ABSTRACT
Given: a column of more than 70 m high has a 20 m diameter at the lower cross section. The wall has 2 rr. thick. Space diagram of loading the column is shown in Fig. 1. The applied ice horizontal load is taken as F = 0.25*109 N = 0.25 GN providing horizontal force on the building is approximately F = 1.0*109 N = 1.0 GN. The vertical thrust load is N = 5.5* 108 N. The bending moment about the lower cross section is obtained as follows:
M = F* H = 0.25 * 10® * 40 * 102 = 1.0 ’* 1012 N*cm
JX s = As,tot /A = 6305.12 / 113.1*104 = 0.005575
i >
Serviceability Offshore 401
SNbi + l N sp,i + £N s,i _ Nbt = SN
T < 0.008*Rb*7C [(r1)3-(r2)3]
T + M * b/c < Mu* b /c + qsr*c*(5
To obtain the new torsion moment T we use the main equation of torsion and bending
T = Mu*b/c + qsr*rs*pM*b/c
CONCLUSIONS
REFERENCES
F < Fb + 0.8 Fsw
