GEOTECH PREBOARD 1
A stratified soil deposit consists of four layers. The thickness of the second, third and fourth layers are
equal to half, one-third and one-fourth, respectively, the thickness of the top layer, while their coefficients
of permeability are respectively twice, thrice and four times that of the top layer. Determine:
1. Equivalent horizontal coefficient of permeability
A. 1.46 k1
C. 1.92 k1
B. 1.03 k1
D. 1.32 k1
2. Equivalent vertical coefficient of permeability
A. 1.03 k1
C. 1.32 k1
B. 1.46 k1
D. 1.92 k1
3. Ratio of equivalent coefficient of permeability
A. 1.32
C. 1.92
B. 1.46
D. 1.03
Solution:
1
h1
2
k2  2k1
1
h3  h1
3
k3  3k1
h2 
1
25
h1 H T  h1
4
12
k4  4k1
h4 
H T K eq / /   hk
25
1
1
1
h1 ( K eq / / )  h1 k1  h1 (2k1 )  h1 (3k1 )  h1 (4k1 )
12
2
3
4
K eq / /  1.92k1
HT
h

K eq 
k
25 h1 /12
K eq 

h1
k1

h1 / 2
2k1

h1 / 3
3k1

h1 / 4
4k1
K eq   1.46k1
ratio 
K eq / /
K eq 

1.92 k 1
 1.32
1.46 k1
The in-site density of a soil mass is to be determined by the core-cutter method. The height and diameter
of the core are 13 cm and 10 cm respectively. The core, when full of soil, weighs 3155 gm, while the selfweight of the empty core is 1250 gm. The natural moisture content and specific gravity of solids are 12%
and 2.66 respectively. Determine the following:
4. Bulk Density in gm/cc
A. 1.87
B. 1.95
5. Dry Density in gm/cc
A. 1.67
B. 1.75
6. Void ratio
A. 0.69
B. 0.59
C. 1.67
D. 1.75
C. 1.87
D. 1.95
C. 0.79
D. 0.89
Solution:
Wt 3155  1250

 1.87 gm / cc
Vol  (10) 2 (13)
4

1.87
d 

 1.67 gm / cc
1  w 1  0.12
G
2.66(1)
 d  s w  1.67 
 e  0.59
1 e
1 e

A sand sample is 50% saturated and has bulk density of 1.75 t/m 3. The specific gravity of solids is 2.65.
Determine the following:
7. Void ratio of the soil
A. 0.85
C. 0.96
B. 0.72
D. 0.65
8. Determine the critical hydraulic gradient
A. 0.85
C. 0.96
B. 0.72
D. 0.65
9. How will the critical hydraulic gradient of the soil change if the soil is compacted to increase bulk
density by 10% when the degree of saturation remains constant.
A. 13.8% increase
C. 16.3% increase
B. 13.8% decrease
D. 16.3% decrease
Solution:
Gs  Se
2.65  0.5(e)
 w  1.75 
(1)  e  0.72
1 e
1 e
G  1 2.65  1
icr  s

 0.96
1  e 1  0.72
G  Se '
2.65  0.5(e ')
 ' s
 w  1.75(1.1) 
(1)  e  0.51
1 e '
1 e '
G  1 2.65  1
icr '  s

 1.09
1  e 1  0.51
1.09
%
 1.135  13.5% increase
0.96

A layer of sand 5 meters deep overlies a 4 m thick bed of clay. Assume sand has saturated unit weight &
dry unit weight of 20.9 kN/m3& 17.4 kN/m3 respectively, while the clay has saturated unit weight of 17.8
kN/m3. Determine the effective stress 9 m below the ground surface if:
10. The water table is at the ground level.
A. 118.96 kPa
C. 87.41 kPa
B. 88.29 kPa
D. 107. 03 kPa
11. The water table is at 2 meters below ground level and the sand above remains saturated with
capillary moisture.
A. 118.96 kPa
C. 87.41 kPa
B. 88.29 kPa
D. 107. 03 kPa
12. If the water table is at the top of the clay.
A. 118.96 kPa
C. 87.41 kPa
B. 88.29 kPa
D. 107. 03 kPa
Solution:
Pe    ' h
Water table is at ground level :
Pe  (20.9  9.81)(5)  (17.8  9.81)(4)  87.41 kPa
2 m below ground level :
Pe  20.9(2)  (20.9  9.81)(3)  (17.8  9.81)(4)  107.03 kPa
Top of Clay :
Pe  17.4(5)  (17.8  9.81)(4)  118.96 kPa
From the figure below:
Assume: 3K1=K2=1.5K3=2K4 and that layer 2 & 3 are of equal height.
AB = 6 cm, BC = 10 cm and CD = 8 cm
13. Find the pressure head at point B
A. 19.3 cm
C. 16.2 cm
B. 24.0 cm
D. 12.0 cm
14. Find the pressure head at point C
A. 19.3
C. 16.2
B. 24.0
D. 12.0
2
15. Find the rate of flow if K1= 3.5 x 10- cm/sec
A. 0.11 cm3/sec
C. 0.08 cm3/sec
3
B. 0.15 cm /sec
D. 0.18 cm3/sec
Solution:
From 3K1  K 2  1.5 K 3  2 K 4
K 2  3K1 K 3  2 K1 K 4  1.5K1
HK eq   hk
4( K 23 )  2(3K1 )  2(2 K1 )
K 23  2.5 K1
H
h

K eq
k
24
6
10
8



K eq K1 2.5 K1 1.5 K1
K eq  1.565 K1
QT  keq iA  1.565 K1 (
12
)(4)(1)  3.130 K1
24
QT  Q1
3.130 K 1  K1 (
24  hb
)(4)(1)  hb  19.305 m
6
QT  Q4
3.130 K1  1.5 K1 (
hc  12
)( 4)(1)  hc  16.173 m
8
QT  3.130 K1  3.130(3.5 x 102  0.11 cm3 / s