# Finding Force and Energy from an indentation into a volume

## Summary:

Finding Force and Energy from an indentation into a volume. Includes a theoretical experiment/problem with multiple known values.
Hello folks, a recent theoretical discussion with a materials engineer has left my head spinning, this sort of thing is beyond me but I am curious if any of the clever folks on here could solve my problem.

I want to find the Force and Energy applied by my Indenter within the following parameters;

Iron block that will be indented;
• Block material - iron, hardness 490 MPa.
• Dimensions, 1 cubic meter.
• Rests on flat stone, will not move from rest upon impact by indenter.
Indenter
• A 8 inch, 0.95 radius solid cone with an area of 26.88 square inches.
• For the sake of argument, the indenter is functionally indestructible, enough so to complete the experiment without itself shattering or shearing.
• The indenter cone indents vertically into the iron block 8 inches (its full length) within 1 second.
The key interest of the experiment is to find Force in newtons required to make the impact, and the energy in joules.

I have been offered this paper, Eq. (15.5) to try and help me but this math is well beyond me, I can just about work out force from pressure/surface area, let alone volume and mechanical calculus and even then, that is with online calculators.

Many thanks for any help with this, props if anyone knows of a calculator that can help me at least part way. I may want to find the force/energy values but for different materials, specifically brass, granite, limestone, sandstone and possibly steel.

jrmichler
Mentor
If I understand correctly, your penetrator cone is like this:

The "semi-vertical angle" is less than 7 degrees, while your linked paper only applies to angles in the range of 50 to 80 degrees. The paper does not apply to your situation. The small angle of your penetrator means that penetration forces will be significantly affected by friction, while the larger angles are chosen to minimize the effects of friction and maximize the effects of material properties.

The paper also assumes that the material will not fracture under worst case plastic strain. That rules out its application to granite, limestone, and sandstone, all of which are brittle materials. Materials like brass and steel have a wide range of properties, depending on exact alloy, strain history, and heat treatment. The paper is about finding those material properties by means of a penetrator test.

Eight inches in one second is not an impact.

A cone of those dimensions capable of being forced 8 inches into brass, granite, limestone, and sandstone would need to be made from unobtanium, the proverbial science fiction material. No currently available material will do that.

All of which raises the question of what you are really trying to do.

If I understand correctly, your penetrator cone is like this:
View attachment 271563

The "semi-vertical angle" is less than 7 degrees, while your linked paper only applies to angles in the range of 50 to 80 degrees. The paper does not apply to your situation. The small angle of your penetrator means that penetration forces will be significantly affected by friction, while the larger angles are chosen to minimize the effects of friction and maximize the effects of material properties.

Yes that looks about right, and fair enough, like I say I did not fully understand it, the person who offered me the paper thought it would help because it talks about plasticity and I knew nothing about it period honestly.

Eight inches in one second is not an impact.
Sorry I don't understand, is not something hitting something an impact all the same?

A cone of those dimensions capable of being forced 8 inches into brass, granite, limestone, and sandstone would need to be made from unobtanium, the proverbial science fiction material. No currently available material will do that.
Well this is theoretical, the material the indenter is made from is not important to my test, lets assume its rigid like metal but is indestructible, I am only interested in the force/energy required for the indenter of said shape to indent 8 inches into the block.

All of which raises the question of what you are really trying to do.
Just discussing some theoretical (not practical) forces and energies. I would like to at least try and understand what math I need to find my results, of a small indenter entering various materials.

Allowing for the indenter being indestructible, is my proposed theoretical experiment possible to calculate for force and energy?

jrmichler
Mentor
Allowing for the indenter being indestructible, is my proposed theoretical experiment possible to calculate for force and energy?
Yes, using a high end FEA program.

Yes, using a high end FEA program.
You know I have never heard of them, a quick search sounds like that could be useful. Any particular one in mind?

Thank you both for the replies in any case.

jrmichler
Mentor
FEA is available in two varieties.

Linear static FEA is the basic FEA used for normal analyses of parts and structures. It is used to find if a part will break. It is simple to use and easy to learn, and gives results that are as good as the analyst builds the model.

Nonlinear dynamic FEA is used where you have impact, dynamic effects, and large plastic deformation. The learning curve is at least an order of magnitude greater than the learning curve of linear static FEA. Building a good model is only a small part of getting good results from nonlinear dynamic FEA. The analyst needs to learn how to use available software tools to get the model to deliver useable results.

I used nonlinear dynamic FEA (Abaqus) to analyze a straighforward sheet metal forming problem. It took me several months of 60 hour weeks to get good results. Some experimental data is necessary in order to know if your model is working properly.

SCP and Lnewqban
FEA is available in two varieties.
Wow, 60 hour weeks, several months? I am not sure my interest in this subject is nearly as close to being worthy of that, I was hoping my problem would be a lot simpler to solve than I thought. I take it, if I estimated the force, e.g.100 newtons, I cannot just do the simple calculation for work (F x Distance in meters, e.g. newton meters) to find joules for this problem? E.g. if I assume the indenter had 50 MN force (not sure if that is a good estimate for this sort of problem?) could I then calculate 50 MN multiplied by 8 inches square to get joules? or is that too simple for this sort of mechanical problem?

Cheers anyway, made me realize how tough this may be, and may not be something I take lightly, I hoped it would be easier.

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Baluncore
A 8 inch, 0.95 radius solid cone with an area of 26.88 square inches.
That is not an indenter, it is a penetrator.
Check the arctangent( friction coefficient ) between the taper and the iron block. Compare that angle with the angle of the tapered cone. Will there be any lubrication. Will you recover the cone, or will the sharp end be left in the iron block.

A sharp penetrator will open a hole, distorting the iron away from the entry axis. Very little material will be pushed inwards as it pierces the bulk. Ductile material may circulate back out at the surface next to the conical penetrator.

Lnewqban
Gold Member
Just to have an idea about the magnitude of those 8 inches of penetration:

https://www.engineeringtoolbox.com/bhn-brinell-hardness-number-d_1365.html

https://en.m.wikipedia.org/wiki/Brinell_scale

Your reference about the block material indicates that you are dealing with pure iron.
The shown value of 490 MPa corresponds to a BHN = Brinell hardness number (BHN) of around 50 Kg-force/mm^2.

A rough calculation shows that 500 Kg-force (4905 Newtons) applied on a steel ball of 10 mm diameter would produce an indentation of around 3.5 mm diameter on the surface of the iron block.

Just to have an idea about the magnitude of those 8 inches of penetration:

https://www.engineeringtoolbox.com/bhn-brinell-hardness-number-d_1365.html

https://en.m.wikipedia.org/wiki/Brinell_scale

Your reference about the block material indicates that you are dealing with pure iron.
The shown value of 490 MPa corresponds to a BHN = Brinell hardness number (BHN) of around 50 Kg-force/mm^2.

A rough calculation shows that 500 Kg-force (4905 Newtons) applied on a steel ball of 10 mm diameter would produce an indentation of around 3.5 mm diameter on the surface of the iron block.
I see, alas my penetrator as Baluncore rightly points out probably cannot be calculated as a ball, as the forces/pressures would be very different as I understand it? What would you say the closest diameter is that I could use for my 80 square inch area cone would be? or rather, if I wanted to convert it as close as possible to a ball?

I took the radius of my Cone (0.95 inches), found the Diameter (48.26 mm) and plugged it into this calculator which gave me a required force of about 17 MN to indent the same diameter into the iron (492 BHN, which is close to my numbers).

Course I feel that does not really describe my problem above.

Baluncore
Course I feel that does not really describe my problem above.
We have no idea of what you are trying to achieve with the cone.
It seems you are trying to harpoon heavy armour, but in slow motion.

To penetrate the iron you must do it very quickly, so that the kinetic energy released on impact melts the iron. That is simple armour piercing technology.

It is unlikely that a pointed cone will survive insertion, let alone extraction for reuse.
How many times will one cone be operated ?
@Physics quest why are you wanting to do this ?
There are better ways of quickly making a conical hole in a block of iron.

We have no idea of what you are trying to achieve with the cone.
Find how much force and energy it requires to make said indentation/hole.

There are better ways of quickly making a conical hole in a block of iron.
Its theoretical, not practical as I say, my only goal and interest is to find required Force and Joules of the operation. The "Cone" in this case does survive, hence why I made it indestructible for this question since the Cone besides its speed and dimensions is mostly irrelevant to my interest, I just want the force required to make an indentation with my penetrator.

It being Iron is just one question, I want to do the same for blocks of other materials, brass, granite, limestone, etc.

Baluncore
It being Iron is just one question, I want to do the same for blocks of other materials, brass, granite, limestone, etc.
Brittle material like granite will shatter, split or spall if confined.

Limestone will behave differently according to moisture content. If it is quite dry it will crush the cavities. If saturated it will split. Marble will behave like granite.

We get back to the problem of friction coefficient. If the penetrator is lubricated, so there is no friction force, then material will move away from the penetrator axis, and not be dragged forward with the penetrator surface. The material pressure at cone contact will push the cone back out.

Lubrication makes a big difference to the analysis. A tapered cone will lock in the material if it has a shallow angle, but will be ejected if it is too steep. That is where the arctan( friction coefficient ) becomes important. It decides the case.

You can make an initial estimate of the force by considering the block is made from many sheets, each with a circular hole being progressively expanded as the penetrator is advanced. Pushing a lubricated cone through a single supported sheet should have a simple mathematical solution where energy is stored in the plastically deformed annular disk about the penetrator, and in the elastically deformed annular disk about that.

You have not yet specified the problem, there are too many unknowns.

Lnewqban
Lubrication makes a big difference to the analysis
Do you mind explaining what you mean by lubrication? Do you mean if its oiled or something? The Cone is dry, e.g. its not been treated in any way. Think of the cone of having similar properties to steel, only its not going to break from the impact for the sake of my interest being more on the force inside the block/applied by the cone, instead of how the cone itself reacts.

plastically deformed annular disk about the penetrator, and in the elastically deformed annular disk about that.
I see, this sounds like progress. I can visualize this at least.

there are too many unknowns.
What more information do you think I need to get a result?

Baluncore
What more information do you think I need to get a result?
We need the friction coefficient between the cone and the materials you will test.

Lnewqban
Gold Member
... What would you say the closest diameter is that I could use for my 80 square inch area cone would be? or rather, if I wanted to convert it as close as possible to a ball?
..
I don’t believe that any ball would do it.
Certain identation is the most that can be achieved.
Note that the measurements of the referenced test becomes irreliable as the diameter of the indentation footprint tends to equal the diameter of the ball.

It is impossible to deeply penetrate blocks of metal without causing destruction or welding.
When too much mechanical energy or work is forced into metals, weird things start to happen, mainly due to generated heat and molecular interaction.

You can watch existing videos of powerful hydraulic presses pushing a metal against another; how they react may surprise you.
Other materials will flow or shatter, according to their degree of brittleness.

Here you can see a reference to the different types of hardness tests that are commonly used:
https://en.m.wikipedia.org/wiki/Indentation_hardness

We need the friction coefficient between the cone and the materials you will test.
I see, so can I estimate that, e.g. the friction coefficient between say, steel (if we assume the cone indenter has similar properties in that area) and Iron, Brass, another piece of steel etc?

It is impossible to deeply penetrate blocks of metal without causing destruction or welding.
Really? So its not possible for my experiment to happen without the entire iron block shattering or something? I chose a cubic block like that because I thought it would be durable enough to take the blow while at least on the outside still looking like a cube (with a 8+ inch hole in it)

You can watch existing videos of powerful hydraulic presses pushing a metal against another; how they react may surprise you.
I think I will.

Here you can see a reference to the different types of hardness tests that are commonly used:
Cheers!

Lnewqban
Baluncore
We need the friction coefficient between the cone and the materials you will test.
I see, so can I estimate that, e.g. the friction coefficient between say, steel (if we assume the cone indenter has similar properties in that area) and Iron, Brass, another piece of steel etc?
If you specify a non-zero friction coefficient, then the sheet model becomes inapplicable.

It is time to stop playing mind games in a field you know almost nothing about.

If you were given an exact numerical answer, what could you possibly do with it in the real world ?

Lnewqban
Gold Member
... Really? So its not possible for my experiment to happen without the entire iron block shattering or something? I chose a cubic block like that because I thought it would be durable enough to take the blow while at least on the outside still looking like a cube (with a 8+ inch hole in it)
The block may keep its external dimensions.
The zone of contact will have different processes happening at once, in the theoretical case that the cone can survive.

As others have explained, it is not only about hardness; your press and the geometry of the cone may be taking the material to its limits of strength, due to pressure and friction and heat.
Besides the penetration, all that huge energy may cause melting, fractures and fusion of both materials.

Boring a 2-inch diameter hole 8-inch deep into that metal block by mechanically removing material with a drill bit requires abundant cooling and lubrication, but much less energy, I believe.
Note that such machining process does not affect the material surrounding the perforation much, while your cone would create a huge profile of pressure and heat radiating away from the zone of penetration (possibly not reaching the exterior walls of your block, as you assume).

By using the penetrating indestructible cone, there would be a huge wedge effect:
https://en.m.wikipedia.org/wiki/Wedge

We need the friction coefficient between the cone and the materials you will test.
Well looking at this source from the engineering tool box, if we take value "clean and dry", because that describes the cone and the block in my example, the coefficient between Steel and iron seems to be 0.4-0.5, brass, another material I was looking into is 0.5.

So could the coefficient be 0.5?

As others have explained, it is not only about hardness; your press and the geometry of the cone may be taking the material to its limits of strength, due to pressure and friction and heat.
Besides the penetration, all that huge energy may cause melting, fractures and fusion of both materials.
So the internal structure of the block may become molten/shatter? Interesting.

By using the penetrating indestructible cone, there would be a huge wedge effect:
I see, is the penetrator/indenter being a cone an issue? would the theoretical idea be better with a cylinder or sphere?

Baluncore
If you were given an exact numerical answer, what could you possibly do with it in the real world ?
@Physics quest
You need to answer that question before you waste more of our time.

Lnewqban
Gold Member
... I see, is the penetrator/indenter being a cone an issue? would the theoretical idea be better with a cylinder or sphere?
I have no idea.
There must be a good reason behind the type of indenters mostly used by the test methods shown in one of my previous links.

The strenght and natural state of solid materials can be challenged with huge amounts of brute force, if available and economical.
For example, these are some of the ways to solder and weld via huge pressure:
https://en.m.wikipedia.org/wiki/Cold_welding

https://en.m.wikipedia.org/wiki/Explosion_welding

You need to answer that question before you waste more of our time.
As i said previously its all theoretical, I never meant to imply I was applying this to the real world, not unless you have any spare cubic iron blocks lying about and a power hydraulic press or something for me to experiment with

I guess I naively thought that this would be easier than it apparently is.

Lnewqban
Gold Member
Some examples of the effect of crushing force from a 150-ton hydraulic press on some materials and shapes:

Physics quest
Some examples of the effect of crushing force from a 150-ton hydraulic press on some materials and shapes:
Very cool! Thank you. And thank all for replies so far, if anyone has any more insight or does this sort of calculation for fun feel free to add to this. I have enjoyed learning a bit more about it, but as I thought, I am out of my depth personally with this sort of thing.

Lnewqban