STRUCTURAL RESPONSE STATISTICS: PART II
189
a — —k\/m\
b=—c\/m\ c = —iû2
d — —2; e = û2à1/2/m
2«i =d + (d2+4c)1/2;
2s2 = d - (d2 + 4c)1/2
2s3 = b + (62 + 4a)1/2;
2s4 = b - (b2 + 4a)1/2
Ci = d(d - 6)/[c(a - c)(d - 6) - (bc - ad)(c + d2 - bd - a)]
C2 = — Ci(c + d2 — bd - a)/(d — b)
C3 — — Ci;
C4 = -(1 + aÛ2)/c
71 = («IG + C2)/(si - $2);
72 - “(s2Ci + C2)/(»i - s2)
73 — (s3C3 + C4)/(s3 - s4);
74 = -(s4C3 + C4)/(s3 - s4)
= -Si, i - 1,2,3,4
Although this intégral solution is exact, it is recalled that the results are
an approximation to reality because of the assumptions inhérent in the mathematical modeL In addition to the single degree of freedom approximation for
the structure, the assumed form for Sp](lu), équation (7.73), does not exactly
replicate ^(lu). Note that as lu —» 0, Spi(iu) —> â, whereas S^(lu) —* 0. However,
these two spectra hâve similar behavior otherwise: both form a single peak and
both approach zéro as u? becomes large.
Example Problem 7.5. Consider the same jackup rig of Example Problem
7.3, for which the characteristic constants m, fci, and Ci are given just after
équation (7.51). Based on the theory of covariant propagation, or the closed
form response results obtained in équation (7.101), compute numerically the
rms deck deflection av. Use the Peirson-Moskowitz wave height spectrum with
a significant wave height of 15 m as the basis for the wave force excitation of
this structure.
The first task is to achieve an approximate fit for à, C an
and (7.73), then
(7.102)
The two terms on the right side of the last équation are known from Example
Problem 7.3: |C(lu)|2 = 4.76 x 108 lb2/ft2, and S,,:.. ) of équation (7.60), which
has its peak at 672.5 ft2-sec/rad at the frequency lu = 0.324 rad/sec. Now match
the peaks for each side of équation (7.102) at lu = lû = 0.324 rad/sec, which
leads to
2
à = (2<)2(4.76 x 10a)(672.5) = 1.28 x 1012< lb2-sec/rad
189
a — —k\/m\
b=—c\/m\ c = —iû2
d — —2; e = û2à1/2/m
2«i =d + (d2+4c)1/2;
2s2 = d - (d2 + 4c)1/2
2s3 = b + (62 + 4a)1/2;
2s4 = b - (b2 + 4a)1/2
Ci = d(d - 6)/[c(a - c)(d - 6) - (bc - ad)(c + d2 - bd - a)]
C2 = — Ci(c + d2 — bd - a)/(d — b)
C3 — — Ci;
C4 = -(1 + aÛ2)/c
71 = («IG + C2)/(si - $2);
72 - “(s2Ci + C2)/(»i - s2)
73 — (s3C3 + C4)/(s3 - s4);
74 = -(s4C3 + C4)/(s3 - s4)
= -Si, i - 1,2,3,4
Although this intégral solution is exact, it is recalled that the results are
an approximation to reality because of the assumptions inhérent in the mathematical modeL In addition to the single degree of freedom approximation for
the structure, the assumed form for Sp](lu), équation (7.73), does not exactly
replicate ^(lu). Note that as lu —» 0, Spi(iu) —> â, whereas S^(lu) —* 0. However,
these two spectra hâve similar behavior otherwise: both form a single peak and
both approach zéro as u? becomes large.
Example Problem 7.5. Consider the same jackup rig of Example Problem
7.3, for which the characteristic constants m, fci, and Ci are given just after
équation (7.51). Based on the theory of covariant propagation, or the closed
form response results obtained in équation (7.101), compute numerically the
rms deck deflection av. Use the Peirson-Moskowitz wave height spectrum with
a significant wave height of 15 m as the basis for the wave force excitation of
this structure.
The first task is to achieve an approximate fit for à, C an
and (7.73), then
(7.102)
The two terms on the right side of the last équation are known from Example
Problem 7.3: |C(lu)|2 = 4.76 x 108 lb2/ft2, and S,,:.. ) of équation (7.60), which
has its peak at 672.5 ft2-sec/rad at the frequency lu = 0.324 rad/sec. Now match
the peaks for each side of équation (7.102) at lu = lû = 0.324 rad/sec, which
leads to
2
à = (2<)2(4.76 x 10a)(672.5) = 1.28 x 1012< lb2-sec/rad