buckling axis, the fabrication route and the flange thickness — a rolled I-section buckling about its minor axis normally takes curve b or c, not curve a. The values used here are therefore a simplification that is optimistic for minor-axis buckling of I-sections. For a code-compliant design, look up the curve from Table 10 and compare.

P_d/N is a utilisation ratio, not a further safety factor. P_d already carries the material partial factor γ_M0 = 1.10 and the buckling reduction χ, so the load has been checked against a factored resistance by the time the ratio is formed. The words "Adequate" and "Marginal" attached to the ratio come from fixed thresholds of 2.0 and 1.0, which effectively ask for a second safety factor on top of the first. A column at a ratio of 1.2 has passed the code check; read the label as a rough measure of how much spare capacity you have, not as a verdict.

What this calculation does not cover. Axial compression only — there is no combined axial and bending interaction, which governs most real columns in frames. No allowance is made for local buckling of slender plate elements, so a thin-walled section may fail locally well before the capacity shown. Torsional and flexural-torsional buckling are not checked, and these can govern for angles, tees and channels. Self-weight is not included, and the column is assumed prismatic and concentrically loaded.

📝 Worked Example 1 — Pinned Steel I-Section Column

Given

A 4.0 m steel column pinned at both ends, so K = 1.0. The section is a 200 mm square I-section with 200 mm flanges 12 mm thick and an 8 mm web. Steel with E = 200 GPa and f_y = 250 MPa, carrying 600 kN.

Step 1 — section properties

h_w = 200 − 2 × 12 = 176 mm
A = 2 × 200 × 12 + 176 × 8 = 6208.0 mm²
I_x = [200 × 200³ − 192 × 176³] / 12 = 46.10 × 10⁶ mm⁴
I_y = 2(12 × 200³/12) + 176 × 8³/12 = 16.01 × 10⁶ mm⁴
The minor axis governs, as it always does for an unbraced I-section.
r_min = √(16.01 × 10⁶ / 6208) = 50.78 mm

Step 2 — slenderness

λ = KL / r = 1.0 × 4000 / 50.78 = 78.8, comfortably inside the limit of 180.
λ_c = π√(2 × 200 000 / 250) = π × 40 = 126
Since 78.8 < 126, the Johnson parabola applies rather than Euler.

Step 3 — critical stress

σ_cr = 250[1 − 250 × 78.8² / (4π² × 200 000)] = 200.88 MPa
P_cr = 200.88 × 6208 = 1247.08 kN — theoretical, not a design value.

Step 4 — IS 800 design resistance

σ_e = π² × 200 000 / 78.8² = 318.1 MPa
λ̄ = √(250 / 318.1) = 0.887
An I-section takes α = 0.21 here, so
φ = 0.5[1 + 0.21(0.887 − 0.2) + 0.887²] = 0.965
χ = 1 / [0.965 + √(0.965² − 0.887²)] = 0.743
f_cd = 0.743 × 250 / 1.10 = 168.8 MPa
P_d = 168.8 × 6208 = 1047.97 kN

Step 5 — the check

N = 600 kN against P_d = 1047.97 kN, so the ratio is 1.75 and the column is adequate. Axial stress is 600 000 / 6208 = 96.65 MPa, well under yield.

Slenderness KL/r = 78.8 against the limit of 180 — PASS

Design resistance P_d = 1047.97 kN against N = 600 kN — PASS

Reduction factor χ = 0.743, so buckling costs about a quarter of the squash capacity

P_d/N = 1.75

Overall verdict — PASS

These are the calculator's default inputs. Press Check Column without changing anything and you should get exactly these figures back. Note how far apart P_cr and P_d sit: 1247 kN against 1048 kN, and the gap widens as the column gets more slender.

⚠ Worked Example 2 — Safety Factor of 17, and Still Not Allowed

Given

A 100 × 100 mm solid steel bar used as a 6.0 m column, pinned both ends, carrying a light load of just 20 kN. E = 200 GPa, f_y = 250 MPa.

Every capacity number looks excellent

A = 10 000 mm², I_min = 8.33 × 10⁶ mm⁴, r = 28.87 mm
P_d = 339.05 kN against an applied 20 kN
P_d/N = 16.95 — nearly seventeen times the load
Axial stress = 2.00 MPa against 250 MPa yield — less than one percent utilised
Demand check reads Adequate.

And the column is not permitted

λ = 1.0 × 6000 / 28.87 = 207.8, against the IS 800 limit of 180.
λ̄ = 2.339 and χ has collapsed to 0.149 — buckling has already destroyed 85 percent of the squash capacity, which is exactly why the code stops you here.

Demand check N = 20 kN ≤ P_d = 339 kN — passes

Safety factor 16.95 — passes

Axial stress 2.00 MPa — passes

Slenderness 207.8 against 180 — FAILS

Verdict — not permitted, and only the slenderness limit says so

Why the limit exists at all

A safety factor of 17 is meaningless for a member this slender. The design equation assumes an initial bow of roughly length over 1000; a very slender column is exquisitely sensitive to that assumption, and a real member with a slightly larger bow, or a load a few millimetres off centre, behaves nothing like the calculation. The limit of 180 is a blunt instrument, and it is there precisely because the capacity equation stops being trustworthy before it stops producing numbers. Brace the column at midheight and λ halves to 104, which fixes the problem without changing the section.

This case is worth running because every intuition points the wrong way. The load is tiny, the stress is negligible, the reported safety factor is enormous — and the member is still inadmissible. A calculator that reported only the demand check would call this column fine.

📐 Inputs, Units and Accepted Ranges

Input	SI unit	Imperial unit	Accepted	Notes
Unbraced length L	m	ft	> 0	Between points of lateral restraint, before K
Effective length factor K	—	—	0.65 – 2.1	Use the recommended value, not the theoretical one
Flange width b	mm	in	> 0, must exceed t_w	Overall width for a rectangular section
Overall depth d	mm	in	> 0, must exceed 2t_f	Rectangular and I-sections
Web thickness t_w	mm	in	> 0	I-section only
Flange thickness t_f	mm	in	> 0	I-section only
Diameter	mm	in	> 0	Solid circular section only
Outside diameter OD	mm	in	> 0	Circular hollow section only
Wall thickness t	mm	in	> 0, must be < OD/2	Circular hollow section only
Elastic modulus E	GPa	GPa	> 0	200 for steel — enter GPa, not MPa
Yield stress f_y	MPa	MPa	> 0	250 for Fe410 mild steel, 345 for Fe490
Applied load N	kN	kip	≥ 0	Factored axial compression; self-weight excluded

The hollow section in this calculator is a circular pipe defined by OD and wall thickness. There is no square hollow section option — a square tube would need to be entered as a rectangular section with the wall subtracted manually, or checked elsewhere.

📚 Reference Tables

Effective length factor K — IS 800 Table 11

End conditions	Theoretical K	Recommended K	Effect on capacity
Fixed both ends, no sway	0.50	0.65	Highest capacity
Fixed one end, pinned other	0.70	0.80	—
Pinned both ends	1.00	1.00	The reference case
Fixed both ends, sway permitted	1.00	1.20	—
Fixed base, free top (cantilever)	2.00	2.10	Lowest capacity

Running the default column at each factor: K = 0.65 gives P_d = 1268 kN, K = 1.0 gives 1048 kN and K = 2.1 gives 359 kN. The end restraint is worth more than three times the capacity, which is why it deserves more thought than it usually gets.

Slenderness limits — IS 800 Table 3

Member	Max KL/r
Compression from dead and imposed loads	180
Compression from wind or seismic only	250
Tension member liable to stress reversal	350
Member normally in tension, reversal under wind only	400

This calculator applies the 180 limit in all cases. If your member falls under one of the higher limits, read the reported KL/r value directly rather than relying on the pass or fail verdict.

Euler–Johnson transition slenderness λ_c = π√(2E/f_y)

Material	E, GPa	f_y, MPa	λ_c	Behaviour below λ_c
Mild steel Fe410	200	250	126	Johnson parabola, inelastic
High strength Fe490	200	345	107	Johnson parabola, inelastic
Fe540	200	410	98	Johnson parabola, inelastic
Aluminium 6061-T6	70	240	76	Johnson parabola, inelastic

Higher strength steel lowers the transition slenderness, so a slender column made of stronger steel gains far less than a stub column does. Above λ_c, capacity depends on E alone — and E is the same for every grade of steel, so upgrading the grade of a slender column buys almost nothing.

Buckling curve imperfection factors used by this calculator

Section in this tool	α	Curve	IS 800 Table 10 reality
I-section	0.21	a	Usually b about the major axis, c about the minor
Solid circular	0.21	a	Broadly reasonable
Circular hollow	0.34	b	a if hot-finished, b if cold-formed
Rectangular solid	0.49	c	Conservative

The assignment here depends only on the shape. Real curve selection also depends on the buckling axis, whether the section is rolled or welded, and the flange thickness. Treat the I-section result as optimistic and check Table 10 for design.

Radius of gyration of common shapes

Shape	r	At 200 mm overall size
Solid rectangle, minor axis	b / √12 = 0.289b	57.7 mm
Solid circle	D / 4	50.0 mm
Circular hollow, thin wall	≈ D / 2√2 = 0.354D	70.7 mm
I-section, minor axis	≈ 0.22 to 0.25 b	≈ 50 mm

A hollow tube gives roughly 40 percent more radius of gyration than a solid circle of the same outside diameter, using a fraction of the material. That is why tubular sections dominate long compression members in trusses and towers.

❓ Frequently Asked Questions

How do you calculate the slenderness ratio of a column?

Slenderness ratio is the effective length divided by the least radius of gyration of the section. Effective length is the actual length multiplied by an end condition factor, which is 1.0 for pinned ends, about 0.65 for fixed ends and 2.0 for a column fixed at the base with a free top. The radius of gyration is the square root of the second moment of area divided by the cross-sectional area, and the smaller of the two principal axis values always governs because the column buckles about its weakest axis.

What is the maximum slenderness ratio allowed for a steel column?

IS 800 limits members carrying compression resulting from dead and imposed loads to a slenderness ratio of 180. Members subject to compression from wind or seismic loading only are permitted 250, and tension members that could see stress reversal are limited to 350. This calculator flags any column above 180 as a failure regardless of how much apparent spare capacity the design equation reports, because the limit is a code requirement independent of the calculated resistance.

What is the difference between the Euler and Johnson buckling formulae?

The Euler formula assumes purely elastic buckling and becomes unrealistic for short columns because it predicts a critical stress above the yield strength. The Johnson parabola replaces it below a transition slenderness, curving smoothly from the yield stress at zero slenderness to meet the Euler curve tangentially at the transition point. That transition occurs at a slenderness of pi times the square root of twice the elastic modulus divided by the yield stress, which is about 126 for mild steel.

Why does a column buckle about its weaker axis?

Buckling capacity depends on the second moment of area, and a column will always fail in the direction that offers least resistance. For an I-section the second moment of area about the minor axis is far smaller than about the major axis, so an unrestrained I-section column buckles sideways out of the plane of its web. This is why the minimum radius of gyration is used in the slenderness calculation, and why adding lateral bracing about the weak axis is often the cheapest way to increase a column's capacity.

What is the effective length factor K for different end conditions?

For a column pinned at both ends the theoretical factor is 1.0. Fixed at both ends gives a theoretical 0.5, though 0.65 is the recommended design value because perfect fixity is never achieved. One end fixed and one pinned gives a theoretical 0.7 with 0.8 recommended, and a cantilever column fixed at the base with a free top gives 2.0 theoretical and 2.1 recommended. Always use the recommended value rather than the theoretical one unless you can demonstrate the restraint really exists.

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Column Buckling Calculator — multicalci.com. Axial compression only, prismatic concentrically loaded members. No combined axial and bending interaction, no local plate buckling, no torsional or flexural-torsional buckling, and buckling curves assigned by section shape alone. Results are indicative and must be verified by a qualified structural engineer against IS 800 Table 10.