Todd,
You're right, the proper unit of measure should have been "in-kips" when
using 17.921. I wasn't
keeping the units straight, and that can give very odd results.
I saw the equation a little differently than the one you quoted, but
only because the structural
who taught me all this stuff used to punch me out if I used letters for
variables that didn't agree
with those in the AISC book. In that book, P represented a point load,
at one point in a span,
while W (upper case) indicated the total load on a beam, and w
(lowercase) represented a
uniformly distributed load per unit of length. Took me quite a while to
understand that W did
not equal w where a structural engineer was concerned.
So in reality, I suppose my formula of (P*L / 4) / Fb = S, is the same
as the one you're using,
and we get the same result. Good exercise.
You're also correct in that the calculation is different if the load
isn't centered. According to the
AISC manual I'm using that would be the equivalent of a simple beam,
with a concentrated load
(P) at any point along the span of the beam. The formula for that,
according to beam diagram #8,
would be P*a*b / L = Mmax at the point of the load. After that the
division by the Fb maximum
of 15,000 ksi would give the required section modulus. The values for a
and b equate to the
distance from the end of beam to the load on one side of the beam
(a=left) and from the load to
the opposite end (b=right).
So, lets see what the deal is. If the same load (426.7#) were located
6' from the left end (a=72"),
and thus 8' from the right end (b=96"), I believe the result would be:
P*a*b / L = 426.7*72*96 / 168 = 2,949,350.4 / 168 = 17,555.657 in-kip =
Mmax.
Then we'd divide 17,555.657 (Mmax) by 15,000 (Fb) = 1.1703 = required
section modulus (S).
This more or less backs up the comment in NFPA-13 that results in a
lesser equivalent length
of the trapeze if the load isn't centered, and the result is for a
lowered section modulus requirement,
even if it's only slightly for the first few inches or so.
Did that work out with what your understanding of the process would be?
PARSLEY CONSULTING
Ken Wagoner, SET
760.745.6181 voice
760.745.0537 fax
[EMAIL PROTECTED] <mailto:[EMAIL PROTECTED]> e-mail
www.ParsleyConsulting.com <http://www.ParsleyConsulting.com> website
IMPORTANT NOTICE: This correspondence is not a Formal
Interpretation issued pursuant to NFPA Regulations. Any opinion
expressed is the personal opinion of the author and does not necessarily
represent the official position of the NFPA or its Technical Committees.
In addition, this correspondence is neither intended, nor should it be
relied upon, to provide professional consultation or services. My lawyer
talks to Todd's and likes this disclaimer a lot.
Todd Williams - FPDC wrote:
Ken,
Something drove me a little crazy because you units weren't working
out. The 17,921.4 in-lbs translates to 17.921 in-kips (1 kip = 1000
lbs, for those who have not run into it before, and even for those who
have). When you divide that by the bending stress of 15 ksi, the units
now work out to the required in **3.
The formula that I used was : s=W*L/4*Z, where "s" is the stress
(load) at the midpoint of a beam (15000psi max, in our case), W is the
weight of the object in pounds, "L" is the length in inches and "Z" is
the section modulus. Solving for the section modulus, we get Z =
W*L/4*s. If you plug in the NFPA max stress, you get a standard
formula of Z = W*L/60,000. The restriction is that the load has to be
at the mid point. When you get off the mid point, the calculation gets
a whole lot scarier.
Disclaimer: This is for information only and by no means is an
endorsement or acceptance of any data generated by the procedure shown
above. When something such as this is encountered and is outside the
prescriptive data in NFPA documents, it should be reviewed and stamped
by a licensed professional engineer with experience in structural
design. My lawyer might even like this statement.
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