Unlock document.
This document is partially blurred.
Unlock all pages and 1 million more documents.
Get Access
175
Chapter 17
17.1 Tube 1: 56.4924.18.50 =−=−= tDD
oi
mm
17.2 Shelby tube A:
Eq. (17.6): 10%6.61% ≤=×
−
=
−
=100
)4.98(
)4.98()6.101(
)100((%)
2
22
2
22
i
io
R
D
DD
A
Shelby tubes A and B: Samples can be considered undisturbed and, therefore,
suitable for consolidation and unconfined compression tests. Grain size
17.3
Depth
(m)
o
σ
′
(kN/m
2
)
5.0−
′
=
a
o
Np
σ
C
N
60
(N
1
)
60
=
C
N
N
60
Depth z
γzσ
o
=
′
2
D
50
N
60
D
r
17.6
)100(
2417
(%)
5.0
60
′
+
=
a
o
r
p
σ
N
D
17.7
34.0
60
1
3.202.12
tan
′
+
=
′
−
a
o
p
σ
N
φ
; p
a
≈ 100 kN/m
2
Depth (m)
o
σ
′
(kN/m
2
) N
60
φ
′
(deg)
178
© 2018 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Design N
60
= (6 + 8 + 11) / 3 =
8.33 ≈ 8
Design
φ
′
(deg) = (35.9 + 36.5 + 38.1) / 3 =
36.8
°
°°
°
b.
22.1
5.1
1
)33.0(133.01 =
+=
+= B
D
F
f
d
17.9 Eq. (17.39) and Table 17.7:
(
)
26.0
pq
ac
2
17.10 Eq. (17.33):
cs
qE 5.3
=
Using q
c
from Problem 17.9:
Depth
(m)
q
c
(kN/m
2
)
E
s
(kN/m
2
)
179
© 2018 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
17.11 Given: N
k
≈ 18.3. Eq. (17.35):
2
kN/m 57.5=
−
=
−
=3.18
)8.17)(5.5(1150
k
oc
u
N
σq
c
17.12 m 35.122
=
×
=
B
Cone penetration resistance and sleeve friction are averaged up to 3 m below the
17.13 Figure 17.13: For the soil type “silty sand to sandy silt,”
(
)
3
60
=
N
pq
ac
17.14 a. Eq. (17.30):
′
+=
′
−
o
c
σ
q
log38.01.0tan
1
φ
Given
3
kN/m 17=
d
γ
. The following table can now be prepared.
Depth (m)
o
σ
′
(kN/m
2
)
c
q
(kN/m
2
)
φ
′
(deg)
Average
°
=
′
41.2
φ
180
b. Eq. (17.33):
cs
qE 5.3
=
Depth
(m)
q
c
(kN/m
2
)
E
s
(kN/m
2
)
17.15 From Eqs. (17.41) and (17.42):
