4) Two branches of different sections and discharges merge and continue to flow in a main channel with trapezoidal section as shown in the figure below. a) Find the discharge in branch A, b) Find the discharge in branch B, c) Find the depth of flow in the main channel. Branch A n=0.025 Branch B 1 SA=0.002 n=0.02 B YA 1.5 m b-25 m n=0.02 B Section A-A. SB-0.0016 n=0.015 ! ilk <->n=0.025 N=0.015- 50 m Tyв=1.15 m -C Section B-B. n=0.011- n=0.015 + 10m + n=0.011 ->n=0.015 50 m Main Channel |yc=? b=30 m Section C-C n=0.02 ▼ n=0.02 3 <->n=0.01 Sc=0.0009 1m

4) Two branches of different sections and discharges merge and continue to flow in a main channel with trapezoidal section as shown in the figure below. a) Find the discharge in branch A, b) Find the discharge in branch B, c) Find the depth of flow in the main channel. Branch A n=0.025 Branch B 1 SA=0.002 n=0.02 B YA 1.5 m b-25 m n=0.02 B Section A-A. SB-0.0016 n=0.015 ! ilk <->n=0.025 N=0.015- 50 m Tyв=1.15 m -C Section B-B. n=0.011- n=0.015 + 10m + n=0.011 ->n=0.015 50 m Main Channel |yc=? b=30 m Section C-C n=0.02 ▼ n=0.02 3 <->n=0.01 Sc=0.0009 1m

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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Concept explainers

Highway Drainage

Highway drainage is the process of extracting and regulating the excess surface water on the pavement by giving the desired slope. The construction of a highway drain helps to divert the water present on the surface, and sub-grade by the means of slope. Highway drainage systems are a necessary part of smooth functioning of traffic, since the highway drainage systems provide proper disposal of unwanted substances from the highways.

Question

Neglect head losses
4) Two branches of different sections and discharges merge and continue to flow in
a main channel with trapezoidal section as shown in the figure below.
a) Find the discharge in branch A,
b) Find the discharge in branch B,
c) Find the depth of flow in the main channel.
Branch A
Branch B
n=0.025-
1
SA=0.002
n=0.02
YA=1.5 m
B
B
b=25 m
Section A-A.
n=0.02
t
SB=0.0016
n=0.015
!
1
R=0.015
50 m
n=0.025
YB 1.15 m
n=0.015
10 m
t
Section B-B.
C
EC
C
n=0.011
+
<->n=0.015
*****
n=0.011
50 m
Main Channel
|yc=?
b=30 m
Section C-C.
n=0.02
n=0.02
Sc=0.0009
1m
n=0.011
3-92
6)
sec

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