Ch04 Solution Manual Soil Mechanics and Foundations
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Exercise 4.1 Prove the following relations:
a) a) d b) b) c) c) d) d)
Gsw
1 e
In the above equations, n is the porosity as a ratio not as a percentage. Solution 4.1 Prove the following: d
(a) (a)
Gsw
1 e
Start with fundamental definitions: d d
(b) (b)
e=
S=
V Vs
; Ws
Vs w G s
; V V s (l e) ; subsitutefor Ws and Vs
w Gs G s w d Vs (l e) le n
1 n
wG s e
Ws
wG s (1 n) n
; ; ; ; (c)
(d)
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
d
;
1 e max
max
Gsw 1 e min
w 1 e
Gs
d
;
Relative density in terms of void ratios is:
Dr
emax
emax
e emin
Solving for the appropriate void ratios and substituting into
e max
G s w d
Dr
1
d
min
d
1
d d = d max
d =
min
1
d
max
dd dd
min
min
max
D r =
d
max
d
=
d
min
d
min
d
min
max
max
min
d
1
min
d
1
d max
1
d d
d
dd
1
d
min
min
= d max d min
max
min
min
d d max
1
1
min
1
min
G s w d
d min
d d d d
d d
1
e
max
ma x
d
1; 1
d Gs w
1
G s w d
Gs w
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: G s = 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% w water ater 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 0.8 o 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 maximum max imum and minimum porosities, and maximum and minimum void ratios. Solution 4.4 Strategy
It is easiest to consider consider that each sphere occupies a unit volume. . If D is the diameter of the sphere, the volume occupied by it in the array is D 3 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: 3 D Vsphere 6
Step 2. Calculate solid volume ratio occupied by sphere sphere
D 3 6 3
VS
6
D
Step 3. Calculate the the porosity n
1
6
0.4764
Step 4. Calculate the void ratio e
n
1
0.4764 n
0.5236
0.91
Dense Array
Step 1. Calculate volume of sphere of diameter D: 3 D Vsphere 6 Step 2. Calculate volume of space occupied by sphere. The height of space occupied is a
tetrahedron. Height of tetrahedron tetrahedron is D D D
3 4
D
2 3
D
3
1 2
2 3
. Space occupied is: is:
D3 6 VS D
3
1
18
2
Step 3. Calculate Porosity n
1
18
0.2595
Step 4. Calculate void ratio e
n
1
0.2595 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 cm3. Determine the specific gravity of the sand. Solution 4.5
Md = mass of dry soil = 100 grams Mwe = mass of water of equivalent volume vo lume to dry soil = volume of water displaced x density of water (1 gram/cm3) = (537.5 – 500) 500) x 1 = 37.5 grams Gs
M d 100 M we
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 = 154.5 154. 5 grams Mass of pycnometer and water = 138.5 grams Solution 4.6
Md = mass of dry soil = 65.8 – 40.1 40.1 = 25.7 grams Mwe = mass of water displaced by b y the soil particles = 138.5 – 154.5 154.5 + 25.7 = 9.7 grams Gs
Md 25.7 M we
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) (a) w
(b) (b) e
Water Content (w) Mw
Ms
MT
Ms
Ms
520
400
400
0.30
30%
Void Ratio (e) Gsw S
2.7 0.30
0.81
Alternatively: 3 (mass density of water) Gs = 2.7, w 1 gm/ cm Vs
Ms
Gss
400 2.7
1
148.1cm3
Because S = 1, Vv = Vw Mw
Vw = Vv = w
e
(c) (c)
3
12 120 0 cm3
0.81
148.1 cm
Saturated Unit Weight (sat)
G sat s 1 (d) (d)
400
1
120 120 cm3
Vv Vs
520
e e
2.7 0.81 9.8 8 19.0 19.0 kN kN/m /m3 w 1 0.81 9.
Effective Unit Weight
( )
sat w 19.0 9.8 9. 9.2 kN / m3
Exercise 4.8
A soil sample has a bulk unit weight of 19.8 kN/m 3 at a water content of 10%. 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) (a) 19.8 d = 18 kN / m3 1 w 1.1 2.7 9.8 G d s w ; e 1 1.47 1 0.47 1 e 18 (b) Degree of Saturation (S) (b) wGs 0.1 0.1 2.7 2.7 0.57 or 57% Se wGs , S e 0.47 (c) % of Air voids (c) Volume of air = 100 – 57 57 = 43 % Alternative solution Assume V = 1 m3 If = 19.8 kN/m3 ; W = 19.8 kN If d = 18 kN/m3 ; Ws = 18.0 kN Vs
Ws Gs w
18 2.7 9.8
0.68m3
V Vs = 1 – 0.68 0.68 = 0.32 m 3 Ws = 19.8 – 18 18 = 1.8 kN Weight of water per m 3 = W – W 1.8 Vw 0.18m3 9.8 Vair V v Vw = 0.32 – 0.18 0.18 = 0.14 m 3 Vv
% of air voids =
Vair Vv
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
Wwater Wsolids
(8 7.5 7.5) 7.5
100% 6.7%
Next, find the dry unit weight: d
Wsolids VT
7.5N 4
4.64 4.64 10 m
3
16.2 kN / m3
Then, using the relationship:
d 1Gs e w , solve it for the void ratio, e. e
Gs d
w
2.70 9.8 kN / m3 1 1 3 16.2 kN / m
0.63
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
Gs w 1 e
=
2.7 9.8 1.63
16.23 kN / m3
Find bulk unit weight d 1 w = 16.23 (1 + 0.235) = 20.04 kN/ m 3 Find porosity n=
e 1 e
0.63 0.39 or 39% 1.63
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 dr y weight of 1.1 N. Determine (a) the degree de gree 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 = wG s /e /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 container is 0.29 kg. The mass of the container is 0.06 kg and its volume is 0.15x10 -3 m3. 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 0.06 = 0.27 kg Mdry + Mcan = 0.29 kg => Mdry = 0.29 – 0.06 0.06 = 0.23 kg Mw = 0.33 – 0.29 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 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; To saturate 0.15x10-3 m3 soil we need
water;
To saturate 1 m3 soil, we need x m3 water.
Exercise 4.13
A sand has a natural water content of 5% and bulk unit weight of 18.0 kN/m3. The void ratios corresponding to the densest and loosest state of this 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) (a)
=
18 kN/ m 3 , Gs = 2.7, emax = 0.87 (loose state), emin = 0.51 (dense state)
Relative Density (Dr)
Dr =
e max e max
e
e min
x 100
Determine “e” by first calculating d
d
e
18 17 17.1 .14 4 kN/ m3 1 w 1.05 Gs
d
w 2.7 2.7 9.8 9.8 1 1 0.543 17.14
Calculate Dr: Dr
(b) S
0.543 100 90.8% 0.87 0.51
0.87
Degree of Saturation 0.05 2.7 25% wGs 0. e
0.543
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) (a)
S = 0.75
(b) (b)
S = 0.95
(c) (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. index . (c) Do you expect a brittle type of failure for this soil? Why? Solution 4.15
(a) Determine (a) LL = 40%
LL
60
) 55 % ( 50 t n e t n 45 o c r 40 e t a W35 30 1
10
25
100
Number of blows (log scale) (b) Liquidity Index (LI) PI =
0.40 – 0.23 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 soil count number (grams) (grams) (grams) (%) Mc Mw Md N w 1 45.3 57.1 54.4 28 29.67 2 43 59.8 56 31 29.23 3 45.2 61.7 57.9 22 29.92 4 45.6 58.4 55.3 18 31.96 Determine the liquid limit. Solution 4.16
(a) Determine LL (a)
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
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 plasticity p lasticity index. If the soil contains 45% clay, calculate the activity. Solution 4.17
75 70
65 60
) % ( 55 t n e t 50 n o c r 45 e t a W40 35 30 1
10
100
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 PL = 58 – 28.3 28.3 = 29.7% Activity (A) =
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 Volume of the container (V1) 32.4 cm3 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) conductedto Atterberg limit tests on soil? Justify your answer. (b) Would Classifyyou thehave soil according USCS, ASTM-CS andthis AASHTO. (c) Is this soil a good foundation material? Justify your answer. Solution 4.19
Coarse-grained
Fine-grained Clay
Sand
Silt
Fine #200
100
Medium #40
Gravel Coarse #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
0.0750.1
0.425
1 Particle size (mm) - logarithmic scale
19.0
10
100
a) a) It is a coarse-grained soil; we don’t need to conduct cond uct Atterberg limit tests. These tests are suitable for fine grained soils. b) 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 P oorly graded sand AASHTO According to left to right elimination process soil is A-1 c) 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
100
10
20
40
100
200
90.8 84.4 77.5 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) 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 PL = 62 – 38 38 = 24 %; A-7-5, A-7-6 PI = 24%
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