# Discussion on A small portion of lab report for Direct Shear

A small portion of lab report for Direct Shear (Strength)
Attached is what I have so far, along with the handout. I just need part 2 of the handout.I just need a very detailed explanation when discussing strength parameters. Using the YouTube link, handout as well as any other research you may conduct, please organize the body of this report to demonstrate your understanding of direct shear and its’ significant elements.

Direct Shear (Strength)

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Report

Sections:

1. Introduction
2. A paragraph on the importance of the direct shear test for Geotechnical Engineering

projects and applications. Show the reader you are knowledgeable on this particular

concept of Geotechnical Engineering.

Direct shear testing is a common, relatively simple method of testing the strength parameters of a soil. This test and these parameters are essential to understanding how soil types deform and become malleable in compression and shearing conditions. The direct shear test consists of packing the direct shear test mold with the soil specimen. The mold has two halves that can move across each other, creating a shear plane for the specimen. The soil is then subjected to a specific stress along the Z axis and left to completely consolidate. After consolidation the soil the mold is pushed, shearing the soil at a constant deformation rate. The force needed to deform the specimen at this constant rate is measured and graphed. Once this is done three or more times at different Z axis stresses, the results are graphed on a Mohr’s Coulomb chart. This plots the Z axis stress on the X axis and the shearing force on the Y axis. These tests for a straight line, the equation of which gives the strength parameters.

In this equation, Φ’ is known as the friction angle and c’ is known as the effective cohesion. These parameters are the same for every consolidation ratio of the same material because these parameters are independent of effective stress.

1. Body (Organize in a reasonable manner)
3. Please review Strength handout from reader
4. Using the aforementioned links as well as any other research you may conduct,

please organize the body of this report to demonstrate your understanding of direct

shear and its significant elements. Be very detailed; especially when discussing strength

parameters.

1. Conclusion

1. Cited references (list them; at least 3)

“CE 326 Mod 12.9a Direct Shear Test.” YouTube, lecture by William A. Kitch, 2015,

Valdes, J. 2020. CIVE 462 Geotechnical Engineering READER, San Diego: CalCopy

STRENGTH
1. The strength of soil arises from two components: friction between grains accompanied
by grain climbing (each other), and bonding between the grains. Before we define
strength, we must understand friction fundamentals.
2. Consider a block with weight W = l00N resting on a tabletop.
The tabletop is perfectly horizontal, and the block is a perfect
square. Is there friction at the interface?
W=
100N
3. The answer to the question above is NO. This may seem odd: friction does not simply
exist at every interface. Friction must be triggered. If friction is not triggered, then it
doesn’t “appear”.
4. Consider the block on the tabletop again. Let us say that we
push the block laterally with a force of lON, but the block
does not slip. Then we push with 30N; no slip. We continue
to increase the pushing force, and we find that the block slips
when we push with 60N. In other words, 60N are required to
make the block slip.
5. You can think of this situation as follows. When the block is not
being pushed laterally, there are 60 soldiers (60 Newtons)
sleeping at the interface between the block and the tabletop.
If we do not push the block laterally, then those 60 soldiers
remain asleep. Therefore, there is no frictional resistance.
W=
100N
,ON ‘sleeping’
6. If we push the block with lON, then 10 soldiers wake up (they mobilize) and resist the
push with a friction force Ff= lON (the remaining 50 are still sleeping). So, when we
push with lON, the mobilizedfriction force is Ff= l0N. That is, friction has now
appeared.
I
W=
100N
F
1= 60N
(no one ‘sleeping’)
BLOCK SLIPS
7. If we push the block with 30N, then 20 additional soldiers mobilize to resist the push,
resulting in a total of 30 soldiers (the other 30 are still sleeping). So, when we push with
30N, the mobilized friction force is Ff= 30N.
so
8. If we further push the block with 60N, then 30 additional soldiers mobilize to resist the
push. That is, all of the available soldiers are resisting when we push with 60N, and so,
the friction force is fully mobilized, i.e., Ft= 60N.
9. Full mobilization of the frictional resistance indicates that we are pushing with the
maximum force that can be sustained by the system. That is, slippage takes place when
friction is fully mobilized. Or in other words, slippage occurs immediately all the soldiers
wake up!
10. Strength is related to failure. Failure is defined by the engineer. For example, a
composites engineer may call the stress required to induce cracking of a composite
cylinder the strength, because the cracking is what he/she identifies as a failure.
11. In the example above, we define failure as slip of the block. Thus, we define the strength
of the block-tabletop interface as the maximum frictional force present when slip
occurs: S = Ft max= 60kN. Clearly, Ft max is a shear force (because it acts parallel to the
interface).
12. For soils, strength is defined as the shear stress at failure. Imagine that we could
fabricate a table composed of unbonded sand grains, and a block of unbonded sand
grains. If we repeated the experiment with the solid block described above, we would
find the same mechanics, with two major exceptions:
a. At the interface, the unbonded grains would not only grind against each other,
but also roll and climb amongst each other. Please … close your eyes and
visualize this.
b. Because of item a (above), the sand block would not slip suddenly as the solid
block would. Instead, the frictional resistance would evolve gradually (soldiers
would gradually wake up), and we would need to push the sand block quite a
ways to mobilize the full frictional resistance (to wake up all the soldiers).
In addition, the strength of the block-tabletop interface, a sand-sand interface in this
case, would be S = Ft max/ A, where A is the area of the interface (i.e., the area of contact
shared by the sand-made block and the sand-made table). We divide by A simply to
render Sa shear stress (as is customary in soil mechanics).
13. Re-imagine the experiment described above, but now with grains of sand that are very
rough and angular (as opposed to smooth and rounded). Clearly, the strength S would
be larger in the case of angular rough grains (there would be more soldiers), because
such grains offer more resistance to grinding, rolling, and climbing. This confirms that
the strength of the soil is dependent on the shape and roughness of the grains.
51
14. Re-imagine the experiment described above, but now with a large weight placed on top
of the block of unbonded sand grains. Clearly, the strength Swould be larger in the case
of a loaded block (vs. unloaded). This is because the stress that acts normal to the
interface is higher, and therefore, a larger force would be required to move the block
laterally. This confirms that the strength of soil is dependent on the stress that acts
normal to the interface in question.
15. Now, you may ask: I can imagine a sand block on a sand table, but where in the real
world are soil-soil interfaces to be found? Look at the drawing below. This is a side view
of a structural foundation (a concrete element; shown shaded) being pushed (loaded)
into the ground. Below the foundation, the soil deforms along interfaces denoted by
dashed and solid curves. These interfaces are the sides of wedges of soil that move
relative to each other. The soil’s ability to sustain the applied load arises from the
mobilization of friction along such interfaces. That is, the resistance arises from the
soldiers that are awaken along those interfaces.
16. So, we know that soil strength depends on grain shape. It also depends on gradation
and density (and water content, for fine grained soils; as you learned from pla·sticity
experiments). The strength of soil also increases if the stress cr’ normal to the interface
in question increases (see item #14 above). Given that strength depends on so many
parameters that can vary, how do we go about parameterizing soil strength?
17. We do so by establishing the relationship between Sand cr’ via lab tests. The direct
shear test is a good lab test for this. It somewhat mimicks the block on a table
experiment. It involves pushing an axially loaded soil mass (confined inside a metal ring)
laterally along the top of another soil mass (confined inside another metal ring). In
reality, the soil specimen begins its life as a single soil mass (the top confined in one ring
and the bottom confined in the other ring). The specimen “breaks” into two subspecimens that grind against each other as the test progresses. Note: the axial load
results in a cr’ that acts normal to the interface.
18. So, how do we define failure for such a case? We define failure with the use of shear
stress vs. displacement data. The shear stress is that which acts on the interface shared
by the two sub-specimens (see item #17). The displacement is simply the relative
displacement between the two sub-specimens.
52
19. When plotted, the shear stress may either (1) peak and then drop to taper off to a
“critical state” (constant shear resistance with increasing displacement), or (2) rise
towards and then taper off to a critical state (that is, no peak).
20. If the shear stress exhibits a peak, then (1) the peak of the curve is identified as a failure
point called “the peak”, with strength= peak shear stress and (2) the onset of the critical
state is identified as another failure point called “the critical state”, with strength=
critical state shear stress. In most cases, soils that exhibit a peak and a post-peak critical
state are dense or overconsolidated.
21. If the data does not exhibit a peak, but rather exhibits resistance that rises towards and
then tapers off to a critical state, then the onset of the critical state is identified as the
failure point (called “the critical state”), with strength = critical state shear stress. In
most cases, soils that exhibit no peak are loose or normally consolidated.
22. We still have an open question: how do we parameterize the strength of a given soil?
Recall that we need a relationship between cr’ and Sfor the soil in question. In fact,
there are two relationships. We proceed as follows:
a. We conduct at least three direct shear tests, each with a with different cr’.
b. We plot the shear stress vs. displacement data for each test.
c. We identify the peak and critical state shear stresses for each test (from the data)
d. We plot the coordinates (cr’, Speak) for each test on a Mohr Circle space, with the
origin at (O, O).
e. We draw a line that best fits the data points obtained in item d.
f. The equation of this line is:
Speak =cr’ tan <I>’peak + c’peak
This is the first relationship that we are looking for. It is called the peak MohrCoulomb failure envelope. In the equation, <I>’peak and c’peak are the peak strength
parameters. <I>’peak is the peak friction angle. It captures the nature of the frictional
interactions amongst grains at the peak. Note also that the slope of the envelope
(the line) is tan <I>’peak- The c’peak is called the peak effective cohesion; it is the yintercept, and is the shear strength of the agent (if any) that bonds the soil grains.
23. You may ask: what about the critical state? We proceed as follows:
a. We plot the coordinates (cr’, Scs) for each test on a Mohr Circle space, with the
origin at (O, O).
b. We draw a line that best fits the data points obtained in item a.
c. The equation of this line is:
53
Scs =er’ tan cj>’cs + c’cs
This is the second relationship we are looking for. It is called the critical state
Mohr-Coulomb failure envelope. In the equation, cj>’cs and c’cs are the critical state
strength parameters. cj>’cs is the critical state friction angle. It captures the nature
of the frictional interactions amongst grains at the critical state. Note also that the
slope of the envelope {the line) is tan cj>’cs, The c’csis the critical state effective
cohesion; it is the y-intercept, and is the shear strength of the agent {if any) that
bonds soil grains.
24. Note from any of the two strength envelope equations, that Sis dependent on cj>’, c’,
and er’. The cl>’ is dependent on the soil (grain shapes, gradation, density). The c’ the
bonding agent’s strength. If c’ is zero, then there is no bonding agent. The er’ is clearly
not a soil parameter. For example, the strength at a point in the field depends on the er’
at the point in question.
25. When asked for the strength of soil, engineers usually convey the strength parameters.
This is because the strength of soil depends on er’, and er’ is not a soil property. The
typical norm is to convey the critical state strength parameters for fine grained soil and
the peak strength parameters for coarse grained soil.
54
NOTE: THESE ARE THE MINIMUM REQUIREMENTS
Direct Shear (Strength)
Report
Sections:
1. Introduction
a. A paragraph on the importance of the direct shear test for Geotechnical
Engineering projects and applications. Show the reader you are
knowledgeable on this particular concept of Geotechnical Engineering.
2. Body (Organize in a reasonable manner)
b. Please review Strength handout from reader
c. Using the aforementioned links as well as any other research you may conduct, please
organize the body of this report to demonstrate your understanding of direct shear and
its’ significant elements. Be very detailed; especially when discussing strength
parameters.
3. Conclusion
4. Cited references (list them; at least 3)

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