LS from tearing off a piece of wooden floor

[StrymULTRA]

In short.

Feat is this, and I tried to calculate it by applying flexural strength to the combined area of the sides at the perimeter, like this calculation.

However, when I have calculated it myself, I have gotten over 632,502 Kg (was halved because the feat was performed by 2 people), and obviously it's unrealistic that a human-sized portion of a thin wooden floor can sustain the combined weight of 100+ adult elephants.

So, is there an alternative to calculate such feats?

 

[Agnaa]

@DontTalkDT @Dalesean027 @Psychomaster35 @Flashlight237

Any of you have thoughts on this?

 

[ByArrow]

I calculated it using another method. The result yield at Class 100.

Gravity Falls Calc: Dipper and Mabel free Stanley from the floor

vsbattles.fandom.com

 

[StrymULTRA]

I calculated it using another method. The result yield at Class 100.

Gravity Falls Calc: Dipper and Mabel free Stanley from the floor

vsbattles.fandom.com

Off to wait which method is better, then.

Though it's still weird for me that a piece of wood that is barely above 6 cm thick can withstand 177 Tons.

 

[ByArrow]

Off to wait which method is better, then.

I think the method I used is better because that was the method used for a similar feat.

Sam Raimi Spider-Man: New Goblin tears a wall

vsbattles.fandom.com

 

[Agnaa]

I calculated it using another method. The result yield at Class 100.

Gravity Falls Calc: Dipper and Mabel free Stanley from the floor

vsbattles.fandom.com

So 3.6x lower

 

[KLOL506]

IDK man, the value seems excessive but at this point feels more like a limitation of the formula itself not being able to account for other criteria like failure points where the whole structure gets weaker if one point fails. But that's just me.

I'll wait and see if DT has anything to say about it.

 

[PhantomØ4]

I am pretty sure these results are not making sense you you lot because the calcs are supposed to be using Flexural Stress instead of shear stress, not because it yields lower results but because when the force is applied to the floor boards it behaves akin to a beam with a load (just opposite direction from gravity) and it'd what would case an elephant to break the floor boards as it would not act like a shear force.

σ = (M*c)/I Taken from here

Set σ equal to the modulus of rupture to find the stress under which the material will give in.
I = Area moment of inertia (Beam's cross section about the neutral axis)
c = Distance from the neutral axis to the outermost fiber.
M = Internal bending moment at a specific cross-section of the beam, there's normally a shorthand for this but since the floorboards broke at two points it cannot be used. It's a very long explanation I spent about 45 minutes (probs more) writing before giving up cuz I don't really wanna do the calc but just look into it.
Once you get that M solved just use algebra to re-arrange the equation to solve for Force.

The annoying part in this feat is that you don't have a single point of force like that since the duo are applying force at Stan and whatever has him glued to the floor is then applying force unto the floor... each board seems to have had two breaking points so you can say those are the points where the flexural load was being applied to each. You'd have to calc the value for each floor board horizontally since they broke horizontally (relative to standing Stan) at different lengths.

Honestly just from trying to explain the idea it's a really big hassle but it should result in much much lower and sensical results (in the sense of forces these floorboards would realistically handle). Up to you lot if a result is worth the effort.

 

[StrymULTRA]

Alright, taking in this image as reference.

I = Area moment of inertia (Beam's cross section about the neutral axis)

Do I take each of the pieces making up the whole thing (I mean separated by the lines) individually, and then apply the weak axis?

c = Distance from the neutral axis to the outermost fiber.

Idk what the latter is.

M = Internal bending moment at a specific cross-section of the beam, there's normally a shorthand for this but since the floorboards broke at two points it cannot be used. It's a very long explanation I spent about 45 minutes (probs more) writing before giving up cuz I don't really wanna do the calc but just look into it.

May DT help us here then.

 

[PhantomØ4]

Alright, taking in this image as reference.

Do I take each of the pieces making up the whole thing (I mean separated by the lines) individually, and then apply the weak axis?

The exact steps depends on if they all bent individually or as a whole/group. Either way you gotta get all of them but if they were bent as a whole then you gotta get the neutral axis of the entire section and compute each piece’s contribution to I about that same neutral axis.

Idk what the latter is.

It's just thickness/2 for this specific feat

If you can't find it on your own DontTalk might be willing to explain it?

 

[StrymULTRA]

The exact steps depends on if they all bent individually or as a whole/group.

Definitely the latter as they're all on the same level here.

Either way you gotta get all of them but if they were bent as a whole then you gotta get the neutral axis of the entire section and compute each piece’s contribution to I about that same neutral axis.

What do you mean by that, assuming that the neutral axis is the full lenght?

 

[Psychomaster35]

It definitely feels like a similar calc I made, but that was with metal and not wood.

 

[Flashlight237]

@DontTalkDT @Dalesean027 @Psychomaster35 @Flashlight237

Any of you have thoughts on this?

I've worked with my dad on floor installation for about two years. I think this is a massive overestimation. My dad did his fair share of floor removals, and he was able to pry glued-down boards out with a hammer and crowbar just fine, though he admits the floor-stripping machines would be way more effective. Of course looking at the boards of this cartoon still here, it looks like these boards were nailed down. In which case, the force needed to remove an individual nail is around 165 pounds of force for a smooth-shank nail:



You would have to turn those boards into pincushions if you want to get anywhere near the force calculated.

 

[Agnaa]

Of course looking at the boards of this cartoon still here, it looks like these boards were nailed down. In which case, the force needed to remove an individual nail is around 165 pounds of force for a smooth-shank nail:

To be fair, they didn't fail at those nails; they cleanly broke, sometimes cleaving a plank in half.

But overall, I think the more important thing is, if this is wrong, then what does that say about other similar calculations? Is there a feasible replacement method?

 

[Flashlight237]

To be fair, they didn't fail at those nails; they cleanly broke, sometimes cleaving a plank in half.

But overall, I think the more important thing is, if this is wrong, then what does that say about other similar calculations? Is there a feasible replacement method?

To answer that last question, unfortunately there is not. Not that I'm personally aware of at least.