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TOPIC 6 LATERAL EARTH PRESSURE. Course : S0705 – Soil Mechanic Year : 2008. CONTENT. RETAINING EARTH WALL (SESSION 21-22 : F2F) RANKINE METHOD (SESSION 21-22 : F2F) ACTIVE LATERAL PRESSURE PASSIVE LATERAL PRESSURE COULOMB METHOD (SESSION 23-24 : OFC) ACTIVE LATERAL PRESSURE

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topic 6 lateral earth pressure

TOPIC 6 LATERAL EARTH PRESSURE

Course : S0705 – Soil Mechanic

Year : 2008

content
CONTENT
  • RETAINING EARTH WALL (SESSION 21-22 : F2F)
  • RANKINE METHOD (SESSION 21-22 : F2F)
    • ACTIVE LATERAL PRESSURE
    • PASSIVE LATERAL PRESSURE
  • COULOMB METHOD (SESSION 23-24 : OFC)
    • ACTIVE LATERAL PRESSURE
    • PASSIVE LATERAL PRESSURE
  • LATERAL PRESSURE DUE TO EXTERNAL LOAD (SESSION 23-24 : OFC)
  • DYNAMIC EARTH PRESSURE (SESSION 23-24 : OFC)
retaining earth wall
RETAINING EARTH WALL

Defined as a wall that is built to resist the lateral pressure of soil – especially a wall built to prevent the advance of a mass of earth/soil

earth lateral pressure
EARTH LATERAL PRESSURE
  • Defined as soil stress/pressure at horizontal direction and a function of vertical stress
  • Cause by self weight of soil and or external load
  • 3 conditions :
    • Lateral Pressure at Rest
    • Active Lateral Pressure
    • Passive Lateral Pressure
earth lateral pressure1

Case I

EARTH LATERAL PRESSURE

Lateral Pressure at rest

earth lateral pressure2

Case II

EARTH LATERAL PRESSURE

Active Lateral Pressure

earth lateral pressure3

Case III

EARTH LATERAL PRESSURE

Passive Lateral Pressure

earth lateral pressure4
EARTH LATERAL PRESSURE

q

Jaky, Broker and Ireland  Ko = M – sin ’

Sand, Normally consolidated clay  M = 1

Clay with OCR > 2  M = 0.95

v =  . z + q

Broker and Ireland

z

Ko = 0.40 + 0.007 PI , 0  PI  40

Ko = 0.64 + 0.001 PI , 40  PI  80

v

h

Sherif and Ishibashi  Ko =  +  (OCR – 1)

  •  = 0.54 + 0.00444 (LL – 20)
  • = 0.09 + 0.00111 (LL – 20)

LL > 110%   = 1.0 ;  = 0.19

At rest, K = Ko

rankine method
RANKINE METHOD

ACTIVE LATERAL PRESSURE

1 = 3 . tan2 (45+/2)+2c.tan (45+/2)

a = v . tan2(45-/2) – 2c . tan (45-/2)

a = v . Ka – 2cKa

Ka = tan2 (45 - /2)

rankine method1
RANKINE METHOD

PASSIVE LATERAL PRESSURE

rankine method2
RANKINE METHOD

PASSIVE LATERAL PRESSURE

p= v . tan2(45+/2) + 2c . tan (45+/2)

rankine method3
RANKINE METHOD

PASSIVE LATERAL PRESSURE

Kp = tan2 (45 + /2)

h = v . Kp + 2cKp

example
EXAMPLE

q = 20 kN/m2

h1 = 2 m

h2 = 8 m

Sheet Pile

1 = 15 kN/m3

1 = 10 o

c1 = 0 kN/m2

2 = 15 kN/m3

2 = 15 o

c2 = 0 kN/m2

h3 = 4 m

  • Questions:
  • Determine the active and passive lateral pressure of sheet pile structure
  • Determine the total lateral pressure
solution
SOLUTION

q = 20 kN/m2

2 m

8 m

Coefficient of Lateral Pressure :

Active ; ka = tan2(45-1/2) = 0.704

Passive ; kp = tan2(45+2/2) = 1.698

Pa1

4 m

Pw1

Pw2

Pq1

Pp1

Pa2

Active Lateral Pressure

Pa1 = ka . 1 . h1 – 2 . c . ka = 0.704 . 15 . 2 – 2 . 0 . 0.704 = 21.12 kN/m2

Pa2 = ka . (1 . h1 + 1’ . h2) – 2 . c . ka = 49.28 kN/m2

solution1
SOLUTION

q = 20 kN/m2

2 m

8 m

Coefficient of Lateral Pressure :

Active ; ka = tan2(45-1/2) = 0.704

Passive ; kp = tan2(45+2/2) = 1.698

Pa1

4 m

Pw1

Pw2

Pq1

Pp1

Pa2

Active Lateral Pressure

Pq1 = ka . q = 0.704 . 20 = 14.08 kN/m2

Pw1 = kw . w . h2 = 1 . 10 . 8 = 80 kN/m2

solution2
SOLUTION

q = 20 kN/m2

2 m

8 m

Coefficient of Lateral Pressure :

Active ; ka = tan2(45-1/2) = 0.704

Passive ; kp = tan2(45+2/2) = 1.698

Pa1

4 m

Pw1

Pw2

Pq1

Pp1

Pa2

PASSIVE LATERAL PRESSURE

Pp1 = kp . 2’ . h3 + 2 . c . kp = 1.698 . 5 . 4 + 2 . 0 . 1.698 = 33.96 kN/m2

Pw2 = kw . w . h3 = 1 . 10 . 4 = 40 kN/m2

solution3
SOLUTION

q = 20 kN/m2

2 m

8 m

Coefficient of Lateral Pressure :

Active ; ka = tan2(45-1/2) = 0.704

Passive ; kp = tan2(45+2/2) = 1.698

Pa1

Pa

za

Pp

4 m

zp

Pw1

Pw2

Pq1

Pp1

Pa2

ACTIVE LATERAL FORCE

Pa = 0.5 . Pa1 . h1 + (Pa1+Pa2)/2 . H2 + Pq1 . (h1+h2) + 0.5 . Pw1 . h2 = 763.52 kN/m

za = 3.56 m

solution4
SOLUTION

q = 20 kN/m2

2 m

8 m

Coefficient of Lateral Pressure :

Active ; ka = tan2(45-1/2) = 0.704

Passive ; kp = tan2(45+2/2) = 1.698

Pa1

Pa

za

Pp

4 m

zp

Pw1

Pw2

Pq1

Pp1

Pa2

PASSIVE LATERAL FORCE

Pp = 0.5 . Pp1 . h3 + 0.5 . Pw2 . h3 = 147.92 kN/m

zp = 4/3 m

slide27
SESSION 23-24COULOMB METHOD

LATERAL PRESSURE DUE TO EXTERNAL LOAD

DYNAMIC EARTH PRESSURE

coulomb method
COULOMB METHOD

ACTIVE LATERAL PRESSURE

  • Assumption:
  • Fill material is granular
  • Friction of wall and fill material is considered
  • The failure surface in the soil mass would be a plane (BC1, BC2 …)

Pa = ½ Ka .  . H2

example1
EXAMPLE

Consider the retaining wall shown in the following figure. Given

- H = 4.6 m

-  = 16.5 kN/m3

-  = 30 o

-  = 2/3 

- c = 0

-  = 0

-  = 90 o

Calculate the Coulomb’s active force

per unit length of the wall

solution5
SOLUTION

Ka = 0.297

Pa = 51.85 kN/m

coulomb method1
COULOMB METHOD

PASSIVE LATERAL PRESSURE

Pp = ½ Kp .  . H2

example2
EXAMPLE

Refer to the following figure. For kv = 0 and kh = 0.3, determine :

  • Pae
  • The location of the resultant, z, from the bottom of the wall

 = 35 o

 = 18 kN/m3

 = 17.5 o

5 m

solution6
SOLUTION

Part a.

Kae = 0.47

solution7
SOLUTION
  • Part b.

Where :

Ka = 0.25

 = 90o

 = 17.5o

 = 0o

Pa = 56.25 kN/m

Pae = Pae – Pa = 105.75 – 56.25 = 49.5 kN/m