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Terramechanics and off-road mobility
Janosi–Hanamoto shear stress
Shear stress mobilised in soil as the shear displacement j grows: it rises towards the maximum strength and reaches 63 % of it when j equals the shear deformation parameter Ks.
- Ref
- TER-13
- Rev
- v1
- Field
- Terramechanics and off-road mobility
- Result
- τ [kPa]
τmax is the maximum shear strength of the soil (from the Mohr–Coulomb relation), j the accumulated shear displacement and Ks the shear deformation parameter fitted to shear-box or shear-ring tests. The stress rises exponentially towards τmax; at j = Ks it has reached 63.2 % of the maximum. It is a monotonic law: it does not model post-peak softening. Displacements and Ks are both in mm here and only their ratio matters.
Shear stress mobilised in soil as the shear displacement j grows: it rises towards the maximum strength and reaches 63 % of it when j equals the shear deformation parameter Ks.
Calculator
- Fraction of the strength mobilised, 1 − e−j/Ks63.21 %
- Displacement for 95 % of the strength, 3Ks60 mm
- Strength still unmobilised, τmax − τ11.37 kPa
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.
- When the displacement equals Ks (20 mm here) the soil has mobilised 63.2 % of its 30.9 kPa strength: 19.53 kPa. At three times Ks (60 mm) it has reached 95 %, 29.36 kPa.
- This is why a wheel or track must slip to develop thrust. The soil at the rear of the contact has sheared much further than the soil at the front, and the average thrust is below the Mohr–Coulomb maximum unless the slip is large.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (kPa) | Local step | Change in result |
|---|---|---|---|---|---|---|
| τmax Maximum shear strength | 5 to 100 kPa | 3.161 | 63.21 | 60.05 | +10% | +10 % |
| j Shear displacement | 0 to 150 mm | 0 | 30.88 | 30.88 | +10% | +5.54 % |
| Ks Shear deformation parameter | 5 to 60 mm | 8.759 | 30.33 | 21.57 | +10% | −5.54 % |
Sweep of Shear displacement
| Shear displacement (mm) | Mobilised shear stress (kPa) |
|---|---|
| 0 | 0 |
| 18.75 | 18.8 |
| 37.5 | 26.16 |
| 56.25 | 29.04 |
| 75 | 30.17 |
| 93.75 | 30.62 |
| 112.5 | 30.79 |
| 131.3 | 30.86 |
| 150 | 30.88 |
Sweep one variable
Variables and how each one contributes
Mobilised shear stress
What the equation gives you. Everything else feeds it.
Maximum shear strength
The maximum shear strength of the soil, in kPa (from the Mohr–Coulomb equation). Directly proportional: it scales the entire curve. Raising it by 10 % (to 33.99) moves the result by +10 %, from 19.53 to 21.49 kPa.
Shear displacement
The accumulated shear displacement of the soil at that point, in mm: the slip distance the wheel or track has travelled over it. Raises the stress quickly at first, then more slowly as it approaches the maximum. Raising it by 10 % (to 22) moves the result by +5.54 %, from 19.53 to 20.61 kPa.
Shear deformation parameter
The shear deformation parameter, in mm: typically 10 to 40 mm for sand and clay and larger for loose or snowy soils. A smaller Ks means the soil mobilises its strength sooner. Lowers the stress at a given displacement, because the soil takes more slip to develop its strength. Raising it by 10 % (to 22) moves the result by −5.54 %, from 19.53 to 18.45 kPa.
Worked example with the values above
- Ratio of displacements: j / Ks = 20 / 20 = 1.
- Mobilised fraction: 1 − e−1 = 1 − 0.3679 = 0.6321.
- τ = 30.9 × 0.6321 = 19.53 kPa.
Assumptions and limits
- A monotonic exponential law, with no peak and softening, and a fixed maximum strength.
- Shear displacement j and Ks in the same units, and the normal stress constant during shearing.
Common mistakes
- Mixing units for j and Ks, such as m for one and mm for the other.
- Using the law for soils that show a strong peak and softening, such as dense sand, where it overestimates the residual stress.
Statistics and uncertainty
Ks is fitted from shear tests and often varies by ±30 to 50 % between tests on the same soil. Since the curve is flat near the top, the stress at large j is only weakly sensitive to Ks (an error of 40 % in Ks at j = 3Ks changes τ by about 5 to 7 %), but at small j the stress is very sensitive to it (at j = 0.5Ks by about 25 to 45 %).
Where it is used
- Predicting the thrust developed by a wheel or track at a given slip, rather than only the ideal maximum.
- Modelling soil shear in wheel–soil and track–soil interaction.
Maths you need
- Exponential function: 1 − e^(−x)
- Shear strength: maximum stress from Mohr–Coulomb