Page 271 - Estimators Piping Man Hour Manual

246 Section Eleven—TECHNICAL INFORMATION

Taking moments about K\ solve for (//-6)". load on H-4 due to the 7/-3-//-4 line.
I5.5C//-6)" = 95(0.25) 4- 1025(8.125) + 453(17.5) 17(//-4)'= 148(0.375)+3667 (2.583)42488(10.708)
= 23,75 + 8328.125 + 7927.5 = 55.5 4 9472.861 +26,641.504
= 16,279.375 = 36,169-865
(//-4>'=-2l28ft>
Total load on //-6 = (//-6)' + (//-6)" = 106 or
1050 = 115(ilb,
(//-4)' = 148 4 3667 4 2488 - 4175 = 21.28Ib

Section 6
Referring to Figure 1 it will be noted that it is
required that there he zero reaction at point marked C.
Draw a sketch to scale as in Figure 12 showing loads
and dimensions.
Consider the forces as acting in two planes which
intersect at 0.
Calculate the load (H-4)" on H-4 due to the section
(H-4) to 0 by taking moments about 0.
Taking moments about (//-('>)" solve for /f], the
load on the flanged tee.
15.5fti = 1025(7.375)495( 15.25)4 140(15.5) -463(2)
= 7559.37541448.7542170-906
==10,272.125
H-4>f—
ft ,=663

= 140 4 95 4 1025 4 453 - 1050 = 663Jb
Section 5
Figure 11 shows the pipe section (//-3)- (//-4) as a
simple team. Solve for the reaction (H-3)" by taking
moments about HA.
I
*663 (FIGURE 10)

2004
IOOO
663
Figure 11
Figure 12
-3)" = 2488(6.292)+3667(14.417)4-148(16.625)
I
= 15,654.496+52,867.13942460,5 7.083(J /-4)" = 285(1.083)4247 (2.125)
= 70.982.135 45418(4.583)4130(6.75)
(ff-3)"==4175lb = 308.6554-524.8754-24,830.694 4-877.5
= 26,541.724
Total load on H-3 = (H-3)' + (H-3)" = 3804 + (//-4)" = 37471b
4175 = 7979lb.
Total load on H-4 - (H-4)' + (H-4)" = 2128 +
Take moments about H-3 and solve for (H-4)', the 3747.- 5875m.