[12.13] A concrete pile 406 mm × 406 mm in cross section is shown in Figure P12.13. Calculate the ultimate skin friction resistance by using the a) a method [use Eq. (12.61) and Table 12.11] Qs = [fpAL = [ac.PAL b) λ method c) ẞ method Use O'R = 20° for all clays, which are normally consolidated. 6.1 m 12.2 m 406 mm FIGURE P12.13 Q, = Σfp ΔL = Σacp ΔL Groundwater table Silty clay Ysat 18.55 kN/m³ Cu = 35 kN/m² Plasticity index, PI = 15 Silty clay Ysat = 19.24 kN/m³ Cu = 75 kN/m² Plasticity index, PI = 20 (12.61)
[12.13] A concrete pile 406 mm × 406 mm in cross section is shown in Figure P12.13. Calculate the ultimate skin friction resistance by using the a) a method [use Eq. (12.61) and Table 12.11] Qs = [fpAL = [ac.PAL b) λ method c) ẞ method Use O'R = 20° for all clays, which are normally consolidated. 6.1 m 12.2 m 406 mm FIGURE P12.13 Q, = Σfp ΔL = Σacp ΔL Groundwater table Silty clay Ysat 18.55 kN/m³ Cu = 35 kN/m² Plasticity index, PI = 15 Silty clay Ysat = 19.24 kN/m³ Cu = 75 kN/m² Plasticity index, PI = 20 (12.61)
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
I need detailed help solving the problem 12.13 about skin friction resistance, please.
![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
TABLE 12.12 Variation of a with cu/
Cu
1
2
3
4
5
6
7
8
9
10
Open-ended pile
0.5
0.4
0.355
0.33
0.31
0.29
0.28
0.26
0.255
0.25
α
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
Closed-ended pile
0.5
0.44
0.41
0.395
0.38
0.365
0.35
0.33
0.32
0.31
1.2
1.0
0.8
0.6
0.4
0.2
0
0.1
TABLE 12.11 Variation of a (Interpo-
lated Values Based on
Terzaghi et al., 1996)
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
Note: Pa = atmospheric pressure
≈100 kN/m²
PI = 55
40
30
20
15
12
PI < 10
1
0.3
1
1
0.5
1.0
α
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
FIGURE 12.26 Variation of a
with c/ for the NGI-99 method
[Eqs. (12.59a) and (12.59b)]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe95c0ed5-6562-4de3-90f0-5232d13c65eb%2F798b6c94-3482-4351-9ed6-9d6bc5f34e38%2Fkz37czt_processed.jpeg&w=3840&q=75)
Transcribed Image Text: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
TABLE 12.12 Variation of a with cu/
Cu
1
2
3
4
5
6
7
8
9
10
Open-ended pile
0.5
0.4
0.355
0.33
0.31
0.29
0.28
0.26
0.255
0.25
α
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
Closed-ended pile
0.5
0.44
0.41
0.395
0.38
0.365
0.35
0.33
0.32
0.31
1.2
1.0
0.8
0.6
0.4
0.2
0
0.1
TABLE 12.11 Variation of a (Interpo-
lated Values Based on
Terzaghi et al., 1996)
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
Note: Pa = atmospheric pressure
≈100 kN/m²
PI = 55
40
30
20
15
12
PI < 10
1
0.3
1
1
0.5
1.0
α
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
FIGURE 12.26 Variation of a
with c/ for the NGI-99 method
[Eqs. (12.59a) and (12.59b)]
![[12.13] A concrete pile 406 mm × 406 mm in cross section is shown in Figure P12.13. Calculate the ultimate skin friction
resistance by using the
a) a method [use Eq. (12.61) and Table 12.11]
Qs = [fpAL = Σ acupAL
b) λ method
c) ẞ method
Use O'R = 20° for all clays, which are normally consolidated.
6.1 m
12.2 m
406 mm
FIGURE P12.13
Qs Efp AL = Σacup AL
table
=
Groundwater
Silty clay
=
Ysat
Cu = 35 kN/m²
Plasticity index, PI = 15
18.55 kN/m³
Silty clay
=
Ysat
Cu = 75 kN/m²
Plasticity index, PI = 20
19.24 kN/m³
(12.61)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe95c0ed5-6562-4de3-90f0-5232d13c65eb%2F798b6c94-3482-4351-9ed6-9d6bc5f34e38%2Fn1ihvl_processed.jpeg&w=3840&q=75)
Transcribed Image Text:[12.13] A concrete pile 406 mm × 406 mm in cross section is shown in Figure P12.13. Calculate the ultimate skin friction
resistance by using the
a) a method [use Eq. (12.61) and Table 12.11]
Qs = [fpAL = Σ acupAL
b) λ method
c) ẞ method
Use O'R = 20° for all clays, which are normally consolidated.
6.1 m
12.2 m
406 mm
FIGURE P12.13
Qs Efp AL = Σacup AL
table
=
Groundwater
Silty clay
=
Ysat
Cu = 35 kN/m²
Plasticity index, PI = 15
18.55 kN/m³
Silty clay
=
Ysat
Cu = 75 kN/m²
Plasticity index, PI = 20
19.24 kN/m³
(12.61)
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