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Q: I-Beam size calaculation for residential construction ( Answered 5 out of 5 stars,   0 Comments )
Subject: I-Beam size calaculation for residential construction
Category: Science > Physics
Asked by: bosarge66-ga
List Price: $50.00
Posted: 05 Sep 2006 05:36 PDT
Expires: 05 Oct 2006 05:36 PDT
Question ID: 762357
I would like to calculate the size of I-Beams needed to carry the load
of a two story house. Outside Dimensions of house: 33'2" X 34'0".
We'll be using (4) I-beams equally spaced running parallel to the
33'2" dimension. The longest span of any beam will be 9'9.3125".
Exterior walls will be 2X6 studs on 16" centers. First and second
floor joists will be 2X12 on 16" centers. 3/4 plywood and hardwood
floors through out.

Request for Question Clarification by redhoss-ga on 05 Sep 2006 06:07 PDT
Is the roof load transfered to the exterior walls or supported by the
interior walls. Since you don't mention anything about the roof in
your question, I would guess that it would be excluded from the beam

Clarification of Question by bosarge66-ga on 05 Sep 2006 10:32 PDT
Sorry....Yes the load would be transferred to the exterior walls.
Pitch of roof (main roof) is 12 12. Roof will be constructed of 3/4
decking with 26ga. 5V-Crimp galvalum.

Subject: Re: I-Beam size calaculation for residential construction
Answered By: redhoss-ga on 05 Sep 2006 14:52 PDT
Rated:5 out of 5 stars
Okay bosarge66, not having to deal with the roof loads is good. We can
use 10 psf dead load and 40 psf live load for the two floors. So, we
have a total of 100 psf on the beams.

The beam formulas for this loading are:

M (maximum bending moment) = wl^2/8
D (deflection @ center of span) = 5wl^4/384 EI
w = (34'/5 spaces) x 100 psf = 6.8' x 100 psf = 680 lb per ft
l = 9.75'
D = l/360 = (9.75 x 12) / 360 = .325"
Where E is a constant for steel = 30,000,000 psi
And I is the moment of inertia

Solving for M:

M = 680 x 9.75^2 / 8 = 8,080 ft lb = 96,960 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 = 96,960/19,800 
                                             = 4.89 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 680 x 9.75^4 / 384 x 30,000,000 x 0.325) x 1728
NOTE: 1728 is a conversion factor to get the proper units for I

I =  14.2 in^4

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

A good choice would be a 6 inch deep Wide Flange beam weighing 8.5 # 
per ft (W6x8.5)
S = 5.08 in^3
I = 14.8 in^4

If you have some other beam you would like to use, or there is
anything that you don't understand, please ask for a clarification and
I will try and help.

Good luck with your project, Redhoss
bosarge66-ga rated this answer:5 out of 5 stars
Thanks for the prompt answers.......Very Helpful!

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