14.8 attached. 14.8 Determine the discharge through the following sections for a normal depth of 5 ft; n = 0.013,and S= 0.2%.a. A rectangular section 20 ft wide.b. A circular section 20 ft in diameter.c. A right-angled triangular section.d. A trapezoidal section with a bottom width of 20 ft and side slope of 1(vertical): 2(horizontal).e. A parabolic section having a top width of 20 ft for a 5-ft depth.8[Hint: A =Ty, P =T+=(y2 /T ), where Tis the top width and y is the depth.][Hint: A =-Ty, P, =T+3W331. in

Answered: 14.8 Determine the discharge through…

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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14.8 attached.

14.8 Determine the discharge through the following sections for a normal depth of 5 ft; n = 0.013,
and S= 0.2%.
a. A rectangular section 20 ft wide.
b. A circular section 20 ft in diameter.
c. A right-angled triangular section.
d. A trapezoidal section with a bottom width of 20 ft and side slope of 1(vertical): 2
(horizontal).
e. A parabolic section having a top width of 20 ft for a 5-ft depth.
8
[Hint: A =Ty, P =T+=(y2 /T ), where Tis the top width and y is the depth.]
[Hint: A =-Ty, P, =T+
3
W
3
31. in

Expert Solution

Step 1

$a) R e \tan g u l a r c h a n n e A = 20 * 5 = 100 f t^{2} P = 20 + 5 * 2 = 30 f t s o, Q = \frac{A^{5 / 3}}{P^{2 / 3}} . \frac{1}{n} . s^{1 / 2} * 1.49 Q = \frac{100^{5 / 3}}{30^{2 / 3}} . \frac{1}{0.013} * {(\frac{0.2}{100})}^{1 / 2} * 1.49 Q = 1143.763 f t^{3} / s$

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