What is A = Asector - Atriangle? Where did they get that?
Principles of Foundation Engineering (MindTap Course List)
8th Edition
ISBN:9781305081550
Author:Braja M. Das
Publisher:Braja M. Das
Chapter7: Settlement Of Shallow Foundations
Section: Chapter Questions
Problem 7.2P: A planned flexible load area (see Figure P7.2) is to be 3 m × 4.6 m and carries a uniformly...
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Question
What is A = Asector - Atriangle? Where did they get that?
Transcribed Image Text:FLUID MECH
CHAPTER EIGHT
Open Channel
FLU
GHYDRAN
510
A = 2 r? (0-sin 0) = 0.03125 (0- sin 0)
Tre
0.25 0
180°
180°
From Eq. (1)
0.03125(0 -sin
sin a)/.
180
= 0.00125
0.25 780 0
(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
![d 1.59 m
d = 5.22 feet
Problem 8-12
e00-mm-diameter concrete pipe is laid on a slope of 1 m per 500 m and is
Solution
IQ = A !
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%2F34c80954-dc4a-4bb0-8fac-8d195efd5bc1%2Facouh1_processed.jpeg&w=3840&q=75)
Transcribed Image Text:d 1.59 m
d = 5.22 feet
Problem 8-12
e00-mm-diameter concrete pipe is laid on a slope of 1 m per 500 m and is
Solution
IQ = A !
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
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