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Module 6 Earth Pressure We have already encountered the concept of Earth Pressure in Module 4. The next four

modules will deal with the issue of Earth Pressure (also known as lateral or horizontal

pressure). We will be investigating three different scenarios that involve earth pressure:

1. Rock anchoring and tied-earth structures [Module 6]

2. Gravity retaining walls [Module 7]

3. Structural retaining walls [Module 8]

To begin, we will investigate the most common use of lateral pressure; to resist horizontal

pressure from earth movements.

Without some form of resistance, unstable cliffs and embankments can collapse, as you can

see in Figure 1.

Figure 1 Holbeck Hall Landslip, UK (from news.bbcimg.co.uk)

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Figure 2 demonstrates the basic processes involved in a landslip event. The earth slides sideways. The slip surfaces are roughly circular.

Figure 2 Landslip process (from www.staffs.ac.uk)

Landslips can be very destructive. On 30 July, 1997, a landslip in Thredbo ski village

destroyed two ski lodges and killed 18 people.

Figure 3 Thredbo landslide (from resources1.news.com.au)

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We will now explore why landslips are so destructive and why the design and construction of

retaining walls is so important.

1.0 Vertical Earth Pressure It is best to start with a simple model of the soil behind the retaining wall. From the basic

properties of the soil, such as the density, we can determine the vertical earth pressure. We

can think of the soil as composed of blocks that are one cubic metre.

1.1 Unit Weight of the Soil

For simplicity, we will assume that the block has a density of 2000 kg/m3. The density of a soil

can be determined on site by a simple density test (refer to AS 1289 Soil Investigation Code).

Using basic Physics that you learned in 300050 Physics 1, Newton’s Third Law,

Force F = Mass m x Acceleration due to Gravity g [Newton] = [kilograms] x [metres/sec2]

abbreviated: [N] = [kg] x [ms-2]

In the same way, the density of the soil can be converted in a “Unit Force” quantity, known as

the Unit Weight. In our example,

Unit Weight γ = Density ρ x Acceleration due to Gravity g

20,000 N/m3 = 2,000 kg/m3 x 10 ms-2

NOTE: 1. The Unit Weight is usually represented in kiloNewtons per cubic metre. Hence, the

answer in our example would be: γ = 20 kN/m3.

2. A more accurate estimate for the Acceleration due to Gravity would be 9.8 ms-2.

However, geotechnics is very uncertain. It is not possible to know with any great

accuracy, the ground conditions that you will encounter. Hence, geotechnical

engineers tend to use very large safety factors, of 3 to 4. It matters little if the

acceleration is taken as 9.8 or 10 ms-2.

The unit weight depends on the mineralogy of the soil and the degree of saturation.

Iron-rich soils (known as mafic) are much heavier than aluminium- or silicon-rich soils (known

as felsic). Similarly, if the soil is inundated with groundwater, the unit weight will increase.

1.2 Vertical Pressure

The vertical earth pressure, which also known as overburden pressure or surcharge,

can be determined from the density ρ (kg/m3) or the unit weight γ (kN/m3).

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If we take the Unit Weight of a block of soil as 20 kN/m3, then a cubic metre block of soil

would weigh:

Weight of soil = Unit Weight γs x Volume of the block

= 20 kN/m3 x 1 m3

= 20 kN (or approximately 2 tonnes)

We can calculate the pressure that the block would place on a surface below. Its plan

dimensions are 1 metre by 1 metre.

From basic physics,

Pressure σv = Force = 20 kN Area 1 m2

= 20 kN/m2 or 20 kPa.

Figure 4 First block

If another block were placed on top of the first, then the vertical pressure below would double

to 40 kPa;

Figure 5 Second block

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After three blocks the vertical pressure would be 60 kPa.

Figure 6 Third block

The vertical pressure is proportional to the depth below the surface. As more overburden is

placed on the base, the pressure will increase.

The overburden pressure can be determined from the following equation:

The overburden (or vertical) pressure, σv, is given by the equation:

σv = γs z

where z is the depth of the soil under investigation

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2.0 Horizontal Pressure If we return to our example of stacked blocks of soil, we can illustrate how horizontal pressure

develops in a soil.

As the second block is placed on the first, the lower block will begin to sag. After the third

block, the sag of the lowest block will increase. If the column of soil blocks were placed

adjacent to a retaining wall, the retaining wall would experience lateral (or horizontal)

pressure.

The bulging of the soil is minimal at the top, hence the lateral earth pressure is small at the

top of the wall. In fact for a granular backfill, the lateral earth pressure will always be zero at

the top of the wall. Since the soil will try to bulge most at the bottom of the wall, the earth

pressure will be greatest at the bottom.

2.1 Analogy of Hydrostatic Pressure

Any material that can flow, like water, (wheat)grain or even soil, exerts a lateral pressure.

The size of the lateral pressure depends on the shearing characteristics of the material.

Lateral pressure is magnified when the material has little shear strength, such as a perfect

fluid like water or air. Water has no resistance to shear. It will flow away from applied

pressure.

As you saw in Module 2, shearing is a way for soil to transfer superstructure loads, sideways

and downwards. A retaining wall only experiences the sideways effect. Some of the

overburden pressure is transferred downwards into the ground.

If we return to the analogy of blocks of soil, as you add more blocks to the stack, the lowest

block of soil will sag further and further. A strong soil (with high shear strength) will only sag a

little with each load; a weak soil (with low shear strength) will sag a lot. The weakest soil, a

slurry, will sag dramatically as the load is applied. Slurries, and indeed water, are easily

sheared.

In slurries and water, the vertical loads are transferred to horizontal pressure (as shown in

Figure 7). The horizontal pressure increases down the tank because of the weight of the water

above. The lateral pressures, σH, are actually as large as the vertical pressures, σV. Using an

example from our industry, soil slurry cannot be held by a retaining wall. Under the action of

its self-weight, it will flow under the wall.

ie. for water or slurry, σH = σV

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Figure 7 Hydrostatic pressure

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2.2 Earth Pressure

The lateral pressure σH is usually expressed as a proportion of the vertical pressure, σV.

σH = K σV

Stiffer (high shear strength) materials have small K values. For instance, sand has K values of

0.33 while gravel has K values of 0.22. Water has a K value of 1.

The way the retaining wall reacts to lateral pressure is also important. The retaining wall can

be forced against the soil, increasing the K value and the lateral pressure (Passive) or it can

move away from the soil, reducing the K value and the lateral pressure (Active) [refer to

Section 3.1 for more detail].

Figure 8 Earth pressure (from Spangler & Handy:p. 530)

For soils, the magnitude of K depends on movement of the retaining wall.

There are three types of earth pressures; Active Earth Pressure; the retaining wall moves away from the earth.

If the retaining wall is able to ‘give’ a little, then the retained soil will move enough for the soil

particles to interlock, increasing their own resistance to shear. The retaining wall has to take

less of the lateral load.

The amount of movement is relatively minor. It is usually about 0.25 % to 1 % of the height of

the wall. A 5 metre wall must rotate 50 mm at the top to reduce the earth pressure to the

active case.

Active earth pressure is actually less than the Earth-at-rest pressure.

Earth Pressure-at-Rest; the retaining wall is restrained from moving

Sometimes, the retaining wall is prevented from rotating by the superstructure. In this case,

the retaining wall must carry all the lateral pressure from the retained soil. The At-rest

pressure can be double the Active earth pressure.

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A builder can also induce the same conditions if he does not provide sufficient construction

joints in the face of the retaining wall. Similarly, if reinforcing bars in the retaining wall are

carried across a construction joint, the wall can become too rigid to flex with the earth

pressure.

Passive Earth Pressure; the wall moves forward into the soil You can actually fail a retaining wall by jacking it. It’s the reverse of active pressure. The soil

isn’t pushing the wall; the wall, through rock anchors, is actually pushing on the soil. There is

a limit to the amount of force you can put on a soil before it starts to move in the opposite

direction. Thankfully, the force is much larger than the active earth pressure, because you

have push all the retained soil up and out of the way. As well as overcoming the shearing

resistance of the soil, you have to impart a pressure large enough to shift the dead weight of

the soil.

The strain needed to mobilise the passive earth pressure is far more than active pressure.

2% to 4% for dense sands

10% to 15% for loose sands

In the case of a 5 metre anchored wall, soil must be pushed backwards 500mm to 750 mm.

There is not much danger of failing a retaining wall, even with loose packing behind it.

Figure 9 Lateral coefficients Ka and Kp (from Bowles, Foundation Analysis & Design, p. 382.)

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2.3 Calculation of Active Earth Pressure

We can use the concept of a stress trajectory to determine the Active Earth Pressure

coefficient, Ka, as shown in Figure 10.

Figure 10 Transformation of vertical to horizontal pressure

As the overburden pressure is placed on the backfill, the “squashing” process produces two

effects:

A. A horizontal pressure is generated on the retaining wall

B. At the same time, some of the overburden pressure is transferred into the foundation

below. The process involves shearing; the maximum shear stresses occur at 45o to

the overburden pressure.

We can show these effects on a Normal Stress versus Shear Stress diagram. Point A

represents the overburden pressure (vertical) while Point C represents the earth pressure

(horizontal) on the retaining wall. In between, the vertical pressure is “turned” into a shear

stress (maximum at 45o to the horizontal and vertical pressures) and then finally back to a

normal stress on the retaining wall.

Figure 11 Stress state behind a retaining wall

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We can combine the stress trajectory diagram with the Mohr-Coulomb diagram to determine

how much shearing a backfill soil can take before it fails.

The active earth pressure is derived from the Mohr - Coulomb failure envelope. The active earth pressure represents the load imposed by the soil on the wall at failure.

Figure 12 Change in active earth pressures

The Mohr-Coulomb diagram, that we encountered in Module 1, can be used to explain what

happens in the soil, as Earth pressure is applied to the wall. We will only investigate the case

of a granular backfill (ie. c = 0). The diagram applies to a point in the soil so A, B and C do not correspond to physical positions in the soil, but rather stress states on different planes at

the point. The radii, ABC, can be considered as stress trajectories. Vertical pressure from the

self-weight (soil) or overburden (buildings, etc) is transformed to horizontal pressure on the

retaining wall; just as we saw with water or slurry in the Section 2.1. For water, A = C, since

water cannot resist shear [the Mohr-Coulomb failure envelope would be flat on the x-axis].

The Mohr-Coulomb diagram is used to calculate the lateral pressure, C3, at failure (using

geometry).

Point C1. corresponds to the earth pressure-at-rest. It is the lateral earth pressure on the

retaining wall, before the backfill has been able to interlock completely.

As the retaining wall “gives”, the lateral pressure on the retaining wall decreases; firstly to C2

and finally C3. When the pressure trajectory reaches the Mohr-Coulomb failure envelope (the

black line on Figure 12), the soil fails. A soil particle cannot experience conditions of normal

and shear stress beyond the Mohr-Coulomb failure envelope without failing.

Notice that the shearing pressure also increases (from B1 to B2 and finally B3), as the lateral

pressure on the wall reduces. The soil backfill has to work harder, as the retaining wall gives.

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The limiting values of Normal and Shear pressure are given by the Mohr-Coulomb failure

envelope. This envelope is derived from the Shear Box test (see Module 1). When a retaining

wall is allowed to give, the value of σH will diminish until part of the stress path reaches the

Failure Envelope.

By geometry (See Das, p. 385-388 for the more general case), we can calculate the limiting

value of the lateral earth pressure (C3):

σH = Ka σv

where Ka = tan2 (45 - φ/2)

σv = γ z

The value of the horizontal pressure decreases until we get to a limiting value Ka.

The failure plane is inclined to the vertical at a slope of θ = 45o - φ 2

The highest values of the horizontal pressure, σH, are at the base of the retaining wall, where

the overburden pressure (or vertical pressure) σv is highest.

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Worked Example

Find the equivalent horizontal force that is applied to the retaining wall, using the data

provided in Figure 13.

Figure 13 Earth pressure data The vertical overburden pressure is given by the equation:

σ = γ z

The vertical overburden pressure is zero at the top of the wall. At the bottom, the pressure is:

σ max = 20 * 10 = 200 kN/m2

Any superimposed loads (from construction materials, buildings, etc) must be added to the

overburden pressure.

At the top of the wall, the vertical pressure produced by the superimposed load is given by:

σV min = 10 kN/m2

The maximum vertical pressure is given by:

σV max = 200 + 10 = 210 kN/m2

The pressure (or stress) distribution is shown in Figure 14.

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Figure 14 Vertical pressure in backfill

The horizontal pressure is determined as the equation:

σH = Ka σ

The horizontal pressure is a simple fraction of the vertical pressure. The horizontal pressure,

σH, is zero at the top of the wall and a maximum at the base.

Ka = tan2(45 - φ/2) = tan2(45 – 25/2) = (tan 32.5o)2 = 0.406

The minimum horizontal pressure is given by:

σH min = 0.406 * 10 kN/m2 = 4.06 kN/m2

The maximum horizontal pressure is given by:

σV max = 0.406 * 210 kN/m2 = 85 kN/m2

The horizontal stress (or pressure) distribution is shown in Figure 15.

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Figure 15 Horizontal pressure The average horizontal pressure on the wall is (4 + 85)/2 = 44.5 kN/m2.

Taking just a one metre length of retaining wall, its surface area would be:

A = 10 m * 1 m = 10 m2.

Since pressure (or stress) can calculated from a force, F, as:

σ = F / A

If we rearrange the equation, the equivalent force on the retaining wall can be determined:

F = σ * A = 44.5 kN/m2 * 10 m2 = 445 kN

To find the “mass” of the earth force, we divide by g, the acceleration due to gravity [10 m/s2].

M = F / 10 = 445 * 103 / 10 = 44,500 kilograms

or 44.5 tonnes.

In other words, a 10 metre retaining wall has the equivalent mass of a “D8” bulldozer locked

up in every metre length of wall [10 metre long wall………10 bulldozers]

………………………….……..waiting to crush an unsuspecting builder!

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2.4 Earth Reinforcement

The simplest form of retaining wall works directly against the lateral earth pressure.

Strips of polypropylene, galvanised or stainless steel are embedded in the backfill. The strips

are structurally connected to facing panels. While the facing panels must be designed to

withstand earth pressures between the ties (reinforcing strips), the system is much simpler

than a cantilever retaining wall [Module 8].

The reinforcing strips resist the movement of the backfill under load. Horizontal forces

are transferred to the strips by friction. Since the strips are tied back into the backfill, the

reinforcement must be torn before the retaining wall can fail. The system has been used on

walls that are 15 metres high.

The fill behind the facing panels must be carefully placed so that earth pressures are

transferred efficiently beyond the slip zone.

The system is marvellously simple. As you can see from Figure 16, the greatest embedment

of the strips (beyond the slip zone) occurs at the base of the wall (where the slip plane is

closest to the facing panels). In other words, the greatest frictional resistance is developed

precisely where the greatest earth pressures are!

Figure 16 Reinforced earth walls

Groundwater drainage can be simple; a geotextile on the inner face to stop fines from

bleeding through the joints in the facing panels. However, some engineers design elaborate

piped collection systems on the inside of the facing panels of free-draining aggregate and

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drainage pipes to collect and dispose of groundwater. We will look at groundwater drainage

systems in the next module.

Reinforced earth walls are characteristically used on highway projects. The strips must be

embedded beyond the slip zone to resist the horizontal movement of the retaining wall.

Building projects usually do not have the clearance behind the retaining wall to embed the

earth ties.

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2.5 Calculation of Passive Earth Pressure

In the passive case, the earth (or lateral) pressures increase beyond the vertical pressure.

In the passive case, the lateral or earth pressure begins at the pressure-at-rest, C1. Wall

jacking can cause the horizontal earth pressure to increase beyond the vertical pressure, A, to

C4 and finally C5. The earth will passively shear at C5. By comparison, the Passive Earth

pressure at failure is much larger than the Active Earth pressure at failure; sometimes, four

times (4x) larger than the Active Earth pressure.

Figure 17 Changes in passive earth pressure

The passive earth pressure at failure can also be determined by geometry.

The pressure, which must be imparted to the soil by a structure to cause it to fail, can also be determined from the Mohr-Coulomb failure criteria.

σH = Kp σv where Kp = tan2 (45 + φ/2)

In this case, the horizontal pressure must be increased beyond the vertical pressure. Hence,

the soil fails at very large shear pressures. The failure surface is also inclined at a shallower

angle to the top of the retaining wall.

The failure planes are inclined to the vertical at a slope of θ = 45o + φ 2

Since the soil in front of a retaining wall, may be clayey, we have to add in an additional factor

to account for the cohesion factor in the soil:

σH = Kp σv + 2 c √ Kp

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2.6 Slope stabilisation

At first glance, you might think that there isn’t any practical application for passive earth

pressure on a building project. However, we will be using the concept of passive earth

pressure for a very conventional sub-structure, sheet piling in Module 7.

In this module, we will investigate the role of passive earth pressure in slope stabilisation. We

will start with a simplified model, that is somewhat practical (indeed, the author has used this

set-up on one occasion).

If we start with an reinforced earth retaining wall AND we add a Deadman (concrete weight or

similar) to the end of the tie, then we dramatically increase the shear resistance of the set-up.

Figure 18 Reinforced earth wall with deadman Only one tendon shown for simplicity; more than one deadman is possible to reduce the bending stresses in the

facing panels

The backfill close to the retaining wall fails actively while the backfill adjacent to the deadman

fails passively. As we demonstrated in the previous section, passive earth pressures can be

up to four times larger than the active earth pressures. Earth in front of the deadman must be

pushed out of the way in order for the deadman to move forward. The passive slip zone can

be much smaller than the active slip zone and still balance the forces on the retaining wall.

A similar principle is used in rock stabilisation. Holes are bored into the rock face. The end of

the hole has an anchor block. High tensile bolts are installed in the holes and grouted in place.

The block at the end of the hole acts like an anchor or deadman. It has to pull the grout out of

the hole before the rock anchor can fail. The grout is held in place by side-wall friction against

the bored hole.

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Figure 19 Rock anchor (from www.williamsform.com)

An expanding nut grips the periphery at the back of the bore hole when the bolt is tightened

with a torque wrench. A special bolt with a forged in-dented end to be used with a chemical

anchorage, is especially suited to soft rocks.

The head of the bolt is fitted with plate washers or lengths of channel to spread the load over

the surface of soft rock.

Figure 20 Rock anchoring machine (from www.mta.info)

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References

Bowles, Joseph E. (1982), Foundation Analysis and Design, Third edition, McGraw-Hill

International Book Company.

Das, Braja M. (1994), Principles of Geotechnical Engineering, Third edition, PWS

Publishing Company, Boston.

Spangler, Merlin G. & Handy, Richard L. (1982), Soil Engineering, Fourth edition, Harper &

Row, New York.

Boral Masonry (2008), Masonry Design Guide: Segmental Block Retaining walls, New South

Wales Book 4, www.boral.com.au/masonry, July.

Created 2004

Modified 2013