DescriptionCIVE 224 Geotechnical Engineering

Final Review Sheet

Spring, 2023

Review Sheet:

The final exam covers the basic knowledge in geotechnical engineering class, there are basic concept questions (B)

and calculation questions (C). Please refer to our mid-term exam for exam format and question types. This list is to

help you focusing on the final exam (comprehensive exam, but more on the new contents after midterm), not an

inclusive list of all the topics we’ve covered throughout the semester.

Chap 6. Soil Compaction

o

o

o

o

Soil compaction introduction (B)

e.g. objectives for compaction, differences between compaction and consolidation, effect of water on

compaction, what is zero-air-void line etc.

Standard Proctor Compaction Test (B, C)

Effect of Compaction Energy (B)

Modified Proctor Compaction Test (B, C)

Study this example:

Chap 7. Permeability

o

o

o

o

o

o

o

o

o

Permeability in Soils (B)

e.g. absolute permeability, constant head vs. falling head test, when to use which one?

Bernoulli’s Equation (B, C)

Darcy’s Law (C)

Hydraulic Conductivity (B, C)

Hydraulic Conductivity Lab Tests (C)

Relationships for Hydraulic Conductivity –granular soils (B, C)

Relationships for Hydraulic Conductivity-cohesive soils (B, C)

Directional (anisotropic) variation of permeability (C)

Equivalent Hydraulic Conductivity

o Theory: stratified soil (C)

o Experimental verification (B)

CIVE 224 Geotechnical Engineering

Final Review Sheet

Spring, 2023

Some important formulas:

2

CIVE 224 Geotechnical Engineering

Final Review Sheet

Spring, 2023

Study these examples:

3

CIVE 224 Geotechnical Engineering

Final Review Sheet

Spring, 2023

4

CIVE 224 Geotechnical Engineering

Final Review Sheet

Spring, 2023

Chap 9. In Situ Stresses

o

o

Stress tensor, stress units (B)

Effective stress, Stresses in saturated soil without seepage (B, C)

5

CIVE 224 Geotechnical Engineering

Final Review Sheet

Spring, 2023

Some important formulas:

Study this example:

6

CIVE 224 Geotechnical Engineering

Final Review Sheet

Spring, 2023

Miscellaneous

o

o

o

Moisture content, dry unit weight, void ratio etc

Relative compaction

Discharge velocity vs. seepage velocity

7

CVIE 224 Soil Engineering

Spring 2023

Dr. Yachi Wanyan, PhD. , PE.

Lecture Slides by Braja M. Das, Lanbo Liu and Yachi Wanyan

¡ Soil compaction introduction

¡ Standard Proctor Compaction Test

¡ Effect of Compaction Energy

¡ Modified Proctor Compaction Test

2

¡ In construction of highway embankments,

earth dams and many other engineering

structures, loose soils must be compacted to

improve their strength by increasing their

unit weight.

¡ Compaction -Densification of soil by

removing air voids using mechanical

equipment;

¡ The degree of compaction is measured in

terms of its dry unit weight.

¡ Increasing the bearing capacity of

foundations;

¡ Decreasing the undesirable settlement of

structures;

¡ Control undesirable volume changes;

¡ Reduction in hydraulic conductivity;

¡ Increasing the stability of slopes.

Compaction: realized by rearrangement of soil particles

without outflow of water. It is realized by application of

mechanic energy. It does not involve fluid flow, but with

moisture altering.

¡ Consolidation: another kind of densification with fluid flow

away. Consolidation is primarily for clayey soils. Water is

squeezed out from its pores under load. (This chapter of the

textbook concentrates on compaction, and the consolidation

will be discussed later in Chapter 11.)

¡

¡ Compaction effort;

¡ Soil type and gradation;

¡ Moisture content;

¡ Dry unit weight (dry density).

•

Compaction is the densification of soil by removal of air, which requires mechanical

energy.

•

Compaction is measured in terms of its dry unit weight.

•

Water acts as a softening agent on soil particles.

•

The dry unit weight at moisture content w = w1 is:

g d ( w= w ) = g d ( w=0) + Dg d

1

•

Initially, the dry unit weight after compaction increases as moisture content

increases.

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly

accessible website, in whole or in part.

¡ In soils, compaction is a function of water

content

¡ Water added to the soil during compaction acts

as a softening agent on the soil particles

§

§

§

§

Consider 0% moisture -Only compact so much

Add a little water -compacts better

A little more water -a little better compaction

Even more water –Soil begins to flow

¡ What is better compaction?

§ The dry unit weight (γd) increases as the moisture content increases TO A

POINT

§ Beyond a certain moisture content, any increase in moisture content

tends to reduce the dry unit weight

¡ Purpose: Find the optimum moisture content at

which the maximum dry unit weight is attained

¡ ASTM D 698

¡ The standard was originally developed to

simulate field compaction in the lab

¡ Equipment: Standard Proctor

§ 1/30 ft^3 mold

§ 5.5 lb hammer

§ 12” drop

§ 3 layers of soil

§ 25 blows / layer

1.

2.

3.

4.

5.

6.

7.

8.

Obtain 10 lbs of soil passing No. 4 sieve

Record the weight of the Proctor mold without the base and the

(collar) extension, the volume of which is 1/30 ft^3.

Assemble the compaction apparatus.

Place the soil in the mold in 3 layers and compact using 25 well

distributed blows of the Proctor hammer for each layer.

Detach the collar without disturbing the soil inside the mold

Remove the base and determine the weight of the mold and

compacted soil.

Remove the compacted soil from the mold and take a sample

(20-30 grams) of soil and find the moisture content.

Place the remainder of the molded soil into the pan, break it

down, and thoroughly remix it with the other soil, plus 100

additional grams of water.

=

It is clear that in the above equation, Gs and Gw are constant, the zero-air-void

density is inversely proportional to water content w.

For a given soil and water content the best possible compaction is represented by the zero-air-voids

curve.

The actual compaction curve will always be below.

For dry soils the unit weight increases as water is added to the soil because the water

lubricates the particles making compaction easier. As more water is added and the

water content is larger than the optimum value, the void spaces become filled with

water so further compaction is not possible because water is a kind like

incompressible fluid. This is illustrated by the shape of the zero-air-voids curve which

decreases as water content increases.

• Aside from moisture content, two

important factors that affect compaction

are soil type and compaction effort (energy

per unit volume).

• Soil type has a great influence on the

maximum dry unit weight and optimum

moisture content.

• Some soils, like sand, exhibit compaction

curves different than a single-peaked bell

curve.

Figure 6.6 Typical compaction curves for four soils

(ASTM D-698) (Note: gw = unit weight of water)

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly

accessible website, in whole or in part.

• If the compaction effort per unit of volume of

soil is changed, the moisture-unit weight curve

also changes.

1. As the compaction effort is increased, the

maximum dry unit weight of compaction

increases

2. Additionally, the optimum moisture content

decreases to some extent

• Note: The degree of compaction is not directly

proportional to the compaction effort.

Figure 6.8 Effect of compaction energy on the

compaction of a sandy clay

© 2022 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly

accessible website, in whole or in part.

¡ Soil Type

§ Grain size

distribution

§ shape of soil

grains

§ Gs

§ Clayey

minerals

Quiz Q 1:

From the graph, what is

the relationship between

Grain size and Optimum

Moisture Content (OMC)?

Compaction curves for different soils with the same compact effort

¡ Compaction Effort

§ Compaction Energy is calculated by

Quiz Q2: Calculate E for Standard Compaction Proctor

Where: Mold volume = 1/30 cubic foot; Compact in 3

layers; 25 blows/layer; 5.5 lb hammer; 12″ drop

¡ With the development of heavy rollers and

their uses in field compaction, the Standard

Proctor Test was modified to better represent

field compaction

¡ As the compaction effort increases:

§ the maximum dry unit weight of compaction

increase

§ The optimum moisture content decreases to

some extend

¡ The modified was developed to simulate larger

compaction effort for more serious loads and

bigger equipment

¡ ASTM D 698

¡ Equipment: Modified Proctor

§ 1/15 ft^3 mold

§ 10 lb hammer

§ 18”drop

§ 5 layers of soil

§ 25 blows / layer

¡ Standard Proctor Test

¡ http://www.youtube.com/watch?v=tqHNK67I

gG4

¡ Standard and Modified Proctor Test

¡ http://www.youtube.com/watch?v=ltmhN4S

Mueg

CIVE 224 Soil Engineering

Spring 2023

Dr. Yachi Wanyan, PhD. , PE.

Lecture Slides by Lanbo Liu, revised by Yachi Wanyan

¡ Stress tensor, stress units

¡ Effective stress, Stresses in saturated soil

without seepage

¡ Stresses in saturated soil with upward

seepage

¡ Stresses in saturated soil with downward

seepage

¡ Seepage force

2

This is a very practical issue, since:

a) Consolidation settlements in soils are strongly

related to increased stresses from applied loads.

b) Shear failure in soils can occur due to applied

loads, and as engineers, we need to be able to

predict whether or not failure will occur.

Stress is the force per unit area on a

surface.

¡ When the force F and the surface

outward normal n are in the same (or

opposite) direction, this stress is called

the normal stress.

¡

¡

When the force and the surface

outward normal are perpendicular

with each other, the stress is the shear

stress.

General mechanics defines tensile stress as

positive, whereas earth science defines the

compressive stress positive, since the

rocks in depth are constantly experiencing

compression.

¡

¡

¡

But the force has directivity, i.e.,

force is a vector and can be

decomposed into components

in 3 orthogonal directions.

Meanwhile, the surface on

which the force is acting on, also

has directivity, it can be facing

any direction.

If we use a unit vector n to

represent a unit area on this

surface, it can also be

decomposed into components

on three orthogonal directions.

So, the stress is a tensor and has

9 components.

¡

Actually, since the force is a

vector, and can act on any

direction, it can be dissolved

into 3 orthogonal directions

(e.g., x, y, z, in Cartesian

coordinate); a surface in or

on the solid can face any

direction, too. The normal of

that surface is also a

function of x, y, z.

Consequently, a complete

stress should be a tensor

with 9 elements.

¡ The unit of pressure:

§ 1 bar = 1 standard atmospheric pressure

§ It also approximately equals to the pressure under

10 meters of water.

¡ Normal stress caused by self-gravity (

overburden)

§ Normal stress= density x gravity x depth = ρgz

¡

In the SI System, the fundamental unit of length is in meters, time in

seconds, and mass in kilograms (basic units) . The unit in SI system for

stress/pressure is in Pascals

¡

A force of one Newton spread out over a square meter is a prettyfeeble

force. Atmospheric pressure (1 bar) is about 100,000 Pascals. A manila

file folder (35 g, 700 cm2area) exerts a pressure of about 5 Pascals.

¡ The normal stress asserted on the solid

matrix minus the pressure asserted on the

pore fluid.

In summary, the total stress is the sum of the effective

stress and the pore pressure (neutral stress).

The effective stress is carried by the soil skeleton. The pore

pressure is carried by pore water.

The effective stress in a soil mass controls its volume change

and strength. Increasing the effective stress (equivalent to

reduce the pore pressure if the total stress is constant)

induces soil to move into a denser state of

packing.

1.

2.

Calculate the total stress, pore

water pressure, and effective

stress at points A, B, and C. b.

If the water table rises to 2 m

below the ground surface, what

would be the change in s9 at A,

B, and C?

CVIE 224 Soil Engineering

Spring 2023

Dr. Yachi Wanyan, PhD. , PE.

Lecture Slides by C.C. Swan, Lanbo Liu and Yachi Wanyan

¡ Permeability in Soils

¡ Bernoulli’s Equation

¡ Darcy’s Law

¡ Hydraulic Conductivity

¡ Hydraulic Conductivity Lab Tests

3

Permeability is the parameter to characterize the ability of

soil to transport water.

¡ Due to the existence of the inter-connected voids, soils are

permeable. The permeable soils will allow water flow from

points of high energy to points of low energy.

¡

¡ It is one of the most important soil properties of

interest to geotechnical engineers

§ Permeability influences the rate of settlement of a

saturated soil under load.

§ The design of earth dams is very much based upon the

permeability of the soils used.

§ The stability of slopes and retaining structures can be

greatly affected by the permeability of the soils involved.

§ Filters made of soils are designed based upon their

permeability.

Knowledge of the permeability properties of soil

is necessary to:

§ Estimating the quantity of underground seepage

(Chapter 8);

§ Solving problems involving pumping seepage

water from construction excavation;

§ Stability analyses of earth structures and earth

retaining walls subjected to seepage forces

1.

environmental engineering:

§ When toxic liquids are retained in holding lagoons,

important questions that need to be answered are:

−−> At what rate is toxic fluid escaping the holding lagoon?

−−> How long might it take the fluid to reach the ground water table?

−−> What can be done to slow down the rate of escape of the

pollutant?

2.

Construction Engineering:

§ When building structures in relatively shallow bodies of water, it is

fairly common construction practice to build a temporary sheet−pile

cofferdam structure around the site and to pump the water out.

Important questions that would need to be answered are:

−−> What will be the rate of water inflow to the site ? (so that the pump can

be sized)

−−> Is it possible that the soil will liquify and endanger construction workers?

¡ Beneath the water table:

§ Fluid pressure is hydrostatic: = !

§ This fluid pressure is really a “gauge” pressure in that the atmospheric

pressure has been subtracted out.

§ Immediately above the water table is the capillary zone:

In the capillary zone, water pressure is less than atmospheric pressure.

−−> Thus, we say that the fluid pressure is negative in this region.

¡

Bernoulli head measures the total energy of a fluid at a fixed spatial point.

¡

The Bernoulli head has three additive contributions:

§

§

z represents the elevation w.r.t. an arbitrary datum. The value is the distance of the point at

which head is being measured above the datum. This number can be either positive if the point

is above the datum, or negative if the point is below the datum.

!⁄

“! represents the pressure head of the fluid and has units of length.

§

# “”

$% represents the kinetic head of the fluid and also has units of length. Since water flowing

in soil typically has very small velocities, the kinetic head is typically negligible compared to

that of the pressure and elevation energies. For this reason, the velocity head is neglected in

soil mechanics.

¡

Therefore, the so−called piezometric head used to measure the elevation and

”

pressure energies of fluids in soils is simply: ℎ = + #

!

¡

The actual value of the piezometric head in fluids is naturally dependent upon

the choice of the reference datum.

¡ The surface of the water column (the head) is the

water table. Water table in an Unconfined Aquifer is

the surface along which the hydrostatic pressure is

equal to the atmospheric pressure.

Water in confined aquifer is separated from air by impermeable

layers known as aquiclude. This type of aquifer forms an artesian

system;

¡ The well drilling into confined aquifer then could be an artesian

well (the water level in the well is above the height of the ceiling

aquiclude).

¡

¡

The dynamic head is usually negligible since the water flow

velocity is usually small. The elevation head is accounted

from the datum to the elevation of the bottom of the well,

and the pressure head is the portion above the well bottom

to the water table.

1.

2.

3.

Fluids flow in soils only when there is a gradient in piezometric

head h. Lack of a gradient in piezometric head implies that fluid

is not flowing

Whenever there is fluid flow in soils, there is energy dissipation.

In soils, fluids always flow down the gradient in piezometric

head. That is, fluids flow from high energy regions to low energy

regions.

¡

At point 1:

¡

At point 2:

¡

The head is identical at points 1 and 2. Thus there is no gradient of

head h in the soil. Therefore, there is no flow of water in this

situation.

¡ Observe that h1 > h2. Therefore water flows

down the head gradient from point 1 toward

point 2.

Hydraulic Gradient:

∆ℎ

=

may exist in fractured rock, stones, gravels,

and very coarse sands

in most soil we found the following relation,

i.e., the water flow velocity in the soil is

proportional to the hydraulic gradient ∝

¡

In 1856, Darcy established an empirical relationship for the

flow of water through porous media known as Darcy’s Law,

which states:

= =

§ q = flow rate (cm3/s)

§ k = coefficient of permeability (cm/s)

§ A = cross-sectional Area (cm2)

§ i = hydraulic gradient

¡

This equation was based primarily on Darcy’s observations

about the flow of water through clean sands.

¡

is the discharge velocity of water based on the gross cross-sectional

area of the soil. However, the actual velocity of water $ (that is, the

seepage velocity) through the void spaces is greater than v.

¡ In the field, the gradient of the head is the head

difference over the distance separating the 2 wells.

( − )

=

=

∆

¡ Darcy’s law states that how fast the

groundwater flow in the aquifer depends on

two parameters:

1. How large is the hydraulic gradient of the water

head (i=dH/dx); and

2. the parameter describing how permeable the

aquifer porous medium –the coefficient of

permeability (hydraulic conductivity) k.

¡ The physical description of groundwater flow

in soil is the Darcy’s law.

¡ The fundamental premise for Darcy’s law to

work are:

1. the flow is laminar, no turbulent flows;

2. fully saturated;

3. the flow is in steady state, no temporal variation.

¡

¡

Hydraulic Conductivity, k, is a measure of soil permeability.

The hydraulic conductivity of soils depends on several

factors:

§ Fluid viscosity,

§ Pore-size distribution,

§ Grain-size distribution,

§ Void ratio,

§ Roughness of mineral particles, and

§ Degree of soil saturation.

In clayey soils, structure plays an important role in hydraulic

conductivity.

¡ Other major factors that affect the permeability of clays are

¡

§ Ionic concentration and

§ Thickness of layers of water held to the clay particles.

¡ The hydraulic conductivity of a soil is also

related to the properties of the fluid flowing

through it by the equation:

¡ In the lab using two methods:

§ Constant-Head Test

§ Falling-Head Test

¡ In-situ:

§ Well−drawdown test

¡ The constant head test is used primarily for

coarse-grained soils;

¡ This test is based on the assumption of

laminar flow where k is independent of i (low

values of i);

¡ ASTM D 2434;

¡ This test applies a constant head of water to

each end of a soil in a “permeameter”.

For Coarse−Grained Soils):

§ Upstream and downstream head

elevations are maintained at

constant levels.

§ The head difference across the

soil is a constant value Δh.

§ The hydraulic gradient i across

the sample is also constant. i =

hydraulic gradient in the soil

(Δh/L)

Q = AvT = volumetric flow through

the soil over an elapsed time T.

¡

q = Q/T = vA = rate of volume flow;

§ v = q/A = the so−called discharge velocity

§ A = the cross−sectional area of the soil sample

¡

Recall from Darcy’s Law that: v = ki

¡

From a constant head test, soil permeability k can be

computed as:

1.Setup screens on the permeameter

2.Measurements for permeameter, (D), (L), H1

3.Take 1000 g passing No.4 soil (M1)

4.Take a sample for M.C.

5.Assemble the permeameter–make sure seals are air-tight

6.Fill the mold in several layers and compact it as prescribed.

7.Put top porous stone and measure H2

8.Weigh remainder of soil (M2)

9.Complete assembling the permeameter. (keep outlet valve closed)

10.Connect Manometer tubes, but keep the valves closed.

11.Apply vacuum to remove air for 15 minutes (through inlet tube

attop)

12.Run the Test (follow instructions in the lab manual) …..

13.Take readings

–Manometer heads h1 & h2

–Collect water at the outlet, Q ml at time t ≈60 sec.

¡ Determine the unit weight;

¡ Calculate the void ratio of the compacted specimen;

¡ Calculate k as:

¡ Calculate

Where η% is viscosity of water at temperature &

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