Crippling Failure of Columns

Nomenclature

2.3.2 Introduction to Crippling Failure of Columns

In the previous sections, the primary failure of simple columns was considered. However, if a column has thin sections, it may fail at a load well below the critical load predicted for primary failure. Thus, a column must, in general, be checked for both primary failure and crippling. Primary failure may be assumed to be independent of crippling effects, in which case, a failure curve such as that shown in Figure 2-63 may be used. The right-hand portion of the curve describes the stress required for primary failure of the column at various effective slenderness ratios. This curve is cut off at the crippling stress level by the flat portion to the left.

If the interaction between crippling and primary failure is to be taken into account, the constant F_co in the Johnson-Euler equation may be set equal to the crippling stress F_cc. Although this procedure is more correct, it also introduces added complications, and only works for columns having a crippling stress less than the primary failure stress as L'/ρ approaches zero. Figures 2-64 and 2-65 show sets of Johnson-Euler curves for various materials. The curve used in a given case is the one that intercepts the ordinate at the value of F_cc for that column.

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2.3.2.1 Crippling Stress of Round Tubes

Steel tubes for which the diameter-to-wall thickness ratio is less than 50 need not be checked for crippling. This gives us some general idea of the thinness required if a tube is to fail by crippling rather than by primary instability. A theoretically correct formula for the crippling stress of a tube is

where r is the mean radius and μ = Poisson's ratio. if μ is taken to be 0.3, as is the case for steel and aluminum alloys, we obtain

This equation, however, is extremely unconservative for small values of t/r and should only be used for approximations in the early stages of design.

More conservative and accurate empirical methods are available for the treatment of tubular columns. If L/r is less than 0.75, use the equation

where the critical stress coefficient, K, is given in Figure 2-66.

If L/r is greater than 0.75, use the equation

where C is given in Figure 2-67.

2.3.2.2 Sample Problem - Crippling Stress of Round Tubes

Given: The tubular aluminum column shown in Fignre 2-68.

Find: The crippling stress if

(a) L = 5 in.
(b) L = 60 in.

Solution: (a) In this case, L/r < 0.75 so we may use Equation (2-22)

$$ {r \over t} = {10 \over 0.02} = 500 $$

From Figure 2-66, K = 11. Substituting this value into Equation (2-22), we find that

$$ F_{cc} = { 2 K \pi^2 r \over L^2 t } = { 2 (11) \pi^2 (10) \over (5)^2 (0.02) } = 4,350 ~\text{psi} $$

(b) If L = 60 in., L/r > 0.75 so that Equation (2-23) may be used

$$ {r \over t} = 500 $$

From Figure 2-67, C = 0.22. Substituting this value into Equation (2-23), we find that

If the theoretically correct formula, Equation (2-20), is used for either of these two columns,

$$ F_{cc} = 0.605 { Et \over r } = 12,100 ~\text{psi} $$

This value may be seen to be much greater than those obtained from the more accurate empirical formulas.

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2.3.2.3 Crippling Stress of Outstanding Flanges

Two idealized cases of edge restraint of long flanges are shown in Figure 2-69. In case (a), the flange is fixed along its edge and the equation for its crippling stress is

The flange shown in (b) is hinged along its edge and the equation for its crippling stress is

A column flange is neither rigidly fixed nor hinged along its edge so its crippling stress lies between those given by Equations (2-24) and (2-25), the latter giving more conservative results.

2.3.2.4 Crippling Stress of Angle Elements and Complex Shapes

The basic design equations for the crippling stress and load of the angle section shown in Figure 2-70 are

and

Here b' is equal to (h+b)/2 as seen in Figure 2-69 and C_e is a constant dependent upon the fixity of the edges, as shown to the right of Figure 2-70.

The area of this angle is given by

Nondimensional plots of Equations (2-26) and (2-27) are shown in Figures 2-71 and 2-72, respectively. These plots may be used to facilitate the solution of angle problems. It must be noted that Equations (2-26) and (2-27) have no significance when F_cc is greater than F_cy. These cutoffs are shown for two alloys in the following figures.

Many complex sections, such as those shown in Figure 2-73, may be treated by considering them to be composed of a number of angles. The crippling stress of these sections may be found by the following procedure: First, break the section up into a number of angles. Secondly, find the crippling load and area of each of the angles. Finally, find F_cc for the entire section from the following equation:

This procedure is illustrated in the following example.

2.3.2.5 Sample Problem- Crippling Stress of a Complex Shape

Given: Column with the cross sectional shape shown in Figure 2-74. It is composed of an aluminum alloy for which E = 10⁷ psi and F_cy = 50 kips.

Find: The crippling stress F_cc·

Solution: This section may be broken into three angle sections as shown by the broken lines above. The calculation of the crippling load and area of each angle section is summarized in the table below:

The values of b and h in this table are as shown in Figure 2-70. The values of P_cc and A may either be found from Equations (2-27) and (2-28) or from Figure 2-72 which shows these equations in graphical form. The crippling stress may now be found from Equation (2-29) to be

Substituting the material properties into the above equation gives

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2.3.2.6 Crippling Stress of I Beams

Figure 2-75 shows an I section.

The crippling stress of a column of this shape is given by

where k_w is given in Figure 2-76.

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