I need the specifications of a steel I-beam to replace a wooden timber
beam.  This room has a 23 foot 4"x6" white oak beam supported on each
end by a 2"x4" stud and headered wall with an 8"x8" wooden post at its
center. Perpendicular to the wooden beam are 4"x6" rough cut unplained
full dimension wooden beams spaced 24" on center and 23 feet long with
the ends resting on 2"x4" wood studded walls and the centers supported
by the main center beam described above.  I would like to replace the
center beam with a steel beam and eliminate the center post in order
to open up the room.  There is also an option of eliminating the
center post and positioning a new steel beam perpendicular to and
supporting the main center beam with supporting poles at each end. 
The area supported by the beam structures is a conventionally
constructed wood frame floor with 1 and 1/2 inches plywood floors, 4
foot knee walls, and a 2 x 8 wood framed attic ceiling, attic, and
roof system with 7 1/8 plywood roof sheathing and fiberglass shingles.
 What size steel beam is required for this project?

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Request for Question Clarification by redhoss-ga on 16 Apr 2006 06:41 PDT

I can do this for you, but I have a couple of questions first. I
assume that the knee walls you describe are not load bearing as
discussed in this article:

http://www.thisoldhouse.com/toh/knowhow/adding/article/0,16417,198322,00.html

"In most cases a knee wall is not a load-bearing partition. If you're
not sure if a knee wall is load-bearing, you should consult a licensed
contractor or structural engineer."

This is very important because, if the knee walls are load bearing,
the beam is supporting some of the roof load and things get much more
complicated. However, if they are not load bearing we don't have to
worry about the roof load since it would be transfered to the walls.
The knee walls themselves are not very heavy and are insignificant in
the calculations.

I don't understand what you mean by, "The area supported by the beam
structures is a conventionally
constructed wood frame floor with 1 and 1/2 inches plywood floors". Is
part of the attic floor 1 inch plywood and the rest 1/2 inch. Also, I
assume when you say "7 1/8 plywood roof sheathing " you mean seven 1/8
inch plies or 7/8 inch thick plywood. Hopefully the knee walls are not
load bearing and this won't be needed. I am sure that these questions
seem stupid to you, but you are very familiar with the construction of
your home and I am not. One other thing. Is there a limit as to how
much head room you are willing to lose. In the case where we would be
adding a steel beam in place of the post this might be important.

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Clarification of Question by waltall-ga on 16 Apr 2006 09:12 PDT

The knee walls are not load bearing.  I rechecked the floor of the
attic area above the beamed system I referred to.  I realize now that
it obviously is not a conventional floor in that it rests on the
beams.  The attic floors are much different than I described. The
original owner first laid down 3/4" plywood, then a layer of 1" white
styrofoam and then another layer of 3/8" plywood. I do want the steel
beam to be as short in height as possible to allow the maximum amount
of head room. I hope this helps.  I look forward to any and all
questions becuase I am truly an amatuer at this.

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Clarification of Question by waltall-ga on 16 Apr 2006 11:19 PDT

The structure is in the Nashville Tennessee area. I do not believe it
is in an earthquake prone area (though Memphis, TN a few hours away
certainly is). I do not mind a larger beam if I can rid myself of the
post in the middle of the room.

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Clarification of Question by waltall-ga on 16 Apr 2006 11:22 PDT

Sorry I missed the last part of the comment question.  No, I do not
know how much load the center post is taking.

Okay waltall, I think we have enough info to get started. First we
need to calculate loading:

Attic live load = 20 PSF
4 x 6 timber = 8.25 #/ft (based on 50# per cubic ft)
3/4 plywood = 2.2 PSF
3/8 plywood = 1.1 PSF
1" styrofoam = 0.25 PSF (based on 3# per cubic ft)

The center beam supports 1/2 (11.5') of the total load. The remainder
is supported by the walls. Calculating the beam loading in lbs. per
ft:

4 x 6 beams @ 24' o/c (say 11 beams) x 11.5' x 8.25 #/ft / 23 ft = 46 ppf
plywood = (2.2 + 1.1) 3.3 PSF x 11.5' = 38 ppf
styrofoam = .25 x 11.5' = 3 ppf
Live load = 230 ppf
Total = 317 ppf (NOTE: weight of steel beam not included)

The formula for maximum bending moment is:

M = wl^2/8
Where:
w = 317 ppf
l = unsupported length of beam = 23'

M = (317 x 23^2) / 8 = 20,962 ft-lbs = 251,544 in-lbs

S (section modulus) = M / s (allowable bending stress = .55 x 36 ksi)
S = 12.7 in^3

The formula for deflection is:

D = 5wl^4/384EI
Where:
E (modulus of elasticity for steel) = 30,000,000 psi
I = moment of inertia required
D = l/360 = (23 x 12) / 360 = 0.76 inches

Solving for I:

I = 5wl^4/384ED

I = [(5 x 317 x 23^4) / (384 x 30,000,000 x 0.76)] x 1728 (conversion factor) 
I = 88 in^4 

Now we are ready to choose a beam. Looking in the AISC tables, a good
beam would be (keeping in mind that we want to reduce headroom as
little as practical):

A wide flange beam 8 inches deep weighing 28 lbs. per ft. looks good:

W8x28 
Depth = 8.06 inches
Weight = 28 ppf
Flange width = 6.54 inches
S = 24.3 in^3
I = 97.8 in^4

You could save weight/money by going with a W10x19 if you don't mind
sacrificing another 2 inches in headroom.

The other option (as you suggested) would be to use an additional
steel beam to support the existing wood beam. I feel that this would
not be a good plan as it would not look very good and severely reduce
your headroom.

Look this over and see what you think. If you have any questions,
please ask for a clarification.

Redhoss

This is perfect. I have located a beam and a contractor to install it.
I appreciate your thouroughness and the formulas add a great deal of
confidence to the answer.  Thank you redhoss-ga.

Building codes may guide or even prevent your project, While a beam of
the same size  s the existing timber will be stronger  you are asking
it to take a heavier load. Do you have an idea of how much load the
center post was taking?
(Be a shame if the beam flexed enough to crack the plaster.)  Earthquake prone area?