4. Delineate the relationship between the total bearing capacity, Qtotal, capacity due to side friction, Qs, capacity due to end bearing, Qp, and settlement at the top, St, of a drilled shaft socketed into soft rock. Assume that 75% of the total bearing capacity comes from the end-bearing portion. You need to draw three curves: Qtotal VS. St, Qp VS. St, and Qs vs. S, on the same graph. Equations and Tables Bearing capacity of mat foundations for saturated clays with p = 0 qu = 5.14c (1+0.195 1952) (1+0.4%) + +9 qu(net) = qu− q = 5.14c (1+0.195! 52) (1+0.42/4) qu(net) qu(net) qa(net) Meyerhof's general bearing capacity equation: qu=c'NcFcs FcaFci+qNqFqsFqaFqi + 0.5yBNyFysFyaFyi FS = Pile: Meyerhof's method Qp = Apqp = Apq'N₁₁ ≤ Apql q₁ = 0.5pa Natano' where pa = atmospheric pressure = 100 kPa The critical depth for skin friction in piles: L'≈ 15D The unit frictional resistance or the unit skin friction: f = Ko'。tand' TABLE 12.6 Interpolated Values of N Based on Meyerhof's Theory TABLE 12.10 Variation of A with Pile Embedment Length, L TABLE 12.11 Variation of a (Interpo- lated Values Based on Terzaghi et al., 1996) Embedment length, L (m) λ C Pa α 0 0.5 ≤0.1 1.00 5 0.336 0.2 0.92 10 0.245 0.3 0.82 15 0.200 0.4 0.74 20 0.173 0.6 0.62 Shape factors by De Beer (1970) Depth factors by Hansen (1970) Df/B≤1 Inclination factors by Meyerhof (1963) & Hanna and Meyerhof (1981) 2 Fcs = 1 + 1+ (-) (~) Fqs = 1 + B Fcd=Fqd 1-Fqd No tan o' +(²) tan o' Fqd = 1 + 2 tan ø′ (1 − sin ')2 Df | Fci = Fqi = (1 Fyi = (1-59)² 1- B° 90° 2 B Fys = 1-0.4 - 0.4 (+) Fyd = 1 Bo Inclination of the load on the foundation with respect to the vertical in degrees Table of bearing capacity factors Nc, Na, Ny for Meyerhof's general bearing capacity equation based on Prandtl (1921), Reissner (1924), Caquot & Kerisel (1953) and Vesic (1973): φ' (°) Nc Na Ny φ' (°) Nc Na Ny 0 5.14 1.00 0.00 1 5.38 1.09 0.07 2345 5.63 1.20 0.15 5.90 1.31 0.24 4 6.19 1.43 0.34 5 6.49 1.57 0.45 11 12 6789012 6 6.81 1.72 0.57 7.16 1.88 0.71 7.53 2.06 0.86 7.92 2.25 1.03 10 8.34 2.47 1.22 8.80 2.71 1.44 9.28 2.97 1.69 13 9.81 3.26 1.97 14 15 45 10.37 3.59 2.29 10.98 3.94 2.65 16 11.63 4.34 3.06 17 12.34 4.77 3.53 18 13.10 5.26 4.07 2222222222231030 15.81 7.07 6.20 16.88 7.82 7.13 18.05 8.66 8.20 24 19.32 9.60 9.44 25 20.72 10.66 10.88 26 22.25 11.85 12.54 27 23.94 13.20 14.47 28 25.80 14.72 16.72 22222222222-23736233 Soil friction angle, ' (deg) 20 21 N₁₂ 25 0.150 0.8 0.54 12.4 30 0.136 1.0 0.48 13.8 35 0.132 1.2 0.42 15.5 40 0.127 1.4 0.40 17.9 50 0.118 24 21.4 1.6 0.38 60 0.113 25 26.0 1.8 0.36 70 0.110 29.5 2.0 0.35 80 0.110 34.0 90 0.110 28 39.7 2.4 2.8 0.34 0.34 29 46.5 Note: Pa = atmospheric pressure 30 56.7 ≈ 100 kN/m² 31 68.2 Coyle and Castello (1981): 81.0 Qp = q'NgAp 96.0 Qs = (Ko' tan 8')PL where 8' = 0.8 ' 34 115.0 35 143.0 168.0 194.0 38 231.0 10 39 276.0 29 27.86 16.44 19.34 40 346.0 20 30.14 18.40 22.40 32.67 20.63 41 420.0 25.99 35.49 23.18 30.21 42 525.0 38.64 26.09 35.19 43 650.0 34 42.16 29.44 41.06 35 46.12 33.30 48.03 44 780.0 36 50.59 37.75 56.31 45 930.0 37 55.63 42.92 66.19 Embedment ratio, L/D T T 20 10 0+ Bearing capacity factor, N 20 40 60 80 100 200 Earth pressure coefficient, K 0.15 0.2 1.0 2 5 0 L 6' = 30° 31° 32° 33° 35° 34° 5 10 40 Embedment ratio, L/D 20 5 36° 20 25 50 38 61.35 48.93 78.02 32° 36° 19 13.93 5.80 4.68 39 67.87 55.96 92.25 20 14.83 6.40 5.39 75.31 64.20 109.41 For saturated clay: Q₁ = Apqp = ApCu N ' = 30° 40° No 9 for = = 0 60 60 T 38° 30- 34° For a method: fav = αcu For method: fλ(σo + 2cu) 70 35 36
4. Delineate the relationship between the total bearing capacity, Qtotal, capacity due to side friction, Qs, capacity due to end bearing, Qp, and settlement at the top, St, of a drilled shaft socketed into soft rock. Assume that 75% of the total bearing capacity comes from the end-bearing portion. You need to draw three curves: Qtotal VS. St, Qp VS. St, and Qs vs. S, on the same graph. Equations and Tables Bearing capacity of mat foundations for saturated clays with p = 0 qu = 5.14c (1+0.195 1952) (1+0.4%) + +9 qu(net) = qu− q = 5.14c (1+0.195! 52) (1+0.42/4) qu(net) qu(net) qa(net) Meyerhof's general bearing capacity equation: qu=c'NcFcs FcaFci+qNqFqsFqaFqi + 0.5yBNyFysFyaFyi FS = Pile: Meyerhof's method Qp = Apqp = Apq'N₁₁ ≤ Apql q₁ = 0.5pa Natano' where pa = atmospheric pressure = 100 kPa The critical depth for skin friction in piles: L'≈ 15D The unit frictional resistance or the unit skin friction: f = Ko'。tand' TABLE 12.6 Interpolated Values of N Based on Meyerhof's Theory TABLE 12.10 Variation of A with Pile Embedment Length, L TABLE 12.11 Variation of a (Interpo- lated Values Based on Terzaghi et al., 1996) Embedment length, L (m) λ C Pa α 0 0.5 ≤0.1 1.00 5 0.336 0.2 0.92 10 0.245 0.3 0.82 15 0.200 0.4 0.74 20 0.173 0.6 0.62 Shape factors by De Beer (1970) Depth factors by Hansen (1970) Df/B≤1 Inclination factors by Meyerhof (1963) & Hanna and Meyerhof (1981) 2 Fcs = 1 + 1+ (-) (~) Fqs = 1 + B Fcd=Fqd 1-Fqd No tan o' +(²) tan o' Fqd = 1 + 2 tan ø′ (1 − sin ')2 Df | Fci = Fqi = (1 Fyi = (1-59)² 1- B° 90° 2 B Fys = 1-0.4 - 0.4 (+) Fyd = 1 Bo Inclination of the load on the foundation with respect to the vertical in degrees Table of bearing capacity factors Nc, Na, Ny for Meyerhof's general bearing capacity equation based on Prandtl (1921), Reissner (1924), Caquot & Kerisel (1953) and Vesic (1973): φ' (°) Nc Na Ny φ' (°) Nc Na Ny 0 5.14 1.00 0.00 1 5.38 1.09 0.07 2345 5.63 1.20 0.15 5.90 1.31 0.24 4 6.19 1.43 0.34 5 6.49 1.57 0.45 11 12 6789012 6 6.81 1.72 0.57 7.16 1.88 0.71 7.53 2.06 0.86 7.92 2.25 1.03 10 8.34 2.47 1.22 8.80 2.71 1.44 9.28 2.97 1.69 13 9.81 3.26 1.97 14 15 45 10.37 3.59 2.29 10.98 3.94 2.65 16 11.63 4.34 3.06 17 12.34 4.77 3.53 18 13.10 5.26 4.07 2222222222231030 15.81 7.07 6.20 16.88 7.82 7.13 18.05 8.66 8.20 24 19.32 9.60 9.44 25 20.72 10.66 10.88 26 22.25 11.85 12.54 27 23.94 13.20 14.47 28 25.80 14.72 16.72 22222222222-23736233 Soil friction angle, ' (deg) 20 21 N₁₂ 25 0.150 0.8 0.54 12.4 30 0.136 1.0 0.48 13.8 35 0.132 1.2 0.42 15.5 40 0.127 1.4 0.40 17.9 50 0.118 24 21.4 1.6 0.38 60 0.113 25 26.0 1.8 0.36 70 0.110 29.5 2.0 0.35 80 0.110 34.0 90 0.110 28 39.7 2.4 2.8 0.34 0.34 29 46.5 Note: Pa = atmospheric pressure 30 56.7 ≈ 100 kN/m² 31 68.2 Coyle and Castello (1981): 81.0 Qp = q'NgAp 96.0 Qs = (Ko' tan 8')PL where 8' = 0.8 ' 34 115.0 35 143.0 168.0 194.0 38 231.0 10 39 276.0 29 27.86 16.44 19.34 40 346.0 20 30.14 18.40 22.40 32.67 20.63 41 420.0 25.99 35.49 23.18 30.21 42 525.0 38.64 26.09 35.19 43 650.0 34 42.16 29.44 41.06 35 46.12 33.30 48.03 44 780.0 36 50.59 37.75 56.31 45 930.0 37 55.63 42.92 66.19 Embedment ratio, L/D T T 20 10 0+ Bearing capacity factor, N 20 40 60 80 100 200 Earth pressure coefficient, K 0.15 0.2 1.0 2 5 0 L 6' = 30° 31° 32° 33° 35° 34° 5 10 40 Embedment ratio, L/D 20 5 36° 20 25 50 38 61.35 48.93 78.02 32° 36° 19 13.93 5.80 4.68 39 67.87 55.96 92.25 20 14.83 6.40 5.39 75.31 64.20 109.41 For saturated clay: Q₁ = Apqp = ApCu N ' = 30° 40° No 9 for = = 0 60 60 T 38° 30- 34° For a method: fav = αcu For method: fλ(σo + 2cu) 70 35 36
Fundamentals of Geotechnical Engineering (MindTap Course List)
5th Edition
ISBN:9781305635180
Author:Braja M. Das, Nagaratnam Sivakugan
Publisher:Braja M. Das, Nagaratnam Sivakugan
Chapter19: Drilled Shaft
Section: Chapter Questions
Problem 19.7P
Related questions
Question
I need detailed explanation solving this exercise from Foundation Engineering, step by step please.
Transcribed Image Text:4. Delineate the relationship between the total bearing capacity, Qtotal, capacity due to side friction, Qs, capacity due to
end bearing, Qp, and settlement at the top, St, of a drilled shaft socketed into soft rock. Assume that 75% of the total
bearing capacity comes from the end-bearing portion. You need to draw three curves: Qtotal VS. St, Qp VS. St, and Qs
vs. S, on the same graph.
Equations and Tables
Bearing capacity of mat foundations for saturated clays with p = 0
qu =
5.14c (1+0.195
1952) (1+0.4%) +
+9
qu(net) = qu− q = 5.14c (1+0.195!
52) (1+0.42/4)
qu(net) qu(net)
qa(net)
Meyerhof's general bearing capacity equation:
qu=c'NcFcs FcaFci+qNqFqsFqaFqi + 0.5yBNyFysFyaFyi
FS =
Pile: Meyerhof's method
Qp = Apqp = Apq'N₁₁ ≤ Apql
q₁ = 0.5pa Natano' where pa = atmospheric pressure = 100 kPa
The critical depth for skin friction in piles: L'≈ 15D
The unit frictional resistance or the unit skin friction: f = Ko'。tand'
TABLE 12.6 Interpolated Values of N Based on
Meyerhof's Theory
TABLE 12.10 Variation of A with Pile
Embedment Length, L
TABLE 12.11 Variation of a (Interpo-
lated Values Based on
Terzaghi et al., 1996)
Embedment
length, L (m)
λ
C
Pa
α
0
0.5
≤0.1
1.00
5
0.336
0.2
0.92
10
0.245
0.3
0.82
15
0.200
0.4
0.74
20
0.173
0.6
0.62
Shape factors by De
Beer (1970)
Depth factors by Hansen (1970)
Df/B≤1
Inclination factors by Meyerhof
(1963) & Hanna and Meyerhof (1981)
2
Fcs = 1 +
1+ (-) (~)
Fqs = 1 +
B
Fcd=Fqd
1-Fqd
No tan o'
+(²) tan
o'
Fqd = 1 + 2 tan ø′ (1 − sin ')2 Df
| Fci = Fqi = (1
Fyi = (1-59)²
1-
B°
90°
2
B
Fys
= 1-0.4
- 0.4 (+)
Fyd
= 1
Bo Inclination of the load on the
foundation with respect to the vertical
in degrees
Table of bearing capacity factors Nc, Na, Ny for Meyerhof's general bearing capacity equation based on Prandtl
(1921), Reissner (1924), Caquot & Kerisel (1953) and Vesic (1973):
φ' (°)
Nc
Na
Ny
φ' (°)
Nc
Na
Ny
0
5.14
1.00
0.00
1
5.38
1.09
0.07
2345
5.63
1.20
0.15
5.90
1.31
0.24
4
6.19
1.43
0.34
5
6.49
1.57
0.45
11
12
6789012
6
6.81
1.72
0.57
7.16
1.88
0.71
7.53
2.06
0.86
7.92
2.25
1.03
10
8.34
2.47
1.22
8.80
2.71
1.44
9.28
2.97
1.69
13
9.81
3.26
1.97
14
15
45
10.37
3.59
2.29
10.98
3.94
2.65
16
11.63
4.34
3.06
17
12.34
4.77
3.53
18
13.10
5.26
4.07
2222222222231030
15.81
7.07
6.20
16.88
7.82
7.13
18.05
8.66
8.20
24
19.32
9.60
9.44
25
20.72
10.66
10.88
26
22.25
11.85
12.54
27
23.94
13.20
14.47
28
25.80
14.72
16.72
22222222222-23736233
Soil friction angle, ' (deg)
20
21
N₁₂
25
0.150
0.8
0.54
12.4
30
0.136
1.0
0.48
13.8
35
0.132
1.2
0.42
15.5
40
0.127
1.4
0.40
17.9
50
0.118
24
21.4
1.6
0.38
60
0.113
25
26.0
1.8
0.36
70
0.110
29.5
2.0
0.35
80
0.110
34.0
90
0.110
28
39.7
2.4
2.8
0.34
0.34
29
46.5
Note: Pa =
atmospheric pressure
30
56.7
≈ 100 kN/m²
31
68.2
Coyle and Castello (1981):
81.0
Qp = q'NgAp
96.0
Qs
=
(Ko' tan 8')PL where 8' = 0.8 '
34
115.0
35
143.0
168.0
194.0
38
231.0
10
39
276.0
29
27.86
16.44
19.34
40
346.0
20
30.14
18.40
22.40
32.67
20.63
41
420.0
25.99
35.49
23.18
30.21
42
525.0
38.64
26.09
35.19
43
650.0
34
42.16
29.44
41.06
35
46.12
33.30
48.03
44
780.0
36
50.59
37.75
56.31
45
930.0
37
55.63
42.92
66.19
Embedment ratio, L/D
T
T
20
10
0+
Bearing capacity factor, N
20 40 60 80 100
200
Earth pressure coefficient, K
0.15 0.2
1.0
2
5
0
L
6' =
30°
31°
32°
33°
35°
34°
5
10
40
Embedment ratio, L/D
20
5
36°
20
25
50
38
61.35
48.93
78.02
32° 36°
19
13.93
5.80
4.68
39
67.87
55.96
92.25
20
14.83
6.40
5.39
75.31
64.20
109.41
For saturated clay: Q₁ = Apqp = ApCu N
' = 30°
40°
No 9 for =
= 0
60
60
T
38°
30-
34°
For a method: fav = αcu
For method: fλ(σo + 2cu)
70
35
36
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 2 steps with 8 images
Recommended textbooks for you
Fundamentals of Geotechnical Engineering (MindTap…
Civil Engineering
ISBN:
9781305635180
Author:
Braja M. Das, Nagaratnam Sivakugan
Publisher:
Cengage Learning
Principles of Foundation Engineering (MindTap Cou…
Civil Engineering
ISBN:
9781305081550
Author:
Braja M. Das
Publisher:
Cengage Learning
Principles of Foundation Engineering (MindTap Cou…
Civil Engineering
ISBN:
9781337705028
Author:
Braja M. Das, Nagaratnam Sivakugan
Publisher:
Cengage Learning
Fundamentals of Geotechnical Engineering (MindTap…
Civil Engineering
ISBN:
9781305635180
Author:
Braja M. Das, Nagaratnam Sivakugan
Publisher:
Cengage Learning
Principles of Foundation Engineering (MindTap Cou…
Civil Engineering
ISBN:
9781305081550
Author:
Braja M. Das
Publisher:
Cengage Learning
Principles of Foundation Engineering (MindTap Cou…
Civil Engineering
ISBN:
9781337705028
Author:
Braja M. Das, Nagaratnam Sivakugan
Publisher:
Cengage Learning