I would like to calculate the size of two I-Beams with a span of 20
feet for a two story residential house addition. I would like to keep
the height of both beams to a minimum by using heavier beams if
possible.
The first beam will carry the second floor 40 psf live and 20 psf
dead. The second beam will be directly above and in the same plane so
the lower beam will not be carrying the load of the ceiling and roof.
The first beam could have a supported at its center so there would be
two 10 foot spans. I would add this center support if it would reduce
the height of the beam.
The second beam will carry the ceiling at 10 psf live and 5 psf dead
and the roof at 20 psf live load and 10 psf dead load. The second beam
will be the full 20? clear span.

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Request for Question Clarification by redhoss-ga on 08 Aug 2006 07:15 PDT

Your description of the loadings is very good. However, you have left
out one important thing. I need to know the size of the area supported
so that we can determine how much tributary area the beams bear. If
you can give me this, we can size the beams.

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Request for Question Clarification by redhoss-ga on 08 Aug 2006 12:37 PDT

When I asked for the size of the area supported, what I need is the
other dimension of the house. You give me the 20' dimension, I just
need the dimension in the other direction.

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Clarification of Question by skierbob-ga on 11 Aug 2006 05:42 PDT

da-- the other dimension is 28'-- it is a 20' x 28' garage with second floor.

Okay skierbob, we have enough info to proceed. In each case the beam
supports 14' of the load. The rest being supported by the walls.

FIRST BEAM 20' SPAN:

The beam formulas for this beam loadings 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 = 20 x 12 / 360 = 0.66 in
w (load per foot) = (40 + 20)psf x 14' = 840 lb per ft
l (beam span) = 20 ft
Where E is a constant for steel = 30,000,000 psi
And I is the moment of inertia

Solving for M:

M = 840 x 20^2 / 8 = 42,000 ft lb = 504,000 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 = 504,000/19,800 
                                             = 25.45 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 840 x 20^4 / 384 x 30,000,000 x 0.66) x 1728
NOTE: 1728 is a conversion factor to get the proper units for I

I =  152.73 in^4

It looks like a good choice for this beam would be:

W8x48  Depth = 8.5"  Flange width = 8.117  Weight = 48 lb per ft

FIRST BEAM WITH CENTER SUPPORT (10 FT. SPANS):

The beam formulas for this beam loadings are:

M (maximum bending moment) = 49 wl^2/512
D (deflection @ center of span) = 0.0092 wl^4/EI
NOTE: Maximum deflection is limited to D = l/360 = 10 x 12 / 360 = 0.33 in
w (load per foot) = (40 + 20)psf x 14' = 840 lb per ft
l (beam span) = 10 ft
Where E is a constant for steel = 30,000,000 psi
And I is the moment of inertia

Solving for M:

M = 49 x 840 x 10^2 / 512 = 8,039 ft lb = 96,469 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 = 96,469/19,800 
                                             = 4.87 in^3

Now we must calculate the required I (moment of inertia):

Solving for I in the above formula for deflection we get:

I = 0.0092 wl^4/ED = (0.0092  x 840 x 10^4 / 30,000,000 x 0.33) x 1728
NOTE: 1728 is a conversion factor to get the proper units for I

I =  13.49 in^4

It looks like a good choice for this beam would be:

W5x16  Depth = 5.0"  Flange width = 5.0 Weight = 16 lb per ft

SECOND BEAM:

The beam formulas for this beam loadings 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 = 20 x 12 / 360 = 0.66 in
w (load per foot) = (10 + 5 + 20 + 10)psf x 14' = 630 lb per ft
l (beam span) = 20 ft
Where E is a constant for steel = 30,000,000 psi
And I is the moment of inertia

Solving for M:

M = 630 x 20^2 / 8 = 31,500 ft lb = 378,000 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 = 378,000/19,800 
                                             = 19.09 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 630 x 20^4 / 384 x 30,000,000 x 0.66) x 1728
NOTE: 1728 is a conversion factor to get the proper units for I

I =  114.55 in^4

It looks like a good choice for this beam would be:

W8x35  Depth = 8.12"  Flange width = 8.027  Weight = 35 lb per ft

If there is any of this you don't understand or you don't like the
beam choices, please ask for a clarification and I will try and
answer.

Good luck with your project, Redhoss

Excellent – this is exactly what I was looking for and the turn around
was a lot quicker than I expected – Thanks