How do you calculate fracturing energy?

[1000TonsofFun]

Trying to figure out how much more energy is needed to fracture a material like granite as opposed to fragmenting it

 

[DarkDragonMedeus]

I recall mentioned that there was a step by step practice related to Mpa among other things. But I think @Ugarik @Spinosaurus75DinosaurFan or @Jasonsith might know more details.

 

[1000TonsofFun]

Ok thank you

 

[Ugarik]

Why do you ask me? Why does it even matter? Our destruction values are 2 to 3 orders of magnitude higher compared to real world and everybody desided to ignore it.
Why can't we just made it up?

 

[Spinosaurus75DinosaurFan]

Depends on the meaning/degree of fracture I guess? Do you mean cutting it in half or making cracks or what?

 

[DarkDragonMedeus]

Ugarik, I do not remember the full details, but I thought your were involved in the fragmentation details. I know our values aren't 100% accurate; some are lowballed or highballed given shape of the object should also be taken into account. But DT said there's no way to go on some mass revision project. But he also has little to no time to visit the wiki ATM.

 

[1000TonsofFun]

Depends on the meaning/degree of fracture I guess? Do you mean cutting it in half or making cracks or what?

I mean Making like one big crack like creating a fault line

 

[Ugarik]

If you want to get a serious responce.

Let's say you have 1 cubic meter of granite. Toughness of granite is about 0.32 J/cc.
The energy to crack the rock in half should be simply "volume times toughness".
So 1000000*0.32 = 320000 J.
Now let's say you hit the rock and break into 16 pieces.
The energy needed should be "toughness times volume times logarithm of number of pieces". Number of pieces can be estimated as average volume of a single piece divided by total volume
1000000*0.32*lg(16) = 1280000 J
But if still an oversimplified model and I don't think this method will be allowed.

 

[1000TonsofFun]

Thank you

 

[Ugarik]

I just want to point out that I derived this formula myself and there is no empirical/experimental prove that it actually works, but the explanation is very simple.

If you apply energy equal to volume times toughness the object will be cracked into two pieces.
If you crack each piece in half you will need the same amount of energy because their combined volume is still the same
If you clack both pieces in half you will end up with 4 pieces. Hence if you need to break an object into 4 parts you'll need twice the energy
Now it's easy to conclude that if you take 3 times the energy, you'll end up with 8 pieces. And you'll end up with 16 pieces if you take 4 times the energy. Hence every time you add the energy equal to toughness times volume, the number on pieces doubles. (this is why you take base 2 logarithm)

 

[DarkDragonMedeus]

Yeah, I wouldn't quite consider cracking something in two or four pieces to be quite fragmentation; it's a bit more than that. But splitting something in half has its own method iirc. And I recall someone saying Violent Fragmentation is that none of the fragments are any larger than a cc or something like that. I know Pulverization is more or less the only destruction method that's mostly correct on most of the methods.

 

[Ugarik]

Yeah, I wouldn't quite consider cracking something in two or four pieces to be quite fragmentation; it's a bit more than that. But splitting something in half has its own method iirc. And I recall someone saying Violent Fragmentation is that none of the fragments are any larger than a cc or something like that. I know Pulverization is more or less the only destruction method that's mostly correct on most of the methods.

You can actually quontify fragmentation, voilent fragmentation or even pulverization using my method. If the original volume of the object was 1 cubic meter and the average volume of fragments is now 1 cc. It means it was fragmentet into one million pieces. In will be abot 20 times higher compared to cracking the object in half

 

[Spinosaurus75DinosaurFan]

If you want to get a serious responce.

Let's say you have 1 cubic meter of granite. Toughness of granite is about 0.32 J/cc.
The energy to crack the rock in half should be simply "volume times toughness".
So 1000000*0.32 = 320000 J.
Now let's say you hit the rock and break into 16 pieces.
The energy needed should be "toughness times volume times logarithm of number of pieces". Number of pieces can be estimated as average volume of a single piece divided by total volume
1000000*0.32*lg(16) = 1280000 J
But if still an oversimplified model and I don't think this method will be allowed.

There's a difference between cracking a large piece of granite in half with a small one even if the size of the crack is equal?

 

[KLOL506]

Making a large crack in the rock is remotely not the same as completely shattering or fragmenting it. So IDK why toughness is even being brought up here.

 

[Ugarik]

Oh, I'm sorry,I missed the point.

The crack propagation is very chaotic and unpredictible. There is no way you can calculate enegry using the total length of those cracks. You can howewer calculate the minimum stress needed create a large crack using stress intensity factor and fracture toughness of the given material. And you can also calculate work needed to cause that stress. The length of those cracks don't really tell us anything

 

[1000TonsofFun]

I might just request for someone else to calculate it because this is getting a bit difficult now. Thanks for your help anyway

 

[Ugarik]

I just pointed out that it cannot be calculated

 

[Flashlight237]

Oh, I'm sorry,I missed the point.

The crack propagation is very chaotic and unpredictible. There is no way you can calculate enegry using the total length of those cracks. You can howewer calculate the minimum stress needed create a large crack using stress intensity factor and fracture toughness of the given material. And you can also calculate work needed to cause that stress. The length of those cracks don't really tell us anything

Would depth be a factor that would make calcing a crack harder? I mean we currently don't have a means to measure crack depth accurately; at least not for the real narrow cracks anyway.

 

[Ugarik]

Cracks suppose to be as deep as the material itself if i'm not mistaken