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

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

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

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

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

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

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

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

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

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

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

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

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

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

Find: sc vc FR=1= vc FR=1= vc critical T flow A g*D d g*dc A dc= b b 2 Q 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

Find: sc vc FR=1= vc FR=1= vc critical T flow A g*D d g*dc A dc= b b 2 Q 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

Find: sc vc FR=1= vc FR=1= vc critical T flow A g*D d g*dc A dc= b b 2 Q 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

Find: sc vc FR=1= vc FR=1= vc critical T flow A g*D d g*dc A dc= b b 2 Q ft3 vc= g Q=210 the area equals, 25.42 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]

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]

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]

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]

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]

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]

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 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]

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]

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]

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]

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]

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]

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]

? 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