90
WAVE FORCES ON STRUCTURES
2.2. In this procedure, the irregular shape is simply replaced by a rectangular
box of the same overall dimensions as the irregular shape, a procedure that
greatly simplifies the calculation of the wave pressure loads. The justification
for this procedure was given by Hogben and Standing (1975), who found that the
total diffraction forces and moments are practically independent of structural
planfonn. The following example, parallel to the analysis of Nataraja and Kirk
(1977), illustrâtes this type of calculation.
Example Problem 4-3.
Calculate Fkx and Mk, the Froude-Krylov horizontal force and overturning moment on a rectangular box caisson resting on
the océan floor. The box has cross section dimensions 2a x 2a and height h,
as shown in Figure 4.2. Assume that there is a single incident wave and that
linear wave theory is appropriate. Relative to the coordinate System of Figure
4.2, the dynamic pressure given in Table 3.1 becomes
,
,
Hcoshfctz + d)
p(x, z, t) = pg— —---- ——— cos (kx - wt)
(4.17)
2
cosh kd
in which k = 2tt/A and w is given by équation (3.16). The value of Fkx is
computed by integrating the pressure différence over the vertical sides of the
box normal to a right-traveling wave, or
r(a—d)
Fi - 2a J
[p(—a, z,t) — p(a, z,t)]dz
(4.18)
With équation (4.17), the last équation is integrated and use is made of trigometric identities, which leads to
„
„
„ sinh kx . ,
,.
Fkx = —2pgaH------ ——sin ka sin ait
(4.19)
kcosh kd
'
1 he overturing moment is the sum of two intégrais. The first is Mkk, the moment
about 0 due to the side pressure forces. The second is Mkv, the moment about
0 due to the pressure forces on the top of the box. Thus
r(a-d)
Alkh = 2a I
\p(—a,z,t)—p(a,z,t)](z + d)dz
(4.20)
J — «f
Mkv = 2a I xp[x, (h — d),t]dx
(4.21)
J —a
where
Mk = Mkh + Mkv
(4.22)
The évaluation of this latter moment in closed form using équations (4.17),
4.20), and (4.21) is a straightforward exercise. With these results, the total
horizontal diffraction force Pll(t) and the overturning moment M0(t) are then
g*wn ’ équations (4.9) and (4.10), using the respective flow coefficients of
équations (4.11) and (4.13). The calculation of Fkz and its corresponding total
verucal load pn(t) is straightforward. See Problems 4.13 and 4.14.
WAVE FORCES ON STRUCTURES
2.2. In this procedure, the irregular shape is simply replaced by a rectangular
box of the same overall dimensions as the irregular shape, a procedure that
greatly simplifies the calculation of the wave pressure loads. The justification
for this procedure was given by Hogben and Standing (1975), who found that the
total diffraction forces and moments are practically independent of structural
planfonn. The following example, parallel to the analysis of Nataraja and Kirk
(1977), illustrâtes this type of calculation.
Example Problem 4-3.
Calculate Fkx and Mk, the Froude-Krylov horizontal force and overturning moment on a rectangular box caisson resting on
the océan floor. The box has cross section dimensions 2a x 2a and height h,
as shown in Figure 4.2. Assume that there is a single incident wave and that
linear wave theory is appropriate. Relative to the coordinate System of Figure
4.2, the dynamic pressure given in Table 3.1 becomes
,
,
Hcoshfctz + d)
p(x, z, t) = pg— —---- ——— cos (kx - wt)
(4.17)
2
cosh kd
in which k = 2tt/A and w is given by équation (3.16). The value of Fkx is
computed by integrating the pressure différence over the vertical sides of the
box normal to a right-traveling wave, or
r(a—d)
Fi - 2a J
[p(—a, z,t) — p(a, z,t)]dz
(4.18)
With équation (4.17), the last équation is integrated and use is made of trigometric identities, which leads to
„
„
„ sinh kx . ,
,.
Fkx = —2pgaH------ ——sin ka sin ait
(4.19)
kcosh kd
'
1 he overturing moment is the sum of two intégrais. The first is Mkk, the moment
about 0 due to the side pressure forces. The second is Mkv, the moment about
0 due to the pressure forces on the top of the box. Thus
r(a-d)
Alkh = 2a I
\p(—a,z,t)—p(a,z,t)](z + d)dz
(4.20)
J — «f
Mkv = 2a I xp[x, (h — d),t]dx
(4.21)
J —a
where
Mk = Mkh + Mkv
(4.22)
The évaluation of this latter moment in closed form using équations (4.17),
4.20), and (4.21) is a straightforward exercise. With these results, the total
horizontal diffraction force Pll(t) and the overturning moment M0(t) are then
g*wn ’ équations (4.9) and (4.10), using the respective flow coefficients of
équations (4.11) and (4.13). The calculation of Fkz and its corresponding total
verucal load pn(t) is straightforward. See Problems 4.13 and 4.14.