Question 11.2: 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. 11-15). 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 standard steel shape (IPN 220) having modulus of elasticity E = 200 GPa and proportional limit \sigma_{p l} = 300 MPa . The total length of the column is L = 8m .
Determine the allowable load P_{\text{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 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 E-2, Appendix E, we obtain the following moments of inertia and cross-sectional area for an IPN 220 column:
I_{1}=3060 \mathrm{~cm}^{4} \quad I_{2}=162 \mathrm{~cm}^{4} \quad A=39.5 \mathrm{~cm}^{2}TABLE E-2
Properties of European Standard Beams 
| Designation | Mass per meter |
Area of section |
Depth of section |
Width of section |
Thickness | Strong axis 1-1 | Weak axis 2-2 | |||||
| G | A | h | b | t_{w} | t_{t} | I_{1} | S_{1} | r_{1} | I_{2} | S_{2} | r_{2} | |
| Kg/m | {cm}^{2} | mm | mm | mm | mm | {cm}^{4} | {cm}^{3} | cm | {cm}^{4} | {cm}^{3} | cm | |
| IPN 550 IPN 500 |
166 141 |
212 179 |
550 500 |
200 185 |
19 18 |
30 27 |
99180 68740 |
3610 2750 |
21.6 19.6 |
3490 2480 |
349 268 |
4.02 3.72 |
| IPN 450 IPN 400 |
115 92.4 |
147 118 |
450 400 |
170 155 |
16.2 14.4 |
24.3 21.6 |
45850 29210 |
2040 1460 |
17.7 15.7 |
1730 1160 |
203 149 |
3.43 3.13 |
| IPN 380 IPN 360 IPN 340 IPN 320 IPN 300 |
84 76.1 68 61 54.2 |
107 97 86.7 77.7 69 |
380 360 340 320 300 |
149 143 137 131 125 |
13.7 13 122 11.5 10.8 |
20.5 19.5 18.3 17.3 16.2 |
24010 19610 15700 12510 9800 |
1260 1090 923 782 653 |
15 14.2 13.5 12.7 11.9 |
975 818 674 555 451 |
131 114 98.4 84.7 72.2 |
3.02 2.9 2.8 2.67 2.56 |
| IPN 280 IPN 260 IPN 240 IPN 220 IPN 200 |
47.9 41.9 36.2 31.1 26.2 |
61 53.3 46.1 39.5 33.4 |
280 260 240 220 200 |
119 113 106 98 90 |
10.1 9.4 8.7 8.1 7.5 |
15.2 14.1 13.1 12.2 11.3 |
7590 5740 4250 3060 2140 |
542 442 354 278 214 |
11.1 10.4 9.59 8.8 8 |
364 288 221 162 117 |
61.2 51 41.7 33.1 26 |
2.45 2.32 2.2 2.02 1.87 |
| IPN 180 IPN 160 IPN 140 IPN 120 IPN 100 |
21.9 17.9 14.3 11.1 8.34 |
27.9 22.8 18.3 14.2 10.6 |
180 160 140 120 100 |
82 74 66 58 50 |
6.9 6.3 5.7 5.1 4.5 |
10.4 9.5 8.6 7.7 6.8 |
1450 935 573 328 171 |
161 117 81.9 54.7 34.2 |
7.2 6.4 5.61 4.81 4.01 |
81.3 54.7 35.2 21.5 12.2 |
19.8 14.8 10.7 7.41 4.88 |
1.71 1.55 1.4 1.23 1.07 |
| IPN 80 | 5.94 | 7.58 | 80 | 42 | 3.9 | 5.9 | 77.8 | 19.5 | 3.2 | 6.29 | 3 | 0.91 |
| Note: Axes 1-1 and 2-2 are principal centroidal axes. | ||||||||||||
Critical loads. If the column buckles in the plane of the figure, the critical load is
P_{\mathrm{cr}}=\frac{\pi^{2} E I_{2}}{(L / 2)^{2}}=\frac{4 \pi^{2} E I_{2}}{L^{2}}Substituting numerical values, we obtain
P_{c \mathrm{r}}=\frac{4 \pi^{2} E l_{2}}{L^{2}}=\frac{4 \pi^{2}(200 \mathrm{~GPa})\left(162 \mathrm{~cm}^{4}\right)}{(8 \mathrm{~m})^{2}}=200 \mathrm{~kN}If the column buckles perpendicular to the plane of the figure, the critical load is
P_{\mathrm{cr}}=\frac{\pi^{2} E I_{1}}{L^{2}}=\frac{\pi^{2}(200 \mathrm{~GPa})\left(3060 \mathrm{~cm}^{4}\right)}{(8 \mathrm{~m})^{2}}=943.8 \mathrm{~kN}Therefore, the critical load for the column (the smaller of the two preceding values)
P_{\mathrm{cr}}=200 \mathrm{~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:
Since this stress is less than the proportional limit (\sigma_{p l} = 300 MPa) , 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 m is the desired factor of safety.
