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.
6.11.2 PROBLEM 2 Filter Design
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
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