Example 10.1.1

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.

Req'd:
(a) The critical load to buckle the column.
(b) If the allowable compressive stress in the Aluminum is 240 MPa, is the column more likely to buckle or yield?
(c) If the factor of safety is F.S.=1.95, what is the allowable buckling load.

Sol'n:
Step 1: The Euler Buckling Formula is given by:

Pcr = p2EI Le2

Where L_e is the effective length of the column. Here, the column is fixed-free in both x- and y-directions. For a fixed-free column, the effective length is:

L_e = 2L = 4.4 m

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

I = (base)(height)³/12

Effective length of a fixed-free column is L_e =2L

For our values of b and d, we have: I_x = 3.84 x 10⁸mm⁴ and I_y = 2.16 x 10⁸ mm⁴

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

Step 2: Will the column Buckle or Yield?

The Critical Buckling Stress is the Euler Buckling Load divided by the area, A=bd. This results in a Buckling Stress of:  

• If s_cr < 240 MPa, the column will buckle (since as the load is applied, the buckling stress is reached first);
• If s_cr > 240 MPa, the column will yield since the yield stress, S_Y is reached first.

Step 3:

With respect to buckling only, the Allowable Load on the column, P_allow, for a Factor of Safety is F.S. = 1.95.

The Factor of Safety is defined as:  F.S. = Failure Load / Allowable Load