Engineering
Euler Column Buckling Load formula: P_cr = PI()^2 * E * I / (K * L)^2
Explore the ideal elastic Euler buckling load and its dependence on effective length, without treating it as a structural design rating.
Calculate an ideal elastic instability load
Euler assumes a straight, prismatic, linearly elastic column under concentric axial compression with idealized supports. Use E in Pa, the relevant bending-axis second moment I in m^4, length L in m and dimensionless effective-length factor K. The result is N before division by 1000.
K=1, 0.5 and 2 describe ideal pinned-pinned, fixed-fixed and fixed-free cases. Welding an end does not by itself establish ideal fixity: joint rotation, frame sway and bracing determine actual restraints. Compare possible axes and modes; the smallest I alone need not govern when effective lengths differ.
These calculations do not check yielding, imperfections, local or torsional buckling, connections or code requirements. Cross-sectional area is not supplied, so the sheet does not calculate slenderness KL/sqrt(I/A) or verify elastic validity. Dividing the Euler result by an arbitrary factor does not establish an allowable working load.
| Column Parameter / Support | Cell / Factor K | Steel Strut Input | Critical Load P_cr (kN) |
|---|---|---|---|
| Steel E & Weak-Axis I | B2 = 200E9 Pa, B3 = 8E-6 m^4 | Flexural rigidity EI = 1.6 MN*m^2 | Governing weak axis |
| Pinned-Pinned (L = 4 m) | B4 = 4 m, B5 K = 1.0 | (1.0 * 4)^2 = 16 m^2 | 986.96 kN |
| Fixed-Fixed Both Ends | K set to 0.5 | (0.5 * 4)^2 = 4 m^2 | 3,947.84 kN (4x ideal Euler result) |
| Fixed-Free Cantilever | K set to 2.0 | (2.0 * 4)^2 = 64 m^2 | 246.74 kN (one-quarter ideal result) |
Euler Critical Buckling Load P_cr in kilonewtons (kN)
The result is the ideal elastic critical load expressed in kN, not the rated capacity of a real member.
=IF(COUNT(B2:B5)<>4,"Invalid inputs",IF(OR(B2<=0,B3<=0,B4<=0,B5<=0),"Invalid inputs",ROUND((((PI() ^ 2) * B2 * B3) / ((B5 * B4) ^ 2)) / 1000, 2)))Illustrative division by a chosen factor
B6 must be at least 1. This arithmetic comparison is not a structural design check or approved safety factor.
=IF(COUNT(B2:B6)<>5,"Invalid inputs",IF(OR(B2<=0,B3<=0,B4<=0,B5<=0,B6<1),"Invalid inputs",ROUND(((((PI() ^ 2) * B2 * B3) / ((B5 * B4) ^ 2)) / 1000) / B6, 2)))Try the example
Use the original values or change them to test your understanding.