Relative strength of equal-area cross sections?

This started out as a seemingly straightforward question but the more I thought about it the less sure I was. Here’s the question: given two bars the same length and the same cross-sectional area, which will be stronger- a bar with a square cross section or an equilateral triangle cross section? IOW, one bar is a square prism and the other is a triangular prism; the bars are the same length and weight, only their shape is different.

I’ll let you figure it out and tell us!

http://www.engineersedge.com/calculators/section_square_case_19.htm

http://www.engineersedge.com/calculators/section_square.htm

Its a very good question. The first thing to think about is what we mean by strength. So let me illustrate with a long bar of metal - say 10 square inches inches in cs area and 4 ft long :

1> Tension (or compression strength) - Imagine using the bar as a piece to hang things of - like a crane that lifts heavy objects. In this case - the triangular section and the circular section will behave the same i.e. they will be able to lift the same weight before failing at a certain weight.

2> Forces with moment - Imagine now using the bar as a bench. (sort of like a bridge between two banks) and sitting or putting something on the middle. In this case - the triangular bar will perform better in terms of deflection under the load and the weight under which it will break.

I have left out torsional stresses (like using the bar in a motor to deliver torque). In most everyday applications, the 2nd case is most prevalent and hence you see structural members shaped to address this (I-beam).

Assuming you are talking about bending (forces with moment), the Moment of Inertia I (also known as the second moment of area) will tell you the resistance to bending of that cross-sectional shape. The higher the I, the better that shape is at resisting bending. Or basically what am77494 said. :slight_smile:

Does it change things if the triangular bar is used pointy edge up or down as a beam?

Compression would be like using a bar as the post in a bar stool? Tension would be hanging something from it vertically (or using it as a tow bar)?

The short answer: the triangular beam is stronger when it comes to bending or compression. Both beams are about the same when it comes to tension.

Long answer: let’s look at the three cases (bending, compression, and tension.)

[Bending] This occurs, for example, in a beam used as the horizontal supports in a bench. Euler-Bernoulli beam theory tells us that the amount which the beam sags is related to it’s second moment of area. The beam whose cross section is an equilateral triangle has a moment that is ~15% larger than a square beam with the same cross-sectional area. Therefore the triangular beam is stronger in the sense that it will flex less than the square one. This is true regardless of orientation (e.g., point up vs. point down). I’m not sure, however, whether it can withstand a greater load; failure should occur when the stress on the top/bottom edges of the beam is large enough to cause mechanical failure, and I haven’t worked out which beam will experience greater stress on the edges. Sounds like a good homework problem!

[Compression] This occurs, for example, in a post in a bar stool. Failure occurs here when the beam buckles, which again is related to the second moment of area due to a formula derived by Euler. So for compression, the triangular beam wins.

[Tension] Tension occurs, for example, when using the beam to tow something. Here it is being pulled length-wise. The strength of the beam here is just related to the area, which is the same for both beams. So they should perform similarly under tension.

I’m gonna say the square one. If they’re the same length and weight and cross sectional area, the triangular one is going to be much bigger and therefore less dense. My guess, though, is that you added an extra constraint that you didn’t mean to and thereby forced the triangle to be too big.

ISTM, the problem would make more sense if you said both bars were to be made of the same material and you dictated their widest dimension (say, they both had to be inscribed in a once inch circle).
Also, keep in mind that the area of a triangle is a third of the area of a rectangle with the same base and height which is why the triangle bar is going to be much bigger if you require them both to have the same cross sectional area, again, I’m curious if that’s what you meant to say. And with that, forcing them to have the same weight is what might make the triangle bar weaker.

I meant given the same amount of material and making the bars the same length, does one cross section have an advantage over the other? The triangular one might have a greater maximum width (and a smaller minimum width) but its not “less dense” than the square one. See this animation of square-to-triangle-and-back dissection.

That’s what I was saying, by adding the constraint of “equal-area cross sections” you changed the problem into something I felt you didn’t intend.

But now that I rethink it, I think I was wrong. If the face of each bar as the same area, they’ll both have the same volume since they’re the same length and since they weigh the same, they both have the same density.

When I pictured the bars along with what you said, it didn’t make sense to me. It wasn’t until I imagined taking two identical chunks of material (say, play-dough) and trying to form these two bars that I realized that it would work just fine.

So, nevermind what I said upthread.

NO.Density and weight changes has nothing to do with the question.

he meant the same amount of material, as in the same cross sectional area and same length… square vs triangular…

He meant the square and triangle are solid , not just a skin… (not a box beam).

One may be better than the other in compression or bending ; certainly an I beam is better than a square beam… a hollow beam is better than a solid beam (of the same weight per unit length…)

The cross sectional shape makes it no better or worse in shear or tension, within reason . (of course if it was as thin as foil, it could be torn at one part and then the tear progresses across easily.)

OK, so let’s look at an example. Say the triangle has a base of 10" and a height of 8.66" (since it is equilateral). The square with an equal cross-sectional are has base and height equal to 6.58". Each has an area of 43.3 in^2

I of the triangle is 1/36bh^3=180.4 in^4
I of the square is 1/12bh^3=156.2 in^4

So assuming my math is correct (it’s been a while since I’ve done these types of calcs!) the triangle has the slight edge for resistance to bending.

Shape has no effect on tensile strength, so they are both the same.
Shape has an effect on compressive strength only if the section is slender enough to buckle before it crushes. So it also depends on the length of the column.

For typical bending loading conditions the triangle would go pointy-side down, but I’m blanking a bit on why exactly that is. I believe it is because the top side is in compression and it’s better to have more of the section at the top to take the compressive bending force. That’s my story and I’m sticking to it…