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Chapter 9
9.1 a. Eq. (9.16):
b. Eq. (9.26):
2
kN/m 000,60)100)(600( as pmE
1.66
)42)(tan206.18)(355.01(2
000,60
c. Eq. (9.36):
9.2 a. Eq. (9.40): L = 15D = (15)(0.46) = 6.9 m
b.
LpKQ os )8.0tan()( 1
Chapter 9
9.3 From Problem 9.1
kN 5563
3
6.64543.52323.5001
p
Q
9.4 a. Eq. (9.17) will apply:
b. = 40; Irr = 50;
93 (Table 9.7).
Chapter 9
d.
m. 72.5381.0)15(15
DL
Let us assume L 6 m.
9.5 The pile tip is 9 m below the ground surface. The average value of N60 for a
distance of 10D above and 5D below the pile tip is:
Chapter 9
© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
kN 4.2732
46.0
9
)5.16)(100)(4.0()46.0(4.0 2
60
D
L
NpA ap
kN 56.1396)]5.16)(100)(4[()46.0()4( 2
60 NpA ap
(USE)
(9 m below ground surface)
From Eq. (9.45):
)](02.0[ 60
NpLpfpLQaavs
9.6
kN 6.1143])5.16)(100)(7.19[()46.0(])(7.19[36.0236.0
60 NpAQ app
9.7 Eqs. (9.35) and (9.33):
6.24433
2000
1600
)347(33347
a
u
rrr p
c
II
Chapter 9
9.8 Eq. (9.18):
kip 5.22
12
15
1000
1600
)9(9
2
pup AcQ
9.9 L = 20 m; = 0.173
Chapter 9
9.10 a.
ft 33.5
12
16
)4( ].[)2(22)1(11
pcLcLpQ uus
b.
2
)av(lb/ft 3.1233
60
)40)(1500()20)(700(
u
c
Chapter 9
© 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
kip301 kip 4.301
)]3.1233)(2(7.2766)[18.0)(60(
12
16
4
1000
1
]2[ )av(av)(
uos cpLQ
2
(av)av(1)
lb/ft 6.28220118
2
1
)20)(tan20sin1(
tan)sin1( c.
oRR
f
9.11 Top Layer:
2
lb/ft 1180)118(
2
20
o
; cu = 700 lb/ft2
Bottom Layer:
2
lb/ft 356012002360
2
40
)4.624.122(20118
o
; cu = 1500 lb/ft2
av(2) lb/ft 4.1155)]3560)(1500[(5.0)(5.0
9.12 Top Layer:
cu = 700 lb/ft2;
Chapter 9
9.13 Eq. (9.68):
= 15.5 kN/m3;
;
323.1)5(72.10;5
2
10 3.1
m
z
9.14 Eq. (9.74):
9.15 Eq. (9.77):
2
(lab)
(design) MN/m16
5
80
5 u
u
q
q
Chapter 9
9.16 Eq. (9.81):
)kN/m 101.2)(46.046.0(
26
)1(
pp
eEA
Eq. (9.82):
Eq. (9.84):
wss
s
ws
eI
E
D
pL
Q
s)1( 2
)3(
9.17 Eq. (9.81):