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Q: Beam size to replace a bearing wall in first level down of a split level home ( Answered,   0 Comments )
Subject: Beam size to replace a bearing wall in first level down of a split level home
Category: Science
Asked by: kjelica-ga
List Price: $50.00
Posted: 07 Aug 2006 09:07 PDT
Expires: 06 Sep 2006 09:07 PDT
Question ID: 753426
I am replacing a bearing wall that supports the second story of my
home.  The roof is supported by W trussed rafters mounted on the
outside walls. The beam will need to support a 24 ft. wide upper floor
for a 16 ft span with no center support.  The supported floor is made
from 2X10 joists with 1 and 1/2 inches of 3/4 in' plywood flooring. 
There is parguet flooring over half of the area.   There are two main
2X4 and sheetrock walls partitioning the second floor, neither is
bearing.  There is foundation at both ends of the proposed span.

What is the calculation? What size beam can I use?  I would like to
use a standard flange if possible, a conservative answer is fine by
Subject: Re: Beam size to replace a bearing wall in first level down of a split level home
Answered By: redhoss-ga on 08 Aug 2006 07:01 PDT
Hello kjelica, I believe we can work out a solution for you. Floor
live load is normally taken to be 40 psf. For the type construction
you have I think a 10 psf dead load is probably a good number. The
beam actually supports a 12' wide portion of the floor and the walls
support the rest.

The beam formulas for your loadings are:

M (maximum bending moment) = wl^2/8
D (deflection @ center of span) = 5wl^4/384 EI
NOTE: Maximum deflection is limited to D = l/360 = 16 x 12 / 360 = 0.53 in
w (load per foot) = (40 + 10)psf x 12' = 600 lb per ft
l (beam span) = 16 ft
Where E is a constant for steel = 30,000,000 psi
And I is the moment of inertia

Solving for M:

M = 600 x 16^2 / 8 = 19,200 ft lb = 230,400 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 = 230,400/19,800 
                                             = 11.64 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 600 x 16^4 / 384 x 30,000,000 x 0.53) x 1728
NOTE: 1728 is a conversion factor to get the proper units for I

I =  55.6 in^4

From my AISC manual it looks like good choices might be one of these
wide flange beams:

W8x17 Depth = 8" Flange width = 5.25" Weight = 17 lb/ft

W10x15 Depth = 10" Flange width = 4.00" Weight = 15 lb/ft

W12x14 Depth = 11.91" Flange width = 3.968" Weight = 14 lb/ft

I just re-read your question and noticed you said "standard beam".
Here is an "S beam":

S8x18.4 Depth = 8" Flange width = 4.001" Weight = 18.4 lb/ft

If there is some other type beam you would rather use, or you have a
beam you would like me to check out just let me know. Please ask for a
clarification if there is something that you don't understand.

Good luck with your project, Redhoss
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