In a deposit of normally consolidated dry sand, a cone penetration test was conducted. Following are the results: Depth (m) Point resistance of cone, q. (MN/m²) 1.5 2.06 3.0 4.23 4.5 6.01 6.0 8.18 7.5 9.97 9.0 12.42 Assuming the dry unit weight of sand to be 16 kN/m², estimate the average peak friction angle, ø', of the sand. Use Eq. (3.53). 1 Pa D, = 0.5 305Q.OCR!8 Pa
In a deposit of normally consolidated dry sand, a cone penetration test was conducted. Following are the results: Depth (m) Point resistance of cone, q. (MN/m²) 1.5 2.06 3.0 4.23 4.5 6.01 6.0 8.18 7.5 9.97 9.0 12.42 Assuming the dry unit weight of sand to be 16 kN/m², estimate the average peak friction angle, ø', of the sand. Use Eq. (3.53). 1 Pa D, = 0.5 305Q.OCR!8 Pa
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Question
Refer to Problem 3.19. Using Eq. (3.51), determine the variation of the relative density with depth. Use Qc = 1.
![In a deposit of normally consolidated dry sand, a cone penetration test was
conducted. Following are the results:
Depth
(m)
Point resistance of
cone, q. (MN/m²)
1.5
2.06
3.0
4.23
4.5
6.01
6.0
8.18
7.5
9.97
9.0
12.42
Assuming the dry unit weight of sand to be 16 kN/m², estimate the average peak
friction angle, ø', of the sand. Use Eq. (3.53).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F71c927ea-e574-4cab-83ed-fe543f0ac29d%2F359324d9-d7d4-4980-852a-fd6624aec083%2F57ie1g_processed.png&w=3840&q=75)
Transcribed Image Text:In a deposit of normally consolidated dry sand, a cone penetration test was
conducted. Following are the results:
Depth
(m)
Point resistance of
cone, q. (MN/m²)
1.5
2.06
3.0
4.23
4.5
6.01
6.0
8.18
7.5
9.97
9.0
12.42
Assuming the dry unit weight of sand to be 16 kN/m², estimate the average peak
friction angle, ø', of the sand. Use Eq. (3.53).
![1
Pa
D, =
0.5
305Q.OCR!8
Pa](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F71c927ea-e574-4cab-83ed-fe543f0ac29d%2F359324d9-d7d4-4980-852a-fd6624aec083%2Fnkb03_processed.png&w=3840&q=75)
Transcribed Image Text:1
Pa
D, =
0.5
305Q.OCR!8
Pa
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