1/ 2
1/ 2
Sin 2 θ
Sin 2 20
K 1 = 1 -
= 1 -
= 0.85
Sin 2 φ
Sin 2 40
2.5
62.4 0.5
12
= 1.1x 0.3 x 1.0 x 1.0 x 10
= 0.79 ft
D30
0.5
102.96 0.85 x 32.2 x 10)
D50 = D30 x (D85 / D15 )1/ 3 = 0.79 x (2)1/ 3 = 1.00 ft
6.12.3 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.
The reference velocity Vr = 6 fps, and the embankment side slope angle θ = 18.4 which
corresponds to a 3:1 side slope. The velocity vector angle with the horizontal λ = 20. If the
embankment is covered with dumped rock having a specific weight Ss = 2.65 and an effective
rock size Dm = 1.0 ft, determine the stability factor.
From Equation 6.16
(
)
0.30 Vr2
(0.30) (6)2
η=
=
= 0.203
(S s - 1) gDm (2.65 - 1) (32.2) (1.0)
This dumped rock has an angle of repose of approximately 35 according to Figure 3.4.
Therefore, from Equation 6.4.
cos λ
cos 20
-1
-1
β = tan
= tan
= 11
2 sin θ + sin λ
2 sin 18.4
+ sin 20
η tan φ
0.203 tan 35
and from Equation 6.6
1 + sin (λ + β)
1 + sin (20 + 11)
η′ = η
= 0.154
= 0.203
2
2
The stability factor for the rock is given by Equation 6.3
6.67
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