Assuming angle of repose φ = 40, the stability number η is obtained from Equation 6.3 or

Figure 6.14 for θ = 20 (η = 0.36). The stone size is then calculated from Equation 6.13.

21τo

21 x 2.0

=

= 1.13 ft

Dm =

(S s - 1) γη (2.65 - 1) 62.4 x 0.36

(b) For the same design shear stress τo = 2.0 psf, determine the stability factor of particle sizes

Dm = 6 inches. From Equation 6.5

21τ o

21 x 2

η=

=

= 0.82

(S s - 1) γDm (2.65 - 1) 62.4 x 0.5

Then, Equations 6.11, 6.10, and 6.9 are used to calculate the stability factor

tan φ tan 40

Sm =

=

= 2.31

tan θ tan 20

ζ = Sm η sec θ = 2.31 x 0.82 sec 20 = 2.02

(

)

Sm

2.31

1/ 2

(ζ 2 + 4)

((2.02 2 + 4)1/ 2 - 2.02) = 0.95 < 1

S.F. =

- ζ) =

2

2

This size fraction (Dm = 6 inches) 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 = 12 fps; y = 10 ft; 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 Sm

0.3 V 2

0.3 x 12 2

=

=

= 1.25 ft

Dm

(S s - 1) g η 1.65 x 32.2 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

6.66

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