Johnson–Euler Column Curve: Critical Stress and Transition

Compare two column models without confusing them with allowable strength

For this teaching example, E=200000 MPa, compressive yield strength Sy=250 MPa and effective slenderness lambda=KL/r=80. The Johnson parabola gives about 199.34 MPa; applying Euler alone gives about 308.43 MPa, above the supplied yield strength. This illustrates why an elastic-only prediction can be unsuitable at smaller slenderness.

The half-yield stress is the tangency level for this particular joined curve. It is not a universal physical proportional limit or proof of Euler validity for every material. Euler assumes an ideal elastic column; Johnson is an empirical intermediate-column approximation. Neither output alone is a code-compliant allowable design stress.

Use consistent material, geometry and end-restraint inputs

B2 and B3 use the same stress unit, here MPa. B4 is the dimensionless effective slenderness KL/r. K represents the appropriate restraint and stability conditions, L is the member length and r=SQRT(I/A) for the bending axis being assessed. L and r must use the same length unit.

Assess all relevant buckling directions: the smallest radius does not by itself identify the governing direction if effective lengths differ. This sheet takes KL/r as an input; it does not infer K or check the geometry. B5 is area in mm2, used only for the optional load conversion.

Calculate the unrounded transition and select the branch

E3 computes Cc=SQRT(2*PI()^2*E/Sy). E4 selects Johnson when lambda<Cc and Euler otherwise. E5 computes Sy*(1-Sy*lambda^2/(4*PI()^2*E)) on the Johnson branch and PI()^2*E/lambda^2 on the Euler branch.

For the initial inputs, Cc is 125.663706…, not exactly 125.66. Both expressions give Sy/2=125 MPa at the exact Cc. Keep E3 unrounded for branching; setting lambda to the displayed 125.66 is close to, but below, the transition.

=IF(E2="OK",IF(B4<E3,B3*(1-B3*B4^2/(4*PI()^2*B2)),PI()^2*B2/B4^2),E2)

Convert critical stress to a model load

E6 multiplies the raw stress by area and divides by 1000 to report kN. Because 1 MPa=1 N/mm2, 600 mm2 with the initial inputs gives about 119.60 kN. This is a model critical load, not a permitted service load. No safety or resistance factor is applied.

E2 rejects missing, text, zero or negative material/slenderness inputs. E6 additionally requires a numeric positive area. The formal lambda→0 limit of the Johnson expression is Sy; a zero-slenderness input is rejected because it does not describe the column case being assessed.

=IF(E2<>"OK",E2,IF(OR(COUNT(B5)<>1,N(B5)<=0),"Invalid area",E5*B5/1000))

Check the transition and the scope of the estimate

Set B4 to150 and verify the Euler branch. Set B4 to =SQRT(2*PI()^2*B2/B3) and verify125 MPa. Compare values just below and above this transition: the joined curve is continuous, with a matching slope. Double the area and only the load output doubles.

Real member strength can be governed by eccentric loading, imperfections, local or torsional buckling, residual stress, connection behavior, section instability or crushing. The worksheet does not check those modes. Use the applicable material/member design rules and actual boundary conditions before treating a result as a capacity.

Further reading

• Autodesk Inventor — Basic calculation of column buckling
• Intermediate-Length Columns with Central Loading — lecture material