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 *t*f. Substitution of Equation 2-29 for *t*f into

Equation 2-28 yields:

' [ exp (&*u * 2) & π *u erfc *(*u*) ]

where

tan(αb )

'

2 ε *t*f

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:

π

π

' 19.4

' 0.96

'*u*

tan(αb )

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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