Prove the following relations:
Exercise 4.1
Gsw
a)
d
b)
c)
( )
d)
1 e
( ))
In the above equations, n is the porosity as a ratio not as a percentage. Solution 4.1
Prove the following: (a)
d
Gsw 1 e
Start with fundamental definitions: d
d
(b)
(c)
(d)
e=
S=
; Ws Vs w G s ; V Vs (l e) ; subsitutefor Ws and Vs w
Gs
Vs (l e)
Gsw le
d
1 n
e
V Vs
n
wG s
;
Ws
wG s (1 n) n
;
; ;
Saturated unit weight (S=1)
therefore; ( )
Exercise 4.2:
Show that
() (()) {()}
Solution 4.2
Dr = = relative density which is usually defined in terms of min, max and current void ratio (e min, emax, e, respectively). The corresponding dry densities are:
d
Gsw
min
;
1 e max
d
max
Gsw 1 e min
w 1 e
Gs
d
;
Relative density in terms of void ratios is:
Dr
emax
emax
e e
min
Solving for the appropriate void ratios and substituting into
e max
G s w d
Dr
1
min
d
min
d
min
d
min
d
min
= d max
min
1
d
max
dd dd
min
min
max
d d d d max
max
d d d d
min
D r =
max
min
G s w
e
1
d
=
d
min
1
d
min
1
1
1
d
max
1
d d
d
dd
max
min
min
d d
max
min
min
= d max d min
1
d
1
d
max
1
Gs w
1
d d
=
d
1;
max
d
d
G s w
d G s w d
1
min
Gs w
d
e min
min
Gs w
d
1;
d
max
d
min
:
Exercise 4.2:
Show that
() (()) {()}
Solution 4.2
Dr = = relative density which is usually defined in terms of min, max and current void ratio (e min, emax, e, respectively). The corresponding dry densities are:
d
Gsw
min
;
1 e max
d
max
Gsw 1 e min
w 1 e
Gs
d
;
Relative density in terms of void ratios is:
Dr
emax
emax
e e
min
Solving for the appropriate void ratios and substituting into
e max
G s w d
Dr
1
min
d
min
d
min
d
min
d
min
= d max
min
1
d
max
dd dd
min
min
max
d d d d max
max
d d d d
min
D r =
max
min
G s w
e
1
d
=
d
min
1
d
min
1
1
1
d
max
1
d d
d
dd
max
min
min
d d
max
min
min
= d max d min
1
d
1
d
max
1
Gs w
1
d d
=
d
1;
max
d
d
G s w
d G s w d
1
min
Gs w
d
e min
min
Gs w
d
1;
d
max
d
min
:
Exercise 4.3
Tests on a soil gave the following results: Gs = 2.7 and e = 1.96. Make a plot of degree of saturation versus water content for this soil. Solution 4.3
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 will equal 1 at about 71.6% water 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.
1.2 1.0 n o 0.8 i t a r u 0.6 t a s , S 0.4
0.2 0.0 0
20
40
60
w, water content (%)
80
100
Exercise 4.4 Assuming soil particles to be spheres; derive equations for the max imum and minimum porosities, and maximum and minimum void ratios. Solution 4.4 Strategy It is easiest to consider that each sphere occupies a unit volume. . If D is the diameter of 3 the sphere, the volume occupied by it in the array is D for the cubic (loose) array and D
3
1 2
for the dense array.
Loose array
Step 1. Calculate volume of sphere of diameter D: V sph ere
D 3
6
Step 2. Calculate solid volume ratio occupied by sphere
D 3 6 VS 3 6
D
Step 3. Calculate the porosity n
1
6
0.4764
Step 4. Calculate the void ratio 0.4764
n e
1
n
0.5236
0.91
Dense Array
Step 1. Calculate volume of sphere of diameter D: V sph ere
D 3
6
Step 2. Calculate volume of space occupied by sphere. The height of space occupied is a tetrahedron. Height of tetrahedron is D D D
3 4
D
2 3
3
D
1 2
2 3
. Space occupied is:
D3 6 VS D
3
1
18
2
Step 3. Calculate Porosity n
1
18
0.2595
Step 4. Calculate void ratio 0.2595
n e
1
n
1
0.2595
0.35
Exercise 4.5 3
A cylinder has 500 cm of water. After a mass of 100 grams of sand is poured into the cylinder and all air bubbles are removed by a vacuum pump, the water level rises to 537.5 3 cm . Determine the specific gravity of the sand. Solution 4.5
Md = mass of dry soil = 100 grams Mwe = mass of water of equivalent volume to dry soil 3 = volume of water displaced x density of water (1 gram/cm ) = (537.5 – 500) x 1 = 37.5 grams Gs
Md M we
100
37.5
2.67
Exercise 4.6
An ASTM D 854 test was done on a sand. The data are as shown below. Calculate the specific gravity. Mass of pycnometer = 40.1 grams Mass of pycnometer and dry soil = 65.8 grams Mass of pycnometer, dry soil and water = 154. 5 grams Mass of pycnometer and water = 138.5 grams
Solution 4.6
Md = mass of dry soil = 65.8 – 40.1 = 25.7 grams Mwe = mass of water displaced by the soil particles = 138.5 – 154.5 + 25.7 = 9.7 grams Gs
Md
M we
25.7 9.7
2.65
Exercise 4.7
The wet mass of a sample of saturated soil is 520 grams. The dry mass, after oven drying, is 400 grams. Determine the (a) water content, (b) void ratio, (c) saturated unit weight, and (d) effective unit weight.
Solution 4.7
Given: S = 1.0, MT = 520g, Ms = 400 g (a)
Water Content (w)
w
(b)
Mw
Ms
MT
Ms
Ms
520
400
400
0.30
30%
Void Ratio (e)
e
Gsw S
2.7 0.30
0.81
Alternatively: Gs = 2.7, 1 gm/ cm 3 (mass density of water) w
Vs
Ms Gss
400 2.7
1
148.1cm3
Because S = 1, Vv = Vw Mw
Vw = Vv = w
e
(c)
Vv Vs
400
1
120 cm3 3
148.1 cm
120 cm3
0.81
Saturated Unit Weight (sat)
G sat s 1 (d)
520
e e
2.7 0.81 3 w 1 0.81 9.8 19.0 kN/m
Effective Unit Weight ( )
sat
19.0 9.8 9.2 kN / m3 w
Exercise 4.8 3
A soil sample has a bulk unit weight of 19.8 kN/m at a water content of 10%. Determine the void ratio, percentage air in the voids (air voids), and the degree of saturation of this sample. Solution 4.8
(a) Void Ratio (e) 19.8 3 = d 18 kN / m 1 w 1.1 2.7 9.8 Gs w ; e d 1 1.47 1 0.47 1 e 18 (b) Degree of Saturation (S) wG s 0.1 2.7 0.57 or 57% Se wG s , S e 0.47 (c) % of Air voids Volume of air = 100 – 57 = 43 % Alternative solution 3 Assume V = 1 m 3
V s
Vv
If = 19.8 kN/m ; W = 19.8 kN 3 If d = 18 kN/m ; Ws = 18.0 kN W s 18 3 0.68m G s w 2.7 9.8
V Vs = 1 – 0.68 = 0.32 m 3
Weight of water per Vw
1.8 9.8
Vair Vv
0.18m
m
3
= W – Ws = 19.8 – 18 = 1.8 kN
3
Vw = 0.32 – 0.18 = 0.14 m 3
% of air voids = V air V v
0.14 0.32
100
44% ( This is slightly higher because of rounding errors)
Exercise 4.9 −4
3
A wet sand sample has a volume of 4.64 × 10 m and weights 8 N. After oven drying, the weight reduces to 7.5 N. Calculate the water content, void ratio, and degree of saturation. Solution 4.9
w
W water W solids
(8 7.5) 7.5
100% 6.7%
Next, find the dry unit weight:
d
W solids VT
7.5N 4
4.64 10
3
m
16.2 kN / m3
Then, using the relationship:
d
e
Gs 1 e
G s d
w
w , solve it for the void ratio, e. 2.70 9.8 kN / m 3 1 0.63 1 3 16.2 kN / m
Finally, use the relationship:
Exercise 4.10
A saturated silty clay encountered in a deep excavation is found to have a water content of 23.5%. Determine its porosity and bulk unit weight.
Solution 4.10
Since the soil is saturated S = 1 Find void ratio Se = wG s ; e = 2.7 x 0.235 = 0.63 Find dry unit weight d
G s w 1 e
=
2.7 9.8 1.63
16.23
3
kN / m
Find bulk unit weight
d 1 w
= 16.23 (1 + 0.235) = 20.04 kN/ m
Find porosity n=
e 1 e
0.63
1.63
0.39 or
39%
3
Exercise 4.11
A soil sample of diameter 37.5 mm and length 75 mm has a wet weight of 1.32 N and dry weight of 1.1 N. Determine (a) the degree of saturation (b) the porosity (c) the bulk unit weight and (d) the dry unit weight. Solution 4.11
Bulk unit weight Dry unit weight
⁄
⁄
You can also calculate the water content then d;
= 20%
⁄
Void ratio
Degree of saturation, S = wGs/e = 0.2 x 2.7/0.99 = 0.54.5 = 54.5% or 55% Porosity
Exercise 4.12
The mass of a wet sample of soil and its container is 0.33 kg. The dry mass of the soil and its -3 3 container is 0.29 kg. The mass of the container is 0.06 kg and its volume is 0.15x10 m . Determine the following. (a) The bulk, dry, and saturated unit weights of the soil. (b) The void ratio and the degree of saturation. (c) How much air void is in the soil? (d) The weight of water required to saturate 1 m3 of this soil. Solution 4.12
Mwet + Mcan = 0.33 kg => Mwet = 0.33 – 0.06 = 0.27 kg Mdry + Mcan = 0.29 kg => Mdry = 0.29 – 0.06 = 0.23 kg Mw = 0.33 – 0.29 = 0.04 kg
a)
To find saturated unit weight, first we need to find void ratio (e);
Now we can find saturated unit weight;
( ) ( )
c) To find air voids in the soil first we need to find the volume of voids in the soil
To find the volume of the voids we will use the equation for porosity;
Now we need to find the volume of the water;
Now we can find the volume of the air;
d) To saturate this soil, we need to fill the air voids with water. To solve this question we will use direct proportion; -3
3
To saturate 0.15x10 m soil we need 3
water;
3
To saturate 1 m soil, we need x m water.
Exercise 4.13 3
A sand has a natural water content of 5% and bulk unit weight of 18.0 kN/m . The void ratios corresponding to the densest and loosest state of th is soil are 0.51 and 0.87. Find the relative density and degree of saturation. Solution 4.13
Given: w = 0.05, (a) Dr =
=
18 kN/ m 3 , Gs = 2.7, emax = 0.87 (loose state), emin = 0.51 (dense state)
Relative Density (Dr ) e max e e max e min
x 100
Determine e by first calculating “
d
e
”
1 w
Gs
w
18 1.05
1
3
17.14 kN /m
2.7 9.8
d
d
17.14
1
0.543
Calculate Dr : Dr
(b) S
0.87 0.543 0.87 0.51
100 90.8%
Degree of Saturation wGs e
0.05 2.7 0.543
25%
Exercise 4.14
The void ratio of a soil is 1.2. Determine the bulk and effective unit weights for the following degrees of saturation: (a) 75%, (b) 95%, and (c) 100%. What is the percentage error in the bulk unit weight if the soil were 95% saturated but assumed to be 100% saturated? Solution 4.14
Given: e = 1.2 Determine bulk () and effective () unit weights for the following degrees of saturation: (a)
S = 0.75
( ) ( ) ( ) ( ) (b)
S = 0.95
( ) ( ) ( ) ( ) (c)
For S = 1,
sat
= 17.4 kN/ m 3
% Error in if S = 0.95 but assumed to be 1.0:
Solution 4.15
The following results were obtained from a liquid limit test on a clay using the Casagrande cup device. Number of blows 6 12 20 28 32 Water content (%) 52.5 47.1 43.2 38.6 37.0 (a) Determine the liquid limit of this clay. (b) If the natural water content is 38% and the plastic limit is 23%, calculate the liquidity index . (c) Do you expect a brittle type of failure for this soil? Why? Solution 4.15
(a) Determine LL = 40%
LL
60
) 55 % ( t 50 n e t n 45 o c r 40 e t a W35 30 1
25
10
100
Number of blows (log scale) (b) Liquidity Index (LI) PI =
0.40 – 0.23 = 0.17
(c) Since
LI is
within the range 0 <
LI < 1,
the soil is plastic and brittle failure is unlikely.
Exercise 4.16
The following data were recorded from a liquid limit test on a clay using the Casagrande cup device. Container Container Water Test Container and dry Blow and wet soil content number (grams) soil count (grams) (%) (grams) 1 2 3 4
Mc
Mw
Md
N
w
45.3 43 45.2 45.6
57.1 59.8 61.7 58.4
54.4 56 57.9 55.3
28 31 22 18
Determine the liquid limit. Solution 4.16
(a) Determine
LL
32.5 32
) 31.5 % ( t n 31 e t n30.5 o c r 30 e t a w29.5 29 28.5 10
25
Number of blows (N) (log scale)
LL = 30%
100
29.67 29.23 29.92 31.96
Exercise 4.17
A fall cone test was carried out on a soil to determine its liquid and plastic limits using a cone of mass 80 grams. The following results were obtained:
Penetration (mm) Water content (%)
8 43.1
15 52.0
80 gram cone 19 56.1
28 62.9
Determine (a) the liquid and plastic limits and (b) the p lasticity index. If the soil contains 45% clay, calculate the activity. Solution 4.17
75
70
65
60
) % ( 55 t n e 50 t n o c r 45 e t a W40 35
30 1
10
Penetration (mm) (log scale) LL = 58% c = 22.97, m = .3024 PL
c(2)m
22.97(2)0.3024
28.3%
PI = LL – PL = 58 – 28.3 = 29.7% Activity (A) =
100
Exercise 4.18
The following results were recorded in a shrinkage limit test using mercury. Mass of container 17.0grams Mass of wet soil and container 72.3 grams Mass of dish 132.40 grams Mass of dish and displaced mercury 486.1 grams Mass of dry soil and container 58.2 grams 3 Volume of the container (V1) 32.4 cm Determine the shrinkage limit. Solution 4.18
( ) ( ) ( )
Exercise 4.19
The results of a particle size analysis of a soil are given in the following table. No Atterberg limits tests were conducted. Sieve No. 9.53 mm (3/8”) 4 10 20 40 100 200 % finer
100
89.8 70.2
62.5 49.8
28.6 4.1
(a) Would you have conducted Atterberg limit tests on this soil? Justify your answer. (b) Classify the soil according to USCS, ASTM-CS and AASHTO. (c) Is this soil a good foundation material? Justify your answer. Solution 4.19
Coarse-grained
Fine-grained
Sand
Silt
Clay
Fine
100
Gravel Coarse
Medium
#200
#40
0.075
0.425
#10
#4
2.0
4.75
Fine
Coarse
#3/8" #3/4"
90 80 70 60
r e n i50 F %
40 30 20 10 0 0.001
0.01
0.1
1
10
19.0
100
Particle size (mm) - logarithmic scale
a) It is a coarse-grained soil; we don’t need to conduct Atterberg limit tests. These tests are suitable for fine grained soils. b) Classify the soil USCS 50% of particles are bigger than 0.075 mm; soil is coarse-grained Sand fraction is bigger than gravel fraction; soil is sand Clay + silt fraction is less than 4%; soil is SW or SP D10 = 0.09, D30 = 0.16, D60 = 0.7
( )
Cc is not between 1 and 3; soil is SP (Poorly graded sand) ASTM-CS Sand fraction is bigger than gravel fraction; soil is sand Clay + silt fraction is less than 4%; soil is SW or SP Soil is SP, gravel is less than 15%; Poorly graded sand AASHTO According to left to right elimination process soil is A-1 c) According to ‘Engineering Use Chart’ soil hasnumber 5 rating which is an average rating.
Exercise 4.20
The results of a particle size analysis of a soil are given in the following table. Atterberg limits tests gave LL = 62% and PL = 38%. The clay content is 37%. Sieve No. 9.53 mm (3/8”)
4
% finer
90.8 84.4 77.5
100
10
20
40
100
200
71.8 65.6 62.8
(a) Classify the soil according to ASTM-CS and AASHTO. (b) Rate this soil as a subgrade for a highway. Solution 4.20
Coarse-grained
Fine-grained Clay
Silt
100
Sand #200
Fine
#40
Medium
Gravel Coarse #10
#4
2.0
4.75
Fine
#3/8" #3/4"
Coarse
90 80 70 60
r e n i50 F %
40 30 20 10 0 0.001
0.01
0.075
0.425
0.1 1 Particle size (mm) - logarithmic scale
10
a) Classify the soil ASTM-CS 50% of particles are smaller than 0.075 mm; soil is fine-grained According to plasticity chart; soil is MH (high compressible silt AASHTO More than 35% passing No. 200; A-4, A-5, A-6, A-7 Liquid limit is bigger than 40; A-5 or A-7 Plasticity index (PI) = LL – PL = 62 – 38 = 24 %; A-7-5, A-7-6 PI = 24%
b) It is poor soil for subgrade
19.0
100
Exercise 4.21
The water contents of soil samples taken at different depths are given in the table below. Depth (m) 1 2 3 4 5 6 w (%) 21.3 23.6 6.1 32.7 41.5 42.0 The ground water level is at the surface. Assume Gs = 2.7. a) Plot on the same graph (a) depth versus water content and (b) depth versus saturated unit weight. b) Are there any questionable water content value(s)? If so, at what depth? c) If the water contents are all correct, what type or types of soils (e.g., clay, sand) are probably at the site? Solution 4.21
sat
Gs e
w; e
1 e Gs Gs w
(a) sat
1 Gsw
Gsw S
w
G sw
1 G s (1 w)
Gsw
1 Gsw
w
G s 2.7 Depth (m)
1
2
3
4
5
6
w (%)
21.3
23.6
6.1
32.7
41.5
42
20.4
20.0
24.1
18.6
17.7
17.6
3
γsat(kN/m
)
w (%); 0
10
20
sat (kN/m
30
3)
40
50
0 1 ) m 2 ( h t 3 p e D 4
water content saturated unit weight
5 6 7 (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 could possibly be a thin sand layer.
Exercise 4.22
A fine-grained soil has a liquid limit of 200% and a plastic limit of 45%. The natural water content of the soil in the field is 60% and the clay content is 63%. a) Calculate the plasticity index, the liquidity index and the activity. b) What is the soil state (e.g. liquid) in the field? c) What is the predominant mineral in this soil? d) This soil is under a rectangular concrete slab, 15 m x 50 m, used as a foundation for a building. A water pipe, 100 mm diameter, is located in a trench 450 mm below the center of the slab. The trench, 300 mm wide and 450 mm deep, running along the length of the slab, was backfilled with the same soil. If this pipe were to leak, what effect would it have on the foundation? Draw a neat sketch of the existing trench and pipe, and show in another sketch how you would mitigate any water related issue related to the pipe and the soil. Explain why your mitigation method is better than the existing construction. Solution 4.22
a) PI= LL – PL = 200 - 45 = 155% LI = (w – PL)/PI = (60 – 45)/155 = 0.096 A = PI/Clay fraction (%) = 155/63 = 2.46 b) Soil is in plastic state PL
Concrete slab
Fine-grained soil
Montmorillonite is also known as swelling or expansive cla y. Because of the water leaking, the soil would swell and this would cause upward movement of the slab. To mitigate this problem you could investigate the solution below with the lowest cost. 1. Put a compressible material (foam) between the soil and the slab. Or a layer of sand/gravel with good drainage. 2. Re-locate the trench and pipeline away from the foundation
Exercise 4.23
An elliptical artificial island is required for a reclamation project. The major axis of the ellipse is 10 km and the minor axis is 7.5 km (Fig. P4.23). A rock breakwater, 100 m thick, forms the edges of the island. The area within the breakwater is to be filled with sand. The sand in its loosest state has a porosity of 50% and the desired porosity, when compacted, is 20%. Assuming an average thickness of the completed island is 200 m, determine the quantity of sand required.
Solution 4.23
First we have to find the volume of the compacted sand, which is equal to; V = (Area of the elliptical artificial island) x (height) Area of the elliptical artificial island =
A B
2A = 7.5 km – 2 x (0.1 km) => A = 3.65 km 2B = 10 km – 2 x (0.1 km) => B = 4.9 km
2
Area of the elliptical artificial island = = 3.14 x 3.65 x 4.9 = 56.187 km Height = 200m = 0.2 km 3
V = 56.187 x 0.2 = 11.237 km
3
At the end of the project, the volume of the compacted sand will be 11.237 km
Now, find the volume of the voids at dense state. We can find it by using porosity equation;
Now find the volume of the sand particles (Vs) at dense state. We can use void ratio equation; For compacted state void ratio will be;
Volume of the sand particles at dense state;
Volume of the sand particles will not change during compaction. Only volume of the voids will change. So at loose state volume of the sands have to be equal to the volume of the sands at dense state. At loose state void ratio will be;
Volume of the voids at loose state;
Exercise 4.24
The highway embankment from Noscut to Windsor Forest, described in the sample practical situation, is 10 km long. The average cross section of the embankment is shown in Fig. P4.24a. The gradation curves for the soils at the two borrow pits are shown in Fig. P4.24b. Pit 1 is located 5 km from the start of the embankment while pit 2 is 3 km away. Estimated costs for various earthmoving operations are shown in the table below. You are given 10 minutes by the stakeholder’s committee to present your recommendations. Prepare your presentation. The available visual aid equipment is an LCD projector.
Operation
Cost Pit 2
Pit 1
Purchase and load borrow pit material at site, haul 2 km round trip, and $10/m spread with 200 HP dozer Extra mileage charge for each km $0.50/m Compaction $1.02/m Miscellaneous $1.10/m
$12/m $0.55/m $1.26 $0.85/m
Solution 4.24
I.
Suitability of Soils
% Fines Cu Cc
PIT 1 5 7 1.3
PIT 2 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
1
Volume of embankment = 13 2.2 2 2.2 4 2.2 10 x 10 3 = 47.96 2
4
10 m
3
Extra mileage charge/m3 10
Pit 1: 8 2n = 190 km n
1
10
Pit 2: 4 2n = 150 km n
1
Pit 1 : 10 190 0.5 1.02 1.1 47.96 10 = $51,374,752 4
Pit 2: 12 150 0.55 1.26 0.85 47.96 10 1.1 = $46,334,156 4
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.
Exercise 4.25
The soil profiles for four boreholes (BH) at a site proposed for an office building are shown in Fig. P4.23. The soils in each borehole were classified using ASTM-CS. Sketch the soil profiles along a diagonal line linking boreholes 1, 2 and 3 and along a line linking boreholes 3 and 4.
Solution 4.25
