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Transcript
Soil Types
 Soil – all unconsolidated material in the earth’s crust
 Soil includes –
 Mineral particles – sand and clay
 Organic Materials – found in topsoil and marsh
 Air
 Water
Mineral Soils
 Result from weathering of rock that forms the solid
crust of the earth
 Physical weathering – due to the action of frost, water,
wind, glaciers, landslides, plant and animal life, and
other weathering agents – that break particles away
from the bedrock
 Particles are often transported by wind, water , or ice

Rounds them and further reduces their size
 Soils created this way are referred granular soils

Grains or particles are similar to the original bedrock
Mineral Soils
 Chemical weathering – occurs when water flows
through rocks and leaches out some mineral
components
 New soil particles are formed from these mineral
 Called clays
 Clay particles are mineral crystals that have very
different properties from those of the original bedrock
Types of Mineral Soils
 Gravel
 Sand
 Silt
 Clay
 Course grained soils – gravel and sand
 Fine grained – silt and clay
 Cobbles – over 75mm or 3 in
 Boulders – over 200mm or 8 in
 Clays are cohesive soils – bonded to each other
Gravels and Sand
 Composed mainly of rounded or cubical grains that
are supported by adjacent grains
 Can carry significant loads
 Loads are spread across many particles through friction
 Fairly easy to compact
 Excellent soils for construction
Clays and Silts




Clays are softer
Do not carry loads very well
Clay grains are small size and flat plate like shape
The mass of the grain as a force is negligible when compared to
the forces resulting from the surface properties of the grain
 Clays have charges on surface – figure 1-1 page 3
 Result of these charges is clay can hold a lot of water




Surface charges attract water molecules
Clays absorb or hold water – permanently unless conditions change
May dry out due to evaporation – or squeeze water out when load is
applied –
Will absorb moisture quickly when re applied
 The plates of clay are attracted to opposites charges
Field Test to Identify Soil types
 Large grains (sand and gravel) are easily to identified
 Organics are also easy
 Silts and clays – are not as easy cause grains are not
visible
 Page 5 table 1-2 differences between silt and clay
Mass-Volume Relations
 Soil sample contains
 Mineral – possibly organic particles
 Water
 Air
 Mass and volume of each phase is usually calculated
 Va = Volume of Air
 Vw = Volume of water
 Vv = Volume of voids (=Va +Vw)
 Vs = Volume of dry soil solids
 V = Total volume (=Va +Vw +Vs)
 Ma = Mass of air (=0 by definition)
 Mw = Mass of water
 Ms = Mass of dry soil solids
 M = Total mass (=Mw +Ms)
Mass and Volume
 Soil sample consisting of 10cm3 of air, 25cm3 of water
(mass = 25g) and 65cm3 of soil solids (mass = 175g)
 Vv = 35cm3
 Va = 10cm3
 Vw = 25cm3
 Vs =65cm3
 V=100cm3
 Mw =25g
 Ms=175g
 M=200g
Mass and Volume
 Relationship between the mass and volume
 Water




Pw = Mw/Vw
Where Pw = density of water
Density of water is 1g/cm3 or 1000kg/m3
Example 1-1 Pw = 25g/25cm3 = 1g/cm3
 Soil solids


Psoil solids = Ms/Vs
Where Psoil solids = density fo the dry soil solids
 Ratio between soil solids and density of water is the relative density
of the solids or specific gravity Gs



RD (relative density) = Psoil solids/Pw = Ms/(Vs*Pw)
Or RD =Ms/(Vs x Pw)
Example 1-1 RD=175g/(65cm3 x 1g/cm3) = 2.69
 Most soils – RD is between 2.6 and 2.8
Properties Calculations
 Density (P)
 Dry Density (Pd)
 Water content (W)
 Void ratio (E)
 Degree of saturation (S)
 Porosity(n)
 Problem 1-1 on page 8
P=m/v
Pp=Ms/V
W=Mw/Ms
e = Vv/Vs
S=Vw/Vv
n=Vv/V
Classification Tests
 Two Tests
 Grain size – to measure grain sizes


Sieve analysis used for sands and gravels
 Grain size distribution graph
 Example 1-9 page 17
Hydrometer used for silts and clays
 Sedimentation test
 Rate at which particles settle
 Strokes law states –that particles in a suspension settle out at a
rate that varies with their size
 Plasticity – to measure grain types
Grain Size Distribution Curve
 Used to help describe and classify a soil
 Shape –


Uniform soil –curve a on page 19
Well graded – curve b on page 19
 Effective size
 10% size is considered effective size – page 19 – sample b
.09mm
 Uniformity coefficient –
 Value gives some indication of the shape of the curve
 Cu=D60/D10
 Coefficient of curvature
 Cc=(d30)2/(d60xD10)
Textural Classification
 American Society for Testing and Materials
 Gravel – larger than 4.75 (no. 4)
 Sand 4.75mm to 0.075mm (no. 4 to No. 200)
 Silt .075mm to .005mm (No. 200 to .005mm)
 Clay – smaller than .005mm
Plasticity Test
 Measures the amount of water that a soil adsorbs
 Plastic limit – soil is roll into a thread(Wp)
 Liquid limit (WL)
 Index of plasticity – range of water contents over which
this soil is plastic

Ip=wl-wp
 Atterberg Limits test
Soil Water
 Type of water found in soil
 Free water or gravitational water – found below
groundwater – free to flow under the laws of gravity
 Capillary water – brought up through the soil pores –
above the groundwater table
 Attached water or held water – moisture film around
soil grains
 The rate of water flow – or permeability of the soil



Darcy’s law
Q=k (h/l)a – page 33
Problem page 34
Soil Strength and Settlement
 Two type of soil failure
 Failures due to shear – grains slide with respect to other grains
 Settlement failures – where a layer of soil is compressed and
becomes thinner under loading
 Forces on soil


Forces acting perpendicular to the plane are normal forces
Forces acting parallel are shear
 Shear strength in most clays is due to cohesion
 T=c were t is sheering pressure and c = cohesion
 Shear strength in granular soils is due to friction

T=o tan o/