View Single Post
  #24   Report Post  
Posted to rec.crafts.metalworking
Stuart Fields Stuart Fields is offline
external usenet poster
 
Posts: 72
Default 8 inch I-beam - how much will it hold?


"Ned Simmons" wrote in message
...
On Wed, 18 Mar 2009 18:25:43 -0700, "Stuart Fields"
wrote:


"Ned Simmons" wrote in message
. ..
On Wed, 18 Mar 2009 08:57:36 -0500, Tim Wescott
wrote:

On Wed, 18 Mar 2009 11:57:21 +0000, _ wrote:

One of the projects involves both re-doing the garage and lifting a
ford
4-cyl motor out of a car. There's a fellow selling an I-beam locally,
13 feet long, 8 inches deep, 4 inches wide (no web thickness
specified).
This would span the garage nicely - could I rely on it to not bend if
I
hook up a hoist to the centre and lift that motor? Supported on the
ends only...

There are engineering tables for that sort of thing. I don't know where
the right place to look is, but that's all been reduced to "look it up
and run a couple of numbers".

The tables are in the AISC Steel Construction Manual, but first you
need to know which table or chart to use and how to apply it
appropriately. There's lots of jargon and abbreviations in the data
that the user needs to interpret and then determine whether any of it
is relevant to the case at hand.

For example, the charts and tables assume a factor of safety for
structural applications which is much lower than the practice for
overhead lifting.


As Lloyd pointed out, mount that sucker so it absolutely positively
cannot turn on it's side or fall off of it's supports.

That's a good start. g

--
Ned Simmons


Ned: Baumeister and Marks Standard Handbook for Mechanical Engineer that
I
have just about ground into unreadable by repeated usage, and I'm not a
mechanical engineer, has a table 2 on page 5-32 that shows a simply
supported beam which means the ends are only restricted vertically and not
welded into a vertical column max deflection is:


f
=
(W* (5* L^3))/(384* E*I) W= weight applied to center of
beam
in pounds

L= length in
inches

E= 29* 10^6 if
it is steel

I = the moment
of inertia of the I beam section. Note the way to calculate this is found
on page 5-38 of the same book. The drawing is a bit messy but the I
factor
does depend on the web thickness. It is a no brainer if you have the
picture
from page 5-38 and all the dimensions from the I- beam


And that's a perfect example of the hazards of putting too much faith
in handbook tables. Look beyond Table 2 and you'll see a section
"Maximum Safe Load on Steel Beams." It deals with load reductions for
beams without lateral bracing, which is quite likely what the OP has
in mind. That's also one of the factors I alluded to that's in the
AISC charts, and it's easy to miss there as well unless you know to
look for it.

BTW, the deflection formula you gave is for a uniformly distributed
load, not a concentrated load like a hoist at the middle of a span
would apply. And in any case, the deflection isn't going to tell you
whether the beam is adequate to support a given load.

Was it Rumsfeld who said it's the stuff you don't know you don't know
that's the most worrisome?

--
Ned Simmons


Oops! Ned you are right. I grabbed the wrong beam condition. BTW I
designed a sail boat trailer for a full keel sail boat using the deflections
as the limiting condition for obvious reasons. Also a spiral staircase, a
hydraulic engine hoist and a building addition, and a helicopter trailer
modification. As well as the deflection calcs for a 30' non standard I beam
with both distributed and concentrated loads. All had some safety factors
added and all have been in use for some time with no failures to perform.
With all that said, the table 4 on page 5-35 does have an easier to use
formula for safe loads. Given that the ends of the I beam are restricted
from rolling or departing from their location, I fail to see that using
deflection, and keeping it small, doesn't yield a safe condition. This
technique was used on the building addition, as well as the 30' non standard
I beam with the concurrance of a P.E.

Stu Fields