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Old 04-09-2010, 03:36 PM   #1
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90 degree corners around beams


Hi guys. I'm trying to plan a conduit layout in the barn.

Does this proposal look ok? It's my understanding that this particular example shows the maximum of 360 degrees for one run between boxes.

What I'm particularly interested in here is whether wrapping around a beam like this to continue on is standard practice? I drew hard angles because I'm lazy, the real corners would be bends, of course.

I do NOT want to drill holes in my beams.

Oh, I'm also curious about the practice of running downstream wiring through each receptable box. My idea is to run 6 circuits through a single 3/4 EMT. I've read that 3/4 accommodates 16 #12 conductors. 6 circuits (120 and 240) would include 12 conductors (8 hots) and a ground for a total of 13. I read that any conduit can accommodate up to 9 hots before having to derate the capacity and use #10.

Does this all sound right so far?

Thanks!
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Last edited by tbeaulieu; 04-09-2010 at 03:40 PM.
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Old 04-09-2010, 03:53 PM   #2
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90 degree corners around beams


<object width="640" height="385"><param name="movie" value="http://www.youtube.com/v/hjrr3ZUcnh4...</param><param name="allowFullScreen" value="true"></param><param name="allowScriptAccess" value="always"></param><embed src="http://www.youtube.com/v/hjrr3ZUcnh4...etailpage&fs=1" type="application/x-shockwave-flash" allowfullscreen="true" allowScriptAccess="always" width="640" height="385"></embed></object>

I would use saddle bends.
Derating begins with 4 or more current carrying conductors. IMO there is no need to derate but if you must know how it's done it can be explained.

Last edited by brric; 04-09-2010 at 04:10 PM.
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Old 04-09-2010, 05:14 PM   #3
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90 degree corners around beams


1. 4, instead of 9? Rats! That changes everything. I need to look that up again so I understand exactly what I'm doing.

2. That sadle bend is interesting. I wonder how big it would be on an 8" beam, though! I suspect that would really stick out a lot.

Thanks!
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Old 04-09-2010, 05:16 PM   #4
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90 degree corners around beams


Would LB fittings facilitate anything here? Might make the push/pull easier?
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Old 04-09-2010, 05:21 PM   #5
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90 degree corners around beams


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Originally Posted by jlmran View Post
Would LB fittings facilitate anything here? Might make the push/pull easier?
The wire will not pull through the lb.
You have to feed the wire throught the lb, but it will make it easier for someone that has never bent a saddle before.
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Old 04-09-2010, 05:33 PM   #6
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90 degree corners around beams


I was just looking at the beam. How about something like this?
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Old 04-09-2010, 05:53 PM   #7
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90 degree corners around beams


The # used for derating is not only the number of hots, it is the number of current carrying conductors (CCC), which includes the neutrals.


4-6 current carrying conductors must be derated to 80% of the value in tables 310.16 thru 310.19 of the NEC.
7-9 @ 80%
10-20 @ 50%

Your initial post would have 12 CCCs, meaning you would have to derate to 50%.
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Old 04-09-2010, 05:57 PM   #8
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90 degree corners around beams


Oh boy, this is NOT what I read in another conversation! They defintely discussed neutrals not counting. Wow. Thanks for cluing me in. I'd sure hate to end up with a mess of conduit. I was hoping for a cleaner layout. Back to the drawing board. Literally.
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Old 04-09-2010, 05:57 PM   #9
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90 degree corners around beams


Derating begins at 4 conductors. However for derating purposes #14 wires don't have a starting current of 15 amps. It is not until you get to 9 conductors that derating drops #14 current below 15 amps.
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Old 04-09-2010, 06:08 PM   #10
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90 degree corners around beams


I haven't found that table online yet, if it is.

My AUDEL books says:

4 - 6 = 80%
7 - 2 = 70%

Furthermore, it has an example of 14#, (I'm planning on all 20A, BTW, so was thinking 12#).

PARAPHRASED QUOTE:

If you have 4 14# wires in the conduit, derated to 80% you get 12A. Since there's no such thing as a 12A breaker, you upgrade to 12#, which handles 20A. Derated to 80%, you get 16A, which is the correct solution.
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Old 04-09-2010, 06:15 PM   #11
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90 degree corners around beams


Quote:
Originally Posted by tbeaulieu View Post
I haven't found that table online yet, if it is.

My AUDEL books says:

4 - 6 = 80%
7 - 2 = 70%

Furthermore, it has an example of 14#, (I'm planning on all 20A, BTW, so was thinking 12#).

PARAPHRASED QUOTE:

If you have 4 14# wires in the conduit, derated to 80% you get 12A. Since there's no such thing as a 12A breaker, you upgrade to 12#, which handles 20A. Derated to 80%, you get 16A, which is the correct solution.
Your derating begins at 25 amps for #14 and at 30 amps for #12
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Old 04-09-2010, 06:34 PM   #12
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90 degree corners around beams


http://www.tnb.com/ps/fulltilt/index.cgi?part=TL291

http://www.tnb.com/ps/fulltilt/index.cgi?part=TL202SC

You could use these to go around the posts but your cost will increase sustantially.

Last edited by brric; 04-09-2010 at 07:22 PM.
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Old 04-09-2010, 09:09 PM   #13
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90 degree corners around beams


Quote:
Originally Posted by williswires View Post
7-9 @ 80%
...OOPS -- typo - correct is

4-6 @ 80%
7-9 @ 70%
10-20 @ 50%
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Old 04-09-2010, 09:38 PM   #14
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90 degree corners around beams


Quote:
Originally Posted by williswires View Post
...OOPS -- typo - correct is

4-6 @ 80%
7-9 @ 70%
10-20 @ 50%
I was going to mention to ya on that one.

Now to OP there is other way as you did make a drawing you can make a double offset kick which I done that pretty often.

However it is little tricky to do it first time but after that you will get a hang of it.

Merci,Marc
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Old 04-10-2010, 07:43 AM   #15
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90 degree corners around beams


Your second picture is the start of a 4 point saddle, repeat that on the other side and you have your full saddle.

I am speaking as a canadian but would think the US would the same on this. Your neutrals can and can not be a current carrying. In a balanced system(2 hots and a neutral) the current on the neutral carries the unbalanced load and does not have to be counted. If your running a hot and a neutral(in this case the proper name would be the indentified conductor) it will carry the same as the hot wire so it has to be counted as current carrying conductor.
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