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G γ s w γ40p6zu91z1c3x7lz71846qd1-wpengine.netdna-ssl.com/wp-content/u… · Solution 3.4...
Transcript of G γ s w γ40p6zu91z1c3x7lz71846qd1-wpengine.netdna-ssl.com/wp-content/u… · Solution 3.4...
Solution 3.1 Prove the following:
(a) e1
G wsd +
γ=γ
Start with fundamental definitions:
sssswsss
d VandWforsubsitute;)el(VV;GVW;V
W+=γ==γ
dws
s
swsd el
G)el(V
GVγ=
+γ
=+
γ=γ
(b) e = n1
n−
∴ S = n
)n1(wGe
wG ss −=
Solution 3.2
d d dr
d d
D min max
max min
γ − γ γ=γ − γ γd
Dr = relative density which is usually defined in terms of min, max and current void ratio (emin, emax, e, respectively). The corresponding dry densities are:
e1G
;e1
G;
e1G ws
dmin
wsd
max
wsd maxmin +
γ=γ
+γ
=γ+γ
=γ
Relative density in terms of void ratios is:
re eD
e emax
max min
−=
−
Solving for the appropriate void ratios and substituting into :
1G
e;1G
e;1G
ed
ws
d
wsmin
d
wsmax
maxmin
−=−=−=γγ
γγ
γγ
s w s w
d dr
s w s w
d d
G G1 1D G G1 1
min
min max
γ γ− − +
γ γ=
γ γ− − +
γ γ
=
maxmin
min
dd
dd
11
11
γ−
γ
γ−
γ
=
maxmin
min
maxmax
max
min
min
min
dd
d
dd
d
dd
d
dd
d
11
11
γγ
γ−
γγ
γ
γγ
γ−
γγγ
=
minmax
minmax
min
min
dd
dd
dd
dd
γγ
γ−γ
γγ
γ−γ
= minmax
minmax
min
min
dd
dd
dd
dd
γ−γ
γγ
γγ
γγ
rD∴ = ⎟⎟⎠
⎞⎜⎜⎝
⎛
γ
γ
γ−γ
γ−γ
d
d
dd
dd max
minmax
min
Solution 3.3 Given Gs = 2.70, w = 0.65, and e = 1.96, Plot S vs. w. Use S = (w Gs) / e and realize that S will equal 1 at about 71.6% water content. Assume that up to this point, the void ratio will remain constant. After this point, the void ratio will increase and the saturation will remain at 1.0. The relationship is linear up to saturation. Notice that there are no data points shown as this is generated data, not observed data.
0.0
0.2
0.4
0.6
0.8
1.0
1.2
0 20 40 60 80 100
w, water content (%)
S, s
atur
atio
n
Solution 3.4 Strategy It is easiest to consider that each sphere occupies a unit volume or else you many calculate the volume of N number of spheres, or if D is the diameter of the sphere, the volume occupied by it in the array is D3 for the
cubic (loose) array and 3 1D2
for the dense array.
Loose array Step 1. Calculate volume of sphere of diameter D:
6
3DVsphereπ
=
Step 2. Calculate solid volume ratio occupied by sphere
66
3
3
ππ
=⎟⎟⎠
⎞⎜⎜⎝
⎛
=D
D
VS
Step 3. Calculate the porosity
4764.06
1 =−=πn
Step 4. Calculate the void ratio
91.05236.04764.0
1==
−=
nne
Dense Array Step 1. Calculate volume of sphere of diameter D:
6
3DVsphereπ
=
Step 2. Calculate volume of space occupied by sphere. The height of space occupied is a tetrahedron.
Height of tetrahedron is2D3
. Space occupied is: 33 2D D D D4 3
× × =12
3
3
D6
VS1 1D2
⎛ ⎞π⎜ ⎟
8π⎝ ⎠= =
Step 3. Calculate Porosity
2595.018
1 =−=πn
Step 4. Calculate void ratio
35.02595.01
2595.01
=−
=−
=n
ne
Solution 3.5 D = 6πrηv
W = s3r
34 γπ
W ′ = Effective weight of particle = )(r34
wsat2 γ−γπ
For equilibrium, D = W ′
( )wsat3r
34vr6 γ−γπ=ηπ
( ) v2
9rwsat γ−γ
η=
[To show that D = vr6 ηπ ]
D = CD γw A2
v2
CD = drag coefficient of sphere = RN
24
NR = Reynold’s number = vdηγ w
d = diameter of sphere
A = 4d 2π
(projected area of sphere)
∴D = vr6vd32vA
N24 2
wR
ηπηπγ ==
Solution 3.6 Md = mass of dry soil = 100 grams Mwe = mass of water of equivalent volume to dry soil = volume of water displaced x density of water (1 gram/cm3) = (537.5 – 500) x 1 = 37.5 grams
ds
we
M 100G 2M 37.5
= = = .67
Solution 3.7 Given: S = 1.0, MT = 500g, Ms = 400 g (a) Water Content (w)
( )T sw
s s
500 400M MMw 0.25 25%
M M 400
−−= = = = =
(b) Void Ratio (e)
e = ( )sG w2.7 0.25 0.675
S= =
Alternatively: Gs = 2.7, 1 gm/ (mass density of water) =ρw
3cm
3ss
s s
M 400V 1
G 2.7 1= = =
ρ ×48.1cm
Because S = 1, Vv = Vw
Vw = Vv = ( ) 3w
w
500 400M100 cm
1
−= =
ρ
3v
3s
V 100 cme 0.675
V 148.1 cm∴ = = =
(c) Saturated Unit Weight (γsat)
ssat w
G e 2.7 0.6759.8
1 e 1 0.675+ +⎛ ⎞ ⎛ ⎞
γ = γ =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠= 19.7 kN/ 3m
(d) Effective Unit Weight )(γ ′
wsat γγγ −=′ = 19.7 – 9.8 = 9.9 kN/ 3m
Solution 3.8 (a) Void Ratio (e)
w1d +γ
=γ = 1.16.19 = 17.8 kN/ 3m
e1G ws
d +γ
=γ ; 8.17
8.97.2e
×=∴ - 1 = 0.49
(b) Degress of Saturation (S)
Se = w , S = sGe
wG s = 49.0
7.21.0 × = 0.55 or 55%
(c) % of Air voids 3
ws
ss m67.0
8.97.28.17
GW
V =×
=γ
=
sv VVV −= = 1 – 0.67 = 0.33 3m
Weight of water per 3m = =×=γ 8.171.0w d 1.78 kN
8.978.1Vw = = 0.18 3m
wvair VVV −= = 0.33 – 0.18 = 0.15 3m
% of air voids = 10033.015.0
VV
v
air ×= = 45%
Solution 3.9 Since the soil is saturated S = 1 Find void ratio Se = ; e = 2.7 = 0.62 sWG 23.0×Find dry unit weight
e1G ws
d +γ
=γ = 3m/kN3.1662.1
8.97.2=
×
Find bulk unit weight ( w1d + )γ=γ = 16.3 = 20.1 kN/( 23.01+ ) 3m
Find porosity
n = 62.162.0
e1e
=+
= 0.38 or 38%
Solution 3.10 The total volume of wet sand, VT is equal to 4.64 x 10-4 m3. The wet weight is 8N and after drying, the dry weight is 7.5N. First, find the water content:
%7.6%100xN5.7
)N5.7N8(www
solids
water =−
=
Next, find the dry unit weight:
334
T
solidsd m/kN2.16
m10x64.4N5.7
Vw
==γ −
Then, using the relationship:
ws
d e1G
γ+
=γ , solve it for the void ratio, e.
64.01m/kN81.9m/kN2.16
70.270.21Ge 33w
d
s =−⎟⎟⎠
⎞⎜⎜⎝
⎛=−γ
γ=
Finally, use the relationship:
sGweS = to obtain the saturation, S = 0.28 = 28%
Solution 3.11 Given: w = 0.10, γ = 18.5 kN/ 3m , Gs = 2.7, emax = 0.87 (loose state), emin = 0.51 (dense state) (a) Relative Density (Dr)
Dr = minmax
max
eeee
−−
x 100
Determine e by first calculating dγ
3d
18.516.8 kN /m
1 w 1.1γ
γ = = =+
s w s w
dd
G G 2.7 9.8e 1 1; e 0.575
1 e 16.8γ γ ×
γ = ⇒ = − = − =+ γ
Calculate Dr: Dr = 0.87 0.575
100 81.9%0.87 0.51
−× =
−
(b) Degree of Saturation
swG 0.1 2.7S 4
e 0.575×
= = = 7%
Solution 3.12 Given: e = 1.2 Determine bulk (γ) and effective (γ′) unit weights for the following degrees of saturation: (a) S = 0.75
3sw
G Se 2.7 0.75 1.29.8 16 kN/m
1 e 1 1.2+ + ×⎛ ⎞ ⎛ ⎞
γ = γ = =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠
3s
sat w
G e 2.7 1.29.8 17.4 kN /m
1 e 1 1.2+ +⎛ ⎞ ⎛ ⎞γ = γ = =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠
wsat γ−γ=γ′ = 17.4 – 9.8 = 7.6 kN/ 3m (b) S = 0.95
2.7 0.95 1.29.8
1 1.2+ ×⎛
γ = ⎜ +⎝ ⎠
⎞⎟ = 17.1 kN/ 3m
wsat γ−γ=γ′ = 17.1 – 9.8 = 7.3 kN/ 3m
(c) For S = 1, satγ = 17.4 kN/ 3m % Error in γ if S = 0.95 but assumed to be 1.0:
% Error = 17.4 17.1100
17.1−
× = 1.8%
Solution 3.13 Sieve size Weight % Rtd Cum.% Rtd % finer
4.75 3.1 12.8 12.8 87.22 5.8 24.0 36.8 63.2
0.85 3.8 15.7 52.5 47.50.25 2.6 10.7 63.2 36.8
0.075 6.8 28.1 91.3 8.7Pan 2.1Sum 24.2 Grain Size Distribution:
Problem 3.13
0
20
40
60
80
100
0.010.1110
Limiting Diameter (mm)
Perc
ent F
iner
(%)
From the grain size distribution, d10 = 0.085 mm, d30 = 0.18 mm, and d60 = 1.70 mm. Strictly speaking, for a sand to be graded SW, it must have a Cu greater than 6 and 1 < Cc < 3.
60u
10
d 1.70C 20.0
d 0.085= = =
2 2
30c
60 10
d 0.18C 0
d d (1.70) (0.085)= = = .22
Hence, by strict application of the classification system, this is an SP. However, there is only about 10% fines and Cu (20) is much greater than 6. Recognize that some gap-grading is happening here. Better to call it an internally unstable well-graded (silty) sand. This soil would be a poor choice to use as a filter.
Solution 3.14
0
20
40
60
80
100
0.010.1110
Limiting Diameter (mm)
Perc
ent F
iner
by
Wei
ght Soil A
Soil BSoil C
Soil A: More than 50% of the sample is greater than 0.075 mm (no. 200 sieve) and all of the soil is smaller than 4.75 mm (no. 4 sieve). Hence, the first letter will be S. For the approximate 37% of the sample that is finer than the no. 200 sieve, there is a ratio of 4:1 clay to silt. This amount of clay should produce significant plasticity in the soil and plot above the A-line. Hence, the soil should be classified as SC, a clayey sand. Soil B: 100% of the sample passes the no. 4 sieve and 0% passes the no. 200 sieve. Hence, the first letter will be S. Look at the grain size distribution. From the grain size distribution, d10 = 0.12 mm, d30 = 0.40 mm, and d60 = 0.75 mm.
60u
10
d 0 75C 6d 0 12
. .
.= = = 25
2 2
30c
60 10
d 0 40C 1d d 0 75 0 12
. .( . ) ( . )
= = = 78
This soil satisfies all criteria for a well-graded sand, SW. Soil C: 100% of the sample passes the no. 4 sieve and 0% passes the no. 200 sieve. Hence, the first letter will be S. Look at the grain size distribution. From the grain size distribution, d10 = 0.10 mm, d30 = 0.19 mm, and d60 = 0.43 mm.
60u
10
d 0 43C 4d 0 10
. .
.= = = 3
2 2
30
60 10
d 0 19Cc 0 84d d 0 43 0 10
. .( . ) ( . )
= = =
This soil does not satisfy Cc (CC = 0.84 which is less than 1) criteria and does not satisfy Cu (Cu = 4.3 < 6) criteria for a well-graded sand and hence, is a poorly-graded sand, SP.
Solution 3.15 . ( Dr. Budhu check if this solution is correct?)
Percent Finer sieve
opening(mm) Soil A B C 4.75 100.00 100.00 100.00
2 (No. 10) 89.89 74.28 97.00 0.85 77.03 63.82 77.40
0.425(No. 40) 56.81 31.64 59.40 0.15 47.75 11.79 23.00
0.075(No. 200) 34.13 0.00 1.20
Step1: % passing No 200 sieve. Soil A: 34.13% passing i.e. < 35%, Soil B: 0% passing i.e. < 35%, Soil C: 1.2 % passing i.e. < 35% Soils A, B and C are granular Step2: Make a table according to Table 3.7
%Passing
Sieve No Soil A B C No. 10 89.89 74.28 97.00No. 40 56.81 31.64 59.40
No. 200 34.13 0.00 1.20LL 23% - - PL 8% - - PL 15% - -
Group Index Soil A: GI = 0.01(F-15)(PI-10) = 0.96 =1 Soil B: GI = 0 (non plastic) Soil C: GI = 0 (non plastic) Step3: Use Table 3.7 with the values in step 2 to classify soils according to AASHTO Soil A: A-2-6 (1) Soil B: A-1-b Soil C: A-3
Solution 3.16 (a) Determine LLLL = 40%
30354045505560
1 10 100
Number of blows
Wat
er c
onte
nt (%
)
25
(b) Liquidity Index (LI) PI = 0.40 – 0.23 = 0.17
LI = 17.0
23.038.0I
ww
p
pl −=
− = 0.88
(c) Since LI is within the range 0 < < 1, the soil is plastic and brittle failure is unlikely. LI
Solution 3.17
y = 22.972x0.3024
R2 = 0.9999
10
100
1 10 100
Penetration (mm)
Wat
er c
onte
nt (%
)
LL = 58% c = 22.97, m = .3024
m 0.3024PL c(2) 22.97(2) 28.3= = = PI = LL – PL = 58 – 28.3 = 29.7%
Solution 3.18
w(%) W γd (kN/m3) 12 17 16.08 13 18 16.87 14 18.7 17.38 16 19.3 17.62 19 18.9 16.82
15.00
16.00
17.00
18.00
19.00
20.00
21.00
10 12 14 16 18 20
water content (%)
dry
unit
wei
ght (
kN/m
3)
zero air voids
Optimum Moisture Content = 15.5%, Maximum dry unit weight = 17.7 kN/m3
e = 5.017.178.97.21G
d
ws =−⎟
⎠⎞
⎜⎝⎛ ×=−⎟⎟
⎠
⎞⎜⎜⎝
⎛γγ
S = sw G 0.155 (2.7)0.84
e 0.5= = = 84%
Zero air voids – use ( )d s w sG /(1 wGγ = γ +
w(%) γd
12 19.98 13 19.59 14 19.20 16 18.48 19 17.49
Solution 3.19 Se = wGs
eembankment = 455.095.0
7.216.0=
×
fill
embankment
fill
embankment
e1e1
VV
++
=
vfill = ( ) 3m2.316151455.01
3.11000,200=
++×
Solution 3.20 (a)
s s ssat w s
s s ssat w w
s s
s
G e G w G w; e G w1 e S 1G G w G (1 w)1 G w 1 G w
G 2.7
+γ = γ = = =
++ +
∴γ = γ = γ+ +
=
Depth (m) 1 2 3 4 5 6
w (%) 21.3 23.6 6.1 32.7 41.5 42
γsat(kN/m3) 20.4 20.0 24.1 18.6 17.7 17.6
0
1
2
3
4
5
6
7
0 10 20 30 40 5w (%); (kPa)
Dep
th (m
)
0
saturated unit weight
water content
(b) The water content at depth = 3m is very low in comparison with the other values and is rather suspicious (c) The soil at depth = 3m is likely a thin sand layer.
Solution 3.21
(a) Fill : 3d m/kN9.15
082.012.17
=+
=γ
Embankment: ( ) 3
d m/kN1.181995.0 ==γ
Minimum fill needed: per 3m of embankment: 9.151.18
= = 1.14 3m
(b) and water in fill = 0.82 = 1.3 kN sW wWW = 9.15× water needed = .11 = 2 kN 1.18×
Additional water = 2 – 1.3 = 0.7 kN per 3m (c) water needed = 0.12 = 2.17 kN 1.18×
additional water = 2.17 – 1.3 = 0.87 kN
(d) Number of truck loads = 400,11m10
m000,1143
3=
(e) Find Cost: Distance: 10 km Extra charge = (10-2) 3(10 2) 0.5 $4 / m− × =
Purchase, load, and spread and compact per 3m = $(15 + 4 + 1.02) = $20.02 Cost = 114,000 × $20.02 = $2,282,280
Solution 3.22 I. Suitability of Soils ITEM PIT 1 PIT 2 % Fines CuCc
5 7 1.3
22 --- ---
Pit 2 contains too many fines (and may not be suitable) Pit 1 has few fines and is well graded ⇒ it will compact to higher densities, which implies it will have higher shear strength and lower compressibility. Pit 1 contains a better soil for the embankment than Pit 2. Cost
Volume of embankment = 3102.242.22122.213 ×⎥
⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛ ×××+× x 10 = 47.96 34 m10×
Extra mileage charge/ 3m
Pit 1: = 190 km (∑=
+10
1nn28 )
)Pit 2: = 150 km (∑=
+10
1nn24
Pit 1 : ( ) = $51,374,752 410 190 0.5 1.02 1.1 47.96 10+ × + + × ×
Pit 2: = $46,334,156 ( ) 412 150 0.55 1.26 0.85 47.96 10 1.1+ × + + × × ×The actual cost is likely to be more because we have calculated the compacted volume. The volume of pit material required is likely to be different for each pit.