American Academy of English Geotechnology Construction Homework for assignment 1 : I need add -introduction -results and discussion -conclusion For

American Academy of English Geotechnology Construction Homework for assignment 1 :

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For assignment 2: I need only (results and discussion) GEOTECHNOLOGY 1
Geotechnology
Student Name
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Date
GEOTECHNOLOGY 2
GEOTECHNOLOGY ASSIGNMENT
Question 1
a) Before the construction of the embankment.
Initial stresses on the clay (z = 4m)
Vertical total stress;
sv = 30 x 4 = 120kPa
Pore pressure;
u = 9.81 x 4 = 39.24kPa
Vertical effective stress;
s´ = sv-u = (120-39.24) kPa = 80.76kPa
Initial stresses on sand (z = 11m)
Vertical total stress
sv = 27×11 = 297 kPa
Pore pressure
u = 9.81×11 = 107.91 kPa
Vertical effective stress
s´= sv – u = 189.09 kPa
GEOTECHNOLOGY 3
b) Immediately after the construction of the embankment.
The construction of the embankment in the surface surcharge:
q = 23×6 = 138 kPa.
The sand is drain and so there is no increase in pore pressure. The clay is undrained and
the pore pressure increases by 138 kPa.
Initial stresses at mid-depth of clay (z = 4m)
Vertical total stress
sv = (30×4) + 138 = 258kPa
Pore pressure
u = (9.81×4) + 138 = 177.24 kPa
Vertical effective stress
s´v = sv – u = 80.76kPa
Initial stresses of sand (z = 11m)
Vertical total stress
sv = (27×11) + 138 = 435kPa
Pore pressure
u = 9.81×11 = 107.91 kPa
Vertical effective stress
s´v = sv – u = 327.09kPa
c) When ground water table lies at surface of ground level and below the foundation
and discuss the change in the stress’s values if any.
GEOTECHNOLOGY 4
In steady state, the hydraulic gradient,
i = Dh/Ds = 6/( 14 + 7 ) = 0.29
Then the effective stresses are:
s´clay= (30×11) – (4×9.81) +( 0.29×9.81) = 293.6 kPa
s´sand = (27×11) – (4×9.81) – (0.29×9.81) = 260.6 kPa
Question 2
Design a square shallow foundation for a residential building of width B and depth
D is carrying design load Q and resting on a soil of unit weight.
Determine the ultimate bearing capacity of the rectangular foundation, 3m x 6m, is to be
founded at a depth of 1.8m below the ground surface of a deep stratum of soft saturated
clay (Unit weight = 28 kN/m3). Undrained and consolidated undrained triaxial tests
established the following soil parameters: Cu=24 kN/m2, ?’ = 25o, c’ = 0 for the following
conditions
i.
Immediately after construction,
GEOTECHNOLOGY 5
As per Coulomb’s eq. ? = c + ? tan ? ——-(A)
Two equations can determine c and ?. • Given: ?1 = 0.37 kg/cm2 ?2 = 0.5 kg/cm2
?1= 24 kg/m2 ?2= 25 kg/m2 Put values in eq. (A) 0.37 = c + 0.75 tan?. ———-(i)
0.5 = c + 1.5 tan?.————(ii)
c = 0.25 kg/cm2 tan ? = 0.1733 ? ? = 9.83o ? 10o some years after construction
ii.
Assume local shear failure. ? Nc ? = 6 , Nq ?= 1.9 N? ? = 0.5 qf = 2/3 c Nc ? + ?D Nq ?
+ 0.5 ? B N? ? 9.
qf = 2/3 x 0.25 x 9 + 1772/109 x 2.5 x 102 x1.9 + 0.5 x 1772/109 x 3.0×102 x0.5
= 1.333 + 0.01069 + 0.001688 = 1.345 kg/cm2 = 13.45 t/m2
Ultimate load = 13.45 x 3 = 40.37 t/m.
Question 3
Determine the lateral earth force at rest per unit length of the retaining wall when
ground water table is acting at 3m below ground surface and find out the location of
the resultant force.
Solution;
We calculate the pressures at the junction of the two soils.
The total pressure distribution will be trapezoidal with top ordinate = 45 kN/m2 and
bottom 27 kN/m2,
GEOTECHNOLOGY 6
the passive pressure at the base, KL = KP1 x ? x 3 = 4.5x19x3.5 = 299.25 kN/m2.
To calculate the pressure in soil 2;
The weight of the first soil = 18 kN/m3x8.5m = 153kN/m2
The already surcharge coming =49 kN/m2
Hence, the total surcharge to be considered for soil 2 (q?) = 153 + 49 = 202 kN/m2
Due to cohesion component of the 3m deep soil due to submerged weight of 3 m deep
soil, hydrostatic pressure in 3 m.
What is foundation settlement and explain how it affects civil engineering
constructions.
A fountain settlement is an architectural design for landscaping with the use of water
(Indraratna, et al. 2016). Most fountains usually eject water into the basis to create a waterfall or
sprays it into the air. The fountains are used in the plumbing industry for the decoration of
buildings and other structures (Das 2017). Mechanical pumps were used to project the water into
the air and recycle.
Nowadays, the fountains use the steam and the electric pumps to access the water source.
The electric pump provides the power to pump the water into the pipes. With the application of
the centrifugal force, the pump is submerged into water reservoir. The pumps have propellers to
force the water into the pumps.
The fountains spray the water through the nozzles. The water sprayed through the
fountain will fall back to the reservoir. The fountain also has the filters that removes the large
GEOTECHNOLOGY 7
solid particles from the water (Yang, et al 2019). The filter is fitted with a pump and designed by
civil engineers to take water to and from the pool.
The fountains have a store room called a plant room. Here, the plumbing controls, filters,
pumps and electrical switches are stored (Roberts 2014). The designs of the plant room are well
constructed to minimize the hazards. It is important to note the founts always use low voltage of
electricity.
Question 4
More water rainfall amounts add weight to the slope. Water is usually denser than air and
contributes to the weight of the soil (Hannigan, et al. 2017). The water will seep into the soil
layers and replaces the air in the pores. Therefore, more force will create stress leading to the soil
slope instability.
Generally, the ground water exists everywhere below the earth’s surface. The pore spaces
are filled with the water within the unsaturated zone above the water table. Therefore, this
changes the water table levels within the soil layers. For instance, the water table level will rise
during the wet seasons to infiltrate the soil and fall during the dry seasons. The shifting in the
water table level leads to the slope instability.
Water is a solvent substance in the soil. The water therefore dissolves the minerals of the
cement. The cements that are made of the gypsum or calcite have high solubility rate (Soltani
and Mirzababaei 2019). Therefore, the infiltration of water into the soil will form the solution of
the mineral’s components of the cement. Moreover, the soil particles will reduce the cohesive
forces that hold the mineral grains of the cement.
GEOTECHNOLOGY 8
The fluid pressure in the water affects the slope stability. The larger soil particles
normally disintegrate into small pieces when buried underground. The soil particles rearrange
themselves into a pact form blocking the water to occupy the pore spaces. The fluid pressure will
be increased to support the weight of the rock mass. Additionally, this will reduce the friction
and the shear strength leading to slope failure.
With increased earth shaking with heavy rainfall leads to the liquefaction. The sediments
of the soil will become loose between grain to grain and hence oversaturation. The result thereby
causes slope instability.
GEOTECHNOLOGY 9
References
Das, B.M., 2017. Shallow foundations: bearing capacity and settlement. CRC press.
Hannigan, P.J., Rausche, F., Likins, G.E., Robinson, B.R. and Becker, M.L., 2016.
Geotechnical Engineering Circular No. 12–Volume II Design and Construction of
Driven Pile Foundations (No. FHWA-NHI-16-010).
Indraratna, B., Nimbalkar, S. and Rujikiatkamjorn, C., 2016. A critical review of rail track
geotechnologies considering increased speeds and axle loads.
Roberts, A., 2014. Geotechnology: an introductory text for students and engineers.
Elsevier.
Soltani, A. and Mirzababaei, M., 2019. Discussion on “Effects of lime addition on
geotechnical properties of sedimentary soil in Curitiba, Brazil” [J Rock Mech
Geotech Eng 10 (2018) 188–194]. Journal of Rock Mechanics and Geotechnical
Engineering, 11(1), pp.214-218.
Yang, S., Leshchinsky, B., Cui, K., Zhang, F. and Gao, Y., 2019. Unified approach toward
evaluating bearing capacity of shallow foundations near slopes. Journal of
Geotechnical and Geoenvironmental Engineering, 145(12), p.04019110.
Geotechnology 1
GEOTECHNOLOGY
Student’s Name
Professor
Course
University
City
Date
Geotechnology 2
Geotechnology
Introduction
Foundation is a critical component of civil engineering design and requires critical
analysis. The nature of the foundation determines the durability of a structure. The design of any
civil engineering structure such as roads, bridges and buildings requires the knowledge of the
foundation system (shallow/deep foundation), loading conditions, and the knowledge of building
codes, the condition and the behavior of soil among other important properties that affect the
stability of a building (Al-Agha, 2015). The knowledge of these geotechnical properties is
guiding the design of the type of foundation and the safety measures to be undertaken.
Different soils possess different loading problems, for instance water table near the
surface or at a distance near the foundation affects the bearing capacity of the soil. Water
increases the hydrostatic pressure on the foundation wall (Das, 2006). Another factor that should
be taken into consideration is settlement. Structures must never be allowed to undergo excessive
settlement. When a structure undergoes excessive settlement, the integrity of the building
elements is affected hence the likelihood of failure (Kent & Becerik-Gerber, 2010). Therefore, an
engineer must assess the properties of the soil and understand the loading conditions and systems
of the proposed structure during the design phase.
Laboratory testing becomes an important aspect in geotechnical design. Some properties
of the soil require sample collection and analysis in the lab. Examples of such properties include,
the particle size distribution, utterbag limits, and the shear strength of soil among others
(Oberlender, 2000). These properties become important in the design of any structure. For
instance, in the design of roads, soils with high plasticity index and high moisture content should
be avoided. In some cases they are cut to spoil while if the areal extent is very large they are
Geotechnology 3
treated with a stabilizing material to improve their engineering properties such as strength
(Thomas, 2013). An example of such materials are lime and cement. In buildings, for
problematic soils such as wet clay, raft foundations are recommended or the use of ground
beams.
Question One
a) Initially, before construction:
Initial stresses at mid-depth of sand (z=3m):
Vertical total stress
?= 24 * 3 = 72 KN/m2
Pore pressure
µ= 9.81 * 3 =29.43 KN/m2
Vertical effective stress
?’ = 72 – 29.43 = 42.57 KN/m2
Initial stresses at mid-depth of clay (z= 10m):
Vertical total stress
Geotechnology 4
?= (24 * 6) + (30 * 4) = 264 KN/m2
Pore pressure
µ= 10 * 9.81 = 98.1 KN/m2
Vertical effective stress
?’ = 264 – 98.1 = 165.9 KN/m2
b) Immediately after the construction of the embankment
The construction of the embankment applies a surcharge:
q= 21 * 8 = 168 KN/m2
The clay is drained, either horizontally or vertically below, hence no increase of pore water
pressure. The sand, however, is undrained and the pore water pressure increases by 168 KN/m2
Initial stresses at mid-depth of sand (z=3m):
Vertical total stress
?= (24 * 3) + 168 = 240 KN/m2
Pore pressure
Geotechnology 5
µ= (9.81 * 3) + 168 =197.43 KN/m2
Vertical effective stress
?’ = 240 – 197.43 = 42.57 KN/m2
(no immediate increase)
Initial stresses at mid-depth of clay (z= 10m):
Vertical total stress
?= (24 * 6) + (30 * 4) + 168= 432 KN/m2
Pore pressure
µ= 10 * 9.81 = 98.1 KN/m2
Vertical effective stress
?’ = 432 – 98.1 = 333.9 KN/m2
(Immediate increase)
c) When ground water table lies at surface of ground level and below the foundation and discuss
the change in the stress values if any.
Geotechnology 6
Stresses at mid-depth of sand (z=3m):
Vertical total stress
?= 24 * 3 = 72 KN/m2
Pore pressure
µ= 9.81 * 3 =29.43 KN/m2
Vertical effective stress
?’ = 72 – 29.43 = 42.57 KN/m2
Stresses at mid-depth of clay (z= 10m):
Vertical total stress
?= (24 * 6) + (30 * 4) = 264 KN/m2
Pore pressure
µ= 10 * 9.81 = 98.1 KN/m2
Vertical effective stress
?’ = 264 – 98.1 = 165.9 KN/m2
The stresses remain the same since the clay is still saturated.
Geotechnology 7
Question Two
a)
?d= 1817 * 9.81/ 1000 = 17.82 KN/m2
By Tezarghi:
qu= 1.3cNc + qNq + 0.4B?N?
qu= qNq + 0.4B?N? (c=0)
qall= qu * FS
qu= Qu/A= (Qall * FS)/A
qu= (2 * 480)/ B2
qu= 960/B2
q=17.82 * 1 = 17.82 KN/m2
From Fig 1, for ? = 380, by interpolation:
Nq= 65.34
Geotechnology 8
N?= 76.96
qu= qNq + 0.4B?N? = (17.82 * 65.34) + (0.4 * B * 17.82 * 76.96)
960/B2 = (17.82 * 65.34) + (0.4 * B * 17.82 * 76.96)
Equation becomes: 548.57088B3 + 1164.3588B2 – 960 = 0
Solving for B yields 0.7769 m
b)
i)
Immediately after construction
For saturated clay soils, initial conditions present undrained conditions for which ? = 00.
From Figure 1,
Nc= 5.14
Nq= 1
N?= 0
qu= 1.3cNc + qNq
qu= 1.3cNc + qNq
Geotechnology 9
Cu= 24 KN/m2
?’= ?sat – ?w = 28 – 9.81 = 18.19 KN/m3
q= Df (?’) = 1.8 (18.19) = 32.742 KN/m2
B= 3m
qu= (1.3 * 24 * 5.14) + (32.742 * 1) = 193.11 KN/m2
ii)
Some years after construction
qu= 1.3cNc + qNq + 0.4B?N?
c=0
qu= qNq + 0.4B?N?
?’= ?sat – ?w = 28 – 9.81 = 18.19 KN/m3
q= Df (?sat – ?w) = 1.8 (18.19) = 32.742 KN/m2
B= 3m
From figure 1, for ? = 250,
Nq= 12.7
N?= 9.7
qu= (32.742 * 18.58) + (0.4 * 3 * 18.19 * 15.7) = 951.0460 KN/m2
Geotechnology 10
Question Three
a) Lateral Earth force at rest and the position of the resultant force
Assume OCR= 1
Coefficient of pressure at rest, Ko is given by;
Ko = 1-sin ?’
K1(sand) = 1 – sin 28 = 0.5305 (4 significant figures)
K2 (clay) = 1 – sin 37 = 0.3982 (4 significant figures)
Geotechnology 11
At z= 0m , ?’o = 0; ?’h = 0
At z = 3m, ?’o = ?’. D = 19 x 3 = 57 KN/m2
?’h = Ko x ?’o = 57 x 0.5305 = 30.2385 KN/m2
Say 30.24 KN/m2
This pressure will be transferred as surcharge into the clay layer (illustrated in the diagram
below)
The pressure is equivalent to
?’h = 57 KN/m2 x 0.3982 = 22.6974 KN/m2
Say 22.70 KN/m2
At z = 5m
There are three components of ??h
?’h1 = 22.70 KN/m2 (Transferred as surcharge)
?’h2 = Pressure due to water= ?w x h= 9.81 x 2 = 19.62 KN/m2
?’h3 = effective weight of clay
?’o = (?sat – ?w) x h
?’o = (19 – 9.81) x 2 = 18.38 KN/m2
?’h3 = 18.38 x 0.3982= 7.3189 KN/m2
Say 7.32 KN/m2
Horizontal pressure diagram on the wall
Geotechnology 12
The total force per unit length of the wall can be determined from the area of the pressure
diagram.
The total force Po is the sum of all the forces of the wall
Po = Area 1+ Area 2 + Area 3 + Area 4
Where Area 1 = P1 ; Area 2 = P2 ; Area 3 = P3 ; Area 4 = P4 and P(s) denote the horizontal forces
on the wall.
P1 = ½ bh = ½ x 30.24 x 3 = 45.36 KN
Geotechnology 13
P2 = bh = 22.70 x 2 = 45.4 KN
P3 = ½ bh = ½ x 7.32 x 2 = 7.32 KN
P4 = ½ bh = ½ x 19.62 x 2 = 19.62 KN
Po = 45.36 + 45.4 + 7.32 + 19.62 = 117.7 KN per meter length of the wall.
The Horizontal force diagram on the wall
Location of the resultant force from the bottom of the wall
? = {P1 (2+3/3) + P2 (2/2) + P3 (2/3) + P4 (2/3)}/Po
? = {45.36(3) + 45.4 (1) + 7.32 (2/3) + 19.62 (2/3)}/ 117.7
?= 1.6945m
Say 1.69m from the bottom of the wall.
Geotechnology 14
b) What is foundation settlement and explain how it affects civil engineering construction.
Foundation settlement is the total vertical displacement of the foundation of a structure
due to the deformation of the soil as a result to the loading of the structure. Structures in different
soils must be designed for a particular settlement. Foundation settlement affects the stability of a
structure and hence foundations can never endure excessive settlement.
The term excessive settlement varies since the degree depends on the condition of the soil
and the type of structure. A plot of the Load per unit areas against the settlement of a foundation
footing gives a specific load at which the foundation fails. This load is known as Qu which is the
ultimate load for the settlement of the foundation. Beyond the load, the foundation will fail.
therefore, all civil engineering constructions must not exceed this limit.
Settlement on a building can be categorized into two which include elastic settlement and
consolidation settlement. Elastic settlement (Se) occurs immediately and is caused by the direct
load of the structure. There are no changes in the water moisture. The amount of the elastic
settlement is a function of the flexibility of the ground whether it is rigid or flexible, and the type
of material that the foundation is resisting. Consolidation involves the change in volume of the
soil because it involves expulsion of moisture and air voids over time. Foundation are designed
considering these types of settlement.
Question Four
Discuss practical applications on how does water in the slope induces instability in
constructions. Permanency of slopes is one of the primary aspects in civil engineering. The
loading instability of slopes is caused by increase in stress and reduction in strength. One of the
factors that causes an increase in stress is increased unit weight of soil by wetting which is a
Geotechnology 15
direct effect of water. One practical example is seepage forces in a granular slope subjected to
rapid draw down. When the water of a submerged slope drains rapidly as compared to the water
level on the slope lags behind, seepage forces are induced leading to stresses in the soil.
Seepage due to rapid down-draw.
In the figure, is a situation where the water level in the river drops suddenly due to tidal
variations. As a result, the water level in the slope lags behind leading to seepage and hence
generation of stresses.
Drainage of Water on Slopes for Road Construction
Water problems especially seepage reduces the durability of roads. In construction of
roads problems of water in slopes is reduced by using discharge channels where water and
discharged down the side of slopes. Embarkment toe ditches can also be used to remove water
from the vicinity of the road embarkment to avoid erosion of the road fills. High water levels in
slopes soaks the subgrade and the improved subgrade layers leading to road failures.
Weight Increase
Geotechnology 16
Addition of water to a slope increases the weight of the slope thus increasing the stress.
This consequently leads to slope instability.
Water in Rock Fractures
Water occupying fractures in medium to hard rock causes a reduction in slope stability.
This is because water pressure acting in a discontinuity within the rock mass reduces the
effective normal stress which causes a reduction in the shear strength along the plane of the
discontinuity. Application of a load on this slope leads to increase in pore pressures. If the load
exceeds the reduced shear strength of the soil, the slope fails. These rock fractures may occur
from blasting during construction.
Bedding Planes
Bedding Planes along a Road
Consider the slope above, the bedding plane to the left is slanted. If water is to enter
along the bedding planes, cohesion is reduced and sliding occurs.
Geotechnology 17
Conclusion
I have achieved learning outcome 1 and 2. I was able to assess the problems of different
soil layers and loading conditions both on foundation and walls. I demonstrated an understanding
of settlement including elastic settlement and consolidation showing their importance to civil
engineering design. I used a problem to demonstrate an understanding of forces on a wall at rest
condition. Further I analyzed a problem on shallow foundations and assessed practical examples
on slope stability problems.
I have found out that the loading conditions and the conditions of the soil (dry or
saturated) influences the analysis adopted for it. The type of soil is a determinant to the stability
of a foundation. Also, water is a major problem to foundations and slope stability and should be
controlled. Through this assignment, I have learn…
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