Quantcast Solution -Cont. - dm2_080127

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Averaging the two values:
IcN + IcP
Ic =
2
8,442 + 8,728
=
2
= 8,585 in4
Gross moment of inertia:
bh3
Ig =
12
18 x 303
=
12
= 40,500 in4
Average moment of inertia of the beams from Equation 5-20
of NAVFAC P-397:
Ig + Ic
Ia =
2
40,500 + 8,585
=
2
= 24,542 in4
(7) From Table 7, KE of a uniformly loaded beam with fixed
ends is:
307 Ec Ia
KE =
L4
307 x 3.8 x 106 x 24,542.5
=
2404
= 8,629.70 lb/in/in
(8) Equivalent elastic deflection is:
ru
XE =
KE
1,065.85
=
8,629.70
= 0.1235 in
(9) Load-mass factor from Table 10 for a plastic range of a
uniformly loaded beam with fixed ends is:
KLM - elastic = 0.77
- elasto-plastic = 0.78
- plastic = 0.66
2.08-111





 


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