What is the Radius of Gyration?

[@carltolentino2044]

6:23 where is the dimension "b"?

[@StructEdOrg]

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

[@momentouscrazynoob1709]

@@StructEdOrg thank you for the clarification :D

[@manmohanbisht69]

What are other practical application. Apart from buckling ?

[@StructEdOrg]

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.

[@gregoryfrank418]

How does the buckling calculation change if this is a tube vs a simple square?

[@StructEdOrg]

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.

[@gregoryfrank418]

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

[@NuwanFernando-mq7kt]

How do you calculate the radius of gyration of a composite section?

[@StructEdOrg]

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.

[@janlopez1453]

This content is well presented. Thank you for sharing. . .

[@jacobrose1664]

Parallel axis theorum???

[@StructEdOrg]

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.

[@sjmy2000]

Iy=2*bt3/12 + (h-2t)*3/12

[@StructEdOrg]

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.

[@cutebonggu]

Great Explanation and Glamorous Voice!😃

[@StructEdOrg]

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

[@nguyenkhanh2101]

Thanks for sharing, I learned one new.

[@user-hu8ih7bn8u]

Thank you so much my friend!

[@muthomimunyua9062]

Thanks.

[@beascott1281]

Great video😀

[@lunditv]

Nice

[@momentouscrazynoob1709]

thank you :D

[@anaishaagarwal3067]

thanksssss

[@thesaver999]

while calculating the moment of inertia for that I-Beam, you forgot steiners theorem, which means adding the distance of the global centre of mass to the particular center of mass, squared, times the Area of that particular area

[@StructEdOrg]

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

[@khangdo4959]

@@StructEdOrg THANK YOU! I learned