EM 1110-2-1100 (Part III)
30 Apr 02
EXAMPLE PROBLEM III-2-10
FIND:
The distance downdrift of a structure at which the shoreline recession is less than or equal to
10 percent of the structure's length before sand begins to naturally bypass the structure.
GIVEN:
A long groin extending a few surf zone widths has been built with no artificial sand fill on
either side. The groin's length, measured from the original shoreline, is Y. Assume that the wave
activity is continuous with breaking angle αb = 5 deg.
SOLUTION:
The structure becomes filled to capacity at time tf. Substitution of Equation 2-29 for tf into
Equation 2-28 yields:
y
' [ exp (&u 2) & π u erfc (u) ]
Y
where
x
x 1
u'
tan(αb )
'
Y π
2 ε tf
Determine the value u (graphically or by iteration), for which y/Y = 0.10. Find u 0.96.
Determine the downdrift distance x (relative to the structure's length Y) using this value for u and
using αb = 5 deg:
x
π
π
' 19.4
' 0.96
'u
tan(αb )
Y
0
tan (5 )
That is, the shoreline recession is equal to or less than 10 percent of the structure's length beyond
approximately 19.4 structure-lengths downdrift. If, for instance, the structure's length was Y =
200 m, the downdrift location at which the shoreline recession is less than 20 m (at the time the
structure is filled to capacity) is (19.4)(200) = 3,880 m downdrift of the structure.
III-2-66
Longshore Sediment Transport
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