& 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 >>> >>> >>> ______________________________**_________________ >>> Sprinklerforum mailing list >>> Sprinklerforum@lists.**firesprinkler.org<[email protected]. org> >>> http://lists.firesprinkler.**org/listinfo.cgi/** >>> sprinklerforum-firesprinkler.**org<http://lists.firesprinkler.org/listinfo.c gi/sprinklerforum-firesprinkler.org> >>> >>> >> >> ______________________________**_________________ >> Sprinklerforum mailing list >> Sprinklerforum@lists.**firesprinkler.org<[email protected]. org> >> http://lists.firesprinkler.**org/listinfo.cgi/** >> sprinklerforum-firesprinkler.**org<http://lists.firesprinkler.org/listinfo.c gi/sprinklerforum-firesprinkler.org> >> >> > > ______________________________**_________________ > Sprinklerforum mailing list > Sprinklerforum@lists.**firesprinkler.org<[email protected]. org> > http://lists.firesprinkler.**org/listinfo.cgi/** > sprinklerforum-firesprinkler.**org<http://lists.firesprinkler.org/listinfo.c gi/sprinklerforum-firesprinkler.org> > _______________________________________________ Sprinklerforum mailing list [email protected] http://lists.firesprinkler.org/listinfo.cgi/sprinklerforum-firesprinkler.org _______________________________________________ Sprinklerforum mailing list [email protected] http://lists.firesprinkler.org/listinfo.cgi/sprinklerforum-firesprinkler.org _______________________________________________ Sprinklerforum mailing list [email protected] http://lists.firesprinkler.org/listinfo.cgi/sprinklerforum-firesprinkler.org
