Calculate Euler critical buckling load for an ideal elastic column. Compare end conditions, weak-axis stiffness, slenderness, and critical stress.
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Worked examples
These are the solver's own reference problems — the answers come from published sources or independent hand computation, never from the solver itself. The solution below is the live solver output for each.
Example 1
SI pin-pin column: L=3 m, E=200 GPa, I=5e6 mm⁴
Given
length
3 m
modulus
200 GPa
moment of inertia
5000000 mm^4
end condition
pin-pin
Assumptions
Ideal Euler column: initially straight, concentric load, linear elastic material, buckling in the elastic range.
K values are the THEORETICAL effective-length factors (design codes recommend modified values for real end fixity, e.g. AISC suggests K = 0.65 instead of 0.5 for fixed-fixed).
Solution steps
Effective length
The end restraint sets the effective-length factor K; the column buckles as an equivalent pin-ended column of length KL.
Le = K·L
K = 1 (pin-pin); Le = 1 × 3 m = 3 m
Euler critical load
The Euler formula gives the axial load at which the ideal elastic column becomes unstable.
Euler formula Pcr = π²EI/(KL)² (NCEES FE Reference Handbook) — pin-pin K=1
verified by
hand-recomputed
derivation
I = 5e6 mm⁴ = 5e-6 m⁴. Pcr = π²·200e9·5e-6/3² = 9.8696044·1e6/9 = 1.09662e6 N = 1096.6 kN.
Example 2
SI fixed-free flagpole with Ix/Iy: weak axis governs
Given
length
3 m
modulus
200 GPa
moment of inertia x
8000000 mm^4
moment of inertia y
5000000 mm^4
end condition
fixed-free
Assumptions
Ideal Euler column: initially straight, concentric load, linear elastic material, buckling in the elastic range.
K values are the THEORETICAL effective-length factors (design codes recommend modified values for real end fixity, e.g. AISC suggests K = 0.65 instead of 0.5 for fixed-fixed).
Solution steps
Governing (weak) axis
A column buckles about the axis with the SMALLER moment of inertia — the weak axis bends most easily.
I(governing) = min(Ix, Iy)
Ix = 8000000 mm^4, Iy = 5000000 mm^4 → governing I = 5000000 mm^4
Effective length
The end restraint sets the effective-length factor K; the column buckles as an equivalent pin-ended column of length KL.
Le = K·L
K = 2 (fixed-free); Le = 2 × 3 m = 6 m
Euler critical load
The Euler formula gives the axial load at which the ideal elastic column becomes unstable.
Euler formula with weak-axis governing I (NCEES FE Reference Handbook) — fixed-free K=2, I = min(8e6, 5e6) mm⁴
verified by
hand-recomputed
derivation
Governing I = 5e-6 m⁴ (Iy < Ix). KL = 6 m. Pcr = π²·200e9·5e-6/36 = 9.8696044e6/36 = 274.156 kN.
Example 3
US pin-pin steel column with area and yield: inelastic-range warning
Given
length
20 ft
modulus
29000 ksi
moment of inertia
100 in^4
end condition
pin-pin
area
10 in^2
yield stress
50 ksi
Assumptions
Ideal Euler column: initially straight, concentric load, linear elastic material, buckling in the elastic range.
K values are the THEORETICAL effective-length factors (design codes recommend modified values for real end fixity, e.g. AISC suggests K = 0.65 instead of 0.5 for fixed-fixed).
Solution steps
Effective length
The end restraint sets the effective-length factor K; the column buckles as an equivalent pin-ended column of length KL.
Le = K·L
K = 1 (pin-pin); Le = 1 × 20 ft = 20 ft
Euler critical load
The Euler formula gives the axial load at which the ideal elastic column becomes unstable.
The Euler formula assumes elastic buckling. The usual validity bound keeps the critical stress at or below half the yield stress; above it the column buckles inelastically and Euler overestimates capacity.
valid while σcr ≤ σy/2
σcr = 49.69 ksi vs σy/2 = 25 ksi
Results
Effective-length factor K
1
Euler critical load Pcr
496.9kip
Slenderness ratio KL/r
75.89
Critical stress σcr
49.69ksi
Notes
σcr exceeds σy/2 — this column buckles in the INELASTIC range, where the Euler formula is not valid and overestimates capacity. Use the Johnson formula or the applicable design-code column curve.