A question for the mechanical engineers here

[BigAl]

Somewhere in my checkered education (including a course called Theoretical Mechanics 101 which introduced me to the wonderful concept of virtual work) I picked up the idea that forces involved with a structure scale as the cube of the dimensions. I've heard hand-waving explanations that sound right.

To what degree is that correct statement? If so, is there a good web page that demonstrates it is terms that someone with no physics might appreciate?

[SpitfireIX]

Somewhere in my checkered education (including a course called Theoretical Mechanics 101 which introduced me to the wonderful concept of virtual work) I picked up the idea that forces involved with a structure scale as the cube of the dimensions. I've heard hand-waving explanations that sound right.

To what degree is that correct statement? If so, is there a good web page that demonstrates it is terms that someone with no physics might appreciate?


Here's one explanation that has to do with nanotechnology.

http://minerva.union.edu/malekis/ESC24/KoskywebModules/sl_nano2.htm

[DavidS]

Somewhere in my checkered education (including a course called Theoretical Mechanics 101 which introduced me to the wonderful concept of virtual work) I picked up the idea that forces involved with a structure scale as the cube of the dimensions. I've heard hand-waving explanations that sound right.

To what degree is that correct statement? If so, is there a good web page that demonstrates it is terms that someone with no physics might appreciate?

How "forces" scale with size depends what forces you're talking about.

For geometrically similar structures of identical materials, weight and buoyancy forces will scale as the cube of any characteristic dimension. That is, pick one dimension (say, width) as the "characteristic dimension", and a geometrically similar structure of twice the width has twice the height (that's what geometric similarity means) and twice the depth (ditto), and therefore 8 times the volume, and therefore eight times the weight (which is just the volume times the identical materials) and buoyancy (which is just the volume times the density difference from the suspending fluid).

Other forces may scale differently.

The "hoop stress" forces in the walls of pressure vessels confining the same pressure scale as the square of the characteristic dimension along with the area against which that pressure acts.

Hydrodynamic drag at a fixed velocity, for example, will scale with the structure's projected *area*, not volume; that is, as the *square* of the dimension (if that fixed velocity is great enough to develop fully-developed turbulent flow about both structures, so their drag coefficients are equal). If you want to scale the velocity, too, you need to specify what sort of scaling you meant; if the velocity is to increase as the dimension, then the drag force will increase as its fourth power because it's proportional to the square of the velocity as well as the projected area.

Surface forces, for example the surface tension keeping your soap bubble as a single sphere instead of two hemispheres, may scale linearly with the characteristic dimension. That's why a little spit on your finger helps pick up a pin but a wet hand just makes a rusty spot on a steel ingot.

[GlennB]

While I'm not a mechanical engineer, the same principles certainly apply in biology (my sphere).

As DavidS explains, mass is a cube function. The strength of a load-bearing member - such as the bone or exoskeleton of an animal, or the stalk or trunk of a plant - is proportional to its cross-sectional area and therefore a square function.

So, wave a magic wand at a stalk of corn to make it 50' high and it would collapse instantly. The mass has risen beyond the increased strength of the stalk. Similarly, if you scaled a mouse up to the size of an elephant, its relatively flimsy bones would snap.

This is one of the (many) reasons why giant radioactive ants will never take over the world ;)

[H'ethetheth]

Think of it this way: Force = mass*acceleration. For identical accelerations, force then scales linearly with mass, which depends on volume and density. As DavidS said, for geometrically similar structures that have the same density, the ratio of the volumes and therefore the masses is given by the ratio of the sizes cubed.

That said, this does not seem to be the right subforum for this question.

[Lurker]

Forces do not scale 3:1 versus dimensions. It varies. I think where your idea came from is from the formula for the Moment of Inertia (I).

I = bh^3/12

is a common formula that is used for bending stresses in beams. b and h are dimensions of the beam (base and height). Note that the height is cubed while the base is to the first power.

So, when the base dimension is doubled the moment of intertia is also doubled. If the height is doubled the moment of inertia is 8x. It depends on the orientation and direction the force is applied.

Lurker (MS Mechanical Engineering)

ETA: Looking at the other responses I wanted to add it depends on what sort of force you are examining. I had assumed you were discussing statically loaded members like I-beams.

[DavidS]

The strength of a load-bearing member - such as the bone or exoskeleton of an animal, or the stalk or trunk of a plant - is proportional to its cross-sectional area and therefore a square function.

So, wave a magic wand at a stalk of corn to make it 50' high and it would collapse instantly. The mass has risen beyond the increased strength of the stalk. Similarly, if you scaled a mouse up to the size of an elephant, its relatively flimsy bones would snap.

Correct, strength of a member scales with the square of its dimension. The (weight) forces on corresponding members in similar structures scale with the cube.

The mouse:elephant analogy neatly demonstrates the difference (similar reasoning demonstrates that sauropod dinosaurs got about as big as they could possibly be and still walk; to support much more weight their legs would have to be so thick they'd have no room to move).

The corn analogy may not be so good, as Lurker describes:

Forces do not scale 3:1 versus dimensions. It varies. I think where your idea came from is from the formula for the Moment of Inertia (I).

I = bh^3/12

is a common formula that is used for bending stresses in beams. b and h are dimensions of the beam (base and height). Note that the height is cubed while the base is to the first power.

So, when the base dimension is doubled the moment of intertia is also doubled. If the height is doubled the moment of inertia is 8x. It depends on the orientation and direction the force is applied.
The distinction here is strength (resistance to forces) versus stiffness (resistance to moments -- bending) and stress (force resistance per unit area). Again, the (weight) forces scale with the cube of size, but the bending moments may not (4th, Lurker?), the local stresses (e.g the tension in the bottom edge of a horizontal beam) may not, and the stiffness does not. GlennB's cornstalk would likely buckle to the bending moment its own weight before it would run out of strength to support that weight in pure compression.

The distinction is of more than academic relevance. Depending on the shape and scale of a structure, a member that's strong enough to handle its forces may not have enough stiffness to be sufficiently rigid against the moments it will experience. I recall reading a tale (sorry, forgot where) about the first attempts to build super-light racing motorcycle frames from titanium; when the structural members were scaled down to the equivalent strength of their steel predecessors, the result was a floppy springy bike that rode like a wet noodle.

Stiffness is also one reason small-ish metal boats are made of aluminum but larger boats are made of steel. Thin steel would be strong enough for the small boat, but to be sufficiently stiff it would have to be thicker. Aluminum isn't as strong as steel, but when it's thick enough to be stiff enough it's still plenty strong and much lighter than the steel. Because the hull's buoyancy increases with the cube of the size, for a big enough boat the weight penalty of steel hurts less than the price penalty of aluminum.

The OP asked about forces, but his stated lack of expertise should have led us to highlight some of these distinctions.

[Dave Rogers]

Just to confuse matters further (well, I am a physicist, after all), there's also the gravitational potential energy stored in a structure to consider. This is an important figure to consider in that it determines what drives the collapse of the structure - see Frank Greening's work for further information. Gravitational potential energy scales as mass times height, and mass scales as the cube of the dimensions of the structure, so GPE scales as the fourth power of the linear dimensions. In other words, double the height of the tower and the GPE goes up 16 times. Since the fracture energy of the beams goes as their cross-sectional area (linear dimension squared), that means that the ratio of energy available to energy needed for collapse has gone up four times, rather than staying the same as some might expect. These are the sort of arguments that demonstrate that physical scale models of the WTC towers are useless for predicting the behaviour of the real towers.

Dave

[Lurker]

Correct, strength of a member scales with the square of its dimension. The (weight) forces on corresponding members in similar structures scale with the cube.

The mouse:elephant analogy neatly demonstrates the difference (similar reasoning demonstrates that sauropod dinosaurs got about as big as they could possibly be and still walk; to support much more weight their legs would have to be so thick they'd have no room to move).

The corn analogy may not be so good, as Lurker describes:

Actually, it is pretty good for members in tension or compression. So between us we covered a lot of the failure modes since mine was looking at bending while yours covered compressive and tensile failure.

I should add that when I used the "b" and "h" vairiables they refer to the cross-section, not the length of the beam at all. The "h" is not the height of the beam, but the height of the cross section (assuming bending around an axis orthogonal to "h" and the length of the beam.

I'd say your post was spot on.

Lurker

[rwguinn]

Actually, it is pretty good for members in tension or compression. So between us we covered a lot of the failure modes since mine was looking at bending while yours covered compressive and tensile failure.

I should add that when I used the "b" and "h" vairiables they refer to the cross-section, not the length of the beam at all. The "h" is not the height of the beam, but the height of the cross section (assuming bending around an axis orthogonal to "h" and the length of the beam.

I'd say your post was spot on.

Lurker

And what these "DE-Debunkers" fail to realize is that before the aircraft plowed through the buildings, "b" and 'h" were about 210 feet. Each.
Afterwards, they were the size of a single beam's cross section, measurable in inches...




Amazing how shear panels work, ain't it...

[peteweaver]

The forces you have to bear in mind are tension torsion shear and compression.

An overloaded beam will bend, with the forces of tension at the top of the beam, and compression on the lower half of the beam...

If an external force acts upon a structure, for instance, in the case of the tacoma narrows bridge: wind, then torsion (twisting) may take place, and cause fatigue in the steel.

Part of an engineers job, is to balance everything, another part is to estimate a safety factor, so that in an estimated worst case scenario, the structure can hopefully survive.

One of the tools an engineer will undoubtably depend on in the design of a structure, is Young's modulus http://en.wikipedia.org/wiki/Young's_modulus (http://en.wikipedia.org/wiki/Young%27s_modulus)

Here's some information about bending moments.
http://www.roymech.co.uk/Useful_Tables/Beams/Shear_Bending.html

[slyjoe]

...
An overloaded beam will bend, with the forces of tension at the top of the beam, and compression on the lower half of the beam...



Could you explain this to me? I realize I am only a lowly EE, but this seems backwards. I was assuming a beam supported at both ends. I would agree with you for a beam with a single point of attachment....am I missing something :confused:

[GregoryUrich]

Just to confuse matters further (well, I am a physicist, after all), there's also the gravitational potential energy stored in a structure to consider. This is an important figure to consider in that it determines what drives the collapse of the structure - see Frank Greening's work for further information. Gravitational potential energy scales as mass times height, and mass scales as the cube of the dimensions of the structure, so GPE scales as the fourth power of the linear dimensions. In other words, double the height of the tower and the GPE goes up 16 times. Since the fracture energy of the beams goes as their cross-sectional area (linear dimension squared), that means that the ratio of energy available to energy needed for collapse has gone up four times, rather than staying the same as some might expect. These are the sort of arguments that demonstrate that physical scale models of the WTC towers are useless for predicting the behaviour of the real towers.

Dave


No, double the height and the GPE is 4 times the original. If you double the height, width and depth the GPE will be 16 times the original.

[Dave Rogers]

No, double the height and the GPE is 4 times the original. If you double the height, width and depth the GPE will be 16 times the original.

OK, I omitted the phrase "and scale all the other dimensions accordingly" - feel free to award yourself a pedant point. The point is that a half scale model will have 1/16th the GPE but require 1/4 the fracture energy per storey to collapse, so energetically a half scale model will be weighted against collapse by a factor of four.

Dave

[GregoryUrich]

OK, I omitted the phrase "and scale all the other dimensions accordingly" - feel free to award yourself a pedant point. The point is that a half scale model will have 1/16th the GPE but require 1/4 the fracture energy per storey to collapse, so energetically a half scale model will be weighted against collapse by a factor of four.

Dave

Now it should be crystal clear for BigAl.