Hello this is Professor Kitch.
Welcome to this webcast on section 12.9.
This is the first of two webcasts covering laboratory shear strength test
This presentation is on the direct shear test When you have finished this presentation you
should be able to describe the direct shear test & draw schematic of testing equipment
In addition, given data from a direct shear test, you should be able to properly plot
the data and determine strength parameters of the soil which was tested
This figure shows a plan view of a typical direct shear test box
The soil specimen used in this test is cylindrical in shape with a cross-sectional are of a-sub-s
And is contained in doughnut shaped rigid metal shear box
To understand the operation of the test, it is best to look at a vertical cross-section
through the shear box.
For the rest of the webcast we will be looking at a vertical section cut through the shear
box at b b-prime as shown.
This figure shows the vertical cross-section b b-prime.
The shear box is split in half at its mid-height by a horizontal plane.
This separates the shear box into two pieces called the upper and lower shear box.
The soil specimen is contained in the shear box and sandwiched between two porous stones,
one above and one below the specimen as shown.
The purposes of the porous stones are to distribute loads evenly to the soil and to allow water
to pass freely in or out of the specimen.
This prevents the buildup of excess pore pressure during testing
The lower shear box is attached to a rigid base and placed in a water bath.
A vertical force f-sub-z can be applied to the top of the upper stone.
This force will generate a normal stress, sigma, in the specimen which will be equal
to f-sub-z divided by the cross-sectional area of the specimen a-sub-s
Since the box is split, the upper box can move separately from the lower box.
By applying a horizontal force f-sub-x to the upper shear box we can generate a shear
stress, tau, in the specimen.
The magnitude of tau will be equal to f-sub-x divided by a-sub-s
Because the box is split in the middle, a shear plane will form along the dashed line
shown in the figure.
A normal stress, sigma, and a shear stress, tau, will be generated along this shear plane
as shown.
During the test, we use dial gages to measure the horizontal displacement of the shear box,
delta-sub-x, and the vertical displacement of the top of the specimen, delta-sub-z
Now let's see how the specimen is sheared.
After the vertical load, f-sub-z, is applied, the horizontal load, f-sub-x, is gradually
increased.
This generates a shear displacement in the specimen.
The dial gages record the horizontal and vertical movements during shearing.
The vertical force and therefore the vertical stress in the specimen are constant during
the testing.
While the horizontal force and shear stress are increasing.
As we just noted, during the test, the vertical or normal stress will remain constant.
In this case we will assume the normal stress for the test is equal to sigma-1
As we increase the applied shear stress, tau, we will record its magnitude as function of
the horizontal displacement, delta-x Depending on the nature of the soil being
tested, the plot of tau versus delta-x will have one of two shapes of curves.
In one case, the applied shear stress increases to a peak level and the decreases with additional
displacement finally reaching a plateau.
In this case the greatest shear stress measured represents the peak strength and the value
of shear stress at the lower plateau represents the residual strength.
This shape of shear-displacement curve is typical of dense sands or over-consolidated
clays.
The interlocking of soil particles in these soils generates relatively high shear strength.
As this strength is exceeded the particles start to slide over one another.
This causes an expansion or dilation of the soil in the shear zone and the strength of
the soil is reduced.
This strength loss eventually levels off as the soil reaches a new equilibrium state.
The second shape of curve shows a monotonically increasing shear stress as the horizontal
displacement increases.
It eventually reaches an asymptotic peak value.
This case is typical of the behavior or loose sands or normally consolidated clays.
In these soils the shearing process causes soil particles to rearrange themselves into
a denser configuration.
This causes contraction or densification of the soil in the shear zone and the soil gains
strength until it reaches its peak value.
This particular figure illustrate one soil, say a certain sand, tested both a dense and
a loose state at the same normal stress.
In this case the peak strength of the loose sand will be equal to the residual strength
of the dense sand.
In both cases the soil will reach the same equilibrium state at large displacements,
because both tests were performed on the same material at the same normal stress.
Remember that the normal stress during a direct shear test remains constant.
As we just described, the soil will usually change volume during the shearing process.
We can measure this volume change in a direct shear test by observing the vertical displacement
during testing.
Because the cross-sectional area of the specimen does not change, any change in volume will
generate a vertical displacement in the specimen.
If the soil is dense, it will dilate or increase in volume during shear and the displacement
at the top of the specimen will be upward as we see here.
During the direct shear test we record the vertical displacement at the top of the specimen
as a function of the horizontal displacement.
Here we see a typical plot of vertical displacement versus horizontal displacement for a dense
sand or over-consolidated clay.
There is a small amount of downward movement of the specimen at the start of shearing,
followed by a larger upward movement indicating, that the specimen is dilating during shear.
The opposite of dilation during shear is contraction, or a decrease in volume.
This occurs when the soil is loose and gets denser during the shearing process.
In this case the displacement at the top of the specimen will be downward as seen here.
Here we compare the previous plot of vertical displacement versus horizontal displacement
for a dense soil with that of a loose soil.
The loose soil is contracting during shear or becoming denser.
Therefore the vertical displacement is downward as we see here.
Thus far we have looked at the result of a single direct shear test.
And explained the difference between contractive and dilative behavior during shear.
We now want to use the results of a direct shear test to determine the strength parameters,
friction angle, phi, and cohesion, c of a given soil.
In order to accomplish this we must run a series of direct shear tests on the same soil
at different normal stress levels.
This figure will present the results of a series of three direct shear test performed
on identical specimens of the same soil.
Here we see, in black, shear stress as a function of horizontal displacement for one test run
at a high normal stress, sigma-1.
A second test, shown in blue, was run at an intermediate normal stress, sigma-2.
And a third test, shown in green, was run at a low normal stress, sigma-3.
For each of the three tests we can determine the peak and residual strengths were as shown.
To determine the friction angle and cohesion of a soil we must develop a mohr-coulomb plot
of shear strength versus effective stress.
In the direct shear test we measure the shear and normal stresses on only one plane, the
failure plane.
Therefore we cannot generate a mohr-circle for each test.
Instead we will simply plot a single point representing the shear and normal stresses
at failure.
The triangles shown, represent the peak shear strength measured from the previous figure.
If we fit a line to these points we can determine the effective peak friction angle, phi-prime-peak,
and the effective peak cohesion, c-prime-peak.
Similarly we can plot the residual strength values shown as squares in this figure, fit
a line to these points and determine the effective residual friction angle, phi-prime-r, and
the effective residual cohesion, c-prime-r.
This particular plot illustrates data that might be measured from testing an overconsolidated
clay soil.
This can be determined by the significant cohesion measured from peak strength.
The residual cohesion is generally very small or near zero as we see here.
If we had been testing a dense sand the plots of strength versus normal stress would have
generated two strength envelopes such as those shown in this figure.
Neither envelop would show any cohesion but we would have a different peak and residual
strength.
If we had been testing the same sand in a loose state we would have measure just one
set of strengths and the single peak strength envelop would appear as shown.
This would be identical to the residual strength envelop obtained for the same sand in a dense
state.
There are a number of limitations to the direct shear test, these include the following
There is no control over drainage in the specimen and we cannot measure pore pressures during
the test Therefore the only way we can know what's
going on is to run the test very slowly so that any excess pore pressures that might
be generated can dissipate.
In this case there will be no excess pore pressures and the effective stress will be
equal to the total stress In this test we force the soil to fail on
a predetermined failure plane which is always horizontal.
This may or may not be the weakest plane in the soil
The actual stress state in the failure zone is much more complicated than the simple model
we assume in our test Finally, because the actual conditions in
the shear zone are complex, we cannot determine the strains during the test.
Therefore while we can determine soil strength, we cannot determine the stress-strain properties
of the soil with this test.
In summary, the direct shear has the following characteristics
The test is easy to setup and run For clay soils the test must be run very slowly
sometimes taking over a day to reach failure.
The data is relatively easy to analyze We can measure the volume change during shear
to determine if the soil dilating or contacting The strength parameters measured in this test
are less accurate than those measured in triaxial test.
This is mostly due to the fact that we constrained the soil to fail on a certain plane rather
than letting the soil fail on whatever plane it would like to.
This concludes this presentation and you should now review our learning objectives.
Now would be a good time to review Example 12.9 in your text.
Try and complete this example before class.
5.