For Questions That Don't Merit Their Own Thread

[JoeSchmuckatelli]

28 minutes ago, JoeSchmuckatelli said:

...

New forum acting weird today 

 

[K^2]

4 hours ago, Aperture Science said:

Yep, since the vertical acceleration is practically the same between them. IIRC there's a Mythbusters experiment for this, even considering there's atmospheric drag and whatnot they pretty much hit the ground at the same time

Yeah. The horizontal velocity has effect of increasing vertical drag slightly, but it won't make a huge difference until you start getting close to terminal velocity in vertical speed, and that would require a lot more height and would mean that bullet is basically dropping at that point. And, well, casing's drag-to-weight is already a bit higher, so the time it takes the two to fall is going to be pretty close. Not identical, for sure, but it's close.

In that Mythbusters episode, they actually did see effect of drag, but they treated it as error. Which is kind of fair, as it's a small effect, but if you do the math for drag properly, the theoretical value ends up closer to their result than if you do the math without accounting for drag.

Drag in ballistics is actually a fun topic. It sounds very complicated, but there are some very interesting formulas that make your life easier. So for the fall of the bullet you usually treat it as free-fall, which is fine, but you also have to account for wind, and that's where things get really exciting. It should be a very hard problem that requires a computer to solve, but snipers and artillerists make wind correction with a single table look-up. What you need to know is time-of-flight over a certain distance and wind speed. You look up time of flight from the table, then multiply by cross wind speed, and you get the correction. The reason this works is because it doesn't matter if you are firing a bullet through air that moves, or at a target that's moving in stationary air! So you do the math for later, and you get an easy answer.

[JoeSchmuckatelli]

1 hour ago, K^2 said:

... . It should be a very hard problem that requires a computer to solve, but snipers and artillerists make wind correction with a single table look-up. What you need to know is time-of-flight over a certain distance and wind speed. You look up time of flight from the table, then multiply by cross wind speed, and you get the correction. The reason this works is because it doesn't matter if you are firing a bullet through air that moves, or at a target that's moving in stationary air! So you do the math for later, and you get an easy answer.

Yeah - us knuckledraggers need all the help we can get!  Truth is, some really smart folks were kind enough to run the numbers and create the tables for us. 

While the tables are good enough for the average soldier or Marine, when we send the Sergeants to the advanced courses and Master Gunner courses they actually learn the numbers and formulae behind the tables.  I've had a lot of fun watching those guys crunch numbers and shoot machine guns and tanks without using the sights, tables or ballistic computers.  Many a six pack has exchanged hands during these events. 

 

The only thing I will add is that (as someone who has used the tables and etc for precision gunnery and long range marksmanship) - aside from the movies or for bragging rights when you feel like you just gotta have a first round shot at extreme range... We can very easily deal with 'that was close' - the Mark 1 Eyeball and a fast reload makes up for minor errors in calculation, and subsequent shots are often successful 

[JoeSchmuckatelli]

1 hour ago, K^2 said:

... .

Drag in ballistics is actually a fun topic. It sounds very complicated, but there are some very interesting formulas that make your life easier. So for the fall of the bullet you usually treat it as free-fall, which is fine, but you also have to account for wind, and that's where things get really exciting. It should be a very hard problem that requires a computer to solve, but snipers and artillerists make wind correction with a single table look-up. What you need to know is time-of-flight over a certain distance and wind speed. You look up time of flight from the table, then multiply by cross wind speed, and you get the correction. The reason this works is because it doesn't matter if you are firing a bullet through air that moves, or at a target that's moving in stationary air! So you do the math for later, and you get an easy answer.

Did you ever play around with really long shots where coriolis played a factor?  That really caused a twist in my noggin when I first tried to understand it. 

[K^2]

2 hours ago, JoeSchmuckatelli said:

Did you ever play around with really long shots where coriolis played a factor?

I haven't. I can do the math for these with a computer, of course, but I don't actually know how you'd stick it into a table for quick lookup. I know that artillery makes corrections for it, but no idea how they do it in practice.

3 hours ago, JoeSchmuckatelli said:

for bragging rights

My only practical use has been in impressing undergrads in a physics classroom. Using spring-loaded cannons, of course, as discharging firearms on campus would get me into trouble, and it's hard to demonstrate the effects at these speeds on classroom scale anyways. But I tell you, when you land a steel ball into a plastic cup someone places on the other side of the classroom using a single shot, the students always look at you like you've summoned the devil to guide the projectile. The fact that they just learned relevant formulae never seems to convince them that the math actually works in the real world. XD I'm pretty sure that the only reason we were teaching labs is to convince kids that we aren't just making up these equations for fun.

[JoeSchmuckatelli]

11 hours ago, K^2 said:

... , the students always look at you like you've summoned the devil to guide the projectile.

 

... .

I LOL'd. 

11 hours ago, K^2 said:

... . The fact that they just learned relevant formulae never seems to convince them that the math actually works in the real world. XD I'm pretty sure that the only reason we were teaching labs is to convince kids that we aren't just making up these equations for fun.

This is the worst thing that we do to students - fail to show how math is actually applicable to the world they live in.   Every kid should have a 'math lab', even if it is only a few days out of every semester. 

For me, math was always a boring grind.  I was too ADD to do the homework for something that had virtually nothing to do with the real world. 

I was a 23 year old construction worker trying figure out how much materials we needed for a roofing project when I stunned myself by using the pythagorean theorum.   It was a head slapping moment.  (Heh - we also use the 3-4-5 thing to ensure foundations, etc are square).  As an adult I found that time and again I needed and benefitted from the math I never learned as a kid.  It's tough to learn this stuff as an old dude - so I really appreciate this forum 

I just wish I could show my kids how to do some of the fun stuff.  (well, other than carpentry - I've got that down). 

[ARS]

So... I've been given a challenge by my teacher. Is it possible to predict the theoretical tensile strength of a composite material solely using mathematical calculation (for example, due to the test cannot be carried out for some reason) if we know the material characteristic of it's individual components? For example, let's say we're gonna find the tensile strength of pykrete. We cannot make the pykrete, but we have the data of ice tensile strength of around 19.0 MPa and it's 18% wood reinforcement is 4.10 MPa. Is it possible to find the theoretical tensile strength of resulting composite solely using mathematical calculation without doing actual test?

Edited October 25, 2020 by ARS

[Shpaget]

You could calculate weighted arithmetic mean, but I wouldn't be too confident with the result. There could be unforeseen interactions that might change the results.

The answer I'd give your teacher would be a "No.".

BTW, that tensile strength for ice you mention seems to be quite off.

[kerbiloid]

Brief version: 
It's greater than both ice's (because they don't build ships out of ice) and wood's (otherwise why did they bother with the pykrete), comparable to bad steel (otherwise they wouldn't even listen him about it),and you can estimate it by dividing the real pykrete ship mass by its walls cross-section, if the teachers brings these values (because the real ship was obviously the biggest what they could build out of the pykrete to impress the sponsors).

Long version.

Spoiler

1. Estimate total energy required to vaporize the material from its solid state.

Roughly you can take the most part of it as the
https://en.wikipedia.org/wiki/Enthalpy_of_vaporization
and round it up.

Properly, you should add (enthalpy of melting) and (heat capacity * (boiling point - melting point)), but as the heat capacity anyway varies on heating, you anyway can't get an accurate answer simply.

Say
Iron (as a chechable example) = 6.09 MJ/kg → let it be 10 MJ/kg
Ice (as the most part of the pykrete, according to wiki) = 2.257 MJ/kg → let it be 3 MJ/kg

So, you need 10 MJ/kg (for iron) or 3 MJ/kg (for ice) to break all bounds between the nodes of the crystal structure.

1 kg of the sample contains (mass/molar_mass * Avogadro_const) of the molecules/atoms, i.e. nodes of the crystal structure.

So,
iron: 1 / 0.056 * 6.022*10²³ ~= 1.1*10²⁵ atom/kg, i.e. crystal structure nodes per mass.
ice: 1 / 0.018 * 6.022*10²³ ~= 3.3*10²⁵ atom/kg, i.e. crystal structure nodes per mass.

To unbound all nodes of the 1 kg sample you need ~10 MJ/kg for iron, ~3 MJ/kg for ice.

iron: 10*10⁶ / 1.1*10²⁵ ~= 9*10^-19 ~ 10^-18 J/node
ice: 3*10⁶ / 3.3*10²⁵ ~= 9*10^-20 ~ 10^-19 J/node

***

1 cubic meter of the material contains (1 * density, kg/m³) of the material mass.

Iron: 1 m³  = 7 800 kg of iron, or 7 800 / 0.056 * 6.022*10²³ ~= 8.4*10²⁸ nodes (atoms)
Ice: 1 m³  = 1 000 kg of ice, or 1 000 / 0.018 * 6.022*10²³ ~= 3.3*10²⁸ nodes (molecules)

So, the average distance between the nodes of the crystal structure is
Iron: (1 / 8.4*10²⁸)^1/3 ~= 2.3*10^-10 m
Ice: (1 / 3.3*10²⁸)^1/3 ~= 3.1*10^-10 m

Average number of the nodes per the sample cross-section area
Iron: 1 / (2.3*10^-10)² ~= 1.9*10¹⁹ nodes (atoms) / m2.
Ice: 1 / (3.1*10^-10)² ~= 1.0*10¹⁹ nodes (molecules) / m2.

***

To break the sample apart in two halves you need to take away from each other at least two neighboring layers of the crystal structure.

So, you should break apart the bounds of ~1.9*10¹⁹ nodes of iron, or 1.0*10¹⁹ nodes of ice.

As you remember, it requires:
iron: ~10^-18 J/node
ice: ~10^-19 J/node

So, in total you should spend:
Iron: (~10^-18 J/node * 1.9*10¹⁹ node/m2 ~= 19 J/m2.
Ice: (~10^-19 J/node * 1.0*10¹⁹ node/m2 ~= 1 J/m2.

Job = force * distance, so 
force = job / distance

The distance is nearly the same as the distance between the nodes.
So, to break apart the sample you ideally have to apply the force:
Iron: 19 J/m2 / (2.3*10^-10) m ~= 80*10⁹ N/m², i.e. stress ~80 GPa.
Ice: 1 J/m2 / (3.1*10^-10) m ~= 3*10⁹ N/m², i.e. stress ~3 GPa.

***

But this is the theoretical upper limit.

While in the dull realm of reality, let me quote the wiki

https://translate.google.com/translate?hl=ru&sl=ru&tl=en&u=https://ru.wikipedia.org/wiki/Теоретический_предел_прочности

Quote

Theoretical tensile strength - the value of the limiting stresses, obtained by calculation, based on the properties of interatomic bonds in the crystal lattice of the material. As a rule, the theoretical strength exceeds the real one (obtained from tests) by several orders of magnitude. For example, the theoretical ultimate strength of iron is 56 GPa, while the practical one is 280 MPa. The main reason for this is the difference in the fracture mechanisms of real crystals from that considered in calculating the theoretical ultimate strength. In real crystals, destruction is associated with the presence and movement of dislocations and other defects of the crystal structure, cracks, etc. When calculating the theoretical limit, it is assumed that the destruction is caused by the breaking of interatomic bonds in the crystal. The approximation of the practical strength of the material to its theoretical limit is possible in two ways: by improving the technological process in order to reduce the number of crystal lattice defects and using the scale effect, that is, increasing the average statistical ultimate strength of the thread (rod) with a decrease in its cross-sectional area. The scale effect is used in fibrouscomposites , for example, fiberglass, which is a frame made of glass filaments bonded with epoxy resin.

(You see, we here roughly estimated 80 GPa for the iron, and the reference value is nearly same, 56 GPa).

The theoretical upper limit is simply  ~= ~1.5 (to include the melting, heating, etc) * enthalpy of vaporization, J/kg * density, kg/m³.
For iron: 1.5 * 6*10⁶ * 7800 ~= 70 GPa (i.e nearly the mentioned 56 and 80).

But irl, you can divide the theoretical values by 200 to get the values for non-ideal matter, and you can just calculate new exact value without knowledge of the sample structure.

(Actually without experimental estimation of this unknown "realism coefficient" you can only guess.)

The ice and wood proportion and these 19 and 4 MPa unlikely is appropriate here, as the wood fibers could strengthen the sample perpendicularly, not directly. 

Edited October 25, 2020 by kerbiloid

[ARS]

42 minutes ago, kerbiloid said:

The ice and wood proportion and these 19 and 4 MPa unlikely is appropriate here, as the wood fibers could strengthen the sample perpendicularly, not directly

So does it means it doesn't factor the orientation of wood fibers inside the ice matrix? If so, should it be assumed as the wood fibers are arranged randomly instead of unidirectional?

[Guest]

4 hours ago, ARS said:

So... I've been given a challenge by my teacher. Is it possible to predict the theoretical tensile strength of a composite material solely using mathematical calculation (for example, due to the test cannot be carried out for some reason) if we know the material characteristic of it's individual components? For example, let's say we're gonna find the tensile strength of pykrete. We cannot make the pykrete, but we have the data of ice tensile strength of around 19.0 MPa and it's 18% wood reinforcement is 4.10 MPa. Is it possible to find the theoretical tensile strength of resulting composite solely using mathematical calculation without doing actual test?

You probably can do this through a finite elements method. I'm not very knowledgeable about structural analysis however so perhaps someone could correct me

Edited October 25, 2020 by Guest

[mikegarrison]

[kerbiloid]

1 hour ago, ARS said:

So does it means it doesn't factor the orientation of wood fibers inside the ice matrix? If so, should it be assumed as the wood fibers are arranged randomly instead of unidirectional?

The fibers may strengthen it along, but can across.
Like in the original clay pots strengthen with wooden twigs which later gave the idea of armored concrete. The arming lattice prevents the concrete stretching, rather than compression.

And the real picture of this is mostly empiric, requiring an experiment to later apply its result on similar structures.

[ARS]

3 hours ago, kerbiloid said:

The fibers may strengthen it along, but can across.
Like in the original clay pots strengthen with wooden twigs which later gave the idea of armored concrete. The arming lattice prevents the concrete stretching, rather than compression.

And the real picture of this is mostly empiric, requiring an experiment to later apply its result on similar structures.

I see. Thanks!

[JoeSchmuckatelli]

3 hours ago, kerbiloid said:

The fibers may strengthen it along, but can across.
Like in the original clay pots strengthen with wooden twigs which later gave the idea of armored concrete. The arming lattice prevents the concrete stretching, rather than compression.

And the real picture of this is mostly empiric, requiring an experiment to later apply its result on similar structures.

Quibble time: organic fibers in clay will burn out during the kilning.  People did use a mix of fibers and mud for things like wattle and daub construction that would not be kilned, but I'm unaware of any kilned pottery that had fibers internal to the clay; it was, however, often used inside the object to be kilned as a support - but it would cook off completely.

 

Now - I will accept being totally wrong, excited even, if you can show me that the ancients did, indeed, strengthen clay pots with fibers - where the fibers remain as part of the internal structure of the body of the pot.

Edited October 25, 2020 by JoeSchmuckatelli

[ARS]

3 hours ago, kerbiloid said:

And the real picture of this is mostly empiric, requiring an experiment to later apply its result on similar structures.

Wait wait... If the previous formula is based on the cross-sectional area of the sample in 1 cubic meter of material mass, then is it possible if:

Assuming that for example, the reinforcement material is around 25% of the total material content (with 75% being the primary matrix), is it possible to take 75% of the force needed to break the matrix and add 25% of the force needed to break the reinforce material (both values are based on the calculation on 75% and 25% of volumes of each material respectively), assuming the reinforce material is distributed evenly throughout the matrix?

[JoeSchmuckatelli]

5 minutes ago, ARS said:

Wait wait... If the previous formula is based on the cross-sectional area of the sample in 1 cubic meter of material mass, then is it possible if:

Assuming that for example, the reinforcement material is around 25% of the total material content (with 75% being the primary matrix), is it possible to take 75% of the force needed to break the matrix and add 25% of the force needed to break the reinforce material (both values are based on the calculation on 75% and 25% of volumes of each material respectively), assuming the reinforce material is distributed evenly throughout the matrix?

I don't think it works that way.  Clearly i'm no scientist, engineer or physicist, but my experience tells me that combining materials does not lead to a strictly additive or even linearly multiplicative effect.

 

Case in point: the concrete we use for construction.  You can add wire to the concrete, and even fiberglass - but the total structural strength of the combo is going to be different than each taken separately.  I don't know the terminology per se; but its one of those 'The whole is greater than the sum of the parts' things.  The method you propose is a good starting place, but experience tells me that in some cases you just gotta test.

[magnemoe]

6 hours ago, ARS said:

So does it means it doesn't factor the orientation of wood fibers inside the ice matrix? If so, should it be assumed as the wood fibers are arranged randomly instead of unidirectional?

Fibers would be randomly oriented so it would be far more like chipboard than planks. You could use wood as reinforcement .
However the thing was designed to be massive with an 10 meter thick hull. Torpedoes would not work well against it. 
An battleship would be an treat but Germany just had an couple and you have a couple in the escort, Heavy bombers is another but Germany could not reach it. 
But they ended up building escort carriers instead who are less cool both ways but much cheaper and faster to build  

[ARS]

32 minutes ago, JoeSchmuckatelli said:

Case in point: the concrete we use for construction.  You can add wire to the concrete, and even fiberglass - but the total structural strength of the combo is going to be different than each taken separately.  I don't know the terminology per se; but its one of those 'The whole is greater than the sum of the parts' things.  The method you propose is a good starting place, but experience tells me that in some cases you just gotta test.

Okay, so the fibers are not additive in terms of total value. But what if the reinforce material is in the form of powder? (such as wood dust) Aka it's not a fiber that acts like a frame inside the primary matrix, but a powder that's integrated into the primary matrix itself. I've already told the lab technician to give me permission to test the sample, but he denied me access because the testing machine is only calibrated for metals, not ice. Trying to run the test would likely give bad results and possibly damaging the machine because of water melting from pykrete

[JoeSchmuckatelli]

11 hours ago, ARS said:

So... I've been given a challenge by my teacher. Is it possible to predict the theoretical tensile strength of a composite material solely using mathematical calculation (for example, due to the test cannot be carried out for some reason) if we know the material characteristic of it's individual components? For example, let's say we're gonna find the tensile strength of pykrete. We cannot make the pykrete, but we have the data of ice tensile strength of around 19.0 MPa and it's 18% wood reinforcement is 4.10 MPa. Is it possible to find the theoretical tensile strength of resulting composite solely using mathematical calculation without doing actual test?

I'll add another thought to what I've written above: I really don't think you can ever use a pure mathematical guess as to an alloy's properties merely by testing or knowledge of the component parts.  I was going to come in here with examples based upon pure iron and various steels... but then it dawned on me... its even simpler than that.

 

Take everything you know about Hydrogen and Oxygen (only in and of themselves)... no amount of testing one or the other alone is going to tell you much about what happens when you mix them together into a new compound.

 

3 hours ago, ARS said:

Okay, so the fibers are not additive in terms of total value. But what if the reinforce material is in the form of powder? (such as wood dust) Aka it's not a fiber that acts like a frame inside the primary matrix, but a powder that's integrated into the primary matrix itself. I've already told the lab technician to give me permission to test the sample, but he denied me access because the testing machine is only calibrated for metals, not ice. Trying to run the test would likely give bad results and possibly damaging the machine because of water melting from pykrete

FIbers are incredibly 'additive' in the performance of the whole.  When I'm pouring a slab for a garage that I know is going to have trucks in it - I want fiber added to the concrete.  You get a heck of a lot more durable foundation that way.  OTOH, when I'm simply making steps or a ramp to a shed, I'll mix the concrete on site and just use wire mesh.

 

The combination you choose has a great effect - and I'm sure you can look at what other's in the industry have written about the benefits of adding fiber to concrete - but whether you can mathematically guess anything by knowing something about fiberglass is unlikely - and beyond my ken.

[kerbiloid]

6 hours ago, JoeSchmuckatelli said:

Quibble time: organic fibers in clay will burn out during the kilning

The armored concrete was suggested by the (clay pots for palms) manufacturer who was strengthening the clay with wooden twigs.
XIX century or so.

The twig-armored clay pots are known and widely used since the stone age. It's not necessary to burn them, usually they just dry it under sun.
It's how the pottery appeared at all, since they started to cover twig baskets with clay.

As you have been to Afghanistan, an adobe should also be familiar for you.

Edited October 26, 2020 by kerbiloid

[kerbiloid]

6 hours ago, ARS said:

If the previous formula is based on the cross-sectional area of the sample in 1 cubic meter of material mass,

It gives the theoretical upper limit of an ideal sample without local defects, of a monocrystal.

But irl the sample is divided by defects (dislocations) of the crystal structure, which decrease this limit by hundreds of times. Because any defect makes a thin place to tear.

6 hours ago, ARS said:

Assuming that for example, the reinforcement material is around 25% of the total material content (with 75% being the primary matrix), is it possible to take 75% of the force needed to break the matrix and add 25% of the force needed to break the reinforce material (both values are based on the calculation on 75% and 25% of volumes of each material respectively), assuming the reinforce material is distributed evenly throughout the matrix?

No, at least because any contact surface between the particles of different materials is a crystal defect itself.
So, the mix on one hand gets weakened by this particle boundaries, on the other hand some materials can either keep the neighboring molecules stronger, or, while weakening the sample in one direction, strengthen it in perpendicular one.

So, the acctual value of the "several hundreds" can be predicted with very low accuracy, and requires an experiment.

(It's normal. Say, aero- and hydrodynamics are almost purely empirical voodoo, full of magic "numbers" of the nature which nobody knows. Even for Space Shuttles.)

So, even if you somehow estimate this  decreasingcoefficient mathematically, its poor accuracy makes it unusable for any engineering.
You can just make experiments, gather results, statistically summarize it, and create empirical formulas.

5 hours ago, ARS said:

But what if the reinforce material is in the form of powder? (such as wood dust)

The same about powder.

Edited October 26, 2020 by kerbiloid

[ARS]

I see. So in other words, it's all just a guess when using mathematical calculation huh? Well anyway thanks for your help, at least it gives me insight about what factors that needs consideration for further testing 

[kerbiloid]

17 minutes ago, ARS said:

So in other words, it's all just a guess when using mathematical calculation huh?

The mathematical calculations are useful when you have either empirical results to build an appropriate mathematical model, or when you are calculating, say, a building, a steam boiler or a nuclear reactor made of materials with well-known characteristics (the maths are monstrous anyway).

But for composite materials like even the simple concrete, it's experiment → statistical processing of its results → building an empiric mathematical model →  mathematical approximation → experiment to check.

A mathematical model of concrete may include about 50 values, afaik.

Edited October 26, 2020 by kerbiloid

[K^2]

20 hours ago, ARS said:

So... I've been given a challenge by my teacher. Is it possible to predict the theoretical tensile strength of a composite material solely using mathematical calculation (for example, due to the test cannot be carried out for some reason) if we know the material characteristic of it's individual components? For example, let's say we're gonna find the tensile strength of pykrete. We cannot make the pykrete, but we have the data of ice tensile strength of around 19.0 MPa and it's 18% wood reinforcement is 4.10 MPa. Is it possible to find the theoretical tensile strength of resulting composite solely using mathematical calculation without doing actual test?

Depends on nature of composite. Something coarse, like reinforced concrete, you can make very decent estimates for. Steel doesn't behave in any fundamentally new ways when encased in concrete, and neither does concrete exhibit any new properties. There are still plenty of nuances. A structure that's just a steel mesh will experience more failures than reinforced concrete, despite the fact that nearly all of the strength is provided by steel. Concrete, however, helps redistribute the stress, meaning you aren't as likely to have a critical failure point. That's a bit tricky to get from a simulation, but if you know what you're looking for, you can build a mathematical model that will work.

Other composites are similar in that regard, but things get rapidly more complex as the mix between fibers and matrix becomes finer. At the extreme range of this you get ceramics suddenly turning superconductive. That is, you suddenly have completely new physical properties not exhibited by any of the components going in. Emergent behavior. You don't get things going quite that badly with typical construction materials, fortunately, but you still can't rely on each material having same mechanical properties in a composite as it does outside of it. For example, when considering strength of glass fiber hulls, you have to keep in mind that resin matrix behaves differently in the composite due to viscous forces. So if you were to model glass fiber composite the same way you would reinforced concrete, you are likely to underestimate material's strength.

In general, the smaller your fibers get, the less predictable material becomes. I'm worried that in case of Pycrete you'll get additional complication of water doing very strange things near the cellulose fibers. Everybody knows that water is weird because it is less dense when frozen. But it's even more weird when it's near freezing. Water at 1°C is less dense than water at 4°C, because while it won't start forming actual ice until it drops to freezing point, the molecules of water start forming large ice-like complexes that stick around long enough to offset average density. This all has to do with strength of hydrogen bonds and is part of the reason why hydrates can get weird. Concrete's actually a good example. I suspect that structure of ice near cellulose fibers in Pycrete is nothing like structure of pure ice. You'll get long chains of water molecules pulling on each other adding to material strength. It is effectively a different phase of water with different physical properties, and you'll never get that from simple mechanical analysis of the composite.

There are entire fields of condensed matter and soft matter physics dealing with these kinds of emergent phenomena in materials. A lot of it is studied in simulation, so it's not a hopeless task. But you do need to know in advance which phenomena you need to account for. If you suspect that viscosity plays a role, you can't use simple rigid body simulation. If you suspect phase transitions are taking place, you'll have to account for the new phase. It's all doable, but it's not trivial, and without doing tests on the actual material, I don't know if you could ever be certain you didn't miss something.