I need the specifications for 2 (or more) steel beams to replace 2
wooden beams barely holding up my garage roof.  The first beam is 211?
long and runs E-W.  It is supported on either end by 4? x 4? posts. 
The only load on this beam presently is the second beam that needs to
be replaced.  This second beam is directly in the center of the span
of the first beam.  At this time this second beam is (inadequately)
2ea 2? x 4? 268? long nailed together running the length of the
building.  The other end of this beam is supported on the south block
wall.  This second beam supports in the center 11ea 2? x 4? 226? long
spaced 24? running perpendicular to the beam.  The ends of these 11ea
2? x 4? are supported on the E and W block walls.  The roof sheathing
is nailed to these 11ea 2? x 4?.  This is a flat roof with a tar paper
upper surface.  If necessary, the 226? long beam could be replaced by
more than one beam.  The critical factor is headroom.  Both the 211?
and the 226? beams need to take up as little vertical space as
possible.

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Request for Question Clarification by redhoss-ga on 23 May 2006 18:42 PDT

I assume that since this is a flat roof that you live in an area that
gets no snow. Is this true. Do you know what your local live load
requirement might be.

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Clarification of Question by zumba-ga on 23 May 2006 20:11 PDT

We live in the southwestern desert (Phoenix).  It snows about every
5-10 years and it stays for an hour or two.  I am sorry but I don't
know the live loads nor where to find them.

Hello zumba, I think that a good live load number for your area would
be 12 PSF. From what you describe a safe dead load would probably be
15 PSF. With a properly designed center beam you will no longer need a
beam in the 211" direction. I would suspect that it was added when the
main beam began to sag at some point in the garage's 60 year life.

The beam formulas for this loading 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 = 268 / 360 = 0.74 in
w (load per foot) = 27 PSF x (113/12 ft) = 254.25 lb per ft
l (beam span) = 268" = 22.33 ft
Where E is a constant for steel = 30,000,000 psi
And I is the moment of inertia

Solving for M:

M = 254.25 x 22.33^2 / 8 = 15,847 ft lb = 190,164 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 = 190,164/19,800 
                                             = 9.6 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 254.25 x 22.33^4 / 384 x 30,000,000 x 0.74) x 1728
NOTE: 1728 is a conversion factor to get the proper units for I

I =  64 in^4

Now we can look for a beam with these minimum properties and a depth
as small as practical.

A good choice would be a 8.14 inch deep Wide Flange beam weighing 20 #
per ft (W8x20)
S = 17.0 in^3
I = 69.4 in^4

Another choice would be a 10.0 inch deep Wide Flange beam weighing 15 #
per ft (W10x15)
S = 13.8 in^3
I = 68.9 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. If you have another beam in
mind, let me know and I will check it out for you.

Good luck with your garage project, Redhoss

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Request for Answer Clarification by zumba-ga on 25 May 2006 09:18 PDT

Thanks for your efforts.  I wish that it was possible to include
sketches.    I think I confused you.  Here is the picture ? the east,
west and south walls are block.  There is no north wall, it is an
opening.  The 211? beam spans this opening, resting on 4? x  4? posts
(unless you want me to use bigger posts, which I will be happy to do).
 The 211? beam is essential since it and the south wall are the only
supports for the 226? long beam that runs N ? S.   The southern end of
the 226? beam rests on top of the south wall and northern end of the
226? beam rests on top of the 211? beam.   The 226? beam probably
would not be necessary if the builder had used something stronger than
the 11ea 2? x 4? (running E ? W, parallel to the 211? beam) to hold up
the roof.  For example, if he had used 16ea 2? x  8? (16? spacing),
the center 226? beam probably would not be necessary.  (It might be
easier just to replace the 11ea 2? x  4? but they are sitting in
openings chiseled into the E ? W walls.  There is no way to replace
them without removing all of the block above the chiseled openings,
unless you know of telescoping steel beams??)   I hope that the
picture is clearer.  Let me explain the headroom issue.  The underside
of the sheathing is at about 85.5? above the slab.  The underside of
the 11ea 2? x  4? are at 82?.   You specified 1ea beam 8? high.  This
means that right down the middle of the garage there will be an
obstruction at 76? height.  This isn?t so good.  It is even worse at
the entrance to the garage (the north) because the new 8? beam rests
on top of the 211? beam.  Let?s say that the 211? beam is 6? high.  
That means that the entrance will be 70?.  Many people going in and
out will bang their heads if they forget to duck and steel should be
particularly painful.  If the 8? 226? beam could instead be 5ea 3?
beams (evenly spaced) and the 211? beam could be a 4? beam that would
make the entrance 77?.  Does any of this make sense?

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Clarification of Answer by redhoss-ga on 25 May 2006 10:30 PDT

Yes, I was confused. Now I understand completely. We will find five
beams to span the 268" length, and one beam to support them at the
open end of the garage.

First the 268" beams:

Solving for M:

M = 84.75 x 22.33^2 / 8 = 5,282 ft lb = 63,384 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 = 63,384/19,800 
                                             = 3.2 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 84.75 x 22.33^4 / 384 x 30,000,000 x 0.74) x 1728
NOTE: 1728 is a conversion factor to get the proper units for I

I =  21.4 in^4

Now for the 211" beam:

Solving for M:

M = 301.5 x 17.583^2 / 8 = 11,652 ft lb = 139,824 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 = 139,824/19,800 
                                             = 7.06 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 301.5 x 17.583^4 / 384 x 30,000,000 x 0.586) x 1728
NOTE: 1728 is a conversion factor to get the proper units for I

I =  36.88 in^4

Here is what I would suggest. Use a W6x20 (6.2" depth x 20# per foot)
for the 211" beam. Do not rest the 268" beams on top of the W6x20, but
clip them to the side. Use W6x12 (6.0" depth x 12# per foot) for the
268" beams. This would give you 82" - 6.2" = 75.8" clear. How does
this sound.

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Request for Answer Clarification by zumba-ga on 25 May 2006 11:54 PDT

Thank you for your help.  Is W6x20 and W6x12 sufficient information to
order the beams?  Are these "I" beams or those hollow tube beams? 
Will the steel company sell the "clips" to clip the long beams to the
short beam or is this a welding job?

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Request for Answer Clarification by zumba-ga on 25 May 2006 13:50 PDT

I found the answers to my questions (except the "clips") on the
Internet.   Would a W4x13 work in place of the 5ea W6x12.  The block
walls are made of 4" high blocks and 4" will fit easier than 6".

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Clarification of Answer by redhoss-ga on 25 May 2006 14:51 PDT

Thanks for the tip. If there is anything else I can help you with, just ask.

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Request for Answer Clarification by zumba-ga on 25 May 2006 16:01 PDT

Hi redhoss - It looks like your calculations are for 3 beams but your
third sentence says 5 beams.  Is it 3 or 5?  Also, what does "clip"
mean?

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Clarification of Answer by redhoss-ga on 26 May 2006 07:53 PDT

The calcs are for 5 beams. The number of beams determines the
tributary area supported by each beam and thus the load per foot. Here
is how I got the 84.75 lb per foot number I used:

226/6 = 37.67" (5 beams = 6 spaces)
(37.67/12) x 27 PSF = 84.75 lb per ft

By clip I mean a vertical plate welded to the 211" beam so that the 5
beams can be attached without having to set the beams on top of the
211" beam. This is a standard method of attaching one beam to another.
It is similar to hanger brackets used in wood construction:

http://www.strongtie.com/products/connectors/HUCQ-HUS-HHUS-HGUS_SCL.html#gallery

Here is a good detail showing how structural beams are prepared:
http://www.nu-waycorp.com/Images/BEAM-SHEET.png

This should show you what you need to do.

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Request for Answer Clarification by zumba-ga on 26 May 2006 10:54 PDT

Thanks very much!!

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Clarification of Answer by redhoss-ga on 26 May 2006 13:34 PDT

You are most welcome. Your garage should last for another 100 years.

Great help.  Thanks.

A steel square tube or I-beam of the same dimensions would likely be
stronger than the existing set up.

Flat roof can be built up over the years to contain weight never designed for.

Termites? or carpenter ants?

Set up sounds a little flakey.  

Can you rent jack posts to hold up the roof while the beams are swithched.

Do building codes apply to an alreddy constructed garage?

Surplus scrap yard might have wood or steel beams.

Regarding Redfoxjumps comments - I don't really have dimensions for
the beam #1 since it wasn't really a beam but a strange 2x6 and 2x4
nailed up mess.  Beam #2 is not holding up well and I hesitate to use
the same dimensions without some logic behind the decision.
The roof only has the standard 3 (?) layers of felt.  The roof has
been scraped clean every time the material has been replaced.
If I remember correctly, southern Arizona has the 2nd (?) highest
concentration of termites per acre in the USA.  It is a constant
battle.
Yes, I can support the roof while the beams are replaced.
This structure was built in 1945 prior to city annexation and was
built under the more relaxed county building codes (if there were any
building codes at all).  It has withstood man and nature for 60 years
but is in need of a little TLC.