SME
Soil Mechanics · Problem type
Cohesion and friction angle from two direct shear tests
Use this for “Two drained direct shear tests fail at τ = 40 kPa under σ = 50 kPa and τ = 98 kPa under σ = 150 kPa. Find c′ and φ′.”
Chart
τ₂ = 0.58σ₂ + 11
Slope m = 0.58: up 0.58 for every 1 across
Intercept c′ = 11: it crosses the y-axis at (0, 11)
τ₂ = 0.58 × 150 + 11 = 98 when σ₂ = 150
This is an example. Type your own values to start a new problem.
m
=
(
τ₂ (kPa)
−
τ₁ (kPa)
)
÷
(
σ₂ (kPa)
−
σ₁ (kPa)
)
φ′
°
=
tan⁻¹(
m
)
c′ (kPa)
=
τ₁ (kPa)
−
σ₁ (kPa)
×
m
Formulas
m
=
τ₂
−
τ₁
σ₂
−
σ₁
0.58
=
98 − 40
150 − 50
φ′
=
tan⁻¹(
m
)
30.1137 = tan⁻¹(0.58)
c′
=
τ₁
−
σ₁
×
m
11 = 40 − 50 × 0.58
c′
=
τ₂
−
σ₂
×
m
11 = 98 − 150 × 0.58
Assumptions
•
Drained tests: the pore pressure stays zero, so each σ is an effective stress.
•
Mohr–Coulomb: the failure points lie on the straight line τ = c′ + σ tan φ′.
Step-by-step
WE KNOW
σ₁ = 50 kPa, τ₁ = 40 kPa, σ₂ = 150 kPa, τ₂ = 98 kPa
FIND
slope (tan φ′) (m), friction angle (φ′), cohesion (c′)
Step 1 · Find slope (tan φ′): m
Use m = (τ₂ − τ₁) ÷ (σ₂ − σ₁)
Rise over run between the two failure points.
m
=
98 − 40
150 − 50
m
=
58
100
m
=
0.58
Step 2 · Find friction angle (φ′)
Use φ′ = tan⁻¹(m)
The line’s slope is tan φ′, so the angle is its inverse tangent.
φ′ = tan⁻¹(0.58)
φ′ = 30.1137°
Step 3 · Find cohesion (c′)
Use c′ = τ₁ − σ₁ × m
Follow the line back from test 1 to σ = 0; what is left of τ there is c′.
c′ = 40 − 50 × 0.58
c′ = 40 − 29
c′ = 11 kPa
Check
Put the numbers back in. Both sides should match.
0.58
=
98 − 40
150 − 50
✓
30.1137 = tan⁻¹(0.58) ✓
11 = 40 − 50 × 0.58 ✓
11 = 98 − 150 × 0.58 ✓
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