soil mechanics handouts

7/26/2019 Soil Mechanics Handouts

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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

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