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Soil and Rock
Soil and rock are the principle components of many construction projects.
Knowledge of their properties, characteristics, and behavior is important to those associated with the design or construction of projects.
Soil and Rock
Soil and Rock
Steel and concrete are construction materials that are basically homogeneous in composition. As such, their behavior can be predicted. Soil and rock are just the opposite. By nature they are heterogeneous. In their natural state, they are rarely uniform.
Soil and Rock
Soil and rock are heterogeneous. They are rarely uniform and work processes are developed by comparison to a similar type material with which previous experience has been gained. To accomplish this, soil and rock types must be classified.
Soil and Rock
GRADATION
Soil gradation is the distribution, in percent (%) by weight, of individual particle sizes.
SOIL TYPES
ORGANIC SOILS
Will usually have to remove before building.
SOIL TYPES
Bulky shaped soil grains
NON-COHESIVE
SOIL TYPES
Small grained #200 Mesh sieve
Platy shaped soil grains
COHESIVE
SOIL LIMITS
Atterburg Limits
LL - Liquid limit
PL - Plastic limit
PI - Plasticity Index
SOIL LIMITSStages of Consistency
Moisture content decreasing
SOIL LIMITS
LL - Liquid limitis the water content of a soil when it passes from the plastic to liquid state.
SOIL LIMITS
LL - Liquid limit
Non-cohesive or sandy soils have low LLs -- less than 20.
Clay soils have LLs ranging from 20 to 100.
SOIL LIMITS
PL - Liquid limitis the lowest water content at which a soil remains plastic.
1/8 inch diameter thread
SOIL LIMITS
PI - Plastic Index
PI = LL - PL
The higher the PI the more clay that is present in the soil.
Volumetric Measure • Bank cubic yards (bcy)
• Loose cubic yards (lcy)
• Compacted cubic yards (ccy)
bcy lcy ccy
COMPACTION
Each soil has its particular optimum moisture content (OMC) at which a corresponding maximum density can be obtained for a given amount of
compactive input energy.
COMPACTION
PROCTOR TESTStandard Proctor
or
AASHTO T-99
Soil sample 1/30 cubic foot3 layers
COMPACTION
COMPACTION
PROCTOR TEST
Modified Proctor
or
AASHTO T-180
Soil sample 1/30 cubic foot5 layers
COMPACTION SPECIFICATIONS
Typically specifications give an acceptable range of water content, OMC ± 2% for example.
COMPACTION SPECIFICATIONS
The specification also sets a minimum density, 95% of max. dry density for a specific test
126.4
Must work in the box.
Soil Weight-Volume Relationships
V
W
volume soil total
soil ofweight totalγweightUnit
V
W
volume soiltotal
solids soil ofweight γweightunit Dry s
d
s
w
W
W
solids soil ofweight
soilin water ofweight ωcontentWater
Equ. 4.1
Equ. 4.3
Equ. 4.2
Water Content
s
w
W
W
solids soil ofweight
soilin water ofweight ωcontentWater
Dry Weight
Dry WeightWet Weight =
Water Content
= = 0.18 or 18%85
8100 5
Soil Weight-Volume Relationships
ω1
γγd
Dry weight is related to unit weight by water content,
and when you move rock and dirt the only thing that stays constant is the weight of the solid particles.
Soil Weight-Volume Relationships
is the weight of the solid particles.
When you move rock and dirt the only thing that stays constant
EXERCISE
• unit weight () of 94.3 pcf• water content () of 8%.
The excavated material has a
EXERCISE
• dry unit weight (d) of 114 pcf• water content () of 12%.
The embankment will be compacted to
EXERCISEThe net section of the embankment is 113,000 cy.How many cubic yards of excavation will be required to construct theembankment?
As material is moved from the excavation
to the compacted fill the only constant is the weight of the solid particles (d).
EXERCISE
EXERCISE Step 1Weight of the solid particles which make up the embankment (fill).
• a dry unit weight (d) of 114 pcf
• 113,000 cy embankment
113,000 cy27 ft
cy114 lb / ft
33
Conversion factor cy to ft3
EXERCISE Step 2Use relationship d - to calculate the dry unit weight of the excavated material.• a unit weight () of 94.3 pcf• a water content () of 8%
d = 87.31 pcf94.3 ft
1 + 0.08
3
EXERCISE Step 3Calculate the weight of the solid particles which make up the excavation (cut).
x cy27 ft
cy87.31 lb / ft
33
EXERCISE Step 4The weights must be equal therefore:
x cy27 ft
cy87.31 lb / ft
33
113,000 cy27 ft
cy114 lb / ft
33
=
Conversion factors cancel out.
EXERCISE Step 4The weights must be equal therefore: x =
x = 147,535 cy
excavated material required
113,000 cy114 lb / ft
87.31 lb / ft
3
3
EXERCISECheck the water requirements.
Will a water truck be needed on the job or will it be necessary to dry the material?
Water ?
Water content () is?
x d = weight of water/cf
Step 1 Water from Cut
147,535 cy
87.31 pcf
27 cf / cy
0.08 = lb of water
Vol. Cut
d
conversion factor()
Step 1 Water from Cut
147,535 cy 87.31 pcf 27 cf / cy 0.08
= 27,825,120 lb waterdelivered with the borrow material
Step 2 Water needed at the Fill
113,000 cy
114 pcf
27 cf / cy
0.12 = lb of water
Vol. Emb
d
conversion factor()
Step 2 Water needed at the Fill
113,000 cy 114 pcf 27 cf / cy 0.12
= 41,737,680 lb waterneeded at the fill
Step 3 Water Deficiency
Needed at the fill 41,737,680 lb
Delivered w/ cut 27,825,120 lb
Water deficiency 13,912,560 lb
PE 2 Step 4 Convert to Gallons
Water deficiency 13,912,560 lb
Water weights 8.33 lb/gal
Need to add 1,670,175 gallons
Step 5 Gallons per cy
Water deficiency 1,670,175 gal
Volume of cut 147,535 cy
Need 11.3 gal/cy
Adding Water
Using sprinklers to add moisture to a foundation fill.
Reducing Moisture
Disking a heavy clay fill to reduce moisture.