## Principle Axis Bending of Single Angles

## Principle Axis Bending of Single Angles

(OP)

Come on, does this ever happen?

AISC Spec F8 has a method for calculating the capacity in these cases. Is there really a practical application where a single angle is bending about its principle axis?

I'm coding a calc sheet and I'm wondering if I should save myself some time and ignore this assuming no one will ever need the capacity of an angle bending about its Z axis.

AISC Spec F8 has a method for calculating the capacity in these cases. Is there really a practical application where a single angle is bending about its principle axis?

I'm coding a calc sheet and I'm wondering if I should save myself some time and ignore this assuming no one will ever need the capacity of an angle bending about its Z axis.

## RE: Principle Axis Bending of Single Angles

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.

## RE: Principle Axis Bending of Single Angles

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.

## RE: Principle Axis Bending of Single Angles

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.

## RE: Principle Axis Bending of Single Angles

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)

## RE: Principle Axis Bending of Single Angles

## RE: Principle Axis Bending of Single Angles

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

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## RE: Principle Axis Bending of Single Angles

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

## RE: Principle Axis Bending of Single Angles

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.

## RE: Principle Axis Bending of Single Angles

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*****

## RE: Principle Axis Bending of Single Angles

## RE: Principle Axis Bending of Single Angles

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.

## RE: Principle Axis Bending of Single Angles

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.

## RE: Principle Axis Bending of Single Angles

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.

## RE: Principle Axis Bending of Single Angles

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

## RE: Principle Axis Bending of Single Angles

## RE: Principle Axis Bending of Single Angles

## RE: Principle Axis Bending of Single Angles

RC

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

Edmund Burke