 Determine the loads on each hinge during rebound phase      Custom Search   ST = 23.24/1.26
= 18.45 in3
Applied moment,
M = Pe
=  5,896 lb/in x 10 in x (2.75 + 1.26) in
=  236,429.6 in-lb or 236.4 in-k
Plastic moment capacity,
Mp = FdyS
= 1.1(50)(4.44)
= 244.2 in-k
Mp > M
O.K.
Consider section as column of length 2'- 3" subject to
combined axial load bending moment (see Equation 2.4-3
AISC Manual).
P/Py + M/1.18Mp < /= 1.0
M < /= Mp
Py = AFdy
= 7.25(50 x 1.1)
= 398.8 k
P = 5,896 x 10
= 59 k
P/Py + M/1.18Mp = 59/398.8 + 236.4/1.18(244.2)
= 0.97 < 1.0
O.K.
Section is O.K. for door reaction.
(5) Determine the loads on each hinge during rebound phase.
From Example 6.a, rebound resistance,
r - = 192.2 psi
Total load on door = 192.2 x 86 x 50
= 826.5 k
This load is picked up by 3 hinges and 3 "locking bolts".
= 137.75 k = 138 k
(6) The stiffeners should be in such a location so as to
locate a stiffener directly underneath a hinge.  The
size of the welds connecting the stiffener plates to
the bent plate should be determined on this basis.
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