Middle East Technical University

CE 363/364
Soil Mechanics

Lab 10
Unconsolidated Undrained (UU) Triaxial Test
merte@metu.edu.tr
 Purpose of the Test 2

• Estimate the undrained shear strength parameters of the

cohesive soils
• Obtain undrained stress-strength behavior of a cohesive soil
• Understand the concept of the triaxial test in general, and be able
to draw the a schematic diagram of the triaxial setup showing
the only active components used in the test
 Purpose of the Test 3

• But before jumping into the test..

• Why do we need to estimate the undrained strength
parameters of the cohesive soils?
• Why do we need to obtain the stress strain behavior of the
soils?
 Purpose of the Test 4

• Assume that you are going to build a structure on top of a clay

layer. Your foundation will overlie this layer. You have to check
that strength of your soil is capable of carrying load transferred
from your superstructure. But …
• Would you use the undrained (short term) shear strength
parameters (cu and Φu) or drained (long term) (c’ or Φ’) to
estimate shear strength for your design?
• Try to think for a minute.

An example sketch for the abovementioned

situation
 Purpose of the Test 5

• As you all know, in a loading case, initially (short term), excess

pore pressure develops (∆σ = ∆u). After a certain time passes
(long term), excess pore pressure dissipates and effective stress
increases.
• Recall shear strength formula => = c + ’tan’
• If the effective stress increases, shear strength increases too. It
means that in the long term, cohesive soils will have higher
shear strength for the loading case.
• You would design your foundation considering undrained
(short term) case for being on the safe side.
• You will obtain undrained shear strength parameters from the
UU test.
 Purpose of the Test 6

• How about stress-strain behavior? Why do we need it?

• The only thing we should consider is not the shear strength.
Also, the settlements are an issue. Settlements should not
exceed the values defined in the design codes.
• You need undrained initial tangent modulus (E0), secant
modulus (E50), etc. to calculate the settlements.
• You will learn how to calculate these parameters from the
stress-strain graph.
 Overall View of The Triaxial Test 7

• Every triaxial test contains two stages.

• First, consolidation stage where the soil specimen is allowed
consolidate
• Second, shear stage where the specimen is sheared in drained
or undrained conditions.
• There are three types of triaxial tests
• Consolidated Undrained (CU)
• Consolidated Drained (CD)
• Unconsolidated Undrained (UU)
• The first latter defines the drainage condition during the
consolidation stage, the second one defines for the shear stage.
 Overall View of The UU Test 8

• We neither consolidate the specimen or let specimen drain its pore

fluid during the shearing stage.
• Therefore, volume change is zero for the UU test.
• Also, excess pore pressure develops during consolidation and shearing
stage. Because we do not let the water out, so, any stress increment
will be carried by the water, not the soil skeleton (assuming if the soil is
fully saturated). But we do not measure the pore pressures in this test.
• Basically, we are going to apply an all around pressure to the specimen
and then shear it.
• It is very fast test compared to CU and CD tests because we do not have
drainage in any part of this tests. It might take long time to consolidate
and drain the cohesive soils since they have low permeability.

https://www.youtube.com/watch?v=hq4UlLm8oIs

Check it out between 9:12 – 11:17

 Equipment 9

Loading frame (fixed)

Loading ram (piston)

Specimen
Load cell: Measures the load
LVDT: Measure the displacement
Data logger: Transfers the digital data to computer
Cell/back pressure units: Apply and control the pressures
Loading unit: It is velocity controlled which pushing the triaxial cell to upwards. Since the loading frame is fixed, the loading
ram is moving relative to down to the specimen, that’s how the specimen is deformed.
 Equipment 10

Drainage lines,
where we apply
water pressure to
soil specimen to
saturate it

Pedestal, on which we will place our specimen on top of it Dashed lines show which drainage line is
As you can see, there are two drainage lines in the connected to which valve. But we don’t use
pedestal where we pressurize the specimen so that we can all of them. You have to distinguish which
saturate the specimen. But in UU test, we don’t saturate one of them we are using in the UU test.
the specimen, we assume it is fully saturated.
 Equipment 11

It is a porous stone, but in UU test we Triaxial base

use impermeable disks.
 Equipment 12

O-ring stretcher

O-rings
 Procedure 13

• Let’s watch this video as it shows how to assemble a clay

specimen into triaxial cell.
• https://www.youtube.com/watch?v=7Hh45k1gqjU
• As you can see in the video, a flexible membrane is placed around
the specimen. Membrane is an impermeable material that
prevents the cell fluid flow into the specimen. However, if you do
not seal it at the bottom and top, with O-rings, pressurized water
will flow into the soil.
• After assembling the specimen, make sure that your loading ram
is in contact with your top cap (see 6:13).
• The next step is filling the triaxial cell with water (make sure that
air vent/bleed valve is open) and apply cell pressure (all-around
pressure).
Procedure 14

Lading ram
(loading piston)

Top cap

Specimen
 Procedure 15

• The next step is shearing the specimen. Do not forget that

drainage valves are not open in any stages of this test. So, there is
no volume change.
• During the shearing stage, we are going to record displacement
and force. Then, you are going to draw stress – strain curve.
• We will set axial strain rate to 0.5% to 2% per minute. So if you
have a specimen with 100 mm height;
• Your strain rate will be between: 100 x 0.5% = 0.5 mm/min -
100 x %2 = 2 mm/min
• We will shear it until we see a visible shear plane or reaching to
15% strain.
 Procedure 16

Example of a visible shear plane

 Calculations 17

• Axial load is converted to axial stress by dividing it by the area of

the specimen. However, the specimen area is enlarged during the
test due to Poisson’s effect. The correct area at each axial strain
can be calculated by assuming the specimen remains a cylinder
and ∆V=0 (undrained behavior), as Ao/(1- εax).
• Here is an example data recorded from the test (not complete
data).

Displacement (mm) Force (N)

0
0.1
0
100
Initial diameter = 50 mm
0.2 200 Initial height = 100 mm
0.3 300
0.4 400
0.5 500
0.6 600
0.7 700
0.8 800
 Calculations 18

• Let’s calculate the axial strain and deviatoric stress for the third
row.
•𝐴 𝜋∗ 1963.5 𝑚𝑚
.
•𝜖 % ∗ 100 0.2 %
.
• 𝐶𝑜𝑟𝑟𝑒𝑐𝑡𝑒𝑑 𝑎𝑟𝑒𝑎 𝑚𝑚 . 1697.4 𝑚𝑚

•𝜎 ∗ 1000 102 𝑘𝑃𝑎

.
Displacement Axial Strain Corrected Area Deviatoric Stress
Force (N)
(mm) (%) (mm2) (kPa)
0 0 0 1963.5 0.0
0.1 100 0.1 1965.5 51
0.2 200 0.2 1967.4 102
0.3 300 0.3 1969.4 152
0.4 400 0.4 1971.4 203
0.5 500 0.5 1973.4 253
0.6 600 0.6 1975.3 304
0.7 700 0.7 1977.3 354
0.8 800 0.8 1979.3 404
 Calculations 19

• The next step is drawing the deviatoric stress vs. axial strain
graph.

After drawing the graph, you have to find the maximum deviatoric stress which is
the stress at the failure.
 Calculations 20

• Then you have to calculate the principle stresses at the failure

and construct the Mohr circle.

𝜎 𝐴𝑙𝑙 𝑎𝑟𝑜𝑢𝑛𝑑 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒

𝜎 𝜎 𝜎

𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑖𝑟𝑐𝑙𝑒 𝜎

 Calculations 21

• But this is only for one test. We need to do at least three UU tests
to construct failure envelope. Therefore, we need to conduct
more UU tests with different 𝜎 .
• Even though the 𝜎 is different for each test, they are expected to
have the same deviatoric stress at the failure. The reason for this
phenomena is that the soil is fully saturated. As mentioned in the
early slides, pore water will carry any additional stress increment
since there is no drainage.
• Increases in total stress 3 don’t affect effective stresses, and
therefore shear strength, resulting in same diameter Mohr circles
If you draw the Mohr-coulomb
failure envelope which is a line
that tangent to all three circles, it
will pass through the top points
of the each circle. Y-intercept is
your undrained cohesion and
slope of the line is undrained
angle of friction, which is zero.
 Calculations 22

• Determine undrained tangent modulus (E0 : initial slope of stress-

strain graph) and undrained secant modulus (E50: slope of a line
from origin to the point on the curve with cu=qu/2)