In a previous post Brian Whatcott says...
Now here's the kicker:
Say we want to replace a wood beam with a steel beam, for strength,
for reduced deflection, or most likely,for reduced depth wasted.

In this situation we DON'T want the low ball safety values, we want
the HIGH ball values for the wood, in order to guarantee the same or
better performance when we match strength or stiffness in steel.

The reason is because the previous (wood) beam may have been a lucky,
strong example at the upper end of the distribution: it is vital not
to degrade values of that rare HIGH end sample.

That has been the other issue: everyone has been so comfortable
with their timber or steel beam program, it was easy to forget that
the brain has to stay engaged, for the safety of the public, whom one
is mandated to keep safe.

This has turned out to be a remarkably civilized thread, in the end.
I salute you all!


Brian:

Here's how I approach this situation if I am unable to determine the
anticipated service loads, but want to have the "new" section at least
as strong and as stiff as the existing section:

1) computer section properties for existing section
2) Using NDS (assumes 3-2x12) and allowable design stresses for the
material, compute an allowable bending moment, then use that to compute
an allowable uniform load.
3) Set a defection limit, typically L/360 for live load, but can be
conservatively used for DL+LL. Using section properties, deflection
limit and "E" for the original material compute an allowable uniform
load.
4) Design of new beam is now governed by the both loads.
5) Since the allowable design stress for Douglas Fir is about 1/20 that
of steel and the "E" for Douglas fir is conservatively about 1/20 that
for steel, you can reasonably design the new steel beam for the lower of
the two loads you computed above.

Now for a really interesting problem, let's say you want to take out a
post and increase the span of the 3-2x12 from 13 feet to 18 feet, but
you must leave the 3-2x12 in place. "Flitch" beams are allowed. Do you
compute combined section properties of wood and steel or do you simply
let the steel carry all the load and the wood becomes "filler"? This
problem has a trick in it that I will not divulge.


--
Bob Morrison, PE, SE
R L Morrison Engineering Co
Structural & Civil Engineering
Poulsbo WA