Material compression

[The Rabid Sloth]

Hello everyone!

Hope this isn't a silly question, or misplaced, but this is my first post, so please excuse me!

Say you are compressing two spheres. The only difference between them being their volume. Would one (ideally, at least) expect the force required to compress the material to have a linear relationship with the volume?

Thanks, in advance!  

[quiet]

On 22/8/2018 at 5:27 AM, The Rabid Sloth said:

Would one (ideally, at least) expect the force required to compress the material to have a linear relationship with the volume?

Hi The Rabid Slot. As you have established, compressing is reducing the volume.

You have compressed the system to a [math]V_{_1}[/math] volume and, to maintain that constant volume, you need to apply an F1 force. Later you want to reduce the volume an infinitesimal more, that is, you want to get to [math]V_{_2} = V_{_1}-dV[/math]. The force required to keep the volume [math]V_{_2}[/math] constant is a little higher than to maintain [math]V_{_1}[/math], that is, [math]F_{_2} = F_{_1} + dF[/math]. This is the case in systems that are opposed to being compressed, for example a bicycle inflator that has the outlet hole blocked. At first it costs little, then it costs more and more. The same happens when you compress a spring.

When you study compression, it is interesting to express that behavior mathematically. When you take a relaxed spring and compress it slightly, you exert force in only one direction and instead of the volume reduction you can measure the length variation of the spring. Hooke's law indicates [math]F_{_2}-F_{_1} = k \left( l_{_2}-l_{_1} \right)[/math], where k is a constant, called the spring constant. Solving the two subtractions as they were defined, remains [math]dF = k \ dl[/math]. Although it is the simplest case, it serves to understand that the variation of size (whether in length or volume) and the variation of force, are related by a mathematical function. Usually the function is expressed as a derivative.

[math] \dfrac{dF}{dl} = k \ \ \ \rightarrow[/math]   (Hooke's law)

[math] \dfrac{dF}{dV} = s \ \ \ \rightarrow[/math] (law of the studied system)

Function [math]s[/math] can be mathematically simple or complicated. And you will not know before studying the system physically. If the system is opposed to being compressed, the only thing you know for sure, is that [math]F[/math] grows when the volume decreases. You can not say that the force [math]F[/math] is inversely proportional to volume, since the relationship depends on the type of system. When the volume decreases, the force [math]F[/math] grows in the form determined by the physical laws that the system is fulfilling and by the conditions of the situation.

Edited September 8, 2018 by quiet

[quiet]

 

On 22/8/2018 at 5:27 AM, The Rabid Sloth said:

Would one (ideally, at least) expect the force required to compress the material to have a linear relationship with the volume?

I have raised the case of an ideal gas. When the collisions are weak and do not modify the internal dynamics of the particles, the behavior of a real monatomic gas approaches that case.

In the statement I have defined the force as the pressure multiplied by the surface of the sphere that encloses the gas. If we were to enclose the gas with a spherical latex balloon, it would be the surface of the balloon.  It is an impractical but acceptable definition. I have arrived at the following result.

[math] \dfrac{ F_{_2} }{  F_{_1} } = \left( \dfrac{ V_{_1} }{  V_{_2} } \right)^{\tfrac{4}{3}}     [/math]

[math]F_{_2} \ \rightarrow [/math] Force when the volume is [math]V_{_2}[/math]

[math]F_{_1} \ \rightarrow [/math] Force when the volume is [math]V_{_1} [/math]

We can symbolize [math] V_{_1}[/math] to the initial volume and [math] V_{_2}[/math] to the volume after the compression.

It is not the ratio proposed by The Rabid Sloth, but in some cases it may differ little from a relationship inversely proportional to the volume. Recall that this result corresponds to an ideal gas. A real gas could deviate from the result. Could any real gas deviate in the necessary way to fulfill what The Rabid Sloth has proposed? I ignore it. Maybe someone can provide information.

Edited September 8, 2018 by quiet

[studiot]

Hello Sloth, I missed this question when you first posted it, but I see you have been back since.

So here is an answer, if you are still interested.

 

Let us start by by noting that by asking about a sphere we are moving into 3 dimensions.

A force is a uniaxial (one dimensional) entity.

 

So the correct question would be is in terms of the stess (pressure) required to compress a sphere.

The answer to that is yes if you use the correct linear-elastic modulus.

This is called the bulk modulus and usually given the symbol K.

[math]{\rm{Bulk Modulus =  K  =  }}\frac{{{\rm{Volume stress}}}}{{{\rm{Volume strain}}}} = \frac{{{\rm{Pressure}}}}{{\frac{{\Delta V}}{V}}}[/math]

The reciprocal of the bulk modulus is called the compressibility.

This finds much use in the theory of sound.

The bulk modulus is not directly measurable. The Bridgman indirect method may be used for some solids.

 

However coming on to another point and addressing quiet's post. (+1 for progress encouragement)

In three dimensions the components in each dimension are not necessarily independent and may interact.

This is most studied in soil mechanics where uniaxial, biaxial and triaxial tests are made on soil and rock samples.

 

Finally forces come in two types.

Body forces eg gravity and contact forces eg via a compression test machine.

Placing a sphere in a compression test machine will not (cannot) generate an even stress distribution.
There will be very high stresses, known as Hertzian stresses, at the points (there must be two) of contact in a uniaxial test.

 

 

 

Edited September 9, 2018 by studiot

[The Rabid Sloth]

Thank you very much for the answers, and of course I'm interested!

 

Suffice to say, this is a little outside my area of expertise, hence the reason for my question.

Out of curiosity, I actually made some compression measurements of these spheres, made up of the exact same material, but with varying diameters. In a plot of compression force as a function of volume, they did indeed fall in a linear line. Note that this (seemingly) linear plot was also at quite high levels of deformation (60 % strain) 

- I guess anisotropy would be less significant than for soil samples/wood etc. (?). As the sphere material is made up, pretty much like this:

(More material concentrated around edges, but otherwise I guess there's no directional structures).

[studiot]

46 minutes ago, The Rabid Sloth said:

- I guess anisotropy would be less significant than for soil samples/wood etc. (?)

Since I can't find where you told us what the material was I can't really comment, but natural wood is stongly anisotropic in mechanical properties.

I can't find any description of you intended use or method of application of your forces either.

So please feel free to extend and develop your question, but also provide more detail so that we can best reply.