Question 15.1: A storm in deep water generates waves which travel towards t......
A storm in deep water generates waves which travel towards the shore, impinging on a breakwater. The breakwater is a vertical wall erected on a rubble mound (Fig. 15.09b). The storm has a wind speed of 10 m s^{-1} and a fetch of 100 km. The base of the vertical wall and the sea bed are respectively 3 m and 8 m below HWL.
Find the maximum force on the vertical wall and the bending moment about A.
As given below, fetch = 100 km and wind speed = 10 m s^{-1}.
JONSWAP spectrum (SEE SECTION 14.9.2)
f_{ m }=3.5 \times \frac{9.81}{10}\left(\frac{9.81 \times 100 \times 10^3}{10^2}\right)^{-0.33}= 0.165 Hz;
T_{ s }=0.777 / f_{ m }=4.7 \,s ;
\alpha=0.076\left(\frac{9.81 \times 100 \times 10^3}{10^2}\right)^{-0.22}= 0.01 ;
H_{ s }=\frac{0.552 \times 9.81}{\pi^2 \times 0.165^2} \sqrt{0.01}=2.01 \, m ;
L_0=\frac{9.81 \times 4.7^2}{2 \pi}=34.5 \,m .
For d / L_0=8 / 34.5=0.23, H / H_0=0.93 \text { and } L / L_0=0.93 (Fig. 14.8), H = 0.93 × 2.01 = 1.87 m and L = 0.93 × 34.5 = 32 m,
\frac{2 \pi d}{L}=\frac{2 \pi \times 8}{32}=1.57 .
p_{\max } on the bed is given by
\rho g\left[d+\frac{H}{\cosh (2 \pi d / L)}\right]=1030 \times 9.81\left(8+\frac{1.87}{2.51}\right)=88.4 \times 10^3 \, Nm ^{-2} .
h_0=\frac{\pi \times 1.87^2}{32} \operatorname{cotanh}(1.57)=0.37 \, mAssume linear distribution:
d+H+h_0=8+1.87+0.37=10.24 \,m .Take y positive downwards from SWL. The pressure on the base of the vertical wall, with the crest of clapotis at the wall, is
The force on the vertical face is
8.63 \times 10^3 \times \frac{5.24^2}{2}=118.5 \times 10^3 \, Nper unit length of the wall.
MOMENT ABOUT A
Because of the linear pressure distribution,
M_{ A }=0.5 \times 8.63 \times 10^3 \times 5.24^2 \times \frac{5.24}{3}= 206.9 Nm
per unit length of the vertical wall.