LATERAL EARTH PRESSURES

Introduction We have seen in shear strength of soils discussion that the soil in a natural deposit is subjected to vertical and horizontal stresses, which are principal stresses on the soil element. We know how to obtain the magnitude of vertical stresses either due to self-weight of the soil or due to externally applied loads. We have yet to arrive at the magnitude of other principal stress or minor principal stress or lateral earth pressure. τ Mohr’s c gin of planes

ircle is a point ori

failure envelope σ σ3=σh σ1=σv We know from basics of fluid mechanics that water exerts a lateral pressure equal to the vertical pressure it is experiencing. We calculate water pressure on dams, walls of water tanks using the same relation. Mohr’s circle will just be a point for any combination of σ1 and σ3. as such σ axis itself is its failure or strength envelope. This is because water doesn’t have any shear strength. Now, soil, we know, have shear strength. We shall imagine a similar situation to that of water i.e. both major and minor principal stresses are equal, as represented by point A in the figure. We shall also draw the strength envelope for the soil on this plot. τ Mohr’s circle C failure envelope σhC < σvC Mohr’s Circle σhB > σvB σ σhC A (σz or σv) σhB

σvA= σhA = σvB = σvC

Now suppose, somehow soil suffers lateral squeeze or compression, either by natural or man-made forces. Then lateral stress will increase. That is lateral stress is increased, vertical stress being unaltered. This new state is represented by Mohr’s circle B, σhB is greater than σvB. Can soil be in such a state physically? Is this a stable state? It is yes for both questions as the Mohr’s circle B is below failure envelope and not touching it. Therefore soil is in stable equilibrium. Now consider the opposite of the above, the soil is allowed to flex or dilate in lateral direction. The soil particles are no longer in a tight fit. As soil particles open up laterally, the horizontal stress that the soil was experiencing decreases. The state of stress is depicted by Mohr’s circle C, σhC is lower than σvC, which remain same. Unlike for water, in which the horizontal stress is always equal to the vertical stress, we see that in soil, the horizontal stress can be less than, equal to or greater than the vertical stress. This is because the soil has shear strength – there is space below the failure envelope for many principal stress combinations providing stable equilibrium. τ (σ1A -σ3A)/2 = [(σ1A +σ3A)/2] sinφ

⎟⎟⎠

⎞⎜⎜⎝

⎛−+

=φφ

σσ

sin1sin1

3

1

A

A also ⎟⎟⎠

⎞⎜⎜⎝

⎛−+

=φφ

σσ

sin1sin1

3

1

B

B

σ σ3A σ3B σ1A σ1B

It is convenient to talk about the lateral earth pressure vis-à-vis the vertical stress. We can easily see a relationship between vertical stress and lateral stress for all possible combinations of principal stresses leading to failure of soil. The two are

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related by a co efficient called co efficient of earth pressure.

σh = Kσv --------- (1) as we can see from the relation, if K is a co efficient, then lateral earth pressure is some factor times vertical pressure in soils. However ‘K’ is not a universal constant. We can see from Mohr’s circle diagrams we used just before that σh varies from a value σhC wherein it is less than vertical pressure and causing failure of soil and σhB wherein it is greater than vertical pressure and causing failure of soil. For the given σv it can vary from σhC to σhB depending upon the lateral deformation. If soil has flexed σh will be lesser magnitude, if soil is squeezed σh will be of larger magnitude and if there is no displacement, called at rest condition, it is somewhere intermediate of these two values. The two extremes and the at rest condition is of importance for soil engineers. Constant of proportionality between horizontal and vertical stress in at rest condition is know as co efficient of earth pressure at rest, K0. The magnitude of K0 can be assessed from the theory of elasticity, for condition of no lateral strain, as follows.

vh σμ

μσ ⎟⎟⎠

⎞⎜⎜⎝

⎛−

=1

⎟⎟⎠

⎞⎜⎜⎝

⎛−

==μ

μσσ

10Kv

h

Experimental evidence, for sands and normally consolidated clays, indicates that the value of K0 can be expressed as being equal to (1- sinφ′). A value of μ = 0.5 gives K0 = 1 and a value of μ = 0.33 gives K0 = 0.5. The former value is applicable for undrained conditions for saturated clay and the later value is more applicable for sand. K0 has been observed to be a function of the angle of shearing resistance, plasticity index and the stress history of soil deposit. For normally consolidated clays and sands K0 varies between 0.4 and 0.6 or can be

computed from K0 = (1- sinφ′). In over consolidated clays K0 is often observed to be more than 1.0. As such it is possible to find soils having, in their naturally existing states, values of K that range from less than 1 to more than 1.

States of failure The figure shows what happens to Mohr’s circle when the soil is pushed more and more into itself. The circle grows bigger and bigger as σh increases. This can continue until it becomes tangent to the strength envelope. Circle B is no longer stable; it has reached a state of failure. This state of failure is called the passive state of failure. At this state

σhB = σp = KpσvB

σp is called passive earth pressure, Kp is called the coefficient of passive earth pressure and is the maximum value that ‘K’ can assume. When soil is not being pushed into itself but laterally relieved, the lateral stress decreases until soil reaches the active state of failure, circle, is no longer stable. At this state

σhC = σa = KaσvC

where σa is active earth pressure and Ka is co efficient of active earth pressure, the minimum value K can assume. The expression for active earth pressure for a general case of c-φ soil can be derived as follows.

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τ Q c φ O′ O C σ σa (σv+ σa)/2 σv

c cotφ (σv+ σa)/2 In triangle O′QC

φsin=OCQC φsin'COQC =⇒

φφσσσσ sincot22

⎟⎠⎞

⎜⎝⎛ +

+=

− cavav

σv - σa = (σv + σa) sinφ +2c cotφ sinφ σv (1 - sinφ) = σa (1+sinφ) + 2c cotφ sinφ

⎟⎟⎠

⎞⎜⎜⎝

⎛+

−⎟⎟⎠

⎞⎜⎜⎝

⎛+−

=φφφ

φφσσ

sin1sincot2

sin1sin1 cva

φφ

φφσσ

sin1sin12

sin1sin1

+−

−⎟⎟⎠

⎞⎜⎜⎝

⎛+−

= cva

σa =Kaσv – 2c√Ka -------(2) similarly for passive earth pressure case the relation can be derived with the help of Mohr’s circle representing the stress condition for passive case given below. τ Q c φ O′ O C σ σv (σp+ σv)/2 σp

c cotφ (σp+ σv)/2

σp =Kpσv + 2c√Kp -------(3) The above expressions for general case of c-φ soils reduces to following forms for special case of sands (c=0) and clays (φ=0) c=0 soils (sands) σa =Kaσv σp =Kpσ τ σa σv σp φ=0 soils (clays) σa =Kaσv – 2c√Ka

σp =Kpσv + 2c√Kp

The following figures gives an idea of the displacement required to achieve active and passive pressures conditions and also the relative magnitudes of them along with their comparison with earth pressure at rest. τ Kaσv K0σv σv Kpσv

Change in lat. Stress to go from ‘at rest to passive condition change in lateral stress to go from at rest to active condition

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Kpσv K0σv Kaσv Away from soil mass into soil mass

0

displacement The following table gives the magnitudes of tilt at the top of a retaining wall away and towards soil for active and passive pressure cases respectively.

Soil Compactness Active case Passive case Coarse grained Dense 0.0005H 0.002H Coarse grained Loose 0.0002H 0.006H

Fine grained Stiff 0.01H 0.02H Fine grained Soft 0.02H 0.04H

The following figures show the possible development of failures surfaces in active and passive failure cases. Major and minor principal planes and the inclination of these planes with horizontal is also indicated in the figures.

Rankine’s theory The expressions for lateral stress developed in the previous section, were developed by W J M Rankine in 1857 . He made the following assumptions while deriving the expressions.

1. The back of the wall is smooth, i.e. no adhesion or friction develops between the wall and the soil.

2. The back of the wall is vertical. 3. Top surface of the backfill is

horizontal. 4. The length of the wall is long when

compared to height so that it can be treated as a two dimensional situation.

5. There is adequate displacement or tilt of the wall to generate the active or the passive case.

The pressure distribution behind the wall retaining the soil for both active and passive cases and for sands and c-φ soils are shown in the following figures. PA=0.5KaγH2/2 H PA=0.5KaγH2/2 H/3 KaγH KpγH 2c 2c z z=2c/γ H H-z Pp=2cH+1/2 γH2 γH+2c Pa=(γH-2c)(H-z)/2 (H-z)/3 For the case of saturated soils, the pressure on the wall is the combination of earth pressure as computed using submerged unit weight and hydrostatic pressure as shown in the figure. KaγH1 KaγH1 H1

H H2 KaγH1+Kaγ′H2 γwH2 KaγH1+Kaγ′H2+γwH2

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The first assumption made by Rankine is never satisfied in that the back of the wall is always rough. Other theories, such as the Coulomb’s theory developed earlier than Rankine’s and discussed in next section, take note of this fact. Theory is widely used because of its simplicity and because its results do not differ much from more realistic theories.

Rankine theory for sloping backfill

Consider a rhombic element of soil. with sides vertical and at an angle β to the horizontal, at depth z in a semi-infinite mass. The vertical stress and the active or passive pressure are each inclined at an angle β to the appropriate sides of element as shown in figure. These stresses are not principal stresses. For the active case, the vertical stress at depth z on a plane inclined at angle β to the horizontal is given by

σz = γz cosβ σz is represented by point A on σ versus τ plot. OA represents σz. at active pressure state this Mohr’s circle touches the failure envelope for the state of plastic equilibrium.

Thus active pressure pa is represented by OB′ on the diagram. When c=0, the relationship between pa and σz, can be derived from the diagram Ka = pa/σz = OB/OA = OB′/OA = (OD-AD)/(OD+AD) Now OD = OC cosβ AD = √(AC2 – CD2 = √(FC2-CD2

= √(OC2 sin2φ - OC2 sin2β

∴ Ka =)cos(coscos)cos(coscos

22

22

φββ

φββ

−+

−−

pa = Kaγz cosβ Pa =1/2 KaγH2 cosβ In passive pressure case σz is represented by the distance OB′ in Mohr’s stress diagram. The passive pressure pp is then represented by OA′ and when c=0 the passive pressure co efficient works out to

Ka =)cos(coscos)cos(coscos

22

22

φββ

φββ

−−

−+

pp = Kpγz cosβ Pp =1/2 KpγH2 cosβ

Coulomb’s theory Almost a century before Rankine, C.A.Coulomb in 1776 had studied this problem of lateral pressures that are applied by soil. He recognised that before we get the soil to the active state or passive state the soil has to undergo lateral deformation. In undergoing lateral deformation the soil also moves in vertical direction.

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When soil moves in vertical direction it develop friction between soil and wall, as walls are not frictionless. On the basis of number of model tests, he determined that when we push a wall into the soil, or the wall tilts outward and relieves the lateral stress in the soil, the failure wedges that are formed behind the wall are not straight lines inclined at angles of (450+φ/2) or (450+φ/2) but are slightly curved as shown in figure. By trial and error he determined the most critical angle of inclination of failure surface. For the active case, the critical angle is the angle that gives the maximum force on the wall, i.e. the soil provides the least help in supporting itself. Conversely, for passive case, the critical angle is the angle that gives the minimum force applied to push into itself to make the soil yield. For simplicity his analysis for a straight failure surface is discussed here. Pa can be expressed in terms of W and the angles in the triangle of forces. Then the maximum value of P, corresponding to a particular value of η, critical failure angle, is given by ∂p/∂θ=0. This leads to Pa = ½ KaγH2 where

2

)sin()sin()sin()sin(

sin)sin(

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−+

++

−

=

βαβφδφδα

αφα

aK

similarly Pp=1/2 KpγH2

2

)sin()sin()sin(sin(

sinsin(

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−++

−−

+

=

βαβφδφδα

αφα

pK

For easy understanding of the theory, the concept as applied to simple cases to most general cases is given here. They are in the order as for c=0 soil backfill horizontal, c=0 and backfill inclined, c-φ soil backfill horizontal and c-φ soil backfill inclined. For the most general case of c-φ soil the forces acting on the failure wedge are

1. The weight of the wedge W. 2. The reaction (P) between the wall

and the soil, acting at an angle δ below the normal.

3. The force due to the constant component of shearing resistance on the wall (Cw =cw x EB).

4. The reaction (R) on the failure plane, acting at an angle φ below the normal.

5. The force on the failure plane due to the constant component of shear strength C=c x BC.

With the forces listed above and indicated in the figures of all the cases the figures are self-explanatory. Knowing the line of action of all the forces and magnitudes of all the forces except P and R they can be determined. With similar exercise passive pressure can also be determined. Case (i) c=0 and horizontal backfill

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Force polygon

Plot of Pa versus η for determining maximum value of PaCase (ii) c=0 inclined backfill

Force polygon

case (iii) c-φ soil with horizontal backfill

Force polygon case (iv) c-φsoil with inclined backfill

for φu =0 (undrained condition) Pa = ½ γ(H2 – z2

0)-2cu (H-z0)√{1+(cw/cu)} For undrained condition φu =0, Ka is unity and it is convenient to introduce a second coefficient Kac, where Kac= 2√{1+(cw/cu)} For fully drained condition in terms of tangent parameters c′ and φ′, it can be assumed that Kac =2√{Ka[1+(cw/c′)]} In general pa= Kaγz-Kacc Where Kac =2√{Ka[1+(cw/c)]} The depth of a dry tension crack (at which pa=0) is given by

z0=a

w

Kccc

γ

+12

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