Question 11.2: A long, slender column ABC is pin-supported at the ends and ......

Use a four-step problem-solving approach.
1. Conceptualize: 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. 11-9c 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. 11-9b for the mode shape of buckling).
Column properties: From Table F-2 obtain the following moments of inertia and cross-sectional area for an IPN 220 column:
\quad\quad\quad\quad I_{1}=3060~{\mathrm{cm}}^{4}\ \ \ \ \ I_{2}=162~{\mathrm{cm}}^{4}\ \ \ \ \ A=39.5~{\mathrm{cm}}^{2}
2. Categorize:
Critical loads: If the column buckles in the plane of the figure, the critical load is
\quad\quad\quad\quad P_{cr}\,=\,{\frac{\pi^{2}\,E\,I_{2}}{(L\,/2)^{2}}}\,=\,{\frac{4\pi^{2}\,E\,I_{2}}{L^{2}}}
3. Analyze: Substitute numerical values to obtain
\quad\quad\quad\quad P_{cr}\,=\,\frac{4\pi^{2}E I_{2}}{L^{2}}\,=\,\frac{4\pi^{2}(200\,\mathrm{GP}{\mathrm{a}})(162 \mathrm{cm}^{4})}{(8\,\mathrm{m})^{2}}=200\,\mathrm{kN}
If the column buckles perpendicular to the plane of the figure, the critical load is
\quad\quad\quad\quad P_{cr}\,=\,{\frac{\pi^{2}E I_{1}}{L^{2}}}\,=\,{\frac{\pi^{2}(200\,\mathrm{GP}{\mathrm{a}})(3060\,\mathrm{cm}^{4})}{(8\,\mathrm{m})^{2}}}\,=\,943.8\,\mathrm{kN}
Therefore, the critical load for the column (the smaller of the two preceding values) is
\quad\quad\quad\quad 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, verify that the critical stresses do not exceed the proportional limit of the material. For the larger critical load, the critical stress is
\quad\quad\quad\quad \sigma_{\mathrm{cr}}={\frac{P_{\mathrm{cr}}}{A}}\,=\,{\frac{943.8\,\mathrm{~kN}}{39.5\,\mathrm{cm}^{2}}}=238\,.9\,\mathrm{MPa}
Since this stress is less than the proportional limit (\sigma_{pl} = 300 MPa), both critical-load calculations are satisfactory.
4. Finalize:
Allowable load: The allowable axial load for the column, based on Euler buckling, is
\quad\quad\quad\quad P_{\mathrm{allow}}={\frac{P_{\mathrm{cr}}}{n}}\,=\,{\frac{200\,\mathrm{kN}}{2.5}}\,=\,79.9\,\mathrm{kN}
in which n = 2.5 is the desired factor of safety.