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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

Therefore:

  • 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?

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  • $\begingroup$ You started with the pin-pin assumption, so the eccentricity will cause instability of the column as the pin support can't restrain the rotation. For a fixed-free or fixed-pin condition, you shall look at the resulting curvature caused by the loads P & M. $\endgroup$
    – r13
    Commented Feb 9, 2022 at 19:19

2 Answers 2

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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.

'

p d

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  • $\begingroup$ Thanks! I'm quite new to this though - could you do a favour and identify what values those represent - codes are full of subscripts and I'm not familiar with them enough to be sure. Bonus request can you show me how it would apply in this concrete example, which will make it absolutely clear. $\endgroup$
    – Stilez
    Commented Feb 10, 2022 at 8:57
  • $\begingroup$ I am flying to Saint Luis Obispo. Later I will answer you concerns. $\endgroup$
    – kamran
    Commented Feb 10, 2022 at 18:44
  • $\begingroup$ I understand the type of bending expected in an eccentrically loaded column, but in not really after the deflection distance. I'm after the critical loading beyond which theoretically, buckling will occur. I've seen a few ways to do something like that - convert bending moment of eccentric load to a maximum stress/strain and add to the vertical load stress/strain, a formula something like 1/X + 1/Y <= 1 (similar to yours but I don't know what all symbols represent and what values to plug in), and a few other versions of it. Symbol clarity and a completed example case would really help? Thanks $\endgroup$
    – Stilez
    Commented Feb 10, 2022 at 19:07
  • $\begingroup$ If eccentricity is very large compared to radius of gyration then elastic bending takes place. $\endgroup$
    – Narasimham
    Commented Dec 3, 2023 at 20:07
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If you're looking for a single formula that includes the effect of eccentricity, I suggest you look up the Perry-Robertson formula. It explicitly includes the effects of eccentricity. It (or similar approaches) are the basis for a number of modern steel design codes (the Eurocode, and I believe Australian Standard AS4100 for example).

Note that for "real" design you will likely need to consider other effects such as initial imperfection of your column, residual stresses from the rolling / fabrication of the column etc. The approach taken by AS4100 (and I believe the Eurocode) is to "hide" allowances for these effects in the code equations, resulting in a number of equations to solve before you get to the buckling capacity.

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