Soil number |
Mass characteristics |
Material characteristics |
Complete name |

1 |
Soil material |
Color |
Organic soils |

2 |
Color |
Shape |
Boulders |

3 |
Consistency |
Texture |
Fine soils |

## Identification of Soil based on Mass and Material Characteristics

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 |

D_{10} = 0.75

D_{30} = 2.25

D_{60} = 6.1

Cu = 6.1/0.75 = 8.133

C_{G} = 2.25^{2}/ (0.75x 6.1) = 1.1066

The soil is therefore well graded since C_{u} is greater than 5 and C_{G} 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/m^{3}

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/m^{2}

stress on 2m of sand below water table

ρgz = 21*9.81*2 = 412.02 N/m^{2}

stress as a result of saturated clay

ρgz = 19*9.81*9 = 1677.51 N/m^{2}

stress as a result of saturated sand

ρgz = 21*9.81*5 = 1030.05 N/m^{2}

pore pressure

ρ_{w}gz = 1000*9.81*16 = 156,960 N/m^{2}

total stresses = 160579.89 N/m^{2}

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/m^{2}

stress on 2m of sand below water table

ρgz = 21*9.81*2 = 412.02 N/m^{2}

stress as a result of saturated clay

ρgz = 19*9.81*9 = 1677.51 N/m^{2}

stress as a result of saturated sand

ρgz = 21*9.81*5 = 1030.05 N/m^{2}

pore pressure

ρ_{w}gz = 1000*9.81*19 = 186,390 N/m^{2}

total stresses = 190,009.89 N/m^{2}

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

D_{10} = 0.75

D_{30} = 2.25

D_{60} = 6.1

Cu = 6.1/0.75 = 8.133

C_{G} = 2.25^{2}/ (0.75x 6.1) = 1.1066

The soil is therefore well graded since C_{u} is greater than 5 and C_{G} 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/m^{3}

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/m^{2}

stress on 2m of sand below water table

ρgz = 21*9.81*2 = 412.02 N/m^{2}

## Particle Size Distribution Analysis and Results

stress as a result of saturated clay

ρgz = 19*9.81*9 = 1677.51 N/m^{2}

stress as a result of saturated sand

ρgz = 21*9.81*5 = 1030.05 N/m^{2}

pore pressure

ρ_{w}gz = 1000*9.81*16 = 156,960 N/m^{2}

total stresses = 160579.89 N/m^{2}

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/m^{2}

stress on 2m of sand below water table

ρgz = 21*9.81*2 = 412.02 N/m^{2}

stress as a result of saturated clay

ρgz = 19*9.81*9 = 1677.51 N/m^{2}

stress as a result of saturated sand

ρgz = 21*9.81*5 = 1030.05 N/m^{2}

pore pressure

ρ_{w}gz = 1000*9.81*19 = 186,390 N/m^{2}

total stresses = 190,009.89 N/m^{2}

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/m^{2}

stress as a result of saturated clay

ρgz = 19*9.81*9 = 1677.51 N/m^{2}

stress as a result of saturated sand

ρgz = 21*9.81*5 = 1030.05 N/m^{2}

pore pressure

ρ_{w}gz = 1000*9.81*19 = 186,390 N/m^{2}

total stresses = 190,127.61 N/m^{2}

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

D_{10} = 0.75

D_{30} = 2.25

D_{60} = 6.1

Cu = 6.1/0.75 = 8.133

C_{G} = 2.25^{2}/ (0.75x 6.1) = 1.1066

The soil is therefore well graded since C_{u} is greater than 5 and C_{G} 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/m^{3}

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/m^{2}

stress on 2m of sand below water table

ρgz = 21*9.81*2 = 412.02 N/m^{2}

stress as a result of saturated clay

ρgz = 19*9.81*9 = 1677.51 N/m^{2}

stress as a result of saturated sand

ρgz = 21*9.81*5 = 1030.05 N/m^{2}

pore pressure

ρ_{w}gz = 1000*9.81*16 = 156,960 N/m^{2}

total stresses = 160579.89 N/m^{2}

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/m^{2}

stress on 2m of sand below water table

ρgz = 21*9.81*2 = 412.02 N/m^{2}

stress as a result of saturated clay

ρgz = 19*9.81*9 = 1677.51 N/m^{2}

stress as a result of saturated sand

ρgz = 21*9.81*5 = 1030.05 N/m^{2}

## Plasticity and Bulk Density Analysis

pore pressure

ρ_{w}gz = 1000*9.81*19 = 186,390 N/m^{2}

total stresses = 190,009.89 N/m^{2}

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/m^{2}

stress as a result of saturated clay

ρgz = 19*9.81*9 = 1677.51 N/m^{2}

stress as a result of saturated sand

ρgz = 21*9.81*5 = 1030.05 N/m^{2}

pore pressure

ρ_{w}gz = 1000*9.81*19 = 186,390 N/m^{2}

total stresses = 190,127.61 N/m^{2}

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 [L^{2}T^{-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 = H_{p)E} * ρ_{w}

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/N_{d}

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/m^{2}

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

? = 6.4^{o}

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

? = 36.9^{o}

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.