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 Solution:
(1) Assume a plate thickness of 3/8 inch.
(2) Determine the elastic and plastic section moduli (per
unit width).
S = bd2/6
= [1 x (3/8)2]/6
= 2.344 x 10 -2 in3/in
Z = bd2/4
= [1 x (3/8)2]/4
= 3.516 x 10 -2 in3/in
(3) Calculate the design plastic moment, Mp.
Mp =
Fdy(S + Z)/2
=
39.6[(2.344 x 10 -2) + (3.516 x 10 -2)]/2
=
39.6 x 2.93 x 10 -2
=
1.160 in-k/in
(4) Calculate the dynamic ultimate shear capacity, Vp, for a
1-inch width.
Vp = FdvAw
= 21.78 x 1 x 3/8
= 8.168 k/in
(5) Evaluate the support shear and check the plate capacity.
Assume DLF = 1.0.
V = DLF x B x L/2
= 1.0(27.0)(15.75)(1)/2
= 212.6 lb/in or 0.2126 k/in
V/Vp = 0.2126/8.168
= 0.0260 < 0.67
No reduction in equivalent plastic moment is necessary.
Note:
When actual DLF is determined, reconsider Step 6.
(6) Calculate the ultimate unit resistance, ru, assuming the
plate is fixed-simply supported.
ru = 12Mp/L2
= 12(1.160)(103)/15.752
= 56.11 psi
2.08-283
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