SME
Soil Mechanics · Problem type
Undrained shear strength from a UU test
Use this for “A saturated clay fails in a UU test at σ₁ = 180 kPa with a cell pressure of 60 kPa. What is sᵤ?”
Diagram
Undrained:
the
line
is
flat
at
sᵤ
=
σ₁ − σ₃
2
=
60
kPa,
the
top
of
the
circle.
The
failure
plane
is
θ
=
45°
+
φ′
2
=
45°
from
the
major
principal
plane;
2θ
on
the
circle.
This is an example. Type your own values to start a new problem.
sᵤ (kPa)
=
(
σ₁ (kPa)
−
σ₃ (kPa)
)
÷
2
Formulas
sᵤ
=
σ₁
−
σ₃
2
60
=
180 − 60
2
Assumptions
•
Unconsolidated undrained (UU): no water leaves the saturated clay, so its strength does not change with σ₃.
•
Total stresses: the pore pressure is not measured, so φ = 0 and the strength line is flat.
•
The strength line touches the circle at its top, so sᵤ is the circle’s radius.
•
An unconfined compression test is the case σ₃ = 0, σ₁ = qᵤ: sᵤ = qᵤ ÷ 2.
Step-by-step
WE KNOW
σ₃ = 60 kPa, σ₁ = 180 kPa
FIND
undrained shear strength (sᵤ)
Step 1 · Find undrained shear strength (sᵤ)
Use sᵤ = (σ₁ − σ₃) ÷ 2
The
circle’s
radius:
half
the
deviator
stress
σ₁
−
σ₃
.
sᵤ
=
180 − 60
2
sᵤ
=
120
2
sᵤ = 60 kPa
Check
Put the numbers back in. Both sides should match.
60
=
180 − 60
2
✓
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Cohesion and friction angle from two direct shear tests
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Bearing capacity
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