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2.12. Struts with unsymmetrical cross-sections

The formulae derived in the preceding paragraphs have assumed that buckling takes place about an axis of symmetry. Loading is then normally applied to produce bending on the strongest or major principal axis (that about which / has a maximum value) so that buckling is assumed to occur about the minor axis. It is also assumed that the end conditions allow rotation in this direction and this is normally achieved by loading through ball ends.

For sections with only one axis of symmetry, e.g. channel or T-sections, the shear centre is not coincident with the centroid and torsional effects are often introduced. These may, in some cases, affect the failure condition of the strut. Certainly, in the case of totally unsymmetrical sections, the loading condition always involves considerable torsion and the theoretical buckling load has little relevance. One popular form of section which falls in this category is the unequal-leg angle section.

Some sections, e.g. cruciform sections, are subject to both flexural and torsional buckling and the reader is referred to more advanced texts for the methods of treatment is such cases.

A special form of failure is associated with hollow low carbon steel columns with small thickness to diameter ratios when the strut is found to crinkle, i.e. the material forms into folds when the direct stress is approximately equal to the yield stress. Southwell has investigated this problem and produced the formula where a is the stress causing yielding, R is the mean radius of the column and t is the thickness. It should be noted, however, that this type of failure is not common since very small t/R ratios of the order of 1/400 are required before crinkling can occur.

Example 2.1

Two 300 mm x 120 mm I-section joists are united by 12 mm thick plates as shown in Fig. 2.14 to form a 7 m long stanchion. Given a factor of safety of 3, a compressive yield stress of 300 MN/m2 and a constant a of 1/7500, determine the allowable load which can be carried by the stanchion according to the Rankine-Gordon formulae.

Examples

12 mm

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