# Ch7 Lateral Earth Pressure Theories (407-440)

November 18, 2017 | Author: rafi | Category: Stress (Mechanics), Soil, Materials Science, Physics & Mathematics, Physics

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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 = K o σ′1 = K o 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

Fig.(7.1): Elastic state of equilibrium in soil.

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

Foundation for Civil Engineers

Chapter 7: Earth Pressure Theories

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).

Failure envelope

Failure (Active state)

Active State Wall movement away from backfill

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

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

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.

Upward movement of wedge B

R.W.

Resisting force

Passive State Wall movement towards backfill

’v = remains the same ’h = ’x = increases till failure occurs

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

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

At rest earth pressure Active earth pressure Passive earth pressure

G.S.

Active Case

At rest Case

Passive Case

(Wall moves away from soil)

(No movement)

(Wall moves into soil)

Fig.(7.4): Wall movement.

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.

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

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. Table (1): Magnitudes of wall movement to reach failure (after NAVFAC DM7.2, 1982).

Soil Type Dense sand Loose sand Stiff clay Soft clay

Value of Y/H* Active

Passive

0.0005 0.002 0.01 0.02

0.002 0.006 0.02 0.04

* Y is the movement of the wall top and H is the height of the wall.

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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: P  K o   h   1  sin  …………………………….….………...…(7.3)  Pv  at .rest  for overconsolidated soils: Ko = (1 –sin ') OCRsin………..………………….……..………..…(7.4)  from elastic analysis: 𝜇 Ko = 1−𝜇 …………………………………………...…….….……...…(7.5)

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

P   K a   h   tan 2 (45  ) 2  Pv active

or

Ka 

1  sin  ..………………(7.6) 1  sin 

P   K P   h   tan 2 (45  ) or 2  Pv Passive

KP 

1  sin  ...….......………(7.7) 1  sin 

 Inclined backfill:

K a  cos 

K P  cos 

cos   cos 2   cos 2  cos   cos 2   cos 2  cos   cos 2   cos 2  cos   cos 2   cos 2 

………..……….……….....….…….(7.8)

……..…………….…….....….…….(7.9)

As shown above for Rankine earth pressure theory:

K p  1 / K a …….…....………..……….….……………………...........(7.10)

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

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

KP 

sin 2 (  )  sin(  ) sin(  )  sin 2 . sin(  ) 1   sin(  ) sin(  )  

2

sin 2 (  )  sin(  ) sin(  )  sin 2 . sin(  ) 1   sin(  ) sin(  )  

2

…………....……….(7.11)

…………....….…....(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.

𝛽

∅ = soil internal friction angle.

 for active state:

R.W.

Backfill

𝑚𝑖𝑛𝑜𝑟 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑠𝑡𝑟𝑒𝑠𝑠

Ka = Ph / Pv = 𝑚𝑎𝑗𝑜𝑟 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑠𝑡𝑟𝑒𝑠𝑠

𝛿

 for passive state: Ka = Ph / Pv =

𝛼

𝑚𝑎𝑗𝑜𝑟 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 𝑚𝑖𝑛𝑜𝑟 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑠𝑡𝑟𝑒𝑠𝑠

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

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

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

Cohesionless soil

Cohesive soil

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).

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Relativly large  Small 

Ka

KP

Ko Movement

Away from backfill

Against backfill

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

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 onethird 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 ∅.

G.S.

G.S.

Resultant

Resultant R.W.

R.W.

H/3

H/3

(a) Levelled backfill

(b) sloped backfill

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.

z

k a ..z

H

PA

R.W.

H/3

k a ..H

Pa .at..base  Ka ..H ….acting at H/3 from base  2

where, K a  tan 2 (45  ) 

1  sin  1  sin 

Case (2): Submerged Backfill (a) Fully Submerged

(b) Partially Submerged W.T.

G.S. H1

H

Pw

PA

R.W.

R.W.

H/3

H/3

k a ..H1

H2

k a . .H  w .H

k a ..H1 k a . .H 2

Pa .at..base  K a . .H   w .H

 w .H 2

Pa .at..base  K a (.H1   .H 2 )   w .H 2

Case (3): Backfill with Uniform Surcharge q/unit area

k a .q

H R.W.

k a ..H The effect of the surcharge of intensity q is the same as that of a fill of height equal to q/ q above the ground surface. K a . .  K a .q  414

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Case (4): Dry Backfill with Sloping Surface Considering the soil element shown in Fig.(7.9) with  and Pa are resultant stresses on two conjugate planes. The principle stresses 1 and 3 are determined from Mohr circle as:

G.S.

PA  R.W. R.W.

H/3

b cos 

 Active condition OA1  Pa

PA

failure envelope

OA 2  

A1

PA

3

B C

A2

1

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

By simple geometry, it can be shown that:   3   3 Pa  OA1  OB  BA1  1 cos   1 sin 2   sin 2  .....(7.13a) 2 2

  OA 2  OB  BA 2  But,



1  3   cos  1 3 sin 2   sin 2  ...…(7.13b) 2 2

.z.b cos   .z. cos  . b

Dividing (7.13a by 7.13b) and substituting for ; gives: Pa  .z. cos 

cos   cos 2   cos 2  cos   cos 2   cos 2 

;

or

Pa  K a ..z

The resultant active thrust on the wall of height H is given by: PA 

where,

1 K a ..H 2 ……..…….……………………......…………………(7.14) 2

K a  cos 

cos   cos 2   cos 2  cos   cos 2   cos 2 

Similarly: K p  cos 

cos   cos 2   cos 2  cos   cos 2   cos 2 

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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. G.S. W.T.

  R.W. R.W.

H/3

PA 

where, K a  cos 

PA

Pw H/3

1 K a ..H 2 ……....……………………..….…..……(7.15a) 2

cos   cos 2   cos 2  cos   cos 2   cos 2  Pw 

1  w H 2 ……....………...….………........…………(7.15b) 2

Case (6): Inclined Back of the Wall (b) Sloped Surface   0

(a) Horizontal Surface   0 W (weight of soil wedge ABC) B

C

G.S.

W

G.S.

B

C

P

P

PA R.W.

R.W.

H/3

PA H/3

A

A

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  3 N  2c N ...……….…………........…………(7.16) 416

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1 2c   where N  tan2 (45  ) 2 N N

or

3 

here,

1  Pv  .z and therefore 3  Ph  .z.Ka  2c Ka

or N 

1 Ka

 2c K a

zo  H R.W.

2c

A

 Ka

B

G.S.

4c

C

 Ka

+

zo = zone of

(net Ph is zero)

z

Pv  .z  1

Ph  3 PA

tension cracks

(Active Case) D

E

F

.H.K a

at z = 0:

Ph  2c K a

when, Ph  0 :

z  zo 

and

2c

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

 Ka

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 2z o  stand with a vertical face up to a depth

4c  Ka 4c  Ka

is zero. This means that a cohesive soil could without any lateral support. Thus the critical

height, Hc of unsupported vertical cut in cohesive soil is given by: For c-  Soil:

H c  2z o 

For Soft Clays   0 :

Hc 

4c  Ka

. ...…..……...…………….….....………(7.18a)

4c ...…...….…….……………..…...…….....………(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: PA 

acts above the wall base at

 1 2c .H.K a  2c K a . H   2  Ka 

1  2c H 3  Ka 

 .  

417

  …..………...…….…...(7.19)  

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Chapter 7: Earth Pressure Theories

(b) Cohesive Soil with Surcharge If there is a uniform surcharge q/unit area, then the lateral pressure is increased by q K a everywhere (diagram AGHF). q / unit.area

A

q z

G

B

Surcharge diagram

C

Kaq

H

q.K a

R.W.

+ E

D

F

H

.H.K a

(PH )active  .HKa  2c Ka  q.Ka …….…………..…..…...(7.20)

Depending on q K a magnitude, the depth of tension cracks is altered. If q K a > 2c K a then there is no tension cracking.

Case (8): Passive Earth Pressure in Cohesive Backfill From Eq.(7.16):

1  3 N  2c N ….……………….…………...…...(7.21) 3  (Pv ) passive  .z and

for passive case:

1  ( Ph ) passive  ( .z.K P )  (2c. K P ) 2c K P

A

G.S.

B

z

Pv  .z  3

Ph  .z.k P  1

P1

H R.W.

P2 (Passive Case) C

D

.H.K P

E

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:

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         

Chapter 7: Earth Pressure Theories

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).

(a) Local active failure

(b) Local passive failure

(c) Mohr stress diagram

Fig.(7.10): Slip planes for active and passive earth pressures.

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:

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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.

C

Wall movement away from soil

B

𝜽−𝜷

𝜷

𝜶+𝜷

Pa

W

𝜶−𝜹 𝟏𝟖𝟎𝒐 − 𝜶 − 𝜽 + ∅ + 𝜹

O

H

O’ Assumed failure surfaces

D N 𝜹 H/3 Pa

Actual failure surface

𝜸 ∅ 𝒄=𝟎

∅ 𝜶

R

R

𝜽

N

W 𝜽−∅

A (a) Retaining wall with single trial wedge.

(b) Polygon of forces.

Fig.(7.11): Coulomb's Active Pressure (Semi-Analytical Solution).

The weight of the soil wedge ABC is 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

[

sin(α + β)

2 sin2 α 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 − ∅)

Pa =

=

W sin(180 − α − θ + ∅ + δ)

W sin(θ − ∅)

……......……………………………...…...(7.22c)

sin(180− α− θ + ∅ + δ)

Combining Eqs. (7.22b) and (7.22c) gives: 420

Foundation for Civil Engineers γ H2

Pa =

Chapter 7: Earth Pressure Theories

[

2 sin2 α

sin(α+β) sin(θ − β)

sin(α + θ)]

sin(θ − ∅)

…....…...(7.22d)

sin(180− α− θ + ∅ + δ)

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: 1

Pa = 2 γ H 2 K a …..……………………………...............................…...(7.22e) where, Ka 

sin 2 (   )  sin(  ) sin(  )  sin 2 . sin(  ) 1   sin(  ) sin(  )  

2

……..................…...(7.22f)

If 𝛽 = 𝛿 = 0 and 𝛼 = 90𝑜 (a smooth vertical wall with horizontal backfill), Eq.(7.22f) simplifies to: Ka 

(1  sin )   tan2 (45  ) ……….............................................…...(7.22g) (1  sin ) 2

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). Graphical determination of Pa (max.)

Pa (max.)

P2 P1

P3

𝜽𝟏 Wall movement away from soil

B

Pa

W

H

𝜶−𝜹 Assumed failure surfaces

D

R

Actual failure surface

𝜸 ∅ 𝒄=𝟎

∅ 𝜽𝟏

𝟏𝟖𝟎 − 𝜶 − 𝜽𝒊 + ∅ + 𝜹 𝒐

O’

𝜶

𝜽𝒊

𝜽𝟏 − 𝜷

O

N 𝜹 H/3 Pa

C3

C2

C1

𝜷

𝜶+𝜷

𝜽𝟑

𝜽𝟐

𝜽𝟐

R

N

A (a) Retaining wall with several trial wedges.

𝜽𝒊 − ∅

(b) Polygon of forces.

Fig.(7.12): Coulomb's Active Pressure (Graphical Solution). 421

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Chapter 7: Earth Pressure Theories

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: γ H2

W=

2 sin2 α

[

sin(α+β)

sin(α + θ)] ….……..…….…....………...…...(7.23a)

sin(θ−β)

The passive force Pp is calculated from force polygon shown in Fig.(7.13b) as: Pp = Setting

𝑑𝑃𝑝 𝑑𝜃

W sin(θ+ ∅) sin(180−α−θ−∅−δ)

.……....…….…………………………...…...(7.23b)

= 0 gives the minimum value of Pp as: 1

Pp = 2 γ H 2 K p …..………………..…………….............................…...(7.23c) where, Kp 

sin 2 (   )  sin(  ) sin(  )  sin 2 . sin(  ) 1   sin(  ) sin(  )  

2

……..................…...(7.23d)

For smooth vertical wall with horizontal backfill (𝛽 = 𝛿 = 0 and 𝛼 = 90𝑜 ) Eq.(7.23d) simplifies to: Kp 

(1  sin )   tan2 (45  ) …...…...............................................…...(7.23e) (1  sin ) 2

which is identical with that of Rankine’s passive earth pressure coefficient. Graphical determination of Pa (min.)

P1 Pp (min.)

𝜽𝟏

Wall movement toward the soil

B

H

𝜽𝟑

𝜽𝟐

C3

C2

C1

𝜽𝒊

𝜽𝟏 − 𝜷

𝜷

𝜶+𝜷

P3 P2

W Assumed failure surfaces

Pp N H/3

∅ 𝜶

𝜽𝟏

Pp

𝟏𝟖𝟎𝒐 − 𝜶 − 𝜽𝒊 − 𝜶 − 𝜹

D

𝜹

𝜽𝟐

N

R

R

𝜸 ∅ 𝒄=𝟎

A (a) Retaining wall.

W 𝜽𝒊 + ∅

(b) Polygon of forces.

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

𝜶+𝜹

Fig.(7.13): Coulomb's Passive Pressure for cohesionless soils. 422

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7.6 COMPARISON OF RANKINE’S EARTH PRESSURE THEORIES

AND COULOMB’S

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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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….). C

B

C4

C3

C2

C1

C5

Tangent

Failure plane

e5 e4

w1

N 𝜹 H/3 Pa

w4

e e2

Pai

w5

e3 H

- line

𝜶−𝜹

𝟏𝟖𝟎𝒐 − 𝜽𝒊 + ∅ − 𝜶 + 𝜹

w3

Culmann's line

e1 PA (Maximum ative pressure)

w2 𝜶

𝜽𝟏

w1

𝜶−𝜹

𝜽𝒊 − ∅

A (a) Gravity retaining wall. Pressure line

C1

B

(b) Polygon of forces.

D

Failure plane

Wi

Ri

C

Tangent

C3

C2

- line

e3 e2 w1 e1

w

w3

e w2

𝜶

w1 𝜽𝟏

𝜶 = 90o

Pressure locus (Culmann's line)

PA (Maximum ative pressure) 𝜶 − 𝜹 = 𝟗𝟎 − 𝛿

A Pressure line

D

(c) Cantilever retaining wall.

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

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.

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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, Vcf parallel to ACF, and take 1/3 distance c  cf from c (see Fig.(7.15ii)).

V

CF

CF B

B

𝜷

𝜶 +𝜷

𝒄

𝜷

𝜶 +𝜷

x C.G.

failure surface

failure surface

𝒄′

C

C

𝜹

𝜹

PA

PA

(i)

𝜽𝒊

C'F 𝜽𝒊

(ii)

A

A

Fig.(7.15): Point of application of active thrust by Culmann's method.

(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)). C1 𝟏𝟖𝟎𝒐 − 𝜽𝒊 − ∅ − 𝜶 − 𝜹

B

𝜷

𝜶+𝜷

H

Pp

𝜽𝒊 − 𝜷

W

𝜶+𝜹

Assumed failure surface

Pp 𝜹

N H/3

D ∅ 𝜶

𝜽𝒊

N

R 𝜽𝒊 + ∅

A (b) Polygon of forces.

(a) Retaining wall.

(c) Geometric relationship.

W

R

(d) Profile of Culmann’s graphical construction.

Fig.(7.16): Passive pressure by Culmann's method for cohesionless soils. 425

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Chapter 7: Earth Pressure Theories

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

PA  ( KCE) 

1 .( KE)( X) …...…...........................................…...(7.24) 2

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. G C A

// to BL

𝜷

Slope line

x

F

// to BL

H

E

X

𝜶+𝜷

O

PA N

𝜹

K H/3

𝜶

B

N

𝝅 − (𝜶 + 𝜹)

Horizontal

Slope line L X

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

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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   22.5 kN/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.

10 kN/m2 B

A 2m

4m

C D

7m

R.W.

45 kN/m2

W.T.

E 135

25 kN/m2

56.25

180 kN/m2

6m

F

G

H

J

K

Solution:

 1 1    37 ; N  tan 2 (45  )  4 ; K a  2 N 4    22.5  10  12.5.kN / m3 ;

Surcharge = 10 kN/m2

Consider 1 m of the retaining wall,

Area diagram

Applicable unit weight

ACD

CFGD

DGH

'

DHJ

w

ABEKJD

Computation 1 1 × 22.5 × 42 × 2 4 1 4 × 22.5 × × 6 4 1 1 × 12.5 × 62 × 2 4 1 2 × 10 × 6 2 1 10 × × 10 4

Lateral thrust (kN)

Point of application below top

45

2.66

135

7.0

56.25

8.0

180

8.0

25

5.0

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

(45)(2.66)  (135)(7)  (56.25)(8)  (180)(8)  (25)(5)  6.98m 𝑍̅  441.25

427

Resultant thrust (kN)

441.25

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Chapter 7: Earth Pressure Theories

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? 17.08 kN/m2

8m

1.98m

3.02m

39.22 kN/m2 26.04

R.W.

W.T.

78.12 kN/m2

3.0m

45 kN/m2

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 30o 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.

36 kN/m2 H

A

18750 kN/m2 W.T. 5m

R.W.

B

C

PA

J 59940 37460 kN/m2 8320 31250 kN/m2

1.87m

D

E

F

G

K

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

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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. Take Ko  1  sin   Solution:

Ko  1  sin    1  sin 30  0.50

 s  2  1.70  3.4 t / m2

At point B,

u0 po  K o s  0.5  3.4  1.7 t / m2

 s  2  1.70  (1.9  1.0)  2  5.2 t / m2

At point C,

po  0.5  5.2  2.6 t / m2 u  2  1  2.0 t / m2 A ∅ = 30o γ = 1.7 ton/m3 B

2m

W.T.

∅ = 30o γsat. = 1.9 ton/m3

R.W.

(1)

2m

(2) (3)

C 1.7m

0.9m

(4) 2.0m

(a) (b) Fig.(b) shows the pressure distribution diagram. The diagram has been divided into four parts. Let P1 , P2 , P3 ,and P4 be the total pressure due to these parts. Thus P1  21  1.70  2  1.7 t

P2  2  1.70  3.4 t P3  21  0.9  2  0.9 t P4  21  2.0  2  2.0 t Total P=8.0 t The line of action of P is determined by taking moments about C, P Z  1.7  2.667  3.4  1.0  0.9  0.667  2  0.667 4.53  3.4  0.6  1.33 Z  1.23 m 8.0

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Problem (7.5): Determine the active earth pressure on the retaining wall shown in the figure (a) below. Solution:

From the equation: For the upper layer, For the bottom layer,

1  sin   1  sin   1  sin 35 Ka   0.271 1  sin 35 1  sin 38 Ka   0.238 1  sin 38

Ka 

 s  2.5  1.7  4.25 t / m2

At point B,

u0 pa  0.271  4.25  1.15 t / m2 Below the interface, pa is given by

pa  0.238  4.25  1.01 t /m2

 s  2.5  1.70  2.5  0.80  6.25 t / m2

At point C,

u  2.5  1  2.5 t / m2 pa  0.238  6.25  1.49 t / m2 Fig.(b) shows the pressure distribution. A

B R.W.

∅ = 35o 2.5m γ = 1.7 ton/m3 W.T.

(1)

∅ = 38o γsat. = 1.8 ton/m3

(2)

2.5m

1.15 ton/m2

(3)

C 1.01

(a)

0.48

(4) 2.5

(b) The forces P1 , P2 , P3 ,and P4 are determined from the pressure distribution diagram. P1  21  2.5  1.15  1.44 t

P2  2.5  1.01  2.53 t P3  21  2.5  0.48  0.60 t P4  21  2.5  2.5  3.13 t Total P=7.70 t Taking moment about C:

Z

1.44  3.33  2.53  1.25  0.60  0.833  3.13  0.833  1.44 m 7.70

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Problem (7.6): Determine the active pressure on the wall shown in the figure below using Rankine's theory.

15o

∅ = 30o γ = 19 kN/m3 15o

R.W.

4m

4/3m

Solution:

From the equation: K a  cos i 

cos i  cos 2 i  cos 2   cos i  cos 2 i  cos 2  

 cos 15

cos 15  cos 2 15  cos 2 30 cos 15  cos 2 15  cos 2 30

 0.373

From the equation:

Pa  21 Ka H 2  21  0.373  19.0  (4)2  56.7 kN 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:

Pa  K a ..Z  2c K a 1  sin12  0.656 1  sin12

where,

Ka 

Thus,

Pa  (0.656 )(1.80 Z)  (2)( 2) 0.656  1.18 Z  3.24

At top Z = 0 :

Pa  3.24.t / m 2

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Chapter 7: Earth Pressure Theories

Pa  1.48.t / m 2

At bottom Z = 4:

From the equation of crack depth: Zc 

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

From the equation of maximum depth of unsupported excavation: Hc 

4c  5.490 m  Ka 3.24

4m

∅ = 12o C = 2 ton/m2 γ = 1.8 ton/m3

+ 1.48

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:

Ka 

1  sin   1  sin 30  Ka   0.333 1  sin   1  sin 30

From the equation: p a  K a Z  2c  K a  0.333  17.5Z  2  5 0.333  5.83Z  5.77 5.77

A

− B

∅ = 20 C = 5 kN/m2 3 5m γ = 17.5 kN/m

0.99m

b

o

4.01m

R.W.

+ C

c

d 23.38

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At top,

Chapter 7: Earth Pressure Theories

Z0 pa  5.77 kN / m2

At point B,

pa  0 5.83Z  5.77  0 Z  0.99 m

At bottom,

Z  5m pa  5.83  5  5.77  23.38 kN / m2

Before the formation of the crack: Negative pressure, P1  21  0.99  5.77  2.86 kN Positive pressure, P2  21  4.01  23.38  46.88 kN Net, Pa  46.88  2.86  44.02 kN Line of action of Pa is determined as under:

Z

46.88 

4.01  2.86  (4.01  0.33) 3  1.14 m 44.02

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

Pa  21  23.38  4.01  46.88 kN act at a height of 4.01/3 m above base. Alternatively, directly from the equation:

Pa   H K a  2cH K a  1 2

2

2  c 

2

 21  17.5  (5)2  0.333  2  5  5 0.333 

433

2(5)2  46.85 kN 17.5

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Chapter 7: Earth Pressure Theories

Problem (7.9): Determine the Rankine’s passive force per unit length of the wall shown in the figure below. Solution:

Kp 

1  sin   1  sin  

K 

1  sin 30  3.00 1  sin 30

1  sin 24  2.37 1  sin 24

For top layer I, p

1

For bottom layer II,

K  p

2

From the equation: Pp  K p . .Z  2c K p

At point A, Z = 0, At point B, Z = 2m, Top layer,

𝑃𝑝 = 0 𝜎̅𝑠 = 2 x 1.6 = 3.2 t /m2 pp  3  3.2  9.6 t /m2

Bottom layer,

pp  3.2  2.37  2  1.0  2.37  10.66 t /m2

At point C,

 s  2  1.6  2  (1.9  1.0)  5.0 t / m2 pp  5  2.37  2  1.0  2.37  14.93 t / m2 u  2  1.0  2 t / m2 A o I: ∅ = 30

B

C =0 γ = 1.6 ton/m3

2m W.T.

II: ∅ = 24

9.6 ton/m2 (1)

o

R.W. C

C = 1 ton/m2 γ = 1.9 ton/m3

2m

(2) (3) 10.66

(4) 2.0

4.27

(a)

(b) Fig. (b) shows the pressure distribution, Total pressure

P  P1  P2  P3  P4  21  2  9.60  10.66  2  21  4.27  2  21  2  2  37.19 t

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Chapter 7: Earth Pressure Theories

Problem (7.10): Determine the Coulomb’s active force on the retaining wall shown in the figure below. Solution:

Ka 

sin 2        sin      sin    i   sin  sin      1   sin      sin    i    sin 2  75  30 

2

2

 sin  30  20  sin  30  15   sin 2 75 sin  75  20  1   sin  75  20  sin  75  15    From the equation: Pa  21 K a H 2

2

 0.548

 21  0.548  1.75  52  11.99 t i=

15o

Pa 20o

R.W.

5m

𝛽

i = 15o ∅ = 30o 𝛿 = 20o γ = 1.75 t/m3 𝛽 = 75o

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. 0.5m γ = 1.9 t/m3 ∅′ = 36o 𝛿 = 24o

4.5m

Pa (2)

(3)

5.7m

24o

(1)

0.4m

0.5m 0.19m

𝛽 = 70o

1.71m

(4)

3.2m 435

0.4m

1.2m 0.7m

Foundation for Civil Engineers

Chapter 7: Earth Pressure Theories

Solution:

From the equation: Pa  21 Ka H 2 where, K a 

sin 2      

 sin      sin    i   sin  sin      1   sin      sin    i    sin 2  70  36 

2

2

 sin  36  24  sin  36  0   sin 2 70 sin  70  24  1   sin  70  24  sin  70  0   

2

 0.417

Therefore, Pa  21  0.417  1.9  5.7 2  12.87 t The total pressure acts inclined at 24o to the normal. Horizontal component, Ph  Pa cos  20  24  9.26 t Vertical component, Pv  Pa sin(20  24)  8.94 t Calculations are shown in the table below. The moments are taken about toe. The clockwise moments are taken as positive. Forces (t) No.

Description

1 2 3 4 5 6

W1  21  5  0.19  2.40 W2  5  0.5  2.40 W3  21  5  1.71  2.40 W4  3.2  0.7  2.40 Pv Ph

1.14 6.0 10.26 5.38 8.94

31.71

Vertical Horizontal

9.26

Lever arm (m)

0.53 0.84 1.66 1.60 2.39 1.90

9.26

Moments about toe Counter Clockwise Clockwise

0.6 5.04 17.03 8.61 21.37 17.59 52.64 17.59 35.05 t-m

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

Fs 

 Rv RH

tan 24 31.72  1.53 (safe) 9.26

The factor of safety against overturning is obtained from equation:

Fo  From the equation: x

 MR 52.64   2.99  Mo 17.59  M 35.05   1.10  V 31.72

436

Foundation for Civil Engineers

Chapter 7: Earth Pressure Theories

From the equation: e  b /2  x  1.60  1.10  0.50 m

e  b /6, there is no tension.

As

The pressure at the base are determined from equation: V  6 e  31.72  6  0.5  2 1   1   19.2 t / m b  b  3.20  3.2  V  6 e  31.72  6  0.5  2  1   1   0.62 t / m b  b  3.20  3.2 

pmax  pmin

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

Fb 

qna 60   3.1 (safe) pmax 19.20

Problem (7.12): Check the stability of the cantilever retaining wall shown in the figure below. The allowable soil pressure is 50 t/m2.

0.4m

𝑖 =15o (5)

γ = 1.8 t/m3 ∅ = 34o 𝛿 = 25o Pv

5m (4)

(1) 0.6m

Pa

6.22m

15o 𝜂

(2)

Ph

𝛽

1.0m 0.6m

0.2m

(3)

3.5m

2.3m 6.22 𝐊 𝐚 γ

Solution:

Let us first ascertain whether Rankin’s theory is applicable to the cantilever retaining wall. From the equation:  sin i      45  i /2    sin 1   2  sin   

 sin 15    45  7.5   17  sin 1    7.9  sin 34  The shear does not intersect the stem. Therefore, Rankin’s theory can be applied. From the equation: Pa  21 Ka H 2 437

Foundation for Civil Engineers

Chapter 7: Earth Pressure Theories

From the equation:

K a  cos i 

cos i  cos 2 i  cos 2   cos i  cos 2 i  cos 2  

 cos 15 Therefore,

cos 15  cos 2 15  cos 2 34 cos 15  cos 2 15  cos 2 34

 0.311

Pa  21  0.311  1.80  (6.22)2  10.83 t Pv  Pa sin 15  2.80 t Ph  Pa sin 15  10.46 t Forces (ton)

No.

Description

1 2 3 4 5 6 7

W1  0.4  5.0  2.4 𝑊2 = (0.2)(5)(2.4)/2 W3  0.6  3.50  2.40 W4  2.3  5.0  1.80 𝑊5 = 0.62)(2.3)(1.8)/2 Pv Ph 

Vertical

Horizontal

4.80 1.20 5.04 20.70 1.28 2.80 35.82

10.46 10.46

Lever arm (m)

1.00 0.73 1.75 2.35 2.73 3.50 2.07

Factor of safety against sliding is:

Fs 

 Rv RH

tan 25 35.82  1.60 (safe) 10.64

Factor of safety against overturning is:

 MR 76.45   3.53 (safe)  Mo 21.65  M 76.45  21.65 x   1.53 m V 35.12

Fo 

e  b /2  x  1.75  1.53  0.22 m  b /6

35.82  6  0.22  2 1   14.12 t / m 3.50  3.50  35.82  6  0.22  2  1   6.34 t / m 3.50  3.50 

pmax  pmin

Factor of safety against bearing capacity failure is: q 50 Fb  na   3.54 (safe) pmax 14.12

438

4.80 0.88 8.82 48.65 3.50 9.80 76.45

Counter Clockwise

21.65 21.65

Foundation for Civil Engineers

Chapter 7: Earth Pressure Theories

PROBLEMS P7.1

Determine the passive pressure per unit run for a retaining wall 4m in height; with i  15,    30,and   1.90 t / m3 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. Take c  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).

Surcharge 50 kN/m2 0.3m

i=

15o

γ = 1.9 t/m3 ∅ = 30o 𝛿 = 20o

(1)

4m

(5)

γ = 18 kN/m3 ∅ = 40o 𝛿 = 25o (4)

𝛽 = 80o

0.45m

4m

(2)

1.9m 1.0m 0.45m

Fig.(7.18)

(3)

2.8m

Fig.(7.19)

439

0.45m

Foundation for Civil Engineers

Chapter 7: Earth Pressure Theories

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.

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