Slenderness Ratio & Radius of Gyration

[bing.katie]

I think I am making this harder than necessary but hopefully someone can help clear this up for me:

Radius of gyration is the square root of the ratio of the moment of inertia to the area

r=√(I/A)

Slenderness ratio is the unbraced length (L) X the end condition factor (k) / radius of gyration (r)

SR = (kL)/r

Question 1:
The inverse relationship between the two is confusing me. I understand (by these equations) that the LARGER the RoG, the SMALLER the SR - so larger radius of gyration leads to lower SR, leads to less likelihood of buckling.

But then, if the area of the cross section goes UP, the RoG goes down; if the RoG goes down, the SR goes UP. How can a larger area lead to a larger SR? Seems like it would be less likely to buckle with a larger area?

Question 2:
See attached image. Is the cross-sectional area in the equation for RoG the surface area or the overall area (i.e., would the area be the same on these top vs bottom images, or would the tubes be larger "areas" (A) because of the holes in the middle spreading their area further)?

THANKS in advance! (I'm totally going to dream in equations again tonight...)

[Pow_Arch]

Radius of gyration is not only affected by the area itself. That is why a member's moment of inertia comes into play, which involves the member's geometry.
Imagine a small ice skater with it's arms spread out, trying to spin around her own axis. And then imagine a "chubby" ice skater with her arms close to her body, trying to spin as well. Even though the chubby skater might have a "larger area", she will still probably be able to spin much faster around her own axis than the small skater.
There are youtube videos that present this issue in a much more dense, much more mathematical way, but I hope my random explanation helped....

[Mike-SE]

STOP! You are not studying for the SE exam. You are so far beyond the knowledge you need to pass the ARE-SS you are confusing yourself. Move on to a different subject.

[Pow_Arch]

Yes sir! haha

[bing.katie]

I'm trying to understand it on a pretty basic level, worried that a question like "If the area of the cross section of a column increases, is it more or less likely to buckle?" will come up. Based on the equations and my limited understanding, I would have to say it's MORE likely to buckle, since when A goes up, r goes down, causing SR to go up.

But maybe that's not a question that would even be asked b/c you need more info to make that call...?

[Mike-SE]

I'm trying to understand it on a pretty basic level, worried that a question like "If the area of the cross section of a column increases, is it more or less likely to buckle?" will come up. Based on the equations and my limited understanding, I would have to say it's MORE likely to buckle, and you would be wrong since when A goes up, r goes down in-correct assumption, causing SR to go up. maybe, maybe not

But maybe that's not a question that would even be asked b/c you need more info to make that call...?

I did not follow the logic of your first post nor you answers in this post. Increasing A reduces fa. KL/r sets Fa. If you reduce fa the change in A must reduce Fa for the column to be more likely to buckle.
Look at column load tables for wood columns. the tables provide values for P passed on kl/r

[bing.katie]

It sounds like I was assuming a relationship was there that isn't there (or not there in the way I assumed...). I broke down the equations more and see (I think) why you say "maybe, maybe not".

ANYWAY - moving on and leaving this one alone! Thanks for humoring me!

[shortkm1]

is a higher or lower slenderness ratio better for a column? Basically, I am wondering if you use an effective length of 2.0, is that less likely to fail than an effective length of 1.0?

[spike]

lower is better . Think of it this way..at least this is how I make sense of it.
for a wooden col SR=L/d where
L=length
d=least lateral dimension.

thinner the col and taller it is ...chances that it will fail by bucking compared to another col with a larger lateral dimension.
compare a broom stick and a pencil of the same length. the broomstick will buckle when a similar axial load is applied.

so for a given length...when you increase the lateral dimension d, the col has better load carrying capacity.... this also results in a decreases of SR

so lower SR better capacity

hope the convoluted explanation help

[shortkm1]

lower is better . Think of it this way..at least this is how I make sense of it.
for a wooden col SR=L/d where
L=length
d=least lateral dimension.

thinner the col and taller it is ...chances that it will fail by bucking compared to another col with a larger lateral dimension.
compare a broom stick and a pencil of the same length. the broomstick will buckle when a similar axial load is applied.

so for a given length...when you increase the lateral dimension d, the col has better load carrying capacity.... this also results in a decreases of SR

so lower SR better capacity

hope the convoluted explanation help

That is great! Thank you so much! So to recap, a K value of 0.5 is better than 1.0 is better than 2.0, right? (in the equation SR=(K*L)/r)

[spike]

yes

[shortkm1]

Thank you!! testing tomorrow!

[spike]

wish you luck! I hope it goes well.