which satisfies requirement 6.24

D15 (Riprap)

6

=

= 16

D15 (Base)

0.38

which satisfies the requirement 6.25

D15 (Riprap)

6

=

=4

D85 (Base) 1.5

which satisfies the requirement 6.26.

The riprap itself satisfies the requirements for the filter so no filter is needed.

The following filter design is taken from Anderson et al. (1968). The properties of the base

material and the riprap are given in Table 6.5. Determine if a filter is needed.

Table 6.5. Sizes of Materials.

Base Material (Sand)

Riprap (Rock)

D85 = 1.5 mm

D85 = 400 mm

D50 = 0.5 mm

D50 = 200 mm

D15 = 0.17 mm

D15 = 100 mm

The riprap does not contain sufficient fines to act as the filter because

D15 (Riprap) 100

=

= 600

D15 (Base)

0.17

which is much greater than 40, the recommended upper limit (requirement 6.25). Also

D15 (Riprap) 100

=

= 67

D85 (Base)

1.5

which is much greater than 5, the recommended upper limit (requirement 6.26).

The properties of the filter to be placed adjacent to the base are as follows:

D50 (Filter )

< 40

(1)

D50 (Base)

D50 (Filter) < (40) (0.5) = 20 mm

so

6.62

Integrated Publishing, Inc. |