How does removing mass make it stronger?

[thetexan]

In the above image if one begins with a long square bar of metal it tends to bend when you apply a force in the middle of the bar. However, if I remove a portion of the bar leaving a L shaped bar the newly created angle bar seems to resist that same force and not bend nearly as much (at least has been my experience).

If this is true how can this be? The angle bar exists within the original bar. In fact you would think that leaving the original bar intact would give even more mass to resist the force. If I use an angle bar to strengthen something wouldn't using the entire bar be even better (strength wise)? Of course, you save on material to use an angle bar but why does it seem to be stronger that the original bar from which it is created?

tex

 

[Khashishi]

I don't believe it is stronger. Did you actually test this quantitatively?

 

[LachyP]

I definitely wouldn't say that removing mass is making it stronger, but when the illusion of "removing mass" is caused by moving at relativistic velocities (when relativistic length applies), this would happen. So maybe, what you mean is that the mass is not being "removed" but merely concentrated/compacted into a particular position while moving in a certain direction? When this mass is concentrated/compacted it would definitely make it "stronger"...

 

[Drakkith]

If this is true how can this be? The angle bar exists within the original bar. In fact you would think that leaving the original bar intact would give even more mass to resist the force. If I use an angle bar to strengthen something wouldn't using the entire bar be even better (strength wise)? Of course, you save on material to use an angle bar but why does it seem to be stronger that the original bar from which it is created?

I'm not sure if it's stronger or not. It was my understanding that removing most of the metal in this manner greatly reduces the weight of the bar yet allows it to keep most of its strength.

I definitely wouldn't say that removing mass is making it stronger, but when the illusion of "removing mass" is caused by moving at relativistic velocities (when relativistic length applies), this would happen. So maybe, what you mean is that the mass is not being "removed" but merely concentrated/compacted into a particular position while moving in a certain direction? When this mass is concentrated/compacted it would definitely make it "stronger"...

Relativity has absolutely nothing to do with this. The OP's is asking about removing the middle of the beam and leaving the edges.

 

[LachyP]

I'm not sure if it's stronger or not. It was my understanding that removing most of the metal in this manner greatly reduces the weight of the bar yet allows it to keep most of its strength.
Relativity has absolutely nothing to do with this. The OP's is asking about removing the middle of the beam and leaving the edges.

I had only mentioned relativity because I do not think that removing mass would make it stronger, so I went into reasoning how it could appear to have been removed but be much stronger. So I had answered in a completely different sense. And to apologise I must say that I had not read the diagram and had just pounced onto a possible conclusion. Sorry about that.

 

[Nugatory]

I'm not sure if it's stronger or not. It was my understanding that removing most of the metal in this manner greatly reduces the weight of the bar yet allows it to keep most of its strength.

That is indeed what is going on.

It's a standard practice in designing aircraft and race cars to do a stress analysis of structural members, then remove material from the lightly stressed parts of the member to reduce the weight but not the strength.

 

[Vanadium 50]

Removing mass does not make it stronger. It makes it weaker. It does (or can), however, make it stronger per unit weight. A rod is stronger than a tube, but pound for pound, a tube is stronger than a rod.

If I flex a rod, one side of the rod gets longer, and so is under tension, and one side gets shorter, and is under compression. There is some spot in the middle that is therefore under neither compression or tension, and so can be removed. My rod is now just as strong, but infinitesmally lighter. Extend this idea to places on the rod that are under only a little tension or a little compression and you see how the tube gets to be almost as strong as a rod, but much lighter.

 

[anorlunda]

Removing mass as in your illustration does not make the object stronger, it makes it stronger per pound.

Let's say a L-shaped steel beam 10 feet long weighs 1000 pounds. That is stronger than a 1000 pound solid rectangular or steel beam 10 feet long.

 

[Chestermiller]

Are you really referring to strength, or are you referring to stiffness. The term strength is usually reserved for conditions under which the member fails and breaks. The term stiffness is used for describing how much the member displaces, bends, or deforms under applied load.

Chet

 

[anorlunda]

I stand corrected Chet. Stiffness

 

[256bits]

Removing mass does not make it stronger. It makes it weaker. It does (or can), however, make it stronger per unit weight. A rod is stronger than a tube, but pound for pound, a tube is stronger than a rod.

If I flex a rod, one side of the rod gets longer, and so is under tension, and one side gets shorter, and is under compression. There is some spot in the middle that is therefore under neither compression or tension, and so can be removed. My rod is now just as strong, but infinitesmally lighter. Extend this idea to places on the rod that are under only a little tension or a little compression and you see how the tube gets to be almost as strong as a rod, but much lighter.

In a roundabout way, this ( and the other posts) why the OP, from his experience, is seeing that a bar with material removed is "stronger" than the solid bar.

An I-beam, due to symmetry, is probably best to use to visualize how, by removing material, the resulting beam can actually be "stronger" with the ability to support more of a load, or with less of a deflection under the same load, as compared to a solid beam.

As mentioned, the outer fibers take up most of the compressive and tensile stress - the inner members less so. If we remove material from the solid, and end up with an I-beam shape, the beam will have one flange in compression, the other in tension, with the web under some stress but mainly serving to separate the two flanges a certain distance. The I-beam should at first glance, be just about as "strong" as the soild, but with much material removed.

Span length is important. A short beam will fail due to shear before it fails in bending, so here a solid beam is preferable.
Thus, we want a long enough beam so that bending stress is important, so that we can compare a solid beam to a cut out beam.

If we place these 2 long beams side by side and simply support them, we should notice something peculiar - the solid beam can actually deflect more the I-beam. And to get the same deflection for both beams we can add some load to the I-beam.The I-beam with material removed really does look to have become stronger than the solid.

So what's going on.
It is simply that the solid beam has to support all of its own dead weight and deflects more, in contrast to the lighter I-beam, (with material removed that wasn't contributing to strength, at least not much anyways ). One could put the removed material and lay it on the I-beam, and both beam deflections should be just about the same.
( Of course, each design of a beam is different with different flange and web thickness, so do it is not an actual hard and fast rule.

This is one of the reasons we try to not build things out of solid material, such as the way they built the pyramids. The dead weight of solid members eventually gets to you, and you cannot build as high, as wide, without extra base area, or extra supporting columns, which again adds to dead weight.

 

[thetexan]

My dad was an aeronautical and mechanical engineer. I'm not sure if this has anything to with it but it seems I remember him talking about a modulus of something regarding this.

If I remember correctly there is something about the width to depth ratio that's important. I. Other words a 10x10 inch solid bar has a different modulus or bending coefficient than a 10x2 inch beam when a force is applied to the narrow edge in the direction of the 10 inch side making the square bar more likely to bend than the beam under the same load.

I don't know but seems like there was something about that.

Tex

 

[Nugatory]

A 10x10 is like five 2x10 beams side by side - it is stiffer (and in this thread, everyone including me who has said "stronger" really means "stiffer") than one 2x10. However, it is much less stiff than a 5x20 would be.

 

[Wes Tausend]

Tex,

One other consideration depends in which cross-section direction the beam is stressed. The angle iron shown above is strong (stiff) when stressed in a 90 degree multiple of lateral or vertical direction.

But if the angle-iron beam is stressed at a 45 degree angle to the "flat" sides, the solid beam is significantly stronger (stiffer) because the angle-iron tends to flatten and collapse. If one then decides to combine two angle-irons together to form a square tube, the combo gets stronger (stiffer) in this 45 stress than a single angle-iron, but can still easily collapse, flatten and fold, from this 45 degree stress. Finally, the square tube is still not as stiff as a similarly 45 degree stressed solid square bar, which resists any collapse and subsequent fold and is actually stronger (stiffer) in this direction than it is squared up with the stress.

For the same "direction-of-stress vs wall" reasons, a 4 inch round frame tube is noodle-y when center-bent in any direction, while a 4 inch square frame tube better resists vertical flexing because the vertical walls are parallel to the up/down force. But the 4" round tube resists torsional twisting stresses far better than the square 4" tube even though the round tube contains less mass. The wall(s) of a round tube are precisely matched to the same direction as torsional stress. That's why most truck driveshafts (and axles) are round rather than square.

Wes
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