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

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

Integrated Publishing, Inc. |