CivilSolve

Statics & Mechanics of Materials

Euler Column Buckling

Critical buckling load Pcr = π²EI/(KL)² for an ideal elastic column with pin-pin, fixed-free, fixed-fixed, or fixed-pin ends (theoretical K = 1, 2, 0.5, 0.7). Give Ix and Iy together to have the weak axis govern automatically; add the area for slenderness KL/r and critical stress, and the yield stress for the inelastic-range validity check. Use for: 'critical/buckling load of the column', 'Euler load', allowable-load-from-buckling problems. Not for: inelastic (Johnson) buckling or code design capacities.

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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

  1. 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

  2. Euler critical load

    The Euler formula gives the axial load at which the ideal elastic column becomes unstable.

    Pcr = π²·E·I/(K·L)²

    Pcr = π² × 200 GPa × 5000000 mm^4 / 3 m² = 1097 kN

    = 1097 kN

Results

Effective-length factor K

1

Euler critical load Pcr

1097kN

Where this answer was checked
source
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

  1. 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

  2. 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

  3. Euler critical load

    The Euler formula gives the axial load at which the ideal elastic column becomes unstable.

    Pcr = π²·E·I/(K·L)²

    Pcr = π² × 200 GPa × 5000000 mm^4 / 6 m² = 274.2 kN

    = 274.2 kN

Results

Effective-length factor K

2

Euler critical load Pcr

274.2kN

Governing (weak-axis) I

5000000mm^4

Where this answer was checked
source
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

  1. 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

  2. Euler critical load

    The Euler formula gives the axial load at which the ideal elastic column becomes unstable.

    Pcr = π²·E·I/(K·L)²

    Pcr = π² × 29000 ksi × 100 in^4 / 20 ft² = 496.9 kip

    = 496.9 kip

  3. Slenderness ratio and critical stress

    With the area known, the radius of gyration gives the slenderness ratio, and the critical load maps to an average critical stress.

    r = √(I/A); slenderness = KL/r; σcr = Pcr/A = π²E/(KL/r)²

    r = 3.162 in; KL/r = 75.89; σcr = 49.69 ksi

    = 49.69 ksi

  4. Euler validity check

    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.
Where this answer was checked
source
Euler formula + σcr = π²E/(KL/r)² validity bound (NCEES FE Reference Handbook) — L=20 ft, E=29000 ksi, I=100 in⁴, A=10 in², σy=50 ksi
verified by
hand-recomputed
derivation
KL = 240 in. Pcr = π²·29000·100/240² = 9.8696044·2.9e6/57600 = 496.91 kip. r = √(100/10) = 3.16228 in; KL/r = 75.895. σcr = 496.91/10 = 49.69 ksi > σy/2 = 25 ksi → inelastic warning must fire.