I've got partial answers but nothing comprehensive found online - most online resources focus on Euler's buckling formula which isn't appropriate for shorter columns.
Data I have collected:
Column is standard 90x90 4mm SHS steel, 2.8m long, treated as pinned-pinned top and bottom, so:
- Young's Modulus E = 2.05 x 1011 N/m2 (205 GPa)
- Yield Strength sy = 4 x 108 N/m2 (400 MPa)
- 2nd Moment of Area I = 1.663 x 10-6 m4 (166.3 cm4) from tables
- Radius of Gyration r = 0.035 m (35 mm) from tables
- Cross section area A = 0.0081 m2 (81 cm2)
- Physical length L = 2.8 m
- End Fixing Factor k = 1 (pinned-pinned)
- Eccentricity e = 40 mm
- Effective length Le = 1 x 2.8 = 2.8 m
- Slenderness S = Le/r = 2.8/0.035 = 80 (intermediate)
- Critical value C = √ (2π2E/sy) = 100.6
- S < C hence use Johnson's formula
I've got the critical load using Johnson's formula without eccentricity allowed for, as
A.sy.(1 – sy. Le2/(4π2r2E)) = 2215 kN
But how do I modify my calculation to work out the critical load with eccentricity?
Also what if the eccentricity means that the vertical line of the load is outside the column (e>45mm), or does that automatically mean the pinned-pinned column buckles?