Wood Shear Strength Inconsistency

[Agnaa]

We attribute different shear strengths to wood in different calculations.

Some of them, such as this calc, use a tensile strength of 70-140 MPa, multiplied by 0.6 to get shear strength,

Some of them, such as this calc, use a shear strength of 7.3774 MPa.

These values are 5.7-11.4x off from each other. Both of them are currently accepted.

We need to find a consistent value to apply to calculations, and we need to update old calcs accordingly.

 

[Arceus0x]

apparently wood shear strength varies from 3 to 15 MPa so i assume that is a wacky part as well

 

[Furudo_Erika]

It also depends on the specific TYPE of wood from the specific species of tree.

For example: Is it balsa wood, or ironwood? Both are "wood", but they have completely different levels of strength.

I would personally use white oak as a default assumption, which has a UTS of 5.31 - 5.50 MPa from what I could find.

 

[ElajRuengies]

It's because it depends on where the direction of force is coming from with respect to the grain/fibers of the wood.

Like, imagine you have a bunch of plastic drinking straws tied together with a rubber band, with all of the straw openings facing up/down. Squishing the bundle of straws perpendicular to the openings is going to be way easier than trying to squish them length-wise.

As a result, there are two strength values for compressive strength listed in @Arceus0x 's pdf-

Five strength properties that are commonly measured for design purposes include bending, compression parallel and perpendicular to the grain, tension parallel to the grain, and shear parallel to the grain. In addition, measurements are sometimes required for tensile strength perpendicular to the grain and side hardness.

(This also means that the compressive strengths of wood at angles between 0 and 90 degrees to the grain are gonna be somewhere between the two values, and can be determined by trigonometry)

We attribute different shear strengths to wood in different calculations.

Some of them, such as this calc, use a tensile strength of 70-140 MPa, multiplied by 0.6 to get shear strength,

Some of them, such as this calc, use a shear strength of 7.3774 MPa.

These values are 5.7-11.4x off from each other. Both of them are currently accepted.

We need to find a consistent value to apply to calculations, and we need to update old calcs accordingly.

The reason why shear strength is inconsistent is because the listed shear strength of wood is measured parallel to the grain, while the listed tensile strength is measured perpendicular to the grain, meaning multiplying the perpendicular tensile strength by 0.6 will (in theory) get the perpendicular shear strength, which is going to be different from the parallel shear strength.

Unlike compressive strength, shear strength of wood perpendicular to the grain is much higher than shear strength parallel to the grain. (Source)

A very limited amount of data suggests that shear strength perpendicular to the grain may be 2.5–3 times that of shear parallel to the grain. Much of the variation in wood properties within and between trees can be attributed to density.

 

[Agnaa]

It's because it depends on where the direction of force is coming from with respect to the grain/fibers of the wood.

Like, imagine you have a bunch of plastic drinking straws tied together with a rubber band, with all of the straw openings facing up/down.

Squishing the bundle of straws perpendicular to the openings is going to be way easier than trying to squish them length-wise.

If so, we should be more clear on which force values correspond to which directions, so we can evaluate whether they're being used correctly or not.

 

[Arceus0x]

so if you have a tree and you hit from the side you use shear and if you cut it in half from top to bottom you use tensile..? I don't get it

 

[ElajRuengies]

so if you have a tree and you hit from the side you use shear and if you cut it in half from top to bottom you use tensile..? I don't get it

Cutting feats are measured as pulverization or compressive strength of a thin slice. (Source- @KLOL506

I think we stopped using that method for cutting calcs altogether.

Now we just use: length of the cut area * width of the cut area * thickness of the blade edge * compressive strength/pulv. energy of the material

Coincidentally, it also works for breaking things in half instead of shattering the entire thing.

Cutting wood from the side (think swinging axe at a tree) will use perpendicular to grain compressive strength

Cutting wood from top-down (think chopping logs on a splitting block) will use parallel to grain compressive strength

 

[Agnaa]

Huh, KLOL said we changed the method in February, but accepted a calc using shear strength in February, was that just because that example was more breaking than cutting?

Also, like that old thread mentioned, if we don't use that anymore, the Calculations Introduction and References page should be updated accordingly.

@KLOL506

 

[Arceus0x]

this needs to be sorted out man, can't add treasure island 1988 without the right formula

 

[KLOL506]

Huh, KLOL said we changed the method in February, but accepted a calc using shear strength in February, was that just because that example was more breaking than cutting?

I believe the cutting calc used shear strength in a different method than how most frag calcs use it.

Also, like that old thread mentioned, if we don't use that anymore, the Calculations Introduction and References page should be updated accordingly.

@KLOL506

I don't think there was ever a method outlined for how to do cutting calcs. But sure, here be the formula:

Blade edge thickness/blade thickness x area of the cut (Basically length x width of the cut but depends on shape) x compressive strength AKA pulv

 

[Arceus0x]

Blade edge thickness/blade thickness

??

 

[KLOL506]

??

Basically, blade edge/blade itself would be the thickness of the super thin shape that you're cutting with a sword.

 

[KLOL506]

Cutting feats are measured as pulverization or compressive strength of a thin slice. (Source- @KLOL506

Cutting wood from the side (think swinging axe at a tree) will use perpendicular to grain compressive strength

Cutting wood from top-down (think chopping logs on a splitting block) will use parallel to grain compressive strength

Aight.

 

[KLOL506]

Cutting wood from the side (think swinging axe at a tree) will use perpendicular to grain compressive strength

Cutting wood from top-down (think chopping logs on a splitting block) will use parallel to grain compressive strength

Now that we figured this one out, we need to find the axe blade edge thickness.

 

[Arceus0x]

Basically, blade edge/blade itself would be the thickness of the super thin shape that you're cutting with a sword.

give me a proper example, like with an actual axe. what would be blade edge and what blade thickness

 

[ElajRuengies]

Now that we figured this one out, we need to find the axe blade edge thickness.

give me a proper example, like with an actual axe. what would be blade edge and what blade thickness

Those were just examples of situations where perpendicular compressive strength or parallel compressive strength would be used, but
[10 minutes of increasingly specific Googling later]

These are battle axes, not wood cutting axes, but I see values from 1/8 of an inch to 1/2 an inch, meaning 0.3175 to 1.27 cm, although the latter was described as "basically a mace strike."

This tactical tomahawk has a blade 0.36 inches thick, or 0.9144, which still seems rather thick.

Also there's a crap ton of different kinds of axes, with felling axes (hitting perpendicular to grain) and splitting axes (hitting parallel to grain) being different in design.

Whatever, let's say you've got an axe 1/8 of an inch thick, and somehow cut a 1 meter thick perfectly circular American White Oak tree in one fell swoop.

• 0.3175 cm * (100/2)^2 * Pi = 2,493.639169 cm3
• 2493.639169 * 9.1 j/cm3 (source) = 22,692.11644 joules, or Wall level

Now let's say you've got a 1 meter tall cylindrical section of that tree on a chopping block, and split the thing in two equal sized halves, again in one hit. (The cross section is gonna be a square this time, not a circle)

• 0.3175 cm * 100 * 100 = 3,175 cm3
• 3175 * ~50 j/cm3 (source) = 158,750 joules, or Wall level

 

[Arceus0x]

What about a hand ~3cm thick instantly cutting through a tree 79cm in radius?

 

[ElajRuengies]

What about a hand ~3cm thick instantly cutting through a tree 79cm in radius?

3 * 79^2 * Pi * 9.1 = 535,262.3572 joules

 

[Arceus0x]

3 * 79^2 * Pi * 9.1 = 535,262.3572 joules

So 9-B+ or so?

 

[KLOL506]

So 9-B+ or so?

9-B+ starts at 10,467,500 joules.

 

[Arceus0x]

9-B+ starts at 10,467,500 joules.

Oh w oops. Well just a nice 9-B character then. Most famous 9-Bs are around the hundreds of thousands of joules so I'll probably have a nice verse for matches

 

[KLOL506]

'Kay, upload the above conclusions in a blog and that should settle that.

Any old calc using the deprecated cutting method should also be likely changed. But I doubt there's enough of those to go around.

 

[Agnaa]

Uhh KLOL, can you be clearer about your resolution to the shear strength inconsistency?

Is it just to use compressive strength instead?

Do you agree with Elaj's method of using different values when cutting with/against the grain?

Are you okay with defaulting to White Oak, or do you want other values to be found too?

And Elaj, you said this earlier

This also means that the compressive strengths of wood at angles between 0 and 90 degrees to the grain are gonna be somewhere between the two values, and can be determined by trigonometry

Can you express what formula would be used when cut at an arbitrary angle such as this?

 

[Agnaa]

Also, we should probably get more input if we're going to be changing a standard.

@Amelia_Lonelyheart @Executor_N0 @Spinosaurus75DinosaurFan @Mr._Bambu @Therefir @DMUA @Damage3245 @DemonGodMitchAubin @Jasonsith @Wokistan @Migue79 @Armorchompy @Psychomaster35 @CloverDragon03 @Dark-Carioca @AbaddonTheDisappointment @Aguywhodoesthings @DontTalkDT

Can you read this thread and give your thoughts on what should be done?

 

[KLOL506]

Uhh KLOL, can you be clearer about your resolution to the shear strength inconsistency?

Is it just to use compressive strength instead?

For cutting feats in general? Yes. Though I doubt the perpendicular/parallel methodology would apply to non-wood materials.

For destroying the entire structure? Just follow the old volume times destruction value.

Do you agree with Elaj's method of using different values when cutting with/against the grain?

Aye, seems good at a glance.

Are you okay with defaulting to White Oak, or do you want other values to be found too?

Don't we have the other values in that chart where we used to find the frag, v. frag and pulv. values of wood? I could've sworn something like that existed.

 

[Agnaa]

For cutting feats in general? Yes. Though I doubt the perpendicular/parallel methodology would apply to non-wood materials.

Cool.

Aye, seems good at a glance.

Neat.

Don't we have the other values in that chart where we used to find the frag, v. frag and pulv. values of wood? I could've sworn something like that existed.

Yeah our destruction values of wood have pulv, listed in J/cc

 

[KLOL506]

Don't we have the other values in that chart where we used to find the frag, v. frag and pulv. values of wood? I could've sworn something like that existed.

Yeah our destruction values of wood have pulv, listed in J/cc

Pretty sure we have a chart linking to it to another website where the perpendicular/parallel stuff is mentioned as well. You could use those for the various types of wood but uh, me no good with table editing formats.

 

[ElajRuengies]

And Elaj, you said this earlier

Can you express what formula would be used when cut at an arbitrary angle such as this?

It's the equation for resolving forces, just reversed.

compressive strength at an angle = sqrt[(cos(cut angle)*perpendicular compressive strength)^2 + (sin(cut angle)*parallel compressive strength)^2]
So, using White Oak Compressive strengths-

Angle of Cut	Compressive Strength at Angle
0 degrees	9.1 MPA
15 degrees	15.64388146 MPA
30 degrees	26.21273545 MPA
45 degrees	35.93612389 MPA
60 degrees	43.53966582 MPA
75 degrees	48.35368624 MPA
90 degrees	~50 MPA

(At 0 degrees, it'll just be the perpendicular compressive strength, while at 90 degrees, it'll just be the parallel compressive strength)

To illustrate it-


(The harder part of cutting at an angle would be getting the area of the cut rather than the compressive strength)

And yeah this can be ignored for materials that don't have axis-specific strengths like wood.

 

[KLOL506]

It's the equation for resolving forces, just reversed.

compressive strength at an angle = sqrt[(cos(cut angle)*perpendicular compressive strength)^2 + (sin(cut angle)*parallel compressive strength)^2]
So, using White Oak Compressive strengths-

Angle of Cut	Compressive Strength at Angle
0 degrees	9.1 MPA
15 degrees	15.64388146 MPA
30 degrees	26.21273545 MPA
45 degrees	35.93612389 MPA
60 degrees	43.53966582 MPA
75 degrees	48.35368624 MPA
90 degrees	~50 MPA

(At 0 degrees, it'll just be the perpendicular compressive strength, while at 90 degrees, it'll just be the parallel compressive strength)

To illustrate it-


(The harder part of cutting at an angle would be getting the area of the cut rather than the compressive strength)

And yeah this can be ignored for materials that don't have axis-specific strengths like wood.

Damn it, image doesn't work for me, imgur link blocked in my place, can you give me some other temporary link to see?

 

[Arceus0x]

I think the question that is in need of being answered more at this point is
Why is elaj not a calc member yet?

 

[Agnaa]

Good question, I'll suggest them for staff.

 

[ElajRuengies]

Damn it, image doesn't work for me, imgur link blocked in my place, can you give me some other temporary link to see?

Alright hold on-

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[ElajRuengies]

I think the question that is in need of being answered more at this point is
Why is elaj not a calc member yet?

Good question, I'll suggest them for staff.

I'm flattered by the sentiment, but my first semester of University starts on Monday- I'm living away from home in a dorm for the first time, and I'm still figuring out the time management skills required to not starve... So I'm not in a place where I'd be able to answer the inevitable Sea of Calc Evaluation Requests that would flood my wall if I did become a Calc Group Member (Not right now at least)

 

[Arceus0x]

my first semester of University starts

Mine's in like 11 days

 

[ElajRuengies]

Mine's in like 11 days

Ayyy, congrats!

 

[Arceus0x]

Ayyy, congrats!

thanks, man, grats' to you too!

 

[Arceus0x]

Good question, I'll suggest them for staff.

Jim Hawkins cuts a tree

vsbattles.fandom.com

i remade it using the suggested formula, is this good now?

 

[Agnaa]

I'd like to wait for more calc group input.

 

[Arceus0x]

I'd like to wait for more calc group input.

k, i just made another calc for the same verse at 9-A so i'll just put that up for evaluation for now

 

[KLOL506]

Alright hold on-

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Thanks.