A closed-end steel pipe pile is driven into thick saturated clay strata. The diameter and length of the pile are 0.4 m and 32 m, respectively. The saturated unit weight and undrained shear strength of the clay are given in the figure. Estimate the side resistance Qs with (a) a method (5%) and (b) λ method (5%), and estimate the tip resistance Qp. 5 m 8 m 24 m Ysat 18.2 kN/m³ Su = 80 kPa Y sat S₁ 3 = 18.7 kN/m = 110 kPa
A closed-end steel pipe pile is driven into thick saturated clay strata. The diameter and length of the pile are 0.4 m and 32 m, respectively. The saturated unit weight and undrained shear strength of the clay are given in the figure. Estimate the side resistance Qs with (a) a method (5%) and (b) λ method (5%), and estimate the tip resistance Qp. 5 m 8 m 24 m Ysat 18.2 kN/m³ Su = 80 kPa Y sat S₁ 3 = 18.7 kN/m = 110 kPa
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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![A closed-end steel pipe pile is driven into thick saturated clay strata. The diameter and length of the pile are 0.4 m
and 32 m, respectively. The saturated unit weight and undrained shear strength of the clay are given in the figure.
Estimate the side resistance Qs with (a) a method (5%) and (b) λ method (5%), and estimate the tip resistance Qp.
5 m
YAV
8 m
24 m
Ysat
18.2 kN/m³
Su = 80 kPa
3
sat 18.7 kN/m
110 kPa
S
U](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F14edd9a8-fad3-4b3f-b234-2144164e926a%2Fb6c666e9-282f-4db7-b423-d59dcce010bd%2F5bjf6n4_processed.jpeg&w=3840&q=75)
Transcribed Image Text:A closed-end steel pipe pile is driven into thick saturated clay strata. The diameter and length of the pile are 0.4 m
and 32 m, respectively. The saturated unit weight and undrained shear strength of the clay are given in the figure.
Estimate the side resistance Qs with (a) a method (5%) and (b) λ method (5%), and estimate the tip resistance Qp.
5 m
YAV
8 m
24 m
Ysat
18.2 kN/m³
Su = 80 kPa
3
sat 18.7 kN/m
110 kPa
S
U
![9net(u) = quq = 5.14c1+ (1+0.4²)
General bearing capacity equation qu= c'NcFesFcaFci + q'NaFas FaaFai +0.5yBNyFysFya Fyi
Shape factors by De Depth factors by Hansen (1970)
Beer (1970)
B No
Fcs = 1+ N
Fas = 1 +
Fys
0
1
2
3
4
5
6
7
8
9
10
B
= 10.4(
Ne
5.14
5.38
5.63
5.90
6.19
6.49
6.81
7.16
7.53
7.92
8.35
tan o'
0.195B
L
TABLE 6.2 Bearing Capacity Factors From Eqs. (6.30), (6.29), and (6.31)
N₁
Na
1.00
2.71
1.09
2.97
1.20
3.26
1.31
3.59
1.43
3.94
1.57
4.34
1.72
1.88
2.06
2.25
2.47
Ne
16.88
18.05
19.32
20.72
22.25
23.94
25.80
27.86
30.14
32.67
35.49
38.64
42.16
46.12
50.59
Fad = 1 + 2 tano (1 — sin y)²:
qd
B
Fyd = 1
Fcd = 1 + 0.4(
Ny
0.00
0.07
0.15
0.24
0.34
0.45
0.57
0.71
0.86
1.03
1.22
7.13
8.20
9.44
10.88
12.54
14.47
16.72
19.34
22.40
25.99
30.22
35.19
41.06
48.03
56.31
=
11
12
13
14
15
16
17
18
19
20
21
TABLE 6.2 Bearing Capacity Factors From Eqs. (6.30), (6.29), and (6.31) (Continued)
φ'
Na
$'
Na
22
7.82
42.92
23
8.66
48.93
24
9.60
55.96
25
10.66
64.20
26
11.85
73.90
27
13.20
85.38
28
14.72
99.02
29
16.44
115.31
30
18.40
134.88
31
20.63
158.51
32
23.18
187.21
33
26.09
222.31
34
29.44
265.51
35
33.30
319.07
36
37.75
37
38
39
40
41
42
43
44
45
46
47
48
49
50
Ne
8.80
9.28
9.81
10.37
10.98
11.63
12.34
13.10
13.93
14.83
15.82
For saturated clay: p
For a method: fav = acu
For method: f = (₁ +2c₂)
Ne
55.63
61.35
67.87
75.31
83.86
93.71
105.11
118.37
133.88
152.10
173.64
199.26
229.93
266.89
Apqp = ApCu Nc
4.77
5.26
5.80
6.40
7.07
Ny
1.44
1.69
1.97
2.29
2.65
3.06
3.53
4.07
4.68
5.39
6.20
(continued)
Ny
66.19
78.03
92.25
109.41
130.22
155.55
186.54
224.64
271.76
330.35
403.67
496.01
613.16
762.89
Inclination
factors
by
Meyerhof (1963) and Hanna
and Meyerhof (1981)
Fci = Fai = (1-2
Fyi = (1 - B²
Bº
Ne 9 for p = 0
TABLE 12.10 Variation of A with Pile
Embedment Length, L
Embedment
length, L (m)
0
5
10
15
20
25
30
35
40
50
60
70
80
90
Coyle and Castello (1981):
Embedment ratio, L/D
0
10
20
30
40
50
60
70
Qp = q'N₁ Ap
10
T
T
T
T
T
e
$'
Bearing capacity factor, N
20
32° 36°
= 30°
34°
λ
0.5
0.336
0.245
0.200
0.173
0.150
0.136
0.132
0.127
0.118
0.113
0.110
0.110
0.110
38°
TABLE 12.11 Variation of a (Interpo-
lated Values Based on
Terzaghi et al., 1996)
40°
Cu
Pa
≤0.1
0.2
0.3
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
2.4
2.8
40 60 80 100 200
Note:
Pa
≈100 kN/m²
Qs
= atmospheric pressure
Embedment ratio, L/D
=
: (Ko'tan 8')pL
0.15 0.2
0
5
10
15
20
25
30
a
35
36
1.00
0.92
0.82
0.74
0.62
0.54
0.48
0.42
0.40
0.38
0.36
0.35
0.34
0.34
Earth pressure coefficient, K
1.0
$'
30°
31°
32°
33°
8=0.80'
2
34°
35⁰
36°
5](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F14edd9a8-fad3-4b3f-b234-2144164e926a%2Fb6c666e9-282f-4db7-b423-d59dcce010bd%2Fo795n34_processed.jpeg&w=3840&q=75)
Transcribed Image Text:9net(u) = quq = 5.14c1+ (1+0.4²)
General bearing capacity equation qu= c'NcFesFcaFci + q'NaFas FaaFai +0.5yBNyFysFya Fyi
Shape factors by De Depth factors by Hansen (1970)
Beer (1970)
B No
Fcs = 1+ N
Fas = 1 +
Fys
0
1
2
3
4
5
6
7
8
9
10
B
= 10.4(
Ne
5.14
5.38
5.63
5.90
6.19
6.49
6.81
7.16
7.53
7.92
8.35
tan o'
0.195B
L
TABLE 6.2 Bearing Capacity Factors From Eqs. (6.30), (6.29), and (6.31)
N₁
Na
1.00
2.71
1.09
2.97
1.20
3.26
1.31
3.59
1.43
3.94
1.57
4.34
1.72
1.88
2.06
2.25
2.47
Ne
16.88
18.05
19.32
20.72
22.25
23.94
25.80
27.86
30.14
32.67
35.49
38.64
42.16
46.12
50.59
Fad = 1 + 2 tano (1 — sin y)²:
qd
B
Fyd = 1
Fcd = 1 + 0.4(
Ny
0.00
0.07
0.15
0.24
0.34
0.45
0.57
0.71
0.86
1.03
1.22
7.13
8.20
9.44
10.88
12.54
14.47
16.72
19.34
22.40
25.99
30.22
35.19
41.06
48.03
56.31
=
11
12
13
14
15
16
17
18
19
20
21
TABLE 6.2 Bearing Capacity Factors From Eqs. (6.30), (6.29), and (6.31) (Continued)
φ'
Na
$'
Na
22
7.82
42.92
23
8.66
48.93
24
9.60
55.96
25
10.66
64.20
26
11.85
73.90
27
13.20
85.38
28
14.72
99.02
29
16.44
115.31
30
18.40
134.88
31
20.63
158.51
32
23.18
187.21
33
26.09
222.31
34
29.44
265.51
35
33.30
319.07
36
37.75
37
38
39
40
41
42
43
44
45
46
47
48
49
50
Ne
8.80
9.28
9.81
10.37
10.98
11.63
12.34
13.10
13.93
14.83
15.82
For saturated clay: p
For a method: fav = acu
For method: f = (₁ +2c₂)
Ne
55.63
61.35
67.87
75.31
83.86
93.71
105.11
118.37
133.88
152.10
173.64
199.26
229.93
266.89
Apqp = ApCu Nc
4.77
5.26
5.80
6.40
7.07
Ny
1.44
1.69
1.97
2.29
2.65
3.06
3.53
4.07
4.68
5.39
6.20
(continued)
Ny
66.19
78.03
92.25
109.41
130.22
155.55
186.54
224.64
271.76
330.35
403.67
496.01
613.16
762.89
Inclination
factors
by
Meyerhof (1963) and Hanna
and Meyerhof (1981)
Fci = Fai = (1-2
Fyi = (1 - B²
Bº
Ne 9 for p = 0
TABLE 12.10 Variation of A with Pile
Embedment Length, L
Embedment
length, L (m)
0
5
10
15
20
25
30
35
40
50
60
70
80
90
Coyle and Castello (1981):
Embedment ratio, L/D
0
10
20
30
40
50
60
70
Qp = q'N₁ Ap
10
T
T
T
T
T
e
$'
Bearing capacity factor, N
20
32° 36°
= 30°
34°
λ
0.5
0.336
0.245
0.200
0.173
0.150
0.136
0.132
0.127
0.118
0.113
0.110
0.110
0.110
38°
TABLE 12.11 Variation of a (Interpo-
lated Values Based on
Terzaghi et al., 1996)
40°
Cu
Pa
≤0.1
0.2
0.3
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
2.4
2.8
40 60 80 100 200
Note:
Pa
≈100 kN/m²
Qs
= atmospheric pressure
Embedment ratio, L/D
=
: (Ko'tan 8')pL
0.15 0.2
0
5
10
15
20
25
30
a
35
36
1.00
0.92
0.82
0.74
0.62
0.54
0.48
0.42
0.40
0.38
0.36
0.35
0.34
0.34
Earth pressure coefficient, K
1.0
$'
30°
31°
32°
33°
8=0.80'
2
34°
35⁰
36°
5
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