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Geotechnical

Triaxial Test — Effective Friction Angle

Interpret a triaxial test at failure to find effective principal stresses and friction angle from a given cohesion, plus the failure-plane angle.

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Worked examples

These are the solver's own reference problems — the answers come from published sources or independent hand computation, never from the solver itself. The solution below is the live solver output for each.

Example 1

CU test with pore pressure (the motivating exam problem)

Given

major principal stress
300 kPa
minor principal stress
100 kPa
pore pressure
40 kPa

Assumptions

  • Mohr-Coulomb failure criterion in effective stresses; pore pressure acts equally on both principal directions.

Solution steps

  1. Effective principal stresses

    Effective stress is total stress minus pore pressure.

    σ1' = σ1 − u; σ3' = σ3 − u

    σ1' = 260 kPa; σ3' = 60 kPa

  2. Friction angle from the failure circle

    With zero cohesion the failure envelope passes through the origin and is tangent to the effective-stress circle.

    sinφ' = (σ1' − σ3')/(σ1' + σ3')

    circle: center 160 kPa, radius 100 kPa → φ' = 38.68°

    = 38.68

Results

Effective friction angle φ' (degrees)

38.68

Effective major principal stress σ1'

260kPa

Effective minor principal stress σ3'

60kPa

Deviator stress σ1 − σ3

200kPa

Failure plane angle from horizontal (45 + φ'/2)

64.34

Where this answer was checked
source
sin φ' = (σ1'−σ3')/(σ1'+σ3') for c' = 0 — σ1 = 300, σ3 = 100, u = 40 kPa
verified by
hand-recomputed
derivation
σ1' = 260, σ3' = 60. sinφ' = 200/320 = 0.625 → φ' = 38.68°. Failure plane 45 + φ'/2 = 64.34°.

Example 2

drained test with cohesion c' = 20 kPa

Given

major principal stress
400 kPa
minor principal stress
100 kPa
effective cohesion
20 kPa

Assumptions

  • Mohr-Coulomb failure criterion in effective stresses; pore pressure acts equally on both principal directions.

Solution steps

  1. Effective principal stresses

    Effective stress is total stress minus pore pressure.

    σ1' = σ1 − u; σ3' = σ3 − u

    σ1' = 400 kPa; σ3' = 100 kPa

  2. Friction angle from the failure circle

    The envelope with intercept c' must be tangent to the effective-stress circle.

    radius = c'·cosφ' + center·sinφ'

    circle: center 250 kPa, radius 150 kPa → φ' = 32.16°

    = 32.16

Results

Effective friction angle φ' (degrees)

32.16

Effective major principal stress σ1'

400kPa

Effective minor principal stress σ3'

100kPa

Deviator stress σ1 − σ3

300kPa

Failure plane angle from horizontal (45 + φ'/2)

61.08

Where this answer was checked
source
radius = c'·cosφ' + center·sinφ' solved in closed form — σ1' = 400, σ3' = 100, c' = 20
verified by
hand-recomputed
derivation
radius 150, center 250, hyp = √(400+62500) = 250.80; φ' = asin(150/250.80) − atan(20/250) = 36.73° − 4.57° = 32.16°. Check: 20·cos32.16 + 250·sin32.16 = 16.93 + 133.1 = 150.0 ✓.

Example 3

US drained: σ1 = 60 psi, σ3 = 20 psi → φ' = 30° exactly

Given

major principal stress
60 psi
minor principal stress
20 psi

Assumptions

  • Mohr-Coulomb failure criterion in effective stresses; pore pressure acts equally on both principal directions.

Solution steps

  1. Effective principal stresses

    Effective stress is total stress minus pore pressure.

    σ1' = σ1 − u; σ3' = σ3 − u

    σ1' = 60 psi; σ3' = 20 psi

  2. Friction angle from the failure circle

    With zero cohesion the failure envelope passes through the origin and is tangent to the effective-stress circle.

    sinφ' = (σ1' − σ3')/(σ1' + σ3')

    circle: center 40 psi, radius 20 psi → φ' = 30°

    = 30

Results

Effective friction angle φ' (degrees)

30

Effective major principal stress σ1'

60psi

Effective minor principal stress σ3'

20psi

Deviator stress σ1 − σ3

40psi

Failure plane angle from horizontal (45 + φ'/2)

60

Where this answer was checked
source
sinφ' = 40/80 = 0.5 — clean 30° case
verified by
hand-recomputed
derivation
sinφ' = (60−20)/(60+20) = 0.5 → φ' = 30.00°; deviator = 40 psi.