1. Find: sc 0.7% 1.1% 1.5% 1.9% d b ft3 Q=210 s b=12 [ft] n=0.025
Find the critical slope, s sub c. [pause] In this problem, ---
b=12 [ft]
n=0.025

2. Find: sc 0.7% 1.1% 1.5% 1.9% d b ft3 Q=210 s b=12 [ft] n=0.025
a rectangular channel flows at a steady 210 cubic feet per second, ---
b=12 [ft]
n=0.025

3. Find: sc 0.7% 1.1% 1.5% 1.9% d b ft3 Q=210 s b=12 [ft] n=0.025
and has a known base width and roughness coefficient. [pause] To find the critical slope, ---
b=12 [ft]
n=0.025

4. Find: sc Q= A * R2/3 * S 1/2 1.49 n d b ft3 Q=210 s b=12 [ft] n=0.025
we’ll start with Manning’s equation, solve for the slope, ---
b=12 [ft]
n=0.025

5. Find: sc Q= A * R2/3 * S 1/2 2 S= 1.49 * A * R2/3 1.49 n Q * n d b ft3
[pause] Then we’ll plug in the values for ---
b=12 [ft]
n=0.025

6. Find: sc Q= A * R2/3 * S 1/2 2 S= 1.49 * A * R2/3 1.49 n Q * n d b
area [ft2]
hydraulic
radius [ft]
ft3 s
Q=210
flowrate, area, hydraulic radius and roughness, during critical flow conditions. [pause] The problem statement ---
ft3
flowrate
roughness s
b=12 [ft]
n=0.025

7. Find: sc Q= A * R2/3 * S 1/2 2 S= 1.49 * A * R2/3 1.49 n Q * n d b
area [ft2]
hydraulic
radius [ft]
ft3 s
Q=210
provides the flowrate and roughness coefficient, ---
ft3
flowrate
roughness s
b=12 [ft]
n=0.025

8. Find: sc Q= A * R2/3 * S 1/2 2 S= 1.49 * A * R2/3 1.49 n Q * n d b
area [ft2]
hydraulic
radius [ft]
ft3 s
Q=210
so we’re left to determine the area and hydraulic radius. [pause] During critical flow, ---
ft3
flowrate
roughness s
b=12 [ft]
n=0.025

9. Find: sc vc FR=1= critical flow g*D d b ft3 Q=210 s b=12 [ft] n=0.025
the froude number equals 1, which equals the critical velocity divided by the square root of the gravitational acceleration times the hydraulic depth, ---
b=12 [ft]
n=0.025

10. Find: sc vc FR=1= critical T flow g*D A A d T b ft3 Q=210 s b=12 [ft]
where the hydraulic depth is defined as the cross-sectional area of the channel, A, divided by the top width of the channel, T.
b=12 [ft]
n=0.025

11. Find: sc vc FR=1= critical T flow A g*D A =dc d T b b ft3 Q=210 s
For a rectangular channel, the hydraulic depth equals the critical depth, lowercase d sub c.
b=12 [ft]
n=0.025

12. Find: sc vc FR=1= vc 1= critical T flow A g*D d g*dc A dc= b b ft3 s
Q=210
Next, we’ll solve for this critical depth which equals, ---
b=12 [ft]
n=0.025

13. Find: sc vc FR=1= vc FR=1= vc critical T flow A g*D d g*dc A dc= b b 2
ft3 s g
Q=210
the critical velocity, squared, divided by the gravitational acceleration. Then we’ll substitute the ---
b=12 [ft]
n=0.025

14. Find: sc vc FR=1= vc FR=1= vc critical T flow A g*D d g*dc A dc= b b 2
ft3 s g
Q=210
area divided by base width, in for the critical depth, and ---
b=12 [ft]
n=0.025

15. Find: sc vc FR=1= vc FR=1= vc critical T flow A g*D d g*dc A dc= b b 2Q
ft3 s
vc= g
Q=210
flowrate divided by area, in for velocity. [pause] If we solve this equation for the area, --- A
b=12 [ft]
n=0.025

16. Find: sc vc FR=1= vc FR=1= vc critical T flow A g*D d g*dc A dc= b b 2Q
ft3 s
vc= g
Q=210
we find the area equals cubed root of the base width times the critical flowrate squared, divided by the gravitational acceleration. A
1/3
b*Qc 2 A=
b=12 [ft] g
n=0.025

17. Find: sc vc FR=1= vc FR=1= vc critical T flow A g*D d g*dc A dc= b b 2Q
ft3
vc= g
Q=210
Plugging in the values for the base width, critical flowrate, and acceleration, --- A s
1/3
b*Qc 2 A=
b=12 [ft] ft g
32.2 s2
n=0.025

18. Find: sc vc FR=1= vc FR=1= vc critical T flow A g*D d g*dc A dc= b b 2Q
ft3
vc= g
Q=210
the area equals, squared feet. [pause] Returning to Manning’s equation, --- A s
1/3
b*Qc 2 A=
b=12 [ft] ft g
32.2 s2
n=0.025
A=25.42 [ft2]

19. Find: sc 2 S= 1.49 * A * R2/3 T Q * n d hydraulic radius [ft] b ft3
the last variable to solve for, is the hydraulic radius, R. The hydraulic radius equals --- s
b=12 [ft]
n=0.025
A=25.42 [ft2]

20. Find: sc 2 S= 1.49 * A * R2/3 A R= P T Q * n d hydraulic radius [ft] b
the cross-sectional area divided by the wetted perimeter, where the wetted perimeter equals --- s
b=12 [ft]
n=0.025
A=25.42 [ft2]

21. Find: sc 2 S= 1.49 * A * R2/3 A R= P P = b + 2*dc T Q * n d hydraulic
radius [ft] R= P b
P = b + 2*dc
ft3
Q=210
the base width, b, plus 2 times the critical depth, d sub c, where the critical depth equals, --- s
b=12 [ft]
n=0.025
A=25.42 [ft2]

22. Find: sc 2 S= 1.49 * A * R2/3 A R= P P = b + 2*dc dc= T Q * n d
hydraulic A
radius [ft] R= P b
P = b + 2*dc
ft3
Q=210
the critical area divided by the base width. [pause] Substituting in the critical area, --- Ac s
dc=
b=12 [ft] b
n=0.025
A=25.42 [ft2]

23. Find: sc 2 S= 1.49 * A * R2/3 A R= P P = b + 2*dc Ac R= dc= b+2 T
Q * n 2 S=
1.49 * A * R2/3 d
hydraulic A
radius [ft] R= P b
P = b + 2*dc
ft3 Ac
Q=210
and base width, we compute the hydraulic radius to be ---- Ac s R=
dc= Ac
b=12 [ft]
b+2 b * b
n=0.025
A=25.42 [ft2]

24. Find: sc 2 S= 1.49 * A * R2/3 A R= P P = b + 2*dc Ac R= dc= b+2T
Q * n 2 S=
1.49 * A * R2/3 d
hydraulic A
radius [ft] R= P b
P = b + 2*dc
ft3 Ac
Q=210
1.566 feet. [pause] Back to Manning’s equation, --- Ac s R=
dc= Ac
b=12 [ft]
b+2 b * b
n=0.025
R=1.566 [ft]
A=25.42 [ft2]

25. Find: sc 2 S= 1.49 * A * R2/3 T Q * n d b ft3 Q=210 s b=12 [ft]
we can now plug the appropriate values, s
b=12 [ft]
n=0.025
R=1.566 [ft]
A=25.42 [ft2]

26. Find: sc 2 S= 1.49 * A * R2/3 T Q * n d b ft3 Q=210 s b=12 [ft]
and the critical slope calculates to --- s
b=12 [ft]
n=0.025
R=1.566 [ft]
A=25.42 [ft2]

27. Find: sc S=0.0105 2 S= 1.49 * A * R2/3 T Q * n d b ft3 Q=210 s
0.0105, --- s
b=12 [ft]
n=0.025
R=1.566 [ft]
A=25.42 [ft2]

28. Find: sc S=0.0105 =1.05% 2 S= 1.49 * A * R2/3 T Q * n d b ft3 Q=210 s
or 1.05%. s
b=12 [ft]
n=0.025
R=1.566 [ft]
A=25.42 [ft2]

29. Find: sc S=0.0105 =1.05% 2 S= 1.49 * A * R2/3 0.7% 1.1% 1.5% 1.9% T
Q * n 2 S=
1.49 * A * R2/3
0.7%
1.1%
1.5%
1.9% d
S=0.0105
=1.05% b
ft3
Q=210
When reviewing the possible solutions, --- s
b=12 [ft]
n=0.025
R=1.566 [ft]
A=25.42 [ft2]

30. Find: sc AnswerB S=0.0105 =1.05% 2 S= 1.49 * A * R2/3 0.7% 1.1% 1.5%T
Q * n 2 S=
1.49 * A * R2/3
0.7%
1.1%
1.5%
1.9% d
S=0.0105
=1.05% b
AnswerB
ft3
Q=210
the answer is B. s
b=12 [ft]
n=0.025
R=1.566 [ft]
A=25.42 [ft2]

31. ? Index σ’v = Σ γ d γT=100 [lb/ft3] +γclay dclay 1
Find: σ’v at d = 30 feet
(1+wc)*γw
wc+(1/SG)
σ’v = Σ γ d d
Sand
10 ft
γT=100 [lb/ft3]
100 [lb/ft3]
10 [ft]
20 ft
Clay
= γsand dsand
+γclay dclay A W S
V [ft3]
W [lb]
40 ft
text
wc = 37% ? Δh
20 [ft]
(5 [cm])2 * π/4