soil mechanics handouts
TRANSCRIPT
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SOIL COMPOSITION
WT=WA+WW+WS ; WA0
WT=WW+WS
VT=VA+VW+VS ; VV=VA+VW
VT=VV+VS
MOIST=DRY(1+w)
SAT=DRY(1+w
SAT)
W =
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SOIL PROPERTIES
Wet Unit Weight(T)
= =
Dry Unit Weight(D)
=
Moisture Content(w)
w =W
W 100%
Void Ratio (e)
=
Degree of Saturation (s)
= 100%
Porosity(u)
=
Example 1:
A sample of soil obtained from a test pit 0.0283m3 in volume has a mass of 63.56kg. The entire sample is
dried in an oven and found to have a dried mass of 56.75kg. Calculate the water content, wet density
and dry density.
Example 2:
Determine the wet density, dry unit weight, void ratio, water content and degree of saturation for a
sample of moist soil which has a mass of 18.18kg and occupies a total volume of 0.009m3 when dried in
an oven, the dry mass is 16.13kg. The specific gravity of the soil solids is 2.70
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Example 3:
A sample soil obtained from a borrow pit has a net weight of 42lbs. The total volume occupied by the
sample when in the ground was 0.34ft3. If small portion of the sample is used to determine the water
content then wet sample mass is 150g and after drying 125g. Determine 1.) Water content 2.) Wet unit
weight 3.) Dry unit weight
Example 4:
Given the ff:
Gs=2.69
e=0.65
w=10%
Determine the dry unit weight and wet unit weight
Example 5/Seatwork:
Given the ff:
VT =150cm3
WT= 250g
S=100%
WS=162g
Determine the dry density, water content, void ratio and specific gravity of soil solids
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NON SATURATED
=1 +
= (1 + )1 +
= (1 + )
=
=
=( + )
1 +
=1 +
=1
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SATURATED SOILS
SUBMERGED SOILS
=
=
= ( ) ; = 1 ,
= ( 1)
=( 1)
+
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=( 1)
+ 1
OR
=
OR
=1
2 ( )
= =1 +
Example 1:
The moist mass of 2.8x10-3m3 of soil is 5.53kg. If the moisture content is 10% and specific gravity of soil
solids is 2.72. Determine the following: T, D, e, n, s
Example 2:
For a given soil, the following are known GS=2.74, moist unit weight (T) 19.8kg/m3 and moisture content
(w) 16.6%. Determine: D, e, n, s
Example 3:
The dry density of a soil is 1750kg/m3. Given GS=2.66. What is the moisture of soil when it is saturated?
Example 4:
In an unsaturated soil formation it is known that the dry unit weight is 18.06KN/m3. The specific gravity
of soil particle is 2.75
a.) What is the saturated unit weight of the sample?
b.) What is the submerged unit weight of the sample?
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INDEX PROPERTIES OF SOIL
Index Properties refers to those properties of soil that indicate the type and condition of soil and
provide relationship to structural properties such as the strength, compressibility, tendency for swelling
and permeability.
A. Consistency in the remolded state and plasticity
At higher water content the soil-water mixtures possesses the properties of a liquid;
At lesser water contents the volume of the mixture is decreased and the materials exhibits the
properties of a plastic;
At still lesser water content, the mixtures behave as a semi-solid and finally solid
ATTENBERG SCALE/LIMITS
Liquid Limit (LL)
The water content indicating the division between the liquid and plastic state.
Plastic Limit (PL)
The water content at the division between plastic and semi-solid state.
Shrinkage Limit (SL)
The water content at the division between solid and semi-solid state.
Plasticity Index (PI)
-indicates the range of water content through which the soil remains plastic.
-use PI as basis for classification of fine grained soils
= % %
Liquidity Index (LI)
-provides indication of soils consistency and/or sensitivity potential.
-value less than 1 indicates the natural water content is less than liquid limit
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-a very low value or near zero indicates that the water content is near plastic limit
-negative values indicates a desiccated or dried soil, hard soil.
=% %
% %=
% %
Example 1:
A fine grained soil is found to have a liquid limit of 70% and plastic limit of 38%. What is the plasticity
index?
Example 2:
A silt soil has a plastic limit of 25 and plasticity index of 30. If the natural water content of the soil is 35%.
What is the liquidity index?
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PARTICLE SIZE DISTRIBUTION
(Mechanical Analysis)
This classification test determines the range of size of particles in the soil and the percentage of particles
in each size of the sizes between maximum and the minimum.
Two methods in particle size distribution:
1. Sieving
2. Sedimentation
Sieve Designation Sieve openings (mm)
#4 4.76
#8 2.38
#10 2.00
#20 0.84
#40 0.42
#60 0.25
#100 0.149
#200 0.074
#270 0.0533
#400 0.037
SHAPE OF PARTICLE SIZE DISTRIBUTION CURVE
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EFFECTIVE SIZE (D10)
-is taken as the particle size corresponding to the 10% passing from the gram size curve. It is this size
that is related to permeability and capillarity.
UNIFORMITY COEFFICIENT (Cu)
-particles comparative indication of the range of particle size in the soil
=
Cu>10 well graded soil
CuCu>5 gap graded soil
Example:
Following are results of sieve analysis. Draw the graph size curve
Sieve #Mass of Soil
Retained
Cumulative Mass
Retained% Retained %Finer
4 0
10 40
20 60
40 89
60 140
80 122
100 210200 56
PAN 12
TOTAL MASS: 729g
Graph:
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COEFFICIENT OF GRADIATION (Cc)
=
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CLASSIFICATION SYSTEMS
Soil Classification systems have been primarily devised to facilitate the transfer of information
between interested parties. In the engineering and construction field the broad general properties of
concern relate to the performances or usefulness for supporting structures and the handling or working
qualities of soil.
Exiting Classification systems
1. US Department of Agriculture
2. AASHTO Classification System
3. Unified Soil Classification System
(USCS ASTM D-2487)
Example: Classify the following soil using USCS
% Finer 1 2 3 4 5 6 7
#4 100 100 100 95 100 99 90#10 98 100 90 82 80 - -
#40 85 96 68 55 78 - -
#200 70 72 30 41 59 57 8
LL 40 58 30 32 32 54 39
PL 26 35 21 20 17 28 31
CC=2.1
CU=3.9
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RELATIVE DENSITY (DR)
The shear strength and resistance to compression are related to the density (unit weight) higher
shear strength and move resistance to compression are developed by the soil when it is in a dense or
compact condition.
=
% =
100%
where
eMAX=void ratio of the soil at loosest condition
eMIN=void ratio of the soil at densest condition
eO=void ratio of the soil at natural condition
% =
1
1
1
1 100%
where
=dry unit weight for loosest condition
=dry unit weight for densest condition
= dry unit weight for natural condition
IN PLANE DENSITY OR FIELD DENSITY
Refers to the volumetric weight usually expressed as KN/m3 or pcf of as soil in the undisturbed
(in situ) condition or in a computed fill.
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METHOD IN DETERMINING IN-PLACE DENSITY
1. Sand Cone Method
2. Rubber Balloon Method
3. Nuclear Equipment
COMPACTIONAny increase in the moisture content tends to reduce the dry unit weight. This is because the
water takes up the spaces that would have been occupied by the solid particles. The moisture content
at which the maximum dry density attained is generally referred to as OPTIMIM MOISTURE CONTENT.
COMMON TEST PROCEDURE
1. Standard Proctor Test
2. Modified Proctor Test
Description of Condition Relative Density %
Loose 85
Example 1:
A granular soil located in a borrow pit is found to have an in plane dry density of 1895 kg/m3. In
the laboratory values of dry maximum and minimum density are determined as 2100 kg/m3 and
1440kg/m3. Calculate the relative density.
Example 2:
In undisturbed sample of sand has a dry weight of 4.20 lbs and occupies a volume of 0.038ft3.
The soil solids have specific gravity of 2.75 Laboratory Test Performed to determine the maximum and
minimum densities indicate void ratios of 0.42 at the maximum density and 0.92 at the minimum
density. Compute relative density of this material.
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MOVEMENT OF WATER THROUGH SOIL
(Permeability, Capillarity)
PERMEABILITY(Hydraulic Conductivity)
Soil deposits are porous and material is considered a permeable material. It should be realized
that flow occurring through the void spaces between particles and not through the particles themselves.
FACTORS AFFECTING FLOW
1. The pressure difference existing between two points where flow is occurring.
2. The density and viscosity of the fluid.
3. The size and shape and number of pore openings.
4. The mineralogical, electrochemical and other pertinent properties of fluids.
DARCYS LAW FOR FLOW
The quantity of water flowing through the soil through a given a period was proportional to the
soil area normal to the direction of flow and the differences in water levels indicated in the piezometer
and inversely proportional to the length of soil between the piezometer through which flow took place.
Q
t
hA
L
U
L = hydraulic gradient (L)
Q
t = k
hA
L
Q
t = q = Av = discharge or flow
PIEZOMETER
h
AREA
q=Q/t
q=Q/t
q=kiA=Av
v=ki
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EMPIRICAL RELATIONSHIPS
For uniform sands (loose condition)
k=10(D10)2
k in mm/s
D10 in mm
For dense or compacted sands
k=3.5(D15)2
k in mm/s
D15 in mm
Soil TypeRelative, Degree of
Permeability
Khyd, Coefficient of
Permeability or
Hydraulic Conductivity
(mm/s)
Drainage Properties
Clean Gravel High 10 to 100 Good
Clean sand, sand and
gravel mixtures
Medium 10 to 10-2 Good
Fine sands, silts Low 10-2 to 10-4 Fair to poor
Sand-silt-clay mixtures
glacial tills
Very low 10-3 to 10-6 Poor to practically
impervious
Homogenous clays Very low to practically
impermeable
10-6 to 10-10
10 -3 to 10-6Poor to practically
impervious
LABORATORY PERMEABILITY TESTS
Constant Head Test
-for coarsed grained soil (sands, gravel)
-volume of flow through the soil is relatively large
-permeability is computed on basis of fluid that passes through the sample
=
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Falling Head Test
-for fined grained soil (silt, clay)
-volume of flow through the soil is very small or zero
-permeability is computed on the basis of fluid flowing into the sample
=
=
2.303
a=area of stand pipe
A=area of soil sample
t1
t2
FILTERS
A= CROSS SECTIONAL AREA
OF THE SAMPLE
h
Q
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Example 1:
Solve for k if given is
L=120mm
A=600mm2
a=200mm2
h1=60cm
h2=55cm
t2-t1=30 minutes
CAPILLARITY
The groundwater table (GWT) or phreatic surface is the level to which the underground water
will rise in an observation well pit or other open excavation in the earth.
Water in Capillary Tubes
Adhesive due to surface tension is equal to the effect of downward pull of gravity in the capillary tube.
2 =
Ts=surface tension
=0.005 lb/ft, 0.064 N/m, 73 dynes/cm
SURFACE TENSION
GLASS TUBE
FREE WATER SURFACE
hC
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=2
=4
=31
,
CAPILLARY RISE IN SOIL
Soil Type Height(m)
small gravel 0.02-0.1
coarse gravel 0.15
fine sand 0.3-0.1
silt 1-10
clay 10-30
Example 1:
A constant head permeability test is performed in a laboratory where the soil sample is 25cm in
length and 6cm2 in cross section. The height of water is maintained at 2ft at the inflow end and 6in at
the outlet end. The quantity of water flowing through the sample is 200ml in 2 minutes.
a. Make a sketch of the describe conditions
b. What is the coefficient of permeability in millimeters/minute?
Example 2:
A falling head permeability test is performed on a fine grained soil. The soil sample has a length
of 120mm and a cross sectional area of 600mm2. The water in the stand pipe flowing into the soil is
0.60m above the top of the sample at the start of the test. It falls 50mm in 30 minutes. The stand pipe
has cross sectional area of 200mm2.
a. Make a sketch of describe condition
b. What is the coefficient of permeability in mm/s?
c. in feet/min?
d. on the basis of the calculated value for khyd, what is the probable soil type?
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HYDRAULIC CONDUCTIVITY IN STRATIFIED SOIL
H1 KH1
H2 KH2
H3 KH3H4 KH4
=1
( + + + )
=( + + + )
FLOW NETS
are pictorial method of studying the path that moving water follows, where water enters and
escapes from a permeable zone of soil by travelling short distance.
FLOW LINES
the path that the water follows
EQUIPOTENTIAL LINES
Lines connecting points of equal total energy head lines
Where:
q= KHW (NF/Nd)(width) (for isotropic soil)
k=coefficient of permeability of soil
hw=elevation difference between at upstream and downstream limits of the flow net
NF= number of flows channels for the flow net
ND=number of equipotential drops for the flow net
width=width of the structure
KV1
KV2
KV3
KV4
H
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Example:
Estimate the equivalent hydraulic conductivity flow in horizontal and vertical direction
KH=KV=10-3cm/sec
KH=KV=2 x 10-4cm/sec
KH=KV=10-5cm/sec
KH=KV=2 x 10-3cm/sec
1m
1m
1m
1m
H=4m
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STEPS IN DRAWING A FLOW NET
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FLOW NETS FOR REPRESENTATIVE SEEPAGE PROBLEMS
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Example 1:
A masonry dam having a sheet pilling cut off at the upstream end is located at a reservoir site, as
indicated by the sketch. Draw a flow net for the subsurface flow and compute the seepage. Also
calculate the uplift force acting on the base and the escape gradient of the water at the downstream tip
of the dam.
SOLUTION:
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COMBINED STRESSES IN SOIL MASSES
= (1)
= (1)
=
SURFACE LEVEL
SOIL DEPOSITS
POINT OF ANALYSIS
major principal plane
minor principal plane
Area=1
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SUMMING FORCES PARALLEL TO N
= +
= +
where:
=1 + 2
2
=1 2
2
=1 + 2
2 +
1 2
2
=+
+
Note:
= 0 = 90
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SUMMING FORCES PARALLEL TO N
=
where:
2 = 2
=2
2
2
2
=
Note:
= 45 = 0
MOHRS CIRCLE
TENSION COMPRESSION
MINOR PRINCIPAL PLANE
MAJOR PRINCIPAL PLANE
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Example 1:
At a point in a stressed material the major principal stress is 120Kpa. (compression) and minor
principal stress is 40Kpa (compression)
a. Determine the maximum shear and normal stress where shear is maximumb. Determine the value of normal and shear stress acting on a plane 30% from the major
principal plane.
SUBSURFACE STRESSES
Stresses caused by soil mass
1. Vertical Stress
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GWT
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2. Horizontal (Lateral) Stresses
=
where:
k=coefficient of lateral earth pressure
Typical values of k
Soil Type k
Granular (loose) 0.5 to 0.6
Granular (dense) 0.3 to 0.5
Clay (soft) 0.9 to 1.1
Clay (hard) 0.8 to 0.9
STRESS CAUSED BY VERTICAL SURFACE LOADING
With increasing depth, the area over which new stress develop will increase, but the magnitude
of the stresses will decrease.
Z1
Z2
Z3
SURFACE LOADING
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BOUSSINES Q STRESS DISTRIBUTION
=3
2( + )
3
2 1 +2
WESTGAARD STRESS DISTRIBUTION
=
1 + 2
SIXTY DEGREE APPROXIMATION
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=( + )( + )
Example 1:
At a point in a stressed material the stress combination acting on a plane is 8000psf
compression and 2000psf shear. While on an orthogonal (perpendicular) plane the stress combination
is 2000psf compression and 2000psf shear.
a. Determine values major and minor principal stress
b. Find the angle between the plane on which the 8000psf compressive stress acts and the
major principal plane.
Example 2:
At a planned construction site substance sampling indicates that the soil unit weight is
19.5Kn/m3
a. Determine the vertical stress at a depth of 4m if the GWT is deep.
b. Determine the vertical stress at depth of 4m, if the GWT is 2m below the surface.
c. Determine the effective vertical stress if GWT is at the ground surface.
Example 1:
Q
z
1/2z 1/2z
2
1
B
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Compute the stress increase resulting 8ft below the center of a 10ft square foundation imposing
500psf.
a. Use 60 approx
b. Use Boussinesq conditions
Example 2:
A circular foundation 12ft in diameter imposes a pressure of 8000psf onto the soil. At the 12ft
depth, determine the vertical stress increase beneath the center and the edge of the loaded area.
a. Westergaard
b. 60 approx