About the Column Buckling Calculator
This column buckling calculator finds the load at which a slender column suddenly bows sideways — the Euler critical buckling load — for the four classic end conditions: pinned–pinned, fixed–fixed, fixed–pinned and fixed–free (a flagpole). Choose a solid round bar, rectangular bar, round tube or enter your own I and area, pick a material, and see the critical load, critical stress, radius of gyration and slenderness ratio.
It is useful for engineering students, designers checking struts, props, jack stands and machine links, and builders getting a feel for how much a post can carry before it buckles. Because Euler’s formula only applies to slender columns, the calculator also compares the slenderness with the transition value and applies the J. B. Johnson parabola for intermediate columns, which keeps the answer from exceeding what the material can actually take.
The results assume a perfectly straight, prismatic column with a centred axial load and ideal end restraint. Real columns have initial crookedness, eccentric loads and imperfect joints, which is why design codes use lower curves and safety factors. For a rectangular bar the weaker axis is used automatically. Treat the output as a preliminary check, not a code-compliant design.
With the default inputs, the euler critical buckling load is 60.81 kN. Change any value above to recalculate instantly.
How to use the column buckling calculator
- 1Choose the end conditions, or enter a custom K.
- 2Enter the unbraced length between lateral supports.
- 3Pick the cross-section and enter its dimensions, or enter I and A directly.
- 4Choose a material, or enter E and yield strength.
- 5Set a safety factor and read the critical and allowable loads.
Formula and method
Euler’s formula gives the axial load at which an ideal, straight, elastic column becomes unstable. The effective length K·L converts any end condition to an equivalent pinned column: fixing both ends halves the effective length (K = 0.5) and multiplies the critical load by four, while a flagpole fixed at the base and free at the top has K = 2 and carries only a quarter of the pinned value. The values used here are the theoretical ones; design codes often recommend slightly larger K for fixed ends because real connections are never perfectly rigid.
Euler’s formula is only valid while the critical stress stays below the material’s proportional limit. The transition slenderness Cc is where the Euler stress equals half the yield strength; below it, the J. B. Johnson parabola gives a more realistic critical load that approaches the squash load A·fy for very short columns. The allowable load divides the governing value by your safety factor.
- Pcr
- Critical buckling load
- E
- Young’s modulus
- I
- Least second moment of area of the section
- K
- Effective length factor for the end conditions
- L
- Unbraced column length
- r
- Radius of gyration
- λ
- Slenderness ratio KL/r
- fy
- Yield (or crushing) strength
Worked examples
3 m pinned steel tube, 60 × 4 mm
The tube has I ≈ 27.73 cm⁴ and A ≈ 7.04 cm², so r ≈ 19.85 mm and KL/r ≈ 151 — above the transition value of about 126 for 250 MPa steel. Pcr = π² × 200,000 × 277,264 ÷ 3,000² ≈ 60.8 kN, or about 30.4 kN with a safety factor of 2.
1.2 m solid 30 mm steel rod as a flagpole (fixed–free)
A free top doubles the effective length to 2.4 m. With r = 30 ÷ 4 = 7.5 mm, KL/r = 320 and Pcr = π² × 200,000 × 39,761 ÷ 2,400² ≈ 13.6 kN — only 5.45 kN allowable at a safety factor of 2.5.
2.4 m timber stud, 38 × 89 mm, pinned
Unbraced, the stud buckles about its weak 38 mm axis: I = 89 × 38³ ÷ 12 ≈ 40.7 cm⁴. With E = 11 GPa, Pcr = π² × 11,000 × 406,967 ÷ 2,400² ≈ 7.67 kN — which is why wall studs rely on sheathing to brace the weak axis.
Short 1 m fixed–fixed steel tube (Johnson range)
With K = 0.5 the slenderness is only about 25, far below Cc ≈ 126. Euler would predict an impossible 2,189 kN; the Johnson parabola gives 7.04 cm² × 250 MPa × (1 − 250 × 25.19² ÷ (4π² × 200,000)) ≈ 172 kN, close to the squash load.
Frequently asked questions
What is the Euler buckling formula?+
Pcr = π²EI ÷ (KL)², where E is Young’s modulus, I the least second moment of area, L the unbraced length and K the effective length factor. It gives the load at which a slender, straight column becomes unstable.
What are the K factors for column end conditions?+
Theoretical values are 1.0 for pinned–pinned, 0.5 for fixed–fixed, 0.7 for fixed–pinned and 2.0 for fixed–free. Design guides such as the AISC commentary recommend slightly higher values (0.65, 0.8 and 2.1) for fixed ends because real connections are not perfectly rigid.
When is Euler’s formula not valid?+
When the slenderness ratio KL/r is below the transition value Cc = √(2π²E/fy), the column yields before it buckles elastically (Cc is about 126 for 250 MPa steel). Use the Johnson parabola or a design-code curve for these intermediate and short columns.
What is the slenderness ratio?+
The slenderness ratio is the effective length divided by the radius of gyration, KL/r. Higher values mean a more slender column that buckles at lower stress; many codes limit compression members to about 200.
Why does a rectangular column buckle about its weak axis?+
A column buckles in the direction of least resistance, which is the axis with the smallest second moment of area. For a 38 × 89 mm stud that is the 38 mm direction, where I is about 5.5 times smaller than in the deep direction.