n = 0.0417 D1/ 6 with D50 in meters n = 0.0417 (0.19 m)
1/ 6
= 0.032
(3.15)
n = 0.0482 D1/06 with D in meters n = 0.0482 (0.19 m)
1/ 6
= 0.037
(3.16)
5
n = 0.0473 D1/56 with D 75 in meters n = 0.0473 (0.23 m)
1/ 6
= 0.037
(3.17)
7
n = 0.046 D1/06 with D 90 in meters n = 0.046 (0.28 m)
1/ 6
= 0.037
(3.18)
9
3.7.4 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
=
1.22 m
S
=
0.00038 m/m
=
0.31mm
D50
D16
=
0.24mm
D90
=
0.46mm
=
0.42mm
D84
The shear stress on the bed, at a single vertical in the cross-section, is 9800 N/m3 x 1.22 m x
0.00038 = 4.543 N/m2
τo/(γs - γ) Ds = 4.543 / (25970-9800) 0.00031 = 0.91
V* Ds/ν = (4.543/1000)1/2 (0.00031)/1.31 x 10-6 = 15.95
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 = 4.543 N/m2; Ds = 3.1 mm we
have:
τo/(γs - γ) Ds = 4.543 /(25970-9800) (0.0031) = 0.091
V* Ds/ ν= (4.543/1000)1/2 (0.0031)/1.31 x 10-6 = 159
This point plots above the incipient motion line in Shields (Figure 3.15) and indicates this
channel bed is in motion.
3.59
Previous Page
