Anonymous at Fri, 21 Jun 2024 18:16:47 UTC No. 16246024

>>16246003
It doesn't. Scienyizts don't know shit. You need to increase length. For more Strengtj.

Anonymous at Fri, 21 Jun 2024 19:29:11 UTC No. 16246112

>>16246003
Get a piece of wood and try it yourself. Honestly it's baffling that you don't already have an intuitive feel for this sort of thing. Are you a woman?

Anonymous at Fri, 21 Jun 2024 19:40:30 UTC No. 16246130

>>16246112
My intuition tells me that beam width would be greater or at least equal to beam depth.

Anonymous at Fri, 21 Jun 2024 19:43:50 UTC No. 16246134

>>16246003
Literally explained in your image.

Anonymous at Fri, 21 Jun 2024 19:45:07 UTC No. 16246137

>>16246003
the wood on the bottom of your pictured beam has more leverage against the pivot point at the weight. It's like using a 12' long wrench on a bolt instead of a 6" one, the 12" wrench only needs half as much force to turn the bolt.

Anonymous at Fri, 21 Jun 2024 19:46:39 UTC No. 16246142

>>16246130
So if you had a very wide and very thin piece of wood, like a sheet of plywood, your intuition tells you that it should be as easy if not easier to bend it along the wide part as the thin part?

Post your tits, dumb whore.

Anonymous at Fri, 21 Jun 2024 19:48:19 UTC No. 16246146

>>16246142
no, she said she can't imagine using a piece of wood that's not square in cross section, all pieces of wood should be square and only vary in length.

Anonymous at Fri, 21 Jun 2024 19:52:09 UTC No. 16246152

>>16246142
>it should be as easy if not easier to bend it along the wide part as the thin part?
You're talking about bending along 2 different axes.
We're talking about how adding to each axis changes how much the beam bends under the same load from the same direction.
It makes sense that doubling the width would halve the deflection, since you're doubling the amount of wood that can distribute the load.
The question is how doubling the depth would reduce the deflection to 1/8th. There has to be some reason for that ratio.

Anonymous at Fri, 21 Jun 2024 19:53:42 UTC No. 16246154

>>16246152
nobody was talking about bending in two different axes

Anonymous at Fri, 21 Jun 2024 20:32:40 UTC No. 16246197

>>16246003
Leveraging. There is a greater distance away from the central fulcrum and therefore more load can be carried dur to that leveraging distance.

Anonymous at Fri, 21 Jun 2024 21:05:29 UTC No. 16246237

>>16246003
Are you stupid? Get a bunch of thin steel rulers, stack them and try to bend them, pretty easy, now try the same with a solid steel bar of equal thickness. You fucking retard, you thought the strength would increase linearly?

Anonymous at Fri, 21 Jun 2024 21:48:26 UTC No. 16246293

>>16246003
Girth, girly doo.

Anonymous at Fri, 21 Jun 2024 21:55:37 UTC No. 16246298

>>16246237
>Are you stupid?
OP asked a good and legitimate question. You're an asshole.

Anonymous at Fri, 21 Jun 2024 22:22:23 UTC No. 16246334

After looking it up a bit, it seems like the deflection of a beam is inversely proportional to its second moment of area (AKA its moment of inertia), which is often called I. And for a rectangular beam, that formula is I = (b*h^3)/12. So that would make sense that doubling the height would divide the flex by 2^3.

But I'm not really sure how moment of inertia comes into play here. It seems like >>16246137
sounds on the right track, seeing how the Wikipedia page talks about torque, but I don't fully understand it.

Anonymous at Fri, 21 Jun 2024 22:40:43 UTC No. 16246362

>>16246334
Since the top compresses, the the bottom expands, I guess you could envision two rotational forces in the beam. Increasing the height of the beam would increase r, which would increase the moment of inertia.

Anonymous at Sat, 22 Jun 2024 05:39:57 UTC No. 16246930

>>16246362
green is my pepper

Anonymous at Sat, 22 Jun 2024 14:52:40 UTC No. 16247485

>>16246137
Yes, that explains why increasing the span would increase deflection. It doesn't answer OP's question though.

Anonymous at Sat, 22 Jun 2024 15:44:23 UTC No. 16247555

>>16246003
Watch the ratios of length to depth.

If you double the span of a beam, you deflect EIGHT times as much
So when you double the depth, think of it as zooming in twice as much, where the ratio of the doubled depth is like cutting the span in half of the original. So doubling the depth of a beam is equal to halving its span; However, cutting the span changes the deflection EIGHT times as much. This is why doubling the depth decreases the deflection by one eighth.

Now, when you double the width of the beam, you only deflect is half as much. Therefore, doubling the width only changes the load and the deflection by half.

Anonymous at Sat, 22 Jun 2024 17:48:49 UTC No. 16247799

>>16246197
>Leveraging. There is a greater distance away from the central fulcrum and therefore more load can be carried dur to that leveraging distance.
OP, this. Look at your picture. DO you see how there is a fulcrum in the center of the beam? Above it there's compression; below, it tension.
They are represented as "cones of arrows": from the fulcrum upwards they are shown as compressing and progressively longer. From the fulcrum downwards, they too get longer but they point outwards.

Now visualize this: if you double the width of the beam, you double the number of arrows, yes?
However, if you double the height of the beam, you have to imagine that the cone of arrows continues to progress up and down from the fulcrum, and so the arrows actually get LONGER if the beam gets taller, do you see that?
For that reason, the total amount of "arrow lengths" is much greater when the cone of arrows is prolonged by doubling the height of the beam, than it is by simply doubting its width.
That's it, leveraging. When you double the height, you double the "leverage" that extends from the center fulcrum of the beam

Anonymous at Sat, 22 Jun 2024 17:50:10 UTC No. 16247801

>>16247799
>as "cones of arrows"
or triangles, or wedges of arrows