The document presents a geotechnical engineering analysis involving calculations related to soil properties such as dry density, void ratio, and degree of saturation using specific parameters including bulk unit weight, water content, and specific gravity. It compares the economic feasibility of extracting soil from two borrow pits for an embankment construction, taking into account excavation and transportation costs. The calculations provide insights into determining dry densities and required volumes from different borrow pits based on given data.

1. Geotechnical Engineering
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2. One cubic metre of wet soil weighs 20 kN/m2. If the specific
gravity of soil particles is 2.60 and water content is 12%, find
the void ratio, dry density and degree of saturation.
STEP I: Given Data
Bulk unit weight = 20 kN/m2
Water Content =12% = 0.12
Specific gravity = 2.6 (unitless quality)
of the soil particle

3. TIP: To find void ratio, we will use relationship between dry unit weight and void ratio.
)
1
( e
G w
d




STEP II: Calculation of dry unit eight
Dry unit weight
3
kN/m
85
.
17
12
.
0
1
20
)
1
(





w
b
d


STEP I: Given Data
Bulk unit weight = 20 kN/m2
Water Content =12% = 0.12
Specific gravity = 2.6 (unitless qualit
of the soil particle
STEP III: Calculation of void ratio e
)
1
( e
G w
d




)
1
(
81
.
9
6
.
2
85
.
17
e



42
.
0

e

4. STEP I: Given Data
Bulk unit weight = 20 kN/m2
Water Content =12% = 0.12
Specific gravity = 2.6 (unitless qualit
of the soil particle
e = 0.42
STEP IV: Calculation of degree of saturation
Degree of saturation,
G
Se
w 
42
.
0
6
.
2
12
.
0 

S
74
.
0

S
Degree of saturation = 74 %

5. A soil has bulk density of 20 kN/m3 and water content of 14%. Calculate
the water content if the soil partially dries to a density of 19 kN/m3 and the
void ratio remains unchanged.
TIP: To find water content, we will use relationship between dry unit weight and water content.
)
1
( w
b
d




STEP I: Given Data
Bulk unit weight
Water content 14
.
0
%
14
1 

w
20
1

b
 19
2

b

?
2 
w
Remember: If the void ratio remains unchanged
while drying takes place, the dry unit weight also
remains unchanged
2
1 d
d 
 

6. )
1
( 1
1
1
w
b
d




2
1 d
d 
 
)
1
( 2
2
1
w
b
d




Step I
Step II

7. )
1
( 1
1
1
w
b
d




Also determine value of void ratio

8. An embankment having volume of 3000 m3 is to be constructed having bulk density of 2 g/cm3 and a water
content of 19%. The soil is to be obtained either from borrow pit A or borrow pit B, which have void ratio of
0.77 and 0.70, respectively and water content 15% and 10% respectively. Taking G= 2.67, for both solids,
determine the volume of soil required to be excavated from each of the pit. If the cost of excavation is $ 4 per
m3 in each pit but cost of transportation is $3 and $4 from pit A and B respectively, find out which which
borrow pit will be more economical for embankment construction ?
TIP: Soil having less dry density will occupy more volume and hence it will be economical. So we will have to determine dry
density of each borrow pit soil and volume using below formula.
w
b
d


1


STEP I: Given data

9. Procedure
1) Determine dry density of Embankment
w
d


1


2) Determine weight of embankment using this dry density Weight = Dry Density X Volume
3) Determine dry density for pit A
)
1
( e
G w
d




4) Determine volume to be dug from pit A, using weight of
embankment
A
pit
Broow
of
density
embankment
of
weight
volume 
5) Determine dry density for pit B
)
1
( e
G w
d




6) Determine volume of pit A using weight of embankment
B
pit
Broow
of
density
embankment
of
weight
volume 
Multiply both Volumes by cost of excavation and transportation and compare.

10. STEP II: Calculation of dry density and weight of soil for embankment
w
d


1


Dry density
3
/
68
.
1
19
.
0
1
2
cm
g
d 



Important: Lets convert this value into kN/m3 using 1 g/cm3 = 9.81 kN/m3. For sake of convenience assume approximate value
1 g/cm3 = 10 kN/m3
3
3
/
8
.
16
/
N
10
68
.
1 m
kN
m
k
d 



Lets determine weight of 3000 m3 embankment using,
Weight = Density X Volume
V
W d
emb 

  kN
Wemb 50400
3000
8
.
16 



11. STEP III: Calculation of dry density and volume for Borrow pit A
For Borrow pit A: As water content for borrow pits are not given we will use different formula to calculate dry densities of pits.
)
1
( e
G w
d




3
/
08
.
15
)
77
.
0
1
(
10
67
.
2
m
kN
d 



 (Assuming γw= 10 kN/m3)
Next volume of borrow pit A required to fill embankment of weight 50400 kN will be given by
A
pit
Broow
of
density
embankment
of
weight
volume  3
1
.
3342
08
.
15
50400
m
VA 


Cost of excavation and transportation = (3+4)x Volume
= (3+4)x3342.1 = $ 23394.7

12. STEP IV: Calculation of dry density and volume for Borrow pit B
For Borrow pit B: we will repeat the same procedure did for pit A
)
1
( e
G w
d



 3
/
70
.
15
)
7
.
0
1
(
10
67
.
2
m
kN
d 




3
2
.
3210
7
.
15
50400
m
VB 


Cost of excavation and transportation = (4+4)x Volume
= (4+4)x3210.2 = $ 25681.6
As cost of excavation and transportation for Borrow pit A is less, it will be more economical.

13. A sample of clay taken from a natural stratum was found to be partially saturated and when tested in the
laboratory gave the following results. Compute the degree of saturation. Specific gravity of soil particles = 2.67 ;
wet weight of sample = 2.60 N; dry weight of sample = 2.20 N ; and volume of sample = 160 cm3.