EXAMPLE PROBLEM V-4-1

FIND:

The overfill factor, *R*A.

GIVEN:

The native and borrow area phi values are:

native beach:

φ05=0.95, φ16=1.31, φ50=1.91, φ84=2.66, φ95=2.90

φ05=1.42, φ16=1.63, φ50=2.49, φ84=3.08, φ95=3.55

borrow area:

The median grain diameter for native and borrow material is 0.27 and 0.18 mm, respectively.

SOLUTION:

The mean sediment diameter in phi units is given in Part III-1 as

(III-1-2)

The standard deviation in phi units is given in Part III-1 as

σφ = (φ84 - φ16) / 4 + (φ95 - φ05) / 6

(III-1-3)

σφn = (2.66 - 1.31) / 4 + (2.90 - 0.95) / 6 = 0.66

σφb = (3.08 - 1.63) / 4 + (3.55 - 1.42) / 6 = 0.72

The sorting ratio from Equation V-4-3 is

σφb / σφn = 0.72 / 0.66 = 1.09

The phi mean differences ratio from Equation V-4-4 is

(*M*φb - *M*φn) / σφn = (2.40 - 1.96) / 0.66 = 0.67

Entering Figure V-4-9 with x = 0.67 and y = 1.09 results in an overfill ratio (*R*A) equal to 2.5.

The finer borrow material may be suitable for use, but it is quite incompatible with the native beach sand.

The value of the overfill ratio suggests that 2.5 units of borrow material will be required to create 1.0

unit of stable native beach material.

Beach Fill Design

V-4-27

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