Quantcast Solution - dm2_080080

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(8) Distance traveled by fragment, Rf = 15 ft
(9) Critical fragment weight: 1.5 oz
Solution:
(1) Weight of explosive = π (1212(40)(0.0558)/4
= 252.4 lbs
W = 1.2(252.4)
= 302.9 lbs.
2
[(13)2 - (121 ](40)(0.281)/4
WC =
= 220.7 lbs.
w/wc = 302.9/220.7
= 1.37
(2) For TNT in mild steel casing, from Table 3:
(2E')l/2 = 6,940.
Then using Equation (21):
(2E')1/2[<W/W,>/(1 + 0.5W/Wc)1/2]
= 6,940 [1.37/(1 + 0.5 x 1.37)]1/2
= 6,258 fps
(3) For TNT in mild steel casing, from Table 3:
B = 0.30.
Then using Equation
(4) For design Confidence Level CL = 0.999 and Equation (29b)
2.08-64





 


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