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20. A trapezoidal canal, 6m at the bottom and having side slopes 2:1 carries water to a depth of 40cm and mean velocity of 0.60 m/s. what length of a rectangular contracted weir 85cm high should be placed in the middle of the canal if the depth of water upstream is to be 1.20m? Given: Find: L =? W.C. H d=1.20m P = 0.85 original w.s 2,4m 0.19 Solution: P 6m *aem- do = 0.40 m T

20. A trapezoidal canal, 6m at the bottom and having side slopes 2:1 carries water to a depth of 40cm and mean velocity of 0.60 m/s. what length of a rectangular contracted weir 85cm high should be placed in the middle of the canal if the depth of water upstream is to be 1.20m? Given: Find: L =? W.C. H d=1.20m P = 0.85 original w.s 2,4m 0.19 Solution: P 6m *aem- do = 0.40 m T

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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How did you come up with 7.6? I mean why you need to add 6+0.8+0.8? To have 7.6 and add this with 6 again?
20. A trapezoidal canal, 6m at the bottom and having side slopes 2:1 carries water to a depth of 40cm and mean velocity of 0.60 m/s. what length of a rectangular contracted weir 85cm high should be placed in the middle of the canal if the depth of water upstream is to be 1.20m? Given: Find: L =? H d=1.20m P = 0.85 do= 0.40 m T B 6m = ½ (6 + 7.6) (0.4) (0.6) H = 1-20 0.85 = 1.63m3/s Q = 1.84 L'H3/2 H = 6.35 1.63 = 1.84 L' (0.35) 3/2 4 L2H 72 L = 4.28 +0.20(0.35) L = 4.35m original W. S 2,4m Solution: L' = 4.28 m Qo = AoVo and 48m- contrations
Transcribed Image Text:20. A trapezoidal canal, 6m at the bottom and having side slopes 2:1 carries water to a depth of 40cm and mean velocity of 0.60 m/s. what length of a rectangular contracted weir 85cm high should be placed in the middle of the canal if the depth of water upstream is to be 1.20m? Given: Find: L =? H d=1.20m P = 0.85 do= 0.40 m T B 6m = ½ (6 + 7.6) (0.4) (0.6) H = 1-20 0.85 = 1.63m3/s Q = 1.84 L'H3/2 H = 6.35 1.63 = 1.84 L' (0.35) 3/2 4 L2H 72 L = 4.28 +0.20(0.35) L = 4.35m original W. S 2,4m Solution: L' = 4.28 m Qo = AoVo and 48m- contrations
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