Lateral Earth Pressure.pdf

ENCE 461 Foundation Analysis and Design

Lateral Earth Pressure Coefficient  x '  K   z '  

K = lateral earth pressure coefficient

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x’ = horizontal effective stress

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    

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        

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      

Mohr’s Circle and Lateral Earth Pressures

Retaining Walls Lateral Earth Pressure Theory

Retaining Walls 

Necessary in situations where gradual transitions either take up too much space or are a re impractical for other reasons

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Retaining walls are analysed for both resistance to overturning and structural integrity

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Two categories of retaining walls 

Gravity Walls (Masonry, Stone, Gabion, etc.)

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In-Situ Walls (Sheet Piling, cast in-situ, etc.)

Development of Lateral Earth Pressure Po b

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1 z 12 K o 2

Groundwater Effects Note Pore Water Effect!

Conditions of Lateral Earth Pressure Coefficient 

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At-Rest Condition 

Condition where wall movement is zero or “minimal”

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Ideal condition of wall, but seldom achieved in reality

Active Condition 

Condition where wall moves away from the backfill

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The lower state of lateral earth pressure

Condition where wall moves toward the backfill

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The higher state of lateral earth pressure

add horizontally

Groundwater Effects 

Steps to properly compute horizontal stresses including groundwater effects: 

Compute total vertical stress

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Compute effective vertical stress by removing groundwater effect through submerged unit weight; plot on P o diagram

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Compute effective horizontal stress by multiplying effective vertical stress by K

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Compute total horizontal stress by directly adding effect of groundwater unit weight to effective horizontal stress

Passive Condition 

subtract vertically

Estimates of At Rest Lateral Earth Pressure Coefficient

Effect of Wall Movement

Jaky’s Equation

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K o 1sin  '  Modified for Overconsolidated Soils

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K o  1sin  '  OCR

sin  ' 

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Applicable only when ground surface is level

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In spite of theoretical weaknesses, Jaky’s equation is as good an estimate of the coefficient of lateral earth pressure as we have

Example of At Rest Wall Pressure 

Given 

Retaining Wall as Shown

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Find 

PA, from At Rest Conditions

Wall Movements Necessary to Achieve Active or Passive States

Development of Passive Earth Pressure

At Rest Pressure Example 

Compute at rest earth pressure coefficient

K o 1sin  '  K o 1sin 30º  0.5 

Compute Effective Wall Force Po

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1 z 12 K o

2 2 120  20  0.5  b 2 Po  lbs  kips 12000 12 b  ft   ft  b

Po

Earth Pressure Theories

h PA

20  6.67 ft. 3

(valid for all theories)

Development of Active Earth Pressure

Rankine Coefficients with Inclined Backfills

Rankine Earth Pressure Equations Level Backfills

Inclined and level backfill equations are identical when  = 0

Example of Rankine Active Wall Pressure 

Given 

Retaining Wall as Shown

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Find 

PA, from At Rest Conditions

Rankine Theory with Inclined Backfills

Rankine Passive Pressure Example 

Compute at rest earth pressure coefficient 2

K P  tan  45º 

 2

Rankine Active Pressure Example 

2

K  A  tan  45º 

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2

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Po

1 3

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1 z 12 K a

2 b 2 120  20  0.333  2 b Po  lbs  kips 8000 8 b  ft   ft 

Po

Po

Summary of Rankine and At Rest Wall Pressures 72,000 lbs.

12,000 lbs.

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Compute Effective Wall Force

1 z 12 K  p

2 2 120  20  3  2 b Po  lbs  kips  72000  72 b  ft   ft  b

2

K  A  tan  45 15 

Compute Effective Wall Force Po



2

K P  tan  45 15  3 

Compute at rest earth pressure coefficient

8000 lbs.

Rankine Passive Pressure Example

2

Example of Coulomb Theory

cos 

K a 2

cos  cos  1

2

sin  sin  cos  cos 



2

cos 

K  p 2

cos  cos  1

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Given 

Wall as shown above

sin  sin  cos  cos 

2

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Find

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KA, KP, PA

Coulomb Theory

Solution for Coulomb Active Pressures 

Compute Coulomb Active Pressure

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KA = 0.3465

Compute Total Wall Force

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PA = 8316 lb/ft of wall

Typical Values of Wall Friction

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Theory of Cohesive Soils

Solution for Coulomb Passive Pressures 

1  sin  1  sin 

   tan 2    4

Compute Coulomb Passive Pressure

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2

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KP = 4.0196

Compute Total Wall Force

Passive Case (Wall Driving) Active Case (Overburden driving)

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Rankine Pressures with Cohesion (Level Backfill) 

Walls with Cohesive Backfill

Active

     3 1 tan 2    2 c tan    4

2

4

2

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 1  H  Overburden Driving 3    2c 2  K  A  tan    tan    1 4 2 4 2  H  

Passive

2









4

2

4

2

Retaining walls should generally have cohesionless backfill, but in some cases cohesive backfill is unavoidable 

 1 3 tan    2 c tan     3 H  Wall Driving 1    2c 2  K    tan    tan   

PA = 96,470 lb/ft of wall

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Cohesive soils present the following weaknesses as backfill: 

Poor drainage

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Creep

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Expansiveness

Most lateral earth pressure theory was first developed for purely cohesionless soils (c = 0) and has been extended to cohesive soils afterward

Example of Equivalent Fluid Method

Comments on Rankine Equations

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Valid if wall-soil friction is not taken in to account

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Do not take into consideration soil above critical height

 H c  

Given 

Wall as shown above

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KA = 0.3465

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KP = 4.0196

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w = 3 degrees

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Find 

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Do not take into consideration sloping walls

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For practical problems, should use equations as they appear in the book 

Forces acting on the wall (both horizontal and vertical)

Example of Equivalent Fluid Compute Equivalent Fluid Unit Weights (Active Case)

G h K a cos  w G h120 0.3465 cos 3º  G h 41.52 pcf  G v  K a sin  w G v 120 0.3465sin 3º  G  2.18 pcf 

2c  K a

Equivalent Fluid Method 

Simplification used to guide the calculations of lateral earth pressures on retaining walls

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Can be used for Rankine and Coulomb lateral earth pressures

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Can be used for at rest, active and passive earth pressures

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Transforms the soil acting on the retaining wall into an equivalent fluid

Example of Equivalent Fluid 

Compute Wall Load (Passive Case)

P p

Example of Equivalent Fluid 

2

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G h H 

Compute Wall Load (Active Case)

Pa

2

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G h H 

b 2 2 P p 481.69  20   96338lb/ft b 2 2 V  p G v H 

b 2 2 P a 41.52  20  8304 lb/ft b 2 2 V  a G v H 

b 2 2 V  p 25.24 20   5048 lb/ft b 2

b 2 2 V  a 2.18 20   436 lb/ft b 2

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Terzaghi Model 

Assumes log spiral failure surface behind wall

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Requires use of suitable chart for KA and KP

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Not directly used in this course, but option in SPW 911

Example of Equivalent Fluid 

Compute Equivalent Fluid Unit Weights (Passive Case)

G h K  p cos  w G h120 4.0196 cos 3º  G h 481.69 pcf  G v  K  p sin  w G v 120 4.0196 sin 3º  G  25.24 pcf 

Homework Set 5 

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Effects of Surface Loading

Reading 

McCarthy: Chapter 16

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Coduto: Chapters 22, 23, 24 & 25

Homework Problems 

McCarthy: 16-1, 16-8, 16-12a, 16-17

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Coduto: 25.3 (Hand and Chart Solutions); 25.5 (SPW 911)

Due Date: 17 April 2002

Questions

Surcharge and Groundwater Loads