Page 270 - Estimators Piping Man Hour Manual

Section Eleven—TECHNICAL INFORMATION 245

11.3« H . 6 = 95(0.25) + 653(5.458) + 146(10.5)
= 23.75 + 3564.074 + 1533
= 5120.824
/?/,.« = 453tb
Taking moments about RH-K solve for (//-?)', the
load on W-7 due to the line KH-&- (H-7).
H.3(//-7)' = 146(0.800) + (>53(5.834) 4- »5(ll..05)
= 116.8 + 3809.602 + 1049.75
= 1976.152
(H-7)' = 44 lib

(//-')' = 146 -f 653 + 95 - 453 - 44ltb
Next consider the section between H-7 and B plus the
reaction RH-T, see Figure 9,

r
Figure 6 h-
Taking moments about /?«.? solvefor (H-6)', the
load on H-Q due to the line (//-(>)-]? w. 7.
H-6
t
LcLfUZL* *- «LZlil ° 792 '
j^^ L. 1J2M

' = !4f>(0.792) + (12(1.563)
= !I5.()32 + 96.906 Figure 9
= 212.538
It is necessary to treat this section as a cantilever
beam. The load on H-7 caused by this section is
(//-7) and can be solved for by summing the forces.
(tf-6)' = 146 + 62 - 102 = 106ft
(H-7)" = 102 + 437 4- 152 + 1318 - f>50 = 13591b
Figure 8 is a free body diagram of section
R«-6~(//-7). Take moments about H-7 to solve for
Total load on H-7 = (H-7)' + (H-7)" = 441 +
1359 = 18001ti.
Section 4
Figure 10 is an elevation view of the section of 6*
pipe tetween H-& and the flanged tee with the reaction
RH-R, which is the load on H-% due to the line
Figure 8