Page 269 - Estimators Piping Man Hour Manual

244 Section Eleven—TECHNICAL INFORMATION

The weight of the 90° bend is shown as 1550lb at the force on H-l due to the line (H-l)--Q.
center of gravity of the bend. Consider this bend as
7.458(//-l)' = 775(3.625) -f 543(6.375)
supported on a beam which passes through the center = 2809.375 + 3460.625
of gravity and rests on the extensions of the tangents to = 6270
the bend. This imaginary beam is shown resting on
(H-l}' = 84 lib
the tangents at a distance D of l'-4^" and the load on
each end of the beam is one half the total load or or
775fc. (H-l)' = 543 + 775 - 477 = 84lib
This latter method of calculating (H-l)' can be used
as a check on the work of calculating the loads by
taking moments, it consists of the sum of the loads
minus the reaction.
Consider line 0-(W-2) as a free body and by taking
moments about H-2 solve for RH-I-
9/? w.i = 790(2) + 775(5.375)
= 1580 + 4165.625
= 5745.625
RH.I = G38tb
By taking moments about RH-I solve for (H-2)' t the
force on H-2 due to the line 0-(//-2).
9(H-2)' = 775(3.625) + 790(7)
= 2809.375 + 5530
= 8339.375
(H-2)' = 9271b

'.» J'-l'^." .-J-I D s'-«'/z" J, z'-o" (H-2)' = 775 + 790 - 638 = 927tb
SL
I T- - For the section of pipe considered, H-l to H-2, we
\t 5jsr5l have reactions as follows:
,^ 4'_
(H-l)' = 84 lib
Figure 2, Horizontal Bend—Plan View (H-2)' = 0271b
K H-\ = 638ft
The distance D is determined trigonometrically or RH.I = 477 fe
from Figure 3 which has been drawn for convenience.
We must now determine the load on H-l due to the
D = 5 X 0.273 - 1.365' or l'-4|" forces between H-l and A. By definition we said that
the force resulting from (H-l )-0 was to be carried by
Now consider the forces between H-l and H-2 acting
H-2, this means that this section is to be balanced by the
in two planes which intersect at 0. There will be two section between H-2 and H-Z so that in calculating the
reactions at 0 which are designated as RH-I and Rn-2- load on A, section (H-l )-0 is considered weightless.
RH-I is the reaction of line Q~(H-2) to be carried on Conversely, section 0~(//-2), which results in
Hanger H-l and RH-Z is the reaction of line (H-l) -0 to
reaction RH,\ at 0, is to be carried by H-l and therefore
be carried on Hanger H-2. Transpose the feet and balanced by section A-(H-l). Section A-0 in this case
inches to their decimal equivalent in feet. Consider
is a simple beam and is solved by taking moments
line (H-l)~Q as a free body and by taking moments about //-! to find the reaction at A.
about H-l solve for RH-I-
3.708A = 4354(2.1,25) + 500(0.417) - 638(7.458)
7.458/i w.2 = 543(1.083) -f 775(3.833) = 9252.25 -f 208.5 - 4758.204
= 588.069 + 2970.575 = 4702.546
= 3558.664 A — 1268tb which is below the allowable load at A
Rn-2 '--- 4771t> of 1500tb
(
By taking moments about /?//-a solve for H-l)', the Taking moments about A solve for ( H - l ) ' ' , the