d 1.59 m d = 5.22 feet Problem 8-12 0-mm-diameter concrete pipe is laid on a slope of 1 m per 500 m and is Solution 1Q = A- R2/351/2] 0.04 = A x Js R2/3 (/2 AR2/3 = 0.011627 A(A/P)2/3 = 0.001627 A5/3/P2/3 = 0.001627 A5/2/P 0.00125 %3D raise both sides to 3/2 %3D > Eq. (1) %3D 0.5 m 0.25 d r 0.25 m 0,25 m From the figure: A= Asector-Atriangle A = 360° -1/2 r2 sin 0
d 1.59 m d = 5.22 feet Problem 8-12 0-mm-diameter concrete pipe is laid on a slope of 1 m per 500 m and is Solution 1Q = A- R2/351/2] 0.04 = A x Js R2/3 (/2 AR2/3 = 0.011627 A(A/P)2/3 = 0.001627 A5/3/P2/3 = 0.001627 A5/2/P 0.00125 %3D raise both sides to 3/2 %3D > Eq. (1) %3D 0.5 m 0.25 d r 0.25 m 0,25 m From the figure: A= Asector-Atriangle A = 360° -1/2 r2 sin 0
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
Why sine was used in the figure
![d 1.59 m
d = 5.22 feet
Problem 8-12
00-mm-diameter concrete pipe is laid on a slope of 1 m per 500 m and is
Solution
IQ = A 2
R2/351/2]
0.04 = A x ds R2/3 (/2
500
AR2/3 = 0.011627
A(A/P)2/3 = 0.001627
A5/3/P2/3 = 0.001627
A5/2/P = 0.00125
%3D
raise both sides to 3/2
> Eq. (1)
%3D
0.5 m
0.25 d
r 0.25 m
0,25 m
From the figure:
A= Agector-Ariangle
A =
360°
- 2 12 sin 0](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F65c63a93-bfcc-4899-9a38-0dc7cdea4ea3%2Fbcaf257a-d74c-478e-8a8e-3ccf0e949ce4%2Fx5uce8s_processed.jpeg&w=3840&q=75)
Transcribed Image Text:d 1.59 m
d = 5.22 feet
Problem 8-12
00-mm-diameter concrete pipe is laid on a slope of 1 m per 500 m and is
Solution
IQ = A 2
R2/351/2]
0.04 = A x ds R2/3 (/2
500
AR2/3 = 0.011627
A(A/P)2/3 = 0.001627
A5/3/P2/3 = 0.001627
A5/2/P = 0.00125
%3D
raise both sides to 3/2
> Eq. (1)
%3D
0.5 m
0.25 d
r 0.25 m
0,25 m
From the figure:
A= Agector-Ariangle
A =
360°
- 2 12 sin 0
Transcribed Image Text:FLUID MECH
510
CHAPTER EIGHT
Open Channel
FLU
GHYDRA
A = 2 r? (0 7 - sin 0) = 0.03125 (0– sin 0)
Tre
0.25 0
180°
180°
From Eq. (1)
0.03125(0 -sin
5/2.
180
= 0.00125
0.25 780 9
(0-sin of/2
180°
= 0.03159
Solve for 0 by trial and error:
0= 140.46°
Then;
cos (6/2) = (0.25 - d)/0.25
cos (140.46/2)(0.25) = 0.25 - d
d 0.1654 m
Problem 8-13 (CE Board November 1988)
on one
over
large
Solution
fi
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