13. FIRST-ORDER WAVE GENERATION
359
Wave Board Force
Instantaneous force on the wave board is given by the integral in Eqn. 7.65
with the exception that the lower limit of integration now becomes the
hinge location rather than the bottom of the wave tank, i.e.,
/
o
p0(z,t)dz
-(h-i)
Substituting p0 from Eqn. 7.63 and integrating yields
(7.92)
Fro (0 —
(sinh
— sinh kl)^ cos at 4(
°° C
\
pa
—-^-(sin knh — sin knF) j sin at 4- y (h - /)2
(7.93)
n=i kn
/
2
for the case of a wave board with water on one side and A and Cn are given
by Eqns. 7.89 and 7.90, respectively.
Wave board forces when there is water of equal depth on both sides of
the board is found by substituting poo from Eqn. 7.64 into Eqn. 7.92 to get
FToo(t) = 2
(sinh kh — sinh/:/)^ cosat +
/
00 Q
\
4-2 I pa
—-(sin knh — sin knF) I sin at
\
*
j
(7.94)
The maximum total force that will occur over a wave cycle on a wave
board with water on one side can be easily found by abbreviating Eqn. 7.93
as
FTo(t) = Fr cos at + Fj sin at 4- ^-(h - /)2
which can be written as
FTo = \JFr + fi cos(at + e) + y (h - I)2
where
(7-95)
(7.96)
(7.97)
e — tan 1 Fr)
The maximum occurs when cos(at 4- c) = 1, i-e->