The buckling stress, using the Rankine and Gordon Equation or the Perry-Robertson Equation, is always less than the yield stress of the material?

  • $\begingroup$ Can you add more information to your question? Maybe adding the equations that you are referencing would help others to guide you to an answer. Buckling is typically dependent on the dimensions of the sample. $\endgroup$
    – hazzey
    Commented Apr 25, 2019 at 13:30
  • $\begingroup$ No. Why do you think it should be? The formula doesn't tell you anything about whether a structure might fail some other way before it buckles. It's hard to see what you are really trying to ask here. $\endgroup$
    – alephzero
    Commented Apr 25, 2019 at 14:10
  • $\begingroup$ Most buckling equations can yield stresses that are higher than the yield stress of a material. However this is not a sensible result, yielding can occur even in compression. $\endgroup$
    – ShadowMan
    Commented Apr 25, 2019 at 15:10
  • $\begingroup$ Hallo guys. The question was phrased the very same way l have just put. $\endgroup$ Commented Apr 25, 2019 at 16:43
  • $\begingroup$ Thank you guys for your ideas. Am trying to get my head around a Structural Analysis course. $\endgroup$ Commented Apr 25, 2019 at 20:56

1 Answer 1


Both formulae cap the column's maximum stress to its yield stress

Both the Rankine-Gordon and Perry-Robertson formulae are means of determining the maximum average stress ($\sigma = P/A$) a given column can support. Due to the fact that short columns will collapse under compression rather than buckling, it's important that these equations cap the result to the material's compressive yield stress. As we will see, this is precisely what both equations do.

Rankine-Gordon Formula

The formula is as follows:

$$\sigma = \dfrac{\sigma_c}{1+a\left(\dfrac{L}{k}\right)^2}$$


  • $\sigma_c$ is the material's compressive yield stress;
  • $a = \dfrac{\sigma_c}{\pi^2 E}$ (where $E$ is the material's Young's modulus), but is usually determined experimentally;
  • $L$ is the column's length;
  • $k = \sqrt{\dfrac{I}{A}}$, the column's least radius of gyration.

So, in the case where the column is infinitely long, we know the column should buckle immediately. This is precisely what happens, since $\sigma$ is inversely proportional to $L$.

Now, if instead we have a very short beam ($L$ tending to zero), the equation simplifies to $\sigma = \dfrac{\sigma_c}{1} = \sigma_c$. Since you can't have $L < 0$, we get that the compressive yield stress is the maximum value possible according to the Rankine-Gordon Formula.

Perry-Robertson Formula

This formula is:

$$\sigma = \dfrac{1}{2}\left(\sigma_c + \sigma_e(1+\theta) - \sqrt{(\sigma_c + \sigma_e(1+\theta))^2 - 4\sigma_c\sigma_e}\right)$$


  • $\sigma_e$ is the average stress ($P/A$) associated with the beam's Euler buckling load
  • $\theta = \dfrac{w_{0,i}c}{k}$
  • $w_{0,i}$ is the initial imperfection
  • $c$ is the maximum distance to the section's centroid (in the relevant direction)

Once again, let's look at the case of an infinitely long column. In this case, we know that $\sigma_e = 0$, so the equation simplifies to

$$\begin{align} \sigma &= \dfrac{1}{2}\left(\sigma_c - \sqrt{\sigma_c^2}\right) \\ &= \dfrac{1}{2}\left(\sigma_c - \sigma_c\right) \\ \sigma &= 0 \end{align}$$

This formula also has infinitely long columns buckling immediately.

Now, for the column with a length approaching zero. This is a more complicated case. Such a beam would have $\sigma_e = \infty$, but let's pretend we can actually play with such numbers. Let's also adopt the simplification suggested by Robertson of assuming $\theta \approx 0.003\lambda = 0.003\dfrac{L}{k}$, which when combined with our zero-length column gives us $\theta = 0$. The equation then simplifies to:

$$\begin{align} \sigma &= \dfrac{1}{2}\left(\sigma_c + \infty - \sqrt{(\sigma_c + \infty)^2 - 4\sigma_c\infty}\right) \\ &= \dfrac{1}{2}\left(\sigma_c + \infty - \sqrt{\sigma_c^2 + 2\sigma_c\infty + \infty^2 - 4\sigma_c\infty}\right) \\ &= \dfrac{1}{2}\left(\sigma_c + \infty - \sqrt{\sigma_c^2 - 2\sigma_c\infty + \infty^2}\right) \\ &= \dfrac{1}{2}\left(\sigma_c + \infty - \sqrt{(\sigma_c - \infty)^2}\right) \\ &= \dfrac{1}{2}\left(\sigma_c + \infty - |\sigma_c - \infty|\right) \\ &= \dfrac{1}{2}\left(\sigma_c + \infty - (\infty - \sigma_c)\right) \\ &= \dfrac{1}{2}\left(\sigma_c + \sigma_c\right) \\ \sigma &= \sigma_c \end{align}$$

We had to play a bit loose with our mathematical concepts, but this demonstrates that even for very short columns, the stress is never greater than the compressive yield stress using the Perry-Robertson Formula.

Given this, we can see the answer to the question is YES, both methods always give results which are lower than the material's yield stress.1

I'd mention, however, that the proper term is just "ultimate stress", not "buckling stress" as in the question. Short columns collapse under compression at stresses far below the stress required for them to buckle. Basically: for short columns, the ultimate stress is roughly equal to the yield stress $\sigma_c$. For slender columns, it's equal to the buckling stress since that's lower.

1 If you want to be a real pedant, you could argue the answer is NO, since both methods give zero-length columns ultimate stresses which are equal to (not less than, as asked by the question) the yield strength. But that would require you to actually consider zero-length columns as a real thing which they obviously aren't, and that you don't accept an implicit "or equal to" after the question's "less than".

  • $\begingroup$ I hope that the questioner at least gives you a copy of the "A" that they get on their homework or exam... : ) $\endgroup$
    – hazzey
    Commented Apr 30, 2019 at 12:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.