Memorandum from Raymond L. Murray and A. C. Menius, Jr. to Dr. Clifford K. Beck
Typescript
2 pp.
August 2, 1951
MurNBweight080251


212 lb/ft³
Aug 2, 1951
NCSC-29
CORRECTED
RM

TO: Dr. Clifford K. Beck
FROM: Raymond L. Murray and A. C. Menius, Jr.
SUBJECT: Weight of Reactor Shield
CC: Newton Underwood and A. W. Waltner

For purposes of estimating needs for material in the reactor shield,
and for crane load considerations, the volumes and weights are computed. Voids
due to various pipes are neglected.
 
 Figure 1

A. Cap. Consider one quadrant as in figure 2.
 

The system can be further subdivided into fourths.
 

If the Level is excluded, the triangular block has a volume

The volume beveled off is-- 1/2 (1.5) (1.5) (5.25 tan 22 1/2°) = 2.45 ft.³,
leaving a volume of 16.19 ft.³ or 64.76 ft.³ total. The correction to ac-
count for the removable top block 1' x 1' x 1' = 1 ft.³, and the overhang
into the reactor cavity of 2.5' x 2.5' x 4" = 2.08 ft.³ is a net 108 ft.³.
Thus, each quadrant of the cap is 65.84 ft.³, with a density of 213 lb./ft³.
The weight is 14,024 lb. = 7.01 tons.

[page 2]

B. The area of this block is
= (4) (3.5) - 1/2 (1.5) (1.5) - 2 (1.5)
= 14 - 1.125 - 3 = 9.875 ft.². Subtract the recess 3" x 6" = .125 ft³,
leaving 9.75 ft.², or a volume of (9.75) (6) = 48.75 58.50 ft.³, a weight of
10,384 lb. 12,460 lb or 5.19 6.23 tons.

C. The area of this block is
= (3.5) (3.5) - (1.5) (1.5) (1) (0.25)
= 12.25 - 2.25 - 0.25 = 9.75 ft.³²
The volume and weight is thus the same as for A.B.
58.5 ft³, 12,460 lb, 6.23 tons

D. The slab may be considered as being two pieces of total volume
5' x 5' x 6" = 12.3 ft.³ plus 5' 6" x 5' 6" x 6" = 15.125 ft.³
or 27.625 ft.³, weight 5,884 lb. or 2.94 tons.

E. Divide the slab into two parts of total volume
5 x 5 x 6" = 12.5 ft.³ plus 5' 6" x 5' 6" x 1' = 30.25 ft.³ or 42.75 ft.³
The weight is 9,106 lb. or 4.55 tons.

In order to determine the total weight of shield, including the per-
manent sides, consider a solid octagon 8' 6", neglecting the bevel.

Since an octagon of diameter has an area .8284 d², the volume of this
port is (8.5)(0.8284)(17)² = 2035.0 ft.³.

Subtract the following voids:a. Bevel. Divide the octagonal rim into 16 parts. the
average length of each is (8.5+7./2) tan 22 1/2° =
3.21 ft; the area is 1/2 (l.5) 1.5) = 1.125 ft.²
the volume is 3.611 ft.³. For 16 of these, the
volume is 57.8 ft.³.
b. Reactor Cavity. 8' 2" x 5' x 5' = 204.2 ft.³.
c. Thermal column cavity. 5 x 5 x 5 = l25 ft.³
plus the ledge 3" x 1' x 5' = 1.25 ft.³, giving
126.2 ft.³.

The sum of the voids is 385.7 ft.³. The complete cap has a total volume (now
including the top block of 4 ft.³) of 4(65.84) + 4 = 267.4 ft.³.

The final shield volume is
2035.0 - 385.7 + 267.4 = 1916.7 ft.³ or 71.0 yds.³

The weight is
408,200 lbs. or 204.1 tons.

Since there are a number of unoccupied spaces--exposure tubes, sample box, locks
and ducts, this is probably high by a few per cent.