Question 9.1: A long, slender column ABC is pin-supported at the ends and ...
A long, slender column ABC is pin-supported at the ends and compressed by an axial load P (Fig. 9-14). Lateral support is provided at the midpoint B in the plane of the figure. However, lateral support perpendicular to the plane of the figure is provided only at the ends.
The column is constructed of a steel wide-flange section (IPN 220) having modulus of elasticity E = 200 GPa and proportional limit \sigma_{pl}= 300 GPa. The total length of the column is L = 8 m.
Determine the allowable load P_{allow} using a factor of safety n = 2.5 with respect to Euler buckling of the column.
Because of the manner in which it is supported, this column may buckle in either of the two principal planes of bending. As one possibility, it may buckle in the plane of the figure, in which case the distance between lateral supports is L /2 = 4 m and bending occurs about axis 2–2 (see Fig. 9-8c for the mode shape of buckling).
As a second possibility, the column may buckle perpendicular to the plane of the figure with bending about axis 1–1. Because the only lateral support in this direction is at the ends, the distance between lateral supports is L = 8 m (see Fig. 9-8b for the mode shape of buckling).
Column properties. From Table E-1, Appendix E (available online), we obtain the following moments of inertia and cross-sectional area for a IPN 220 column:
I_1 = 3060 cm^4 I_2 = 162 cm^4 A = 39.5 cm^2
Critical loads. If the column buckles in the plane of the figure, the critical load is
P_{cr} = \frac{4\pi ^2 EI_2 }{(L/2)^2} = \frac{4\pi ^2 EI_2 }{L^2}
Substituting numerical values, we obtain
P_{cr} =\frac{4\pi ^2 EI_2 }{L^2} = \frac{4p^2(200 GPa)(162 cm^4)}{(8 m)^2} = 200 kN
If the column buckles perpendicular to the plane of the figure, the critical load is
P_{cr} =\frac{\pi ^2 EI_1 }{L^2} =\frac{p^2(200 GPa)(3060 cm^4)}{(8 m)^2}
Therefore, the critical load for the column (the smaller of the two preceding values) is
P_{cr} = 200 kN
and buckling occurs in the plane of the figure.
Critical stresses. Since the calculations for the critical loads are valid only if the material follows Hooke’s law, we need to verify that the critical stresses do not exceed the proportional limit of the material. In the case of the larger critical load, we obtain the following critical stress:
\sigma _{cr}=\frac{P_{cr}}{A} =\frac{943.8 kN}{39.5 cm^2} =238.9 MPa
Since this stress is less than the proportional limit (\sigma_{pl} = 300 MPa), both critical load calculations are satisfactory.
Allowable load. The allowable axial load for the column, based on Euler buckling, is
P_{allow}=\frac{P_{cr}}{n} =\frac{200 kN}{2.5} =79.9 kN
in which n = 2.5 is the desired factor of safety.
