SOIL MECHANICS SAB1713 · “ Soil mechanics is the application of laws of mechanics and hydraulics...
Transcript of SOIL MECHANICS SAB1713 · “ Soil mechanics is the application of laws of mechanics and hydraulics...
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SOIL MECHANICS SAB1713
DR. HETTY
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INTRODUCTION
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SOIL MECHANICS
-Concerned solely with soils
-Concerned with the deformation and strength of bodies of soils
-Concerned with the interaction of structures with their
foundation materials (STABILITY & SERVICEABILITY)
“ Soil mechanics is the application of laws of mechanics and hydraulics to engineering problems dealing with sediments and other unconsolidated accumulations of solid particles produced
by the mechanical and chemical disintegration of rocks regardless of whether or not they contain an admixture of
organic constituent” (Terzaghi, 1948)
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What is Soil???
Un-aggregated or un-cemented deposits of mineral and/or organic particles or fragments covering large portion of the earth’s crust.
An uncemented aggregate of mineral grains and decayed organic matter (solid particles) with liquid and gas inempty spaces between solid particles. (Das)
Any uncemented or weakly cemented accumulation of mineral particles formed by the weathering of rock, the void space between the particles containing water and air. (R.F.Craig)
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The Source of Soils?????
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The Differentiation between SOIL & ROCK
SOIL – removed without blasting
ROCK – required blasting
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ROCKS The Source of Soils Soil particles are the result of weathering (disintegration and decomposition) of rocks and decay of vegetation.
Some soil particles may, over a period of time, become consolidated under the weight of overlying material and become rock.
Figure2: Rock Cycle
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Figure 3:Schematic diagram of a typical residual weathering soil profile
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V
II
I
III
IV
VI Residual
soils
Completely
decomposed
Highly
decomposed
Moderately
decomposed
Slightly
decomposed
Fresh
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Weathering Process
DISENTEGRATION (Mechanical weathering)
- Disintegrates rocks into small particles by temperature changes, frost action, rainfall, running water, wind, ice, abrasion, and other physical phenomena.
DECOMPOSITION
(Chemical weathering)
- Chemically decomposed rocks, which can drastically change it physical and chemical characteristics.
- Results from reactions of rock minerals with oxygen, water, acids, salts and…..
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Rock Classification
Igneous
Sedimentary
Metamorphic
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Igneous Rock
Formed by the cooling of molten magmas, or by the recrystallization of the older rocks under heat and pressure.
2 main classes of igneous rock:
- Intrusive (cooling and solidifying slowly beneath the earth) – coarse grained
- Extrusive (poured out at the surface and cooling rapidly) – fine grained
Granite and Basalt
Hard, Dense and durable
Good construction materials
Have high bearing capacity – good for foundation
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Granite
Basalt
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Sedimentary Rock
Results from weathering process that being transported, redeposited, and partly or fully consolidated or cemented into a new rock.
Sediment = loose particle material (clay, sand. Gravel, etc)
Sediment becomes sedimentary rock through lithification, which involves:
1) Compaction
2) Cementation
3) Recrystallization
Can be identified easily when their layered is observable
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Shale Siltstone
Layered gypsum Quartz Sandstone
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Metamorphic Rock
Produced when sedimentary or igneous rocks literally change their texture and structure as well as mineral and chemical composition, as a result of heat, pressure and shear.
Granite metamorphose to gneiss Shale Slate Schist
Limestone Marble
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gneiss
schist
schist
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limestone
marble
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BASIC PHYSICAL SOIL PROPERTIES
SOIL COMPOSITION
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SOIL COMPOSITION
Generally soil can be divided to three components in natural occurrence
This separation know as three phase systems or three phase diagram
It consists of soil solids, water and air
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Soil components in natural occurrence
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Volume-Weight relationships
As a result from 3 phase diagram several relationship were produce
This relationship known as volume weight relationships.
Refer to 3 phase diagram
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Relationship equations
V = Vs +Vv ( eq. 1) = Vs + Vw + Va
where Vs = volume of soil solids Vv = volume of voids Vw = volume of water in the voids Va = volume of air in voids
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Relationship equations
W = Ws + Ww ( eq. 2) where, Ws = weight of soil solid Ww= weight of water Wa = zero (negligible)
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Void ratio, e
The ratio of the volume of voids Vv to the volume of soil solid Vs
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Porosity, n
The ratio of the volume of voids, Vv to total volume, V expressed as either a decimal or a percentage.
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Relationship between void ratio and porosity
e = Vv = Vv = (Vv/V ) = n Vs V – Vv 1 – (Vv/V) 1- n n = e 1 + e
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Degree of saturation, Sr
The ratio of the volume of water, Vw to the volume of voids, Vv. It’s commonly in percentage
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Water content@ Moisture content
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Unit weight, W
The weight of soil per unit volume
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Relationship between dry unit and bulk unit weight
d = /(1 +w)
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Specific Gravity
Specific gravity can be determined as a ratio of the weight of the soil solids to weight of water of equal volume
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Relationships between Unit weight, Moisture Content and Specific gravity
let us consider a volume of soil in which the volume of the soil solids is one
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Derivation of equations
Therefore Gs will be substituted as
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Derivation –cont’d
= W = WS +WW =GS W + wGs W = (1+w) Gs w
V V 1+e 1+e
d= Ws = Gs w
V 1+e
S= Vw = wGs Vv e
Therefore
How Vw become wGS???
Vw = Ww = wGs w =wGs w w
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sat = W = WS +WW = Gs w + e w = (Gs+e) w
V V 1+ e 1+e
So if S = 1 (100%)
Se = wGs
e = wGs
Beside bulk unit weight, dry unit weight and saturated unit weight
there is another type of unit weight that is submerge unit
weight, ’= sat- w
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Table 1: Void ratio, moisture content, and dry unit weight for some typical soil
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Derivation –cont’d
So what is the different between unit weight and density?
Unit Weight is influence by gravity forces while mass isn’t (Unit Weight=Weight/ Volume while Density= Mass/Volume)- Weight = Mass x Gravity acceleration
So we can use the similar equations on related calculation if there is only density
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For example: If the equations use unit weight you must use weight in the calculation but if you are only given density you must substituted weight with mass or vice versa.
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The air content, A can be expressed as
A = e- wGs / 1+ e
or A = e/ 1+e ( 1- wGs/e)
Since n = e/1+e
Therefore, A = n (1-Sr)
Beside using symbol A as air content sometimes authors use Na
Exercise