Calculation of critical column buckling load using Johnson's formula, but with eccentricity?

Question

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 s_y = 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

Therefore:

• Effective length L_e = 1 x 2.8 = 2.8 m
• Slenderness S = L_e/r = 2.8/0.035 = 80 (intermediate)
• Critical value C = √ (2π²E/s_y) = 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.s_y.(1 – s_y. L_e²/(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?

Answer 1

You have to apply the off-center moment to the column and let it bend. Then apply the axial load but with an offset of $\Delta$

So basically you can calculate the deflection of a cantilever column with its L half of the length of your column Under the moment $M=P*40mm$, due to symmetry this will be the deflection at both ends that we call $\Delta$

Now you apply the axial load that you already calculated at this offset.

It creates both moment and compression stress. we use $e= \frac{M_u}{P_u}$ and look up the charts or use code formula

$\frac{M_u}{S*\sigma_{allo}}+ \frac{ P_u}{A*\sigma_{axial}}\leq 1$

Edit

After OP's comment

When you have a column under both axial load and moment on way to handle a case like this is to assume a cantilever column or beam with a length half of the effective length of the column is subjected to moment and deflects. This deflection called delta is now plugged in to calculated the moment of load P applied at this offset. Why half of L, because in a pin, pin, supports the column will bend symmetrically about the center and the max deflection is at the center. we call that delta, and the situation is called the $P\Delta$ effect.

In the figure, the column has bent with max deflection delta at the middle.

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