Euler critical load, critical stress and slenderness ratio for any end condition, with a yield check and the Johnson parabola for short columns.
Column
Use the least moment of inertia (weak axis), unless the weak axis is braced. Steel shapes: take A and I from a section table.
Results
Euler buckling
Critical load Pcr = 987 kN
Critical stress σcr = 197.4 MPa
Slenderness KL/r = 100
K = 1, KL = 4 m
Section
A = 5,000 mm²
I = 8,000,000 mm⁴
r = √(I/A) = 40 mm
Squash load Fy·A = 1,250 kN
Transition slenderness Cc = 125.7
KL/r = 100 is below Cc = 125.7, so the Euler stress is above Fy/2 and the column buckles inelastically. Euler overestimates the capacity here; the Johnson parabola governs.
σcr = Fy − (Fy/2π)²(KL/r)²/E = 170.8 MPa
Pcr = 854.2 kN
Critical stress (MPa) vs slenderness KL/r
Theoretical buckling loads with no safety factor. Design codes (AISC 360 Chapter E, Eurocode 3) apply resistance factors and residual stress effects.
Euler Buckling Formulas
Pcr=(KL)2π2EIσcr=APcr=(KL/r)2π2Er=AI
End conditions
Theoretical K
Design K
Fixed, fixed
0.5
0.65
Fixed, pinned
0.7
0.8
Pinned, pinned
1.0
1.0
Fixed, free
2.0
2.1
K converts the real length into the length of an equivalent pinned-pinned column: the distance between points of zero moment on the buckled shape. A fixed-fixed column buckles like a pinned column half its length, so it carries four times the load.
Short Columns: Yield and the Johnson Parabola
The Euler load grows without limit as the column gets shorter, but no column can carry more than its squash load Py=FyA. The transition slenderness
Cc=Fy2π2E
separates the two regimes. Below it, the calculator also reports the Johnson critical stress:
σcr=Fy−(2πFy)2E1(rKL)2
Worked Example
A pinned-pinned steel column (E = 29,000 ksi, Fy = 50 ksi) is 20 ft long with A = 10 in² and a least I = 100 in⁴. Then r = √(100/10) = 3.16 in and KL/r = 240/3.16 = 75.9.
That is above Fy/2 = 25 ksi (KL/r is below Cc = 107), so Euler is unconservative here. Johnson gives σcr = 50 − (50/2π)²(75.9)²/29,000 = 37.4 ksi, a critical load of 374 kip. Try these numbers in the calculator in US units.
Pcr = π²EI/(KL)², where E is the elastic modulus, I the least moment of inertia, L the unbraced length and K the effective length factor. Dividing by the area gives the critical stress σcr = π²E/(KL/r)².
What are the effective length factors for different end conditions?
Theoretical values are K = 0.5 for fixed-fixed, 0.7 for fixed-pinned, 1.0 for pinned-pinned and 2.0 for fixed-free (a flagpole). Design practice uses 0.65, 0.8, 1.0 and 2.1 because real connections are never perfectly fixed.
What is the slenderness ratio of a column?
KL/r, the effective length divided by the radius of gyration r = √(I/A). Long, slender columns (high KL/r) fail by elastic buckling; short, stocky columns fail by yielding or inelastic buckling. AISC recommends KL/r ≤ 200 for compression members.
When is the Euler formula not valid?
When the Euler critical stress is above the proportional limit, roughly half the yield stress for steel with residual stresses. That happens when KL/r is below Cc = √(2π²E/Fy), about 126 for A36 steel and 107 for 50 ksi steel. Below Cc, use the Johnson parabola or the design code column curve.
What is the Johnson parabola?
An empirical curve for intermediate columns: σcr = Fy − (Fy/2π)²(KL/r)²/E. It equals Fy at zero length and meets the Euler curve tangentially at KL/r = Cc, where both give Fy/2.
Which moment of inertia should I use?
The smallest one, about the weak axis, because the column buckles in the direction with the least resistance. If the weak axis is braced at closer spacing, check both axes with their own unbraced lengths and use the lower load.