Ch7 Lateral Earth Pressure Theories (407-440)

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CHAPTER

7

LATERAL EAERTH PRESSURE

THEORIES

7.1 STATES OF STRESS

7.1.1 ELASTIC STATES OF EQUILIBRIUM

Referring to Fig.(7.1), if no movement of the wall takes place, the soil is said to be in "elastic equilibrium" or at rest condition. The corresponding earth pressure under this condition is known as earth pressure at rest.

For this state of stress, the vertical and horizontal effective stresses acting on any element of soil such as A or B are:

σ′

z= σ′1 = γ′ z ……….……….(7.1) PA = σ′3 = Ko σ′1 = Ko γ′z ………..……….(7.2) Since no point of soil is on verge of failure, the Mohr’s circle for the at rest stress state stays within the failure surface boundaries. 6

7.1.2 PLASTIC STATES OF EQUILIBRIUM

The soil is said to be in "plastic equilibrium" if every point of it is on the verge of failure (failure is about to occur simultaneously at all points in the mass). This state of equilibrium is classified into active and passive states. When the soil is in active state of plastic

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equilibrium, the lateral pressure is known as active earth pressure. Whereas when it is in passive state, the pressure developed is known as passive earth pressure.

The active state of plastic equilibrium in soil behind a retaining wall with horizontal ground surface is shown in Fig.(7.2). Since the wall is moved away from element A and towards element B, the effective horizontal stress in element A will reduce but the effective vertical stress will remain constant. Therefore, the Mohr’s circle for active stress state will expand until it touches the failure surfaces in Fig.(7.2).

Fig.(7.3) shows the passive state of plastic equilibrium in soil behind a retaining wall with horizontal ground surface. Since the wall is moved towards B, its effective horizontal stress will increase but the effective vertical stress will remain constant. Hence, the Mohr’s circle will first contract and then expand.

’v = remains the same

’h = ’x = increases till failure occurs

Fig.(7.3): Passive state of plastic equilibrium in soil.

Upward movement of wedge

Passive State R.W.

Wall movement towards backfill Resisting force

B

Fig.(7.2): Active state of plastic equilibrium in soil.

Failure(Active state)

Failure envelope

’v = z (remains the same)

’h = K0’v = K0z (decreases till failure occurs)

Wall movement away from backfill Active State

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7.2 CATEGORIES OF LATERAL EARTH PRESSURE

There are three categories of lateral earth pressure. The magnitude of each type depends upon the retaining wall movement relative to the backfill as shown in Fig.(7.4), type of backfill and the value of the vertical pressure

P

_v which affects the state of stress. The three categories are:

 At rest earth pressure

 Active earth pressure

 Passive earth pressure

The at rest pressure develops when the wall experiences no lateral movement. This typically occurs when the wall is restrained. The active pressure develops when the wall is free to move outward such as a typical retaining wall and the soil mass stretches sufficiently to mobilize its shear strength. On the other hand, if the wall moves towards the soil, then the soil mass is compressed which also mobilizes its shear strength and the passive pressure develops. This situation might occur along the section of wall that is below grade and on the opposite side of the retained section of fill. Some engineers might use the passive pressure that develops along this buried face as additional restraint to lateral movement, but it often is ignored.

In order to develop the full active pressure or the full passive pressure, the wall must move a sufficient amount; otherwise the full active or full passive pressure will not develop. The wall movement effect on development of the active or passive earth pressure is shown in Fig.(7.5). Note that the at rest condition is shown where the wall rotation is equal to 0, which is the condition of zero lateral strain.

Fig.(7.4): Wall movement. Active Case

(Wall moves away from soil)

At rest Case (No movement)

Passive Case (Wall moves into soil) G.S.

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From Fig.(7.5) it is evident that:

 As the wall moves away from the soil backfill, the active condition develops and the lateral pressure against the wall decreases with wall movement until the minimum active earth pressure force (Pa) is reached.

 As the wall moves towards (into) the soil backfill, the passive condition develops and the lateral pressure against the wall increases with wall movement until the maximum passive earth pressure force (Pp) is reached.

Thus the intensity of the active / passive horizontal pressure, which is a function of the applicable earth pressure coefficient, depends upon the degree of wall movement since the movement controls the amount of shear strength mobilized in the surrounding soil. Table (1) shows the movement of a retaining wall top necessary to reach minimum active or maximum passive pressure developed by tilting or lateral translation.

Soil Type Value of Y/H*

Active Passive

Dense sand 0.0005 0.002

Loose sand 0.002 0.006 Stiff clay 0.01 0.02 Soft clay 0.02 0.04

Fig.(7.5): Effect of movement on wall pressure (after NAVFAC DM-7, 1971).

Table (1): Magnitudes of wall movement to reach failure

(after NAVFAC DM7.2, 1982).

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7.3 EARTH PRESSURE COEFFICIENTS

(a) At Rest Earth Pressure Coefficient:

If a mass of soil is deposited by either natural or artificial process, the coefficient K will be equal to Ko (the coefficient of earth pressure at rest):

 for normally consolidated soils:

          1 sin P P K rest . at v h o ……….….………...…(7.3)

 for overconsolidated soils:

Ko = (1 –sin ') OCRsin………..……….……..………..…(7.4)  from elastic analysis:

Ko = 𝜇

1−𝜇 ………...…….….……...…(7.5)

(b) Active and Passive Earth Pressure Coefficients for Rankine Theory:

 Level backfill: ) 2 45 ( tan P P K 2 active v h a            or      sin 1 sin 1 K_a ..………(7.6) ) 2 45 ( tan P P K 2 Passive v h P            or      sin 1 sin 1 K_P ...…...………(7.7)  Inclined backfill:             2 2 2 2 a cos cos cos cos cos cos cos K ………..……….………...….…….(7.8)             2 2 2 2 P cos cos cos cos cos cos cos K ……..……….……...….…….(7.9)

As shown above for Rankine earth pressure theory:

a p 1/K

K  …….…....………..……….….………...(7.10)

where, Ka = active coefficient of earth pressure. KP = passive coefficient of earth pressure, and

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(c) Active and Passive Earth Pressure Coefficients for Coulomb Theory:

2 2 2 a ) sin( ) sin( ) sin( ) sin( 1 ) sin( . sin ) ( sin K                              …………....……….(7.11) 2 2 2 P ) sin( ) sin( ) sin( ) sin( 1 ) sin( . sin ) ( sin K                              …………....….…....(7.12)

Note: Unlike the Rankine earth pressure coefficients, KP ≠ 1/ Ka where,

𝛼 = angle of inclination of back face of wall with horizontal.

𝛽 = angle of inclination of backfill or ground surface with horizontal. 𝛿 = soil-wall-friction angle.

∅ = soil internal friction angle.

 for active state: Ka = Ph / Pv=

𝑚𝑖𝑛𝑜𝑟 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 𝑚𝑎𝑗𝑜𝑟 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑠𝑡𝑟𝑒𝑠𝑠

 for passive state: Ka = Ph / Pv=

𝑚𝑎𝑗𝑜𝑟 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 𝑚𝑖𝑛𝑜𝑟 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑠𝑡𝑟𝑒𝑠𝑠

Typical values of earth pressure coefficients are shown in Table (2). They depend relatively on density of soil, the process by which the deposit is formed and on the overconsolidation ratio (O.C.R.).

To change the values of (K) for a mass of sand from Ko to Ka or KP, it is necessary to give the entire mass of soil an opportunity either to stretch or to be compressed in a horizontal direction. Pv is unaltered, but Ph = K. Pv decreases if the soil mass stretches (Active

Rankine Case) and it increases if the soil mass compresses (Passive Rankine Case); see Fig.(7.7).

Earth pressure coefficient

Cohesionless soil Cohesive soil

Dense sand Loose sand Stiff clay Soft clay

Ka 0.4 0.6 1 2

Ko 0.33 0.22 0.4 0.8

KP 3 14 1 0.5

Table (2): Usual range of earth pressure coefficients.

Fig.(7.6): Retaining wall with inclined back face and sloped ground surface.

𝛼

𝛽

𝛿

Backfill

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7.4 RANKINE'S EARTH PRESSURE THEORY

This theory was developed in 1857 for frictionless soil and then it was extended by Bell (1915) for cohesive soils. Rankine's theory assumes that:

1. The retained soil is cohesionless, homogeneous, isotropic (similar stress-strain properties in all directions), semi-infinite (wall is very long and soil goes back a long distance without bends) and well drained to avoid consideration of pore pressures.

2. The back surface of the wall is smooth (i.e., there is no adhesion or friction between the wall and soil;  = 0).

3. Lateral pressure is limited to vertical walls with horizontal or inclined ground surface. Thus the resultant force must be parallel to the backfill surface as shown in Fig.(7.8). 4. Lateral pressure varies linearly with depth and the resultant pressure is located

one-third of the height (H) above the base of the wall.

5. The wall yields about its base and therefore it satisfies the deformation condition for plastic equilibrium.

6. Failure (in the backfill) occurs as a sliding wedge along an assumed failure plane defined by ∅.

Small 

Relativly large 

Against backfill Away from backfill

L a tera l ea rth p ress u re co ef ficient Movement a K P K o K

Fig.(7.7): Variation of K- coefficient versus relative movement.

R.W.

(a) Levelled backfill Resultant R.W. (b) sloped backfill Resultant   G.S. G.S.

Fig.(7.8): Resultant of earth pressure for smooth wall.

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7.4.1 SPECIAL CASES OF RANKINE'S EARTH PRESSURE THEORY

(EXCLUSIVELY FOR SANDY SOILS):

Case (1): Dry or Moist Backfill with no Surcharge.

H . . K base .. at .

P_a  _a  ….acting at H/3 from base where, ) 2 45 ( 2 tan a K        sin 1 sin 1

Case (2): Submerged Backfill

(a) Fully Submerged (b) Partially Submerged

Case (3): Backfill with Uniform Surcharge

The effect of the surcharge of intensity q is the same as that of a fill of height equal to q/ above the ground surface. K_a.q.K_a.q

 H z H/3 A P H . . k_a  z . . k_a  R.W. H H/3 H . . k_a  W.T. H . w  R.W. H/3 Pw PA H . w H . . a K base .. at . a P    H1 1 a. .H k  _w.H₂ H2 1 a. .H k  2 a. .H k  R.W. G.S. 2 H . w ) 2 H . 1 H . ( a K base .. at . a P     H H . . k_a  q/unit area q . k_a R.W.

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Case (4): Dry Backfill with Sloping Surface

Considering the soil element shown in Fig.(7.9) with  and

P

_a are resultant stresses on two conjugate planes. The principle stresses 1 and 3 are determined from Mohr circle as:

By simple geometry, it can be shown that:

               1 3 1 3 2 2 1 1 a sin sin 2 cos 2 BA OB OA P ...(7.13a) ...…(7.13b) But,   .z.cos b cos b . z . .

Dividing (7.13a by 7.13b) and substituting for ; gives:

             2 2 2 2 a cos cos cos cos cos cos cos . z . P ; or

P

_a



K

_a

.



.

z

The resultant active thrust on the wall of height H is given by: 2 a A K . .H 2 1 P   ……..…….………...………(7.14) where,             2 2 2 2 a cos cos cos cos cos cos cos K Similarly:             2 2 2 2 p cos cos cos cos cos cos cos K                 1 3 1 3 2 2 2 2 sin sin 2 cos 2 BA OB OA

Fig.(7.9): Active earth pressure for bakfill with sloping surface.

   Active condition   2 OA a P 1 OA A1  failure envelope B C A2 1  3  G.S. R.W.   A P     A P b cos R.W. A P   H/3

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Case (5): Submerged Backfill with Sloping Surface

If the backfill is submerged, the lateral pressure due to submerged weight of soil will act at  with horizontal, while lateral pressure due to water will act normal to the wall.

2 a A K . .H 2 1 P   ……....………..….…..……(7.15a) where,             2 2 2 2 a cos cos cos cos cos cos cos K 2 w w H 2 1 P   ……....………...….………...…………(7.15b)

Case (6): Inclined Back of the Wall

(a) Horizontal Surface 0_{(b) Sloped Surface}0

The resultant pressure P is a vector sum of PA and W.

Case (7): Active Earth Pressure in Cohesive Backfill (a) Cohesive Soil without Surcharge

From soil mechanics, the relationship between₁ and ₃ at failure is given by:

     ₁ ₃N 2c N ...……….…………...…………(7.16) A B _C A P   H/3 G.S. R.W. W P A P H/3 G.S. R.W.

W(weight of soil wedge ABC)

P A B C R.W. A P   H/3 W.T. w P H/3 G.S. R.W.

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417 or       N c 2 N 1 3 where ) 2 45 ( tan N 2  or a K 1 N

here,



₁



P

_v





.

z

and therefore ₃ P_h .z.K_a2c K_a

at z = 0: P_h 2c K_a and when, P_h 0: a K c 2 o z z  

 (depth of tension cracks) ……..…..…...…(7.17)

Tension cracks usually develop in soil at top of wall and decrease to zero at depth zo. The total net pressure up to a depth of

a K c 4 o z 2 

 is zero. This means that a cohesive soil could

stand with a vertical face up to a depth

a K

c 4

 without any lateral support. Thus the critical height, Hc of unsupported vertical cut in cohesive soil is given by:

For c- Soil: a K c 4 o z 2 c H    . ...…..……...……….…...………(7.18a)

For Soft Clays 0:

 4c

H_c ...…...….…….………..…...……...………(7.18b) Due to tension cracks, it is usual to neglect the negative pressure diagram (ABC) and consider the positive diagram below zo. Therefore, the resultant thrust is:





_            a K c 2 H . a K c 2 a K . H . 2 1 A P …..………...…….…...(7.19)

acts above the wall base at _         a K c 2 H 3 1 . H a K c 2  R.W. a K c 2 o z   ₋ + G.S. a K c 4  E D A B F A P (net Ph is zero) a K . H .  C zo = zone of tension cracks z Pv.z1 3 h P  (Active Case)

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(b) Cohesive Soil with Surcharge

If there is a uniform surcharge q/unit area, then the lateral pressure is increased by qK _a everywhere (diagram AGHF).

a a a active H) .HK 2c K q.K P (    …….…………..…..…...(7.20)

Depending on qK magnitude, the depth of tension cracks is altered. If q_a K > _a 2c K_a then there is no tension cracking.

Case (8): Passive Earth Pressure in Cohesive Backfill From Eq.(7.16):      ₁ ₃N 2c N ….……….…………...…...(7.21)

for passive case: ₃ (P_v)_passive .z and ₁(P_h)_passive(.z.K_P)(2c. K_P)

7.5 COULOMB'S EARTH PRESSURE THEORY

This theory was developed in 1776. It can be used for different boundary conditions such as inclined walls, walls with a break, inclined uniform or non-uniform slopes, under concentrated and/or distributed surcharge loads. Coulomb's theory assumes that:

H area . unit / q R.W. − + E D A B F a K . q a K . H .  C z q q K_a H G Surcharge diagram H P K c 2 R.W. G.S. D C A B E 2 P P K . H .  z Pv .z3 1 P k . z . h P   (Passive Case) 1 P

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 The retained soil is cohesionless, isotropic and homogeneous.

 The surface of the wall is rough (i.e., the soil-wall-friction angle δ ≠ 0). Note that δ ranges from ∅/2 - 2∅/3 and δ = 2∅/3 is commonly used.

 The resultant force is not necessarily parallel to the backfill surface because of the soil-wall friction value δ.

 A condition of limit equilibrium is satisfied in the soil mass retained behind the wall (i.e., the wall deforms to produce active or passive condition in soil).

 The limit equilibrium describes the state of a soil mass that is on the verge of failure (i.e., the applied stresses are equal to the available strength along the slip plane).

 The retained soil mass will slip along a failure plane inclined at an angle θ to the horizontal

 The slope of the slip surface failure plane is planer.

 The critical slip plane gives the maximum lateral pressure on the wall.

 Failure is a plane strain problem with always two sets of slip planes - one for positive shear stress and the other for negative shear stress as shown in Fig.(7.10).

 Soil constants have definite values (i.e.,  , c and  are constants and their values are known).

7.5.1 COULOMB'S ACTIVE EARTH PRESSURE

FOR COHESIONLESS SOIL

Figure (7.11a) shows a retaining wall of height H with its back face inclined at 𝛼 with horizontal, retaining a soil of friction angle 𝜙 that slopes at 𝛽 with horizontal. If 𝛿 be the angle of wall friction and under active pressure the wall will move away from the soil mass.

 Semi-Analytical Solution

To find the active force, assume the failure surface in the soil mass to be a plane such as AC inclined at an angle (θ= 45 + ∅/2) with horizontal and a possible soil failure wedge such as ABC. Then forces acting on the wedge ABC per unit length of the wall are as follows:

(a) Local active failure

(b) Local passive failure (c) Mohr stress diagram

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1. Weight of the wedge, W acting through the center of gravity (O′) of ABC.

2. The reaction of soil against the wedge, R, inclined at an angle 𝜙 to the normal drawn to the failure surface AC.

3. The active force Pa, inclined at an angle 𝛿 to the normal to the back face of the wall.

For equilibrium, these three forces must meet at a point. Since their directions and the magnitude of W are known, R and Pa can be determined from force polygon.

The weight of the soil wedge ABCis calculated from Fig.(7.11a) as:

Area of wedge ABC = 1/2 (AC) (BD) ….…..…….…….………...…...(7.22a) where BD is drawn perpendicular to AC.

From the Law of Sines: AC = AB sin(α + β)

sin(θ − β) , BD = AB sin(α + θ), AB = H sin α Substituting into Eq.(7.22a) and simplifying gives:

W = γ A(1) = γ H2 2 sin2_α[

sin(α + β)

sin(θ − β) sin(α + θ)]…...………...…...(7.22b)

The active force Pa is calculated from force polygon shown in Fig.(7.11b) as:

Pa sin( θ − ∅)

=

W sin(180 − α − θ + ∅ + δ) P_a= W sin(θ − ∅) sin(180− α− θ + ∅ + δ)……...………...…...(7.22c)

Combining Eqs. (7.22b) and (7.22c) gives:

Fig.(7.11): Coulomb's Active Pressure (Semi-Analytical Solution). 𝜷 A B Pa W R _N 𝜹 𝜽 O’ O D 𝜶

Assumed failure surfaces

∅ N

(a) Retaining wall with single trial wedge. (b) Polygon of forces.

Pa 𝜶 − 𝜹 𝜽 − ∅ 𝟏𝟖𝟎𝒐_{− 𝜶 −}_𝜽₊_∅_{+ 𝜹} 𝜸 ∅ 𝒄 = 𝟎

Wall movement away from soil

W R

𝜶 + 𝜷

𝜽 − 𝜷

H

Actual failure surface

H/3

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421 P_a = γ H2 2 sin2_α

[

sin(α+β) sin(θ − β)

sin(α + θ)

]

sin(θ − ∅) sin(180− α− θ + ∅ + δ)…....…...(7.22d)

where, 𝛾 is unit weight of soil, parameters 𝛾, ∅, 𝛿, 𝛼, 𝛽, 𝐻 are constants, 𝜃 is a variable coressponding to assumed failure surface AC .

Setting 𝑑𝑃𝑎

𝑑𝜃

=

0 gives the maximum active wall force Pa as:

Pa =1₂γ H2 Ka …..………...…...(7.22e) where, 2 ) sin( ) sin( ) sin( ) sin( 1 ) sin( . 2 sin ) ( 2 sin a K                            ……...…...(7.22f)

If 𝛽 = 𝛿 = 0 and 𝛼 = 90𝑜 (a smooth vertical wall with horizontal backfill), Eq.(7.22f) simplifies to: ) 2 45 ( 2 tan ) sin 1 ( ) sin 1 ( a K        ………...…...(7.22g)

which is identical with the Rankine’s coefficient for active earth pressure.

 Graphical Solution

Several trial wedges are selected such as ABC1, ABC2, ABC3,..corresponding to assumed failure surfaces AC1, AC2, AC3,..that makes an angles of 𝜃1, 𝜃2 , 𝜃3 ,…with the horizontal.

Initially for each trial wedge, the active force is determined using the force polygon as shown in Fig. (7.12b) or using Eq.(7.22d) with specified 𝜃𝑖. Then the maximum value of Pa determined is the Coulomb’s active force as shown at the top part of Fig.(7.12a).

Fig.(7.12): Coulomb's Active Pressure (Graphical Solution). C3 𝜷 A B C1 C2 Pa W R _N 𝜹 𝜽𝟏 𝜽𝟐 O’ O D 𝜶

∅ N

(a) Retaining wall with several trial wedges. (b) Polygon of forces.

Pa 𝜶 − 𝜹 𝜽𝒊− ∅ 𝟏𝟖𝟎𝒐_{− 𝜶 − 𝜽} 𝒊+∅+ 𝜹 𝜸 ∅ 𝒄 = 𝟎 Pa (max.)

Wall movement away from soil

W R P1 P2 P3 𝜽𝟏 𝜽𝟐 _𝜽 𝒊 𝜽𝟑 𝜶 + 𝜷 𝜽𝟏− 𝜷

Graphicaldetermination of Pa (max.)

H

Actual failure surface

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7.5.2 COULOMB'S PASSIVE EARTH PRESSURE

FOR COHESIONLESS SOIL

Following similar method used in obtaining the active earth pressure, the passive earth pressure Pp can be derived and expressed by the following equations:

The weight of the assumed failure wedge ABC is calculated from Fig.(7.13a) as:

W = γ H2 2 sin2_α

[

sin(α+β)

sin(θ−β)

sin(α + θ)

]

….……..…….…....………...…...(7.23a) The passive force Pp is calculated from force polygon shown in Fig.(7.13b) as:

Pp = _{sin(180−α−θ−∅−δ)}W sin(θ+ ∅) .……....…….………...…...(7.23b)

Setting 𝑑𝑃𝑝

𝑑𝜃

=

0 gives the minimum value of Pp as:

Pp =1₂γ H2 Kp …..………..………...…...(7.23c) where, 2 ) sin( ) sin( ) sin( ) sin( 1 ) sin( . 2 sin ) ( 2 sin p K                            ……...…...(7.23d)

For smooth vertical wall with horizontal backfill (𝛽 = 𝛿 = 0 and 𝛼 = 90𝑜_{) Eq.(7.23d)} simplifies to: ) 2 45 ( 2 tan ) sin 1 ( ) sin 1 ( p K         …...…...…...(7.23e)

which is identical with that of Rankine’s passive earth pressure coefficient.

C3 𝜷 A H B C1 C2 Pp W R N 𝜹 𝜽𝟏 𝜽𝟐 D 𝜶

∅ N (b) Polygon of forces. Pp 𝜶 + 𝜹 𝜽𝒊+ ∅ 𝟏𝟖𝟎𝒐_{− 𝜶 − 𝜽} 𝒊− 𝜶 − 𝜹 𝜸 ∅ 𝒄 = 𝟎 Pp (min.)

Wall movement toward the soil

H/3 R W P1 P2 P3 𝜽𝟏 𝜽𝟐 _𝜽_𝒊 𝜶 + 𝜷 𝜽𝟏− 𝜷

Graphicaldetermination of Pa (min.)

Fig.(7.13): Coulomb's Passive Pressure for cohesionless soils.

(a) Retaining wall.

𝜽𝟑

 for semi-analytical solution single trial wedge is needed with 𝜽𝟏= 𝟒𝟓 − ∅/𝟐.

 for graphical solution several trial wedges are needed.

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423

7.6 COMPARISON OF RANKINE’S AND COULOMB’S

EARTH PRESSURE THEORIES

The results obtained from Rankine’s and Coulomb’s earth pressure theories are identical under the same conditions (smooth wall surfaces, level grounds, and homogeneous cohesionless soils) though the two theories are quite differently based.

 Coulomb’s theory is derived according to the principle of force equilibrium. As a result, there is only one failure surface, which is a plane, assuming the wedge between the failure surface and the retaining wall is rigid. Whereas, Rankine’s earth pressure theory is based on the principle of plastic equilibrium of the strained soil (as a result there are infinite failure surfaces within the failure zone).

 Coulomb’s theory is applicable to more complicated conditions than Rankine’s theory,

though it is difficult to obtain a theoretical solution.

 Coulomb’s wedge theory calculates less earth pressure than Rankine’s theory for a level back slope whereas the values converge under back slope conditions when δ = β.

 Coulomb’s theory calculates a unique failure angle for every design condition whereas the application of Rankine’s theory to reinforced soil structures fixes the internal failure plane at (45 + ∅/2).

 Coulomb’s earth pressure theory gives an upper bound estimate or an unsafe solution because it is based on a limit equilibrium analysis which always results in a failure load greater than the true failure load. The main reason for this is that the soil will always be able to choose a failure mechanism that is more efficient than the assumed failure mechanism (shape and location of slip plane). Whereas Rankine’s theory gives a lower bound estimate or safe solution of lateral earth pressure due to it is based on plastic equilibrium states of stresses which usually results in a failure load smaller than the true failure load.

 Coulomb’s active wedge theory and a calculated failure plane is favored by the National Masonry Concrete Association (NCMA). While, the application of Rankine’s "state of stress" earth pressure theory and fixed failure plane is favored by the transportation agencies (AASHTO and FHWA).

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424

7.7 GRAPHICAL METHODS FOR EARTH PRESSURE

CALCULATION OF COHESIONLESS SOIL

7.7.1 CULMANN'S CONSTRUCTION

(a) Active Case:

1. Draw the retaining wall, backfill, etc….. , to a convenient scale, as shown in Fig.(7.14). 2. From point A (the base of the wall) draw two lines; the first line AC inclined at  to the

horizontal, and the second line AD inclined at an angle () or () to AC line; where  is the angle between the backside of the wall and a horizontal line, and



is the angle of wall friction.

3. Draw some possible failure wedges, such as ABC1, ABC2, ABC3 and so on.

4. Compute the respective weights of wedges ABC1, ABC2, ABC3 as w1, w2, w3 and so on.

5. Using a convenient weight scale along line AC, lay off the respective weights of wedges locating points (w1, w2, w3, etc….).

6. Through each point of (w1, w2, w3, etc…), draw a line parallel to the line AD intersecting the corresponding lines AC1, AC2, AC3 at points e1, e2, e3 and so on. Triangle Aw1e1 represents the triangle of forces for the trial wedge ABC1 and w1e1 is the pressure Pa1 on the wall from this wedge.

Fig.(7.14): Active pressure by Culmann's method for cohesionless soils.

Failure plane ∅ 𝜶 − 𝜹 = 𝟗𝟎 − 𝛿 C A C1 𝜽𝟏 C3 C2 D ∅ - line Pressure line e3 e2 e1 w1 w2 w3

PA (Maximum ative pressure)

Tangent e B w 𝜶

w1 𝜶 = 90o

(c) Cantilever retaining wall. Pressure locus (Culmann's line) (a) Gravity retaining wall.

w1 Failure plane ∅ 𝜶 − 𝜹 C A B C1 C2 Pa 𝜹 𝜽𝟏 𝜶 N H H/3 C4 C5 C3 D ∅ - line Pressure line e5 e4 e3 e2 e1 w1 w2 w3 w4 w5 Tangent e (b) Polygon of forces. Pai 𝜶 − 𝜹 𝜽𝒊− ∅ 𝟏𝟖𝟎𝒐_{− 𝜽} 𝒊+∅− 𝜶 + 𝜹 Wi Ri Culmann's line PA (Maximum ative pressure)

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425

7. Repeat steps 1-6 for different planes AC2, AC3, etc. and connect e1, e2, e3, etc. by a smooth curve (Culmann’s line). Through points of intersection determined in step (6), trace a tangent to the curve parallel to AC. Then, the distance PA shown in Fig.(7.14) to the chosen scale represents the active thrust on the wall and the real surface of sliding lies on AeC.

Point of Application of Active Thrust

(i) No Concentrated Load: from the center of gravity (C.G.) of the failure wedge in Fig.(7.15i) draw a line parallel to AC till intersecting the wall face at the point of application.

(ii) Concentrated Load: draw Vc parallel to AC,

V

c



_f parallel to ACF, and take 1/3 distance

c





c

_f



from c (see Fig.(7.15ii)).

(b) Passive Case

The method is the same as that for active case except that the slope line AC is drawn at an angle  below the horizontal (see Fig.(7.16)).

Fig.(7.15): Point of application of active thrust by Culmann's method. 𝜷 A B CF 𝒄′ 𝜹 𝜽𝒊 failure surface ∅ C 𝜶 + 𝜷 x C.G. PA (i) 𝜷 A B CF 𝒄′ V 𝜹 𝜽𝒊 failure surface ∅ C 𝜶 + 𝜷 C'F PA (ii)

(d) Profile of Culmann’s graphical construction. (c) Geometric relationship.

Fig.(7.16): Passive pressure by Culmann's method for cohesionless soils.

(a) Retaining wall.

𝜽𝒊 + ∅ (b) Polygon of forces. Pp 𝜶 + 𝜹 𝟏𝟖𝟎𝒐_{− 𝜽𝒊 − ∅ − 𝜶 − 𝜹} W R 𝜷 A H B C1 Pp W R N 𝜹 𝜽𝒊 D 𝜶 Assumed failure surface ∅ N H/3 𝜶 + 𝜷 𝜽𝒊 − 𝜷

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426

7.7.2 REBHANN'S CONSTRUCTION

Active Case: (for cohesionless backfill, Fig.(7.17))

1. Draw BG at  to the horizontal.

2. Draw BL at

  









or () to BG. 3. Draw AF parallel to earth pressure line BL. 4. Draw semi-circle on BG.

5. Draw a perpendicular to BG from F to meet the semi-circle in (X).

6. With center B and radius BX, draw an arc to meet BG in E. Through E draw a line parallel to BL to meet the ground surface in C. Join BG then BC is the surface of rupture.

7. With E as center and EC as radius, draw an arc to cut BG in K, join CK. 8. Then total active pressure on the wall

) X )( KE .( 2 1 ) KCE ( P_A     …...…...…...(7.24) where  is the unit weight of backfill.

9. Locate the point of application of PA by drawing parallel to final rupture plane from center of gravity of wedge to cut the wall surface at required point.

𝜷 A H B PA X N 𝜹 𝝅 − (𝜶 + 𝜹) E 𝜶 Horizontal ∅ N Slope line H/3 𝜶 + 𝜷

Fig.(7.17): Rebhann's active Pressure for cohesionless.

soils. L F G C // to BL // to BL O x K X Slope line

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427

SOLVED PROBLEMS

Problem (7.1):

Compute the total active pressure per meter length of a retaining wall 10 m high retaining sand having 37and 22.5kN/m3_{up to its top. The backfill has a horizontal top with a} uniform surcharge of 10 kN/m2 _{and the water table is located 4m below it.}

Solution:    37 ; ) 4 2 45 ( tan N 2   ; 4 1 N 1 K_a    3 m / kN . 5 . 12 10 5 . 22     ; Surcharge = 10 kN/m2

Consider 1 m of the retaining wall,

Area diagram

Applicable

unit weight Computation

Lateral thrust (kN) Point of application below top Resultant thrust (kN) ACD  1 2× 22.5 × 42× 1 4 45 2.66 441.25 CFGD  4 × 22.5 ×1 4× 6 135 7.0 DGH ' 1 2× 12.5 × 62× 1 4 56.25 8.0 DHJ w 1 2× 10 × 62 180 8.0 ABEKJD 10 ×1 4× 10 25 5.0

Point of application of resultant thrust below top of the wall:

𝑍̅ 6.98m 25 . 441 ) 5 )( 25 ( ) 8 )( 180 ( ) 8 )( 25 . 56 ( ) 7 )( 135 ( ) 66 . 2 )( 45 (       4m 10 kN/m2 R.W. 6m F G H J K B A 2m 7m _{C D} _E 45 25 180 135 56.25 W.T. kN/m2 kN/m2 kN/m2

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428

Problem (7.2):

A retaining wall has a vertical back of 8 m height. The soil is sandy loam of 17.6 kN/m3 unit weight, and an angle of internal friction of 20o_{. If water table builds up behind the wall to a} level 3 m above the bottom of the wall, calculate the magnitude of the resultant thrust on wall per linear meter. Neglect effect of wall friction and take a horizontal top fill. What is the maximum likely depth of tension cracks that may develop?

Answer:

PA = 179.2 kN/m run of wall at point of application 1.88 m above base.

Problem (7.3):

A retaining wall of 5 m height, has a smooth vertical back, the backfill has a horizontal surface with the top of wall. There is a uniformly distributed surcharge load of 36 kN/m2_{. The} density of the backfill is 18 kN/m3_{, its angle of shear resistance is 30}o_{and the cohesion is} zero. Water table is located at mid height of the wall; calculate the magnitude and point of application of active thrust per meter length of wall.

Answer:

PA = 155.7 kN/m length of the wall at point of application 1.87 m above base.

R.W. 8m 3.02m 17.08 kN/m2 39.22 kN/m2 45 kN/m2 W.T. 3.0m 1.98m 78.12 kN/m2 16.95 26.04

−

36 kN/m2 R.W. 5m D E F G K H A B C J 18750 kN/m2 59940 8320 W.T. Active

P

A 1.87m 31250 kN/m2 37460 kN/m2

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429

Problem (7.4):

Determine the at rest lateral earth pressure per unit length of the wall shown in the figure (a) below. Then, determine the location of the resultant of earth pressure. TakeK_o  1 sin

Solution: K_o 1 sin 1 sin 30 0.50 At point B, 2 2 2 1.70 3.4 / 0 0.5 3.4 1.7 / s o o s t m u p K t m           At point C, 2 2 2 2 1.70 (1.9 1.0) 2 5.2 / 0.5 5.2 2.6 / 2 1 2.0 / s o t m p t m u t m             

Fig.(b) shows the pressure distribution diagram. The diagram has been divided into four parts. Let P P P₁, , ,and ₂ ₃ P₄be the total pressure due to these parts. Thus

1 1 2 2 1 3 2 1 4 2 1.70 2 1.7 2 1.70 3.4 0.9 2 0.9 2.0 2 2.0 Total P=8.0 P t P t P t P t t               

The line of action of P is determined by taking moments about C, 1.7 2.667 3.4 1.0 0.9 0.667 2 0.667 4.53 3.4 0.6 1.33 _1.23 8.0 P Z Z m              (a) 2m A B ∅ = 30o γ = 1.7 ton/m3 C ∅ = 30o γsat.= 1.9 ton/m3 W.T. 2m (1) (2) (3) (4) 1.7m 0.9m 2.0m (b) R.W.

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430

Problem (7.5):

Determine the active earth pressure on the retaining wall shown in the figure (a) below. Solution:

From the equation: 

      1 sin 1 sin a K

For the upper layer,   

  1 sin 35 0.271 1 sin 35 a K

For the bottom layer,   

  1 sin 38 _0.238 1 sin 38 a K At point B, 2 2 2.5 1.7 4.25 / 0 0.271 4.25 1.15 / s a t m u p t m        

Below the interface, pais given by

_{0.238 4.25 1.01 /}  2 a p t m At point C, 2 2 2 2.5 1.70 2.5 0.80 6.25 / 2.5 1 2.5 / 0.238 6.25 1.49 / s a t m u t m p t m            

Fig.(b) shows the pressure distribution.

The forces P P P₁, , ,and ₂ ₃ P₄ are determined from the pressure distribution diagram.

1 1 2 2 1 3 2 1 4 2 2.5 1.15 1.44 2.5 1.01 2.53 2.5 0.48 0.60 2.5 2.5 3.13 Total P=7.70 P t P t P t P t t               

Taking moment about C:

1.44 3.33 2.53 1.25 0.60 0.833 3.13 0.833 _1.44 7.70 Z         m (a) 2.5m A B ∅ = 35o γ = 1.7 ton/m3 C ∅ = 38o γsat. = 1.8 ton/m3 W.T. 2.5m (1) (2) (3) (4) 1.01 0.48 2.5 (b) R.W. 1.15 ton/m2

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431

Problem (7.6):

Determine the active pressure on the wall shown in the figure below using Rankine's theory.

Solution:

From the equation:

                       2 2 2 2 2 2 2 2

cos cos cos

cos

cos cos cos

cos15 cos 15 cos 30

cos15 0.373

cos15 cos 15 cos 30

a

i i

K i

i i

From the equation:

P_a ₂1K H_a 2  1₂ 0.373 19.0 (4)  2 56.7kN The pressure acts at a height of 4/3 m inclined at an angle 15o with horizontal.

Problem (7.7):

Determine the stresses at the top and bottom of the cut shown in the figure below. Also determine the maximum depth of potential crack and the maximum depth of unsupported excavation.

Solution:

From the equation:

P_a K_a..Z2c K_a where, 0.656 12 sin 1 12 sin 1 K_a       Thus, P_a (0.656)(1.80Z)(2)(2) 0.656 1.18Z3.24 At top Z = 0 : P_a 3.24.t/m2 4m ∅ = 30o γ = 19 kN/m3 R.W. 4/3m 15o 15o

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432

At bottom Z = 4: P_a 1.48.t/m2

From the equation of crack depth:

    2  2 2.0 2.745 1.8 0.656 c a c Z m K

From the equation of maximum depth of unsupported excavation:

   4 5.490 c a c H m K

Problem (7.8):

A 5m high retaining wall is shown in the figure below. Determine the Rankine's active pressure on the wall for the following cases:

a. Before the formation of the crack. b. After the formation of the crack. Solution:               1 sin 1 sin 30 _0.333 1 sin 1 sin 30 a a K K

From the equation:

2 0.333 17.5 2 5 0.333 5.83 5.77 a a a p K Z c K Z Z           4m ∅ = 12o C = 2 ton/m2 γ = 1.8 ton/m3 − + 3.24 1.48 − + 5.77 23.38 5m ∅ = 20o C = 5 kN/m2 γ = 17.5 kN/m3 R.W. A B C b c _d 4.01m 0.99m

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433 At top, 2 0 5.77 / a Z p kN m    At point B, 0 5.83 5.77 0 0.99 a p Z Z m     At bottom, 2 5 5.83 5 5.77 23.38 / a Z m p kN m     

Before the formation of the crack:

Negative pressure, P₁  1₂ 0.99 5.77 2.86  kN Positive pressure,

Net, P_a 46.88 2.86 44.02  kN Line of action of Pa is determined as under:

4.01 46.88 2.86 (4.01 0.33) 3 _1.14 44.02 Z      m

After the formation of the crack:

After the formation of the crack, the negative pressure is eliminated. The pressure distribution is given by the area bcd

1

2 23.38 4.01 46.88

a

P     kN act at a height of 4.01/3 m above base.

Alternatively, directly from the equation:

 

2 2 1 2 2 2 1 2 2 2 2(5) 17.5 (5) 0.333 2 5 5 0.333 46.85 17.5 a a a c P H K c H K kN                  1   2 2 4.01 23.38 46.88 P kN

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434

Problem (7.9):

Determine the Rankine’s passive force per unit length of the wall shown in the figure below. Solution:        1 sin 1 sin p K For top layer I,

 

     1 1 sin 30 _3.00 1 sin 30 p K For bottom layer II,

 

_  _   2 1 sin 24 2.37 1 sin 24 p K From the equation:

P_p K_p..Z2c K_p At point A, Z = 0, 𝑃_𝑝 = 0 At point B, Z = 2m, 𝜎̅_𝑠 = 2 x 1.6 = 3.2 t /m2 Top layer, p_p  3 3.2 9.6 / t m2 Bottom layer, p_p 3.2 2.37 2 1.0    2.37 10.66 / t m2 At point C,                 2 2 2 2 1.6 2 (1.9 1.0) 5.0 / 5 2.37 2 1.0 2.37 14.93 / 2 1.0 2 / s p t m p t m u t m

Fig. (b) shows the pressure distribution, Total pressure                 1 2 3 4 1 1 1 2 2 9.60 10.66 2 2 4.27 2 2 2 2 37.19 P P P P P t (a) 2m A B ∅ = 30o C = 0 γ = 1.6 ton/m3 C 2m (1) (2) (3) (4) 10.66 4.27 2.0 (b) R.W. 9.6 ton/m2 I: ∅ = 24o C = 1 ton/m2 γ = 1.9 ton/m3 II: W.T.

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435

Problem (7.10):

Determine the Coulomb’s active force on the retaining wall shown in the figure below. Solution:











 





 













_

 

_{ }



_

                                  _{  } _{  }      _  _           2 2 2 2 2 2 sin sin sin sin sin 1 sin sin sin 75 30 0.548 sin 30 20 sin 30 15 sin 75 sin 75 20 1 sin 75 20 sin 75 15 a K i i

From the equation:

       2 1 2 2 1 2 0.548 1.75 5 11.99 a a P K H t

This will act at a height of 5/3 m, inclined at 20o to normal, in the direction shown in the figure. The reader should note that the direction of Pa is equal and opposite to that on the

wedge.

Problem (7.11):

Check the stability of the gravity retaining wall shown in the figure below, if the allowable soil pressure equals to 60 t/m2_.

γ = 1.9 t/m3 ∅′ = 36o 𝛿 = 24o 5.7m 4.5m Pa 0.5m 0.5m 1.2m 0.7m 1.71m 3.2m 0.4m 0.19m 𝛽 =70o (1) (2) (3) (4) 0.4m 24o 5m i = 15o ∅ = 30o 𝛿 = 20o γ = 1.75 t/m3 𝛽 = 75o R.W. i= 15o 20o 𝛽 Pa

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436 Solution:

From the equation: P_a ₂1K H _a 2

where,

















































                                  _{  } _{  }        _{  } _{  }      2 2 2 2 2 2 sin sin sin sin sin 1 sin sin sin 70 36 0.417 sin 36 24 sin 36 0 sin 70 sin 70 24 1 sin 70 24 sin 70 0 a K i i Therefore, P_a  1₂ 0.417 1.9 5.7  2 12.87t The total pressure acts inclined at 24o to the normal. Horizontal component, P_h P_acos 20



   24



9.26t Vertical component, P_v P_asin(20   24 ) 8.94t

Calculations are shown in the table below. The moments are taken about toe. The clockwise moments are taken as positive.

No. Description

Forces (t) Lever arm

(m)

Moments about toe Vertical Horizontal Clockwise Counter

Clockwise 1 W₁   1₂ 5 0.19 2.40 1.14 0.53 0.6 2 W₂  5 0.5 2.40 6.0 0.84 5.04 3 W3   21 5 1.71 2.40 10.26 1.66 17.03 4 W₄ 3.2 0.7 2.40  5.38 1.60 8.61 5 Pv 8.94 2.39 21.37 6 Ph 9.26 1.90 17.59  31.71 9.26 52.64 17.59 35.05 t-m

Neglecting passive resistance, the factor of safety against shear is given by the equation:

  tan 24 31.72 1.53 (safe) 9.26 v s H R F R

The factor of safety against overturning is obtained from equation:

     52.64 _2.99 17.59 R o o M F M From the equation:

     35.05 _1.10 31.72 M x V

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437

From the equation:

      /2 1.60 1.10 0.50 /6, there is no tension. e b x m As e b

The pressure at the base are determined from equation:

       _  _ _  _            _  _ _  _     2 max 2 min 6 31.72 6 0.5 1 1 19.2 / 3.20 3.2 6 31.72 6 0.5 1 1 0.62 / 3.20 3.2 V e p t m b b V e p t m b b

The factor of safety against bearing capacity failure is given by the equation:

max 60 3.1 (safe) 19.20 na b q F p   

Problem (7.12):

Check the stability of the cantilever retaining wall shown in the figure below. The allowable soil pressure is 50 t/m2_.

Solution:

Let us first ascertain whether Rankin’s theory is applicable to the cantilever retaining wall. From the equation:









1 1 sin 45 /2 sin 2 sin sin 15 45 7.5 17 sin 7.9 sin 34 i i              _ _            _ _    

The shear does not intersect the stem. Therefore, Rankin’s theory can be applied. From the equation: P_a ₂1K H _a 2

γ = 1.8 t/m3 ∅ = 34o 𝛿 = 25o 6.22m 5m Pa 0.4m 1.0m 2.3m 3.5m 0.6m 0.2m (2) (1) (4) (3) 15o 0.6m 𝛽 𝑖 =15o 𝜂 _P_h Pv (5) 6.22 𝐊𝐚 γ

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438

From the equation:

2 2

cos cos cos

cos

cos cos cos

cos15 cos 15 cos 34

cos15 0.311

cos15 cos 15 cos 34

a i i K i i i                        Therefore, 1₂ 0.311 1.80 (6.22)2 10.83 sin 15 2.80 sin 15 10.46 a v a h a P t P P t P P t            No. Description

Forces (ton) Lever arm

(m)

Moments about toe Vertical Horizontal Clockwise Counter Clockwise 1 W1 0.4 5.0 2.4  4.80 1.00 4.80 2 𝑊₂ = (0.2)(5)(2.4)/2 1.20 0.73 0.88 3 W3 0.6 3.50 2.40  5.04 1.75 8.82 4 W4 2.3 5.0 1.80  20.70 2.35 48.65 5 𝑊₅ = 0.62)(2.3)(1.8)/2 1.28 2.73 3.50 6 Pv 2.80 3.50 9.80 7 Ph - 10.46 2.07 - 21.65  35.82 10.46 76.45 21.65

Factor of safety against sliding is:

  tan 25 35.82 1.60 (safe) 10.64 v s H R F R

Factor of safety against overturning is:

    76.45 _{3.53 (safe)} 21.65 R o o M F M      76.45 21.65 _1.53 35.12 M x m V e b /2 x 1.75 1.53 0.22  m b /6 2 max 2 min 35.82 ₁ 6 0.22 _{14.12 /} 3.50 3.50 35.82 ₁ 6 0.22 _{6.34 /} 3.50 3.50 p t m p t m     _  _       _  _  

Factor of safety against bearing capacity failure is:

   max 50 _{3.54 (safe)} 14.12 na b q F p

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439

PROBLEMS

P7.1 Determine the passive pressure per unit run for a retaining wall 4m in height; with

 

 _{15 ,}  _{30 ,and} _{1.90 /} 3

i t m as shown in Fig.(7.18). The back face of the wall

is smooth and vertical.

P7.2 For the retaining wall of problem (P7.1), determine the active pressure per unit run.

P7.3 Determine active and passive pressures, using Coulumb’s theory, on the wall shown in Fig.(7.18).

P7.4 A retaining wall has a vertical back of 8m height. The back face of the wall is smooth and the upper surface of the fill is horizontal. Determine the thrust on the wall per unit length. Takec1.0 /t m2, 1.8 /t m3 and  20 . Neglect tension.

P7.5 A retaining wall with a vertical smooth back face of 8m height. The wall supports a cohesionless soil( 1.90 /t m3, 30 ). The surface of the soil is horizontal. Determine the thrust on the wall.

P7.6 Check the overall stability of the cantilever retaining wall shown in Fig.(7.19). 4m γ = 1.9 t/m3 ∅ = 30o 𝛿 = 20o i= 15o 𝛽 = 80o Fig.(7.18) γ = 18 kN/m3 ∅ = 40o 𝛿 = 25o 4m 0.3m 1.0m 1.9m 2.8m 0.45m 0.45m (2) (1) (4) (3) 0.45m (5) Surcharge 50 kN/m2 Fig.(7.19)

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440

REFERENCES

Bell, A. L. (1915). “The lateral pressure and resistance of clay and the supporting power of clay foundations”, in A Century of soil mechanics, ICE, London, pp. 93-134.

Bjerrum, L. and Andersen, K. (1972). “In-situ measurement of lateral pressures in clay”, in Proc. 5th European Conference SMFE, Madrid, Vol.1, Spanish Society SMFE, Madrid, pp. 11–20.

Caquot, A., and Kerisel, J. (1949).“Tables for the calculation of passive pressure, active pressure, and the bearing capacity of foundations”, Gauthier-Villars, Paris.

Janbu, N. (1957). “Earth pressures and bearing capacity calculations by generalized procedure of slices”, in Proc., IV Int. Conf. Soil Mech. Found. Eng., London, 2, 207. Kerisel, J. and Absi, E. (1990). “Active and passive earth pressure tables”, 3rd. edition, A.A.

Balkema, Rotterdam.

Mazindrani, Z. H., and Ganjali, M. H. (1997),“Lateral Earth Problem of Cohesive Backfill with Inclined Surface,” Journal of Geotechnical and Geoenvironmental Engineering, ASCE, Vol. 123, No. 2, 110 –112.

NAVFAC (1982a). “DM-7.2, Foundations and earth structures”, U.S. Department of the Navy, Naval Facilities Engineering Command, 200 Stovall Street, Alexandria, VA 22332, P. 7.2-209.