Foundation Engineering_Earth Pressure

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      010Foundation Engineering(3-1-0-4)

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    Course instructor 

    Dr. Trudeep N. DaveInstitute of Infrastructure Technology Research and Management

    E-mail: [email protected]

    Class timings:

    Monday: 11:00 to 12:00Tuesday: 10.00 to 11.00

    Thursday: 11.00 to 12.00

    Retaining wall for embankment

    Box culvert for vehicular traffic

    Retaining wall with Box culvert

    Basement wall

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    Bridge AbutmentPedestrian Underpass

    At-rest / Active / Passive Earth Pressures

    Granular Soils

    smooth wall

    Wall moves away

    from soil

    Wall moves

    towards soil A 

    B

    Earth pressure at rest

    zσv

    σh = K o σv

     A

    B

    If wall  AB remains sta tic – soi l

    mass will be in a state of elas tic

    equil ibrium – horizontal strain is

    zero.

    Ratio of horizontal stress to vertical 

    stress is called coefficient of earth

    pressure at rest, K o, or 

    v

    ho

      K  

     z  K  K  ovoh  

    LATERAL EARTH

    PRESSURE

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    Soil with Cohesion and Friction

    Mohr’s Circle of

    Stress

    c

    c

    f

    Soil fails when

    Mohr’s circle

    touches theselines

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    σX = Ko σz

    σz

    ACTIVE EARTH PRESSURE (RANKINE’S)

    (in simple stress field for c=0 soil)

    Active Earth Pressure

    - in granular soils

    v’ 

    decreasing h’ 

    Initially (K0 state)

    Failure (Active

    state)

     As the wall moves away from the soil,

    active earth

    pressure= pa

    σzKo σzσx’Aø

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    PASSIVE EARTH PRESSURE (RANKINE’S)

    (in simple stress field for c=0 soil)

    σX = Ko σz

    σz

    Passive Earth Pressure- ingranular soils

    v’ 

    Initially (K0 state)

    Failure (Active

    state)

     As the wall moves towards the soil,

    increasing h’ 

    passive

    earthpressure

    = pa

       S   h  e  a  r  s   t  r  e  s  s

    Normal

    stress

    f tanc f   

    C

    D

    D’

    O A σpKoσv

    b

    a

    σv

    f

    c

    Mohr’s circlerepresenting

    Rankine’spassive state.

    Passive Earth Pressure 

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    Based on the diagram :

     pressureearthactives Rankine' of t coefficien  Ratiov

    a

     

    a K  (K a is the ratio of the effective stresses)

    Therefore :

    f

    f

      sin1

     sin-1  )

    2 (45 -tan K  2

    v

    aa

    It can be shown that :

    aa

    2

    a

     K 2c- K  z

     )2

     (45 -tan2c- )2

     (45 -tan zf

     

    Active Earth Pressure 

    LATERAL EARTH

    PRESSURE

    aa K 2c- K  z

    z o

    a K 2c-

     Active pressure distribution

    Active Earth Pressure 

    a K 2c-

     K za

    LATERAL EARTH

    PRESSURE

     Active pressure distribution

    Active Earth Pressure 

    Based on the previous slide, using

    similar triangles show that :

    a

    o K 

    c z g 

    2  where z o is depth of tension

    crack

    For pure cohesive soil, i.e. when f = 0 :

    c z 

    o

    LATERAL EARTH

    PRESSURE

    For cohesionless soil,

    c = 0

    aava  K  z K   

     Active pressure distribution

    Active Earth Pressure 

     K za

    LATERAL EARTH

    PRESSURE

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    Earth pressure at rest may be obtained theoretically from thetheory of elasticity applied to an element of soil,

    remembering that the lateral strain of the element is zero.

    Various researchers proposed empirical relationships for K 0,

    EARTH PRESSURE THEORIES

    The magnitude of the lateral earth pressure is evaluated

    by the application of one or the other of the so-called ‘lateralearth pressure theories’ or simply ‘earth pressure theories’.

    Theories given by Coulomb (1776) and Rankine (1857) stood the

    test of time and are usually re fer red to as the “Classical earth

    pressure theories”.

    These theories have been developed originally to apply to

    cohesionless soil backfill, since this situation is considered to be

    more frequent in practice and

    since the designer will be on the safe side by neglecting cohesion.

    RANKINE’S THEORY

    Rankine (1857) developed his theory of lateral earth pressure

    when the backfill consists of dry, cohesionless soil. The theory

    was later extended by Resal (1910) and Bell (1915) to be

    applicable to cohesive soils.

    The following are the important assumptions in Rankine’s theory:

    (i ) The soil mass is semi infinite, homogeneous, dry and

    cohesionless.(ii ) The ground surface is a plane which may be horizontal or

    inclined.

    (iii ) The face of the wall in contact with the backfill is vertical

    and smooth. In other words, the friction between the wall and

    the backfill is neglected (This amounts to ignoring the presence

    of the wall).

    (iv ) The wall yields about the base sufficiently for the active

     pressure conditions to develop; if it is the passive case that is

    under consideration, the wall is taken to be pushed sufficiently

    towards the fill for the passive resistance to be fully mobilised .

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    Plastic Equilibrium of Soil—Active and Passive RankineStates

     A mass of soil is said to be in a state of plastic equilibrium if 

    failure is incipient or imminent at all points within the mass.

    This is commonly referred to as the ‘general state of plastic

    equil ibrium’ and occurs only in rare instances such as when

    tectonic forces act.

    Most of the times only in a small portion of the mass such as

    that produced by the yielding of a retaining structure in the soil

    mass adjacent to it. Such a situation is referred to

    as the ‘local state of plastic equilibrium’.

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    Rotation of Frictionless Wall About the Bottom

    Rankine vs. Coulomb Theory

    Coulomb’s Earth Pressure Theory

    Coulomb’s wedge theory of earth pressure is based on the concept of a slidingwedge whichis torn off from the restof the backfill on movementof the wall.

    Assumptions:

    The backfil l is dry, cohesionless, homogeneous, isotropic and elast icallyundefomable but breakable.

    Theslip surface is plane which passess throughthe heel of thewall.

    The sliding wedge itself acts as a rigid body and the value of earth pressure isobtained by considering the limiting equil ibrium of the sliding wedge as awhole.

    Theposition and direction of theresultant earth pressure areknown.

    The back of the wall is rough and a relat ivemovement of the wall and the soildevelops frictionalforcesthat influencethe direction of the resultantpressure

    Coulomb: Active Case

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    Coulomb: Passive Case

    Rebhann’s Graphical Method for Active Pressure

    1. Draw the ground line and f  – line at angles b and f, respectively to meetin pointD.

    2. Draw semi-circle on BD as diameter.

    3. Through B,draw a l ine BH at an angle ψ with BD. Line BH is called theearth pressure lineor ψ – line.

    4. ThroughA, draw lineAE parallel to theψ – line.5. Draw EF perpendicular to BD, to meet thesemi-circlein F.

    6. WithB ascentre,and BF asradius,draw an arc to cut BDin G.

    7. Through G, draw GC parallel to the  ψ  –  line. BC then represents the slipline.

    8. With G asthe centre and GCas radius, draw an arc to cut BD in L.JoinCL.

    9. Calculate the total active earth pressure from therelation:

    Pa = g(∆CLG)= ½ g (LG) * x

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    Culmann’s Graphical Solution

    A graphical solution of Coulomb’s earth-pressure theory by Culmann(1875).

    Culmann’s solution can be used for any wall fr iction, regardless o f  

    irregularity of backfill and surcharges.

    Steps in Culmann’s solution of active pressure with

    granular backfill

    Step 1: Draw the features of the retaining wall and the backfill to a convenient scale.

    Step 2: Determine the value of  y  (degrees) = 90 –  q  –  d ’, where, q   the inclination of 

    the back face of the retaining wall with the vertical,and  d ’ = angle of wall friction.

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    Step 3: Drawa line BDthatmakesan angle f‘ with the horizontal.

    Step 4: Drawa line BE thatmakesan angle y with lineBD.

    Step 5: To consider some trial failure wedges, draw lines BC1, BC2, BC3, . . . ,BCn.

    Step 6: Find the areas ofABC1,ABC2,ABC3, . . . ,ABCn.

    Step 7: Determine the weight o f soil, W, per unit length of the retaining wall ineach of the trial failure wedges as follows:

    W1 = (Area of ABC1) × (g) × (1)

    W2 = (Area of ABC2) × (g) × (1)

    W3 = (Area of ABC3) × (g) × (1)

    Wn = (Area of ABCn) × (g) × (1)

    Step 8:Adopt a convenient load scale and plot the weights W1, W2, W3, . . ., Wndetermined from step 7 on line BD. (Note: Bc1 = W1, Bc2 = W2, Bc3 = W3, .. . , Bcn = Wn.)

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    Step 9: Draw c1c1’, c2c2’, c3c3’, . . . , cncn’ parallel to the line BE. (Note: c1’,c2’, c3’, . . . , cn’ are located on lines BC1, BC2, BC3, . . . , BCn, respectively.)

    Step 10: Draw a smooth curve through points c1’, c2’, c3’, . . . , cn’, Thiscurve is called the Culmann line.

    Step 11: Draw a tangent B’D’ to the smooth curve drawn in Step10. B’D’ isparallel to line BD. Let c’a be the point of tangency.

    Step 12: Draw a line ca c’a parallel to the line BE.

    Step 13: Determine the active force per unit length of wall asPa = (Length of cac’a) × (Load scale)

    Step 14: Draw a line Bc’aCa .ABCa is the desired failure wedge.

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