A 4-inch schedule 40 main is routed parallel in-between two beams and requires a trapeze hanger. The main is 2 ft. from one beam and 8.5 ft from the other for a total span of 10.5 ft. What is the smallest schedule 10 pipe capable of supporting this main as a trapeze hanger? a. 2 inch b. 2-1/2 inch c. 3 inch d. 3-1/2 inch Solution | Posted 10/22/18
10 Comments
KC
10/18/2018 08:15:24 am
C
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Edw H
10/18/2018 09:44:18 am
c. 3 inch
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Ed
10/18/2018 01:51:28 pm
From T 9.1.1.7(a) required 0.84
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TG
10/18/2018 10:30:19 am
c. 3 inch
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Seatexans@gmail.com
10/18/2018 10:49:19 am
B. Since it isn't loaded at the midspan, the max moment is different. M_max=803 ft-lb. S=M_max/15ksi=0.642 in^3. Per Table 9.1.1.7.1(b), that's 2-1/2" sched 10. Unless I over thunk it.
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Jimmy
10/18/2018 11:14:27 am
Look at A.9.1.1.7: the equivalent trapeze length is L=4ab/(a+b) = 6.47'. easier calculation then my ramblings above. But S_min = 0.64 in^3 just like the alternate method above... This is the easier sub-6 minute solution for test day! Good luck all!
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Pari
10/18/2018 10:54:27 am
c
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Peter
10/18/2018 12:50:54 pm
SeaTexan and Jimmy are correct. The section modulus minimum is to limit the bending stress on the trapeze member. Bending stress Sigma= MC/I where M is the bending moment, C is the distance from the neural axis to the outer-most fiber of the member cross-section, and I is the 2nd moment of area. Section Modulus, S= I/C; so, Bending stress = M/S. The intent of NFPA 13 section 9.1.1.6. is to hold S to a minimum (high) value in order to limit bending stress. Unfortunately, the way 9.1.1.6 is written, section modulus values cannot be interpolated from table (a). One can, however, find an equivalent span based on whether the trapeze member is centrally located, or located to one side. This can be accomplished by Leq= 4ab/(a+b) where a is the distance from the load to one point of attachment of the trapeze member to the structure, and b is the distance from the load to the other point of attachment. Overhang of the member has no bearing on the bending stress. In this example Leq= (4*2*8.5)/10.5= ~6.48 From Table (a) select a 7'-0" span for 4" schedule 40 gives S= 0.69 in^3. From Table (b) 0.69 coincides with 2 1/2" schedule 10.
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Brian
10/18/2018 03:34:01 pm
b. SFPE 5th pg 1442 or 13-A.9.1.1.7 pg 13-349:
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Ron
10/21/2018 03:02:13 pm
"B" 2-1/2" Sch 10 Pipe
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