n = 0.388 D1/56 with D 75 in ft. n = 0.0388 (0.75 ft)1/ 6 = 0.037
(3.17)
7
n = 0.038 D1/06 with D 75 in ft. n = 0.038 (0.92 ft)1/ 6 = 0.037
(3.18)
9
3.8.4 PROBLEM 4 Beginning of Motion
Bed Material Movement Using Shields Figure
To establish if the bed material of a channel is in motion the Shields' relationship can be
used. Recall that τo = γRS and V* = τo / ρ .
(a) For the following conditions determine if the bed material in a sand bed channel is in
motion.
R
=
4.0 ft
S
=
0.00038 ft/ft
=
0.31mm = 0.00102 ft
D50
=
0.24mm = 0.0008 ft
D16
D90
=
0.46mm = 0.0015 ft
=
0.42mm = 0.0014 ft
D84
The shear stress on the bed, at a single vertical in the cross-section, is 62.4 lb/ft3 x 4.0 ft x
0.00038 = 0.0948 lb/ft2.
τo/(γs - γ) Ds = 0.095/ (165 - 62.4) (0.00102) = 0.92
V* Ds/ν = (0.095/1.94)1/2 (.001)/ 1.41 x 10-5 = 15.69
This point plots above the incipient motion line in Shields (Figure 3.15) and indicates the bed of
the channel is in motion.
(b) For the gravel sized material with the same values of τo = 0.095 lb/ft2; Ds = 3.1 mm =
0.0102 ft we have:
τo/(γs - γ) Ds = 0.095/ (165 - 62.4) (0.0102) = 0.091
V* Ds/ ν= (0.095/1.94)1/2 (0.0102)/ 1.41 x 10-5 = 160
This point plots above the incipient motion line in Shields (Figure 3.15) and indicates this
channel bed is in motion.
Critical Velocity for Beginning of Bed Material Movement
(a) Sand Size Bed Material
Given depth y is 12 ft and bed material size D50 is 0.31 mm = 0.00102 ft, what is the critical
velocity Vc?
Vc = 11.25 y1/ 6 D1/ 3 = 11.25 x 121/ 6 x 0.001021/ 3 = 1.72 ft / s
3.64
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