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Q: Sizing I or H beam to support 2"x12" lumber floor joists at a 25' span ( Answered 5 out of 5 stars,   0 Comments )
Subject: Sizing I or H beam to support 2"x12" lumber floor joists at a 25' span
Category: Science > Physics
Asked by: bad67deuce-ga
List Price: $40.00
Posted: 02 Apr 2006 08:35 PDT
Expires: 02 May 2006 08:35 PDT
Question ID: 714614
I am building an addition onto my house. The area in question has a 25'
x 32' dimension. I would like to reduce the 32' span with an
unspecified I or H beam to carry the load of the floor. The beam would
run parallel to the 25' length and be set 13' from one end of the 32'
span. There would be 2" x 12" lumber spaced 16" on center spanning the
13' and the 19' distances. 3/4"decking will be used, and 3/4" hardwood
flooring on top of that. This area will be used  a kitchen on one half
of the room, and a living room on the other half.
   What I or H beam would be adequate to support this layout? Thank you
Subject: Re: Sizing I or H beam to support 2"x12" lumber floor joists at a 25' span
Answered By: redhoss-ga on 03 Apr 2006 07:48 PDT
Rated:5 out of 5 stars
Hello bad67deuce, you did a very good job describing your problem.
Here is the info we need to do the calculations:

Live Load = 40 psf
Deflection limitation = L/360 = (25 x 12)/360 = 0.83 inch
2 x 12 weight = 4.1 # per ft
3/4 plywood = 2.13 # per sq ft
3/4 hardwood flooring = 3 3 per sq ft

Dead load calculation:
Quantity of 2 x 12 = (25 x 12)/16 (minus 1) = 18
Tributary area (area which beam supports) = 13/2 + 19/2= 16 ft x 25 ft
Weight of 2 x 12 = (18 x 4.1 x 16)/25 = 47.2 #/ft
Weight of flooring & plywood = (2.13 + 3) x 16 = 82.1 # per ft
Total dead load = 47.2 + 82.1 = 129.3 # per ft
Live load = 40 psf x 16 = 640 # per ft
Total load (w) = 129.3 + 640 = 769.3 # per ft

The beam formulas for this loading are:

M (maximum bending moment) = wl^2/8
D (deflection @ center of span) = 5wl^4/384 EI
Where E is a constant for steel + 30,000,000 psi
And I is the moment of inertia

Solving for M:

M = 769.3 x 25^2 / 8 = 60,102 ft lb = 721,224 in lb

The allowable bending stress for structural steel (s) = 0.55 x 36,000 psi
                                                  = 19,800 psi

The section modulus of the required beam (S) = M/s = 721,224/19,800 
                                             = 36.42 in^3

Now we must calculate the required I (moment of inertia):

Solving for I in the above formula for deflection we get:

I = 5wl^4/384 ED = (5 x 769.3 x 25^4 / 384 x 30,000,000 x 0.83) x 1728
NOTE: 1728 is a conversion factor to get the proper units for I

I =  272 in^4

Now we can look for a beam with these minimum properties.

A good choice would be a 16 inch deep Wide Flange beam weighing 26 #
per ft (W16x26)
S = 38.3 in^3
I = 300 in^4

Of course, you can use a heavier beam if there is one more available.

Please ask for a clarification if there is anything you don't
understand and I will try to explain.

Good luck with your project, Redhoss

Request for Answer Clarification by bad67deuce-ga on 03 Apr 2006 14:05 PDT
Looks great Redhoss. So I need 16" I beam, wow that's much bigger than
I was thinking. I'm glad I asked this question. The only thing that I
did'nt see was the beam thickness?? And also, in your opinion is this
the best material to use for this job, or is there something better
that you can suggest??
Thank you Redhoss, I will be rating this answer as a 5 star when I
receive your follow up.

Request for Answer Clarification by bad67deuce-ga on 03 Apr 2006 14:09 PDT
One more thing. If your calculations were for lets say 3/8" thick I
beam at 16" deep. Could we go with 1/2" web thickness to reduce the
16" dimension. If I could use a shallower depth beam this would
increase my headroom. Thanks again, your the man Redhoss.

Clarification of Answer by redhoss-ga on 03 Apr 2006 16:37 PDT
There are many beam that would satisfy the I and S minimum values we
calculated. I selected what would be the best choice for price. When
it comes to structural members pounds per foot determines how many
dollars you will spend. The W16x26 which I chose has these dimensions:
Depth = 15.65"
Flange width = 5.5"
Flange thickness = 0.345"
Web thickness = 0.250"

If you drop down to a 14 inch depth beam, you would need a W14x30,
going on down to a 12 inch requires a W12x36, 10 inch requires a
W10x54, and so on. You can see that after 12 inches the weight goes up
drastically. So, if you want to spend a few more dollars to gain
headroom the W12x36 might be the best bet. And yes, I do think that a
steel beam is the best material choice.
bad67deuce-ga rated this answer:5 out of 5 stars
Redhoss is a total professional. Thank you

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