MIL-STD-209K APPENDIX C
= 1.5 x T
= 1.5 x 26,667 lb = 40,000 lb
Aft U
= 1.5 x T
= 1.5 x 53,333 lb = 80,000 lb
= 1.5 x T
= 1.5 x 26,667 lb = 40,000 lb
C.4.1.2 Minimum required vertical design and ultimate loads. The vertical inertia force acting through the CG is:
2 x GW = 2 x 20,000 lb = 40,000 lb
It should be assumed that this force will be restrained by the vertical force compo- nents of all four tiedown provisions against upward movement of the item.
1V 2V 3V 4V
= 40,000 lb
For this example, the provisions are not located symmetrically about the CG so the proportionate share of the load that is applied to each provision is not equal and is as follows:
S 2
L3
T = (S
+ S ) x (L
+ L )
x (T + T + T + T )
1V 1 2
10
1V = (5 + 10) x
S1
1 3
10
(15 + 10)
L4
x 40,000 = 10,667 lb
T = (S
+ S ) x (L
+ L )
x (T + T + T + T )
2V 1 2
5
2V = (5 + 10)
2 4
10
x (15 + 10)
x 40,000 = 5,333 lb
S 4
L1
T = (S
+ S ) x (L
+ L )
x (T + T + T + T )
3V 3 4
10
3V = (5 + 10) x
1 3
15
(15 + 10)
x 40,000 = 16,000 lb
Source: https://assist.dla.mil -- DCow-4nloaded: 2014-09-28T23:10Z Check the source to verify that this is the current version before use.
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