Hello cpopetz, you seem to have done a very good job of describing
your problem. I think we can get some answers.
(a) the load that beam is bearing with
one floor
NOTE: What you have here is a continuous beam with two equal spans.
However, to make things a little easier we can consider it as a simple
beam. The answer will be a little conservative, but that is probably
good and these formulas are not 100% accurate anyway.
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 = 12 x 12 / 360 = 0.40 in
w (load per foot) = (30 + 20 + 10)psf x 11' = 660 lb per ft
l (beam span) = 12 ft
Where E is a constant for steel = 30,000,000 psi
And I is the moment of inertia
Solving for M:
M = 660 x 12^2 / 8 = 11,880 ft lb = 142,560 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 = 142,560/19,800
= 7.2 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 660 x 12^4 / 384 x 30,000,000 x 0.40) x 1728
NOTE: 1728 is a conversion factor to get the proper units for I
I = 25.66 in^4
This tells us what we already know, that the existing beam was more than adequate.
(b) what it bears with the second floor and extra ceiling
The only difference in these calcs is that we now add in another 30 +
20 psf. So, w now becomes: 110 x 11' = 1210 lb per ft
Solving for M:
M = 1210 x 12^2 / 8 = 21,780 ft lb = 261,360 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 = 261,360/19,800
= 13.2 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 1210 x 12^4 / 384 x 30,000,000 x 0.40) x 1728
NOTE: 1728 is a conversion factor to get the proper units for I
I = 47.04 in^4
(c) whether the beam is strong enough to handle that
I checked your numbers for the existing S 7x20 and you are exactly
right. So, we are comparing our calculated required value of 47.04
in^4 to 42.4 in^4. I would say that your existing beam is adequate. I
have a beam program that will handle the actual end conditions that
you have. I will check out the beam using that program and let you
know what results I get. If there is any of what I have posted that
you don't understand, please ask for a clarification.
Back with you soon, Redhoss 
Request for Answer Clarification by
cpopetzga
on
18 Jul 2006 12:48 PDT
Thanks so much! That all makes sense to me.
The inspector is also concerned about the column itself. (He hasn't
seen it yet, he's going by my descriptions.)
There are actually two columns, although they are placed so close
together that I have assumed one was used to install the other, or
that one was installed after the other for some reason (though since
the steel beam isn't likely to sag, I can't imagine the reason.)
Anway, the column at 12' looks like a cast concrete pier, 6" diameter
octagonal, with two peices of rebar running vertical in it. It's on a
24" square pad, but I don't know how deep. The column 1 foot away
from it is a telescoping steel column, on another 24" square pad.
Both columns are in good shape, on rust on the steel, no
cracking/crumbling on the concrete.
Any thoughts on this? I realize this is a lot more vague than the
previous question.
