6.10.4 PROBLEM 4 Riprap Design on an Abutment
o
(a) Consider a spill through abutment with side slope θ = 18.5 (3:1 side slope). The flow
o
m/s and angle with the horizontal λ = 20 . The specific gravity of the available rock is 2.65.
Determine the size of riprap in this area required to resist the erosive force of water.
The spill slope riprap rock size is obtained by iteration from assuming Dm and calculating
successively Equations 6.16, 6.4, 6.6, and 6.3 until the stability factor, S.F. equals 1.5. As a
first approximation, a stone size of 0.091 m is used. From Figure 3.4, φ ≅ 40.
From Equation 6.16,
0.3 x 1.83 2
η=
= 0.68
(2.65 - 1) x 9.81x .091
From Equation 6.4,
cos 20
-1
β = tan
= 32.8
2 sin 18.5
+ sin 20
0.68 tan 40
From Equation 6.6,
1 + sin (20 + 32.8)
η′ = 0.68
= 0.61
2
From Equation 6.3,
cos 18.5 tan 40
S.F. =
= 1.02
0.61 tan 40 + sin 18.5 cos 32.8
The procedure is repeated with increasing stone size until S.F. = 1.5. A riprap size of 0.23 m
has a S.F. slightly over 1.5.
m/s and an angle with the horizontal λ of 0 degrees. Determine the size of riprap to protect
the toe of the abutment.
The riprap size is determined either by the same iterative procedure used in (a) or by using
Equations 6.11, 6.12, and 6.16.
Using the latter method:
Equation 6.11
6.60
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