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.
At 04:14 PM 5/20/2008, you wrote:
Randy,
I'm not sure I agree with a couple of the responses you've received,
so I'll weigh in with my own.
The values shown in Table 9.1.1.6.1(b) in the '02 edition of NFPA-13
are based on the required
section modulus for a load created by 15'-0" of pipe filled with
water, plus 250 pounds, and a
maximum allowable bending stress of 15.0 ksi. (This is from note #2
of that table). So far so good.
The formula they use for this simple beam calculation is directly
from the AISC manual, beam diagram
#7, and shows the calculation for the maximum moment of inertia
(Mmax) as the result of PL/4, where
P is the load in pounds, and L is the length of the beam in
inches. In the case of what you're describing
the P is approximately 426.7# (based on weight of 4" sch 10 as 11.78
lb/ft x 15' and 250#), and the
value for L is 168" (14' x 12" per foot).
The resulting maximum moment of inertia (Mmax) is then 17,921.4
inch-pounds, or 17.921 kips.
The maximum allowable bending stress (Fb) is established by NFPA-13
as 15.0 ksi (or 15,000 pounds
per square inch), so you would simply divide Mmax by Fb, or 17.921
kips by 15.0 ksi, resulting in a
minimum required section modulus of 1.1947 in^3. So you're
assumption is basically correct.
For proof, just do the math on the same weight (426.7#) with a 120"
(10') span, and you'll come up
with a required section modulus of at least 0.8534 in^3, which is
awfully close to the value shown in
NFPA-13.
As for Ray's comment, I'm in agreement, if you can figure out a way
to use the trapeze with the load
not centered you can reduce the equivalent length of the span using
the formula in A.9.1.1.6,
which uses 4AB/A+B to reach an equivalent length if the load is not
centered. In the trap you're
describing if the load was 11'-0" from one end and 3'-0" from the
other this would work out to an
equivalent length of about 9.4', as a result of (4x11'x3')/14'
I'm completely in agreement with Todd on this one though, I'd be
sure and get a structural PE to
sign off on the math. You're not required to have a PE behind your
name to think these thoughts
or even to do the math, however it would be a prudent course of
action to protect yourself by
getting one to carefully review it and sign off.
Just some thoughts on your question. Hope that helps.
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.
-----Original Message-----
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Matthew
J. Willis
Sent: Tuesday, May 20, 2008 12:25 PM
To: [email protected]
Subject: RE: Trapeze lengths beyond 10'-0"
Can you apply the formula and get an equivalent modulus based on not
center loading the members?
R/
Matt
Matthew J. Willis
Living Water Fire Protection, LLC.
1160 McKenzie Rd.
PO Box 877
Cantonment, FL. 32533
850-937-1850 Voice
850-937-1852 Facsimile
[EMAIL PROTECTED]
----- >-----Original Message-----
----- >From: [EMAIL PROTECTED]
----- >[mailto:[EMAIL PROTECTED] On
----- >Behalf Of Randy Knutson
----- >Sent: Tuesday, May 20, 2008 11:05 AM
----- >To: [email protected]
----- >Subject: Trapeze lengths beyond 10'-0"
----- >
----- >Forumites,
----- >
----- >
----- >
----- >I searched the archives and didn't find a discussion on
----- >this topic.
----- >
----- >Here's my question?
----- >
----- >
----- >
----- >Can I extrapolate out beyond the 10'-0" length on the
----- >section modulus required for trapeze in table
----- >9.1.1.6.1(a) '07ed.?
----- >
----- >
----- >
----- >A fitter called this morning wanting 14' trapeze hangers
----- >for 4" pipe. He was placing the main in the wrong
----- >location but it brought up an interesting question. I
----- >noticed that the section modulus has a linear increase
----- >with trapeze length.
----- >
----- >For 4" Sch 10 y=.085x, where y=section modulus &
----- >x=Trapeze length. So, for a 14' trapeze for 4" sch 10,
----- >would a 1.19 section modulus work?
----- >
----- >Do I have to have the initials PE after my name to think
----- >these thoughts?
----- >
----- >
----- >
----- >Thanks,
----- >
----- >
----- >
----- >Randy Knutson
----- >
----- >Shilo Automatic Sprinkler, Inc.
----- >
----- >_______________________________________________
----- >Sprinklerforum mailing list
----- >[email protected]
----- >http://lists.firesprinkler.org/mailman/listinfo/sprinklerforum
----- >
----- >To Unsubscribe, send an email
----- >to:[EMAIL PROTECTED]
----- >(Put the word unsubscribe in the subject field)
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Todd G. Williams, PE
Fire Protection Design/Consulting
Stonington, Connecticut
www.fpdc.com
860.535.2080
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