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Q: steel spans ( Answered,   0 Comments )
Question  
Subject: steel spans
Category: Science > Physics
Asked by: blisstout-ga
List Price: $10.00
Posted: 08 Feb 2006 08:22 PST
Expires: 10 Mar 2006 08:22 PST
Question ID: 443116
Hello!  
I am thinking of building a balcony off the back of my house. The
total dimensions will be 6' x 30' with the longest unsupported span
being 15'.  I would like to use 3"x6"x3/16" rectangular steel tube
laid out on one of the following schedules A) 24" centers(best) B) 18"
centers(better) or C) 16" centers(good).  Could you tell me what the
deflection would be?

Request for Question Clarification by redhoss-ga on 08 Feb 2006 08:34 PST
I can do this for you, but first we need to consider loadings.
Normally live load for this type of construction is considered to be
100 psf because of the potential for having many people on the
balcony. Depending on the material you plan to use for the deck
(floor) you would expect for a dead load of something in the range of
5 to 10 psf. Of course the deflection expected will vary greatly
depending on what live/dead loads you want to use. The actual
defection calcs are not difficult once we decide the loading. Can you
help me.

Request for Question Clarification by redhoss-ga on 08 Feb 2006 08:44 PST
Sorry to bother you again, but I just looked at some info that changes
what I just said in the previous request. The figure I gave you of 100
psf dead load was for balconies in public buildings. I see that values
used in residential construction are in the 50 to 60 psf range.

Clarification of Question by blisstout-ga on 08 Feb 2006 09:43 PST
Thanks for getting back to me.  The dead load will be less then
5psf(Trex brand decking) and I think that Residential Building Code
for Seattle requires that it be built to hold 60psf of live load but
I'm not sure...
Answer  
Subject: Re: steel spans
Answered By: redhoss-ga on 08 Feb 2006 12:27 PST
 
Okay blisstout, here we go. The formula comes from my "AISC Manual of
Steel Construction" for a cantilever beam:

Deflection = wl^4 / 8EI

Where:
w = (live load + dead load) x spacing x (1/144)  
NOTE: 1/144 is a factor to keep the units correct.

l = length of beam in inches = 72

E = constant for steel = 30,000,000 psi

I = moment of inertia for 3x6 tubing = 14.0 in^4 (from AISC)

-------------------------------------------------------------------------------
A. For 24 inch centers
w = 65 x 24 x (1/144) = 10.83 # per inch. 
Deflection = 10.83 x 26,873,856 / 8 x 30,000,000 x 14.0 = 0.087 inches

-------------------------------------------------------------------------------
B. For 18 inch centers
w = 65 x 18 x (1/144) = 8.125 # per inch
Deflection = 8.125 x 26,873,856 / 8 x 30,000,000 x 14.0 = 0.065 inches

-------------------------------------------------------------------------------C.
For 16 inch centers
w =   65 x 18 x (1/144) = 7.22 # per inch
Deflection = 7.22 x 26,873,856 / 8 x 30,000,000 x 14.0 = 0.058 inches

If there is something that you don't understand, please ask for a clarification.

Redhoss

Clarification of Answer by redhoss-ga on 09 Feb 2006 11:56 PST
I was looking back at this question and I am having second thoughts. I
solved the problem as if the 3x6 beams are cantilevered out from the
house. Now I notice the 15' dimension you mention. Are you using
columns. Do the beams run parallel with the house wall. If so the
numbers I gave you are wrong. I will be glad to recalculate. Just let
me know.
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