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XII. STRESSES IN SOILS Soil Mechanics
STRESSES IN SOIL Contents
• TOTAL STRESS
• PORE WATER PRESSURE
• EFFECTIVE STRESS
• STRESS DISTRIBUTION
STRESS Assumptions
Assumptions
Continuous material
Homogeneous (Engineering properties are
same in all locations)
Isotropic (Modulus and n are equal in all
directions) Linear-elastic stress-strain properties
STRESSES IN SOIL Stress Concept
xz
x
z
s Normal Stresses
xsxs
xz
zx
zx
zs
zs
Shear Stresses
Ten (-), Comp (+)
Clock (-), CC (+)
STRESSES IN SOIL Strain Concept
dL
P
L
dL
Normal Strain
L
g Shear Strain
g = shear strain [radians]
g
STRESSES IN SOIL Stress vs. Strain
sE
g
G
pa
p
STRESSES IN SOIL
1. Geostatic Stresses
Due to soil’s self weight
2. Induced Stresses
Due to added loads (structures)
3. Dynamic Stresses
e.g., earthquakes
0.248811
0.241673
ASi
g
81.950 ti
0 10 20 30 40 50 60 70 80 900.4
0.2
0
0.2
0.4
XII.1. STRESSES IN SOILS Soil Mechanics Effective Stress Concept
EFFECTIVE STRESS Introduction
The pressure transmitted through grain to grain at the contact points
through a soil mass is termed as intergranular or effective pressure. It is
known as effective pressure since this pressure is responsible for the
decrease in the void ratio or increase in the frictional resistance of a soil mass.
If the pores of a soil mass are filled with water and if a pressure induced
into the pore water, tries to separate the grains, this pressure is termed as
pore water pressure or neutral stress. The effect of this pressure is to increase the volume or decrease the frictional resistance of the soil mass.
EFFECTIVE STRESS Introduction
EFFECTIVE STRESS Introduction
If the valve V provided in the piston is opened, immediately there will be
expulsion of water through the hole in the piston. The flow of water
continues for some time and then stops.
The expulsion of water from the pores decreases the pore water pressure
and correspondingly increases the intergranular pressure. At any stage
the total pressure Q/A is divided between water and the points of
contact of grains. A new equation may therefore be written as,
Total pressure =𝑄
𝐴= Intergranular pressure + Pore water pressure
𝜎 = 𝜎′ + 𝑢
XII.2. STRESSES IN SOILS Soil Mechanics
Effective Stress with and without Seepage
EFFECTIVE STRESS Stresses in Saturated Soil without Seepage
EFFECTIVE STRESS Stresses in Saturated Soil with Upward Seepage
Pore water pressure increases,
effective stress decreases
EFFECTIVE STRESS Stresses in Saturated Soil with Upward Seepage
Note that ℎ/𝐻2 is the hydraulic gradient i caused by the flow, and so
If the rate of seepage and thereby the hydraulic gradient are gradually
increased, a limiting condition will be reached, at which point
where 𝑖𝑐𝑟 is the critical hydraulic gradient. In such a situation, the stability of the soil will be lost. This is generally referred to as boiling, or quick
condition.
EFFECTIVE STRESS Stresses in Saturated Soil with Downward Seepage
Pore water pressure decreases,
effective stress increases
ILLUSTRATIVE PROBLEMS Stresses in Saturated Soil with and without Seepage
Revisit Problems 1 and 2 of Problem Set 8 (Class Note 11)
CASE I: Assume NO FLOW.
1.2 m
A
‘0’
‘4’
1.8 m
Sand
γsat = 20 kN/m3
11.772
47.772 29.43 18.342
ILLUSTRATIVE PROBLEMS Stresses in Saturated Soil with and without Seepage
Revisit Problems 1 and 2 of Problem Set 8 (Class Note 11)
CASE II: Assume DOWNWARD FLOW.
1.2 m
A
‘0’
‘4’
1.8 m
Sand
γsat = 20 kN/m3
11.772
47.772 53.658 -5.886
ILLUSTRATIVE PROBLEMS Stresses in Saturated Soil with and without Seepage
Revisit Problems 1 and 2 of Problem Set 8 (Class Note 11)
CASE III: Assume UPWARD FLOW.
1.2 m
A
‘0’
‘4’
1.8 m
Sand
γsat = 20 kN/m3
11.772
47.772 41.202 6.57
EFFECTIVE STRESS Stresses in Saturated Soil with Seepage
Effective force (no flow)
Effective force (upward flow)
Decrease of effective total force due to
upward flow
Seepage force per unit volume of soil
Problem Set 9 A soil profile is shown in the figure. Calculate the total stress, pore water pressure,
and effective stress at points A, B, C, and D.
Problem 1
Problem Set 9 An exploratory drill hole was made in a saturated stiff clay. It was observed that the sand
layer underlying the clay was under artesian pressure. Water in the drill hole rose to a height of H1 above the top of the sand layer. If an open excavation is to be made in the
clay, how deep can the excavation proceed before the bottom heaves? We are given H = 8 m, H1 = 4
m, and w = 32%.
Problem 2
Problem Set 9 A 10-m-thick layer of stiff saturated clay is underlain by a layer of sand. The sand is under
artesian pressure. If H is 7.2 m, what would be the minimum height of water h in the cut so that the
stability of the saturated clay is not lost?
Problem 3
Problem Set 9 The figure shows a layer of granular soil in a
tank with an upward seepage by applying
water through the valve at the bottom of
the tank. The loss of head caused by the upward seepage between the levels of A
and B is 0.70 m and between levels A and C
is 0.28 m. The void ratio of the soil is 0.52 and
its specific gravity is 2.72. 4.1 Compute the effective stress at C.
4.2 Compute the critical hydraulic gradient for zero
effective stress.
4.3 Compute the upward seepage force per unit
volume.
Problem 4
1 m
5 m
2 m
0.70 m
0.28 m
Problem Set 9 The figure shows a granular soil in a tank
having a downward seepage. The water
level in the soil tank is held constant by
adjusting the supply from the top and the
outflow at the bottom. Void ratio of soil is
0.47 and specific gravity is 2.68. 5.1 Compute the effective stress at C.
5.2 Compute the effective stress at B.
5.3 Compute the critical hydraulic gradient.
5.4 Compute the downward seepage force per unit
volume.
Problem 5
1 m
6 m
4 m
0.90 m
0.6 m
Problem Set 9 A dense silt layer has the following properties: void ratio = 0.40, effective size D10 = 10 μm,
capillary constant C = 0.20 cm2. Free ground water level is 8.0 m below the ground
surface.
6.1 Find the height of capillary rise in the silt.
6.2 Find the vertical effective stress in kPa at 5 m depth. Assume unit weight of solids = 26.5 kN/m3 and
that the soil above the capillary action rise and ground surface is partially saturated at 50 %.
6.3 Find the vertical effective stress in kPa at 10 m depth. Assume unit weight of solids = 26.5 kN/m3 and
that the soil above the capillary action rise and ground surface is partially saturated at 50 %.
Problem 6
XII.3. STRESSES IN SOILS Soil Mechanics
Effective Stress due to Capillary Rise
EFFECTIVE STRESS Effect of Capillary Rise in Soils
EFFECTIVE STRESS Effect of Capillary Rise in Soils
EFFECTIVE STRESS Effect of Capillary Rise in Soils
As the effective grain size decreases, the size of the voids also
decreases, and the height of capillary rise increases. A rough estimation
of the height of capillary rise can be determined from the equation,
𝒉𝒄 =𝑪
𝒆𝑫𝟏𝟎
in which 𝑒 is the void ratio, 𝐷10 is Hazen's effective diameter in centimeters, and 𝐶 is an
empirical constant which can have a value between 0.1 and 0.5 𝑐𝑚2.
EFFECTIVE STRESS Effect of Capillary Rise in Soils
The general relationship among total stress, effective stress, and pore water
pressure is,
𝝈 = 𝝈′ + 𝒖 The pore water pressure u, with the atmospheric pressure taken as datum, at
a point in a layer of soil saturated by capillary rise is equal to
𝒖 = −𝜸𝒘𝒉
Where ℎ is the height of the point under consideration measured from the
groundwater table.
If partial saturation is caused by capillary action, it can be approximated as
𝒖 = −𝑺
𝟏𝟎𝟎𝜸𝒘𝒉
Where 𝑆 is the degree of saturation in percent.
Problem Set 9 Given the following soil profile:
Problem 7
Depth Soil type g (kN/m3) gsat (kN/m3) Ko
0 to 2 m ML 16 17 0.60
2 to 5 m CH 18 19 0.55
5 to 10 m SM 20 21 0.50
The water table is at 3 m depth. Calculate and plot the following:
7.1 Total vertical stress
7.2 Pore water pressure
7.3 Effective vertical stress
7.4 Effective horizontal stress
7.5 Total horizontal stress