10a-ex. Buckling Examples
Ex 10a.1
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Example 10a.1

Given: An aluminum (E = 70 GPa) column built into the ground has length, L = 2.2 m, and is under axial compressive load P. The dimensions of the cross-section are b = 210 mm and d = 280 mm.

(a) Determine the critical load to buckle the column.
(b) If the yield strength of the aluminum is 240 MPa, is the column more likely to buckle or to yield?
(c) If the factor of safety is F.S.=1.95, what is the allowable buckling load.

Step 1. The Euler Buckling Formula is:

where Le is the effective length of the column. The column is fixed-free for buckling about both x- and y-directions. For a fixed-free column, the effective length is:

Le = 2L = 4.4 m

The column may buckle about the x- or y- axis. The moment of inertia for a rectangle is:

The effective length of
a fixed-free column
is Le =2L


Here:   Ix = 3.84 x 108mm4 and Iy = 2.16 x108 mm4

The smaller moment of inertia governs since it results in the smaller Euler Buckling load.
This load is:

Pcr = 7711 kN

Step 2. Will the column Buckle or Yield?

The Buckling Strength is the Euler Buckling Load divided by the area A=bd:  

scr = 131.1 MPa
  • If scr < Sy = 240 MPa, the column will buckle (as compressive stress s increases from zero, the buckling strength is reached first);
  • If scr > Sy, the column will yield (s reaches Sy first).

In this our case, the column will buckle.

Step 3
. Allowable Load Pallow against buckling with a factor of safety of F.S. = 1.95.

The factor of safety is:  F.S. = (Failure Load)/(Allowable Load):

Pallow = Pcr / F.S. = (7711 kN)/(1.95)

Pallow = 3954 kN

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Updated: 05/22/09 DJD