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In the buckling of columns we know that:

$$P = \dfrac{n^2\pi^2EI}{L^2}$$

The smallest value of P occurs when $n=1$ which gives a simple buckling shape (one wave):

$$P_{cr} = \dfrac{\pi^2EI}{L^2}$$

However for $n > 1$, as shown below the buckling shape is more complex and has many waves:

Buckling shapes

My question is do the buckling mode shapes for $n > 1$ ever occur in reality? If the column begins to buckle as per the shape for $n = 1$ then wouldn't it just continue to buckle like this until failure? How would the other buckling modes ever occur?

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Whether or not buckling modes with $n>1$ exists depends on how you look at the structure.

As @hazzey notes in his answer, columns with bracing may display buckling modes with $n>1$. These buckling modes, however, are simply equivalent to the $n=1$ modes of the individual segments that compose the column. To be clear, this doesn't mean that the segments behave independently (you'll never have two consecutive unbraced lengths buckling to the same side), only that any $n>1$ mode can be composed by a series of continuous $n=1$ modes for the unbraced lengths.

So, if you have a column with a single bracing which buckles, do you consider that an $n>1$ mode for the entire column or an $n=1$ mode for each of the unbraced lengths? Both? Your call.

enter image description here

To paraphrase @starrise's comment on @hazzey's answer, this can be demonstrated by looking at the buckling equation: \begin{align} P &= \left(\dfrac{n}{L}\right)^2\pi^2EI \\ P_{column,\,n=2} &= \left(\dfrac{2}{L}\right)^2\pi^2EI \\ P_{segment,\,n=1} &= \left(\dfrac{1}{\frac{L}{2}}\right)^2\pi^2EI = \left(\dfrac{2}{L}\right)^2\pi^2EI \\ \therefore P_{column,\,n=2} &= P_{segment,\,n=1} \end{align}

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If you are looking at the column as being supported at the ends, you are correct that the n=1 mode gives the lowest buckling load.

The other modes (n=2,3,...) aren't useless though. Long columns are often braced at regular intervals to reduce the unbraced length of the column. For a given length of column, these braces force the column to buckle under a different mode (n=2,3,...) with the corresponding increase in buckling load.

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  • $\begingroup$ I didn't realize that the mode shapes referred to bracing of the columns but that really makes sense now that I think about it. $\endgroup$
    – pauloz1890
    Commented Oct 26, 2015 at 1:11
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    $\begingroup$ But wouldn't the load for the column's global $n>1$ mode be equal to the $n=1$ mode of one of its unbraced segments? This means that whether $n>1$ modes exist depends on how you look at the structure. If you look at it from a global perspective, then yes, $n>1$ modes are possible. If, however, you look at the local segments that compose the structure, then only $n=1$ modes exist. @pauloz1890 $\endgroup$
    – Wasabi
    Commented Oct 26, 2015 at 1:43
  • $\begingroup$ @Wasabi Yes i think that's what has confused me you're right. $\endgroup$
    – pauloz1890
    Commented Oct 26, 2015 at 1:47
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    $\begingroup$ As @Wasabi noted, only $n=1$ modes exist when considering bracing. To see why, note that in the $n=2$ case, $L_{segment}=L_{global}/2$. Then $P=4\pi^2 EI/(4L_{segment}^2)=\pi^2 EI/L_{segment}^2$ which is identical to the $n=1$ case but for a shorter column. The same naturally applies for any $n$. This should all make sense since the top and bottom of the original global column can be said to be braced in the same sense (at least with these boundary conditions). $\endgroup$
    – do-the-thing-please
    Commented Oct 26, 2015 at 2:11
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    $\begingroup$ @SamWatkins, indeed, the cases are not independent. They couldn't be, since we are talking about a single monolithic column with bracing. If both sections buckled to the same side, then there'd be a discontinuity in the deformation angle of the column, which is impossible. The statement that $n>1$ modes are actually just a series of mode 1 is not meant to imply that each of the mode 1's is independent, but rather that an $n>1$ mode only happens in the real world if it can be composed by a series of continuous mode 1's. $\endgroup$
    – Wasabi
    Commented Oct 26, 2015 at 9:56

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