Soil and Rock Soil and rock are the principle components of many construction projects. Knowledge of...

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.