topic 6 lateral earth pressure

TOPIC 6 LATERAL EARTH PRESSURE

Course : S0705 – Soil Mechanic

Year : 2008

Bina Nusantara

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)

Bina Nusantara

SESSION 21-22

RETAINING EARTH WALL

RANKINE METHOD

Bina Nusantara


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

Bina Nusantara


Bina Nusantara

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

Bina Nusantara

Case I


Lateral Pressure at rest

Bina Nusantara

Case II


Active Lateral Pressure

Bina Nusantara

Case III


Passive Lateral Pressure

Bina Nusantara

EARTH LATERAL PRESSUREq

v

h

z v = . z + q

v

hK

At rest, K = Ko

Jaky, Broker and Ireland Ko = M – sin ’

Sand, Normally consolidated clay M = 1

Clay with OCR > 2 M = 0.95

Sherif and Ishibashi Ko = + (OCR – 1)

= 0.54 + 0.00444 (LL – 20)

= 0.09 + 0.00111 (LL – 20)

LL > 110% = 1.0 ; = 0.19

Broker and IrelandKo = 0.40 + 0.007 PI , 0 PI 40Ko = 0.64 + 0.001 PI , 40 PI 80

Bina Nusantara

RANKINE METHOD

Ka = tan2 (45 - /2)

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

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

a = v . Ka – 2cKa

ACTIVE LATERAL PRESSURE

Bina Nusantara

RANKINE METHOD

PASSIVE LATERAL PRESSURE

Bina Nusantara

RANKINE METHOD

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


Bina Nusantara

RANKINE METHOD

Kp = tan2 (45 + /2)

h = v . Kp + 2cKp


Bina Nusantara

EXAMPLE

h1 = 2 m

h2 = 8 m

h3 = 4 m

1 = 15 kN/m3

1 = 10 o

c1 = 0 kN/m2

2 = 15 kN/m3

2 = 15 o

c2 = 0 kN/m2

q = 20 kN/m2

Questions:

1. Determine the active and passive lateral pressure of sheet pile structure

2. Determine the total lateral pressure

Sheet Pile

Bina Nusantara

SOLUTION

2 m

8 m4 m

q = 20 kN/m2

Coefficient of Lateral Pressure :

Active ; ka = tan2(45-1/2) = 0.704

Passive ; kp = tan2(45+2/2) = 1.698

Pa2

Pa1

Pp1Pw1Pw2

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


Pq1

Bina Nusantara

SOLUTION

2 m

8 m4 m

q = 20 kN/m2

Pa2

Pa1

Pp1Pw1Pw2

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

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

Pq1


Active ; ka = tan2(45-1/2) = 0.704

Passive ; kp = tan2(45+2/2) = 1.698


Bina Nusantara

SOLUTION

2 m

8 m4 m

q = 20 kN/m2

Pa2

Pa1

Pp1Pw1Pw2

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


Pq1


Active ; ka = tan2(45-1/2) = 0.704

Passive ; kp = tan2(45+2/2) = 1.698

Bina Nusantara

2 m

8 m4 m

q = 20 kN/m2

Pa2

Pa1

Pp1Pw1Pw2

ACTIVE LATERAL FORCE

Pq1

Pp

Pa

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

za = 3.56 m

za

zp

SOLUTION


Active ; ka = tan2(45-1/2) = 0.704

Passive ; kp = tan2(45+2/2) = 1.698

Bina Nusantara

SOLUTION

4 m

q = 20 kN/m2

Pa2

Pa1

Pp1Pw1Pw2

PASSIVE LATERAL FORCE

Pq1

Pp

Pa

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

zp = 4/3 m

za

zp

2 m

8 m


Active ; ka = tan2(45-1/2) = 0.704

Passive ; kp = tan2(45+2/2) = 1.698

Bina Nusantara

RANKINE EARTH PRESSURE FOR INCLINED BACKFILL

22

22

coscoscos

coscoscoscosKa

22

22

coscoscos

coscoscoscosKp

Ka.H..Pa 22

1

Kp.H..Pp 22

1

Bina Nusantara

SESSION 23-24 COULOMB METHOD

LATERAL PRESSURE DUE TO EXTERNAL LOAD

DYNAMIC EARTH PRESSURE

Bina Nusantara

COULOMB METHOD

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

2

2

2

)sin().sin()sin().sin(

1sin.sin

)(sinKa

ACTIVE LATERAL PRESSURE

Bina Nusantara

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

Bina Nusantara

SOLUTION

2

2

2


1sin.sin

)(sinKa

Ka = 0.297

aa KHP ...21 2

Pa = 51.85 kN/m

Bina Nusantara

COULOMB METHOD

2

2

2


1sin.sin

)(sinKp

Pp = ½ Kp . . H2


Bina Nusantara

COULOMB METHOD WITH A SURCHARGE ON THE BACKFILL

Bina Nusantara

LATERAL EARTH PRESSURE DUE TO SURCHARGE

222

2

ba

ba.

nH

q2

222

2

ba

ba.

nH

q4

22b16,0

b203,0.

H

q

a > 0,4

a 0,4

Bina Nusantara

LATERAL EARTH PRESSURE DUE TO SURCHARGE

2cos.sinH

q

12H90

qP

12

122

H2

H'a30,57QRHz

H

'btan 1

1

H

'b'atan 1

2

22 90'b'aR

12 90'bQ

Bina Nusantara

LATERAL EARTH PRESSURE FOR EARTHQUAKE CONDITIONS

aev2

21

ae Kk1H..P

2

2

2

ae

sin'sin'sinsin

1'sin.sin'.cos

'sinK

v

h1

k1

ktan'

Bina Nusantara


Bina Nusantara


ae

ae

P

PaH

PHz

36.0

PaPP aeae

Bina Nusantara


pev2

21

pe Kk1H..P

2

2

2

pe

sin'sin'sinsin

190'sin.sin'.cos

'sinK

Bina Nusantara


Bina Nusantara

EXAMPLE

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

a. Pae

b. The location of the resultant, z, from the bottom of the wall

= 35 o

= 18 kN/m3

= 17.5 o5 m

Bina Nusantara

SOLUTION

Part a.

aev2

21

ae Kk1H..P

Kae = 0.47

mkNP

P

ae

ae

/75.105

47.0.015.18. 221

Bina Nusantara

SOLUTION

• Part b.

aa KHP ...21 2

Where :

Ka = 0.25

= 90o

= 17.5o

= 0o

Pa = 56.25 kN/m

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

m

P

PaH

PHz

ae

ae

29.23

6.0

topic 6 lateral earth pressure

Documents

lateral pressure of

lateral earth pressure

active ka

passive kp

solution4 mq

h2 pq1

external load session

f2frankine method session