I have an engineering competition coming up and one of the things I need to build is structure with both compression and tension members. The structure is scored based on the load held/weight ratio, and I already have the design I'm looking for but I need ways to reduce weight.

I was thinking about comparing the strength of a hollow rectangular section and a solid one with the same cross-sectional area. I think that I understand that for members under tensional stress, the strength of the two should be equal because the cross section of the two members is the same.

Would this be different for compression members? i.e., would one be able to carry more weight if one used a hollow member instead of a solid one?

EDIT: Another question: If I'm making this in a structural simulation, does the dimensions of the member matter if the cross-section is still the same? I guess you can assume here that the member is undergoing pure compression parallel to the length of the member.

  • $\begingroup$ Start with thinking about why bridges and viaducts have pillars to support them and are not solid.... $\endgroup$
    – Solar Mike
    Commented Mar 4, 2021 at 18:26
  • $\begingroup$ I tried looking it up but I couldn't really find anything about it - I guess it would be more stable? But I'm trying to see if there's maybe a mathematical thing behind it $\endgroup$
    – Lucas Wu
    Commented Mar 4, 2021 at 18:54

3 Answers 3


You are correct on both counts.

For members under tension, both cross-sections should behave equally, since the limit to the allowable tension force is simply:

$$P = f_y A$$

Where $f_y$ is the material's yield strength and $A$ is its area. So if both cross-sections have the same material and area, they'll resist the same load.

However, compression is different. Elements under compression can fail either by being crushed or by buckling.

The force needed to crush an element has the same equation as tension:

$$P = f_y A$$

($f_y$ may or may not be the same in compression and tension, it depends on the material). So as far as crushing is concerned, your two cross-sections will also behave identically.

The force needed for an element to buckle, however, is harder to calculate. Buckling is a behavior which "slender" members face, wherein they might suddenly get out of shape (i.e. collapse) at much lower loads than the crushing force (think of squeezing a plastic straw from both ends; you apply just a bit of force and it suddenly bends at the middle).

The theoretical buckling force is known as Euler's critical load, and is given by

$$P = \dfrac{\pi^2 EI}{(KL)^2}$$

where $E$ is the material's modulus of elasticity (basically how "unsquishy" it is), $I$ is the cross-section's moment of inertia, $L$ is the member's length, and $K$ is a coefficient having to do with how the member is supported.

If you look at those variables, the only thing that could change between the two cross-sections is the moment of inertia $I$. The moment of inertia is basically a measure of how good a cross-section is at resisting bending.

Now, I could show you how to calculate the moments of inertia for these different cross-sections, but if we just want to know which one will have a higher moment of inertia, we just need to use our intuition.

Think of a rectangular popsicle stick. Think of how easily it bends if you push it when it is horizontal, and how it basically doesn't budge if you push it when it's vertical. From this example, we can already see that the moment of inertia is very dependent on a cross-section's height and not so much on its width.

The reason why is a bit complicated to explain right now, but this means that when it comes to bending (and therefore buckling), you want to put as much of your cross-section as far away (height-wise) from its center as possible.

So, if we think of things this way, it becomes clear that the hollow section is better than the full one. After all, a hollow section doesn't "waste" material by putting it near the center and instead throws everything to the edges (specifically, it will be taller than the full section).

You could therefore imagine spreading that area into a super-thin-walled, super-tall cross-section, but don't take things too far. If the walls get too thin and the section too tall, you can start having local buckling effects which are way too complicated to discuss here. Just keep things reasonable and the hollow section will take you a long way.

  • $\begingroup$ Thanks. This was really helpful! $\endgroup$
    – Lucas Wu
    Commented Mar 4, 2021 at 19:19

There is a fundamental difference between tension and compression. Bars under compression are prone to buckling.

enter image description here

Compared to tensile strength which depends on the cross-sectional area A (Compressive strength also depends on A), buckling of compressed members depends on

  • the length L
  • the modulus of elasticity E
  • the second moment of area I

The second moment of Area is a geometric constant of the cross-section which shows how the cross-sectional area is distributed in space (not just the surface area but its shape and distance). As a result the compressive behaviour of long members under compression is better for hollow cross-sections.

An example, assume a solid bar (SOLID) of square cross-section 1x1 (assume its meters although any other unit of length measurement would be ok also).

Assume a hollow section of wall thickness 0.05[m]. In order for the crosssection to be equal you'd need to have an breadth of about 5[m] (more precisely $b_{ext}=$5.05m external and precisely $b_{int}=$4.995 internal).

In both cases the cross-sectional areas are the same $A_{solid}= A_{hollow}$. However because the second moment of area of a solid bar is given by $$I_{solid} = \frac{1}{12}b^4 = 0.0833 [m^4]$$

and the hollow section I can be calculated by subtracting the second moment of for the external and the internal, you get

$$I_{hollow} =I_{ext} -I_{int} =\frac{1}{12}b_{ext}^4- \frac{1}{12}b_{int}^4$$ $$I_{hollow} =2.0834[m^4]$$

As you can see the ratio $\frac{I_{ext}}{I_{int}}=25$

So although, you have the same total cross-section and therefore tensile behaviour, (for sufficiently long members) you can get significant improvements when you are using hollow sections.

  • $\begingroup$ This is interesting. Thanks! $\endgroup$
    – Lucas Wu
    Commented Mar 4, 2021 at 19:20

In compression, except for very short members the member fails by buckling under Euler's formula:

$$P_c= n\frac{ π^2 E I}{ L^2} $$

Where n is a factor based on the column's end connections.

Therefore we see that with the same area of a section if we can increase its I, second area moment, we increase the compression force the member can safely take.

The I of hollow square or rectangular or circular section is very close to a section with the same size. Alternatively, the same section area if used in a hollow section creates a lrger I than a solid section.

Lets use a square of sides a as an examole

$$I=a^4/12 $$

but for a hollow section where the void is a square with sides a/2$$ I=a^4/12-a/2^4/12= \ 15/16 *a^4/12$$

Removing 25% of section area leads to only 1/16 less I.


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