I currently have a project to frame in and finish the inside of my
24'x30' "pole barn" construction shop. I need some help with sizing a
rectangular steel tube to span the 23' 1/4" (inside dimension) width &
help support an insulated sheetrocked ceiling. The beam in question
would fit in the 3" void between two wooden trusses notched into
either side of the 6 in square support poles (spaced on 10' centers
along the length of the building), and would have to be tied into the
poles & through bolted to the existing trusses. The wooden trusses
have 2x6 top & horizontal members with a 2x4 "W" bracing. 
Perpendicular to the trusses would be 2x6 headers space 24" on center
and 5/8 sheetrocked. There would be no storage above this ceiling, so
a 10# live load  & a 5# dead load would only be required. With a 19'
2" length, 23'1/4" width, and the beam centered in the width at 8' 7",
I calculated the total load of the beam to be just over 3300". If
possible include information on bolt size, grade, and spacing to
attach to the trusses & poles.

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Request for Question Clarification by redhoss-ga on 24 Feb 2006 07:20 PST

I have read your question several times and drawn a sketch of what I
think your pole barn looks like. I guess that you have 8 wooden
trusses and that the rectangular steel tube would be 3 inches x some
height (to be determined) and would be a straight section with
connection plates welded on both ends for through bolting to the
columns. There are 4 columns down each 30' side so there would be 4 of
these wood truss/steel tube beams. If this is correct, then I would
say that the 2 inner beams would support 10 x 24 x 15 = 3600#. The 2
end beams would support 5 x 24 x 15 = 1800#. The end beams could be
different (less depth) than the center beams if you wish. Does this
accurately describe your situation. If so, I can have you an answer
back soon after your reply.

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Clarification of Question by akmudman-ga on 24 Feb 2006 12:04 PST

This is definitely hard to describe having no drawing, so I can see
your difficulty. There are two trusses spanning across on each of the
two inside poles and a single truss on each end wall. Each of the end
walls are framed with the proposed headers resting on these. There is
also a framed room 10'x11.5'in the back corner that has an end wall
that is supporting half the distance of one of the two sets of the
double trusses. Across the other half the distance there are 2-2x10's
sandwiched inside the trusses (nailed & 1/2" grade 5 bolts every 2')
and supported by a steel bracket on the outside end. The 2-2x10's by a
beam chart would support 2850# across a 12' span. If this is adequate,
I would leave these in place. The steel beam in question would fit
inside the the other set of double trusses with 2x6's perpendicular,
and attached with hangers to the reinforce or supported double trusses
on one end and the framed outside wall on the opposite side.  I hope
this clarifies this for you.

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Clarification of Question by akmudman-ga on 24 Feb 2006 13:38 PST

After reading my reply there appears to be some of your questions that
were not answered. Yes there would be attaching plates to welded on to
both through the columns for any 3" wide steel beams used. Available
height for the beams is 14". The reinforced double truss not supported
by the room wall would have to support roughly 1800# or 1/2 the area
of 12'x 20' and the side supported by the room is 1/2 of 10'x12'.
There are 4 poles along the lenth of the side as you assumed with
double trusses on each of the two interior poles (square 6"x6"). There
is a 4" concrete rebar reinforced floor throughout the building

Okay akmudman, I understand exactly what you are building. Here are
the calcs for your beam:

First we will calculate the maximum bending moment and required section modulus:

M = wl^2/8
Where w = 15 PSF x 10 ft = 150 #/ft
M = 150 x 24^2 / 8 = 10,888 ft# = 129,600 in#

S (section modulus) = M /s
Where s is the allowable bending stress = 36,000 psi x .55 = 19,800 psi
S = 129,600 / 19,800 = 6.54 in^3

Next we will calculate the moment of inertia required to limit the
deflection to (beam length)/360 = 24 x 12 / 360 = 0.8 inch

I (req'd moment of inertia) = (5wl^4/384 ED) x 1728
Where E is a constant for steel = 30,000,000 psi
D is deflection
1728 is a constant to keep the units in inches

I = (5 x 150 x 331,776 / 384 x 30,000,000 x 0.8) x 1728 = 47 in^4

Here is a chart of rectangular tubing properties:
http://www.chicagotube.com/stockbook/structure/elements_rect.html

Looking for a tube that has the I and S we need and also has a width
of 3 inches we find:

9x3x1/4  S= 11.4  I= 51.1
8x3x3/8  S= 12.7  I= 51

Either of these would work. The 9x3x1/4 would be the better choice
since it weighs less per foot.

Now for the through bolts:

The shear load per end is 3,600# / 2 = 1,800#

A 1/2 inch bolt has a shear area of 0.196 sq in

The allowable shear for an ASTM A307 bolt is 10 ksi.
The allowable shear load for one bolt is 10 x .196 = 1,960#

So, one bolt would actually be enough. However, two would make for a
better connection and we need one top and bottom to prevent the beam
from twisting. The plate itself could be made from 1/4 in flat bar or
plate.

I think that this answers your question, but if you need any further
explanation or info please ask for a clarification.

Good luck with your building, Redhoss

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Request for Answer Clarification by akmudman-ga on 25 Feb 2006 13:00 PST

This looks to be just the information that I was looking for, but
there are a couple of details that need some clarification. Are the
ASTM A307 bolts a grade that can be requested at the local fastener
store or does this cross over to something in the range of a grade 5
or 8 bolt? Also, are the stress & bending calculation inclusive of the
weight of the beam itself? Thank you for getting back so quickly and
appreciate your help in this matter.

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Clarification of Answer by redhoss-ga on 26 Feb 2006 07:38 PST

You caught me in a little bit of laziness, which I will admit that I
am prone to. No, I didn't include the weight of the beam in the
calculations. Normally when a designer calculates roof members they
would use a nominal dead load that includes an allowance for the
members themselves. In your case I should have probably added more
because your 5 PSF dead load only includes the actual ceiling and
framing materials. However, the weight of the beam itself is somewhat
insignificant if you consider that the 150 #/ft load that I used would
become about 170 #/ft if we included the beam. The percentage
difference being 150/170 = 88%. I hate to destroy your faith in the
entire engineering profession, but most of the standard formulas used
are not 100% accurate. That is the reason that the term "safety
factor" exists. The safety factor in these calculations is that we
only used 55% of the published yield strength for the beam material
(.55 x 36,000 psi = 19,800 psi). To be more exact you can use what is
called an iterative solution. Since we had no idea what beam we would
be using when we started we could have assumed a value, solved the
equations, and then plugged the beam weight we calculated back into
the equations again until the answer came out more exact. I hope all
this hasn't bored you too much, but I just wanted to explain why I
wasn't too worried about the actual beam weight. However, I will solve
the equations using 170 #/ft just to see how much difference it makes.

First we will calculate the maximum bending moment and required section modulus:

M = wl^2/8
Where w = 15 PSF x 10 ft = 170 #/ft
M = 170 x 24^2 / 8 = 12,240 ft# = 146,880 in#

S (section modulus) = M /s
Where s is the allowable bending stress = 36,000 psi x .55 = 19,800 psi
S = 146,880 / 19,800 = 7.41 in^3
NOTE: This would make no difference since both possible beam choices
have an S value greater than 7.41

Next we will calculate the moment of inertia required to limit the
deflection to (beam length)/360 = 24 x 12 / 360 = 0.8 inch

I (req'd moment of inertia) = (5wl^4/384 ED) x 1728
Where E is a constant for steel = 30,000,000 psi
D is deflection
1728 is a constant to keep the units in inches

I = (5 x 170 x 331,776 / 384 x 30,000,000 x 0.8) x 1728 = 53 in^4
NOTE: We would still choose the same two beams.

So, now I feel better and hope that you do also. Now on to the bolt
question. ASTM A307 is equivalent to Grade 2. Very likely the bolts
you buy at the hardware store will be either Grade 2 or 5.

Thanks for keeping me honest, Redhoss

This was just the information that I needed. Redhoss answered in a
timely manner and clarified details of the answer. I decided to use
this forum primarily after viewing some of his answers.