1. 1
•
4
•
5
•
Example 1):
Plot the variation of total and effective stresses, and pore water pressure
with depth for the soil profile shown.
Solution:
At point (1)
0σt = , 0u = and 0σ =′
At point (2)
2
tt kN/m71.24*17.8hγσ ===
0u =
(kPa)kN/m2.71σσ 2
t ==′
At point (3)
2
sattt kN/m108.22*18.571.2hγhγσ =+=+=
2
w
kN/m19.629.81*2hγu ===
(kPa)kN/m88.5819.62-108.2σ 2
==′
At point (4)
2
t kN/m186.219.5*4108.2σ =+=
2
w
kN/m58.99.81*6hγu ===
(kPa)kN/m127.358.9-186.2σ 2
==′
At point (5)
2
t kN/m281.219*5186.2σ =+=
2
w
kN/m107.99.81*11hγu ===
(kPa)kN/m173.3107.9-281.2σ 2
==′
Example 2):
For the subsoil condition shown in figure below, calculate the total,
neutral and effective stress at 1m, 3m, and 6m below the ground level.
Solution:
3ws
d(sand) kN/m569.18
4.01
81.9*65.2
e1
γG
γ =
++
==
2
•
•
3
•
2. 3
w
s
sat(sand) kN/m372.21
4.01
81.9*)4.065.2(
γ
e1
G
γ =
+
+
+
+
==
e
At point A
kPa18.5691*18.569hγσ dt ===
0u =
kPa569.18σσ t ==′
At point B
kPa61.3132*21.37218.569σt =+=
2
w
kN/m19.629.81*2hγu ===
kPa41.69319.62-61.313σ ==′
At point B
kPa121.31320*361.313σt =+=
2
w
kN/m49.059.81*5hγu ===
kPa72.26349.05-121.313σ ==′
Example 3):
For the system shown below, calculate the
effective stress at points (1) and (2)
1) Immediately after loading,
2) After very long time of loading.
Solution:
1) Immediately after loading
At point (1)
kPa100.55262.5*19.51.5*17.2σt =++=
kPa50.525269.81*2.5hγu w
=+==
kPa50.02550.525-100.55σ ==′
At point (2)
kPa176.554*19100.55σt =+=
kPa89.765269.81*6.5hγu w
=+==
kPa86.78589.765-176.55σ ==′
2) After very long time of loading
At point (1)
kPa100.55σt =
kPa24.5259.81*2.5u ==
kPa76.02524.525-100.55σ ==′
At point (2)
kPa176.55σt =
kPa63.7659.81*6.5u ==
kPa112.78563.765-176.55σ ==′
Example 4):
2
kN/m26q =
3
d kN/m17.2γ =
3
sat kN/m19.5γ =
3
sat kN/m19γ =
2
•
1
•
3. For the soil profile shown below, calculate and plot the variation of
total, effective stresses and pore water pressure with depth.
Example 5):
On a certain site surface layer of silty sand (γ = 18.5 kN/m3
) 5m thick
and overlies a layer of peaty clay (γ = 17.7 kN/m3
) 4m thick, which is
underlain by impermeable rock. Calculate the σt, σ' and u with depth for
the following conditions:
a) Water table at the surface, and
b) Water table at a depth of 2.5m with the silty sand above the water
table saturated with capillary water.
Solution: a)
Depth σt kN/m2
u kN/m2
σ'= σt-u kN/m2
0 0 0 0
5 5*18.5 = 92.5 5*9.81 = 49.05 43.45
9 92.5+4*17.7 = 163.3 9*9.81 = 88.29 75.01
b)
4. Depth σt kN/m2
u kN/m2
σ'= σt-u kN/m2
0 0 -2.5*9.81 = -24.525 24.525
2.5 2.5*18.5 = 46.25 0 46.25
5 5*18.5 = 92.5 2.5*9.81 = 24.525 67.975
9 92.5+4*17.7 = 163.3 6.5*9.81 = 63.765 99.535
Example 6):
The water table in a certain area is at a depth of 4m below the ground
surface. To a depth of 12m, the soil consists of very fine sand having an
average void ratio of 0.7. Above the water table the sand has an average
degree of saturation of 50%. Calculate the effective stress at a depth of
10m below the ground surface. What will be the increase in the effective
stress if the soil gets saturated by capillary up to a height of 1m above the
water table? Take Gs = 2.65.
Solution:
3
w
s
t kN/m312.17
7.01
81.9*)]7.0*5.0(65.2[
γ
e1
seG
γ =
+
+
+
+
==
3
w
s
sat kN/m331.19
7.01
81.9*)]7.0*1(65.2[
γ
e1
seG
γ =
+
+
+
+
==
1)
kPa185.2346*19.3314*17.312σt =+=
kPa58.869.81*6hγu w
===
kPa126.37458.86-185.234σ ==′
2)
kPa187.2537*19.3313*17.312σt =+=
kPa58.869.81*6hγu w
===
kPa128.39358.86-187.253σ ==′
Therefore, the increase in kPa2.019126.374-128.393σ ==′
Example 7):
The water table in a deposit of sand 8m thick is at a depth of 3m below
the surface. Above the water table, the sand is saturated with capillary
water. The bulk unit weight of sand is 2.0 g/cm3
. Calculate the effective
stress at 1m, 3m and 8m below the surface.
Solution:
At 1m
5. 2
g/cm400100)*2*(-1-100*2*1σ ==′
At 3m
600300*2σσ t ===′ 2
g/cm
At 8m
2
g/cm1100500*1-800*2σ ==′
Example 8):
Calculate the effective stress for a soil element at depth 5m in a uniform
deposit of soil as shown in figure below.
Solution:
1.35
0.6
2.7*0.3
S
ωG
e s ===
Unit weight above water table
3
ws kN/m652.1481.9*7.2*
35.11
3.01
γG
e1
ω1
γ =
+
+
=
+
+
=
Below water table
1.08
1
2.7*0.4
S
ωG
e s ===
3
w
s
sat kN/m17.8289.81*
1.081
1.082.7
γ
e1
eG
γ =
+
+
=
+
+
=
kPa82.788828.17*314.652*2σt =+=
kPa29.439.81*3hγu w
===
kPa53.35829.43-82.788σ ==′
6. 2
g/cm400100)*2*(-1-100*2*1σ ==′
At 3m
600300*2σσ t ===′ 2
g/cm
At 8m
2
g/cm1100500*1-800*2σ ==′
Example 8):
Calculate the effective stress for a soil element at depth 5m in a uniform
deposit of soil as shown in figure below.
Solution:
1.35
0.6
2.7*0.3
S
ωG
e s ===
Unit weight above water table
3
ws kN/m652.1481.9*7.2*
35.11
3.01
γG
e1
ω1
γ =
+
+
=
+
+
=
Below water table
1.08
1
2.7*0.4
S
ωG
e s ===
3
w
s
sat kN/m17.8289.81*
1.081
1.082.7
γ
e1
eG
γ =
+
+
=
+
+
=
kPa82.788828.17*314.652*2σt =+=
kPa29.439.81*3hγu w
===
kPa53.35829.43-82.788σ ==′
