Quantcast Solution -Cont. - dm2_080128

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KLM for plastic mode deflections; from paragraph
7.6(1)(c) of Section 1.
0.77 + 0.78
KLM = [() + 0.66]/2
2
= 0.72
(10) Natural period of the beam from Equation 6-15 of NAVFAC
P-397 is:
TN = 2[pi](KLM m/KE)1/2
Where m is the mass of the beam plus 20 percent of the
slab's span perpendicular to the beam:
m = w/g
150
1,0002
= (30 x 18 + 2 x 8 x 102 x 0.20) x x
123
32.2 x 12
= 194,638.50 lb-ms2/in2
0.72 x 194,638.5 1/2
TN = 2[pi] ( )
8,629.70
= 25.3 ms
(11) Find Xm/XE, ductility ratio from Figure 6-7 of NAVFAC
P-397.  From Figures 38b ad 38c:
T/TN = 60.7/25.3 = 2.40
B = (18 + 2 x 102) x 7.2
= 1,598.40 lb/in
1,598.40
B/ru =
1,065.85
= 1.50
Xm/XE = 17.0
(12) From Table 8 support rotation is:
L tan[theta]
Xm =
2
Xm = Xm/XE x XE
= 17.0 x 0.1235
= 2.10 in
2.08-112





 


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