Natural slopes tend to slide or fall to their lower side, given long enough time. Depending on the saturation degree of soil, the slope angle, overall 3D geometry, and the soil material itself, this sliding or falling can happen in very different ways and speeds, some can be very slow and virtually harmless, and some can be very sudden and catastrophic.

Slope failure can be in different forms. It can be in the form of:

- an avalanche or continuous flow of soil,
- just the creep of soil with much slower speeds when the soil just spreads laterally,
- sudden rockfalls and toppling off of both soil and rock,
- a landslide which looks like it is rotating around a point,
- sliding along a straight line.

Significant loss of life and property can occur, if these landslides are not predicted and precautions are not taken. Although it is difficult to predict landslides, still the slope can show some hints, even before any slope engineering analysis is done. For example, evidence of displacements, such as tilted or warped soil strata, bowed trees can be a hint. Slopes with sensitive clays, or the ones formed with the cutting action of eroding rivers carry high landslide risks. The projects should be designed to minimize slopes and excavations (not just for soil stability but economical reasons too) and drainage should be provided at the top and toe of slopes.

Natural slopes are vulnerable to failure after heavy rains, which makes the soil saturated, and increase pore water pressure, and decrease effective stress, as per the formula we saw before,

σ_{T
}= σ’ + u ,

which means, reduction of resistance of the soil to failure. Because when effective strength decreases, shear strength also decrease, as we saw before. And when shear strength is less, which as we mentioned before is the key to soil strength, failures happen.

That is why after heavy rains, sometimes slopes that are on hills or mountains fail. Same condition can occur, after a rapid drawdown of water near an earth embankment dam, which leaves the slope saturated and vulnerable to failure with same logic.

For the slopes that are close enough to human life and property, depending on soil type, depth of water table, and slope angle and geometry, different precautions must be taken to keep the slopes stable.

Slope stability analyses are made to determine the nature of the slope, to assess short and long term slope stability including under seismic effects.

For a slope to be stable, the driving forces must be less than the strength of slope, and not only that it must be smaller by a certain factor, the factor of safety. Here by strength we basically mean the shear strength of that soil along the failure plane.

So for example, if we call the driving
shear force as τ_{d} and the shear strength
as τ_{f}, upon performing our
slope stability analysis, we can find a factor of safety as:

This FS is what we aim to find, in a slope stability analysis.

Driving force is a result of the weight of the slope itself, that tends to slide or fall down plus any additions such as rainwater or structures built on it or even water seepage. Driving force, creates shear stress along a failure surface, which is resisted by the shear strength of the soil. Note that for slope analysis we ignore any plants and trees that are often present in slopes, their roots provide natural anchorage in some cases which affects the depth of actual slide surface, but they can also just move as a whole with a sliding slope.

We can classify slopes broadly into two, finite and infinite slopes.

Now let’s consider the infinite slope first.

Infinite slopes has length much greater than their height, so that their height is not even comparable to their length, and therefore they can be assumed for our slope analysis purposes that they continue indefinitely. The figure below shows the main components of an infinite slope and how we approach to its analysis.

Soil strength 24

Although we will not present the formula for analysis of this type of slope, we will list the items that goes into the equation. In other words, what factors affect the stability of such as slope as seen above, and how. These factors are:

- Cohesion of the soil (c’): As we have
seen, cohesion is one of the main ingredients of the general formula for shear
strength, which was:

Since we said that analysing slopes is all about comparing shear strength of ailure surface versus the destailizing forces, we must know the shear strength, and hence, cohesion of the soil. Again as usual we put apostroph sign near c, as only the effective cohesion is what matters, a concept we explained before. As explained before, cohesion is only present in clays. Not sands. So for sandy soils, this term doesn’t apply, unless there is considerable amount of clay in sand.

- Angle of internal friction (Ø’): For the same reasons above, since angle of internal frictoin is a main ingredient in shear strength formula, the slope stability equation depends on this angle. And for the same reasons as above, we use effective angle.

- Slope angle with the horizontal (β): As also can be seen intuitively from figure above, the steeper the slope, the less stable it would be. So the stability equation is dependent on this angle beta as well.

- Thickness of the soil layer (H): This effects the weight of the block as seen in the figure and therefore goes into stability equation. How this H is determined is beyond the level of this text.

- Unit weight of soil. So this time let’s ask before explaining… What can be the reason to include unit weight of the soil? …. We include it because of two reasons… To calculate the weight of the soil block and also the vertical stress in the soil at the failure plane. And do you think we must take the total (saturated) unit weight or the effective? We must take the saturated unit weight, because the driving force of the slope is the whole weight of the soil block, including whatever water it has in it. That is why, the more water, the heavier the soil and the more driving force. That is why slopes fail more often after heavy rains. (Rain also reduces effective stress in the soil and thus shear strength, so it has a negative effect to stability in two ways both in terms of increasing detsbilizing force and also reducing stabilizing force)
- Any water seepage. Seeping water exerts pressure in soil. Therefore its effects must be considered, if it is fast enough as it can exert reasonable force on soil particles.

Now let’s mention another important point, before we move on to finite slopes…

We already discussed shear strength of soil before, which was given in general form as,

After we divide driving stress, by the shear strength, for sands we reach an important conclusion. Taking cohesion as zero for sands, and after some simple mathematics, we see that for sands, an infinite slope is stable, if, the slope angle β (as also seen in figure above) is less than the angle of internal friction Ø of sand.

In other words, if,

β < Ø

the slope is stable, when it is made of sand.

As can be seen, this is an important conclusion and it is totally independent of the height of the slope or other dimensions or the soil unit weight.

This is another significant use of the internal friction angle. *(and this is the situation where internal
friction angle means angle of repose, as we talked about before, which is the
angle a sand soil can keep its natural posture)*

Now let’s take a look at finite slopes…

A finite slope is where, the height of the failing soil amounts to a considerable portion of the whole slope length. The general shape of a finite slope is as below and it can fail in different modes as shown:

Soil strength 25a

here for different cases, an arc can be drawn and its center can be established. After that, the analysis is done as if the arch is rotating around the center point. Cohesion, height of the slope, angle of internal friction, distance to base rock are determining factors for stability. Again we compare the driving forces such as the weight of the slope that tries to rotate the arc around its center point, versus stabilizing force, such as the shear strength of the soil along the failure surface (one of the arcs as shown above).

Below, we see the situation where bedrock is at reasonably close distance to the failing slope. Then the arc can only form upto the bed rock, as can be seen in the figure.

Soil Strength 25b

An alternate method when analysing finite slopes is, dividing the failure arc into vertical pieces, and analyse the stability of pieces, and then adding their effect overall and come up with a result of driving and stabilizing forces. It is shown in the figure below. The width of slices can be taken as convenient and need not be the same. The important thing is that we reasonably divide the arc into slices and analyse and add the driving forces and stabilizing forces of each piece together, to come up with an overall result. The forces here are not shown for clarity of the figure but are drawn very similar to what we drew in figure 24, meaning, each slice has the weight and the resisting shear force at the bottom and the reaction force from below and its sides. which are used in the analysis. As always, angle of internal friction and cohesion plays important role in determining stability, as they make up the shear strength.

Soil strength 26

Finally, when we also consider earthquake effects, that force must also be taken into account as a destabilizing force and will make slope analysis considerably more complex.

In the next post of this series, we will discuss “Critical State of Soil”

]]>As we wrote before, settlement of soils in general has three stages:

- The immediate settlement as the load is applied, which is elastic settlement,
- the gradual, slow consolidation settlement while water is escaping through soil pores,
- and the secondary compression due to rearrangement of soil skeleton structure only.

Sands settle elastically, which we have talked about before, and for them consolidation settlement is negligible. For clays on the other hand, apart from elastic settlement consolidation settlement is also important, which are the second and third items in this list. The consolidation process we will talk about right below is the second one, which is the settlement that takes long time, while the water is squeezed out of clay soil which has very low permeability.

When a heavy load is placed over fine grained soils such as clay, water in the soil gradually, slowly exits the pores in the soil, which causes those pores to contract, which causes overall decrease in soil volume. As the soil can not move laterally, the volume decrease can be only because of reduction of the soil height, in other words, settlement. It is similar to pushing a piston which has a spring and water in it.

Soil strength 21

In this figure, the water in the piston represents the water in soils pores, and the spring represents the force that the soil skeleton takes. At first the piston is in balance. Then, an extra load is placed on it. At first, the extra load is entirely taken by water. But then, as water escapes from the soil slowly, which is represented by water exiting the open valve at the top of the piston, more and more of the extra load is is carried by the spring, and, the spring deforms, up to a point that all of the extra load is now taken by soil skeleton. So, when new balance is reached, the soil volume is now less than the original, in other words, settlement of soil has occurred. If we were to place further extra load on the piston, more water would exit, and further soil settlement would occur, and so on.

This consolidation process may take months or even years. To make the process faster, sometimes gravel or prefabricated drains can be placed in these soils, so that water can escape quicker.

Through consolidation equations we can determine the amount of settlement under given load for a certain soil and also the time it takes for a certain amount of settlement to occur. But we will not go into the details of how to calculate those settlements here, as that would be too detailed for the purpose of this book.

And what if you were to take away that extra load? Then the spring would push back the piston upwards, which means, the soil volume will increase again. But this can not bring back the soil to its original height again, it can recover only part of it. This is because some permanent particle rearrangement has already taken place when the extra load was first applied and we let the full consolidation take place under that load. This permanent rearrangement of particles can not be recovered, and that is why the piston can not go up to its original height even if the extra load was to be entirely removed. This partial recovering of soil of its existing height is called heaving of soil.

Soil strength 22

But in real life when can this load removal occur? It can happen when we excavate an area. After we complete our excavation, the new excavated surface, which had more soil on top, or load in the past, was in balance, as it had completed all of its consolidation for previous load. Now, as the soil on top of it is gone, the soil will push back, which causes heaving (or swelling) of soil. The soil in such state is called “Overconsolidated”. The degree of overconsolidation, called Overconsolidation Ratio, shortly written as OCR, is a parameter frequently used in soil mechanics.

Overconsolidation ratio is defined as:

OCR = maximum effective stress in the past / current effective stress

Writing with proper symbols:

Where

σ’_{o} stands
for maximum effecitive stress in the past.

σ’ stands for current effective stress.

Note that in soil mechanics, to indicate effective stress parameters, we add the apostrophe sign near the symbols (such as the stress symbol or the angle of internal friction symbol). We use effective stress, which is the soil skeleton stress, because that is what matters for strength and settlement after water is gone. Water just takes on the pressure temporarily, before it escapes through pores.

σ’_{o} is also
called “preconsolidation pressure”, which means the same thing. So it
is the stress existed for the soil, prior to the excavation or removal of
stress above the soil.

We will not go into further details of overconsolidation here, but the reader should be aware that overconsolidation is an important concept in soil mechanics, and it causes significantly different or even reverse behaviour in soils.

The most fundamental graph when discussing consolidation settlement would of course be the plot of void ratio versus effective stress, which is e vs. log σ’. We just plot effective stress σ’ by taking its logarithm for convenience, as the graph would be difficult to draw and read otherwise, so do not be confused by it, in the figure below. A typical graph looks as below:

Soil strength 23a

Void ratio, e, is a volume related property as you remember from section “phase relationships” that we introduced before. So this e vs. log σ’ graph basically shows the relationship of volume change with increasing or decreasing stress on the soil. So in other words, since consolidation is all about settlement of clay soil when we apply load on it, and since settlement also means decrease of soil volume, when we express void ratio versus effective stress in soil, we are basically showing, describing, what is happening to the soil during a consolidation process. This is why this graph is the first thing we plot and look, whenever we talk about consolidation in soils.

As can be seen in the figure above, at first the stress increase does not cause much volume change, and the curve goes flatter. In other words, a lot of stress increase is needed, to cause a little volume change at first. This is a good thing, from construction point of view, when we see that the loads we add do not cause much volume change. The reason that curve is flat here, is because, clay soils have “memory”, and if it had been subjected to higher stresses in the past, it does not change volume easily when a stress lower than in the past is applied on it. It first “waits” that the stress increase reach to that highest point in the past. This is exactly the overconsolidated condition, that we described in previous page.

Then gradually, every increase in stress cause more and more volume change, and finally it becomes a flat, down sloping line, which means it goes linearly now, in direct proportion now. That flat line is called the virgin compression line. The point where virgin curve starts, is where our new stress reaches the maximum past stress. And as soon as we go beyond the maximum past stress, now the curve is flat and steeply sloping downward, which means and every increase in stress cause a lot of volume change. But how can a past stress be more than the current stress in the past? It can happen by human acitivity such as excavation of soil above. It can happen by geological processes, where in the past there was thicker soil on top, and now a thinner layer exis, such as glaciers or water eroding the soil on top, or seismic movements. It canalso happen due to sampling process of soil specimen from field, when we take out a core of soil from natural ground, it does not have the overburden anymore, that existed on it for many millions of years.

Now let’s look at what happens to the curve, if we were to release the load we introduced, and then reloaded the sample and then release it again, and then reload it again, and so on (by the way all this can be done in an oedometer test in the lab, so we obtain the graphs above and below in the lab with oedometer test, also called consolidation test).

The figure below is the same thing with the previous one, except, in this one we also show releasing of load and the loading it back. So this is how the curve looks, when we perform such a cycle:

Soil strength 23b

As can be seen in the figure above, this figure is basically the repetition of the previous figure in fact. First we start loading as in the previous figure, first soil settles less, then it reaches virgin line, and starts to deform a lot. Then, different than the preivous figure, we do something extra here, and release the load. As you can see, the volume goes up again (soil swells), and the stress falls. But does it go up all the way back to original? No. It can only recover some of it. Then we start loading again and a next cycle begins, exactly looking like the preivous one in shape, with the only difference that the stresses are higher here and volume is smaller. So again, at first the curve is flat like before, not a lot of deformation since we had higher stress from previous cyce, but, the moment we reach to that previous maximum stress level, now the curve joins to virgin curve and it just continues on it, as we now compress the soil for the first time with this much stress. Then we release the load again, so the soil swells a little again, and we strart loading, again the deformation is not much at first, but as soon as we reach the preivous maximum stress, again the soil enter back into virgin line and the cycles can be continued like this, up to a certain practical limit.

In the next post of this series, we will discuss “Slope Stability”

]]>Settlement of soils in general has three stages:

- The immediate settlement as the load is applied, which is elastic settlement, (for all soils)
- the gradual, slow consolidation settlement while water is escaping through soil pores, (for cohesive soils only)
- and the secondary compression due to rearrangement of soil skeleton structure only (for cohesive soils only)

In this section, we will cover the first one, elastic settlement, which applies to all soils. For sands, this is the only settlement that matters, and the second and third one are only for clay soils. So when we study elastic settlement, we also study sand settlement. Because for sandy soils which have coarse grains, permeability is large enough, so that water can be assumed to dissipate immediately when the extra load is applied, so second and third items do not apply. The settlement of sand soils due to applied loads from above occurs mainly because of compression of soil skeleton, and it is called elastic settlement. Elastic settlement requires more force than consolidation settlement, for the same amount of settlement, and therefore sands are stronger than clays, as we also have seen before.

Other things that facilitate settlement in sand is vibration or submerging or soaking in water, together with applied loads from above. The vibrating motion can naturally be produced by earthquakes too. This is why loose sandy soils are vulnerable to earthquakes, especially if they are loose, and water is present close to surface, as in liquefaction that we described previously in this book .

So whenever you see construction work where a vibrating equipment is used to compact soil, you can understand that it is sandy soil, and you will probably notice that some water also has been poured into it, to make compaction easier.

In addition to particle rearrangement as described above as a result of loading, sandy soils can also further settle due to crushing of particles, if extremely heavy loads are applied on them as we described before, such as at the bottom parts of earth embankment dams.

The more angular the sand particles, the stronger the soil due to interlocking of particles and higher internal friction. And needless to say, denser soils are also stronger than loose soils.

In some situations, there can be a stronger sandy soil on top, but under that a weaker soil of loose sand or even clay can exist. So a settlement analysis of the top soil layer is not enough in that case. Lower layers must also be analyzed for the given loads.

In practice, to calculate settlements from dead loads for most ordinary shallow foundation systems, Meyerhof’s bearing capacity method, and Terzaghi’s works usually form the basis of designs. International Building Code Chapter 18 also provides more information on this subject.

As far as earthquake effects, seismic load duration is very small in relation to gravity loads, and it reverses a number of times in a short time, so for dense soils, seismic loads are not expected to contribute a lot to settlement. Regardless of that they must be accounted for and this can be done by including them in design load combinations by increasing soil bearing capacity requirement by a certain factor (i.e. %30).

The equations for calculating settlement amounts will not be given here as it goes beyond an introductory level text, and as always, interested readers are guided to countless of engineering texts that are available in the books or on internet. But we can at least list the following about calculating elastic settlement:

- Elastic settlement is calculated by using soil’s modulus of elasticity, E, (which can vary with depth), Poisson’s ratio (ratio of how a material converts vertical stress to horizontal and vice versa), dimension of the foundation, the stresses in soil (which is calculated using principles given in previous section).
- The settlement amount of rigid versus flexible foundation assumptions make a difference and they can be converted to each other using equations.
- Elastic settlement on clays and sands are calculated differently.
- Settlement on sand can also be estimated from the results of a commonly used tests, called Standard Penetration Test (SPT) and Pressuremeter Test (PMT), as the data are correlated with extensive research. We will talk briefly about these tests in the following sections.

In the next post of this series, we will discuss “Consolidation Settlement”

]]>We have already seen total and effective stresses that exist in natural soil a few sections before in this book. But when we introduce new loads to the soil such as by building structures on it, the stresses increase beyond those existing natural levels. This new stress increase will cause a certain settlement in the soil.

In other words, we must know how much stress at what depth and what horizontal location increased in soil below as a result of new loads that we introduce. There are various methods to estimate stress increase in soils.

Mathematical solutions exist for estimating stresses increase in soil, based on point loading, strip (linear) loading and area loading, which go beyond an introduction level text. The purpose though, is always the same: Given a certain new loading, find the stress increase in soil at a certain depth, and horizontal distance relative to the loading.

To illustrate the point, we will only present area loading, which means, uniformly loading of soil from a square or circular footing.

The most straightforward of them, called 2:1 method, is shown below:

In this method, the stress below a footing gets distributed gradually over a larger area which widens as 2 vertical to 1 horizontal distance below the footing. This means, the vertical stress on the soil decreases with increasing depth.

There are other more complex methods, developed by researchers. One of the most commonly known methods, developed by Boussinesq in 1883, mathematically estimates the stresses at a certain depth and horizontal location. For these estimates to be valid, he assumed that the soil medium is perfectly elastic (no permanent deformations upon stress removal), homogeneous and isotropic (transmits stress equally in all directions). Although we will not present the equations here, the graphical representation of those equations looks similar to the figure below. Note that this graph is drawn just to present the concept to show that as we go on a further curve, the percentage of q reduces dramatically, and it not specifically given here the shape of loading:

Soil strength 20

The curves represent the percentage of stress q that is imposed on the soil by the footing. For example at the imaginary line where the foundation touches soil, that line would be a straight line and its value would be 100% of q. Then as soon as we start to go deeper, the percentage would decrease and the lines start to be more and more curved. So each curve represents the equal stress points given as a percentage of q. In other words, if you just traveled on one curve, there would always be same stress increase due to foundation stress q.

As can be seen in the figure above, the equal vertical stress curves resemble bulbs, that is why they are called stress bulbs (they are also called pressure isobars). Also you can see that as soon as we get further away from a foundation horizontally, the effect of that load reduces dramatically and reduces to zero after a very short distance. For each shape of foundation and type of loading such as point or linear or area loading, different stress bulbs exist.

There are also charts that engineers use, such as influence charts, and other graphs, based on extensive research, to make this process simple and estimate stresses in soil. *(Actually in our age, it is mostly left to software, but that is also based on theory)*. After finding stresses in soil, the settlement amounts can be estimated as we will look into below…

In the next post of this series, we will discuss “Elastic Settlement”

]]>Now we are ready to start seeing how these are actually used, when designing foundations.

In this subsection we will now learn about bearing strength, which is a direct result of shear strength, which in turn is a direct result of angle of internal friction, that we had introduced.

Soil under footings can fail in two ways:

- bearing failure.
- excessive settlement failure,

First, the
bearing capacity is checked, but even if it is found to be sufficient, the
settlement must also be checked regardless *(which,
in practice actually, is the governing criteria in majority of the cases)*.
In other words, a soil can bear a stress applied by a foundation, but it may do
so by so much settlement, which could be over acceptable limits, and this can still
be considered as failure. If that is the case, then the bearing capacity of the
soil or the stresses from foundation on that soil must still be adjusted, so
that the settlement will be within acceptable limits *(there is not just such thing to totally eliminate settlement, at least
theoretically, so in practice, we limit it to what we consider acceptable)*.
So we will cover bearing strength in this subsection, and soil settlement in
the following subsections.

The bearing capacity of the soil means how much load can the soil below the foundation can totally carry, without undergoing shear failure and large settlements. Basically, the total load applied by the column above is divided by the area of the footing, to obtain applied stress, as stress equals: load divided by area, as we have seen before.

σ=N/A

If this stress value is more than the bearing strength of the soil, then,

- either the load coming form the column must be decreased,
- or the area of the footing must be increased which means a bigger footing,
- or, the soil bearing capacity must be improved.

Bearing failure occur differently, depending on soil strength. The figure below shows the general shear failure and it occurs in soils with greater strength, such as dense sands, strong clays such as clays that were not subject to larger loads in the past (normally consolidated clays, which we will explain), or even some rocks:

In the figure above, it can be seen how shear strength and angle of internal friction is used, to find bearing strength. You can also see in first figure that, the footing is not on ground surface but at a certain depth. This is a good thing… Because, the soil above the footing presses the soil wedges on sides of the footing, that would otherwise fail and heave, as in figure 15b.

This type of failure as in the figure above happens in strong soil only. If the soil is loose or weak, the footing just sinks, as in figure below, as it does not have the strength to form wedges like above. This type of failure is called punching shear failure:

For soils in between, the shape of failure is something in between the two cases we showed, and that one is called local shear failure.

As we saw previously under the section Structural Basics > Strength of Materials, every material has an ultimate strength, and also an allowable limit that is actually used for design. This applies also to soils. The bearing failure (shear failure) figures you see above are not actually the state to which we design to. The figures above are the ultimate shapes, where total collapses occur, when the ultimate bearing capacity is reached. But the actual value we design to, which is called allowable bearing capacity, is far less than these ultimate states. We divide the ultimate value by a safety factor (which varies between 2-3), to reach to an allowable bearing capacity value. That allowable capacity value, is what we use as our criteria, when deciding if a soil is safe against bearing failure or not. Because before reaching figures above, there are already some local failure, which means bearing failure still. To be consistent with the level of this book, we will not go into further details of bearing capacity calculations here.

When horizontal loads, such as earth pressure, wind, earthquake etc.. acts on a structure, a turning effect (moment) is created on the foundation, where it can decrease the load or even create an uplift on one side of the foundation.

So, the foundation must be designed such that no part of it will loose contact with the soil, because if it happens, it means our foundation is now suddenly smaller. So then the same vertical load that always exist, must now be carried by a smaller soil area that is still in contact with foundation. As there is less area to resist the same load, this means increase in load per area, in other words, increase in stress. This can now mean that the bearing capacity of the soil be exceeded. So to prevent this loose of contact situation, the foundation itself must be wide and heavy enough.

In this figure, you can see that one side of the foundation is under more pressure. In fıgure 17a, the right side is still under pressure here, which is the desired situation. The design must be made such that the pressure will not fall to zero anywhere at the bottom of foundation. To achieve this, M<PB/6 and e<B/6 must be satisfied. “e” is the eccentricity of the resultant force relative to foundation geometric center.

The pressure may fall to zero, or worse, the zero point may even move leftward, which further reduces foundation contact area and increases stress on soil as in figure 17b above.

Of course we showed eccentricity in one plane, but it can also happen in two perpendicular planes or even as inclined loading in real life, which would have different equations but the main logic is that the footing must not loose contact with the soil.

In the next post of this series, we will discuss “Stress Increase in Soils”

]]>We saw the relationship

τ = c + σ’sin Ø

Now,

if σ’ = 0

this means,

τ = c

Sticking action is a result of surface attraction forces between particles, and therefore applicable to only very fine particles, such as clay and some silt. Cohesive soils, as we have talked about before, are fine grained soils. They get their strength by this sticking action, but they are not big enough to generate frictional resistance between them, such as, internal friction.

So, in other words,

a purely cohesive, fine grained soil, should have zero internal friction (Ø=0), and its strength results only from cohesion as:

τ = c

and,

a perfectly coarse grained soil such as sand, should have zero cohesion (c=0), and its strength results from internal friction only as:

τ = c + σ’sin Ø

In real life however, most soils have characteristics from both sides and they are called c-Ø soils.

So, the shear strength of a soil, has two components, the component coming from cohesion, and the component coming from friction.

At the start of this section, we said in soil mechanics, cohesion means, shear strength when applied normal stress is zero. But how do we measure strength of something when we do not apply load? For this, we apply loads, and draw Mohr’s circles, and then we obtain the c value by intersecting the failure envelope tangent line by the vertical axis, as in figure Soil Strength 12, in the previous post of this series. In other words, cohesion c is an inferred value, and not measured directly.

In the next post of this series, we will discuss “Bearing Strength of Soil”

]]>Angle of repose is for sands only. Clays can not be just spilled on a surface and take this kind of shape. We cannot speak about angle of repose for clays, unless we submerge it into water, so that all cohesive bonds are broken (remember that clay is cohesive), to measure it.

In the next post of this series, we will discuss “Cohesion of Clay”

]]>In this test, a sample of cylindrical volume is used, and it is confined from its sides, and compressed. The cylindrical specimen must be carefully obtained form the site, with least possible disturbance. This is usually not possible with sands but it is with clays. After extracting the specimen, it is wrapped in a rubber membrane and placed in cylinder. Around the specimen is filled with water and the pressure of that water can be controlled, which applies an all round pressure of σ_{3} at the specimen, from everywhere, including top. We say σ_{3 }because saying σ_{2 }would be redundant as it is also coming from side of the specimen. In addition, with a loading ram from top, a vertical stress is applied, which is σ_{1}. Now that we can control all horizontal and vertical stresses, and also the pore water pressure in the soil (u), we can measure the strength properties of the sample.

So in direct shear test, we could not draw the Mohr’s circle, as we knew the failure plane, we dictated it though the plane which we slid the two pieces, and therefore it was not possible to draw a circle. But in triaxial test, we have a cylindrical specimen that can fail from any plane, in other words, it “decides” where it will fail, so we can draw a Mohr’s circle. Each circle here represents a different combination of sigma 1 and sigma 3, in other words σ_{1} and σ_{3}, which are axial and confining stresses respectively. This is a major advantage to determine angle of internal friction. Because we “force” the failure plane in direct shear test, we usually obtain a little higher values for Ø. This is because in nature, things always tend to take the easiest path possible, and in triaxial test a lower, more natural value is possible.

Like we saw in Mohr’s circle subject before, under structural section, nothing can exist above the failure envelope in the graph, in other words, no combination of sigma 1 and sigma 3 and shear stress can happen above this line, for a particular material. If you try to do it, the material will fail and can not take that combination, before you ever reach to that point. In other words, this sloped line is the failure “limit” of that material. Below this line however, any point can be obtained with different combination of sigma 1 and 3. So all points below this line, it means that particular material is able to handle without failure.

The equation of the failure envelope is:

In the graph above, we see everything in this equation, except c. What is c? It means cohesion and for sands they are often zero, so the graph above stared at origin. But for clays c is not zero, in fact it is a major component of their shear stress. So for clays, it can look like this:

Another main advantage of triaxial test over direct shear test is that we can control whether the water can escape or not. This makes a big difference. When we do not let water to escape, it is called undrained test. When we let water to escape, it is called the drained test. Both versions make big difference in strength properties. Both can exist in nature. For example, when we consider loading in a short time, we consider undrained conditions, where water did not have time to escape and so we must make undrained test in the lab to simulate those conditions. When we consider long term, where water has time to escape, we consider drained conditions and perform the test accordingly. Drained tests take much, much longer time than undrained tests, as permeability of clay soils are low. Therefore, there are used less often.

Also it is useful to note that at very high stresses, such as at the bottom of dams, sand particles crush and the slope of failure envelope will decrease as in the figure above.

In the next post of this series, we will discuss “Angle of Repose”

]]>Direct shear test is a very simple but common test. In this test, as you can see, the failure plane is fixed as horizontal. So we can not draw a Mohr’s circle here, because as you have seen before in our previous post for Mohr’s Circle, that circle represented different failure planes. (That post is under structural section but it makes no difference, principles for drawing Mohr’s circle are the same. Therefore, not controlling failure plane is the major limitation of direct shear test.

And then, for various values of N and T, we can obtain a simple graph as follows:

This angle here, is the angle of internal friction of this sand, depending on its initial density. This graph is for dense sand. The reason we drew two lines here is because, for the same force N, the force T to shear the sample varies between beginning and final stages, after large amount of sliding. For dense sand, at first, we spend a lot of effort, to start pushing, and that is why, the Ø in the beginning is Ø_{peak }but in later stages, pushing is easier, once it starts to move. That angle then is called Ø_{residual }or_{ }Ø_{ultimate}. This is similar to trying to push a heavy block on the floor. At first you spend more effort, but once it starts moving, then you can continue pushing with less effort.

This means for example, when you analyze a soil slope stability problem, if you see that the soil was dense but it deformed greatly, you must use the residual, lower value of Ø. If you use the initial higher value, your design will not be safe. If however no deformation or very little deformation has occurred, it is okay to use the higher angle value, which results in more strength value for soil.

And in this graph above, you see how differently dense and loose sands behave when we continue to push them in shear. The graph for dense sand describes just the same thing with previous figure. First we need a lot of effort, shear stress, and then after a peak is reached, we need less stress to deform it more. Loose sands however, need low stress to deform them at first, but then, ultimately, they reach to same stress level with dense sands. On the graph at the bottom, we can see that dense sands volume keeps increasing, and loose sands volume keeps decreasing, until they reach to critical state. The critical state, which we will see later in a little more detail, is the ultimate state of soil, a region where the soil’s own characteristic properties come into equation, irrelevant of how much dense or compact they were in the beginning. For example, critical void ratio, as seen in the figure, is a material property. In other words, it is a defining characteristic of that material, as it does not depend on density of material anymore. In other words, it is a characteristic of material. Critical state soil mechanics is a relatively newer subject in comparison to decades old of established soil mechanics theory but very useful. We will describe of briefly in the coming sections.

If the initial density of soil is more, this graph would be steeper, which means a higher Ø, which means a stronger soil. There is one exceptional case however. When the sand is extremely dense, under very high confining pressures, the sand particles are actually crushed on to each other, under heavy stresses. When this happens, the particles are now less angular and more round, which decreases the overall amount of friction, and consequently, Ø. This may be the case, for instance, at the bottom of a high earth embankment dam. The vertical stress from above is so high, and the confining horizontal pressure is so high, that the particles can be crushed, and the Ø at the bottom of the dam may be lower than the Ø at the upper parts of the dam. This difference can amount to considerable difference in calculations, as Ø affects all foundation calculations.

In the next post of this series, we will discuss “Triaxial Test”

]]>Let’s consider a block resting on a surface, where there is its weight and the table has friction.

In the figure above, when we try to push this block, we must apply a certain horizontal force, shown here as T. If we apply less than that, we can not push it. Again, let’s think about the ratio of horizontal to vertical as we just showed.

This angle Ø, is called, “Angle of Internal Friction”.

And if we applied the concept for soil loading it looks like this:

So, this is the angle where the soil fails, when stresses are applied in all directions. Here σ_{2} would be perpendicular to the page you are looking at and equal to σ3, thus, it is usually not shown. This is also how we test the strength of soils in the lab, in triaxial test, which we will talk about shortly.

Rearranging the equation above, we reach to the famous equation in geotechnical engineering, which relates shear stress to normal stress as:

And if we put it in graph form, it will look like:

The larger this angle of internal friction, the stronger the soil. It determines how large friction a soil can generate inside. The more friction is generates, the more it can resist loads from above, and the less it converts that vertical stress to horizontal stress. In other words, angle of internal friction is the ability to withstand shear stress. Remember the example in previous section. We said we can not walk on water, because it’s angle of internal friction is zero. We had left it at that. Now we can say why. The angle of internal friction of water is zero, because, water has no resistance to shearing. But soil has resistance to shearing. That is why, when we step on soil, we can walk on it. Unlike water, soil doesn’t convert all of the vertical stress to horizontal stress without resistance, because it has shear resistance / strength. It resists, as in K value we saw in previous section, and only some of it is converted to horizontal stress. So again, shear resistance is the key, to a soil’s strength.

Both clay or sandy soils have internal friction angle – especially sands. It characterizes their strength. For preliminary calculations, a value of 30 degrees may be assumed for angle of internal friction of sand, before any further data can be obtained. For clays, in undrained conditions, (when the water did not have enough time to escape under loading) the friction angle is approximately 0 as long as clay remains saturated. In drained conditions however, (when water had enough time to escape after loading) the friction angle for clays at failure (in the critical state) is somewhere between 20 and 30 degrees, which is considerably less than sands, but still provides an amount of resistance.

In the next post of this series, we will discuss “Direct Shear Test”

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