Now then, for floors, deflection s/b < 1/360 of the span  of 144 in

in this

case, .4"  The load @ 60lbs/lineal ft = 720 lbs, soooo

if deflection = (5(weight, evenly dist'd)(span in inches^3))/(((384
E)((beam width )(beam height^3))/12))

then I should get .21", which is well within acceptable limits.

Did I screw any of this up? I gotta try this w/5" plate, maybe that

will

work.  Did I mention there isn't much going on at work today??

"Kevin R. Walsh" wrote:

This is the gotcha I warned you about.  Ceiling beams are not evenly
distributed loads. For a house with a peaked roof, you should find most
ceiling beams in tension, actually.   Floor beams are considered evenly
distributed because they are live loads.  I suppose even a roof beam
could be evenly distributed when calculating for something like snow
loading.  In all likelyhood, it will be fine the way you have it.  If
you want the real answer, get yourself a copy of a first course in
mechanics.  I use Crandall/Dahl/Lardner because that's what my
professor had us buy back when.

As a quick catchup-
Cantilever beam, point load at the end
d=FL^3/3EI

Cantilever beam, uniform load
d=WL^4/8EI  W=load per unit length

Simply supported beam, point "a" from one end, "b" from the other
d=Fb(L^2-b^2)^(3/2)/9LEI*sqrt(3)

Simply supported beam, evenly distributed load
d=5WL^4/384EI

Have some fun engineering at Holiday time!

=====
Kevin Russert Walsh
Process & Equipment Development Engineer

Even if this has nothing to do with rotaries it is an
education in structural mechanics never-the-less :-)

Thanks Kevin.

Paul Lamar
 
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