Lecture 12- Earth Pressure and Sturucture Rankine Theory [1]

Earth pressure and structure (Rankine theory)

Dr. Md Mizanur Rahman

School of Natural and Built EnvironmentsCIVE 3008-Lecture 12: Earth pressure and structure (Rankine theory)

Acknowledge: Dr. D. A. CameronPrevious course coordinator

Earth pressures on retaining structures


Two Methods


- to estimate the earth pressures on structures

1. RankinePlausible stress states

2. CoulombPlausible failure mechanisms

Relative merits of approaches?

Rankine earth pressures


A lower bound estimate

Effective horizontal stress,

H = Kz

where, K = earth pressure coefficient

z = effective vertical stress

Earth pressure states (retaining walls)


“At rest” wall is not moving, so the soilan intermediate state

PassiveActive

Both are failure states

Earth pressure at rest


“AT REST” PRESSURE

The intermediate state

K = Ko = fn(soil type, density, OCR)

The soil is unable to move laterally - can’t expand, OR contract

e.g soil confined in a large body of soil

Active state (stress relaxation)


Normal stress

Shear stress

3f 3o 1

At rest stateActive state

failure envelope

Passive state (stress intensification)


Normal stress

Shear stress

1f3o 1

Passive stateAt rest state

failure envelope

3f

All three states


Normal stress

Shear stress

1o

Active state

Passive state

At Rest failure envelope

Note:Active state: stress relaxation

Passive state: stress intensification

The 3 States (consider a vertical retaining wall)


H/z

Wall movement

Kp

Ka

NB: Passive needs LARGE strains

KO

Equations for Rankine States


(Can be derived from Geometry of Mohr’s circles)

For ACTIVE STATE

Case 1A: c = 0

H = Kaz

and Ka =

[Ka max 0.333 for loose sand]

)sin(1

)sin(1

Active state


Normal stress

Shea

r str

ess

NB: Active state = a failure state

Failure, f , nf (

1 - 3 )/2

(1 + 3)/2

Active state


31

31sin

)sin1()sin1( 31

)sin1(

)sin1(

1

3

aK

From the geometry,

Active state (with cohesion)


Case 2A: c 0

H = Kaz - 2cKa

Notes:

• the 2nd term is a constant!

• z = (z) + z

i.e. stress due to self weight + extra due to surface load

Can soil undergo tension?


If z = 0, then H 0

Now if z = z,

At what depth will H = 0?

H = Kaz - 2cKa

This depth is called the depth of

cracking, zc, & defines the potential

tension zone

z

Depth of Cracking


By definition:

At zc, H = 0

Therefore,

H = 0 = Ka zc - 2cKa

Therefore,

zc = [2cKa][ Ka ]

Or a

cKγ'

c2z

z

The tension zone


The pulling power of cohesive soil is ignored in calculations of pressures behind retaining walls over the depth zc because:

- tension is unsustainable

i.e. short term only!

However, no compressive pressures exist in this zone = a dead zone

Evidence of a tension zone


How can unsupported, vertical-sided trenches be cut to metres depth in clay soils?

What depth is possible?

What happens if it rains?

Warning: people laying pipes have died in collapsed trenches!

OH&S???

Summary of active state


• Stresses relaxedcommon retaining wall situation

• Ka = (1 – sin)/(1+ sin )clean sand, Ka 0.33 usually

• Theoretical tension or crack zone from cohesive strength (c)

may be applied to slope stability

Passive state


Again, from Geometry of Mohr’s circles

Case 1P: c = 0

H = Kpz

and Kp =

[Kp min 3 for loose sand]

aK

1

)sin(1

)sin(1

Passive state (with cohesion)


Case 2P: c 0

H = Kpz + 2cKp

Note:

1. the 2nd term provides greater constant passive pressure component

Orientation of Failure Planes


From Mohr’s circlesActive state:

(45 + /2) to horizontalPassive state:

(45 - /2) to horizontal

Orientation of Failure Planes


Sliding surfaces?

ACTIVE

PASSIVE


Tension is ignored!

ACTIVE-2cKa

zc

+2cKp PASSIVE

Typical Lateral stresses, c 0

The Influence of Pore Water


Steady state (and seepage) pressures add to lateral stresses on walls

Should Ka be applied to the pore water pressure?

NO WAY!

Hydrostatic means K = 1



No Water

c = 0

H = H = Ka z

Water

c = 0

H = Ka z u = wz

+

TOTAL LATERAL STRESS



In the previous example

uniform soil, no surface load and with or without a Water Table at

ground level,

Almost twice the total lateral pressure is

experienced with the high Water Table

Effective lateral stresses are halved, BUT full pwp

is exerted!

Importance of Drainage for Retaining Walls (Drains, Filters & Weep holes)


Weep holes

Granular zone or geofabric drain

Examples


From Whitlow - modified

Find the total resultant thrust and its point of action behind vertical-backed retaining walls of height, 12 m, resulting from earth and water pressures given the following situations

1. Surface horizontal; no surcharge; single soil layer, c = 0, = 30, = 18 kN/m3

2. Surface horizontal; uniform surcharge of 10 kPa; single soil layer: c = 0, = 30, = 18 kN/m3

3. Surface horizontal; no surcharge; two soil layer:

0-5 m depth, c = 0, = 30, = 18 kN/m3

> 5 m depth, c = 0, = 36, = 20 kN/m3

Examples


Thrust = lateral pressure x area

= average pressure x height over which it acts, per m length of wall

9 m

40 kPa

360 kN

20 kPa

360 kN

60 kPa3 m

Examples


Resultant Thrust = Resultant, Pa = (all thrusts)

Point of action found by summing moments about a point and dividing by Pa

10 kPa12

m

360 kN

60 kPa4 m

120 kN

X mLocation of

resultant force

Examples


AnswerResultant, Pa = (120 + 360)

Pa = 480 kN per m length of wall

Point of action found by summing moments about the base and dividing by Pa

120 x 6 + 360 x 4 = Pa x X

X = (720 +1440)/480 = 4.5 m

Examples


Q1. c = 0, = 30, = 18 kN/m3

Ka = 0.333

72 kPa

12 m

432 kN

8 m

Examples


Q2. As for 1 but 10 kPa surcharge

Ka = 0.3333.33 + 72 kPa

432 kN

8 m

40 kN

10 kPa

ANSWER 472 kN/m, 4.17 m

Examples


Q3. Two granular soils

Ka1 = 0.333

12 m

5 m

Ka2 = 0.26

At z = 5 m, z = 90 kPaAt z = 12 m, z = 140 kPa

ANSWER

366 kN/m, 4.15 m

Example


From Whitlow, cont’d6. Surface horizontal; no surcharge; single soil layer, cu = 45 kPa, u = 0,

= 18 kN/m3

7. Surface horizontal; no surcharge; single soil layer, c = 15 kPa, = 20, = 18 kN/m3

11. Surface horizontal; no surcharge; two soil layer,

0-4 m depth, c = 0, = 30, = 19.6 kN/m3

> 4 m depth, c = 25 kPa, = 15, = 18.2 kN/m3

ANSWERSQ6 441 kN/m, 2.33 mQ7 408 kN/m, 3.21 mQ11 458 kN/m, 3.61 m

Example


Q7. c = 15 kPa, = 20, = 18 kN/m3

(0.49x216 - 300.49 kPa

OR 84.8 kPa

12 m

408 kN

zc

Ka = 0.49

zc = 2.38 m 21 = 0.49x18xzcANSWER: 408 kN/m, 3.21 m

Limitations of Rankine


1. Vertical backs of walls only

2. Backfill surface must be regular

– a solution exists for a sloping backfill, provided slope angle, <

– BUT pressures act parallel to the slope - theoretically wrong!

3. Backfill loads / surcharge effects approximated

4. Wall friction ignored!

– friction is beneficial

Summary


1) Earth pressures are needed for design of retaining walls & excavations

2) Three major states: at rest, active and passive

─ Last 2 are failure states

3) Earth pressure coefficients are based on effective stresses

4) Water pressures are important

− total lateral stresses

5) Cohesion leads to potential cracked zone for Active state

Excavation Bracing


Trench

Support systems:soldier beams (vertical)& shuttering between them or steel sheeting

Strut

Possiblefailure shape

Steelsheeting

Wale

PLAN

Example


Design of Bracing


Earth pressures are not simple

- propping forces from struts

- progressive construction

Empirical design earth pressures

- struts designed for thrust

Refer to Notes for guidance

Information Only

Lecture 12- Earth Pressure and Sturucture Rankine Theory [1]

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