The resulting power equation is:
v/0.738 = 1.37 (y/yo)0.24
Shear Stress Analysis
Using the above information calculate the shear stress on the bed using the several methods given
in Section 2.4.5.
(a) Average Shear Stress on the Bed
Calculate the average shear stress on the bed using Equation 2.97
τ0 = γ R Sf = 9800 x 2.38 x 0.000206 = 4.805 N/m2
Using R = yo (for a wide channel)
(b) Local Shear Stress on the Bed Using Single Point Velocity near the Bed
Using the above information and the logarithmic velocity distribution equation derived from Equation
2.75 (v/V* = 2.564 ln (y/0.009)) the shear stress on the bed using a single point velocity near the
bed is:
ρv2
1000 (0.518)2
= 5.11N / m 2
τ0 =
=
2
2
y
0.152
2.564 ln
2.564 ln
0.009
0.009
(c) Local Shear Stress on the Bed Using Two Point Velocities near the Bed
For this determination use Equation 2.99
(
)
(
)
2
2
ρ v1 - v 2
1000 0.633 - 0.518
= 4.37 N / m 2
τ0 =
=
2
2
y1
0.305
5.75 log
5.75 log
0.152
y2
(d) Shear Stress on the Bed Using Average Vertical Velocity
Using Equation 2.100 and the average depth 2.38 m, average velocity 0.738 m/sec, and a value
for ks determine the average shear stress on the bed.
Taking ks as (30.2) y' = 0.27 m
()
(
)
2
2
ρV
1000 0.738
= 3.98 N / m 2
τ0 =
=
2
2
12.27 y 0
12.27 x 2.38
5.75 log
5.75 log
0.27
ks
2.73
Previous Page
