Question 11.1: 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. 11-14). 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 (W 8 × 28) having modulus of elasticity E = 29 × 10³ ksi and proportional limit \sigma_{pl} = 42 ksi. The total length of the column is L = 25 ft.
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 = 12.5 ft and bending occurs about axis 2-2 (see Fig. 11-8c 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 = 25 ft (see Fig. 11-8b for the mode shape of buckling).
Column properties. From Table E-1, Appendix E, we obtain the following moments of inertia and cross-sectional area for a W 8 × 28 column:
I_{1}=98.0 in.^{4} I_{2}= 21.7 in.^{4} A= 8.25 in.^{2}
Critical loads. If the column buckles in the plane of the figure, the critical load is
P_{cr}=\frac{\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{4\pi^{2}(29\times10^{3} ksi)(21.7 in.^{4})}{}=276 kIf the column buckles perpendicular to the plane of the figure, the critical load is
P_{cr}=\frac{\pi^{2}EI_{1}}{L^{2}}=\frac{\pi^{2}(29\times10^{3} ksi)(98.0 in.^{4})}{}=312 kTherefore, the critical load for the column (the smaller of the two preceding values) is
P_{cr}=276 kand 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:
Since this stress is less than the proportional limit (\sigma_{pl} = 42 ksi), both critical-load calculations are satisfactory.
Allowable load. The allowable axial load for the column, based on Euler buckling, is
in which n = 2.5 is the desired factor of safety.
