Angle Braket Loading Calculations

[nonsinusoidal]

Hello Guys:

I am in need of assistance. My task is to build a support made out of Steel Angle Brackets as a means for supporting a duct system. So far I have not advance to a point that I can proudly proceed with my task. Can anyone help me determine the allowable loading for a given Steel Angle Bracket. Lets say that I would like to utilize a L 5" x 3 1/2" x 3/8" bracket to support a vertical load laying evenly on top of it. The steel angle bracket is to be supported at both ends by means of other two sets of angle brackets (vertical positioned) which are to be attached to I-Beams. I have available to me the eight edition of a manual of steel construction book. Can anyone be so kind and provide me guidance on how to utilize the values given in such manual. I also do not clearly understand the given data in that books. Please advice.

 

[desertfox]

Hi nonsinusoidal

If a beam as symmetry in that plane then the neutral axis occurs at the centre of the shape.
So for instance a rectangle, square and a standard "I" beam will have the neutral axis at the centre of the shape.
Look up neutral axis in your text books.

regards

desertfox

 

[rb1957]

desertfox,

do you really think that those two (ok, four) angles are going to bend together like a beam ? i guess, obviously, you do.

i'd rather see each angle as an independent beam, carrying 1/4 of the load.

nonsinusoidal,

you're asking very basic questions, and i don't think you should rely on this forum to completely answer them all. i think you need to either find a structures guy where you work, or 2nd best, find a structures text to help you. as desertfox has said, "This is only a small portion of the calculations your faced with".

 

[desertfox]

hi rb1957

Yes I believe a pair of angles as shown in my uploaded file will act as a composite beam due to the connecting rods.
Whilst I appreciate they might be some small local bending between the vertical rods I think this will be small, I likened it to castellated beams in buildings where they burn out the centre web to reduce weight which is also based on the assumption that all the bending stress is carried by the beam flanges so I am led to believe.
Your method would be very conservative and theres nothing wrong with that.

desertfox

 

[nonsinusoidal]

Desertfox,

Thank you for all of your responses.

RB1957,

Unfortunately, structural engineers are not a part of my organitation. However, my original question has not been answered (It was diverted). My original question was as follows:

"I have available to me the eight edition of a manual of steel construction book. Can anyone be so kind and provide me guidance on how to utilize the values given in such manual. I also do not clearly understand the given data in that books."

Please advice as appropriate.

 

[rb1957]

i agree your OP wasn't answered, but the thread wasn't "diverted", rather other questions that have to be answered were posed. going back to your post of 3rd July, the running load is 15*10 = 150 lbs/ft for conduit and 7*10 (or 7*30) lbs/ft for cables. so each support would react 8ft worth of load = 1760 lbs (or 2980 lbs), and at each support you've got two angles (no?). the maximum moment in the structure looks to be something like 220(or 360)*64/16 = 880ft.lbs; i would say each angle reacts 220 ft.lbs, you have to check the angle section (crippling is probably the critical failure mode).

my advice, pick one of the following ...
1) find a structures textbook that will explain all the things you need to look into to solve this problem;
2) maybe a building code will explain what they want checked (and i think they also say how to check); or
3) if there are structures people in a different part of your organisation, go find them.

 

[rb1957]

oops finger trouble ... Mmax = wL^2/6 = 220*64/6 about 1 ton.ft

 

[desertfox]

Hi nonsinusoidal

You need to specify which properties your asking about, it is unlikely that we can give you a detailed answer to all the things you would like to understand however we might be able to point you in the right direction.

desertfox

 

[trainguy]

rb1957,

1) Since when is 880 / 2 angles equal to 220?
2) I would use simple support, so wl^2/8, not wl^2/6

Agree that crippling is the failure mode, as long as by that you mean lateral/torsional buckling. I'd use the full length as unsupported. Loads are so low, it may still work.


Desertfox:

The angles are linked by vertical ties, and NO DIAGONALS OR SHEAR WEBS, so there is no composite action. Just take the total load per support assembly and divide by 2 angles. Better yet, I'd assume one angle could take maybe 60-65% of the total, and sleep better.

tg

 

[gwolf2]

> it gave the formula below that you provided:
"?= M*y/(I).................. and the rest.

OMG!

no

OMFG!!!


:-((((((

 

[nonsinusoidal]

Trainguy,

The two horizontal angles are indeed supported by two vertical angles. Since my expertise do not fall withing this subject matter, I would like to know if you can explain what you stated below:

"The angles are linked by vertical ties, and NO DIAGONALS OR SHEAR WEBS, so there is no composite action. Just take the total load per support assembly and divide by 2 angles. Better yet, I'd assume one angle could take maybe 60-65% of the total, and sleep better."

Thank you for your assistance in this matter.