8. Design of Coastal Structures
271
Problem IV
A vertical wall 3 m high in a water depth of ds = 2.5 m on a near shore
slope of 1:20 (m = 0.05), Hb — 2.8 m for 6 sec wave, Hb = 3.2 m for
10 sec wave. Find the reduced force and overturning moment because of
the reduced wall height.
Solution
For T - 6 sec, Hb = 2.8 m
From previous problem,
Rm = 309 kN/m,
and for T — 10 sec, Hb = 3.2
Rm = 194 kN/m,
Mm = 772 kN-m/m,
Mm = 485 kN-m/m.
For the breaker with T = 6 sec, the height of the breaker crest above the
bottom is
(d„ + y) = (2-5 +
= 3'9 “
The value of b' as defined in Fig. 8.13 is 1.9 m, b' is obtained as (Hb minus
the height obtained by subtracting the wall crest élévation from the breaker
crest élévation).
i.e. h' = Hb - \(d3 Hb
2
— wall élévation = 2.8- [3.9-3] = 1.9 m.
b'
1.9
— = —— = 0.679, from Fig. 8.13, rm = 0.83
Hb
2.8
R'm = rm Rm = 0.83(309) = 256 kN/m
b'
From Fig. 8.14 with —— — 0.679
Hb
2a/Hb = 0.57
Hence,
w
0.57(2.8)
Hence a =----- -—2
= 0.8 m
— Rm['ï'm(ds “F ®)
M'm = 309[0.83(2.5 + 0.8) - 0.8]
M'm = 600 kN-m /m.
For T = 10 sec, rm = 0.79, a = 0.86
R'm = 153 kN/m, M'm = 348 kN-m/m.