Geotechnical Engineering Worksheet

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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