SME
Soil Mechanics
Shear strength
Use this for “A silty sand (c′ = 10 kPa, φ′ = 30°) is sheared drained at σ′₃ = 100 kPa. Find σ′₁ at failure and the failure plane’s angle.”
Diagram
The line τ = c′ + σ′ tan φ′ touches the circle from σ′₃ = 100 to σ′₁ = 335 kPa: the soil fails.
The
failure
plane
is
θ
=
45°
+
φ′
2
=
60°
from
the
major
principal
plane;
2θ
on
the
circle.
This is an example. Type your own values to start a new problem.
θ
°
=
45°
+
φ′
°
÷
2
σ′₁ (kPa)
=
σ′₃ (kPa)
×
tan(
θ
°)²
+
2
×
c′ (kPa)
×
tan(
θ
°)
Δσ
Deviator stress
calculated · kPa
Formulas
θ
=
45°
+
φ′
2
60
=
45°
+
30
2
σ′₁
=
σ′₃
×
tan
(
θ
)
2
+
2
×
c′
×
tan(
θ
)
334.641
=
100
×
tan
(60°)
2
+
2
×
10
×
tan(60°)
Δσ = σ′₁ − σ′₃
234.641 = 334.641 − 100
Assumptions
•
Mohr–Coulomb failure: the soil fails where shear stress reaches τ = c′ + σ′ tan φ′.
•
Drained, so the pore pressure stays zero and every stress is effective (σ′).
•
θ is measured from the major principal plane, horizontal in the triaxial cell.
Step-by-step
WE KNOW
c′ = 10 kPa, φ′ = 30°, σ′₃ = 100 kPa
FIND
failure plane angle (θ), major stress at failure (σ′₁), deviator stress (Δσ)
Step 1 · Find failure plane angle (θ)
Use θ = 45° + φ′ ÷ 2
The
circle
touches
the
strength
line
at
2
θ
=
90°
+
φ′
from
σ′₁,
so
halve
it.
θ
=
45°
+
30
2
θ
=
45°
+
15
θ
=
60°
Step 2 · Find major stress at failure (σ′₁)
Use σ′₁ = σ′₃ × tan(θ)² + 2 × c′ × tan(θ)
The
biggest
circle
the
line
allows:
σ′₃
grown
by
tan²
θ
,
plus
the
part
cohesion
adds.
σ′₁
=
100
×
tan
(60°)
2
+
2
×
10
×
tan(60°)
σ′₁
=
100
×
1.7321
2
+
2
×
10
×
1.7321
σ′₁ = 300 + 34.641
σ′₁ = 334.641 kPa
Step 3 · Find deviator stress (Δσ)
Use Δσ = σ′₁ − σ′₃
The extra vertical stress the piston adds at failure.
Δσ = 334.641 − 100
Δσ = 234.641 kPa
Check
Put the numbers back in. Both sides should match.
60
=
45°
+
30
2
✓
334.641
=
100
×
tan
(60°)
2
+
2
×
10
×
tan(60°)
✓
234.641 = 334.641 − 100 ✓
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Cohesion and friction angle from two direct shear tests
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More problem types
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