topic 6 lateral earth pressure
DESCRIPTION
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 - PowerPoint PPT PresentationTRANSCRIPT
TOPIC 6 LATERAL EARTH PRESSURE
Course : S0705 – Soil Mechanic
Year : 2008
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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)
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SESSION 21-22
RETAINING EARTH WALL
RANKINE METHOD
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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
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RETAINING EARTH WALL
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RETAINING EARTH WALL
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RETAINING EARTH WALL
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RETAINING EARTH WALL
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RETAINING EARTH WALL
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RETAINING EARTH WALL
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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
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Case I
EARTH LATERAL PRESSURE
Lateral Pressure at rest
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Case II
EARTH LATERAL PRESSURE
Active Lateral Pressure
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Case III
EARTH LATERAL PRESSURE
Passive Lateral Pressure
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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
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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
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RANKINE METHOD
PASSIVE LATERAL PRESSURE
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RANKINE METHOD
p= v . tan2(45+/2) + 2c . tan (45+/2)
PASSIVE LATERAL PRESSURE
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RANKINE METHOD
Kp = tan2 (45 + /2)
h = v . Kp + 2cKp
PASSIVE LATERAL PRESSURE
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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
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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
Active Lateral Pressure
Pq1
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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
Coefficient of Lateral Pressure :
Active ; ka = tan2(45-1/2) = 0.704
Passive ; kp = tan2(45+2/2) = 1.698
Active Lateral Pressure
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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
PASSIVE LATERAL PRESSURE
Pq1
Coefficient of Lateral Pressure :
Active ; ka = tan2(45-1/2) = 0.704
Passive ; kp = tan2(45+2/2) = 1.698
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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
Coefficient of Lateral Pressure :
Active ; ka = tan2(45-1/2) = 0.704
Passive ; kp = tan2(45+2/2) = 1.698
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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
Coefficient of Lateral Pressure :
Active ; ka = tan2(45-1/2) = 0.704
Passive ; kp = tan2(45+2/2) = 1.698
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RANKINE EARTH PRESSURE FOR INCLINED BACKFILL
22
22
coscoscos
coscoscoscosKa
22
22
coscoscos
coscoscoscosKp
Ka.H..Pa 22
1
Kp.H..Pp 22
1
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SESSION 23-24 COULOMB METHOD
LATERAL PRESSURE DUE TO EXTERNAL LOAD
DYNAMIC EARTH PRESSURE
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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
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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
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SOLUTION
2
2
2
)sin().sin()sin().sin(
1sin.sin
)(sinKa
Ka = 0.297
aa KHP ...21 2
Pa = 51.85 kN/m
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COULOMB METHOD
2
2
2
)sin().sin()sin().sin(
1sin.sin
)(sinKp
Pp = ½ Kp . . H2
PASSIVE LATERAL PRESSURE
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COULOMB METHOD WITH A SURCHARGE ON THE BACKFILL
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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
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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
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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'
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LATERAL EARTH PRESSURE FOR EARTHQUAKE CONDITIONS
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LATERAL EARTH PRESSURE FOR EARTHQUAKE CONDITIONS
ae
ae
P
PaH
PHz
36.0
PaPP aeae
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LATERAL EARTH PRESSURE FOR EARTHQUAKE CONDITIONS
pev2
21
pe Kk1H..P
2
2
2
pe
sin'sin'sinsin
190'sin.sin'.cos
'sinK
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LATERAL EARTH PRESSURE FOR EARTHQUAKE CONDITIONS
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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
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SOLUTION
Part a.
aev2
21
ae Kk1H..P
Kae = 0.47
mkNP
P
ae
ae
/75.105
47.0.015.18. 221
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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