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 
Clarification of Answer by
redhossga
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
