Chapter 2 Manual for... · 2019. 10. 27. · sat(clay) 18.55 kN/m 1 (0 3478)(2. 74) (1 )...
Transcript of Chapter 2 Manual for... · 2019. 10. 27. · sat(clay) 18.55 kN/m 1 (0 3478)(2. 74) (1 )...
1
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Chapter 2
2.1 d. 3kN/m 17.17
)05.0)(1000(
)81.9)(5.87(
c. 3
kN/m 14.93
15.01
17.17
1 wd
a. Eq. (2.12): 0.76
e
ee
G ws ;1
)81.9)(68.2(14.93 .
1
b. Eq. (2.6): 0.43
76.01
76.0
1 e
en
e. Eq. (2.14): 53%
)100(
76.0
)68.2)(15.0(
e
Gw
V
VS s
v
w
2.2 a. From Eqs. (2.11) and (2.12), it can be seen that,
3
kN/m 16.48
22.01
1.20
1 wd
b. e
G
e
G sws
1
)81.9(
1kN/m 16.48 3
Eq. (2.14): ).)(22.0( ss GwGe So,
2.67
s
s
s GG
G;
22.01
81.948.16
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Chapter 2
2
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2
2.3 a. 0.55
e
ee
wG ws ;1
)12.01)(4.62)(65.2(119.5 .
1
)1(
b. 0.355
55.01
55.0n
c. 57.8% 10055.0
)65.2)(12.0(
e
GwS s
d. 3
lb/ft 106.7
12.01
5.119
1 wd
2.4 a. w
eGs . 0.97
ee
e
e
w
ew
d ;1
)4.62(36.0
85.43 .1
)(
b. 0.49
97.01
97.0
1 e
en
c. 2.6936.0
97.0
w
eGs
d. 3
lb/ft 115.9
97.01
)4.62)(97.069.2(
1
)(sat
e
eG ws
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Chapter 2
3
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2.5 From Eqs. (2.11) and (2.12): 3lb/ft 108
08.01
64.116
d
Eq. (2.12): 53.0 ;1
)4.62)(65.2(108 ;
1
e
ee
G wsd
Eq. (2.23): 0.94
max
max
max
minmax
max ;44.0
53.082.0 e
e
e
ee
eeDr
3lb/ft 85.2
94.01
)4.62)(65.2(
1 max
(min) e
G wsd
2.6 Refer to Table 2.7 for classification.
Soil A: A-7-6(9) (Note: PI is greater than LL 30.)
GI = (F200 – 35)[0.2 + 0.005(LL – 40)] + 0.01(F200 – 15)(PI – 10)
=
=
(65 – 35)[0.2 + 0.005(42 – 40)] + 0.01(65 – 15)(16 – 10)
9.3 9
Soil B: A-6(5)
GI =
=
(55 – 35)[0.2 + 0.005(38 – 40)] + 0.01(55 – 15)(13 – 10)
5.4 5
Soil C: A-3-(0)
Soil D: A-4(5)
GI =
=
(64– 35)[0.2 + 0.005(35 – 40)] + 0.01(64 – 15)(9 – 10)
4.585 ≈ 5
Soil E: A-2-6(1)
GI = 0.01(F200 – 15)(PI – 10) = 0.01(33 – 15)(13 – 10) = 0.54 ≈ 1
Soil F: A-7-6(19) (PI is greater than LL 30.)
GI =
=
(76– 35)[0.2 + 0.005(52 – 40)] + 0.01(76 – 15)(24 – 10)
19.2 ≈ 19
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Chapter 2
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4
2.7 Soil A: Table 2.8: 65% passing No. 200 sieve.
Fine grained soil; LL = 42; PI = 16
Figure 2.5: ML
Figure 2.7: Plus No. 200 > 30%; Plus No. 4 = 0
% sand > % gravel – sandy silt
Soil B: Table 2.8: 55% passing No. 200 sieve.
Fine grained soil; LL = 38; PI = 13
Figure 2.5: Plots below A-line – ML
Figure 2.7: Plus No. 200 > 30%
% sand > % gravel – sandy silt
Soil C: Table 2.8: 8% passing No. 200 sieve.
% sand > % gravel – sandy soil – SP
Figure 2.6: % gravel = 100 – 95 = 5% < 15% – poorly graded sand
Soil D: Table 2.8: 64% passing No. 200 sieve
Fine grained soil; LL = 35, PI = 9
Figure 2.5 – ML
Figure 2.7: % sand (31%) > % gravel (5%) – sandy silt
Soil E: Table 2.8: 33% passing No. 200 sieve; 100% passing No. 4 sieve.
Sandy soil; LL = 38; PI = 13
Figure 2.5: Plots below A-line – SM
Figure 2.6: % gravel (0%) < 15% – silty sand
Soil F: Table 2.8: 76% passing No. 200 sieve; LL = 52; PI = 24
Figure 2.5: CH
Figure 2.7: Plus No. 200 is 100 – 76 = 24%
% gravel > % gravel – fat clay with sand
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Chapter 2
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2.8 43.01117
)4.62)(68.2(1e ;
1
d
wswsd
G
e
G
Eq. (2.37):
43.01
43.0
63.01
63.0
22.0;
1
1
3
3
2
2
32
1
31
2
1
k
e
e
e
e
k
k; k2 = 0.08 cm/s
2.9 From Eq. (2.41):
n
nn
e
e
e
e
k
k)6316.0(1667.0 ;
9.1
2.1
2.2
9.2
1091.0
102.0or ;
1
16
6
2
1
1
2
2
1
898.31996.0
778.0
)6316.0log(
)1667.0log(
n
cm/s 10 0.075 6-
)10216.0(9.1
9.0
1
10216.02.1
)2.2)(102.0()1(
6998.3
3
6
998.3
6
1
11
e
eCk
e
ekC
n
n
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Chapter 2
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6
2.10 The flow net is shown.
/m/sm 1017.06
36-
8
)4)(25.5(
10
105.6
So, m. 25.575.17 cm/s; 105.6
2
4
21max4
q
HHhk
2.11 a.
7825.03
2
7825.03
210
6.01
6.0)2.0(4622.2
)1(4622.2
e
eDk cm/s 0.041
b. 32.2
6.0332.2
106.0
3
)2.0(2.0
4.0
6.01
6.0)35()(
135
DC
e
ek u cm/s 0.171
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Chapter 2
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2.12 3
dry(sand) kN/m 16.8455.01
)81.9)(66.2(
1
e
G ws
3
sat(sand) kN/m 81.0248.01
)48.066.2(81.9
1
e
eG wws
3
sat(clay ) kN/m 18.55)74.2)(3478.0(1
)3478.01)(81.9)(74.2(
1
)1(
s
ws
wG
wG
At A: σ = 0; u = 0; σ = 0
At B: σ = (16.84)(3) = 50.52 kN/m2
u = 0
σ = 50.52 kN/m2
At C: σ = σB + (20.81)(1.5) = 50.52 + 31.22 = 81.74 kN/m2
u = (9.81)(1.5) = 14.72 kN/m2
σ = 81.74 – 14.72 = 67.02 kN/m2
At D: σ = σC + (18.55)(5) = 81.74 + 92.75 = 174.49 kN/m2
u = (9.81)(6.5) = 63.77 kN/m2
σ = 174.49 – 63.77 = 110.72 kN/m2
2.13 Eq. (2.54): Cc = 0.009(LL – 10) = 0.009(42 – 10) = 0.288
Eq. (2.65):
mm 87.2
110
155log
82.01
mm) 10007.3)(288.0(log
1 o
o
o
ccc
e
HCS
2.14 Eq. (2.69):
mm 56.69
128
155log
82.01
)3700)(288.0(
110
128log
75.01
mm) 3700(5
288.0
log1
log1 c
o
o
cc
o
c
o
csc
e
HC
e
HCS
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Chapter 2
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8
2.15 a. Eq. (2.53): 0.392
150
300log
792.091.0
log1
2
21
eeCc
From Eq. (2.65): o
o
o
ccc
e
HCS
log
1
Using the results of Problem 2.12,
2kN/m 87.88)81.955.18(
2
5)81.981.20(5.1)84.16)(3( o
953.0)74.2)(3478.0( so wGe
mm 194.54
87.88
5087.88log
953.01
mm) 5000)(392.0(cS
b. Eq. (2.73): 2H
tCT v
v . For U = 50%, Tv = 0.197 (Table 2.11). So,
days 609 sec 105262;cm) 500(
1036.9197.0 4
2
4
tt
2.16 a. Eq. (2.53): 0.377
120
360log
64.082.0
log1
2
21 eeCc
b. 0.736
2
2
1
2
21 ;
120
200log
82.00.377 ;
log
eeee
Cc
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Chapter 2
9
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2.17 Eq. (2.73): 2H
tCT v
v . For 60% consolidation, Tv = 0.286 (Table 2.11).
Lab time: min 6
49 min 8
61 t
/minin. 0788.0;)5.1(
6
49
286.0 2
2
v
v
C
C
Field: U = 50%; Tv = 0.197
days 6.25 min 9000;
2
1210
)0788.0(197.0
2
t
t
2.18 5.060
30U
tt
H
tCT
v
v6
221
)1(
)1( 102
2
10002
))(2(
tt
H
tCT
v
v6
222
)2(
)2( 108
2
10001
))(2(
So, Tv(1) = 0.25Tv(2). The following table can be prepared for trial and error
procedure.
Tv(1) Tv(2) U1 U2
UHH
HUHU
21
2211 (Figure 2.22)
0.05
0.10
0.125
0.1125
0.2
0.4
0.5
0.45
0.26
0.36
0.40
0.385
0.51
0.70
0.76
0.73
0.34
0.473
0.52
0.50
So, Tv(1) = 0.1125 = 2 10-6
t; t = 56,250 min = 39.06 days
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Chapter 2
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10
2.19 Eq. (2.84): .2H
tCT cv
c tc = 60 days = 60 24 60 60 sec;2
2H m = 1000 mm.
0415.0)1000(
)60602460)(108(2
3
cT
After 30 days: 0207.0)1000(
)60602430)(108(2
3
2
H
tCT v
v
From Figure 2.24 for Tv = 0.0207 and Tc = 0.0415, U = 5%. So
Sc = (0.05)(120) = 6 mm
After 100 days: 069.0)1000(
)606024100)(108(2
3
2
H
tCT v
v
From Figure 2.24 for Tv = 0.069 and Tc = 0.0415, U 23%. So
Sc = (0.23)(120) = 27.6 mm
2.20
N
S1tan
Normal force, N (lb) Shear force, S (lb)
NS1tan (deg)
50
110
150
43.5
95.5
132.0
41.02
40.96
41.35
From the graph, 41
2.21 Normally consolidated clay; c = 0.
245tan2
31 ; 38
;
245tan309630 2
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Chapter 2
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2.22
245tan2
31 ; 30
;
245tan204020 2
2.23 c = 0. Eq. (2.91): 2kN/m 387.8
2
2845tan140
245tan 22
31
2.24 Eq. (2.91):
245tan2
24531 c
245tan2
245tan140368 2 c (a)
245tan2
245tan280701 2 c (b)
Solving Eqs. (a) and (b), = 24; c = 12 kN/m2
2.25 25
1332
1332sinsin 1
31
311
31
311sin
23
23 lb/in. 5.75.513 ;lb/in. 5.265.532
34
5.75.26
5.75.26sin 1
Normally consolidated clay; c = 0 and c = 0
2.26
245tan2
31 . 22
1 kN/m 305.92
2045tan150
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Chapter 2
12
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12
2kN/m 61.9
u
u
u ;
2
2845tan
150
9.305 ;
245tan 22
3
1
2.27 a. )log(6.14.01026 50DCD ur
30.7 )]13.0)[log(6.1()1.2)(4.0()53.0)(10(26
b. bae
1
327.209.0
21.0097.0101.2097.0101.2
15
85
D
Da
081.0)327.2)(398.0(845.0398.0845.0 ab
33.67
081.0)68.0)(327.2(
1
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