Design of an Angle as a Beam 3

[astructurale]

How do I go about designing an angle as a beam (a short one of course)? My angle is loaded on the inside face of one of the legs (not an upside down L where stiffeners could be added) so it is not stiffened, and it is not braced the full length. I've looked through the LRFD (2nd Ed) and it has equations and tables for Channel sections and I-beams, but not angles. Can anyone point me in the right direction? Do the same equations apply?

 

[UcfSE]

For the AISC LRFD 2nd edition, look on page 6-277 of volume 1 for the "Specification for Load and Resistance Factor Design of Single Angle Members". The design equations for single angles are different from those we use for other members. There is also an ASD version in the 9th ASD edition if you are interested in that.

 

[astructurale]

Thank you UcfSE!!

I missed it, and it's even in BOLD LARGE TYPE.

 

[UcfSE]

A couple of years ago I asked someone the same question ;-)

 

[capiere]

"A couple of years ago I asked someone the same question"
- the best answer I have ever read in this forum.

 

[astructurale]

UcfSE,

On 6-284 (LRFD 2nd ed) it defines "b" as the full width of angle leg with tip in compression. Of course it does not show this pictorally, so I am a bit confused.

If I have a 6x4x3/8" angle with the load applied on the inside face of the 4" leg with the 6" leg pointing up
(loaded as an "L" instead of an upside down L), what is my "b" dimension? I would guess that b=6", since it is used in the b/t equation and t is the thickness of the angle.

Likewise if my angle were upside down b would be 4" right?

 

[astructurale]

I think I've got it right. My design seems reasonable, and I have logical value choices for "b" based on LTB & LB.

 

[UcfSE]

Note the headings for 5.1.1 and 5.1.2. You will use different equations depending on whether the tips are in compression or tension. Also, if your angle is not constrained externally to bend about the geometric axes, then you will also need to check the provisions of 5.3, and specifically 5.3.2 for an L6x4.

That is a lot of the problem with angles. The geometric axes do not coincide with the principal axes. If you have a leg pointing vertically and a load pointing parallel to the leg, the angle will still deflect out of plane, even though there is no load out of plane. That complicates the design equations, as you can see in section 5.3.

 

[astructurale]

Yes, I did notice 5.1.1 and 5.1.2, and I have checked 5.3 & 5.3.2 also.

It was a tough go the first time, but I think I've got the hang of it. Now I'm just trying to create a MathCad worksheet to be able to easily do these calculations in the future which I think I've just about got done.

Thanks for all of your help!

 

[astructurale]

Arghhh...

Using my "b" b/t is always a large number, which does not make since given the way the equations are written (comparing it to decimals of numbers).

So my "b" cannot be right.

I had, b=6"

for 6x4x3/8 with 6" leg up and in compression. Guess I do still need an explanation after all. Why can't they just illustrate the dimensions instead of trying to describe them? It's frustrating.

If you know of some place that shows the dimension pictorally, or how to better describe "b" so that I can understand please let me know.

 

[astructurale]

FYI - I know the decimals are multiplied by sqrt(E/Fy0\). My Q reduction factor is negative though...

Q=1.34-0.761*(b/t)*sqrt(...

OOPS! I think I just found my mistake writing this out. I had the last part as "sqrt(E/Fy)" but for this ONE case it's sqrt(Fy/E). How sneaky!!!

It's OK now. Guess I did not need help afterall.

 

[UcfSE]

Let's start here. L6x4x3/8. b/t = 6/.375=16. Looking at b/t limiting ratios, say 0.382sqrt(E/Fy) in eq 5-1a. 0.382sqrt(E/Fy)=0.382sqrt(29000/36)=10.84. The numbers I used represent units of stress in ksi.If you were using a 3x3x3/8 then b/t would be 8. Those are in the range of the limits shown. Try checking your units or values used for E and Fy.

When the angle is oriented such that it looks like an upright "L" in profile and the load is directed vertically downward (parallel to the vertical leg toward the bottom), the angle will deflect down and to the left. This puts the angle tips in compression and the heel, or corner in tension. If the load were reversed and pointed upward then the angle would try to deflect up and to the right putting the tips in tension and the heel in compression. You have to remember that angles deflect perpendicular to the minor principal axis, not about the geometric axes. If you think in terms of this and picture your loading it helps to understand whether the tip is in compression or tension.

 

[14159]

Download the 2005 AISC Spec. I think it has updated information for your question.

DBD