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 Solution:
(1) Determine the equivalent static load:
DLF = 1.65 (reusable)
w = DLF x p
= 1.65 x 4.30 x 144
= 1,021.7 lb/ft2
(2) Determine required ultimate moment capacities:
Mup = Mun = wL2/12
= 1,021.7(4.5)2/12
= 1,724.1 lb-ft/ft
(3) Determine required section moduli:
Fdy = 1.1 x 40,000 = 44,000 psi
S+ = S -
= (1,724 x 12)/44,000
= 0.47 in3/ft
Select a UKX 18-18, 1-1/2 inches deep.
(4) Determine actual properties of selected section.
From manufacturer's guide:
S+
=
0.472 in3/ft
S -
=
0.591 in3/ft
I
=
0.566 in4/ft
w
=
4.8 psf
(5) Compute maximum unit resistance, ru.
Mup = (44,000 x 0.472)/12
= 1,730 lb-ft/ft
Mun = (44,000 x 0.591)/12
= 2,167 lb-ft/ft
ru
= 4(2Mup + Mun)/L2
= 4(2 x 1,730 + 2,167)/(4.5)2
= 1,111.8 lb/ft2
(6) Determine equivalent static stiffness:
KE = ru/XE = EIru/0.0062ruL4
= (30 x 106 x 0.566)/[0.0062 x (4.5)4 x 144]
= 46,380.3 lb/ft
2.08-179
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