ca write up

8/3/2019 CA write up

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Continuous AssessmentEG-321 GeomechanicsElliot Lawrence Lowe : 521706


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Elliot Lawrence Lowe EG-321Geomechanics 24/11/2010

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Background knowledgeRetaining structures

There is a wide variety of structures used to retain soil for both temporary works and permanent

structures. Some of these retaining structures we see every day, but may be unaware of their existence

or there important to structural design and safety in everyday life.

Mass concrete or masonry walls rely largely on their weight for stability against overturning andsliding much like a mass dam. They are unreinforced so their height must be limited to ensure internalstability of the wall in bending and shear when subjected to the lateral stresses. They are typically no

more than about 3 m high.

Reinforced concrete walls are more economical with the reinforcement enabling the stem and basesections to be designed as cantilevered structural elements. Overall stability is provided by an adequatebase width and the weight of backfill resting on the base slab behind the stem.

Embedded wallsExcavations are supported by reinforced concert diaphragm walls, contiguous bored pile walls or secant

bored pile walls. These walls are made to aid

construction typically and are not usually used

as a permanent solution to a retaining problem.These types of retaining walls are constructed

around the basement perimeter before excavationcommences, occupying minimal space but support

to the soil and groundwater both in the temporary

condition during excavation and construction ofthe basement and in the permanent condition asthe final structural basement wall. They may also provide support to vertical loads such as the external

columns and walls of a building.

Victoria Square Shopping Center

An example of retaining walls being used to allow construction would be the Victoria squareshopping center (2008) in central Belfast. This site is a basement excavation where 63 000 m2 of retail

floor space is spread over three floors which are below ground level. Huge ground anchors are set at55 from the vertical as the site is tightly set in a heavily developed city center (2). The retaining walls

are sheet pile design these are especially designed to high standards due to their loads and are made

out of high modulus steel. The large tree level piled basement has a footprint of approximately 21 500m2 this involved 24 000 lorry loads of excavation 4000 tons of steel sheet pile and 550 aroundanchors.

Figure (2) shows a side on view of how the retaining walls are used to allow the huge excavation.Before any excavation was carried out the dewatering wells were constructed to reduce the water table

Figure (2)

Figure (1)


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so the excavation would

not reach the water table.This excavation was done

horizontally to remove

any chance of a land slip

or slide as most of theexcavation was soft

sediment and sleech.Figure (3) shows theHigh-modulus sheet steel

piles that were used to

construct the perimeterwall, stiffened bywelding beams to the

face of the sheet piles

prior to installation. Themain function of the

beams was to stiffen the

piles and reduce ground

movements anddeflections. The sheet piles were installed using a vibrating hammer. Here you can see the angle in

which they are being applied due to the close proximity of a well built up area of a city Centre. Hereyou see the importance of the retaining wall and the vast amount of time effort any money being spent

before even the foundations and the structure its self can be erected and build.

Lateral Earth PressuresIn order to design embedded walls the assessment of earth pressures is needed so the intended

structure can withstand these loads. The assessment of earth pressures is based generally on the

effective stress analysis, effective stress on an embed wall is caused by the horizontal and verticalpressures of the retained soil. So the lateral earth pressuresare those imparted by soils onto vertical or

near-vertical structures such as the embedded wall or retaining wall. They may include both normal andshear pressures, these pressures are especially important in the design of retaining walls, retaining

walls as shown above are common civil engineering projects.

Embedded Wall Question

Figure (3) is a geometry diagram of the question the embedded wall is a simple retained wall with aground anchor located close to ground level. The labeled parameters are known apart from the passivedepth d this is unknown and will be calculated at a later question. From simple analysis of Figure (4)

it is clear that the wall will experience both the active and passive condition.

Figure (4)

Figure (3)


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Eq set (1)

Single anchor embedded wall

From research the problem given is known as a single anchor or propped embedded wall, its design issaid to be capable in many different ways but I will be following the gross pressure methods and the

method laid out in the lecture notes for Lateral Earth Pressures.

The overall stability of the wall is considered by taking moment about the Anchor level in this case

level Ho, of the active and passive thrusts Paand Ppassuming a free-earth support condition. In thiscondition the depth of embedment is not sufficient to prevent rotation at the toe of the wall, as in the

fixed-earth condition. However, rotation of the toe is necessary for the consideration of overallstability in terms of moments about the Anchor level. Design methods for this problem are given inEC7 Euro Codes 7.

Distribution of Effective StressesAs mentioned before the wall is subject to active and passive stresses so both of these will need to becalculated in order to plot a distribution of the two. Using the lecture notes on lateral earth pressures the

coefficient of lateral earth pressure is deduced with this the horizontal and vertical effective stresses can

be calculated for both active and passive sides of the wall.

Gross pressure methodIn the question the only perimeter that is unknown is the depth of the embedment d this is obtained by

equating moments about the point O of the full value of the active thrust pa but balanced by a reduced

value ofpp. Moments are taken about the point O near the toe of the wall for the cantilever mode andabout the level of the anchor for the propped wall.

Then increased by 20% to give the full depth

of embedment and to ensure that the passiveresistance below O represented by the

reaction R will be obtained this increase is

not an additional factor of safety therecommended vale of safety is given in the

question as m = 1.5 this figure is used to

divied by in respect to every soil property,Characteristic soil properties / m = Designsoil properties . Figure (4) shows the

effective active and passive vertical stresses

as you can see the active stress is on theright hand side this is because the active

stress is due to the movement of the wallaway from the embedment. The passive sidethe left hand side only comes into account

due to the depth d this passive distribution of

stress aids in reducing the active amount of stress and acts like a canter lever the anchor acts to stop

the canter lever situation. The active and passive thrust are located 2/3 of the soil height this is due tothe location of the center of force due to the force shape being triangular.

Figure (5)


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Eq set (2)

Penetration depth calculation

The following equations are used in reference to figure (5) to find the optimum depth of embedment dand the resultant anchor force R once d is found.

Using the following equations the residual moment and anchored force can be calculated to gain anunderstanding of where the real depth of the embedded wall lies. This is done as the equation for thedepth o the embedded wall becomes a nonlinear algebraic cubic equation this cannot be solved in anynormal numerical manner this can be solved with the Bisection method the Scent method or the

Newton-Rapson method.

The Newton-Rapshon Method

Is said to be one of the most widely used root-allocating formulas it works by using an initial guessthen a tangent is extended to the second point where this point crosses the x axis usually represents an

improved estimate of the root. Although the Newton-Raphson method is often very efficient, there aresituations whet performs poorly. Multiple roots can cause problems with this method However, even

when dealing with simple roots, difficulties can also arise.

The Newton-Raphson formula: The Secant Method

One of the problems also with the Newton-Raphson method is evaluating the derivative, this can be

ok for many functions but some of the derivatives can be extremely difficult to evaluate. Instead ofthis method the backward finite difference method can be used the Sceant method.

The Secant Formula: The Bisection Method

Is a increment search method in which the interval is always divided into half this is due to thefunction if a interval changes sign then the root must lie between these two points. The midpoint isevaluated this seems very crude but if there are

enough intervals then the error becomes very small

to improve this further the process is repeated tofind a refined estimate. Figure (6) graphical

depiction of the Bisection method

Figure (6)


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MatLab Script

function Geomechanics_CA_521706

clc;close all

%%%assign values to a function% initial conditions%the display and input function is used to enter the parameters of the%question or alternatively the parameters are already assigned below forthis type of question

disp('Input the Parameters to the problem')Ho=input('Depth to Anchor (Ho) = ');H=input('Height (H) = ');phi=input('Value of Angle of Internal friction (phi) = ');Gamma_b=input('The Bulk Density Value = ');d=input('Guess the Depth of Embedment = ');

% % initial conditions% H=5; % tie rod depth (m)% Ho=1; % (m)% phi=20; %KN/m% Gamma_b=20; % Angle of internal shearing (degrees)% d=5; %passive depth (m) unknown at this point

%%%%%%%%%%%%%%%%%%% Q 1 %%%%%%%%%%%%%%%%%%%display('-------------------- Q 1 --------------------')%%coefficient of lateral earth pressure K is obtained for active andpassive%side of the wall.display(' Lateral Earth Pressure K')ka=(1-sin((2*pi/360)*phi))/(1+sin((2*pi/360)*phi)),display('ka < 1')%ka =1

k_check=(ka*kp);for k_check=1display('ka and kp check ok')end%total depth of the embedded wallz=-(H+d):0.01:0;%matrix assignment for the effective vertical and horizontal stress.EVa=zeros(size(z))';EVp=zeros(size(z))';EHa=zeros(size(z))';EHp=zeros(size(z))';%for the active side of the embedded wall the depth of active vertical%stress is from the datum to the depth h+d the depth are below ground level%so are assigned a negative value.i=0;for z=-(H+d):0.01:0

i=i+1;EVa(i)=Gamma_b*-z;EHa=EVa*ka;%for the passive side of the embedded wall only the depth d is required so%once the depth has reached a value of h the wall is now at the d sectionofthe wall so the passive vertical stress is calculated.if -z>=HEVp(i)=Gamma_b*(z+H);EHp=EVp*kp;endend


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%now the values of effective stress have been calculated for a guessedvalue of d% the graphical representation will require the displayed position of the% ground level and the passive depth dz=-(H+d):0.01:0;ground_level=zeros(size(z));passive_depth=ones(size(z))*(-H);

%plotting the effective vertical stress for both the active and passivefigure(1)plot(EVa,z,'r');hold onplot(EVp,z,'b');hold on%plotting the straight lines for the ground level and the passive deth dplot(EVa,ground_level,'k','LineWidth',2)hold onplot(EVp,passive_depth,'k','LineWidth',2)grid onhold offxlabel('Effective Stress KPa');ylabel('Depth (m)');title('Effective Vertical Stress Distribution');legend('Active Side','Passive Side','Ground Level','Passive Depth (d)');axis([(min(EVp)-5) (max(EVa)+5) (min(z)-1) max(z)+1]);

% plot effective horizontal stress passive and activefigure(2)plot(EHa,z,'r');hold onplot(EHp,z,'b');hold onplot(EHa,ground_level,'k','LineWidth',2);hold onplot(EHp,passive_depth,'k','LineWidth',2);grid onhold offxlabel('Effective Stress KPa');ylabel('Depth (m)');

title('Effective Horizontal Stress Distribution');legend('Active Side','Passive Side','Ground Level','Passive depth (d)');axis([(min(EHp)-5) (max(EHa)+5) (min(z)-1) max(z)+1]);

%%%%%%%%%%%%%%%%%%% Q 2 %%%%%%%%%%%%%%%%%%%display('-------------------- Q 2 --------------------')%to find the value of d for which the equilibrium of moments satisfies rand%values of will be entered in the formulas for force with respect to d and%the anchored force R where these two lines cross will be the values of d%that satisfies the two equationsj=0; % initial iterationfor D=0:0.1:5j=j+1;Zp(j)=H-Ho+((2*D)/3);

Pp(j)=((kp*Gamma_b)/2)*D^2;Za(j)=(2*(H+D))/3 - Ho;Pa(j)=((ka*Gamma_b)/2)*(H+D)^2;F(j)=Pa(j)*Za(j)-Pp(j)*Zp(j);R(j)=Pa(j)-Pp(j);end%plot residual over embedmentD=0:0.1:5;figure(4)plot(D,F,'bs-')xlabel('Depth of Embedment (m)');ylabel('Residual(KNm/m)');


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Real_depth_d=(root);display('The Embedded Depth d')display(root)

%plot log residual over iteration graphfigure(6);plot([0:iter],log(residual_vec(1:iter+1)),'bo-');

grid on;title('convergence');xlabel('iteration');ylabel('Log(residual)');

%%%%%%%%%%%%%%%%%%% Q 4 %%%%%%%%%%%%%%%%%%%%here equation set (2) is called again to find the real anchor force as therela depth of the embedment has been calculated.display('-------------------- Q 4 --------------------')Pa=((ka*Gamma_b)/2)*(H+Real_depth_d)^2;

Pp=((kp*Gamma_b)/2)*(Real_depth_d^2);

REAL_R=Pa-Pp;

display('The Anchored force'),display(REAL_R)

%%%%%%%%%%%%%%%%%%% Q 5 %%%%%%%%%%%%%%%%%%%

display('-------------------- Q 5 --------------------')

% using method of section the embedded wall is broken into three sectionsas described in equation set (3)% the first cut is from the ground level to the anchor depth%the second is from the anchored depth to depth H%the third is from the depth H to the embedded depth

%%%%%%%%%%%%%%%% section 1 %%%%%%%%%%%%%%%%

V in order to do this the data is feed into a for loop for the give sectionof distance z each section is give a specific depth for instance depthsection one is called z1 and so on.%firstly the matrixes are calculated so the data can be filled within thesematrices so later that can be called again to a plot function%%set depth for function

z1=0:0.1:Ho;n=0;% matrix creation for both BM and ShearV1=zeros(size(z1));BM1=zeros(size(z1));

% here is the for loop needed to calculate the BM and Shear for section onefor z1=0:0.1:Ho;n=n+1;V1(n)=-((ka*Gamma_b*(z1^2))/2);BM1(n)=-((ka*Gamma_b*(z1^3))/6);end% the simple for loop enters all the possible increments from 0 to 1 meterand calculates all the values for BM and Shear%here the steps are completely the same as section one the only differenceare the names for depth 1 to 3 and the BM and Shear for 1 to 3.


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%%%%%%%%%%%%%%%% section 2 %%%%%%%%%%%%%%%%z2=Ho:0.1:H;n=0;V2=zeros(size(z2));BM2=zeros(size(z2));for z2=Ho:0.1:H;n=n+1;

V2(n)=REAL_R-((ka*Gamma_b*(z2^2))/2);BM2(n)=-((ka*Gamma_b*(z2^3))/6)+(REAL_R*(z2-Ho));end

%%%%%%%%%%%%%%%% section 3 %%%%%%%%%%%%%%%%z3=H:0.1:(H+Real_depth_d);n=0;V3=zeros(size(z3));BM3=zeros(size(z3));for z3=H:0.1:(H+Real_depth_d);n=n+1;V3(n)=REAL_R-((ka*Gamma_b*(z3^2))/2)+((kp*Gamma_b*((z3-H)^2))/2);BM3(n)=-((ka*Gamma_b*(z3^3))/6)+(REAL_R*(z3-Ho))+((kp*Gamma_b*((z3-H)^3))/6);end

%%%%%%%%%%%%%%%%%%% Graphical plot %%%%%%%%%%%%%%%%%%%% here the data is plotted for simplicity the depth of the embedment andthe anchor are plotted as black lines to show the different sections and toassess their behaviour after the three significant points

z1=0:0.1:Ho;z2=Ho:0.1:H;z3=H:0.1:(H+Real_depth_d);VL=-80:10:100;figure(7)plot(V1,(-z1),'Rs-'),hold on,plot(V2,(-z2),'Ro-'),hold on,plot(V3,(-z3),'Ro-')hold on,plot(VL,-1,'kx','LineWidth',2),hold on,plot(VL,-5,'kx','LineWidth',2)hold off,title('Shear Force Distribution'); grid onxlabel('Shear Force (KN/m)');ylabel('Depth (m)');

axis ([-80 100 -9 0])VL=-50:10:200;figure(8)plot(BM1,(-z1),'bo-'),hold on,plot(BM2,(-z2),'bo-'),hold on,plot(BM3,(-z3),'bo-')hold on,plot(VL,-1,'kx','LineWidth',2),hold on,plot(VL,-5,'kx','LineWidth',2)hold off,title('Bending Moment Distribution'); grid onxlabel('Bending Moment (KN.m/m)');ylabel('Depth (m)');axis ([-20 200 -9 0])display(' see Graphical Data ')

%%%%%%%%%%%%%%%%%%% Q 6 %%%%%%%%%%%%%%%%%%%display('-------------------- Q 6 --------------------')display(' see Graphical Data ')

% matrix allocation for all three sections for the wall%the leght of the wall as it moves through the earthDis_z=[z1 z2 z3]';%the recalling of the previous bending momentDis_BM=[BM1 BM2 BM3]';int_BM=zeros(size(Dis_z,1),1);EI=1e4;for i=size(Dis_z,1)-1:-1:1


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int_BM(i)=int_BM(i+1)+(Dis_z(i)-Dis_z(i+1))/2/EI*(Dis_BM(i)+Dis_BM(i+1));endDis_p=zeros(size(Dis_z,1),1);

for i=size(Dis_z,1)-1:-1:1Dis_p(i)=Dis_p(i+1)+(Dis_z(i)-

Dis_z(i+1))/2*(int_BM(i)+int_BM(i+1));

end

VL=-0.37:0.01:0.05;figure(9);plot(-Dis_p,-Dis_z,'rs');hold onplot(VL,-1,'kx','LineWidth',2),hold on,plot(VL,-5,'kx','LineWidth',2)hold off,title('Embedded Wall Deformation'); grid onxlabel('Displacement (m)');ylabel('Depth (m)');axis([-0.37 0.05 -9 0])

%%%%%%%%%%%%%%%%%%% Q 7 %%%%%%%%%%%%%%%%%%%display('-------------------- Q 7 --------------------')display(' see Graphical Data ')display('---------------------------------------------')

%here the question is asking for the data ran in q3 and q4 to be recalledand ran for a varying anchor depth. The changing variable here will becalculated depth of the embedment calculated using the help file function.In order to do this the code written for q3 and q4 is recalled and placedwithin a for loop.%initial the data was not assigned to a matrix and the graph was plottedvery slowly then using increments and vectors the graph was plotted at afaster pace.

%setting out the length of wall interested in placing the anchor I havechosen 5 and I have also plotted a end magnitude of 3 to see how the anchorforce changes past the point calculated as the real depth d at around 3.65this figure was taken from the third plotted graph in question 2

Ho=0:0.1:5;%matric allocationD=zeros(size(Ho));R=zeros(size(Ho));i=0;

%start of the for lop where q3 and q4 have been placed so the data can beobtainedfor Ho=0:0.1:5

ka=(1-sin((2*pi/360)*phi))/(1+sin((2*pi/360)*phi));kp=(1+sin((2*pi/360)*phi))/(1-sin((2*pi/360)*phi));

%%%%%%%%%%%%%%%%%%% Q 7(3) %%%%%%%%%%%%%%%%%%%% here q3 and q4 have been copied in to the loop for simplicity

i=i+1;tolerance=1.0e-5;maxiter=100;residual_vec=zeros(maxiter);[root]=root_finder(tolerance,maxiter);%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%D(i)=(root);


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%%%%%%%%%%%%%%%%%%% Q 7(4) %%%%%%%%%%%%%%%%%%%Pa(i)=((ka*Gamma_b)/2)*(H+D(i))^2;Pp(i)=((kp*Gamma_b)/2)*(D(i)^2);

R(i)=Pa(i)-Pp(i);

end

%graphical representation of the data calculated within the for loopHo=0:0.1:5;figure(10)plot(Ho,R,'rs-')title('Anchor Force with respect to varying Anchor depth'); grid onxlabel('Depth of Anchor (m)');ylabel('Anchor Force (KN/m)');

figure(11)plot(Ho,D,'bo-')title('Embedment depth with respect to varying Anchor depth'); grid onxlabel('Depth of Anchor (m)');ylabel('Embedment Depth (m)');

%sub functions taken from the help functions%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%functions taken from help file%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%function [D]=residual(x)

%here the equation is changed to the appropriate one here the equation is%changed to the equilibrium of moments w.r.t O: PaZa-PpaZp=0D=(((ka*Gamma_b)/2)*(H+x)^2)*((2*(H+x))/3 - Ho)-(((kp*Gamma_b)/2)*x^2)*(H-Ho+((2*x)/3));end

function [Dy]=stiffness(x)%here the equation is changed to derivative of the original function this%calculation is performed in a separated function using the diff functionDy=(Gamma_b*ka*(H+x)^2)/3 - (Gamma_b*kp*x^2)/3 + (Gamma_b*ka*(2*H +2*x)*((2*H)/3 - Ho + (2*x)/3))/2 - Gamma_b*kp*x*(H - Ho + (2*x)/3);endfunction [root]=root_finder(tolerance,maxiter)residual_vec=zeros(maxiter);%first guessiter=0;root=4;res=residual(root);residual_vec(iter+1)=abs(res);%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% fprintf('Iteration %g, Root %g, Residual %g\n',iter,root,res);%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%K=stiffness(root);while abs(res)>tolerance && iter


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Output Data

Numerical Data

Data displayed within the command window

-------------------- Q 1 --------------------

Lateral Earth Pressure K

ka =

0.4903

ka < 1

kp =

2.0396

kp > 1

ka and kp check ok-------------------- Q 2 --------------------

See Graphical Data

-------------------- Q 3 --------------------

The Embedded Depth (d)

root =

3.6511

-------------------- Q 4 --------------------

The Anchored force

REAL_R =

95.0514

-------------------- Q 5 --------------------See Graphical Data

-------------------- Q 6 --------------------

See Graphical Data

-------------------- Q 7 --------------------See Graphical Data

---------------------------------------------

Analysis of data

Q1: the values of Ka and Kp are sufficient that they meet there domains and pass the check this provesthat the MatLab code is correct.

Q3: the depth presented is close to that read of the graph in question 2 the closets zoom in on figure

(3) was to a value of 3.65. this agrees with the rough calculations from question 2

Q4: from the close up of Figure (3) you cal scroll up the image to read off a value of the anchor at thispoint of 3.65 m to that of a anchor force of about 95KN/m. this shows that the numerical data is

sufficient and acceptable and I am happy with it.


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Figure (1)

Figure 1 shows the effective vertical stress in the passive and active side. The expected graph would

shoe the passive effective stress growing linear with depth as it is a straight multiplication formula

this is shown in the graph. The passive side will not produce a figure until the depth of the y axis hasreached that of the initial point of d 5 meters here the value of d is guessed and to be cautions I havechosen that value to be the same as the height of the wall 5m. This is shown in the graph. The graph

plots the expected data and if the dimensions of the embedded wall were to change the lines shownwould only grow or shrink depending on the change in the wall.

Figure (2)

Figure 2 shows the horizontal stress which is a function of the vertical stress so the graphs are related

notice the passive horizontal stress is now greater than the active the figures are similar and opposite.


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Figure (3)

This graph is used to show the relation between the anchor force and the residual f of the given depthof the embedment the data interesting me is the data on the right hand side of the center line this

represent the a true physical solution to the embedded depth.

This graph is used to read off a rough value for the depth of the embedment this is done by reading thedepth at the point in which the residual is equal to zero reading the graph a rough value is 3.65 meters


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Figure (8)

Figure 8 shows as predicted from the previous graph the maximum bending moment occurs at a depth

of 4.5 meters this figure is 185 KN/m all these figures are reasonable and most commercial steel that

is readily available will with stand these forces sufficiently so I believe the wall design is possible

Figure (9)

Figure 9 shows the deflection curve of the beam along its depth this is calculated using three sections

and three sets of equations. The plotted curve shape is acceptable and I predicted this would be its

shape deflection steadily growing towards the ground level. This is due to the type of embedded wall

the single anchor is simply a canter lever with a tie rod anchor around the top to stop seriousdeflection and a canter lever will deflect more the further you move away from the fixed end in this


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case the fixed end is the embedded section as the passive and active thrust oppose each other fixing

the wall into place.

Figure (10)

This graph simply shows the force needed from the anchor to oppose the moments will become graterthe deeper the anchor is set this shows how more force is placed on the anchor as the anchor is placedat a deeper depth due to the moment length of the anchor being reduced the closer I moves towards

the origin this is logical as the factor the force is being times my in the equilibrium equation is being

reduced so the force will need to be increases as to oppose the opposite forces and maintain

equilibrium.

Figure (11a)


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Figure (11b)

I have chosen to produce two plots of Figure 11 to show how the behavior of the embedded depthchanges with respect to the placement of the anchor. As you can see as the anchor is placed past 3.6

meters the embedded depth becomes a negative value this means the depth of the embedment is

removed and I place above that of the anchor to oppose the force being produced by the anchor at thispoint the embedment is not working in its proper manor and the design of the wall completely

changes. The further down the anchor move the less depth of embedment is needed as the anchor it

taking the force that the embedment is used to oppose the anchor is taking the embedments job. Thisis not the function of the anchor the anchor is used to reduce any significant deflection in he wall as

the design can be thought as a large cantilever.

References

For the theoretical understanding of the question and how to answer it appropriately I have used

printed text other then my own work or class notes.

1) Geotehnical Engineering Principles and Practice: Donald P. Coduto: pages 581-600: Used inBackground knowledge.

2) Soil Mechanics Principles and Practice: G E Barnes: Pages 348-535: Used in Backgroundknowledge.

3)

ICE Civil Engineering: November 2009: Volume 162: issue CE4: Paper 08-00050: pages162-170: Used in Background knowledge.

4) EG-321 Geomechanics lateral Earth Pressure: Dr. Antonio J. Gil:5) Numerical Methods For Engineers: Chapra Canale: pages 116-150: used in Q3 and Back

ground knowledge.

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