Good point John. I don't recall anything in the prescription that says you
can use less than 1/2" rod on a trapeze supporting 6" pipe. Although I am
hanging from more than one rod and I could calculate the distributed loads
between them wouldn't that need to be engineered. Or have I missed a
provision that allows me to do such a thing while remaining prescriptive?


On Wed, Jun 5, 2013 at 2:18 PM, John Denhardt
<[email protected]>wrote:

> Rahe - I am so confused.  This is a requirement of the standard already.
>  Are you requiring that they comply with the standard?  That would be a
> great starting point!
>
> John
>
> John August Denhardt, P.E., FSFPE
> Strickland Fire Protection Incorporated
> 5113 Berwyn Road
> College Park, Maryland 20740
> Office Telephone Number:  301-474-1136
> Mobile Telephone Number:  301-343-1457
> FIRE SPRINKLERS SAVE LIVES - Can you live without them?
>
>
> -----Original Message-----
> From: [email protected] [mailto:
> [email protected]] On Behalf Of Rahe Loftin
> - 7PCE
> Sent: Wednesday, June 05, 2013 5:04 PM
> To: SprinklerFORUM
> Subject: Re: required section modulus (it don't get no easier)
>
> Ken,
> You might mention that if one is trying to support a 6 inch pipe, requiring
> 1/2 inch all thread rod with a trapeze using 3/8 inch all thread rod, as
> the 6 inch pipe is moved from the middle of the span, one of the 3/8 ATRs
> will begin to be supporting more of the 6 inch pipe load.  At some point,
> the 3/8 inch rod will no longer share the load properly leaving one of the
> rods trying to support too much of the load.  Even if the section modulus
> is correct, I always have the contractor use 1/2 inch ATR to support the
> trapeze if they are trying to support 6 inch pipe.
>
> *
> *
> *Thanks*
> *
> *
> *Rahe Loftin, PE *
> *Fire Protection Engineer (AHJ)*
> *General Services Administration*
> *Region 7 - TX, OK, NM, AR, LA*
> *Design & Construction 7PCE*
> *
> *
> *c - 817-371-3102*
> *o - 817-978-7299*
>
>
> On Mon, Jun 3, 2013 at 9:59 AM, ParsleyConsulting <
> [email protected]
> > wrote:
>
> > Whoops, forgot that to get to 1.09 for the required section modulus, you
> > first have to divide PL by 4, then by 15,000.  See how easy it is for
> even
> > a smart aleck like me to make a mistake? The answer was correct, but the
> > explanation was flawed, that's why my teachers always asked to show the
> > work.
> >
> >     65,696.4divided by 4= 16,421 and that divided by 15,000 = 1.094
> >
> > Forgetting a step like that got me much lower grades.
> > *
> >
> > PARSLEY CONSULTING
> >
> > *
> >
> > Ken Wagoner
> >
> > 760.745.6181 voice
> >
> > 760.745.0537 fax
> >
> > website: www.ParsleyConsulting.com <http://www.ParsleyConsulting.**com<
> http://www.ParsleyConsulting.com>
> > >
> >
> > On 6/3/2013 7:51 AM, ParsleyConsulting wrote:
> >
> >> Brad,
> >>     Part 1 - Gibberish.  Pure and simple.  The tables in NFPA 13 didn't
> >> reduce the "force" and "distance" by half in their calculations.  It
> simply
> >> used the formula to calculate the maximum moment at the point of the
> load
> >> for a simple beam, based on the maximum length of various pipe sizes
> >> between hangers.  That's why the figures for 1" and 1-1/4" are based on
> 12'
> >> between hangers.
> >> The maximum moment is the "worst case", and was what the committee saw
> as
> >> providing system designers with enough information to allow them to not
> use
> >> trapeze members which would have excessive loads placed on them.
> >>     The formula to determine maximum moment is PL/4, with P=weight of
> >> pipe being supported in pounds (including 250#), and L=the length of the
> >> span of the trapeze in inches. The moment is then divided by the maximum
> >> allowable bending stress, which for steel is 15,000.
> >>     The resulting equation looks like this:  (PL/4)/15,000 = required
> >> section modulus. So, for 15' of 1-1/2" sch 40 pipe, the value of P would
> >> equal 15' x 3.61 = 54.15 + 250 = 304.15#.  For an 18' span, the value
> of L
> >> = 216".  Multiply PxL = 65,696.4, and then divide that value by 15,000,
> and
> >> the result is a required section modulus for the worst case location of
> the
> >> load (mid span) of 1.09 in³. That is just slightly less than doubling
> the
> >> value for a 9'-0" span, however using that method would result in a
> larger
> >> required section modulus, so it would work out in the end.
> >>     The 4ab/a+b you reference is a method to calculate the equivalent
> >> length of the trapeze, in order to reduce the value of "L" in the
> original
> >> equation. Going back to your 18' span, if the load were 3' from one end,
> >> and 15' from the opposite end, that results in the equation
> >> 4x180x36/180+36, which equals 120", or 10'.  The value of L in the above
> >> equation is changed from 216"to 120", reducing the required section
> >> modulus. PL/4 now becomes 304.15 x 120 = 36,498which is divided by 4,
> >> resulting in 9,124.5, which is further divided by the allowable bending
> >> stress of 15,000 for steel, with a final required section modulus of
> 0.6083
> >> in³. Round that up to 0.61, and it matches the value in NFPA 13, Table
> >> 9.1.1.7.1(a) (2013).
> >> No reduction by half anywhere in those equations.
> >> Part 2 - Relatively correct regarding doubling the require section
> >> modulus for a 9' span, however as the structural engineers in the room
> when
> >> this table was expanded to include spans up to 16'-0" notedyou can keep
> >> increasing the span to the point where the weight of the trapeze itself
> now
> >> becomes an issue in the overall capacity to resist bending, and that
> sort
> >> of calculation was one they were reluctant to provide without a
> significant
> >> amount of additional information.
> >> Please note that this is my opinion only, and does not reflect that of
> >> the NFPA 13 Hanging and Bracing Committee.
> >> *
> >>
> >> PARSLEY CONSULTING
> >>
> >> *
> >>
> >> Ken Wagoner
> >>
> >> 760.745.6181 voice
> >>
> >> 760.745.0537 fax
> >>
> >> website: www.ParsleyConsulting.com <http://www.ParsleyConsulting.**com<
> http://www.ParsleyConsulting.com>
> >> >
> >>
> >> On 5/31/2013 7:36 PM, Brad Casterline wrote:
> >>
> >>> saw an e-mail the other day about 18 foot trapeze spans.
> >>>
> >>> wished someone had asked me.
> >>>
> >>> i would have said---either:
> >>>
> >>> section modulus is moment divided by stress. moment is half the
> force(lb)
> >>> times half the distance(in).
> >>>
> >>> the force is the weight of the water-filled pipe plus 250. the distance
> >>> is
> >>> 4ab/a+b.
> >>>
> >>> the stress is 15,000.
> >>>
> >>> or: double the required section modulus for a nine foot span.
> >>>
> >>>
> >>> Brad Casterline, NICET IV
> >>>
> >>> Fire Protection Division
> >>>
> >>>
> >>> FSC, Inc.
> >>>
> >>> P: 913-722-3473
> >>>
> >>> [email protected]
> >>>
> >>> www.fsc-inc.com
> >>>
> >>>
> >>> Engineering Solutions for the Built Environment
> >>>
> >>>
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-- 
Ron Greenman
(Пушечное ядро Ивана 6 дюймов)
Instructor
Fire Protection Engineering Technology
Bates Technical College
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