American Academy of English Geotechnology Construction Homework for assignment 1 :

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Geotechnology

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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 stresss 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 Coulombs 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 earths 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 minerals 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. 12Volume 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) 188194]. 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

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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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