(
)
Sm
2.31
1/ 2
(ζ 2 + 4)
((2.04 2 + 4)1/ 2 - 2.04) = 0.94 < 1
S.F. =
- ζ) =
2
2
This size fraction (Dm = 0.15 m) is unstable.
(c) Determine riprap size Dm for a side slope. The side slope angle θ = 20; very angular rock
with angle of repose φ = 40; Vss = 3.66 m/s; y = 3.05 m; S.F. = 1.1 and D85 / D15 = 2.0. From
Equations 6.11, 6.12, and 6.16:
tan φ tan 40
Sm =
=
= 2.31
tan θ tan 20
Sm - (S.F.)2
2
2.312 - 1.12
η=
cos θ =
cos 20 = 0.66
2
1.1x 2.312
(S.F.) x S m
0.3 V 2
0.3 x 3.66 2
=
=
= 0.38 m
Dm
(S s - 1) g η 1.65 x 9.81 x 0.66
(d) Compare the size calculated in (c) with a riprap size calculated using the U.S. Army Corps
of Engineers equation (Section 6.5.4). From Equations 6.18 and 6.20:
2.5
γ 0.5
V
= S f C s C v C T y
γ - γ (K g y)0.5
D30
s
1
1/ 2
1/ 2
Sin 2 θ
Sin 2 20
K 1 = 1 -
= 1 -
= 0.85
Sin 2 40
Sin 2 φ
2.5
9800 0.5
3.66
= 1.1x 0.3 x 1.0 x 1.0 x 3.05
= 0.24 m
D30
0.5
16,175 0.85 x 9.81x 3.05)
D50 = D30 x (D 85 \ D15 )1/ 3 = 0.24 x (2)1/ 3 = 0.30 m
6.10.3 PROBLEM 3 Stability Factors for Riprap Design
For flow around spill through abutments the angle between the horizontal and the velocity
vector can be large (Figure 6.13). The draw down as the flow goes around the upstream
end of the abutment can be very large. The draw down angle can range from 0 to 35
degrees and the reference velocity Vr in the vicinity of the riprap can be very large (Lewis
1972, Richardson et al. 1975, 1990). The following problem addresses the design of riprap
for the protection of the spill through embankment.
6.57
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