QUESTION 1
PROPERTIES OF THE SOIL INDEX
Source: Soil Mechanics Book (pages 11-12)
By: Hadi Y. Ce
Question:
At the moment of drilling, it turns out that tanha is found to be saturated with corrosine oil.
Sat from the ground = 2,4 grams/cm3. Specify e and d when Gsolid = 2,65 and Goil= 0.98
Solutions:
Known:
sat = 2,4 gram/cm3
Ws = 1 gram
Asked:
Po pore (e)?
When Gsolid =2,65 and Goil= 0.98 then, wet volume weight (d )?
Answered:
Since the soil is saturated, then all the pores are filled with oil.
Take the volume of oil juice in the ground = Vo.
Ws diambil 1 gram.
Ws
1. Vs =
S
1
Vs =
G× W
= 12.65
= 0.377 cm3
2. Ws = 1 gram
= 0.001 kg
W of oil (Wo) = Goil × Vo
= 0.89 × Vo
= 0.89 Vo
3. Wtotal = Ws + Wo
= 1 + 0.89 Vo
4. Vtotal = Vs + Vo
= 0.377 + Vo
1
, W total
5. sat =
V total
1+ 0,89V o
2.4 =
0,377 +V o
1 + 0.89 Vo = 2.4 Vo + (2.4) × (0.377)
1.51 Vo = 0.0952
V o = 0.063 cm3
Vv Vo
1. e = =
Vs Vs
0,063
=
0,377
= 0.167
Looking for d means dried first.
So Vo = 0.063 will be filled with air (Wo = 0)
2. Wtotal = 1 + 0
3. Vtotal = 0,377 + V
= 0,377 + 0,063
= 0.44 cm3
Ws
1. d =
V s +V v
Ws
=
V total
1
=
0,44
= 2,272 gram/cm3
So, the calculation results are obtained:
Pori number (e) = 0,167%
Wet volume weight (d ) = 2.271 grams/cm3 = 2270.64 kg/m3
2
, QUESTION 2
SOIL CLASSIFICATION
Source: Wesley, (2019). Soil Mechanics. (pages 21-37)
Question:
Known soil sieve data as follows:
Number of Retained Weight Retained Weight
Details Diameter
Sieve (grams) (grams) Percentage (%)
4 4,76 50 10
10 2 75 15
20 0,84 15,7 31,4
40 0,42 23 4,6
100 0,25 130 26
200 0,14 40 8
Pan 0,075 25 5
S - 500 100
The following data are obtained
PL: 28 PI: 12 D30: 0,36
LL: 40 D10: 0,24 D60: 1,4
Asked:
Classify the land using AASHTO and USCS methods!
Solutions:
Retained Weight
Retained
No Details (grams) Cumulative
Weight % Passes
Sieve Diameter Restrained (%)
(grams)
Percentage (%)
4 4,76 50 10 10 90
10 2 75 15 25 75
20 0,84 15,7 31,4 56,4 43,6
40 0,42 23 4,6 61 39
100 0,25 130 26 87 13
200 0,14 40 8 95 5
Pan 0,075 25 5 100 -
S - 500 100
3
PROPERTIES OF THE SOIL INDEX
Source: Soil Mechanics Book (pages 11-12)
By: Hadi Y. Ce
Question:
At the moment of drilling, it turns out that tanha is found to be saturated with corrosine oil.
Sat from the ground = 2,4 grams/cm3. Specify e and d when Gsolid = 2,65 and Goil= 0.98
Solutions:
Known:
sat = 2,4 gram/cm3
Ws = 1 gram
Asked:
Po pore (e)?
When Gsolid =2,65 and Goil= 0.98 then, wet volume weight (d )?
Answered:
Since the soil is saturated, then all the pores are filled with oil.
Take the volume of oil juice in the ground = Vo.
Ws diambil 1 gram.
Ws
1. Vs =
S
1
Vs =
G× W
= 12.65
= 0.377 cm3
2. Ws = 1 gram
= 0.001 kg
W of oil (Wo) = Goil × Vo
= 0.89 × Vo
= 0.89 Vo
3. Wtotal = Ws + Wo
= 1 + 0.89 Vo
4. Vtotal = Vs + Vo
= 0.377 + Vo
1
, W total
5. sat =
V total
1+ 0,89V o
2.4 =
0,377 +V o
1 + 0.89 Vo = 2.4 Vo + (2.4) × (0.377)
1.51 Vo = 0.0952
V o = 0.063 cm3
Vv Vo
1. e = =
Vs Vs
0,063
=
0,377
= 0.167
Looking for d means dried first.
So Vo = 0.063 will be filled with air (Wo = 0)
2. Wtotal = 1 + 0
3. Vtotal = 0,377 + V
= 0,377 + 0,063
= 0.44 cm3
Ws
1. d =
V s +V v
Ws
=
V total
1
=
0,44
= 2,272 gram/cm3
So, the calculation results are obtained:
Pori number (e) = 0,167%
Wet volume weight (d ) = 2.271 grams/cm3 = 2270.64 kg/m3
2
, QUESTION 2
SOIL CLASSIFICATION
Source: Wesley, (2019). Soil Mechanics. (pages 21-37)
Question:
Known soil sieve data as follows:
Number of Retained Weight Retained Weight
Details Diameter
Sieve (grams) (grams) Percentage (%)
4 4,76 50 10
10 2 75 15
20 0,84 15,7 31,4
40 0,42 23 4,6
100 0,25 130 26
200 0,14 40 8
Pan 0,075 25 5
S - 500 100
The following data are obtained
PL: 28 PI: 12 D30: 0,36
LL: 40 D10: 0,24 D60: 1,4
Asked:
Classify the land using AASHTO and USCS methods!
Solutions:
Retained Weight
Retained
No Details (grams) Cumulative
Weight % Passes
Sieve Diameter Restrained (%)
(grams)
Percentage (%)
4 4,76 50 10 10 90
10 2 75 15 25 75
20 0,84 15,7 31,4 56,4 43,6
40 0,42 23 4,6 61 39
100 0,25 130 26 87 13
200 0,14 40 8 95 5
Pan 0,075 25 5 100 -
S - 500 100
3
