(

)

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

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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