What is rT, radius of gyration of section?

[MrFurleyEIT]

I am trying to find out how to calculate the term "r subscript T". rT is used in the determination of Lu, which is the maximum unbraced length of compression flange for which Fb=0.60 Fy. In PPI's Civil Engineering Reference Manual 8th Ed on Page 59-3, rT is defined as the radius of gyration of a section comprised of the compression flange plus one third of the compression web area, taken about an axis in the plane of the web. It says [Sec. F1] which I take to be a reference to the AISC Manual.

It also states that, "rT is tabulated in the AISC Manual for each beam, but is approximately 1.2ry."

I tried looking for rT in Manual, but am not sure where it is. Does anyone know whereabouts is this in the Manual, 9th or 13th editions? Then, I thought that maybe rT refers to either rx or ry, but if rT is approx. 1.2ry, then it cannot be ry, can it?

I guess, in the strictest sense, I do not have to know rT, since Lc and Lu are provided in the AISC Manual, 9th edition, at least, but it is good to know what rT is.

BTW, how do you find Lc and Lu in the AISC Manual, 13th edition? I think I am seeing Lp and Lf, and am not really what these are.

Would appreciate help or advice from anyone familiar with rT, or have come across this problem, and found a resolution for it. Thanks for the heads up in advance.

 

[JAE]

For a wide flange, you take the full compression flange and include with it 1/3 of the portion of the web that is above the neutral axis of the member.

This gives you a "T" section. The full flange is the top of the "T" and the small portion of web is the vertical stem of the "T".

Turn the section 90 degrees and draw a neutral axis line through the shape (for a WF then it would be at the centerline of the now-horizontal web portion.)

|
==| ----- neutral axis
|

Calculate the I of the sideways "T" about that neutral axis.
Calculate its area, A

rT = sqrt (I/A)

 

[YoungTurk]

This is defined, and tabulated, in Machinery's Handbook.

"The radius of gyration with reference to an axis is that distance from the axis at which the entire mass of a body may be considered as concentrated, the moment of inertia, meanwhile, remaining unchanged."

Google gives:

http://www.uoregon.edu/~struct/courseware/461/461_lectures/461_lecture30/461_lecture30.html

 

[rb1957]

a query on JAE's post ... why wouldn't you take the axis with the minimum I ?

 

[MrFurleyEIT]

Thanks for all the responses, people. YoungTurk, were you talking more about radius of gyration in general?

For JAE, when you said to include the 1/3 of the portion of the web that is above the neutral axis of the member, did you mean only 1/3 of the top half of the web. The N/A is gnerally at the mid-point of the web, say. So, only the top half of the web is in compression. Therefore, 1/3 of the top 1/2 of the web is just 1/6 of the total web area, right?

Also, why do we need to turn the T-section 90 degrees? Is there any reason for this?

Say, we turn the T-section 90 degrees, how do we determine what I is? I is generally bh^3/12. What are my values for b and h, say, for a W section. With the T-section rotated 90 degrees, per your sketch (good one!), is my h then going to be the bf? What about b? Do I take tf, or some other value?

Thanks for your response, and hope you could elaborate some more on this.

 

[shin25]

You can find it out in the AISC-ASD book. See under the worked out examples related to plate girders.

 

[UcfSE]

rb, r[sub]T[/sub] is used to calculate the lateral-torsional buckling strength of the beam (I's and C's). The "T" section shown is braced in its weak direction by the rest of the web and the tension flange, so it won't buckle in that direction. When a beam buckles due to LTB, it moves about the strong axis of the "T".

That logic applies to "normal" proportions for typical beams used in construction. I'm sure you could probably come up with an "I" shape for which that isn't true by exaggerating its proportions, but for the most part that's how it works.

Furley, 1/6, yes. For your other answers, think about the actual deformation of the beam when it fails due to LTB. See above also. As far as calculating the I, A and R[sub]T[/sub], that's basic undergrad mechanics of materials. You should be able to do that.

 

[nutte]

The reason it's "one third of the compression area of the web" and not "one sixth of the web" is because you could have unequal top and bottom flanges, which would make the neutral axis not at the middle of the web, which would give you different areas of web in compression than in tension.

You take the ry of the T-section because this is a lateral torsional buckling failure. The flange buckles sideways, away from the web, so you look at the radius of gyration in that direction.

This procedure is found in the green 9th edition ASD Manual of Steel Construction. AISC quit doing it that way in the 2nd edition and 3rd edition LRFD manuals. In those, the lateral torsional buckling check used the torsional properties of the shape (J and Cw) to calculate the lateral torsional buckling capacity.

However, it looks like the 13th edition manual now uses an "rts" value, which is approximately equal ro the old rT. This new "rts" is tabulated in the 13th edition manual, and it is pretty close to the old rT. They quit tabulating the rT value in the 9th edition manual.

 

[YoungTurk]

I read too quickly and thought you weren't familiar with the concept of radius of gyration; sorry. If I understand from the above posters correctly, the question revolves around what portion of the web to use for the T section. Pardon the aero structures guy here!

Am I right in assuming JAE only rotates the section to make the neutral axis "horizontal"? If so, you could logically skip this step and calc ry of the T section, equivalent to rx of the rotated T section.

 

[JAE]

So, only the top half of the web is in compression. Therefore, 1/3 of the top 1/2 of the web is just 1/6 of the total web area, right?

Yes...but per the great comments above - don't get in the habit of using 1/6 of total web, but stay with 1/3 of the compression web for cases where NA is not in the center height.

YoungTurk - Yes - I only rotate it for my I calculations as a convenience for me to better visualize it.

The I of the "rotated" section about the NA would be the following:

I[sub]T[/sub] = b[sub]wc[/sub](t[sub]w[/sub])[sup]3[/sup] + t[sub]f[/sub](b[sub]f[/sub])[sup]3[/sup]
---------- ------------
12 12

where:
b[sub]wc[/sub] is 1/3 of the compression web length
t[sub]w[/sub] is the web thickness
t[sub]f[/sub] is the thickness of the flange
b[sub]f[/sub] is the width of the flange

A[sub]T[/sub] is of course the area of the "T".

UcFSe's and nutte's comments on the concept I agree with..nicely put.

 

[rulljs]

What if I wanted to calculate the value of rt for a wide flange (W6x15) with a channel attached to the flange (C6x13), what would rt be?

My guess is 1.61in+2.13in=3.74in, which is rt of the W6x15 plus the moment of inertia about the x-x axis of the C6x13, would this be correct?

 

[JLNJ]

For combined shapes use the same method. The WF or I shape capped with a channel would be one of those cases where the neutral axis is not at mid-depth and 1/3 the compression area is NOT 1/6 or the total web area.

The calculation is not that tough by hand.

 

[rulljs]

Agreed, not that tough, still learning though, thanks for the info.