bending rigidity of a rod

[engineheat]

Hi,

Say you have a metal rod with a square cross section (say 10cm by 10cm) of a certain length. You'd like to make it less prone to bending. Is it better to:

1. keep it the way it is

or

2. remove material so that it has a T shaped cross section

I wonder if removing material would actually make it more rigid from a bending perspective simply due to the T shaped cross section.

Thanks

[FreddieChopin]

You could cut this rod to 4 plates and then weld them in cross shape - that would be preferable from strength standpoint.

[Domagoj T]

"Area moment of inertia" or "second moment of area" is the term you want to look up.

[wraper]

How removing material is supposed to make it more rigid? Although removing material can decrease bending under it's own weight due to lower weight to support.

[edy]

And check out this SlideShare show....

https://www.slideshare.net/Navazhusen/shear-stresses-on-beam-mechanics-of-solids

[ajb]

The basic rule is that for a given amount of cross sectional area, the farther the material within that area is from the neutral axis on average the more rigid the shape will be against bending.  The neutral axis is  where the material, when subject to a bending load, transitions from being in compression to being in tension along its length axis (see edy's diagrams for some examples).  Based on this rule, it becomes fairly obvious that for a given amount of material, something like an I-beam or truss structure is going to be the strongest, and a wide, short cross section (like a horizontal plate) is going to be the weakest.  There is no situation in which removing material from within a shape will make a shape that is stronger than what you started with (but it will reduce the weight of the beam itself, reducing the load on it, which may be just as good in certain cases).  However, by removing material near the neutral axis that has a relatively low contribution to the strength of the structure relative to the amount of weight it contributes, you can improve the strength to weight ratio of the structure.  Hence way lattice or truss structures are quite common (think of them like I-beams where strategic holes are cut in the web but the top and bottom plates remain intact).

[MosherIV]

neutral axis is  where the material, when subject to a bending load, transitions from being in compression to being in tension along its length axis

You could cut the rod in half and then join the 2 halves together by welding along the seem.
Or pin/bolt the 2 halves together.
This has the effect of converting the compression and tension forces into shear forces.
It is a lot harder to shear through a meterial.
I learned this when working with composite materials, ie a glass or carbon fibre rod gets most of its stiffness by converting bending forces to shearing force by the strands of glass/carbon fibre trapped in the resin

[ajb]

neutral axis is  where the material, when subject to a bending load, transitions from being in compression to being in tension along its length axis

You could cut the rod in half and then join the 2 halves together by welding along the seem.
Or pin/bolt the 2 halves together.
This has the effect of converting the compression and tension forces into shear forces.

(emphasis mine)

This already happens within the structure.  There are horizontal shear planes running parallel to the neutral axis due to the differential compressive/tensile stresses at different levels of the material as you move between top and bottom.  Slicing and then pinning the beam back together would only concentrate that horizontal shear into the pins rather than it being distributed through the entirety of the original contiguous material.

[MosherIV]

Slicing and then pinning the beam back together would only concentrate that horizontal shear into the pins

Precisley my point!
You can make a rod appear to be stiffer (not bend) by pinning 2 halves together.
Now, either the pins get sheared or the holes around the pins get distorted.

[coppercone2]

its like armor, you can rearrange the molecules to be thicker along a certain bend axis so you can try to concentrate force to an area that is 'thicker' in the direction applied but there is always a trade off some where with volume etc.

[engineheat]

The basic rule is that for a given amount of cross sectional area, the farther the material within that area is from the neutral axis on average the more rigid the shape will be against bending.  The neutral axis is  where the material, when subject to a bending load, transitions from being in compression to being in tension along its length axis (see edy's diagrams for some examples).  Based on this rule, it becomes fairly obvious that for a given amount of material, something like an I-beam or truss structure is going to be the strongest, and a wide, short cross section (like a horizontal plate) is going to be the weakest.  There is no situation in which removing material from within a shape will make a shape that is stronger than what you started with (but it will reduce the weight of the beam itself, reducing the load on it, which may be just as good in certain cases).  However, by removing material near the neutral axis that has a relatively low contribution to the strength of the structure relative to the amount of weight it contributes, you can improve the strength to weight ratio of the structure.  Hence way lattice or truss structures are quite common (think of them like I-beams where strategic holes are cut in the web but the top and bottom plates remain intact).

Thanks, I'm not looking for the strongest bending strength for a given amount of material, but just the two options I listed. I also cannot alter the shape much beyond that (such as cutting it in certain way and welding them together). It's either remove material to give it a I beam shaped or T shaped or keep it the way it is.

From the response, it seems I should just keep it a square profile for the maximum absolute strength, not strength to weight ratio. Weight is not relevant for my application. But size/space constraint is.

Thanks for the introduction to 2nd moment of inertia. It mathematically shows it never make sense to remove material as long as one is trying to improve the absolute resistance to bending, if weight is not a concern.

Thanks

[T3sl4co1l]

Right. The only remaining thing you can do is choose a material with higher elastic (Young's) modulus.  Going from aluminum (~70 GPa modulus) to steel (~200 GPa) for example is a huge improvement.  Besides its strength and hardness, this is what really sets carbide (~600 GPa) apart from the rest, for machine tools.  Not just the obvious application, using the hardness for cutting bits -- but also sprung elements like boring bars.

It's very handy that, because stiffness of a beam goes as width squared, the stiffness to weight ratio goes up considerably with light materials -- aluminum may be ~1/3 the stiffness of steel, but it's 1/3 the weight, so a member can be made 3 times thinner and 3 times wider, having about the same overall strength but 3 times the stiffness.

But since you aren't going that way, you can't take any advantage from it, so all you can do is pack in more material -- or use a stiffer material in the first place.

There aren't actually many materials any better than tungsten carbide; and they're all molecular exotica (nanotubes, graphene) or worse (diamond).

Ed: Also, I'm not sure if there's confusion over stiffness versus strength.  Stiffness is deflection under load.  Everything bends, even if imperceptibly so.  It's only when the bending is so extreme that the material tears itself apart, that we also ask how strong something is.  If you need strength but don't mind if there's a lot of deflection, you may find a special steel does a better job, or even carbon fiber.  If you need low deflection for any load, consider carbide.  If you need both, carbide is still a darn good choice, but it isn't very forgiving, obviously.

Tim