arth Pressure and Retaining Structure Notes

Page 1 of 9

Lateral Earth Pressure Theories The Rankine Theory The Rankine theory is a theory of plastic failure which considers the stresses acting within a soil element and assesses the likelihood of plastic failure.

By adopting this approach Rankine developed an expression which allows lateral earth pressures to be calculated if the vertical earth pressure is known.

The theory involves the application of various coefficients of lateral earth pressure and results assume the soil to be in a state of plastic failure.

The Mohr-Coulomb Failure Criterion and Mohr’s Circle The Rankine theory requires the application of a failure criterion to assess whether plastic failure will occur. The failure criterion used is the Mohr-Coulomb criterion, which states that failure occurs when the shear stress on any given plane, , is equal to the following: tan c

Where: c is the soil cohesion

is the soil angle of internal friction

is the normal stress on the plane When this criterion is satisfied, plastic failure begins to develop by sliding on the plane.

Mohr’s circle is a geometric construction which allows the shear and normal stresses on any given plane within a physical element to be plotted in order to determine whether failure has occurred.

Mohr’s plot basically consists of a vertical shear stress and a horizontal normal stress axis. A failure envelope can be plotted which has a gradient equal to and an intercept of c. Any points that plot outside this envelope constitute soil that has failed.

In order to develop his theory, Rankine considered a square element of soil at some depth, z, below the ground surface in the vicinity of a smooth vertical wall.

Moveable Smooth

Wall z

h

z

Page 2 of 9

In this case the vertical stress acting on the soil element will be a principal stress, meaning that there is zero shear stress on the plane on which this acts.

Principal stresses have a number of features in respect of the Mohr’s diagram, as follows:

When plotted in two dimensions there will be two principal stresses, known as the major and minor values

Principal stresses always plot on the horizontal axis of the Mohr’s circle diagram

If a circle is drawn on the diagram with a diameter equal to the difference between the two principal stresses, this will indicate the greatest values of shear and normal stress that occur within that element

Given that the wall is smooth, there will be no shear stresses acting on it and consequently the horizontal force must also be a principal stress.

Now, if the wall is gradually moved away from the soil element under consideration, the horizontal stress will necessarily fall.

At some point, the horizontal stress will fall to a low enough level such that a failure will be generated in the soil element.

On the point of failure, the Mohr circle will be as shown in the diagram below:

After Craig Fig 6.2

Page 3 of 9

In this case the vertical stress on the soil element is the major principal stress, 1, and the horizontal stress is the minor principal stress, 3.

What Rankine essentially did was to develop an expression for 3 in terms of 1 from the above diagram. In this expression 1 is the vertical stress, or overburden, and is known. 3 is the horizontal stress, i.e. the horizontal earth pressure.

Rankine’s solution was:

sin1sin12

sin1sin1

13 c

Then by writing Ka

sin1sin1

And realising that the vertical effective stress, 1, is equal to the soil overburden, which is z, the lateral earth pressure, pa, can be calculated as:

pa = Kaz – 2c√Ka

The value of Ka in solution obtained is known as the coefficient of active earth pressure, and applies where the soil fails by the wall moving away from it (i.e. by lateral expansion).

Alternatively, if the wall were pushed into rather than moved away from the soil, the horizontal stress would increase rather than decrease. The Mohr Coulomb plot for this case would be as before, but in this case the horizontal stress will be larger than the vertical stress, so that 1 is now the horizontal stress and 3 the vertical stress.

Rankine’s solution here was:

sin1sin12

sin1sin1

13 c

In which case, we can say:

Kp

sin1sin1

and pp = Kpz + 2c√Kp

The value of Kp here is known as the coefficient of passive earth pressure, and applies where the soil fails due to the wall moving into it. (i.e. by lateral compression).

Page 4 of 9

Lateral Earth Pressure – Sloping Ground The coefficients described thus far relate only to cases where the ground is horizontal. This does not, however, cover all the cases that we might be interested in.

Rankine’s theory can be extended to allow for the case where the ground is sloping rather than horizontal. The procedure is generally similar to that for horizontal ground, but in this case the construction is more complex because in this case neither the vertical nor horizontal soil stresses are principal stresses.

The case considered is shown in the diagram below and the results obtained for a purely granular soil – i.e. no cohesion – are as follows:

Ka

22

22

coscoscos

coscoscos

The above value is for a pressure acting parallel to the slope, so the horizontal pressure on the back of the wall in this case is given as:

pa = Kaz cos

A similar approach can also be adopted to find the following values for passive pressures:

Kp

22

22

coscoscos

coscoscos

and the horizontal pressure on the back of the wall is given as:

pp = Kpz cos

Mobilisation of Lateral Earth Pressures As should be clear from the foregoing, provided that the vertical stress remains constant, the horizontal stress in the soil depends critically on how much horizontal movement takes place and in which direction this occurs.

Moveable Smooth

Wall

z

Pa

z

Page 5 of 9

The diagram below illustrates the relationship between the coefficient of lateral earth pressure and the movement that has taken place.

The diagram clearly shows that much greater strains are required to mobilise significant passive than significant active stresses. As a guideline, laboratory tests indicate that the strain required to fully mobilise the active force is in the order of 0.25% for dense sand and 1% for loose sand. This compares to values of 2–4% and 10–15% respectively for full passive resistance to be mobilised.

After Craig Fig 6.10

Earth Pressure at Rest The above diagram also shows a lateral earth pressure coefficient, Ko, which lies between the full active and passive values, Ka and Kp. This is the coefficient of earth pressure at rest, which is in theory the earth pressure that exists in a soil deposit where no horizontal movement has taken place, i.e. a virgin soil.

The coefficient of earth pressure at rest is frequently quoted as being equal to the value derived by Jaky, which is:

Ko sin1

In fact, this is something of an over-simplification as Jaky derived his value from a soil that was placed in a very specific manner.

The actual value of Ko will depend on a number of factors, perhaps most import of which is the stress history, the way in which the soil has previously been loaded.

In many cases, naturally occurring soil deposits will have been subjected to much higher stresses than they are currently, most typically through having previously been overlain by deposits which have now eroded away. Overconsolidation can also occur through construction processes such as compaction.

Overconsolidation can cause the actual Ko value in a soil deposit to be considerably higher than predicted by Jaky’s equation. This is illustrated in the following Figure and Table below:

Page 6 of 9

Soil Ko

Dense Sand 0.35

Loose Sand 0.6

Normally Consolidated Clays (Norway) 0.5 – 0.6

Clay, OCR = 3.5 (London) 1.0

Clay, OCR = 20 (London) 2.8

After Craig

After Craig Fig 6.11

The Coulomb Theory Although the Rankine Theory produces a number of useful results, it does have some limitations, as touched upon above.

As an alternative, it is also possible to derive expressions for coefficients of lateral earth pressure using the Coulomb Theory.

The Coulomb Theory is a procedure that also looks at the situation adjacent to a retaining wall, but in this case the assumption is that a failure “wedge” will form behind the wall.

By looking at the forces acting on the wedge, it is possible to derive generalised expressions for the earth pressure coefficients.

The main differences between this and the Rankine approach are that:

The Coulomb Theory is a theory of limiting equilibrium

It works with forces rather than stresses

It can take account of wall friction

It can deal directly with conditions such as sloping soil surfaces and walls where the back is non-vertical

For simple cases the results obtained by the Coulomb theory are similar to those obtained by Rankine. One difference, however, is that Rankine directly predicts a hydrostatic soil pressure distribution whereas, although Coulomb agrees with this, it is not predicted directly but only by successive applications for different wall heights.

The figure below illustrates the general condition considered by the Coulomb theory for the case where there is no cohesion:

Page 7 of 9

In the general case illustrated, Coulomb suggests the following values for active earth pressure coefficients:

2

)sin()sin()sin()sin(

sin)sin(

Ka

And for the passive case:

2

)sin()sin()sin()sin(

sin)sin(

Kp

W

P R

h

Page 8 of 9

Cohesive Soils and Tension Cracks As noted above, the Rankine Theory predicts the lateral earth pressure for a soil that exhibits cohesion to be:

pa = Kaz – 2c√Ka

The consequence of this is that the lateral earth pressure at the ground surface is predicted to be less than zero – i.e. the soil will, in theory apply a tension to the wall. The profile of lateral earth pressure will be as shown below:

The profile indicates that the lateral earth pressure only becomes positive below a depth of zo below the ground surface. In practice, the soil cannot resist tension in this way and as a result, a tension crack develops. The depth of the tension crack will be zo, which can be calculated from the above formula as:

Kaczo

2

Where a tension crack occurs in the soil behind a retaining wall, this will affect the lateral earth pressure acting on the wall. The clearest way to demonstrate the effect of tension cracking is by reference to the generalised Coulomb method for a soil which exhibits cohesion, which is shown below. It is clear from the diagram that:

The depth of the tension crack increases with the value of cohesion

From reference to the force polygon, the higher the value of cohesion adopted, the lower the value of the lateral force on the wall

Tension = 2c√Ka

zo

Ground Level

Page 9 of 9

After Smith Fig 6.16

Coefficient of active earth pressure for cohesive soils It is possible to write an equation for the generalised solution of the Coulomb Theory in the form:

pa = Kaz – cKac

where Kac will depend on both the value of the soil cohesion and the cohesion between the soil and the wall, cw and will be given as:

ccKK w

aac 12

Page 1 of 9

Gravity Retaining Walls In general, retaining walls will be required at any location where there is a change in ground level and it is not possible to accommodate this by constructing a stable soil slope. The stability of gravity retaining walls is due to their self weight, with the wall being proportioned such that this is sufficient to maintain stability.

Retaining wall design will comprise two processes:

1. Design of the wall in accordance with the geotechnical requirements of the site

2. Structural design of the wall

The following sections deal exclusively with the first of these processes. Potential Failure Modes of Gravity Walls Walls may be subject to failure either through loss of serviceability or collapse. Serviceability Failures These would typically include:

1. Excessive deformations of the soil or the wall These may also adversely affect adjacent structures

2. Adverse seepage effects, erosion or leakage through the wall Detailed design must give adequate consideration to the effects of groundwater

This category would also include failure of any structural element of the wall or a combined wall/structural failure, neither of which will be taken into account in the following analyses. Collapse Failure Mechanisms As noted above, the wall is required to be proportioned such that it can resist the loads acting on it, which will largely be due to the horizontal earth pressure generated in the retained fill. In practice, four major collapse failure mechanisms are considered in design of retaining walls, these being as follows: Failure by deep-seated rotation of the soil mass

This occurs as shown in the diagram to the left. It constitutes what must be thought of as a generalised ‘global’ failure and needs to be checked for by slip circle analysis. Failures of this type occur wholly outside the structure and consequently cannot be prevented by minor changes in the wall geometry.

Page 2 of 9

One way of preventing these failures is by installing piles (often driven sheet piles) to extend across the critical failure plane. In addition to global failures of this type, three more localised failures must be considered. Because they are localised, each of these can generally be prevented by changing the geometry of the wall, although sometimes such changes may be so extensive that it might not be economical to do this and a revised design might be required. Failure by Sliding This mechanism occurs when the lateral earth pressure on the back of the wall is so large that it causes the wall to slide forward. In general terms, the tendency of the wall to slide is resisted both by friction at its base and passive force generated in any soil that is present in front of it. If the horizontal force due to the retained soil is greater than the combined friction on the base and passive resistance due to the soil in front of it, the wall will fail by sliding forward.

One way to prevent failure by sliding is to include a heel or a toe cast under the base of the wall. This forces any failure through the base deeper and consequently increases the sliding resistance here.

Failure by Overturning

Failure is again due to the influence of the lateral earth pressure behind the wall, but in this case the wall overturns about its toe. The lateral pressure due to the retained soil generates an overturning moment about the toe of the wall which is resisted by a restoring moment about the same point due to

Wall with Heel Wall with Toe Basic Wall

Page 3 of 9

the weight of the wall. If the weight of the wall is insufficient, the wall will overturn about this point. The simplest way to increase the resistance of the wall to overturning is to amend its geometry to increase either its weight and/or thickness, which will increase the restoring moment. Bearing Failure The soil beneath the base of the wall will be loaded due to the weight of the wall. In the absence of any retained fill, the bearing pressure would simply be equal to the weight of the wall divided by the area of the base. However, where fill is retained, the pressure distribution beneath the base is no longer uniform – the action of the fill causes it to increase under the toe.

If the bearing pressure is greater than the bearing resistance of the soil beneath the toe, this will result in a bearing failure where the toe of the wall ‘punches’ into the foundation soil. It should be noted that a bearing failure is quite different to an overturning failure.

Gravity Wall Design Procedure The first stage of the design procedure is to check for a global failure. This requires the use of slip circle analysis and will not be illustrated in the following examples. The second stage is to derive Ka and Kp values for the retained fill and the fill in front of the wall and then to check for the three failure mechanisms described above. Provided that the wall is shown to be safe for the three failure mechanisms shown, the wall design is safe. If not, either the wall geometry or the material used to backfill the wall needs to be changed and the wall stability re-checked.

Page 4 of 9

Factors of Safety in Geotechnical Design General In designing any structure, it is vital to ensure that it will be and will remain safe throughout its lifetime. Consequently, the design must include allowance for an appropriate “Factor of Safety”.

Although the choice of an appropriate factor of safety and the assessment of the actual factor for a given structure may appear to be a straightforward process, this is not necessarily the case. There are, in fact, a number of considerations that must be taken into account when determining a structure’s factor of safety.

Geotechnical design has for many years adopted a procedure which is known as the “Lumped Factor” approach. However, in recent years there have been moves away from this towards a “Factor on Strength” approach, which is considered to be more compatible with the limit state approach that is now used in structural engineering.

Although the Factor on Strength approach will become increasingly prevalent in UK practice, not least because it is the method that constitutes the basis of the new Eurocode for geotechnical design, the Lumped Factor approach is still quite commonly used and is likely to be so for some time. Consequently, it is important that both approaches are understood.

The basis of the Lumped Factor approach is explained in the following section. The Factor on Strength Method is not considered here, but will be explained fully later in the course.

The Lumped Factor of Safety Approach This approach can be summarised in terms of the following elements:

The design uses the best assessment of the actual soil strength

Using this soil strength, the value of the total action (usually a force or moment) causing instability of the structure is calculated

The value of the total action (again a force or moment) stabilising the structure is calculated on the same basis

The overall factor of safety is then given as:

Moment or Force ingDestabilisMoment or Force gStabilisinF Safety, of Factor

The design then requires that the structure is proportioned in order to ensure that a minimum factor of safety is achieved

It should be noted that the Lumped Factor approach only considers total failure of the structure – i.e. the Ultimate Limit State. No account is taken of any serviceability limits.

Page 5 of 9

Lumped Factor of Safety – Required Minimum Factor of Safety It is important to appreciate that the Lumped factor approach does not actually require that one single factor of safety will be appropriate for all different design conditions.

The actual factor of safety required for a check against any given failure mechanism may well be different from that required for another mechanism, as will become apparent in the following sections.

In practice, the required factor of safety for a design check against a particular failure mechanism will generally depend on knowledge drawn from previous design experience. This will be influenced by factors such as:

The type of structure

The type of soil present

How well the ground conditions are understood (e.g. the extent of ground investigation that has been carried out)

The consequences of failure (e.g. a lower factor of safety may be accepted for sliding of a 0.5m high garden wall than for a 20 metre high wall retaining a major highway)

Lumped Factors of Safety Required for Gravity Retaining Walls As noted above, the actual factors required will be influenced by site-specific conditions and guidance should usually also be sought from previous design experience and relevant codes of practice (e.g. BS8002 Retaining Structures), but the following gives some indication of the relevant order of required factors of safety.

Factor of Safety against Overturning Typically this should be in the order of 2.0 and this value is used in all of the following examples given for this course.

Factor of Safety against Sliding Typically this should be in the order of 1.5 – 2.0.

A value of 1.5 is used in all of the following examples given for this course.

Factor of Safety against Bearing Failure The Lumped Factor design approach generally requires that the imposed bearing stress is calculated based on the actual soil strength. The bearing capacity check then requires that:

Calculated Maximum Imposed Bearing Stress ≤ Allowable Soil Bearing Pressure

This assumes that the allowable soil bearing pressure will itself have been calculated using a lumped factor approach by dividing the ultimate foundation bearing pressure by a lumped factor of safety, which is usually 3.0. A second check is also required to ensure that there is no tension under the base, which could not be resisted and would cause uplift under the base of the wall, so:

Calculated Minimum Imposed Bearing Stress ≥ 0

Page 6 of 9

Gravity Wall Design – Example 1 The diagram on the following sheet shows a gravity retaining wall for which: Bulk density of retained fill = 18 kN/m3 ’ for the retained fill of 40° Surcharge Load = 10 kN/m2 Bulk density of concrete = 23.5 kN/m3 The angle of friction on the base of the wall, δ, is equal to 0.75’. Ka = (1 – sin ’) / (1 + sin ’) = 0.3572 / 1.6428 = 0.217

Stage 1 – Sliding Check: Sliding will occur if the total lateral force on the rear of wall due to the retained soil and the surcharge is greater than the frictional resistance between the base of the wall and the soil beneath it.

Working in Forces per metre width:

Element Force (kN)

(1) – Force due to Surcharge 0.217 x 10 x 3.0 = 6.51 kN

(2) – Force due to Retained Fill 0.217 x 18 x 32 / 2 = 17.577 kN

H = 24.09 kN

(3) – Weight of wall section 3.00 x 1.80 x 23.5 = 126.9 kN

V = 126.9 kN

Ratio of δ on base to = 0.75

Sliding Resistance = 126.9 x tan (0.75 x 40) = 73.27 kN

Factor of Safety on Sliding = 73.27 / 24.09 = 3.04

Factor of Safety vs. Sliding ≥ 1.5 O.K.

Surcharge – 10kN/m2

(1) (2) 3.00 m

1.80 m

(3)

A

Page 7 of 9

Stage 2 – Check on Overturning: Given that the lateral thrust behind the wall acts at some height above the base, this will result in an overturning moment on about the base of the wall. This moment will be resisted by the action of the weight of the wall. Should these moments lead to overturning of the wall, this will develop about point A and in order to check on failure by overturning moments must consequently be considered about this point. Note that the forces given in the table below have already been obtained from Stage 1, above.

Working in Forces per metre width:

Element Force (kN) Lever Arm about point A Moment about A (kNm)

(1) – Surcharge 6.51 3.0 / 2 = 1.5m 9.765

(2) – Retained Fill 17.577 3.0 / 3 = 1.0m 17.577

Overturning Moment = 27.342

(3) – Wall section 126.9 1.8 / 2 = 0.9m 114.21

Restoring Moment = 114.21

Factor of Safety on Overturning = 114.21 / 27.342 = 4.18

Factor of Safety vs. Overturning ≥ 2.0 O.K. Stage 3 – Check on Bearing Pressure: One effect of the horizontal thrust on the rear of the wall is to change the distribution of bearing pressures beneath the base of the wall. This results in an increase in bearing pressure under the toe and a decrease under the back of the wall as shown below:

e

max =

B6e1

BV min =

B6e1

BV

Base of Wall

Distribution of Bearing Stresses

Centreline of Foundation

Page 8 of 9

A

Thrust due to Fill

Weight of Wall

Thrust under base

L

The eccentricity of the base resultant can be found by once again considering moments about point A, at the toe of the wall as shown on the right: From this moment balance and noting that the thrust due to the fill causes the overturning moment and the weight of the wall causes the restoring moment calculated during Stage 2, above:

Wallof WeightMoment Overturn Moment RestoringL

And: L2Be

So that the maximum and minimum bearing stresses can then be calculated as follows:

Working in Forces per metre width and taking values from previous calcs:

Overturning Moment = 27.342 kNm L = 0.685m

Restoring Moment = 114.21 kNm B = 1.80m

Weight of Wall = 126.9 kN e = 0.215m

Stress under Base =

B6e1

BV max 121.0 kN/m2

min 20.0 kN/m2

For the Lumped Factor approach, the required conditions will be as follows:

Maximum Bearing Pressure ≤ Foundation allowable bearing pressure Minimum Bearing Pressure ≥ Zero (i.e. no tension)

The second of these is clearly satisfied; the first condition requires the imposed bearing pressure to be compared with the allowable bearing pressure, which will be calculated separately.

Page 9 of 9

The full calculation may be tabulated as set out below:

Calculations all per metre width – Taking Moments about Toe of wall

Element Force (kN) Lever Arm (m) Moment (kNm)

(1) 0.217 x 10 x 3.0 = 6.51 kN 3.0 / 2 = 1.5 m 9.765 kNm

(2) 0.217 x 18 x 32 / 2 = 17.577 kN 3.0 / 3 = 1.0 m 17.577 kNm

H = 24.09 kN MH = 27.342 kNm

(3) 3.00 x 1.80 x 23.5 = 126.9 kN 1.8 / 2 = 0.90 m 114.21 kNm

V = 126.9 kN MV = 114.21 kNm

ΣM = MV – MH 86.868 kNm

Dist. of base resultant from Toe of Wall = ΣM / V = 0.685 m ; e = (1.80 / 2) – 0.685 e = 0.215 m


Sliding Resistance = 126.9 tan (0.75 x 40) = 73.27 kN Factor of Safety vs. Overturning

Factor of Safety on Sliding = 73.27 / 24.09 = 3.04 = MV / MH = 114.21 / 27.342 = 4.18

Stress under Base =

B6e1

BV Max Stress = (126.9/1.80) x (1+ 6 x 0.215/1.80) = 121.0 kN/m2

Min Stress = (126.9/1.80) x (1– 6 x 0.215/1.80) = 20.0 kN/m2

Page 1 of 8

Retaining Walls – Part 2 Layered Soils All of the examples encountered so far have investigated the case of walls constructed within a single type of soil. In practice, it is not unusual for walls to be constructed in layered soils. Where this is the case, the wall design procedure is essentially identical to that for a single soil type except that separate Ka and Kp values have to be derived for each soil layer and applied appropriately. It is also important to ensure that the correct parameters are adopted for the soil that is actually present beneath the base of the wall when it comes to calculating the sliding resistance. The following example illustrates the procedure:

Where:

Soil 1 ’ = 30º; c’ = 0; bulk = 17 kN/m3 Soil 2 ’ = 35º; c’ = 0; bulk = 18.5 kN/m3 Unit Weight of Concrete = 23.5 kN/m3

Hence, for Soil 1, Ka = (1–sin 30) / (1+Sin30) = 0.333

for Soil 2, Ka = (1–sin 35) / (1+Sin35) = 0.271



(1) 0.333 x 10 x 0.5 = 1.665 kN 2.5 + 0.5 / 2 = 2.75 m 4.58 kNm

(2) 0.333 x 17 x 0.52 / 2 = 0.706 kN 2.5 + 0.5 / 3 = 2.67 m 1.89 kNm

(3) 0.271 x 10 x 2.5 = 6.775 kN 2.5 / 2 = 1.25 m 8.47 kNm

(4) 0.271 x 17 x 0.5 x 2.5 = 5.759 kN 2.5 / 2 = 1.25 m 7.20 kNm

(5) 0.271 x 18.5 x 2.52 / 2 = 15.667 kN 2.5 / 3 = 0.83 m 13.00 kNm

H = 30.572 kN MH = 35.14 kNm


(1) (2)

2.50 m

3.00 m

1.70 m

(3) (4)

Soil 1

2.50 m

Soil 2 (5) [A]

[B]

Page 2 of 8

[A] 1.7 x 3.0 x 23.5 = 119.85 kN 0.8 + 1.70 / 2 = 1.65 m 197.75 kNm

[B] 0.5 x (2.50– 1.70) x 3.0 x 23.5 = 28.2 kN

2 x 0.8 / 3 = 0.53 m 14.95 kNm

V = 148.05 kN MV = 212.70 kNm

ΣM = MV – MH 177.56 kNm

Dist. of base resultant from Toe of Wall = ΣM / V = 1.20 m ; e = (2.50 / 2) – 1.20 m e = 0.05 m




Stress under Base =

B6e1


Min Stress = (148.05/2.50) x (1- 6 x 0.05/2.50) = 52.1 kN/m2

General Design Considerations There are a number of factors that must generally be taken into account when designing any retaining wall. These are discussed below: Surcharge Load Depending on its exact location, a surcharge load imposed on the fill retained behind the wall can tend to have a destabilising effect on the wall. This effect must be taken into account in the wall design and this is conventionally done by considering a standard nominal surcharge of 10 kN/m2 applied to the surface of the retained fill in such a way as to have the greatest effect on the wall stability. This surcharge is deemed to allow for any unexpected temporary loading in this area such as traffic movements, short term storage of materials etc. In applying this surcharge, care must be taken to ensure that it is applied in the correct location, as shown below. It is also possible in certain cases that it may be inappropriate to apply a surcharge as it is not possible for a load to actually be applied here, for example in the case of an extremely steep slope or where the surface of the fill is not actually accessible (this may, for example, happen on the bank seat of a bridge)

Page 3 of 8

Surcharge applied incorrectly – stabilises wall

Surcharge applied correctly – destabilises wall

Surcharge applied incorrectly – stabilises wall

Surcharge applied correctly – destabilises wall

May not be appropriate to apply surcharge load on steep slope

Allowance for Over-excavation in front of the Wall Many retaining walls depend to some extent on the passive resistance from the soil above the base and in front of the wall. The difficulty with the contribution of this element of wall stability is that it can be unreliable. In particular, if someone comes along and excavates a trench in front of the wall, for example to install wall drainage, then this passive resistance will be lost. In order to account for this possibility, retaining walls are generally designed with an allowance made for an excavation taking place in front of them. In accordance with British Standards, this allowance should be either 10% of the wall height or 0.5 m, whichever is less. Factoring of the Passive Resistance As noted above, many retaining walls rely at least in part on the passive resistance generated at the front of the wall.

H 0.1H

or 0.5m

Page 4 of 8

It must, however, be appreciated, that the development of passive resistance is much less reliable than the development of the active thrust, primarily due to the fact that significantly greater strains are required to mobilise passive resistance than active thrust. In order to take account of the unreliability of the passive resistance, it is generally factored when designing wall, with a factor of 2 usually applied, so that: Design Passive Resistance = Theoretical Passive Resistance / 2.0 Effects of including allowance for wall friction The Rankine Theory considers the case of lateral earth pressures where the wall is frictionless. This is unrealistic in practice, as there will always be friction between the wall and the retained fill. In terms of retaining wall design, friction on the rear of the wall will generally be advantageous to wall stability in that it will act in a direction such as to oppose the tendency of the wall to overturn, as shown in the adjacent diagram. Consequently, any efficient design will be required to take account of wall friction. In practice, the two effects of wall friction are:

1. To reduce the horizontal component of lateral earth pressure on the back of the wall

2. To introduce an additional restoring moment about the toe of the wall

In calculating the effects of wall friction, the angle of friction between the wall and the soil (the angle of wall friction) for a purely frictional soil is generally taken as a proportion of the shear strength of the soil, with the ratio being dependent on the wall material. BS 8002, the code of practice for retaining walls suggests that in the absence of test data, the angle of wall friction, , should be taken as: Design tan = 0.75 Design tan ’ For the BS8002 approach, where design strengths are factored, this equates approximately to:

Representative value of = 2/3 x Representative value of ’ In the case of a soil which exhibits effective cohesion, or when considering the undrained condition, it is also possible to make an allowance for wall adhesion, cw, where:

Design cw = 0.5 x Representative cohesion

Effect of Wall Friction

Page 5 of 8

Design for Effects of Groundwater Should a situation occur where groundwater can be impounded behind the wall, such as a combination of heavy rain and poor wall drainage, then the effects of the presence of the groundwater will need to be taken into account in the design. Irrespective of whether the wall design will be required to take account of the presence of groundwater, allowance will need to be made in the design to permit drainage of water from the retained fill. Typical examples of how this is achieved are shown in the following diagram:

After Smith

Where the wall design does need to take account of the presence of groundwater in the retained fill, the approach will depend on the nature of the wall drainage. Where the wall drainage is provided by means of a vertical drain at the rear of the wall, the design will need to consider the water pressures that will be induced by the flow net shown below: The design adopts a wedge failure mechanism approach, but allowance must be made for an additional force due to the water pressure acting on the inclined face of the soil wedge as shown.

Page 6 of 8

After Smith

As an alternative, if an inclined drain is provided as shown below, so that the drain lies below the level of the critical failure wedge, the water pressure acting on the inclined face of the wedge will be zero. Under these circumstances, no additional water pressure will act on the failure wedge and the lateral pressure acting on the rear of the wall will be exactly equal to that where there is no water behind the wall.

After Smith

In practice, retaining walls are generally designed to include an appropriately sized wedge of free-draining granular fill directly behind the wall face together with weep holes to allow drainage through the face or a toe drain behind the wall. In this way, any groundwater which might collect behind the wall will automatically be drained away and the lateral pressures on the rear of the wall will be the same as those which would be generated for a dry fill.

Free – draining structural granular fill required here

45°

Example – Highways Agency Specification for Retaining Structures

Note: Drainage also required – not shown

Page 7 of 8

Effects of Compaction Wall construction necessitates the use of compaction equipment, typically rollers, to compact the fill placed behind the wall. It has been suggested that the use of compaction plant at the rear of the wall can be modelled as being equivalent to a line load, as shown in the diagram below.

Simple elastic theory can be used to calculate the values of the vertical and lateral stresses due to the line load. For a roller placed immediately at the rear of the wall, the theory predicts that;

z π

2Qz

Where; Q = load/unit width

z = depth

Ingold suggested that if the lateral stress due to this value of z is maintained when the roller is removed, then the soil could fail vertically due to this high stress (i.e. the lateral stress behind the wall will tend towards the passive value, Kpz). Based on this assumption, Ingold suggested that the result will be that a maximum lateral soil stress, Pmax will develop at a depth, zc, where;

π2QγPmax and

πγ2QKazc

This is shown on the left below

After Craig

Wall fill will be placed in layers and as a result of this, the lateral pressure in a given soil layer will be maintained at Pmax until a sufficient number of layers of soil have been placed above this layer such that the value of the active earth pressure, Kaz,

Q/m

z

x

x

z

Page 8 of 8

exceeds the “locked in” pressure. Once this happens, the lateral earth pressure will revert to the conventional hydrostatically distributed active value. This assumption results in the profile of lateral earth pressures shown on the right hand diagram in the figure above, where;

πγ2Q

Ka1za

Page 1 of 12

Gravity Retaining Walls Design Assuming a Virtual Back of a Wall As noted previously, the lateral earth pressures on the rear of a gravity retaining wall may be calculated using either the Rankine or Coulomb Theory as appropriate. In practice, structures that resist loads purely due to their own self weight can be somewhat inefficient. A reinforced concrete cantilever wall is, for example, more efficient than a gravity wall because it is proportioned in such a way that the weight of part of the fill material acts to stabilise it, as shown below. In this case, the lateral earth pressures tending to destabilise the wall are those that act on a “virtual surface” some distance behind the wall rather than those actually acting on the back of the wall. In addition to the design of cantilever walls, a similar approach may also be adopted for walls where the rear face slopes or for similar forms of construction where it is clear that a “block” of soil will contribute to the stability of the wall. The latter category will, for example, include reinforced earth retaining walls, as illustrated below: Wall with Sloping Rear Face Reinforced Earth Wall

Weight of soil here acts to stabilise wall

W

Lateral earth pressure

calculated on “virtual surface”

here



here



here

Page 2 of 12

Effective Wall height for design

The use of a virtual back in wall design can result in there being options for the method of design to be used. For example, in the case of a wall with a sloping back, design may follow one of two procedures:

Procedure (a): Using this approach, which considers the forces which actually act on the rear of the wall, a Coulomb wedge analysis is utilised. The advantages of this approach are that the effects of wall friction can be considered. The disadvantage is that it can be more complex to calculate the lever arm for the forces on the back of the wall (particularly the wall friction)

Procedure (b)

Where a virtual back of the wall is assumed in the design, the effective height for design may be greater than the actual wall height (as shown). The wedge of soil behind the virtual back of the wall will also contribute to wall stability and the lateral earth stresses will be calculated on the basis of this height. Calculation of lateral stresses can be carried out using a Rankine approach as there will be no wall friction on the virtual plane considered.

In theory, the more critical of the two mechanisms (i.e. the one with the lowest factor of safety) will govern. In practice, either of the two approaches would generally be considered acceptable, although procedure (b) would usually be followed if the design is to be completed by hand calculation. Procedure (b) might also be expected to give a lower factor of safety as the effect of wall friction is not taken into account. Gravity Wall Design – Example The diagram on the following sheet shows a gravity retaining wall for which: Bulk density of retained fill = 18 kN/m3 ’ for the retained fill of 40° Surcharge Load = 10 kN/m2 Bulk density of concrete = 23.5 kN/m3 The angle of friction on the base of the wall, δ, is equal to 0.75’. Ka = (1 – sin ’) / (1 + sin ’) = 0.3572 / 1.6428 = 0.217

Page 3 of 12

The required calculation is then as set out below:



(1) 0.217 x 10 x 5.0 = 10.85 kN 5.0 / 2 = 2.5 m 27.125 kNm

(2) 0.217 x 18 x 52 / 2 = 48.825 kN 5.0 / 3 = 1.667 m 81.39 kNm

H = 59.68 kN MH = 108.52 kNm

Stem 0.40 x (5.0 – 0.5) x 23.5 = 42.3 kN 3.00 – 1.50 – 0.20 = 1.30 m

54.99 kNm

Base 3.00 x 0.50 x 23.5 = 35.25 kN 3.00 / 2 = 1.50 m 52.875 kNm

Soil over Base 1.50 x (5.00 – 0.50) x 18 = 121.5 kN 3.00 – 1.50 / 2 = 2.25 m 273.375 kNm

V = 199.1 kN MV = 381.24 kNm

ΣM = MV – MH 272.72 kNm

Dist. of base resultant from Toe of Wall = ΣM / V = 1.37 m ; e = (3.00 / 2) – 1.37 e = 0.13 m




Stress under Base =

B6e1


Min Stress = (199.1/3.00) x (1– 6 x 0.13/3.00) = 49.1 kN/m2


(1)

(2)

3.00 m

1.50 m

0.50 m

0.40 m 5.00 m

Page 4 of 12

Factors of Safety in Geotechnical Design As noted previously, the design given above is based on a Lumped Factor approach with the following required factors of safety:

Factor of Safety vs. Sliding ≥ 1.5

Factor of Safety vs. Overturning ≥ 2.0

Maximum Bearing Pressure ≤ Foundation allowable bearing pressure

Minimum Bearing Pressure ≥ Zero (i.e. no tension)

As also noted previously, the wall design could also have been undertaken using a Factor on Strength approach.

The following sections describe the Factor on Strength approach and repeat the wall design example given above using this method.

The Factor on Strength Approach The Factor on Strength Approach has been developed as an alternative to the Lumped Factor Approach for defining the design factor of safety for geotechnical structures.

In the Factor of Safety on Strength Approach, the structure is design using a Soil Design Strength which differs from the Actual Soil Strength.

The Soil Design Strength is defined as:

FStrength Soil ActualStrength Design Soil

Where: F is the Factor on Strength

In this approach the factor of safety for the structure arises from the fact that it is being designed assuming a soil strength that is less than the actual soil strength

This is directly analogous to the use of partial safety factors on material strengths in structural engineering.

Note that this approach is significantly different to the Lumped Factor Approach.

Using Factors of Safety on Soil Strength When adopting this approach, it must be borne in mind that the strength of any soil will generally have two components – a frictional component [’ or u] and a cohesion component [C’ or Cu]

The intention of this approach is actually to directly factor the shear strength of the soil, rather than these two individual components. Consequently, the factors of safety are applied as described below. It is also important to appreciate that the partial factors applied to each of the two strength components are likely to be different.

Page 5 of 12

Frictional Component of Soil Strength

The shear strength of a soil is actually given as tan ’ or tan u.

As a result, the design strength is given as:

friction

actualdesign F

' tan' tan

or

friction

actualdesign F

tan tan u

u

Cohesion Component of Soil Strength

The cohesion values, C’ and Cu give a direct measure of the soil shear strength.

As a result, the design strength is given as:

cohesion

actualdesign F

C'C' or

cohesion

actualUdesignU F

CC

Partial Factors on Applied Loads One further element of the Factor on Strength Method is that it will generally require that externally applied loads must also be factored by a partial load factor.

This allowance is made to account for any uncertainty in the magnitude of the applied loads and for the fact that there is no lumped safety factor included in the final stability assessment.

These partial factors are effectively directly analogous to the partial load factors used in structural design.

Where partial load factors are applied, this is done in accordance with the following equation:

Load Applied x ActualF Load AppliedDesign load partial

The Factor on Strength Design Process This approach can be summarised as follows:

An assessment is made of the actual soil strength

The actual soil strength is then factored using the appropriate strength factor to give the design soil strength

Any external applied loads are multiplied by the appropriate partial factor to give the applied design loads

Using the design soil strength and design loads, the value of the total action (usually a force or moment) causing instability of the structure is calculated

The value of the total action (again a force or moment) stabilising the structure is calculated on the same basis as for the destabilising action

The structure is then acceptable providing that:

Moment or Force ingDestabilis Moment or Force gStabilisin

This is the only check that is required at this stage and no further safety factor(s) needs to be considered in the design.

Page 6 of 12

Values of Partial Factors to be used in Geotechnical Design As with other aspects of geotechnical design, the partial factors that are used in any actual design will depend on a number of factors related to the specific component of the design being considered.

Guidance on the use of appropriate partial factors to be used in any given case will usually be obtained by reference to the relevant design standard.

For design of retaining walls, BS: 8002 suggests the use of the following Factors on Soil Strength (which are described in the code as “Mobilisation Factors”):

Parameter Factor on Strength

[Mobilisation Factor] Value used for design

Cu 1.5 1.5CU

C’ 1.2 1.2C'

tan ’ 1.2 1.2

' tan' tan design

Page 7 of 12

Design to Eurocode 7

As noted previously, Eurocode 7 adopts a limit state approach with partial factors applied to actions, material properties and resistances.

British practice follows design approach 1 when considering the GEO and STR Limit states. This requires two load combinations to be considered as follows:

Load combination 1 (A1+ M1 + R1)

Load combination 2 (A2 + M2 + R1)

Now, considering the previous design again, the first point to consider is that Eurocode 7 does not require a surcharge to be applied, so that the problem becomes simplified to:

We then need to identify the actions, which will be:

The force due to the weight of the soil plus the weight of the wall

The thrust due to the retained backfill

The material properties will be:

The ’ value for the retained granular fill

The bulk density of the retained fill,

In this case there are no resistances to be considered. We can then establish the relevant partial factors and carry out the sliding and resisting checks for each of the two required load combinations as follows:

3.00 m

1.50 m

0.50 m

0.40 m 5.00 m

(1)

Page 8 of 12

Load Combination 1 (A1 + M1 + R1) For the sliding check:

Action Type Partial Factor Case A1

Force due to the weight of soil plus wall Permanent – favourable G, stb = 1.0

The thrust due to the retained backfill Permanent – unfavourable G, dst = 1.35

Material Property Partial Factor Case M1

Coefficient of shearing resistance (tan ’) ’ = 1.0 Weight Density = 1.0

Resistance Partial Factor Case R1

Bearing R; v = 1.0 Sliding R; h = 1.0

For ’ = 40º and ’ = 1.0: tan ’ design = tan 40º / 1.0 and ’ design = 40º As the angle of friction on the base of the wall, δ, is equal to 0.75’,

design = 0.75 x ’ design = 30º Hence:

Ka design = (1 – sin 40) / (1 + sin 40) = 0.3572 / 1.6428 = 0.2174 For bulk density of retained fill = 18 kN/m3 design = 18 / 1.0 = 18 kN/m3 For bulk density of concrete = 23.5 kN/m3 design = 23.5 / 1.0 = 23.5 kN/m3 For failure by sliding, the sliding resistance should be factored by R; h = 1.0 We should also then re-assess the partial factors for failure by overturning. The factors on actions and material properties will clearly be the same as the same actions will be favourable or unfavourable. However, there will be no actual soil resistance to overturning, so no R factor will be necessary The calculations are then as follows:

Page 9 of 12

3.00 m

1.50 m

0.50 m

0.40 m 5.00 m

(1)

Calculations all per metre width – Load Combination 1 Note that all partial factors are shown in bold and underlined


(1) 1.35 x 0.2174 x 18 x 52 / 2 = 66.035 kN 5.0 / 3 = 1.667 m 110.08 kNm

H = 66.035 kN MH = 110.08 kNm

Stem 0.40 x (5.0 – 0.5) x 1.0 x 23.5 = 42.3 kN 3.00 – 1.50 – 0.20 = 1.30 m 54.99 kNm

Base 3.00 x 0.50 x 1.0 x 23.5 = 35.25 kN 3.00 / 2 = 1.50 m 52.875 kNm

Soil over Base 1.50 x (5.00 – 0.50) x 1.0 x 18 = 121.5 kN 3.00 – 1.50 / 2 = 2.25 m 273.375 kNm

V = 1.0 x 199.1 = 199.1 kN MV = 381.24 kNm


Sliding Resistance = 199.1 tan (0.75 x 40) = 114.95 kN

Check on Sliding:

Sliding Resistance > H 1.0 x 114.95 > 66.035 OK

Check on Overturning:

MV > MH 381.24 > 110.08 OK

Bearing Pressure Calculation is not included here – the calculated bearing pressure will need to be checked against a factored bearing resistance where the factor used will be R; v = 1.0

Page 10 of 12

Load Combination 2 (A2 + M2 + R1) For the sliding check:

Action Type Partial Factor Case A2

Force due to the weight of soil plus wall Permanent – favourable G, stb = 1.0

The thrust due to the retained backfill Permanent – unfavourable G, dst = 1.0

Material Property Partial Factor Case M1

Coefficient of shearing resistance (tan ’) ’ = 1.25 Weight Density = 1.0

Resistance Partial Factor Case R1

Bearing R; v = 1.0 Sliding R; h = 1.0

For ’ = 40º and ’ = 1.25: tan ’ design = tan 40º / 1.25 = 0.6713 and ’ design = 33.87º As the angle of friction on the base of the wall, δ, is equal to 0.75’,

design = 0.75 x ’ design = 25.4º Hence:

Ka design = (1 – sin 33.87) / (1 + sin 33.87) = 0.4427 / 1.5573 = 0.2843 For bulk density of retained fill = 18 kN/m3 design = 18 / 1.0 = 18 kN/m3 For bulk density of concrete = 23.5 kN/m3 design = 23.5 / 1.0 = 23.5 kN/m3 For failure by sliding, the sliding resistance should be factored by R; h = 1.0 We should also then re-assess the partial factors for failure by overturning. Again, the factors on actions and material properties will clearly be the same as the same actions will be favourable or unfavourable. However, there will be no actual soil resistance to overturning, so no R factor will be necessary The calculations are then as follows:

Page 11 of 12

3.00 m

1.50 m

0.50 m

0.40 m 5.00 m

(1)

Calculations all per metre width – Load Combination 1 Note that all partial factors are shown in bold and underlined


(1) 1.00 x 0.2843 x 18 x 52 / 2 = 63.97kN 5.0 / 3 = 1.667 m 106.64 kNm

H = 63.97 kN MH = 106.64 kNm

Stem 0.40 x (5.0 – 0.5) x 1.0 x 23.5 = 42.3 kN 3.00 – 1.50 – 0.20 = 1.30 m 54.99 kNm

Base 3.00 x 0.50 x 1.0 x 23.5 = 35.25 kN 3.00 / 2 = 1.50 m 52.875 kNm

Soil over Base 1.50 x (5.00 – 0.50) x 1.0 x 18 = 121.5 kN 3.00 – 1.50 / 2 = 2.25 m 273.375 kNm

V = 1.0 x 199.1 =199.1 kN MV = 381.24 kNm


Sliding Resistance = 199.1 tan (0.75 x 33.87) = 94.55kN

Check on Sliding:

Sliding Resistance > H 1.0 x 94.55 > 66.035 OK

Check on Overturning:

MV > MH 381.24 > 106.64 OK

Bearing Pressure Calculation is not included here – the calculated bearing pressure will need to be checked against a factored bearing resistance where the factor used will be R; v = 1.0

Page 12 of 12

Parameter Symbol EQU GEO/STR - Partial factor set A1 A2 M1 M2 R1 R2 R3 Permanent action (G) Unfavourable γG, dst 1.1 1.35 1.0 Favourable γG, stb 0.9 1.0 1.0 Variable action (Q) Unfavourable γQ, dst 1.5 1.5 1.3 Favourable - - - - Accidental action (A) Unfavourable γA, dst 1.0 1.0 1.0 Favourable - - - - Coefficient of shearing resistance (tan ') γφ' 1.25 1.0 1.25 Effective cohesion (c') γc' 1.25 1.0 1.25 Undrained shear strength (cu) γcu 1.4 1.0 1.4 Unconfined compressive strength (qu) γqu 1.4 1.0 1.4 Weight density (γ) γγ 1.0 1.0 1.0 Bearing resistance (Rv) γRv 1.0 1.4 1.0 Sliding resistance (Rh) γRh 1.0 1.1 1.0 Earth resistance (Rh) γRe 1.0 1.4 1.0 Design Approach 1 Combination 1 (A1+M1+R1) Piles & anchors: (A1+M1+R1) Combination 2 (A2+M2+R1) (A2 + (M1 or M2) + R4) Design Approach 2 (A1+M1+R2) Design Approach 3 (A1or A2) +M2+R3

Table 1 – Eurocode 7 Partial Factors Reference: Smith’s Elements of Soils 8th Edition Pg 268

Cantilever Sheet Pile Walls In cantilever sheet pile wall construction, heavy steel sheet piles are driven into the ground prior to excavation taking place. When excavation is carried out, the soil behind the sheets is retained by their cantilever action. Cantilever sheet pile walls tend to be prone to movement and are generally only used for temporary structures or for permanent walls of low height in sands and gravels.

Figure 1 After Craig – Figure 6.22

Additional Design Requirements In addition to the above, there are also two other important criteria that are adopted in practical design of sheet pile walls as follows:

1. An additional depth is allowed at the front of the wall to allow for the possibility that an accidental excavation may occur here. BS 8002 (Earth Retaining Structures) suggests that this allowance should be 10% of the retained height up to a maximum allowance of 0.5 m.

2. A surcharge load is applied behind the wall to allow for loading in this area. The applied surcharge will depend on the load that might be expected, but general practice allows for a minimum nominal surcharge of 10 kN/m2.

Design of Cantilever Walls by Moment Balance Design of cantilever walls is carried out by finding a moment balance about the assumed point of fixity, which is shown as point C in Figure 1(c). As has been discussed previously, generation of full active and passive pressure requires a certain strain to develop in the soil. Generation of full passive resistance will, in particular, require quite significant strains to develop. Several different design methods have been proposed for cantilever retaining walls depending on different assumptions of precisely what earth pressures are mobilised. The four main design methods are described below:

Gross Pressure Method (CP2 Method)

This method commences with the assumption that full active pressure will develop behind the wall and full passive pressure in front of the wall. The approach requires that there is a factor of safety against failure and this is obtained by dividing the calculated moment due to the passive earth pressure by a factor, Fp, and then equating this to the moment due to the active earth pressure plus the net moment due to any water pressure acting on the wall.

moment PressureNet Water Pressure Active fromMoment F

Pressure Passive fromMoment

p

Net Pressure Method (British Steel Method) This method was developed by British Steel and adopts the pressure distribution shown bounded by the solid lines in the diagram opposite, where the net pressures acting on the rear and front of the wall are calculated by subtracting the pressures acting on one side of the wall from those on the other side, with the effects of water pressure included in these calculations. As for the Gross Pressure method, a factor of safety is applied to the moment due to the passive pressure in order to ensure that the wall design is safe. In this case:

Water)(Earth Pressure ActiveNet fromMoment F

Water)(Earth Pressure PassiveNet fromMoment

np

As a supplier of steel sheet sections, British Steel contributed significantly to the literature on steel sheet pile design and produced a Piling Handbook, which went through a number of editions. The part of the business dealing with sheet piles was subsequently sold to Arcelor, whose web site includes a lot of data relevant to sheet pile design. In particular, see www.arcelor.com/sheetpiling under documentation, where a copy of the current Piling Handbook can be downloaded.

Net Available Passive Resistance Method (Burland and Potts) This method was developed by Burland and Potts and adopts the pressure distribution shown bounded by the solid lines in the diagram opposite. The active pressure is assumed to reach a maximum value at the level of the ground in front of the wall. This value of active pressure is then maintained to the base of the wall while the net active pressure is calculated as the gross passive pressure less the drop in active pressure below the theoretical hydrostatic value. As for the previous methods, a factor of safety is incorporated in the moment balance in order to ensure that the design is safe, as follows:

moment PressureNet Water Pressure Active Modified fromMoment F

Pressure Passive Modified fromMoment

r

Factor of Safety on Soil Strength Method This method differs from the previous three methods in that the factor of safety for the wall is applied to the soil strength rather than to the moments induced by the soil. Using this method for a granular soil and effective stress analysis, the soil strength for design will be determined as:

tan(’design) = (tan measured’) / Fs

The unfactored gross soil pressures based on the factored soil strength are then used in the analysis, so that:

moment PressureNet Water Pressure Active fromMoment Pressure Passive fromMoment Factors of Safety to be used in the above Analyses The appropriate factors of safety for the above analyses will depend on the circumstances of the particular problem. The following table summarises the relevant factors for a number of cases:

Method Effective Stress Analysis

’ ≤ 20° 20° < ’ ≤ 30° ’ > 30°

Factor on Strength, Fs 1.2 1.2 1.2

Gross Pressure Factor, Fp 1.5 1.5 – 2.0 2

Net Pressure Factor, Fnp 2 2 2

Net Available Resistance, Burland and Potts, Fr

2 2 2

Example – The Design Procedure – Gross Pressure Method Design follows a set procedure in which the following steps need to be taken:

1. Adopt the simplified model shown in Figure 1(c), above 2. Find the appropriate design values of Ka and Kp 3. Determine the lateral earth pressures in front of and behind the wall, making

allowance for groundwater as necessary. Note that the design is almost always carried out on the basis of an effective stress approach

4. Use a moment balance to find the depth to the point of fixity 5. Increase the proposed embedment of the pile by 20% to provide the necessary

force, R, at the rear of the wall 6. Find R 7. Check that the increase in depth allowed in step 5 is adequate to provide R

Examples: The following two examples illustrate the procedure. Example 1 is a basic illustration and excludes the standard assumptions for overdig and surcharge. Example 2 is a more general example, including these and the presence of groundwater. This is, however, a special case where the groundwater levels in front of and behind the wall are the same, in which case the water pressures on the front and the back of the wall are equal and can be ignored in the calculation.

Example 1 Consider the case of a wall with a 2 metre retained height of fill constructed in a soil for which ’ = 30° and bulk = 18kN/m3

The groundwater table is well below the level of the base of the wall and ignoring any allowance for accidental overdig or surcharge load but taking a factor of safety of 2.0 on the passive pressure (Gross Pressure Method)

Assuming there is no wall friction:

Ka = (1 – sin(1 + sin) = 0.333

Kp = (1 + sin(1 – sin) = 3.0

Taking moments about point C

Moment due to active force:

Mmt. = 0.5 Ka bulk (h+d)2 x (h+d)/3

= 0.5 x 0.333 x 18 x (2 + d)3/3

= 8 + 12 d + 6 d 2 + d 3

Moment due to passive force:

Mmt. = 0.5 Kp bulk d2 x d/3

= 0.5 x 3.0 x 18 x d3/3

= 9 d 3

Then, equating the active and passive moments and allowing for a factor of safety of 2.0 on the passive gives:

4.5 d 3 = 8 + 12 d + 6 d 2 + d 3

Which can be solved by trial and error to give d = 3.07 m

The final embedment is then taken as 3.07 x 1.2 = 3.68 metres

It is then required to carry out a force balance to calculate the required value of R for horizontal stability:

Active force = 0.5 Ka bulk (h+d)2 = 0.5 x 0.333 x 18 x 5.072 = 77.11 kN

Passive force = 0.5 Kp bulk d 2/2 = 0.5 x 3.0 x 18 x 3.072 /2 = 127.24 kN

Hence, R = 127.24 – 77.11 = 50.13 kN

This force must be generated in the additional length of pile below point C

The two force elements generated will be due to the passive load on the back of the wall and the active force on the front of the wall (see figure 1(b), above). In each case there will be two elements – one due to the effective surcharge imposed by the fill above the level of point C and one due to the soil self weight below level C. Overall, the lateral effective stress profile will be as shown below:

Passive Pressures:

Force 1, due to overburden surcharge = Kp x bulk x (h+d) x 0.2 d = 168.1 kN

Force 2, due to soil self weight = 0.5 x Kp x bulk x (0.2 d) 2 = 10.18 kN

Active Pressures:

Force 3, due to overburden surcharge = Ka x bulk x d x 0.2 d = 11.31 kN

Force 4, due to soil self weight = 0.5 x Ka x bulk x (0.2 d) 2 = 1.13 kN

Available force = Passive – Active = 168.1 + 10.18 – 11.31 – 1.13 = 165.84 kN

Required force = 50.13 kN < Available Force of 165.84 Hence, OK

R C

2 3

4

0.2 d 1

Active Pressure Passive Pressure

Example 2 A sheet pile wall is to be installed to support a 3m high cut in sand of bulk density 18kN/m3 and ’ 30°. If the saturated bulk density of the sand is 20kN/m3 and a factor of safety of 2 is to be applied to the passive earth pressure, find the required length of sheet piles required if the groundwater level is 4 m below the existing ground level. In carrying out the design, the following points should be considered:

10% of the wall height is 0.3m, so the design height will be 3.3m

A 10kN/m2 surcharge must be applied behind the wall

The effective vertical stress due to the soil below the water table can be taken as (sat – water), i.e. 20–9.8 = 10.2kN/m3

In the absence of other data, take:

Ka = [(1-sin)/(1+sin)] = 0.333 Kp = [(1+sin)/(1-sin)] = 3.0

The required wall and the effective earth pressures acting on it will then be as shown below, where d is the depth of embedment below the water table and is what we are required to determine: Figure 2: Design Example 2 – Configuration and Active and Passive Earth Pressures By considering the earth pressures shown in Figure 2 and by taking moments about point C, the following values are obtained:

3.30m

0.70m

d

1 2

3

4

5

6 7

R

Surcharge = 10kN/m2

W.T.

C

Passive Earth Pressures Active Earth Pressures Retaining Wall Configuration

Calculations all per metre width

Force (kN) Lever Arm (m) Moment (kNm)

(1) 0.333 x 10 x (d + 4.0) = 3.33d + 13.33 d/2 + 4.0/2 1.665d2 + 13.32d + 26.64

(2) 0.5 x 0.333 x 18 x 4.02 = 47.95 d + 4.0/3 47.95d + 63.94

(3) 0.333 x 18 x 4.0 x d = 23.976d d/2 11.988d2

(4) 0.5 x 0.33 x 10.2 x d2 = 1.689d2 d/3 0.566d3

(5) – (0.5 x 3.0 x 18 x 0.72) / 2 = – 6.62 d + 0.7/3 – 6.62d – 1.544

(6) – (3.0 x 18 x 0.7 x d) / 2 = – 18.9d d/2 – 9.45 d2

(7) – (0.5 x 3.0 x 10.2 x d2) / 2 = – 7.65d2 d/3 – 2.547 d3

Note – densities below the groundwater table are taken as (sat – water) in order to obtain soil effective vertical and lateral stresses. Table 1: Design Example – Calculated Bending Moments Balancing the moments then gives:

– 1.981 d3 + 4.203 d2 + 54.65 d + 89.036 = 0

Giving: d3 = 44.94 + 27.59 d + 2.12 d2

Which is best solved by trial and error to find d = 7.0 m

Depth of Embedment As noted above, the depth of embedment should be increased by 20%

In solving the problem, the actual depth of embedment is 0.70 m + d = 7.7 m.

i.e. actual required embedment = 1.2 x 7.7 = 9.24 metres (1.54 m increase in length)

Required Value of R R is the force generated by the passive soil pressure on the back of the wall below point C, where the wall tends to rotate back into the soil.

For horizontal equilibrium:

Total passive Force = Total Active Force + R

So R = Total Calculated Passive Force – Total Calculated Active Force

Hence, R = – ([(5)+(6)+(7)] – [(1)+(2)+(3)+(4)])

Substituting d = 7.0 into the values calculated in Table 1, allowing for the necessary changes of sign then gives:

R = [6.62 + 132.30 + 374.85] – [36.63 + 47.95 + 167.83 +83.22] = 178.13 kN This force has to be generated on the back of the sheet piles by the extra 20% length added above – i.e. between point C and a point 1.54 metres below this.

The situation is shown in Figure 3, where forces 8, 9 and 10 will be passive earth pressures and 11 and 12 will be active pressures Figure 3: Design Example2 – Active and Passive Earth Pressures below point of fixity

Force (kN)

(8) 3 x 10 x 1.54 = 46.2

(9) 3 x [(4.0 x 18) + (d x 10.2)] x 1.54 = 662.508

(10) 0.5 x 3 x 10.2 x 1.542 = 36.29

(11) 0.333 x [(0.7 x 18) + (d x 10.2)] x 1.54 = 43.12

(12) 0.5 x 0.333 x 10.2 x 1.542 = 4.03

Note: In this case no reduction factor is applied to the passive earth pressures as owing to where they are generated they are more reliable R available = 46.2 + 662.508 + 36.29 – 43.12 – 4.03 = 697.85 kN > 178.13 kN OK

3.30m

0.70m

d

1 2

3

4

5

6 7 R

Surcharge = 10kN/m2

W.T.

C

10 11

12

0.2 d 8 9

Eurocode 7 Check Considering design example 2 as above but adopting a Eurocode 7 analysis using a factor on strength approach.

Firstly:

Eurocode 7 does not require any allowance for surcharge

Eurocode 7 requires the same allowance for over-excavation in front of the wall

Hence, in this case there is no surcharge to be considered but the overall depth of excavation for design will again be 3.30 metres (3.0 metres + 10%)

Figure 4: Design Example 2 – Configuration for Eurocode 7 Design As before:

The bulk density of the sand above the water table is 18kN/m3

The saturated bulk density of the sand is 20kN/m3

And the angle of internal friction for the sand, ’ is 30°.

We now consider the two load combinations to be considered for Eurocode 7 Design approach 1 as follows:

Load combination 1 (A1+ M1 + R1)

Load combination 2 (A2 + M2 + R1)

3.30m

0.70m

d

1

2

3

4

5 6 R

W.T.

C

Passive Earth Resistances Active Earth Pressures Retaining Wall Configuration

Load Combination 1 (A1+ M1 + R1)

Action Type Partial Factor

Case A1 The thrust due to the retained

backfill Permanent – unfavourable G, dst = 1.35

Material Property Partial Factor

Case M1 Coefficient of shearing resistance (tan ’) ’ = 1.0

Weight Density = 1.0

Resistance Partial Factor

Case R1 Earth Resistance R; e = 1.0

For ’ = 30º and ’ = 1.0: tan ’ design = tan 30º / 1.0 and ’ design = 30º Hence:

Ka design = (1 – sin 30) / (1 + sin 30) = 0.5 / 1.5 = 0.3333

Kp design = (1 + sin 30) / (1 - sin 30) = 1.5 / 0.5 = 3.0 For bulk density of retained fill = 18 kN/m3 design = 18 / 1.0 = 18 kN/m3 For bulk saturated bulk density of fill = 20 kN/m3 design = 20 / 1.0 = 20 kN/m3 Effective unit weight of soil below water table = (20 – 9.8) = 10.2 kN/m3 So, as = 1.0, again for this value: design = 10.2 kN/m3 The partial load factor for the action due to the active earth pressure behind the wall, G, dst = 1.35, will be applied to the force due to the active earth pressure behind the wall. The partial factor for resistance R; e = 1.0, which will be applied to the force due to the passive earth pressure in front of the wall. The calculations are then as follows:



(1) 1.35 x 0.5 x 0.333 x 18 x 4.02 = 64.8 d + 4.0/3 64.8d + 86.4

(2) 1.35 x 0.333 x 18 x 4.0 x d = 32.4d d/2 16.2d2

(3) 1.35 x 0.5 x 0.333 x 10.2 x d2 = 2.295d2 d/3 0.765d3

(4) – 1.0 x (0.5 x 3.0 x 18 x 0.72) = – 13.23 d + 0.7/3 – 13.23d – 3.087

(5) – 1.0 x (3.0 x 18 x 0.7 x d) = – 37.8d d/2 – 18.9 d2

(6) – 1.0 x (0.5 x 3.0 x 10.2 x d2) = – 15.3d2 d/3 – 5.1 d3

Which gives: 64.8 d + 86.4 + 16.2 d2 + 0.765 d3 = 13.23 d + 3.087 + 18.9 d2 + 5.1 d3

Which simplifies to: 83.313 + 51.57 d – 2.7 d2 – 0.765 d3 = 0, from which d = 3.82 m

The pile penetration is then increased by 20 % to ensure stability Actual pile penetration is taken as (d + 0.7) metres = 4.52 m Required penetration length calculated from actual pile length but increased by 20% = 1.2 x 4.52 = 5.42 m Overall pile length required = 3.0 + 5.42 = 8.42 metres

Load Combination 2 (A2+ M2 + R1)

Action Type Partial Factor

Case A2 The thrust due to the retained

backfill Permanent – unfavourable G, dst = 1.0

Material Property Partial Factor

Case M2 Coefficient of shearing resistance (tan ’) ’ = 1.25

Weight Density = 1.0

Resistance Partial Factor

Case R1 Earth Resistance R; e = 1.0

For ’ = 30º and ’ = 1.25: tan ’ design = tan 30º / 1.25 and ’ design = 24.8º Hence:

Ka design = (1 – sin 24.8) / (1 + sin 24.8) = 0.5805 / 1.4195 = 0.409

Kp design = (1 + sin 24.8) / (1 - sin 24.8) = 1.4195 / 0.5805 = 2.445 For bulk density of retained fill = 18 kN/m3 design = 18 / 1.0 = 18 kN/m3 For bulk saturated bulk density of fill = 20 kN/m3 design = 20 / 1.0 = 20 kN/m3 Effective unit weight of soil below water table = (20 – 9.8) = 10.2 kN/m3 So, as = 1.0, again for this value: design = 10.2 kN/m3 The partial load factor for the action due to the active earth pressure behind the wall, G, dst = 1.0, will be applied to the force due to the active earth pressure behind the wall. The partial factor for resistance R; e = 1.0, which will be applied to the force due to the passive earth pressure in front of the wall. The calculations are then as follows:



(1) 1.0 x 0.5 x 0.409 x 18 x 4.02 = 58.896 d + 4.0/3 58.896 d + 78.528

(2) 1.0 x 0.409 x 18 x 4.0 x d = 29.448 d d/2 14.724 d2

(3) 1.0 x 0.5 x 0.409 x 10.2 x d2 = 2.086 d2 d/3 0.695 d3

(4) – 1.0 x (0.5 x 2.445 x 18 x 0.72) = – 10.78 d + 0.7/3 – 10.78d – 2.515

(5) – 1.0 x (2.445 x 18 x 0.7 x d) = – 30.807 d d/2 – 15.404 d2

(6) – 1.0 x (0.5 x 2.445 x 10.2 x d2) = – 12.47 d2 d/3 – 4.16 d3

Which gives: 58.896 d + 78.528 + 14.724 d2 + 0.695 d3 = 10.78d + 2.515 + 15.404 d2 + 4.16 d3

Which simplifies to: 76.013 + 48.116 d – 0.68 d2 – 3.465 d3 = 0, from which d = 4.27 m

The pile penetration is then increased by 20 % to ensure stability Actual pile penetration is taken as (d + 0.7) metres = 4.97 m Required penetration length calculated from actual pile length but increased by 20% = 1.2 x 4.97 = 5.96 m Overall pile length required = 3.0 + 5.96 = 8.96 metres Hence the design is governed by Load Combination 2 and the overall required pile length will be 8.96 metres

Tied Back Sheet Pile Walls Tied back walls can offer advantages over cantilever walls due to reduced embedment lengths and lower required section modulus due to the lower bending moments generated in the pile. The major disadvantage is that additional construction is required, often at some distance behind the wall, in order to generate sufficient resistance to provide the force in the tie. Design of Tied Back Walls Two distinct approaches have been suggested for tied back wall design, which are known as the free earth and the fixed earth method. The primary difference between these methods lies in the assumed behaviour at the toe of the wall. In the free earth method, it is assumed that the toe of the wall is free to translate laterally without wall failure resulting. In the fixed earth method, the toe of the wall is assumed to be fixed and cannot translate horizontally (the same assumption that is made in cantilever wall design). As might be imagined, the required wall length calculated using the free earth method is shorter than that using the fixed earth method. Consequently, the fixed earth method is seldom used in practice. A description of the fixed earth method is given here for the sake of completeness, but all subsequent wall design is undertaken using the free earth method.

The Fixed Earth Method for Tied back Wall Design

Reference: CIRIA Report C580

The above diagram summarises the design assumptions for the fixed earth approach. In effect, the key assumption is given in Figure (d), that the toe of the pile

does not translate. As indicated, this requires a simplified model where two lateral forces, P and Q, are required to stabilise the wall in addition to the lateral earth pressures.

The Free Earth Method for Tied back Wall Design

Reference: CIRIA Report C580

The basic assumption of the free earth method is that, given that the top of the wall is restrained by the applied force, the wall can fail only be rotation about this point of fixity. The design method then consists of:

1. Finding a moment balance about the point of application of the force at the top of the wall

And then;

2. Carrying out a horizontal force balance to find out the required value of P to satisfy horizontal force equilibrium

In carrying out the design, as for the design of cantilever walls, it is possible to make different assumptions about the soil lateral earth pressures. In fact, all four design methods used for cantilever walls may also be used for tied-back walls, i.e.

1. Gross Pressure Method

2. Burland and Potts

3. Net Pressure (British Steel) Method

4. Factor on Soil Strength

Once the required tie back force has been calculated, it is then necessary to design an appropriate system to ensure that this force can be provided.

Example 1 Find the required length and tie-back force for a 2 metre high wall provided with a tie back 0.5 metres below the top of the wall assuming that wall is to be constructed in a sand of ’ 30° and bulk 18 kN/m3. The groundwater level is significantly below the base of the wall and no allowance is to be made for either accidental overdig or for surcharge loading and the analysis is to be carried out using a Gross Pressure approach with a factor of safety of 2.0 on the passive pressure.

Assuming there is no wall friction:

Ka = (1 – sin(1 + sin) = 0.333 Kp = (1 + sin(1 – sin) = 3.0

Taking moments about T:

Moment due to active force:

Mmt. = 0.5 Ka bulk (h+d)2 x [2/3 x (h+d) - 0.5)]

= 0.5 Ka bulk x [2/3 x (h+d)3 - 0.5(h+d)2)]

= 0.5 x 0.333 x 18 x [2/3 x (2+d)3 - 0.5 x (2+d)2)]

= 9.99 + 18 d + 10.5 d 2 + 2 d 3

Moment due to passive force:

Mmt. = 0.5 Kp bulk d2 x [2 x d/3 + h - 0.5]

= 0.5 x 3.0 x 18 x 2 x d3/3 + 0.5 x 3.0 x 18 x d 2 x 1.5

= 18 d 3 + 40.5 d 2

Then applying a factor of safety of 2.0 gives:

Design Passive moment = 9 d 3 + 20.25 d 2

Equating the active and passive moments then gives:

9.99 + 18 d + 10.5 d 2 + 2 d 3 = 9 d 3 + 20.25 d 2

Which can be solved by trial and error to give d = 1.33 metres

Calculation of required Tie Force:

The required Tie Force is then calculated as the difference between the total active and passive forces as follows:

Total Active Force = 0.5 Ka bulk (h+d)2 = 0.5 x 0.333 x 18 x 3.332 = 33.27 kN

Total Passive Force = 0.5 Kp bulk d2 / 2 = 0.5 x 3.0 x 18 x 1.332 / 2 = 23.88 kN

[Note: the factor of safety of 2.0 on the passive force is included here]

Hence; T = 33.27 – 23.88 = 9.39 kN

h = 2.0 m

0.5 m

d

T

Again, since all of the other calculations have been carried out per metre width, this value of T is the tension required per metre width of wall.

Example 2 Tied back walls are frequently used adjacent to watercourses, such as for rivers or for sea walls. The following provides a typical design example in such a situation adopting a Gross pressure approach with a factor of safety of 2.0 on the passive resistance.

A 6.5 metre high wall is to be constructed in a soil for which ’ 35°, the bulk unit weight is 18 kN/m3 and the saturated unit weight 20 kN/m3. The tie is to be 1.0 m below the top of the wall, the depth of water in front of the wall is to be 1.5 m and the water level behind the wall is at the same as that in front of the wall.

From the above data, the wall height is 6.5 m, so an accidental overdig allowance of 0.5 m should be allowed (the lesser of 0.5 m or 10% of the wall height). This must be allowed for on the lower side of the wall, so that the actual design height of the wall will be 7.0 m with a total water depth of 2.0 m at the front of the wall, as shown below.

In accordance with standard practice, a 10kN/m2 surcharge must be applied behind the wall, as also shown.

Tied Back Retaining Wall – Example 2

As before, the first stage of the analysis is to calculate the relevant Ka and Kp values and then carry out a moment balance about the point of application of the tie force, T.

5.0 m

d

1

2

3 4

5

T

Surcharge = 10kN/m2

W.T.

C

2.0 m

1.0 m

Passive Earth Pressures Active Earth Pressures Retaining Wall Configuration

1 Smith (2006) “Smith’s Elements of Soil Mechanics” 8th Edition, Blackwell, Oxford

Assuming that the wall is frictionless


Also, the effective vertical stress due to the soil below the water table can be taken as (sat – water), i.e. 20–9.8 = 10.2kN/m3



(1) 0.271 x 10 x (d + 7.0) = 2.71d + 18.97 d/2 + 7.0/2 – 1.0 1.35d2 + 16.26d + 47.42

(2) 0.5 x 0.271 x 18 x 5.02 = 60.975 2/3 x 5.0 – 1.0 142.275

(3) 0.271 x 5.0 x 18 x (2.0 + d) =

48.78 + 24.39 d

0.5 x (d + 2) + (5.0 – 1.0)

12.195d2 + 146.34d + 243.9

(4) 0.5 x 0.271 x (20 – 9.8) x (2.0 + d)2 =

1.382 d2 + 5.53 d + 5.53

2/3 x (d + 2.0) + (5.0 – 1.0)

0.92d3 + 11.08d2 + 33.20d + 29.49

(5)+ (0.5 x 3.69 x 10.2 x d2) / 2 = 9.41d2 2/3 x d + (7.0 – 1.0)

6.27d3 + 56.46d2

Note – densities below the groundwater table are taken as (sat – water) in order to obtain soil effective vertical and lateral stresses. + Factor of safety of 2.0 applied to passive pressure

Equating the moments due to active and passive pressures gives:

6.27d3 + 56.46d2 = 0.92d3 + 24.61d2 + 195.76d + 463.07 Which can be solved by trial and error to find d = 4.95 m. Calculation of Required Tie Force: The required Tie Force can be given by substituting the relevant value of d into the expressions for forces (1) to (5) above: T = (1) + (2) + (3) + (4) – (5)

From which;

T = 32.38 + 60.975 + 169.51 + 66.77 – 230.57 = 99.07 kN per metre width

Clearly, the design value of Tie Force should include some factor of safety. It should be noted that the above value does implicitly allow for a factor of safety of 2.0 on the passive force. Any increase in the actual value of passive force above the value allowed for in the design will lead to a reduction in the Tie Force. On this basis, it is not very clear what the precise factor of safety is and an element of judgement must be exercised in finalising the design Tie Force. Various authors make recommendations as to what should be done – Smith1, for example, suggests that the value calculated above should be increased by 25% in order to allow for flexibility in the piling and arching in the soil.

Irrespective of the factor of safety adopted to derive a design Tie Force, it should be borne in mind that any structural elements design to carry the tie force will, in themselves, be required to incorporate appropriate factors or safety to ensure that the required force can be carried safely. As noted above, design of tied back walls can be carried out using a number of assumptions regarding the distribution of lateral earth pressures on the wall. The following example considers the same problem as in Example 2 but using a factor on strength approach. Example 3 Considering Example 2, but using a factor of safety of 1.25 on soil strength and 1.3 on surcharge (with all other partial factors equal to 1.0): ’ = 35°, so that ’des = tan-1[tan 35°/1.25] = 29.26°

As before, assuming that the wall is frictionless;


The bulk unit weight is 18 kN/m3 and the saturated unit weight 20 kN/m3. The tie is to be 1.0 m below the top of the wall, the depth of water in front of the wall Repeating the previous calculation gives:



(1) 0.343 x 10 x 1.3 x (d + 7.0) =

4.459d + 31.213

d/2 + 7.0/2 – 1.0 2.23d2 + 26.75d + 78.03

(2) 0.5 x 0.343 x 18 x 5.02 = 77.175 2/3 x 5.0 – 1.0 180.075

(3) 0.343 x 5.0 x 18 x (2.0 + d) =

61.74 + 30.87 d

0.5 x (d + 2) + (5.0 – 1.0)

15.4352 + 185.22d + 308.7

(4) 0.5 x 0.343 x (20 – 9.8) x (2.0 + d)2 =

1.749 d2 + 7.00 d + 7.00

2/3 x (d + 2.0) + (5.0 – 1.0)

1.166d3 + 13.99d2 + 42.0d + 37.33

(5)+ (0.5 x 2.912 x 10.2 x d2) = 14.851 d2 2/3 x d + (7.0 – 1.0)

9.90d3 + 89.11d2

Note – densities below the groundwater table are taken as (sat – water) in order to obtain soil effective vertical and lateral stresses. + Full passive pressure taken – factor of safety applied to soil strength

Equating the moments due to active and passive pressures gives:

9.90d3 + 89.11d2 = 1.166d3 + 31.655d2 + 253.97d + 604.135 Which can be solved by trial and error to find d = 4.22 m.

Calculation of Required Tie Force: The required Tie Force can be given by substituting the relevant value of d into the expressions for forces (1) to (5) above: T = (1) + (2) + (3) + (4) – (5), so that;

T = 50.03 + 77.175 + 192.01 + 67.69 – 264.47 = 122.44 kN per metre width

Note that the above Tie Force value will already include a factor of safety due to the fact that the soil strength has been factored, but that this must also be taken into account in any structural design in order to ensure that the structural and geotechnical factors are compatible.

Summary of Design Outputs The following table summarises the results of the analyses carried out as Examples 2 and 3, which are alternative ways to design a similar wall.

Design Method Required Wall Embedment

(m)

Tie Force (kN per metre

width)

Notes

Gross Pressure 4.95 99.07 Factor of safety of 2.0 on Passive. Tie Force value not directly factored, but will be affected by this

Factor on Soil Strength 4.22 122.44

Factor of safety taken on soil strength – Tie Force will also be factored

It can be seen from the summary table that the two methods, which are both nominally similarly acceptable, produce different answers for both the required wall embedment and tie force. Consequently, there will necessarily be an element of judgement in the design process both in identifying the most appropriate method of analysis and understanding the use to which the results will be put.

Anchor Design – Providing the Tie Force The Tie Force for tied back walls is usually provided by means of tie rods or cables.

The tie rods or cables are usually anchored either using concrete blocks or steel plates (individual anchors), or using beams or sheet piles (continuous anchors).

One of the most important factors in the design of anchor blocks is the need to design them so that they bear on undisturbed ground which is sufficiently far back from the wall in order to ensure that the distribution and attenuation of stresses through the soil does not affect the performance of the wall itself. The basic geometric requirements are indicated in the second of the two figures shown below.

Reference: Smith

Basic Geometric Requirements for Anchor Blocks

In designing anchor plates the active zone is defined by a line which originates from the base of the wall and is inclined at an angle of 45°+ /2 to the horizontal as shown in the above diagram. The boundary of the passive zone caused by the anchor block will originate from the base of the block and extend backwards from it at an angle of 45°– /2 to the horizontal.

45°+ /2

45°– /2 Active Zone due to wall Passive Zone due

to anchor plate

Tie Rod Anchor Plate

Where the tie rod or cable is anchored to a row of steel sheets driven into the ground, the situation will be as shown below:

Pullout of the anchor is resisted by the passive earth pressure acting on the front of the sheet piles. However, the active earth pressure acting on the back of the sheet piles will tend to push the anchor forward, thus increasing the possibility of the anchor pulling out. Hence, the net earth pressure resisting pullout of the anchor will be equal to the passive earth pressure acting on the front of the sheets minus the passive earth pressure acting on the back of the sheets.

On the basis of the above reasoning, we can write expressions for the net pressure resisting pullout of the anchor and hence for the force resisting pullout of the anchor, F, which is given as:

F = 0.5 x (Kp – Ka) x da2

Anchor Block Design

Continuous Beams

Where a continuous beam anchor is used to resist the load in the ties, the passive pressure is assumed to act over the full depth of the anchor (da) only provided that the depth of the beam is greater than half of the depth of the anchor,

i.e. h ≤ da

In practice the anchor block should be proportioned so that this requirement is always satisfied.

Individual Anchor Blocks

As an alternative to continuous beams, individual anchor blocks may be used as shown in plan below.

Passive Pressure on front of anchor plate

Active Pressure on rear of anchor plate

da

da

h

In this case the anchor block spacing is shown above as S and the width of each block as L.

One advantage of this approach is that a separate passive soil failure wedge is generated for each plate and in addition to the simple passive soil resistance, the shearing resistance of each side of this wedge can also be counted as contributing to the plate pullout resistance.

Shearing resistances on these surfaces may also be taken into account

Passive soil failure wedge behind individual anchor block

S

L

L

Documents

arth Pressure and Retaining Structure Notes