Triaxial Test Mohr Circle Calculator
Builds the Mohr circle of stress from a single triaxial test's principal stresses σ1 (axial) and σ3 (cell/confining), returning the deviator stress, circle centre and radius (maximum shear), and — if a friction angle is supplied — the normal and shear stress on the theoretical failure plane. Used by geotechnical engineers interpreting triaxial (UU/CU/CD) laboratory results.
Enter Values
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How to use this calculator
- Enter the major principal stress σ1 (axial stress at the stage of interest, kPa).
- Enter the minor principal stress σ3 (the cell / confining pressure, kPa).
- Optionally enter the effective friction angle φ (°) to get the failure-plane stresses.
How it works
In the geotechnical convention (compression positive) a triaxial stress state plots as a circle on the normal-stress axis. Its centre is s = (σ1 + σ3)/2 and its radius is t = (σ1 − σ3)/2, which equals the maximum shear stress τ_max. The deviator stress σ1 − σ3 is the circle's diameter.
If the friction angle φ is known, the theoretical failure plane makes an angle θ = 45° + φ/2 with the major-principal plane. Rotating twice that angle around the circle gives the stresses on that plane: σn = s + t·cos(2θ) and τ = t·sin(2θ).
Worked example
σ1 = 300 kPa, σ3 = 100 kPa, φ = 30°. Deviator = 200 kPa; centre s = 200 kPa; radius t = 100 kPa = τ_max. With φ = 30°, θ = 60°, so σn = 200 + 100·cos120° = 150 kPa and τ = 100·sin120° = 86.6 kPa on the failure plane.
Common mistakes
- Swapping σ1 and σ3 — σ1 is always the larger (axial) stress; entering σ1 < σ3 is rejected.
- Confusing this with the structural Mohr's-circle tool, which transforms a general σx/σy/τxy state rather than principal stresses.
- Reading the radius as the deviator stress — the radius is half the deviator (t = (σ1 − σ3)/2).
Frequently asked questions
What is the deviator stress in a triaxial test?
The deviator stress is σ1 − σ3, the axial stress applied in excess of the confining (cell) pressure. It equals the diameter of the Mohr circle and drives shear failure.
Why is the failure plane at 45° + φ/2?
The Mohr–Coulomb envelope is tangent to the circle at the point rotated 2θ from the major-principal plane, which corresponds to a physical plane at θ = 45° + φ/2 — steeper than 45° because friction rotates the critical plane.
What does the radius represent?
The radius t = (σ1 − σ3)/2 is the maximum shear stress τ_max, acting on planes at 45° to the principal directions.
Should I use total or effective stresses?
Use whichever matches your envelope: total stresses for a total-stress (cu) interpretation, effective stresses (σ' = σ − u) for a drained/effective-stress interpretation. Keep σ1 and σ3 consistent.
How is this different from the structural Mohr's circle tool?
This tool takes principal stresses σ1/σ3 directly (the triaxial framing). The structural tool starts from a general in-plane state σx, σy, τxy.
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