Single angles will tend to bend or buckle about their principal axes unless restrained in some way.  

Restrained in some way could mean they are continuously connected to a deck or something which forces it to bend about it's geometric axis.   

Thanks for the quick response Josh.

Interesting, you just blew my mind. I have to completely change my approach now.

I see in the commentary (p.16.1-280) ....

"When bending is applied about one leg of a laterally unrestrained single angle, the angle will deflect laterally as well as in the bending direction. It's behavior can be evaluated by resolving the load and or moments into principal axis components...."

But now I'm wondering what the intent of the code is now because the next sentence of the comentary says "Section F10.2(i) is provided to simplify and expedite the calculations for this common situation"

So it seems that the spec accounts for this bending effect anyway in the geometric axis bending equations, but only for equal leg angles. For unequal leg members then I'm assuming I having to do it manually by resolving the load to each axis and summing the effects and that must be why they provide the principle axis bending equations. This makes sense now. That explains why this once seemingly ludiscris equation is provided. I thought that there may be instances where an angle is loaded primarily about its principal axis.


ASSIDE: Shouldn't it be princiPLE? not PrinciPAL? Principals are people who manage grade schools and companies.



 

Yes the equations account for principle axis bending for equal leg single angles - the "bending about the geometric axis" wording is a bit misleading, it has been changed to "bending moment about the geometric axis" in 2010 Spec to indicate it is for cases where the applied moment is about the geometric axis, but the actual bending is still about the principal axis.

Unequal leg angles are complicated.  You need to resolve loads into components about each principal axis.  Then go through the LTB equation to determine the strength for the component through the major principal axis.  The strength of the minor principal axis is the elastic moment (FySw) about the minor principal axis.  Then you have to use Chapter H interaction equations to combine it all.  You technically need to do this for each tip and heel of the angle to make sure you have the worse case depending on the loading.     

Thanks WillisV,

So is the yeild moment that is suggested ->  0.8 x Yield Moment calculated using geometric section modulus -> just a quick and dirty way to get an interaction of the yield moments about the 2 principle axis.

0.8 x My(geometric) = My(major principal) x sin(a) + My (minor principal) x cos(a)


 

The word in this case is indeed "principal".

And it's ASIDE, not ASSIDE...

If you "heard" it on the internet, it's guilty until proven innocent. - DCS

http://www.eng-tips.com/supportus.cfm

If anyone has tried doing these calcs both ways using the provisions of F10-4 and then trying it by resolving the moments and using F10-5, have you had success getting the same solutions?

I'm about 20% off. I'm combining the moments about the major and minor principal axis using chapter H -> Muz/Mnz+Muz'/Mnz' yaddayaddayadda
 

Why doesn't AISC publish My_z and My_w values to make this whole process easier?  Or even if they would publish the maximum "c" values (as in My = Fy*I/c) for the heals and toes of the angles.  When you include the fillets at the toes, this is very difficult to calculate.

I have been drawing the angles in autocad, rotating the shape alpha degrees, and measuring these distances.  It would be much easier if this were tabulated somewhere.

I get these C values using a spreasheet. Its pretty easy and it took me like 10 min,..

1) Plot the angle in its geometric axis by getting the coordinates of each corner of the angle. Should be 6 corners -> 6 coordinates.

2) Adjust these coordinates so that the centroid is at (0,0) by subtract Yave from the Y's and Xave from the X's.

3) Then you can rotate this shape by any degree by...

a) Convert the coordinates to polar which consist of a vector (Angle, Radius) where Angle=tan-1(y/x) and Radius = sqrt(X^2+Y^2)

b) Add the angle that you want to rotate the shape to each angle of the coordinates

c) Convert the polar coordinates back to cartesian coordinates.


Now the max negative Y and max positive Y coordinates are your respective C+ and C- values.


********Be carefull about signage when the angle enters a new quadrant, you need conditional statements to manually adjust the sign for the 3rd quantrant since tan-1(-x/-y)=tan-1(x/y). Remember that from trig*****
 

Do you account for the fillets at the toes?  If so, how?

No, but if I were to then it would only affect the moments of inertia. I would just subtract the moment of inertia of the missing material at the toe which is provided on P. 17-40 of the AISC 13th. "Parabolic Fillet in Right Angle"

I = (11/2,100) t ^4

And then use parallel axis theorum to move that moment of inertia to the centroid.

Although there is no way to adjust the new Ixy (which you need in order to get the rotated Ix and Iy). I might just take the area of these fillets =(t/(2sqrt(2)) and multiply them by the x and y coordites to the centroid of the section (also given in the manual) to get an approximated Ixy that you can subtract from the non fillet sections Ixy.
 

Why do you need Ixy?  Ix + Iy = Iz + Iw (using AISC's notation for principal and geometric axes).  Iz is given as rz^2 * A.

Assuming square ends at the toes will give larger "c" values at the toes and so would yield a smaller S at the toes.

Thanks, by the way, for the polar coordinate method of calculating the distances.  It worked great!  I never would have thought of that myself.  I am very stuck on the cartesian system.

Oh yeah, I forgot about that little short cut, although that too might be an aproximation. I've been using Ixy for the full transformation equation since I'm starting from scratch.

I'm not sure exactly how fillets work, I always assumed that they are an inverse quarter-circle taken off the tips of the angle, but if the very tips of the angle are still intact, then wouldn't the c values be the same? Unless you're taking c as the distance to the base of the fillet and not to the outer fiber.

I would think that the fillets reduce "I" which reduces S.


 

I want to make sure I'm following this correctly:

The Sw can be found by dividing the Iw (used AutoCAD to find principal axis properties) by the largest distance from centroid to rotated tip of toe?

Thanks in advance.

This is a rather confusing short spec.

RC
All that is necessary for the triumph of evil is that good men do nothing.
    Edmund Burke

 

I have recently made a post to the freely downloadable AISC steeltools.org where Mathcad spreadsheets resolve these issues, the more complete being a Biaxial beamcolumn unequal legs angle. Here you have a printout with most calculations collapsed but if you download it you can see all within.

• http://files.engineering.com/getfile.aspx?folder=3319ee22-a26d-4a53-8b26-53

http://www.steeltools.org/STEELTOOLS/STEELTOOLS/Resources/ViewDocument/Default.aspx?DocumentKey=f174a7b0-6a9b-4bb7-a024-dd6933645621
 

No mathcad here in the office. Thx though.

RC
All that is necessary for the triumph of evil is that good men do nothing.
    Edmund Burke