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THE EARTH PRESSURE PROBLEMEARTH PRESSURES UNDER ADDITIONAL LOADS

excavation CONDITIONS OF PLASTIC EQUILIBRIUMA. CONCEPT OF FAILURE

B. EQUATIONS OF PLASTIC EQUILIBRIUM

failure surfaces3Zemini ya basarak (daha zor) yada ekerek (daha kolay) yenebiliriz. DEVELOPMENT of ACTIVE EARTH PRESSURE

s1 =rzs3 =?tss1s3=s0fH0.001H-0.003H :required movementKo= coefficient of earth pressure at-restKa=active earth pressure coefficients3=Karzs3=Korzs1=r.zat rest circleActive condition is reached by reduction in lateral stress (extension)ACTIVE EARTH PRESSURE

sh= r.z.KaPa acts at H from base=r.zDEVELOPMENT of PASSIVE EARTH PRESSURE

s1s1s3s3ss1s3=sps3=sofH0.02H-0.2H: required movement to reach plastic equilibriumKo=at-rest earth pressureKp=passive earth pressure coefficientts3=Kprzs3=Korzs1=rz

at rest circle moving to rightplastic equilibriumPASSIVE EARTH PRESSURE sh = r . z . Kp

THE SIGNIFICANCE OF MOVEMENTSON THE DEVELOPMENT OF EARTH PRESSURE

8Kp ye gelebilmek iin Ka nn 4 kat bir hareket gerekiyorRANKINES THEORY OF EARTH PRESSURE (Lower bound in that the condition of equilibrium and yield is satisfied).Rankines theory considers the state of stress in a soil mass when the condition of plastic equilibrium has been reached, that is when the soil is at the point of failure throughout the mass. This condition is best illustrated using Mohr circles. The initial stress conditions in a soil prior to the construction of a retaining wall is taken asThis is a non failure condition, hence it is represented on a Mohr diagram by a circle which does not touch the Mohr Coulomb failure line. In the case of wall movement away from the soil, it is assumed that the vertical stress does not change and that the horizontal stress can be reduced until the stresses on some plane in the soil mass reaches the yield stress. That condition is represented on the Mohr diagram by a circle which touches the Mohr Coulomb failure line. At that condition, the limiting horizontal stress is called the active earth pressure pa (which is the minimum earth pressure that can act on the wall) and its relationship to the vertical stress sz is got by geometry (as shown on the board) to be:-The coefficient of active earth pressure is and is equal to Note that the failure surfaces are at an angle of 45+f/2 with the horizontal.Similarly, if the wall is pushed towards the soil, the vertical stress is assumed to remain constant and the horizontal stress increases. This can increase until yield is reached within the soil mass which is represented on the Mohr diagram by a circle which touches the failure line. This limiting horizontal stress is called pp and can be determined from the geometry (as shown on board) to bewhere the coefficient of passive earth pressure is Note that the failure surfaces are at angles of 45-f/2 with the horizontalTension cracksVertical cut in cohesive soil -

PLASTIC EQUILIBRIUM in SAND

ACTIVEPASSIVE

qfK0

9Kazya balaynca yanal gerilme giderek azalr. Zemin aktif olarak plastik dengeye ulatnda yatayla 45+f/2 lik alarla krlma dzlemleri oluur.

PRESSURE DISTRIBUTION in ACTIVE STATE(SAND)

HH

PaH/3

FailurewedgeSand

PRESSURE DISTRIBUTION FOR PASSIVE STATE (SAND)HH

PpH/3

FailurewedgeSandEQUILIBRIUM IN SOIL CAN BE DESCRIBED BY THREE STATES

ELASTIC EQUILIBRIUM CONDITION : K0

PLASTIC EQUILIBRIUM- ACTIVE STATE :Ka caused by extension

PLASTIC EQUILIBRIUM- PASSIVE STATE: Kp caused by compression/shortening ACTIVE AND PASSIVE PRESSURES of SAND (SUBMERGED)Soil pressure at any depth

Total thrust (submerged)

ACTIVE STATE

PASSIVE STATE

13EFFECT of DISTRIBUTED LOAD (q)

use equivalent height h'or

ACTIVE PRESSURE of CLAY

plastic equilibrium equations

15Zeminde c varsa bundan dolay bir ekme gerilmesi var denilebilir. Yani belli bir yere kadar bu zemin kendini desteksiz tutabilir. Kil z0 derinlie kadar ekme gerilmesi alyor dalays ile 2 z0 derinlie kadar teorik olarak kazlabilir. Kil kumdan farkl deildir aslnda. Kili Makarna gibi uzatp sallarsak kopmuyor buna c diyorlar. Ama uzun vadede c kayboluyor ve kum gibi bir hale geliyor. Kohezyon ne kadar byk olursa olsun buna gvenmemeliyiz. at the surface

in tension zone

The distribution of pressures in c-f conditions:+-z=0z=Hs30 CRITICAL EXCAVATION DEPTH IN CLAY

Theoretically, it should be possible to excavate to a depth where the active thrust becomes equal to the passive resistance : 2 z0If f =0, then PASSIVE RESISTANCE of CLAY

DistributedLoad qcohesionmassEXAMPLE 1:Determine the active lateral earth pressure on the frictionless surface shown in figure. Calculate the resultant force and its location from the base of the wall.

H=5mSTEP-1: Calculate Ka

STEP-2: Calculate the vertical effective stress (sz)

STEP-3: Calculate the lateral effective stress (sx)a

EXAMPLE 1: (continued)STEP-4: Sketch the lateral earth pressure distributionsH=5m17 kPa49 kPaLateral Earth PressureHydrostatic PressureSTEP-5: Calculate the active thrust

where (Pwater pressure) is the due to the pore water and (Pearth pressure) is the lateral force due to the soil

STEP-6: Determine the location of the resultantSince both lateral pressure and the pore water (hydrostatic)pressure distributions are triangular over the whole depth, the resultant is at the centroid of the triangle, that is;

from the base of the wallEXAMPLE 2:Determine the active earth pressure on the frictionless surface shown

H=4mSTEP-1: Calculate Ka

STEP-2: Calculate the vertical effective stress (sz)

STEP-3: Calculate the lateral effective stress at the base (sx)a

GWTSTEP-4: Calculate the lateral effective stress due to external load at the base(sq)a

15 kPaSTEP-5: Calculate the lateral tension effect due to cohesion (sc)a

EXAMPLE 2: (continued)STEP-6: Sketch the distributions of lateral earth pressureH=4m14.34 kPa39.24 kPaLateral Earth PressureHydrostaticSTEP-7: Calculate the total lateral force(thrust)

GWTq=15 kPa5.85 kPaLoadCohesion-10 kPa(+)(+)(+)(-)EXAMPLE 3:Determine the active earth pressure on the frictionless surface shown.

H=3STEP-1: Calculate Ka values

GWT10 kPaSWSP

H=3

H=3GWT10 kPaSWSP

H=3Lateral Earth PressureSWHydrostatic PressureLoadIII12IIIIV34VIV65Lateral Earth PressureSPPRESSURES

FORCES

EXAMPLE 3: (continued) 3.33 16.98 9.50 29.43PROBLEM 4SW sand is filled into the timber box with shown dimensions. Calculate thrust on the door(a)at rest (b)active in dry state (c)passive force required when the box is full of water. Results of shearbox tests are supplied.

Results from Shearbox testsTest No121503001132291 ) Calculate shearing resistance of sand:

AktifDurum

Pasif DurumPROBLEM 5Calculate the force F required to keep the door in active equilibrium.

PRESSURES

FORCES(THRUST)

MOMENTS

rn=18rd=20rd=22f=35f=30PROBLEM 6 HOMEWORKConstruct the active pressure diagram on vertical plane XX and calculate the thrust Pa.rk=19 kN/m3f=300rd=22 kN/m3f=450rd=18 kN/m3f=100c=10 kPa

pressures

FORCES

SPGWYASS33212345678IIIIIIIVVCHVIVIIVIIIDry sandSubmerged CHcohesion CHsubmerged GW waterXX( - )