Assignment 1 Solution
(1)
For the 1 (m) above the ground water, you have two assumptions that you can make the first is to
assume the sand above the ground water level is completely dry (then you have to calculate the dry unit
weight of Sand which will be 14.5 kN/m3) or to assume that the sand layer is saturated due to capillary
rise effect then you can use directly the saturated unit weight given in the problem.
Now assuming that the sand layer is dry and using unit weight of 14.5 kN/m3
(a) At the middle of the clay layer:
′
1 15
3 19
∆
4 19
4 21
7 10
78
84
78 84
0.15 8
log
78
1 0.85
0.206
21
(b) Dividing the clay layer into equal layers
After calculating the effective stresses at the center of each layer:
0.15 4
60 84
log
1 0.85
60
0.15 4
96 84
log
1 0.85
96
0.212
(c) The rebound of the clay if the surcharge was removed:
0.85
1
0.15 log
78
84
78
78
0.04 8
log
162
0.8024
0.8024
0.056
6
21
(2) Shear Strength of Soil:
300 tan 35
210
The principle stresses could be obtained either by drawing the Mohr –circle using geometrical procedure
or solving the following equations simultaneously:
cos
210 (1)
=300 (2)
Solving (1) and (2)
703
191
(3)
1.2
1
Void ratio (e)
0.8
0.6
0.4
0.2
0
0
0.5
1
1.5
2
Stress (σ)
2.5
3
3.5
4
From the above graph:
The recompression index Cr : 0.08
The virgin compression index Cc : 0.42
The preconsolidation pressure = 178 kpa
To check if the clay is under overconsolidation or normal consolidation
125 + 160 > 178 Æ Overconsolidated Clay
Therefore the settlement is calculated as follow:
178
0.42 6
log
160
1 0.959
Swap
225
0.08 6
log
160
1 0.959
ht = 0.67m;
nf = 8;
ne = 4;
equipotential drop: (2-0.67)/ne = (2-0.67)/4 = 0.3325m
k = 5x10-5 m/s
Thus: Q ≈
nf
ne
Ht k =
8
(2 − 0.67) x(5 x10−5 ) = 1.33x10−4 m3 / s / m
4