Something else to remember / revise in your calculations is that the minimum

Yield Strength of almost all steel pipe today is 30,000 psi, some going to
33 or 35k.

The 15,000 used in the "red book" (how long since it's been red?) is quite
out of date.

Jeff Garrison
Project Manager
C 586-871-3705
Detroit Automatic Sprinkler Co.
28501 Ryan Rd., Suite C
Warren, MI 48092
P 586-574-0100
F 586-574-0101
[email protected]




-----Original Message-----
From: [email protected]
[mailto:[email protected]] On Behalf Of Brad
Casterline
Sent: Wednesday, June 05, 2013 5:44 PM
To: [email protected]
Subject: RE: required section modulus (it don't get no easier)

& Ken I wish your PL/4 would have occurred to me sooner (I was re-reading
your post last night). Since the trapeze member has two supports I see each
one handling half the force time half the distance, or, .5P x .5L, which is
equivalent to your PL/4. I would have simply asked you to give that some
thought...


-----Original Message-----
From: John Denhardt [mailto:[email protected]]
Sent: Wednesday, June 05, 2013 4:18 PM
To: [email protected]
Subject: RE: required section modulus (it don't get no easier)

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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