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Terramechanics and off-road mobility

Shear deformation length from one mobilisation point

Deformation length K of the monotonic shear curve τ = τmax(1 − exp(−j/K)) from one point (j, τ) at the same normal stress and an independently determined strength τmax.

Ref
TER-38
Rev
v1
Field
Terramechanics and off-road mobility
Result
K [m]

The Janosi–Hanamoto curve τ = τmax(1 − e^(−j/K)) rises from zero at zero displacement towards τmax, and K is the displacement scale of the rise. Inverting it for one measured point gives K = −j/ln(1 − τ/τmax). The inversion is ill-conditioned: near the plateau the logarithm blows up and near zero stress the ratio of two small numbers is noisy, so use points between about 30 % and 80 % of τmax.

Deformation length K of the monotonic shear curve τ = τmax(1 − exp(−j/K)) from one point (j, τ) at the same normal stress and an independently determined strength τmax. The result is ill-conditioned near zero stress or at the plateau.

Calculator

Solve for
m
m
0.0010.2
Pa
109,900
Pa
10,0001,000,000
  • Mobilised fraction, τ/τmax50 %
  • Displacement for 95 % of τmax, −K ln(0.05)0.08644 m
  • Initial shear stiffness, τmax/K346,600 Pa/m

What drives the result

Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.

Deformation length (m) current value 0.02885 m
τShear stress at that displacementKey driver
0.0043420
τmaxAsymptotic shear strength
3.99
jShear displacement
0.001440.289
  • The deformation length is 28.85 mm; the soil reaches 63 % of its strength at j = K and 95 % at j = 86.4 mm.
  • The measured point is at 50 % of τmax, which is inside the well-conditioned range.
Chart data as tables

Ranked variables

VariableTypical rangeLowest resultHighest resultSwing (m)Local stepChange in result
τ Shear stress at that displacement10 to 9,900 Pa0.00434319.9919.99+10%−13.2 %
τmax Asymptotic shear strength10,000 to 1,000,000 Pa0.028853.993.961+10%+14.4 %
j Shear displacement0.001 to 0.2 m0.0014430.28850.2871+10%+10 %

Sweep of Shear stress at that displacement

Shear stress at that displacement (Pa)Deformation length (m)
1019.99
1,2460.1503
2,4830.07009
3,7190.04301
4,9550.02923
6,1910.02072
7,4280.01473
8,6640.009937
9,9000.004343

Sweep one variable

Deformation length (m)Hover, touch, or use the arrow keys on the chart
05101520 2000400060008000 Shear stress at that displacement (Pa) now 0.02885

Variables and how each one contributes

K

Deformation length

Result · m

What the equation gives you. Everything else feeds it.

j

Shear displacement

m · typical 0.001 to 0.2

The shear displacement of the measured point, in m. Raises K in direct proportion for a fixed mobilisation. Raising it by 10 % (to 0.022) moves the result by +10 %, from 0.02885 to 0.03174 m.

τ

Shear stress at that displacement

Pa · typical 10 to 9,900

The shear stress at that displacement, in Pa. A higher stress means the soil mobilises faster, so K falls. Raising it by 10 % (to 5,500) moves the result by −13.2 %, from 0.02885 to 0.02505 m.

τmax

Asymptotic shear strength

Pa · typical 10,000 to 1,000,000

The asymptotic shear strength, in Pa, determined independently. A larger strength means the measured stress is a smaller fraction of it, so K rises. Raising it by 10 % (to 11,000) moves the result by +14.4 %, from 0.02885 to 0.033 m.

Worked example with the values above

  1. Fraction: τ/τmax = 5,000 / 10,000 = 0.5.
  2. Logarithm: ln(1 − 0.5) = −0.69315.
  3. K = −0.02 / −0.69315 = 0.028854 m.

Assumptions and limits

  • The same soil state, plate or ring geometry and loading rate for each point.
  • A power-law pressure–sinkage law, a monotonic exponential shear curve and a linear Mohr–Coulomb envelope in the tested range.

Common mistakes

  • Using a point near the plateau, where the logarithm blows up.
  • Using the displacement from the start of the test rather than from the start of shear.
  • Using τmax from a different normal stress.

Statistics and uncertainty

Soil strength and stiffness parameters from two-point methods carry the full measurement scatter of two points, which is typically 10 % to 30 %, and give no estimate of the fit uncertainty. Use a least-squares fit when more tests are available.

Where it is used

  • Fitting Bekker and Janosi shear parameters from bevameter and ring-shear tests.
  • Getting first-estimate strength parameters from a pair of simple field tests.

Maths you need

  • Bekker relation: p = (kc/b + kφ) zⁿ
  • Janosi–Hanamoto curve: τ = τmax(1 − e^(−j/K))
  • Mohr–Coulomb strength: τ = c + σ tan φ

Related formulas

Practise this formula