Introduction
Column buckling is a curious and unique subject. It is perhaps the only area of
structural mechanics in which failure is not related to the strength of the material.
A
column buckling analysis consists of determining the maximum
load a column can support before it collapses. But for long columns,
the collapse has nothing to do with material yield. It is instead governed by the
column's stiffness, both material and geometric.
This page will derive the standard equations of column buckling using two approaches.
It will first cover the usual development of the equations,
i.e.,
Euler Buckling Theory. This is the
derivation found in text books and presented in engineering courses. But I have
never liked it. Not because it is incorrect
(it is correct), but because I don't think it satisfactorily presents the physical
mechanisms governing the buckling process. That is why a second
derivation of the buckling equations will also be presented.
Curiously, objects are referred to as
columns when they are loaded axially in compression,
as is the case here, but they are referred to as
beams when they are loaded transversely.
Nevertheless, beam bending theory is central to column buckling analyses, so it is recommended
that the reader review this
beam bending page.
Euler Buckling Theory
Euler Buckling Theory is the classical theory presented in textbooks and classrooms.
It begins simply by noting that the internal bending moment in a loaded and deformed
column is \(P \, y\) where \(P\) is the compressive load and \(y\) is the column
deflection. So insert \(P \, y\) in for \(M\) in the
beam bending equation,
\( E \, I \, y'' = M \).
\[
E \, I \, y'' = M = P \, y
\]
This produces the following differential equation
\[
E \, I \, y'' + P \, y = 0
\]
which has the solution
\[
y = A \sin \left( \sqrt{{P \over E \, I}} \; x \right) + B \cos \left( \sqrt{{P \over E \, I}} \; x \right)
\]
where \(A\) and \(B\) are constants determined from the boundary conditions.
The boundary conditions are \(y = 0\) at \(x = 0\) and \(x = L\).
The first boundary condition, \(y = 0\) at \(x = 0\), leads to the conclusion that \(B = 0\).
And this leaves
\[
y = A \sin \left( \sqrt{{P \over E \, I}} \; x \right)
\]
So far, so good. But it's at this point that the classical derivation tends
to leave physical intuition behind and become overtly mathematical...
Things become very interesting with the 2nd boundary condition because,
as we will see, it does not lead to determination of the unknown
constant, \(A\). To see this, insert the second boundary condition as follows.
\[
y(L) = 0 = A \sin \left( \sqrt{{P \over E \, I}} \; L \right)
\]
There are basically two possibilities here. In the first case, \(A = 0\),
but this is boring because it leads to the result that
all
displacements are zero. This is just the nonbuckled solution. Before
the column buckles, its lateral displacements are simply zero.
The second case is the interesting one, and the one directly related
to column buckling. The second method of satisfying the boundary condition
is to note that \(\sin(\pi) = 0\). Therefore the way to satisfy
the boundary condition is to require that the argument in the equation,
\(\left( \sqrt{{P \over E \, I}} \; L \right)\) must equal \(\pi\).
Doing so gives
\[
\sqrt{{P \over E \, I}} \; L = \pi
\]
and solving for \(P\) gives
\[
P_{cr} = { \pi^2 \, E \, I \over L^2 }
\]
This is the classical Euler buckling theory result. It gives the
critical value of load \(P\), called \(P_{cr}\), above
which, the column will buckle.
This result is perfectly legit. However, as should be evident by now,
it is very mathematical in nature, and provides little physical
insight as a result.
The upcoming derivation below will present an
alternative method of arriving at the same equation that I believe
provides a much more direct physical connection to the buckling process
than the above derivation did.
Buckling vs Yielding
As stated at the outset, classical buckling analysis is independent of a material's yield strength.
This is evident in the above derivation because at no time was stress or strain discussed or
compared to a material's strength.
But in fact, yielding considerations should never be totally ignored. Once one obtains an estimate of
\(P_{cr}\) from the above equation, one should always divide it by the column's crosssectional area,
\(A\), to obtain a stress
\[
\sigma_x = { P_{cr} \over A }
\]
and compare this value to the material's yield strength
to determine if yielding will occur before buckling. This is critical for short columns since
they have inherently high \(P_{cr}\) values because \(L^2\) is in the denominator of the buckling
equation.
End Constraints in Buckling
Though not the focus of this page, it is important to recognize that end constraints
are critical to buckling analyses because they alter the value of \(P_{cr}\). For example,
consider that the critical buckling load of the column shown here is given by
\[
P_{cr} = { \pi^2 \, E \, I \over 4 \, L^2 }
\]
Note that this value is 1/4 that given in the earlier equation for \(P_{cr}\).
But it is easy to see why. The sketch shows that the buckling condition here
is exactly equivalent to the buckling of a column of twice the length and having
the same boundary conditions as in the above derivation. And this leads to
\[
P_{cr} = { \pi^2 \, E \, I \over ( 2 \, L)^2 } = { \pi^2 \, E \, I \over 4 \, L^2 }
\]
PhysicallyBased Buckling Derivation
This section will present an alternative method of determining
critical buckling loads that I believe is more physically intuitive
than classical Euler buckling theory.
Its key feature is the
comparison of the internal bending moment arising from the internal stress
distribution to the external bending moment resulting from the
load applied to the column. (The previous sentence is critical.)
This approach picks up approximately where it was
noted earlier that the classical Euler approach becomes overtly
mathematical. The first step is to assume a deformed shape.
We know that \(y = 0\) at both ends of the column, and that
the shape follows a \(\sin()\) function based on the above
differential equation. So the most logical choice is
\[
y(x) = \delta_{max} \sin ( {\pi x \over L} )
\]
where \(\delta_{max}\) is the lateral displacement at the
midpoint of the column. Its value is unknown, but it is
known to be greater at the midpoint than at any other point
on the column, hence the \(max\) subscript. The choice
of \(\pi x / L\) as the argument of the \(\sin()\) function
ensures that the displacements are zero at \(x = 0\) and
\(x = L\). (We will talk about other assumed shapes shortly.)
If you are wondering about the \(\sqrt{P/EI}\) term in the
\(\sin()\) function from the earlier Euler solution, don't.
It arose from the solution of the differential equation,
but we are pretending to know nothing about the details
of that solution here (exception alert!).
We only need a function that can resemble a bowed column
with zero displacements at its ends.
Using \(\sin ( {\pi x \over L} )\) accomplishes this.
The "exception alert" is present above because we are
in fact taking advantage of one piece of knowledge
about the earlier analysis. It is that trig functions
are the solutions of the differential equation.
Therefore, sines and cosines should be used
whenever possible to describe the deformed shapes because
they will lead to the most accurate estimates
of \(P_{cr}\).
Recall from
beam bending theory
that the bending moment, \(M\), is related to the deflection
of the column by
\[
M = E \, I \, y''
\]
Though not critical, it is helpful to remember that this relationship
came from calculating the moment in the crosssection due to the
stress distribution. The relationship shows that we need the
second derivative of the assumed displacement function.
\[
\begin{eqnarray}
M \; = \; E \, I \, y'' \; & = & \; E \, I \, {d^2 \over dx^2 } \left\{ \delta_{max} \sin ( {\pi x \over L} ) \right\} \\
\\
\\
& = & \;  {\pi^2 \, E \, I \, \delta_{max} \over L^2 } \sin ( {\pi x \over L} ) \\
\\
\\
& = & \;  {\pi^2 \, E \, I \, y(x) \over L^2 }
\end{eqnarray}
\]
This is the internal bending moment in the column due to the stress distribution
within it, which is in turn due to the fact that the column is bent.
Here comes an important thought... This bending moment can be thought of as
the column's internal resistance to bending, or the strength with which it
tries to straighten back out.
The simple nextstep is to equate this internal resistive bending moment to
that resulting from the external load. That amount is simply
\(M = P \, y(x)\). Equating the two gives
\[
M \; = \; P \, y(x) \; = \;  {\pi^2 \, E \, I \, y(x) \over L^2 }
\]
It is not hard to see at this point that this approach is leading to the
same expression for \(P_{cr}\) as the classical Euler buckling theory did.
But along the way, it has provided much more insight into the physical
process of buckling than the former theory. Namely...
 We have arrived at this relationship by equating the internal
bending moment (due to the internal stresses arising from the column's bending)
to the external bending moment resulting from the external load, \(P\).
It should be clear that buckling occurs when \(P\) is large enough
to satisfy the equation. Any value less, and \(P \, y(x)\) will
be less than the "resisting bending moment."

The fact that \(y(x)\) appears on both sides of the equation, and will
therefore be cancelled out, means that when buckling does occur, it
does so simultaneously throughout the length of the column.
(A fascinating result that is not evident in the Euler theory.)
Anyway, cancelling out \(y(x)\) from both sides of the equation gives
\[
P_{cr} = { \pi^2 \, E \, I \over L^2 }
\]
again, but with much more physical insight this time.
Fixed End Example
This time, assume a deformed shape of
\[
y = \delta_{max} ( 1  \cos ( { \pi x \over 2 L }) )
\]
Calculate the bending moment due to this.
\[
M \; = \; E \, I \, y'' \; = \; {\pi^2 \, E \, I \, \delta_{max} \over 4 \, L^2 } \cos ( {\pi x \over 2 L} )
\]
The bending moment due to the external load is \(M = P ( \delta_{max}  y(x) )\). Equating these two
and simplifying gives the familiar result.
\[
P_{cr} = { \pi^2 \, E \, I \over 4 L^2 }
\]
Fixed End Example with Different Assumed Shape
This example will demonstrate that alternative functions can be assumed for the deformed shape of the
column, and that the resulting formula for \(P_{cr}\) will not be significantly different from the
exact solution.
This time, assume a deformed shape of
\[
y = \delta_{max} \left( { x \over L } \right)^2
\]
Calculate the bending moment due to this.
\[
M \; = \; E \, I \, y'' \; = \; { 2 \, E \, I \, \delta_{max} \over L^2 }
\]
The bending moment due to the external load remains \(M = P ( \delta_{max}  y(x) )\). Equating these two gives
\[
P ( \delta_{max}  y(x) ) = { 2 \, E \, I \, \delta_{max} \over L^2 }
\]
It's clear from the equation that the minimum value of \(P\) will occur at \(x = 0\)
because it's at this point that \(y(x)\) is a minimum (zero, in fact) and therefore
\((\delta_{max}  y(x))\) is a maximum. Setting \(y(x)\) to zero and cancelling
\(\delta_{max}\) from both sides gives
\[
P_{cr} = { 2 \, E \, I \over L^2 }
\]
This result is obviously not equal to the exact solution above. The difference is that
the exact solution contains \(\pi^2 / 4 = 2.467\), while this approximate solution contains 2,
a 23% difference. Significant, but not large like, say, a factor of 2, or an order of magnitude.
Note also that the approximate solution is conservative since it gives a critical buckling load
less than that of the exact solution.
It is interesting to note that this solution based on the quadratic shape, led to a lower critical
buckling load and a concentration of the buckling failure at the base of the column, \(x = 0\).
In contrast, the exact solution consisting of the trig function, produced an equal buckling
tendency all along the length of the column, and a corresponding higher \(P_{cr}\).
A key factor here is that the assumed quadratic deformed shape is not an exact solution of the
governing differential equation (though it is close). Trig functions are.
Textbooks