|
Soil number |
Mass characteristics |
Material characteristics |
Complete name |
|
1 |
Soil material |
Color |
Organic soils |
|
2 |
Color |
Shape |
Boulders |
|
3 |
Consistency |
Texture |
Fine soils |
Q 2
|
Size mm |
Material retained g |
Percentage retained % |
Percentage passing % |
|
37.500 |
0 |
0 |
100 |
|
28.000 |
37 |
5.27 |
94.73 |
|
20.000 |
24 |
3.42 |
91.31 |
|
14.000 |
73 |
10.40 |
80.91 |
|
10.000 |
9 |
1.28 |
79.63 |
|
6.300 |
108 |
15.38 |
64.25 |
|
5.000 |
41 |
5.84 |
58.41 |
|
3.350 |
80 |
11.4 |
47.01 |
|
2.000 |
60 |
8.55 |
38.46 |
|
1.180 |
97 |
13.82 |
24.64 |
|
0.600 |
66 |
9.40 |
15.24 |
|
0.425 |
29 |
4.13 |
11.11 |
|
0.300 |
23 |
3.28 |
7.83 |
|
0.212 |
21 |
2.99 |
4.84 |
|
0.150 |
17 |
2.42 |
2.42 |
|
0.063 |
17 |
2.42 |
2.42 |
D10 = 0.75
D30 = 2.25
D60 = 6.1
Cu = 6.1/0.75 = 8.133
CG = 2.252/ (0.75x 6.1) = 1.1066
The soil is therefore well graded since Cu is greater than 5 and CG is between 0.5 and 2.0, which is 1.1
The liquid limit is taken at 20mm penetration which is 68.5%
Using the chart, the soil will be classified to have high plasticity, CL
Q 5
Bulk density
The bulk density will be (Gs+ Sr * e)/ (1+e) ?w
= (2.70 + 0.82* 0.738)/(1+0.738) * 1000
= 3.30516/1.738 *1000
= 1901.70 kg/m3
Water content
Sr = mGs/e
0.82= m*2.70/0.738
M = (0.82*0.738)/2.70
M= water content= 0.22413
Q 7
The stress in each soil is given by the formula ρgz
Where ρ is the unit weight of the soil
g is the gravitational pull
z height of the soil
The following stresses are available when the water table is 3 meters below the surface of the soil
Stress on 3m of sand above water table
ρgz = 17*9.81*3 = 500.31 N/m2
stress on 2m of sand below water table
ρgz = 21*9.81*2 = 412.02 N/m2
stress as a result of saturated clay
ρgz = 19*9.81*9 = 1677.51 N/m2
stress as a result of saturated sand
ρgz = 21*9.81*5 = 1030.05 N/m2
pore pressure
ρwgz = 1000*9.81*16 = 156,960 N/m2
total stresses = 160579.89 N/m2
Immediately after the rise of water table, the sand on 3m from the top will not be saturated. Nevertheless, the pore pressure will increase.
Stress on 3m of sand above water table
ρgz = 17*9.81*3 = 500.31 N/m2
stress on 2m of sand below water table
ρgz = 21*9.81*2 = 412.02 N/m2
stress as a result of saturated clay
ρgz = 19*9.81*9 = 1677.51 N/m2
stress as a result of saturated sand
ρgz = 21*9.81*5 = 1030.05 N/m2
pore pressure
ρwgz = 1000*9.81*19 = 186,390 N/m2
total stresses = 190,009.89 N/m2
Q 1
|
Soil number |
Mass characteristics |
Material characteristics |
Complete name |
|
1 |
Soil material |
Color |
Organic soils |
|
2 |
Color |
Shape |
Boulders |
|
3 |
Consistency |
Texture |
Fine soils |
Q 2
|
Size mm |
Material retained g |
Percentage retained % |
Percentage passing % |
|
37.500 |
0 |
0 |
100 |
|
28.000 |
37 |
5.27 |
94.73 |
|
20.000 |
24 |
3.42 |
91.31 |
|
14.000 |
73 |
10.40 |
80.91 |
|
10.000 |
9 |
1.28 |
79.63 |
|
6.300 |
108 |
15.38 |
64.25 |
|
5.000 |
41 |
5.84 |
58.41 |
|
3.350 |
80 |
11.4 |
47.01 |
|
2.000 |
60 |
8.55 |
38.46 |
|
1.180 |
97 |
13.82 |
24.64 |
|
0.600 |
66 |
9.40 |
15.24 |
|
0.425 |
29 |
4.13 |
11.11 |
|
0.300 |
23 |
3.28 |
7.83 |
|
0.212 |
21 |
2.99 |
4.84 |
|
0.150 |
17 |
2.42 |
2.42 |
|
0.063 |
17 |
2.42 |
2.42 |
Q 3
D10 = 0.75
D30 = 2.25
D60 = 6.1
Cu = 6.1/0.75 = 8.133
CG = 2.252/ (0.75x 6.1) = 1.1066
The soil is therefore well graded since Cu is greater than 5 and CG is between 0.5 and 2.0, which is 1.1
The liquid limit is taken at 20mm penetration which is 68.5%
Using the chart, the soil will be classified to have high plasticity, CL
Q 5
Bulk density
The bulk density will be (Gs+ Sr * e)/ (1+e) ?w
= (2.70 + 0.82* 0.738)/(1+0.738) * 1000
= 3.30516/1.738 *1000
= 1901.70 kg/m3
Water content
Sr = mGs/e
0.82= m*2.70/0.738
M = (0.82*0.738)/2.70
M= water content= 0.22413
The stress in each soil is given by the formula ρgz
Where ρ is the unit weight of the soil
g is the gravitational pull
z height of the soil
The following stresses are available when the water table is 3 meters below the surface of the soil
Stress on 3m of sand above water table
ρgz = 17*9.81*3 = 500.31 N/m2
stress on 2m of sand below water table
ρgz = 21*9.81*2 = 412.02 N/m2
stress as a result of saturated clay
ρgz = 19*9.81*9 = 1677.51 N/m2
stress as a result of saturated sand
ρgz = 21*9.81*5 = 1030.05 N/m2
pore pressure
ρwgz = 1000*9.81*16 = 156,960 N/m2
total stresses = 160579.89 N/m2
Immediately after the rise of water table, the sand on 3m from the top will not be saturated. Nevertheless, the pore pressure will increase.
Stress on 3m of sand above water table
ρgz = 17*9.81*3 = 500.31 N/m2
stress on 2m of sand below water table
ρgz = 21*9.81*2 = 412.02 N/m2
stress as a result of saturated clay
ρgz = 19*9.81*9 = 1677.51 N/m2
stress as a result of saturated sand
ρgz = 21*9.81*5 = 1030.05 N/m2
pore pressure
ρwgz = 1000*9.81*19 = 186,390 N/m2
total stresses = 190,009.89 N/m2
after several years f the rise of water table, the pore pressure will increase and the whole sand at the top will be saturated
Stress on 5m of saturated sand layer
ρgz = 21*9.81*5 = 1030.05 N/m2
stress as a result of saturated clay
ρgz = 19*9.81*9 = 1677.51 N/m2
stress as a result of saturated sand
ρgz = 21*9.81*5 = 1030.05 N/m2
pore pressure
ρwgz = 1000*9.81*19 = 186,390 N/m2
total stresses = 190,127.61 N/m2
Q 1
|
Soil number |
Mass characteristics |
Material characteristics |
Complete name |
|
1 |
Soil material |
Color |
Organic soils |
|
2 |
Color |
Shape |
Boulders |
|
3 |
Consistency |
Texture |
Fine soils |
Q 2
|
Size mm |
Material retained g |
Percentage retained % |
Percentage passing % |
|
37.500 |
0 |
0 |
100 |
|
28.000 |
37 |
5.27 |
94.73 |
|
20.000 |
24 |
3.42 |
91.31 |
|
14.000 |
73 |
10.40 |
80.91 |
|
10.000 |
9 |
1.28 |
79.63 |
|
6.300 |
108 |
15.38 |
64.25 |
|
5.000 |
41 |
5.84 |
58.41 |
|
3.350 |
80 |
11.4 |
47.01 |
|
2.000 |
60 |
8.55 |
38.46 |
|
1.180 |
97 |
13.82 |
24.64 |
|
0.600 |
66 |
9.40 |
15.24 |
|
0.425 |
29 |
4.13 |
11.11 |
|
0.300 |
23 |
3.28 |
7.83 |
|
0.212 |
21 |
2.99 |
4.84 |
|
0.150 |
17 |
2.42 |
2.42 |
|
0.063 |
17 |
2.42 |
2.42 |
Q 3
D10 = 0.75
D30 = 2.25
D60 = 6.1
Cu = 6.1/0.75 = 8.133
CG = 2.252/ (0.75x 6.1) = 1.1066
The soil is therefore well graded since Cu is greater than 5 and CG is between 0.5 and 2.0, which is 1.1
The liquid limit is taken at 20mm penetration which is 68.5%
Using the chart, the soil will be classified to have high plasticity, CL
Q 5
Bulk density
The bulk density will be (Gs+ Sr * e)/ (1+e) ?w
= (2.70 + 0.82* 0.738)/(1+0.738) * 1000
= 3.30516/1.738 *1000
= 1901.70 kg/m3
Water content
Sr = mGs/e
0.82= m*2.70/0.738
M = (0.82*0.738)/2.70
M= water content= 0.22413
The stress in each soil is given by the formula ρgz
Where ρ is the unit weight of the soil
g is the gravitational pull
z height of the soil
The following stresses are available when the water table is 3 meters below the surface of the soil
Stress on 3m of sand above water table
ρgz = 17*9.81*3 = 500.31 N/m2
stress on 2m of sand below water table
ρgz = 21*9.81*2 = 412.02 N/m2
stress as a result of saturated clay
ρgz = 19*9.81*9 = 1677.51 N/m2
stress as a result of saturated sand
ρgz = 21*9.81*5 = 1030.05 N/m2
pore pressure
ρwgz = 1000*9.81*16 = 156,960 N/m2
total stresses = 160579.89 N/m2
Immediately after the rise of water table, the sand on 3m from the top will not be saturated. Nevertheless, the pore pressure will increase.
Stress on 3m of sand above water table
ρgz = 17*9.81*3 = 500.31 N/m2
stress on 2m of sand below water table
ρgz = 21*9.81*2 = 412.02 N/m2
stress as a result of saturated clay
ρgz = 19*9.81*9 = 1677.51 N/m2
stress as a result of saturated sand
ρgz = 21*9.81*5 = 1030.05 N/m2
pore pressure
ρwgz = 1000*9.81*19 = 186,390 N/m2
total stresses = 190,009.89 N/m2
after several years f the rise of water table, the pore pressure will increase and the whole sand at the top will be saturated
Stress on 5m of saturated sand layer
ρgz = 21*9.81*5 = 1030.05 N/m2
stress as a result of saturated clay
ρgz = 19*9.81*9 = 1677.51 N/m2
stress as a result of saturated sand
ρgz = 21*9.81*5 = 1030.05 N/m2
pore pressure
ρwgz = 1000*9.81*19 = 186,390 N/m2
total stresses = 190,127.61 N/m2
Flow and equiptential lines
There are 17 equiptential lines are market with 0, 1, 2, 3,4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15 and 16
There are 5 flow lines are running perpendicular to equiptential lines and are marked with numbers 0, 1, 2, 3 and 4.
Total head in upstream and downstream
Total head upstream h, = water head + head from the datum
= 12+17
= 29m
Total head downstream = total head upstream – head loss
= 29 – 12+ 1
= 18m
Seepage per m width
Using the flow nets to calculate the seepage,
q = (m/n)KH [L2T-1]
where:
q = seepage per unit width
m= number of flow channels
n= number of equipotential drops
h = total head loss in flow system
K = hydraulic conductivity
q = (4/16)5 x10-6 *11
= 1.375 x10-5 m/s per meter width
Pore pressure at point E
Pore pressure at point E = Hp)E * ρw
Pressure head at point e = pressure head at E + elevation head at E
Elevation head of E = 15.5m from the datum
The elevation head of E = total head – elevation head of E
Total head at point E= total head at upstream head – head loss at point E
(12+17- head loss at E)
Head lose at E = number of equipotential drops * H/Nd
Total head at E = {(12+17)- 11 *11/17}
= 29 -7.12= 21.88m
Pressure head at E = total heat at E – elevation head at E
= 21.88 – 15.5 = 6.38m
Pore pressure at E = pressure head at E * unit weight of water
= 6.38 * 9.807
= 62.56866 Kn/m2
Purpose of cut-off wall
The cut-off in gravity dams helps to prevent piping, reduce the exit gradient and also reduce the amount of seepage under hydraulic structures.
Q 17
a.
t= = radius of the circle
= center of the circle
|
Test |
Cone pressure |
Peak deviator stress |
s |
t |
|
1 |
300 |
185 |
242.5 |
57.5 |
|
2 |
450 |
290 |
370 |
80 |
|
3 |
700 |
520 |
610 |
90 |
a’= c’cos?
a’ = 28
tanα can be found from the slope of the graph
taking a graph of t-s
the slope will be (90-80)/ (610-520) = 0.11111
since tanα =sin?
we find arcsin of 0.11111 to find
? = 6.4o
the value of c =a’/cos6.4
= 28/cos6.4
C’ = 28.18
b.
|
Test |
Cone pressure |
Ultimate deviator stress |
s |
t |
|
1 |
300 |
210 |
255 |
45 |
|
2 |
450 |
235 |
342.5 |
107.5 |
|
3 |
700 |
310 |
505 |
195 |
a’= c’cos?
a’ = 140
tanα can be found from the slope of the graph
taking a graph of t-s
the slope will be (195-45)/ (505-255) = 0.6
since tanα =sin?
we find arcsin of 0.6 to find
? = 36.9o
the value of c =a’/cos36.9
= 140/cos36.9
C’ = 175.07
Cohesion depends on the amount of pressure exerted, therefore impossible little or no pressure will lead to near zero cohesion in ultimate state.
c.
pore water pressure parameter A
for test 1 the value of parameter A =
change in pore pressure = B* change in cone pressure
210-200 = B* 185-145
=B = 10/40 = 0.25
Change in deviator stress – BA (change in cone pressure)
185-145 = BA* (185-145)
40 = 0.25A*40
A = 4
for test 2 the value of parameter A =
change in pore pressure = B* change in cone pressure
250-235 = B* 290-285
=B = 15/5 = 5
Change in deviator stress – BA (change in cone pressure)
290-285 = BA* (290-285)
5 = 5A*5
A = 5
for test 3 the value of parameter A =
change in pore pressure = B* change in cone pressure
315-310 = B* 520-485
=B = 5/35 = 0.143
Change in deviator stress – BA (change in cone pressure)
520-485 = BA* (520-485)
5 = 0.143A*35
A = 1
The value of A is dependent on different types of soils. That is the main reason for the different changes with some values less than 1 while others are above 1.
d.
for effective design of a retaining structure using this material, I will highly concentrate on the cohesion and frictional angle. The angle is high meaning the failure of the structure will be high. The cohesion factor will help to increase the resistance against failure for the structure.
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