CIV3247 Summary Notes

Semester 2, 2020 Contents

1 Soil Consolidation 1 1.1 Sources and Causes of settlement ...... 1 1.2 Understanding Consolidation ...... 1 1.3 Oedometer Test Analysis ...... 3 1.4 Normally and Overconsolidated Clays ...... 4 1.5 Finding Preconsolidation Pressure and Over-Consolidation Ratio ...... 5 1.5.1 Casagrande Method ...... 5 1.5.2 Simpliﬁed Method ...... 6 1.5.3 Over-Consolidation Ratio ...... 7 1.6 Compression, Swelling indices and other parameters ...... 7 1.7 Calculation of Primary Consolidation Settlement ...... 8 1.7.1 NC Clays ...... 9 1.7.2 OC Clays ...... 10 1.8 Secondary Consolidation Settlement ...... 11

2 Time Rate of Consolidation 12 2.1 Terzaghi’s Equation ...... 12 2.1.1 Derivation ...... 12 2.1.2 Analytical solution ...... 13 2.1.3 Maximum Drainage Path ...... 14 2.1.4 Average degree of consolidation ...... 14 2.1.5 Graphical Solution ...... 14 2.2 Average Degree of Consolidation ...... 15 2.3 Coeﬃcient of Consolidation ...... 16 2.3.1 Logarithm of Time Method ...... 16 2.3.2 Square root of time method ...... 17 2.4 Primary consolidation settlment under a foundation ...... 18

3 Shear Strength of Soils and the Consolidated Drained Test 20 3.1 Soil Strength ...... 20 3.2 Mohr-Coulomb Failure Envelope ...... 20 3.3 Shear Strength Measurement: Triaxial Test ...... 21 3.4 Types of triaxial tests ...... 23 3.5 Consolidated-Drained (CD) Triaxial Test ...... 23 3.5.1 Stage 1: Saturation state ...... 24 3.5.2 Stage 2: Isotropic consolidation process ...... 24 3.5.3 Stage 3: Deviatoric Loading Phase ...... 25 3.5.4 Steps/notes for analysis ...... 26 3.6 Typical results from CD Tests ...... 26 3.7 Advantages/Disadvantages of the CD Test ...... 27

4 Consolidated Undrained and Unconsolidated Undrained Triaxial Tests 28 4.1 Consolidated-Undrained (CU) Triaxial Test ...... 28 4.1.1 Stage 1: Saturation state ...... 28 4.1.2 Stage 2: Isotropic consolidation ...... 28 4.1.3 Stage 3: Deviatoric loading phase ...... 28 4.2 Typical results obtained from the CU test ...... 29 4.2.1 Skempton’s Pore Pressure at Failure ...... 30 4.3 Advantages/Disadvantages of the CU Test ...... 30 4.4 Unconsolidated Undrained (UU) Test ...... 31

i 4.4.1 Stage 1: Conﬁning phase ...... 31 4.4.2 Stage 2: Deviatoric Loading Phase ...... 31 4.5 Typical results from the UU Test ...... 32 4.6 So many tests... which one to use??? ...... 32

5 Critical State Soil Mechanics 34 5.1 Stress Paths on the Shear Plane ...... 34 5.1.1 Cambridge Deﬁnition ...... 34 5.1.2 Stress path in isotropic loading ...... 35 5.1.3 Stress path for deviatoric loading ...... 35 5.1.4 Stress Paths in Triaxial Tests ...... 35 5.1.5 Critical State Line (CSL) ...... 36 5.2 The Volumetric Plane ...... 37 5.3 Drained Triaxial Tests in CSSM ...... 38 5.3.1 NC Clays and loose sands ...... 38 5.3.2 Heavily OC Clays and dense sands ...... 39 5.4 Undrained Triaxial Tests in CSSM ...... 40 5.4.1 NC Clays and loose sands ...... 40 5.4.2 Heavily OC Clays and dense sands ...... 41 5.5 Soil Yielding ...... 42 5.5.1 Yield surfaces (locus of yield points) ...... 43 5.5.2 Evolution/expansion of the yield surface ...... 43 5.6 State Boundary Surfaces (SBS) ...... 46

6 Modiﬁed Cam-Clay Model 47 6.1 Yield Surface Equation ...... 47 6.2 Elastic Strain ...... 48 6.2.1 Iteration ...... 49 6.3 Plastic Deformation ...... 50 6.3.1 Iteration ...... 51 6.3.2 But what does η mean?? ...... 52 6.4 Model Advantages and Limitations ...... 53

7 Slope Stability Analysis 54 7.1 Factor of Safety ...... 54 7.2 Stability of Inﬁnite Slopes ...... 55 7.3 Factor of Safety with Seepage ...... 56 7.4 Limit Equilibrium Method ...... 57 7.5 Total Stress Analysis (Undrained stability of slopes) ...... 58 7.5.1 Determining the type of Slope Failure ...... 59 7.5.2 Toe circle failure ...... 60 7.5.3 Midpoint circle failure ...... 61

8 Slope Stability Analysis: Eﬀective Stress Analysis 62 8.1 Fellenius Method ...... 63 8.2 Bishop’s Simpliﬁed Method ...... 64 8.2.1 Iteration using Bishop’s Method ...... 65 8.3 Solving problems with the Methods of Slices ...... 66 8.4 Limitations ...... 67

ii 9 Rock Slope Stability: Stereographic Analysis 68 9.1 Types of Rock Failure ...... 68 9.2 Some terms relating to discontinuities and planes ...... 69 9.2.1 Notation ...... 69 9.3 What is a stereonet? ...... 70 9.3.1 Setting up the stereonet ...... 70 9.3.2 Plotting planes ...... 70 9.3.3 Plotting Dip Vectors and Strikes ...... 73 9.3.4 Plotting poles ...... 74 9.3.5 Intersections between planes ...... 75 9.3.6 Friction Circle ...... 76 9.4 Planar Failure - Kinematic analysis ...... 77 9.5 Stability Analysis of Planar Failure ...... 78 9.5.1 Surcharge loads ...... 78 9.5.2 Eﬀect of groundwater ...... 79

10 Rock Slope Stability: Wedge and Toppling failure 80 10.1 Wedge Failure (Kinematic analysis) ...... 80 10.2 Wedge Failure (Stability Analysis) ...... 81 10.3 Toppling failure (Kinematic Analysis) ...... 82 10.4 Toppling Failure (Stability Analysis) ...... 83 10.4.1 Sliding Failure ...... 83 10.4.2 Toppling Failure ...... 84

11 Lateral Earth Pressure 85 11.1 Earth pressure at rest ...... 85 11.1.1 Thrust due to Earth pressure ...... 86 11.2 Active and Passive Earth Pressure ...... 87 11.3 Rankine’s Pressure theory ...... 88 11.3.1 Active pressure ...... 88 11.3.2 Passive Pressure ...... 90 11.3.3 Sloped backﬁlls ...... 91 11.4 Coulomb’s Earth Pressure Theory ...... 92 11.4.1 Active Case ...... 92 11.4.2 Passive Case ...... 92

12 Earth Retaining Structures 93 12.1 Retaining Walls ...... 93 12.1.1 Overturning failure ...... 94 12.1.2 Sliding Failure ...... 95 12.1.3 Bearing Failure ...... 96 12.2 Sheet Pile Walls ...... 98 12.2.1 Simplified Analysis ...... 98 12.2.2 Simplified analysis (feat. water table) ...... 100 12.2.3 Not-so-simplified analysis ...... 101 12.2.4 Not-so-simplified analysis (feat. water table) ...... 103 12.2.5 Anchored sheet pile walls ...... 103

Appendix 1: Derivation of relationship between σ1 and σ3 on MC 105

iii 1 Soil Consolidation

When load is placed on soil, it gets compressed, as the void space reduces. Consolidation occurs when pore water is pushed out of soil voids, causing soil to reduce in volume. This can cause buildings to sink (e.g. the Leaning Tower of Pisa).

1.1 Sources and Causes of settlement Construction activity can place loads on soils, resulting in settlement. The effects of this are different with different types of soils, which behave/react differently to loading: Sandy Soils: Due to its high permeability, pore pressure increases and dissipates rapidly. Clay soils: Pressure dissipates much slower due to low permeability. Thus, consolidation occurs over a very long period of time (i.e. several years) Before we can understand the mechanics of consolidation, we need to know some definitions 1. The total stress in soil can be expressed as:

0 σt = σ + u Where:

σt is the total stress σ0 is the eﬀective stress, which is the stress expreienced by the soil solids u is the pore water pressure, which is the stress taken by the water in between the soil particles. This is also known as the Bishop’s eﬀective stress.

1.2 Understanding Consolidation The behaviour of clayey soils can be likened to a piston/spring model. A piston is placed on top of a container with water, which represents the pore water in between the soil particles. It is held up by a spring, which represents the soil particles.

The valve starts out closed. Since water is incompressible, The spring will not be compressed, and all the load will be taken by the water. Once we open the valve, water will squirt out through the hole, due to the high hydrostatic pressure. As a result, the piston will lower, and the load will be transferred to the spring, rather than the water. Eventually, all of the water will exit the system and the piston will be pushed all the way down.

1From CIV2242!

1 Now how has this got anything to do with soils? Well, it turns out, the process of consolidation/settlement in soils over time is very similar to this. When we place a load on top of clay, this happens:

As we can see, there are some stages to this:

Stage 1: This is before the load is applied. The soil is at its initial eﬀective stress and pore water pressure. This is the same as when the valve on the piston is closed.

Stage 2: This occurs as soon as the load is applied on the soil. All the stress is taken by the pore water. The eﬀective stress remains the same. In the previous model, this occurs at the very moment that the valve is opened.

Stage 3: After the load is applied, the volume gradually reduces. Going back to the piston model, this is the period after the valve is opened.

2 Stage 4: At the very end of all this (t = ∞), all excess water is drained and load is carried solely by soil grains. In the spring model, this is what happens when all the water has been pushed out of the system.

There are two types of consolidation settlement that we are concerned with:

Primary settlement: Occurs as a result of excess pore water pressure dissipating as the soil is squashed.

Secondary settlement: Occurs over time after primary consolidation, due to creeping behaviour of soil particles.

1.3 Oedometer Test Analysis In order to understand the behaviour of soils in the ﬁeld, lab testing must be done to estimate soil parameters. The Oedometer test can be used to obtain consolidation properties of soils.

To analyse data and estimate parameters of soil following an oedometer test, we need to look at the change in void ratio as pressure increases. Usually, the pressure and change in height are given. We need to ﬁnd the initial void ratio, knowing that:

From this, we can calculate the initial void ratio as:

Vv HvA Hv e0 = = = Vs HsA Hs Where:

Ws is the weight of solids, or the dry weight (be careful with units!!) A is the cross sectional area of the sample

Gs is the speciﬁc gravity of the solids

3 γw is the unit weight of water = 9.8kN/m

3 1.7.2 OC Clays

0 0 For clays that START OFF overconsolidated (that means σ0 < σc), there are two diﬀerent cases we need to consider. The clay can either stay overconsolidated, or become normally consolidated and begin to behave diﬀerently.

0 0 0 If σ0 + ∆σ < σc, or in other words, the clay remains overconsolidated, we can simply calculate the primary consolidation settlement as the clay goes from point A to B:

OC, because it's on the flat part and

0 0 CsH σ0 + ∆σ0 Sp = log 0 1 + e0 σ0

0 0 0 0 Now, if σ0 +∆σ > σc, it’s a bit awkward because the clay will pass σc, and become normally consolidated, Thus, there will be two parts to the equation. One considers when the clay goes from point A to point C, then the other considers when it goes from point C to point B’.

OC, because it's on the flat part and

0 0 0 CsH σc CcH σ0 + ∆σ Sp = log 0 + log 0 1 + e0 σ0 1 + e0 σc

10 3 Shear Strength of Soils and the Consolidated Drained Test

When we are putting buildings (among other things) on the ground, they will place a load on the soil, causing it to consolidate and experience stress increases. While consolidation can be bad, there can be much worse consequences when soil failure occurs. Failure of soil can cause buildings to fall, slopes to collapse and landslides to occur. Therefore, we need to understand shear stength of any soil we are working with.

3.1 Soil Strength In geotechnical engineering, we are concerned with the ability of soil to resist shear. Soils can resist shear due to frictional forces between its grains. The normal (vertical) stress on the soil also helps hold it together, and prevents it from failing.

The shear strength of a soil is the resistance per unit area that the soil can provide to resist failure. Depending on the normal stress applied, diﬀerent amounts of shear will be required to bring a soil to failure.

3.2 Mohr-Coulomb Failure Envelope The amount of shear and normal force required to fail a soil can be plotted, and will form a linear plot with equation:

τ = c0 + σ tan φ Where: c0 is the cohesion of the soil sample. This is the frictional resistance that prevents shear failure when the soil is not held down by a normal force. σ is the normal force acting on the sample. φ is the friction angle of the soil.

20 5 Critical State Soil Mechanics

So far, we have seen volumetric responses of soil (e.g. consolidation) as well as shear behaviour. We have treated them as separate things that occur separately. However, in reality, they occur together. As soil is sheared, it also exhibits volumetric changes.

In order to consider them together, we need to ﬁnd a common point where they behave in the same way.

While dense and loose sand, OC and NC clays behave diﬀerently when considering volumetric and shear strains, they ALL reach a point where they converge to a constant value, and stop changing. This occurs regardless of the type of testing. This is known as critical state. Critical State Soil Mechanics (CSSM) is the framework for analysing this behaviour.

Knowing this, we can analyse volumetric and shear behaviour together. This can be done using:

The shear plane, which is a plot of deviatoric stress q against mean stress p.

The volumetric plane, which is a plot of speciﬁc volume ν against mean stress p.

5.1 Stress Paths on the Shear Plane 5.1.1 Cambridge Definition Previously, we saw the MIT definition for ∆p and ∆q. In this definition, when calculating the mean stress ∆p, σ2 is ignored, as it is assumed to be the same as σ3.

With the Cambridge definition, ∆p considers σ2. Unlike the MIT definition, ∆q is the deviatoric stress, not the shear stress. σ + 2σ p = 1 3 3 q = σ1 − σ3 = σd Mean effective stress is equal to:

p = p − ∆uw The stress path is a representation of changes in stress states of a soil specimen as it is loaded. This is can be plotted on the shear plane and volumetric plane.

34 5.2 The Volumetric Plane Along with deviatoric stresses, we can also plot the speciﬁc void ratio ν = 1 + e against ln p. ν is used instead of e, unlike when plotting oedometer test data. This is another way to visualise behaviour of soils as they are loaded to critical state.

On the volumetric plane, we can plot:

Normally Consolidated Line (NCL): Represents points where the soil specimen is normally consolidated.

Critical State Line (CSL): Represents points where the soil reaches critical state (and fails).

Unloading-Reloading Line (url)12: When a specimen is overconsolidated, it follows this line until it reaches σc and joins the NCL. There can be multiple of these (e.g. if a clay is unloaded and reloaded more than once).

Both the NCL and CSL have the same slope, which is given as λ. The unloading-reloading line has a slope of κ. Here’s what it might look like:

The equations for the NCL and CSL are:

νNCL = Nν − λ ln p

νCSL = Γν − λ ln p

Where Γv and Nv are the intercept of the lines (reference values at p = 1kP a). These are material properties of the soil.

While being isotropically loaded, soil specimens can:

When NC: Start on the NCL and move along the NCL.

When OC: Start on the OCL/url and move along this line. If it is isotropically loaded past the pre- consolidation pressure, it will join the NCL and continue moving along it.

When an NC soil is unloaded, a new url is formed.

Movement along the OCL/URL is elastic and movement along the NCL is plastic. (More on this later on!)

When a specimen is deviatorically loaded, it will move towards the CSL either horizontally or at a slope, depending on the loading condition.

12Also known as over consolidated line or OCL

37 Similarly, when a clay is isotropically loaded until it’s lightly OC, then deviatorically loaded in drained conditions (i.e. a CD test), the specimen moves to point A as it is deviatorically loaded. It moves up to point B where it yields. From here, the yield surface starts to expand until the soil fails at point D on the CSL.

Notice that on the volumetric plane, the soil follows the URL while being elastically loaded. It only follows the drained loading path once it yields.

44 When an undrained triaxial loading is performed (don’t consider isotropic loading), the behaviour is slightly diﬀerent:

NC soils that start right on the pre-consolidation pressure will deform plastically from point C to point A on the shear plane.

Heavily OC soils will move vertically up on the shear plane until it reaches its peak, before moving back down to the critical state line. The vertical movement is because the load is all taken by the pore water pressure.

Lightly OC soils will move up above the initial yield surface, before reaching a peak, then moving down onto the CSL.

Note that for both of these tests, the load path on the volumetric plane is always a horizontal line. In these cases, the yield surface only expands along path CD, as this results in plastic deformation.

45 6 Modiﬁed Cam-Clay Model

Now that we know how stress paths work, we can better model how clay soils behave under isotropic and deviatoric loading. Apart from volumetric and shear changes, we need to understand how soil samples deform when loaded. This is where the Cam-Clay Model comes in.

The Cam-Clay model allows us to use maths to calculate strain increments in soils that occur as a result of elastic and plastic deformation. There are a few assumptions made:

Clays are assumed to be homogeneous/isotropic and undergoing elasto-plastic behaviour during loading. This means that they behave the same in tension and compression, and elastic and plastic deformation occurs together after yielding.

Total deformation can be separated into elastic and plastic components (just like steel).

The yield locus is assumed to be an ellipse and the critical state line passes through its maximum point.

Remember that elastic strain is recoverable, while plastic strain is not. So that means when we load a material past its yielding point, then unload it, the yielding point will be moved. This is similar to what happens when we load a soil past its pre-consolidation pressure (as seen here).

Strain is represented by its volumetric and shear strain components where:

e δεv δεe = e δεs

0 δεv is the volumetric strain increment caused by the normal stress p .

δεe is the shear strain increment caused by the deviatoric load p.

6.1 Yield Surface Equation From before, we assume that the yield surface is an ellipse. We’ve seen it in many of the diagrams before, and we know how it behaves (expands/contracts) as soil is loaded while normally consolidated. But still, we don’t know how to plot it. This begs the question, what is the equation of this ellipse and how do we use it to tell if a soil is yielded or not??

The size and shape of the yield surface is controlled by M, which is the slope of the CSL on the q − p0 plane, 0 along with the pre-consolidation pressure p0 (or pc or whatever you want to call it). The equation is:

2 2 0 0 0 f = q − M [p (p0 − p )] = 0

47 8.3 Solving problems with the Methods of Slices To analyse slopes using methods of slice, we need to determine the coordinates of each slice.

The coordinates along the bottom of the failure surface (’b’ coordinates) can be determined using the equation of a circle:

2 2 2 (x − x0) + (y − y0) = R q 2 2 yi,b = − R − xi,b The coordinates at the midpoint of a segment can be calculated in a similar manner. The x coordinate can be calculated as the average of x at the start and end of the segment. q 2 2 yi,m = − R − xi,m For the sloped portion of the top (’a’ coordinates) can be found using the equation. In the above image, (xend,b, yend,b) would be equal to (x9,b, y9,b).

yi,a = yend,b + (xi,a − xend,b) tan β

The angle θi of each segment can be found using:

−1 O1 · O2 θi = cos |O1||O2| The inclination angle α of a segment can be calculated using: −y α = cos−1( i,m ) i R

The length li of each segment can be approximated using:

li = 2 × R sin θi

The weight from pore water uw,i for segment i is:

uw,i = γw × bi × hw

Once all the coordinates of the sections are known, the area Ai of each segment can be calculated. For the more complicated shapes, try splitting the area into diﬀerent components. For example, segment 5 in the previous diagram can be split into:

66 0 350 10 340 20

330 30

320 40

310 50

300 60

290 70

280 80

270 90

260 100

250 110

240 120

230 130

220 140

210 150

200 160 Trend = 186 degrees 190 170 180

The angle between two lines can be calculated by ﬁnding the plunge of each line, then measuring the angle between them (similar to ﬁnding the trend.)

9.3.6 Friction Circle When we know the friction angle φ of the soil, we can represent that by drawing a circle20 on the plane, with radius of 90 − φ. φ is measured inwards from the north, east, south and west. This is what a friction circle corresponding to φ = 40◦ would look like:

0 350 10 340 20

330 30

320 40

310 50

300 60

290 70

280 80

Φ = 40 deg 270 90

260 100

250 110

240 120

230 130

220 140

210 150

200 160 190 170 180

20To draw a circle, rotate the paper, marking points oﬀ, then join the points. Or just cheat using a compass.

76 The stress distribution for cohesive soils can be drawn as follows. The total force per unit length of the wall can be calculated as the total area of the distribution:

1 p P = (γHK − 2c0 K ) × H a 2 a a The top part of the distribution is in tension, thus, tension cracks can form up to a maximum depth of:

2c0 Z0 = √ γ Ka The total force considering tension cracks is given by:

1 p P = (γHK − 2c0 K )(H − Z ) a 2 a a 0 1 2c2 = γH2 − 2c H + u 2 u γ When groundwater and surcharges are applied to the soil, the pressure diagram can be adjusted to reﬂect this. The total force can be calculated as the area (as usual).

1 1 P = K qH + K [γH2 + 2γH H + γ0H2] + γ H2 a a 2 a 1 1 2 2 2 w 2

89 12 Earth Retaining Structures

Earth retaining structures provide permanent or temporary lateral support to soil slopes. Examples of these include gravitational retaining walls, braced cuts and sheet pile walls.

There are two phases in the design of reatining walls:

Geotechnical design: Determination of lateral earth pressure and checking for failure by overturning, sliding and bearing. Consolidation settlement (see section 1) needs also to be considered.

Structural design: Checks the strength of each structural component.

Retaining walls should be proportioned as follows. This makes it easier to divide it into sections for safety checks.

12.1 Retaining Walls In order to apply Rankine’s theory to retaining walls, we need to draw a vertical line AB and apply Rank- ine’s active25 pressure along it. The area enclosed by the line and the wall will be considered part of the wall.

The components of the active force Pa will pass through A.

25Because the soil will push the wall outwards

93 12.2.3 Not-so-simpliﬁed analysis The net lateral earth pressure diagram considers all possible earth pressures acting on the sheet pile wall. This means we can’t ignore the pressure below point O. This pressure diagram will look more like:

Combining this will result in a non-linear stress distribution:

Simplifying this into easy-to-calculate areas (i.e. triangles), we get this distribution. We are trying to solve for the required depth (d = z0 + z2).

The point of zero net stress z0 is calculated by setting the net stress equal to 0:

0 0 0 σnet = σa − σp = 0

101 12.2.4 Not-so-simpliﬁed analysis (feat. water table) When the water table is involved, the pressure diagram will change slightly.

The location of zero net stress is also slightly diﬀerent:

0 Ka(γmh1 + γ h2) z0 = 0 (Kp − Ka)γ 0 The depth z2 is found using the same equation as on the previous page, but diﬀerent σk:

4 3 2 z2 + A1z2 − A2z2 − A + 3z2 − A4 = 0 0 0 0 σk = Kp(γmh1 + γ h2) + (Kp − Ka)γ z0

The required depth is simply equal to d = z0 + z2, as usual.

12.2.5 Anchored sheet pile walls When the excavation gets deeper (h > 6m), the loads on the sheet pile walls increase signiﬁcantly, thus, we need to implement additional measures to prevent the piles from falling over. Sheet piles can be anchored to a weight using a tie rod. Here’s an example:

As usual, the required installation depth for the sheet pile can be obtained using moment equilibrium about the location of the tie rod: X M0 = 0

103