I have a room that is 20' x 25', the peak runs the 25' direction. The
ht. of the  peak is 12' the walls are 8'. The board at the peak is a 2
x 10. The roof rafters are 2 x 8, spaced 16 on center. So the rise
from the top of the wall to the peak is 4' and from the wall to the
center of the peak horizontally it is 10'(half of the 20' width of the
room).  I've put a 6 x
16 Pine beam in the peak of the room and removed all of the ceiling
joists. The beam is supported at either end of the room (again a 25'
span) by 6 x 6 Pine posts ( . I wanted to quietly (unpermitted) raise my
ceiling to be a cathedral style ceiling in this room, a sort of
studio.  Someone called Building & Saftey on me. So my question is how
would the calculation for this beam span in this size of a room look?
Note: I've added steel hangers to all of the rafters both at the peak
and the wall.

I'm not a wiz with math, so laymen's descriptions along with the
equations would really help.
Please help and Thank You!!

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Request for Question Clarification by redhoss-ga on 24 Jul 2006 11:51 PDT

I think that I can probably help you. Do you know what the live
load/snow load is for your area. Also, what type of roof do you have
or do you know the dead load.

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Clarification of Question by pizlou-ga on 24 Jul 2006 12:28 PDT

Yeah, this is from the Los Angeles Co. Residential building reqs.

ALLOWABLE SPANS FOR DF #2 ROOF RAFTERS
Light Dead Load (up to 15 psf)/ Max. Roofing Load: 6 psf (Asphalt Shingles)
Live Load: 20 psf

ALLOWABLE SPANS FOR DF #2 CEILING JOISTS
Dead Load: 10 psf/ Live Load: 10 psf

But the ceiling Joists have been removed in this case.

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Clarification of Question by pizlou-ga on 24 Jul 2006 12:30 PDT

Sorry, and my roof is asphalt shigle

Hello pizlou, I believe that I have a good understanding of your
problem. First, here is a very good place to find information for wood
beams:

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

You say your beam is pine and it looks like the numbers we need would be:
Fb (allowable bending stress) = 1105 psi
E (modulus of elasticity) = 1,400,00 psi

You will also notice that the deflection should be limited to L/240 or
in your case:
D = (25 x 12) / 240 = 1.25 in

We also need to know what I (moment of inertia) and S (section
modulus) are for your 6 x 16 beam.

I = bd^3/12 = (6 x 16^3)/12 = 2048 in^4

S = bd^2/6 = (6 x 16^2)/6 = 256 in^3

Keeping this as simple as possible the load on the beam is:
(Using a live load of 20 psf and a dead load of 16 psf)

w = 10 x 36 psf = 360 lb per ft

The beam formulas for this loading are:

M (maximum bending moment) = wl^2/8
D (deflection @ center of span) = 5wl^4/384 EI
l (beam span) = 25 ft

Solving for M:

M = 360 x 25^2 / 8 = 28,125 ft lb = 337,500 in lb

The section modulus of the required beam (S) = M/Fb = 337,500/1105
                                             = 305 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 360 x 25^4 / 384 x 1,400,000 x 1.25) x 1728
NOTE: 1728 is a conversion factor to get the proper units for I

I =  1808 in^4

Here are the numbers we need to compare:

S (for your beam) = 256 
S (required)      = 305

I (for your beam) = 2048
I (required)      = 1808

Your 6 x16 pine beam is probably adequate. It lacks a little in
allowable yield strength (Fb), but what we have used might not be the
actual value for your beam. If you look at the chart, you will see
that if your beam is a better grade with Fb = 1580 psi the required S
would then be:

S = M/Fb = 337,500/1580 = 213 in^3 which looks good (213 less than 256).

The fact that your beam is more than strong enough to satisfy the
deflection criteria is most important.

If you can get a more accurate value for the Fb of your actual beam,
we can possibly make you feel a little more safe.

Please ask for a clarification if there is anything you don't understand.

Redhoss