Design of flange section -Cont.

Custom Search

Thus the effective mass of the element is:
me = mKLM
= 449.30 x 0.65
= 292.04 lb-ms2/in2
and the natural period of vibration is equal to:
TN = 2[pi](me/KE)1/2
= 2[pi](292.04/68.88)1/2
= 12.9 ms
T/TN = 43.9/12.9
B/r = 1.1/1.03
= 3.40
= 1.07
Entering Figure 6-7 of NAVFAC P-397 with these ratios;
Xm/XE = 5.5 No Good; section must remain
elastic.
Increase flange thickness to 3 inches and use two layers
of welded wire fabric, 6 X 6 - W1.4 X W2.0 in top layer
d = 3 -.625-(0.159/2)
= 2.25 in
a = 0.04 x 71,5000/(.85 x 6250 x 12)
= 0.045 in
Mu = 0.04 x 71,500(2.25 - 0.045/2)/12
= 531 in-lb/in
ru = 2 x 531/(21.125)2
= 2.38 psi
Available resistance
ravail = 2.38 - 0.26 -0.10
= 2.02 psi
Ia = 0.6 x 33/12
= 1.35 in4/in
KE = 8(4,286,825.8)(1.35)/(21.125)4
= 232.47 psi/in
m = (150/123 x 3)/(32.2 x 12 x 10 -6)
= 673.96 lb-ms2/in2
me = 0.65 x 673.96
= 438.07 lb-ms2/in2
TN = 2[pi](438.07/232.47)1/2
= 8.6 ms
T/TN= 43.9/8.6
B/r = 1.1/2.02
= 5.10
= 0.54
2.08-258