Y = 15.7 kN/m³ d' = 32° c' = 0 3 m Groundwater table . Ysat = 18.2 kN/m³ 4' = 32° c' = 0 3 m = 19.2 kN/m3 Ysat o' = 40° c' = 0 15 m 381 mm Figure P9.4 f = f;=L' (9.42)
Y = 15.7 kN/m³ d' = 32° c' = 0 3 m Groundwater table . Ysat = 18.2 kN/m³ 4' = 32° c' = 0 3 m = 19.2 kN/m3 Ysat o' = 40° c' = 0 15 m 381 mm Figure P9.4 f = f;=L' (9.42)
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
A driven closed-ended pile, circular in cross section, is shown in Figure P 9.4.
Calculate the following.
a. The ultimate point load using Meyerhof’s procedure.
b. The ultimate point load using Vesic’s procedure. Take Irr = 50.
c. An approximate ultimate point load on the basis of parts (a) and (b).
d. The ultimate frictional resistance Qs. [Use Eqs. (9.40 (L' ≈ 15 D)) through (9.42), and take K = 1.4 and ẟ' = 0.6 Φ'.]
e. The allowable load of the pile (use FS = 4).
![Y = 15.7 kN/m³
d' = 32°
c' = 0
3 m
Groundwater
table .
Ysat = 18.2 kN/m³
4' = 32°
c' = 0
3 m
= 19.2 kN/m3
Ysat
o' = 40°
c' = 0
15 m
381 mm
Figure P9.4](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fde2410ed-e82e-4041-abed-831ba768d43e%2F946603c4-032e-4096-a624-f36b95c17075%2Fkbyaon_processed.png&w=3840&q=75)
Transcribed Image Text:Y = 15.7 kN/m³
d' = 32°
c' = 0
3 m
Groundwater
table .
Ysat = 18.2 kN/m³
4' = 32°
c' = 0
3 m
= 19.2 kN/m3
Ysat
o' = 40°
c' = 0
15 m
381 mm
Figure P9.4
![f = f;=L'
(9.42)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fde2410ed-e82e-4041-abed-831ba768d43e%2F946603c4-032e-4096-a624-f36b95c17075%2F8j6cow_processed.png&w=3840&q=75)
Transcribed Image Text:f = f;=L'
(9.42)
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