Sorry for the late reply! Yes, there are standards that prohibit use of those local-buckling-prone shapes, particularly when high reliability is required or when predictable yielding behavior is needed. Seismic and blast-resistant design are the two that immediately spring to my mind, but I suspect the Nuclear Regulatory guys have similar requirements (just my suspicion, that's not a sector I'm familiar with).

looks to be about 0.25!

Thanks for the interest! I'm looking to have this be primarily used in tension, and in that configuration you're right that horizontal wouldn't be a big deal, but I'm also looking to perhaps make some compression tests as well, in which case thin materials could be unduly influenced by gravity.

Thanks for the interest Wink, I have this project tabled at the moment, a casualty of moving and studying for the SE exam, but picked up a milling machine at auction and some other tooling recently, so hoping to get back into it in the spring! If all goes well, I'll drop some parts lists and Send Cut Send design files for anyone looking to build along

Hamed, good tip, this one has fallen a bit into the weeds as I moved and have been studying for the SE exam, but hope to get back to it soon with Send Cut Send files linked for anyone who wants to copy it!

Nuwan, if you're fully-composite and all of one material, then exactly the same way. Fully-composite and multiple materials, you'll need to transform the section using the modular ratio, if this is what you're interested in, please let me know what shape you're looking for and I'll shoot a quick vid. Otherwise, partially-composite get a bit more complex.

Jacob, at 5:45 I mention that this is all centered on the weak axis, so we have no need for the A * d^2 component from parallel axis theorem, as there's no offset distance to be squared. If we were working strong axis moment of inertia then we would need to either utilize the parallel axis theorem, or subtract the I-value of the negative space from the overall bounding rectangle.

No problem Juan! It's buried in the whole q = V * Q / I equation. Q is the area outboard of the section you're cutting through times the distance from the centroid of that area to the centroid of the overall shape, and it's the only part that changes as we march along the shape. The overall shear V and the overall shape I-value are constant everywhere. When you're in the horizontal flanges, the distance centroid-to-centroid remains constant, it's just half the height of the overall shape minus half the thickness of the flange, so the only part that's increasing is the area, which goes up linearly as we move along the flange. In the vertical web, two things are going into that increasing Q. We've got the area increasing at a linear rate again as we slide down the web, but also have the centroid-to-centroid distance shifting, which multiplies by our coordinate again, giving that second order, or parabolic, distribution. Does that clarify it? Happy to explain further or shoot a video expanding on that if it's at all unclear!

@@StructEdOrg thank you so much for replying. I understand it more now, but I think I’m still struggling to visualize it a little bit. I will try to go through it again to see if I can make more sense of it. Thanks again!

@@StructEdOrg thanks a lot!

Thanks Fedor! So unless you load at the shear center, the torsion is going to always be there, and your best defense is to somehow incorporate tubular shapes into your structural member. Otherwise, anything you can do to either try to get the load applied closer to the shear center, or to move the shear center closer to the load will help to limit the eccentricity and therefore your torsion on your section. Even a tiny little tube applied somewhere on your shape helps tremendously! If you have a specific design question, definitely feel free to email and we can discuss more! [email protected]

@@StructEdOrg wow. This is so detailed explanation. At uni, they taught how to get shearcentre but they didnt teach what to do after getting the shear centre (maybe they dis but I am the one who forgot) Thank you

😂 no, just cursed to have some microphones not work well for me

More of a forgetting the thickness of the 2x4 was going to be somewhat more than the length of the shaft from the motor. Planning a 3-piece bolted steel design for the top, but ended up moving and getting bogged down with other projects so far this summer.

ProBel, you're right that I've confused the issue here, but it's actually I.y = 2*t*b^3/12 + (h-2t)*t^3/12. I'm going to go try to add one of those little notes into the video now for that section of the video, unfortunately KZbin doesn't let me go back and edit an already-posted video for actual content.

You're absolutely right, that was a silly mistake to make right at the end of the video! I'll drop a pop-up card on there with the correction once I get my desktop fired up. Thank you for paying such close attention, and for letting me know!

missing lol. I had already decided in my head that b was going to be equal to h, but hadn't said it yet and foolishly substituted in the h

@@StructEdOrg thank you for the clarification :D

unfortunately still in the making. Life has gotten crazy and I haven't made it to the steel yard yet, though I have most of the other components on-hand. Also working on a CNC router to help finish the job, but the plans for that I purchased have been absolutely horrible, so it's taking much longer than expected.

@@StructEdOrg hope to see it built soon.

You got it! I seem to have misspoken 🙈

No, I wouldn't say that. You may be able to take more accurate measurements on a ruler with millimeters than with inches, depending on the gradations (1/16" is about 1.59 mm, 1/32" is about 0.79 mm). But if you're measuring with, say, a digital calipers capable of displaying in either unit, the accuracy should be equivalent.

😊 thanks! Quite certain no one has ever called my voice "glamorous" before!

Any time! And happy to make any videos on topics you could benefit from!

You are very welcome

Rasmus, remember that we're calculating the weak axis moment of inertia, so in this case the distance term of Steiner's would be zero, all three rectangles are located on the neutral axis. Rewatching, looks like I did accidentally have "h" instead of "b" in the flange terms, but that didn't end up changing the outcome as a couple lines down when I plugged in numbers I had h=b=4", so that's my bad and confusing

@@StructEdOrg THANK YOU! I learned

Good question! Lots of things, like the shear center if a channel, can be located outside the shape, but the PNA has to be located within it. Remember that we're looking for where the full force resulting from complete yield of the material above and below the axis is balanced. If the PNA is above the top flange, then the whole section would be yielded in just tension or just compression, with no material in the opposite stress state to balance it out.

@@StructEdOrg thank you so much

Anytime! I should add that by "within the shape", I mean within the bounding dimensions of the shape. The actual "plastic centroid" (not sure if that's a real term or not) where your x and y-PNAs intersect could be out in space for parts like a channel shape with returns on the flanges, but it will always be within the bounding box of the shape.

The radius of gyration would still be the square root of I.y/A, but you'll get a larger radius of gyration than for the solid members. For both I.y and A, we can find the values of the larger shape, the outside of the tube, and subtract off the values of the smaller shape, the inside of the tube.

@@StructEdOrg That's what I thought, just wanted to make sure, thanks!

Buckling of beams and columns, and bracing to prevent it, is all I've seen this concept used for. Apparently there's a similar concept in polymer chemistry, but I didn't really follow what it is.