I was wondering if someone can help me analyze my need for a
structural beam in my remodel project.  I will probably hire an
engineer later I just want to see if I can get an idea about what I
will need.
First this room is the entire second floor of a 2 story house built in
the 1920?s.  The 2nd floor is currently 24?x15? split down the middle
with a wall.  The plan is to remove the wall and open up the room. 
The ceiling joists are full dimension 2?x4? spliced in the middle,
overlapped about 4? running the 24? direction.  The rafters slope up
from all four sides creating a short peak in the center.  What I would
like to do is run an engineered beam across the top and perpendicular
to the ceiling joists and use hangers to hang the 2?x4? for support. 
Or worst case a beam below the ceiling joists. I would like to know
how to calculate the load the beam must carry, and what kind of beam I
would need to carry that load.  I?m hoping I can use two 2?x8? nailed
together since I already have the lumber.

Information:
Room dimension -  24?x15?
Span of Beam ? 15?
For this, let?s assume 40 lb/sf for roof loading.  I?m still trying to
figure out the UBC but as far as I can tell that is the requirement.

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Request for Question Clarification by redhoss-ga on 26 Apr 2005 15:22 PDT

I think that I understand what you have described. The 2x4's that run
in the 24' direction support only the dead load of the ceiling
(assuming I understand correctly). The 40 PSF that you call "roof
loading" seems to me would actually be your ceiling load. You say the
rafters slope up from all four sides and I assume that they would
carry the roof load (transfering it to the outer walls). Do you plan
on inhabiting the attic area or storing items there. If not, the 40
PSF is very conservative. Let me know if you agree with what I have
said and I can help you size a beam.

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Clarification of Question by gvjeep-ga on 26 Apr 2005 16:40 PDT

I?m really just looking for a little help understanding the problem. 
I thought that seeing someone else doing the problem would be a good
way to understand.  I?m not going to use the result to get a permit. 
I really don?t even need an answer just a method. Like I said in the
original question I?ll probably hire an engineer once I decide exactly
what I want to do.

To clarify the original question, the crawl space is only about 4 feet
tall at the highest point and won?t be used for storage.  The house is
in Sacramento, CA so I doubt that there is a snow load requirement. 
The rafters do slope up from all four sides.

Hello gvjeep, I have worked for years in the metal building industry
and am very familiar with this type of problem. What you describe is a
simply supporter beam with a uniformly distributed load. The length is
15 feet and it supports 12 foot of the 40 PSF load. The formula for
the maximum bending moment, which occurs at the center of the span, is
given by:

M = (w x L^2)/8   where w = 40 PSF x 12 FT. = 480 LB./FT.

M = (480 x 225)/8 = 13,500 FT.-LB. = 162,000 IN.-LB.

You don't say what type of wood you plan to use, so we will assume
"Western white pine" which has an allowable bending stress of 800 PSI.

We will try two 2x8's as you suggest. The section modulus for a single
2x8 is 13.14 IN.^3 or doubled we have 26.28 IN.^3.

Bending stress = Max. moment / section modulus

               = 162,000 / 26.28 = 6,164 PSI

Comparing 6,164 to our allowable bending stress of 800 PSI it is
obvious that we will have to use a larger beam. Solving the above
formula for section modulus we get:

Section modulus = Max. moment / Allowable bending stress
                = 162,000 / 800 = 202.5 IN.^3

The formula for section modulus is:

S = (b x d^2)/6

Assuming a 6 inch wide beam and solving for d we get:

d = sqrt(s) = 14.23 IN. 

So you can see that we are talking about a fairly large beam. There
are many other factors to be considered when dealing with a wood beam.
However, this will get you in the ballpark. One thing you might want
to consider is using a glulam beam. Also, since the beam will be
hidden (if you hang the 2x4's underneath) you could use a steel beam.
I hope that this hasn't confused you too much and you are able to
follow each step. If you have further questions, please ask for a
clarification and I will try and answer.

Good luck, Redhoss

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Request for Answer Clarification by gvjeep-ga on 27 Apr 2005 07:32 PDT

I?m able to follow your calculation for the most part.  One question I
have is about you first moment calculation.  How do you decide how
much of the 40 lb/sf load the beam caries?  You say w = 40 psf X 12ft
= 480 ft/lb.  Where do you come up with the 12 ft?  The way I was
looking at it was:

40 lb/sf  x (15 ft x 24ft) = 14,400 lb total load from the roof.  Then
trying to figure out how much of that load each wall carries.  Maybe
I?m missing something.

Thanks for the answer it really helped out.

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Clarification of Answer by redhoss-ga on 27 Apr 2005 11:31 PDT

First, I noticed a typo in the first paragraph of my answer
"supporter" should read "supported". The two end walls each carry half
the distance from the end wall to the center or 6 foot. This leaves
the center support (wall or beam) to carry the remaining 12 foot (2 x
6 FT.). If you draw yourself a picture of what is happening, I think
you can see that the center support does carry 12 foot of the load. In
my question to you about whether your attic space was to be used for
storage or living area I was questioning the 40 PSF loading. If you
are not using it for storage, as you say, the 40 PSF sounds excessive
to me. I found a website with some info that gives some numbers for
different UBC live load requirements:

http://www.awc.org/technical/spantables/tutorial.htm

Looking at this it seems that a more reasonable number for live load
might be 10 PSF (no storage) plus a dead load of 10 PSF for a total of
20 PSF. Of course when you consult with an engineer in your area I am
sure that they will be able to tell you better than me what is
normally done locally.

Here is a website with a good explanation and drawing of what I tried
to explain on how I determined the 12 foot number (see bottom of the
page):

http://64.78.42.182/sweethaven/BldgConst/Carpentry01/lessonMain.asp?iNum=030106

Let me know if you need more explanation or information and I will try to help.

Thanks!! just what I was looking for

Most engineers will be reluctant to answer this question for a few reasons.

First of all, the Professional Engineers act of many states makes it
unlawful to give out such advice unless you are a PE.  Next, there are
quite a few liabilities associated with this matter.

I suggest you consult a local registered professional civil engineer
for information on this project.  The scope seems simple enough, but
there are way too many other factors that need to be taken into
account, such as local conditions, live loads, and so on that change
from region to region.

For such project, a professional engineer probably wouldn't charge
much more than a couple hundred dollars, which isn't that much more
than the $40 you are offering here.  You'd be getting peace of mind
knowing the work were done correctly.

If you are insistent on doing the calculations yourself, then you
should go to the local library and reference a copy of the current UBC
along with a textbook on Timber Design, along with any applicable
local building codes.

Just so you know, the 40 psf number you gave is a live service load
number only.  You also need to add a number of other loads, such as
the roof dead load, snow load, seismic load, among others

I?m able to follow your calculation for the most part.  One question I
have is about you first moment calculation.  How do you decide how
much of the 40 lb/sf load the beam caries?  You say w = 40 psf X 12ft
= 480 ft/lb.  Where do you come up with the 12 ft?  The way I was
looking at it was:

40 lb/sf  x (15 ft x 24ft) = 14,400 lb total load from the roof.  Then
trying to figure out how much of that load each wall carries.  Maybe
I?m missing something.

Thanks for the answer it really helped out.