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Terramechanics and off-road mobility
Work per area to mobilise soil shear
Shear work per unit contact area, from integrating the monotonic Janosi–Hanamoto shear stress over accumulated displacement j at constant normal stress, with asymptotic strength τmax and deformation length K.
- Ref
- TER-47
- Rev
- v1
- Field
- Terramechanics and off-road mobility
- Result
- E/A [J/m²]
Integrating the Janosi–Hanamoto shear stress τ = τmax(1 − e^(−j/K)) over displacement from 0 to j gives the shear energy per unit area, E/A = τmax[j − K(1 − e^(−j/K))]. It is the area under the stress–displacement curve, in J/m², and it is the energy the wheel or track loses to shear in each square metre of contact, one part of the slip loss.
Shear work per unit contact area, from integrating the monotonic Janosi–Hanamoto shear stress over accumulated displacement j at constant normal stress, with asymptotic strength τmax and deformation length K.
Calculator
- Rigid-plastic upper bound, τmax j1,000 J/m²
- Share of the upper bound that is dissipated63.28 %
- Shear stress at the end of the displacement18,360 Pa
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.
- Shearing the soil through 50 mm takes 632.8 J per square metre, 63.3 % of the rigid-plastic upper bound.
- The shortfall is the energy not dissipated while the stress is still rising, and it shrinks as j grows past a few K, here 2.5 K.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (J/m²) | Local step | Change in result |
|---|---|---|---|---|---|---|
| τmax Asymptotic shear strength | 100 to 1,000,000 Pa | 3.164 | 31,640 | 31,640 | +10% | +10 % |
| j Accumulated shear displacement | 0 to 0.5 m | 0 | 9,600 | 9,600 | +10% | +14.7 % |
| K Deformation length | 0.002 to 0.2 m | 115.2 | 960 | 844.8 | +10% | −4.35 % |
Sweep of Accumulated shear displacement
| Accumulated shear displacement (m) | Shear work per unit area (J/m²) |
|---|---|
| 0 | 0 |
| 0.0625 | 867.6 |
| 0.125 | 2,101 |
| 0.1875 | 3,350 |
| 0.25 | 4,600 |
| 0.3125 | 5,850 |
| 0.375 | 7,100 |
| 0.4375 | 8,350 |
| 0.5 | 9,600 |
Sweep one variable
Variables and how each one contributes
Shear work per unit area
What the equation gives you. Everything else feeds it.
Asymptotic shear strength
The asymptotic shear strength, in Pa. Raises the work in direct proportion. Raising it by 10 % (to 22,000) moves the result by +10 %, from 632.8 to 696.1 J/m².
Accumulated shear displacement
The accumulated shear displacement, in m. Raises the work, almost linearly once j is well above K. Raising it by 10 % (to 0.055) moves the result by +14.7 %, from 632.8 to 725.6 J/m².
Deformation length
The deformation length, in m. A longer deformation length means the soil mobilises more slowly, which lowers the work for the same displacement. Raising it by 10 % (to 0.022) moves the result by −4.35 %, from 632.8 to 605.3 J/m².
Worked example with the values above
- Exponential term: 1 − e^(−j/K) = 1 − e^(−2.5) = 0.91792.
- Bracket: j − K(…) = 0.05 − 0.02 × 0.91792 = 0.031642 m.
- E/A = 20,000 × 0.031642 = 632.83 J/m².
Assumptions and limits
- A rigid, flat track patch or cone in homogeneous soil, with a linear pressure profile where stated.
- Constant soil strength and deformation length over the test, and the same rate and cone for comparisons.
Common mistakes
- Using the linear pressure formula beyond the one-sixth limit on the load offset.
- Mixing the left and right sign convention for yaw moment.
- Including rod friction in the cone force.
Statistics and uncertainty
Pressure cells and penetrometers carry 5 % to 10 % uncertainty, and the penetration or shear work depends on rate and soil state. Compare only tests run under the same conditions.
Where it is used
- Checking track pressure distribution, thrust sharing and steering demand.
- Comparing soil strength and shear energy between test beds.
Maths you need
- Linear pressure distribution: p = W/A (1 ± 6e/L)
- Work as an integral: W = ∫ F dz
- Moment balance: the sum of moments about the vehicle centre is zero