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Terramechanics and off-road mobility
Bekker moduli from two plate widths
Cohesive modulus kc from the pressure–sinkage coefficients B1 and B2 fitted at two plate widths b1 and b2 with a common exponent, using B = kc/b + kφ.
- Ref
- TER-36
- Rev
- v1
- Field
- Terramechanics and off-road mobility
- Result
- kc [Pa m^(1−n)]
Plates of two widths give pressure–sinkage coefficients B1 and B2 in p = Bzⁿ. With Bekker’s form B = kc/b + kφ, subtracting the two relations gives kc = (B1 − B2)/(1/b1 − 1/b2), and kφ = B1 − kc/b1. The cohesive modulus kc captures the effect of plate width, and the frictional modulus kφ the part independent of it.
Cohesive modulus kc from the pressure–sinkage coefficients B1 and B2 fitted at two plate widths b1 and b2 with a common exponent, using B = kc/b + kφ. The frictional modulus kφ = B1 − kc/b1 is in the derived rows. Keep the width convention (rectangular width, circular radius) and units with the data.
Calculator
- Frictional modulus, kφ = B1 − kc/b1100,000 Pa/m^(n)
- Coefficient predicted for a 0.3 m plate, kc/0.3 + kφ166,700 Pa/mⁿ
- Cohesive share of the first plate’s coefficient, kc/(b1 B1)66.67 %
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.
- The cohesive modulus is 20,000 and the frictional modulus 100,000, so the cohesive term makes up 66.7 % of the narrow plate’s coefficient.
- The two-plate method subtracts two similar numbers, so a 5 % error in B1 or B2 can change kc by far more than 5 %; use more widths when you can.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (Pa m^(1−n)) | Local step | Change in result |
|---|---|---|---|---|---|---|
| B1 Fitted coefficient, plate 1 | 1,000 to 1e7 Pa/mⁿ | −39,800 | 1,960,000 | 2,000,000 | +10% | +30 % |
| B2 Fitted coefficient, plate 2 | 1,000 to 1e7 Pa/mⁿ | −1,940,000 | 59,800 | 2,000,000 | +10% | −20 % |
| b1 Width of plate 1 | 0.02 to 0.19 m | 2,222 | 380,000 | 377,800 | +10% | +22.2 % |
| b2 Width of plate 2 | 0.2 to 1 m | 11,110 | 20,000 | 8,889 | +10% | −8.33 % |
Sweep of Fitted coefficient, plate 2
| Fitted coefficient, plate 2 (Pa/mⁿ) | Cohesive modulus (Pa m^(1−n)) |
|---|---|
| 1,000 | 59,800 |
| 3,162 | 59,370 |
| 10,000 | 58,000 |
| 31,620 | 53,680 |
| 100,000 | 40,000 |
| 316,200 | −3,246 |
| 1,000,000 | −140,000 |
| 3,162,000 | −572,500 |
| 1e7 | −1,940,000 |
Sweep one variable
Variables and how each one contributes
Cohesive modulus
What the equation gives you. Everything else feeds it.
Width of plate 1
The width of the narrower plate, in m. A narrower first plate widens the 1/b spacing and gives a better conditioned estimate. Raising it by 10 % (to 0.11) moves the result by +22.2 %, from 20,000 to 24,440 Pa m^(1−n).
Width of plate 2
The width of the wider plate, in m. A wider second plate does the same, so well-separated widths are better. Raising it by 10 % (to 0.22) moves the result by −8.33 %, from 20,000 to 18,330 Pa m^(1−n).
Fitted coefficient, plate 1
The fitted coefficient from the first plate, in Pa/mⁿ. Raises kc in direct proportion to the difference B1 − B2. Raising it by 10 % (to 330,000) moves the result by +30 %, from 20,000 to 26,000 Pa m^(1−n).
Fitted coefficient, plate 2
The fitted coefficient from the second plate, in Pa/mⁿ. Lowers kc in direct proportion to the same difference. Raising it by 10 % (to 220,000) moves the result by −20 %, from 20,000 to 16,000 Pa m^(1−n).
Worked example with the values above
- Width term: 1/b1 − 1/b2 = 10 − 5 = 5 1/m.
- Coefficient difference: B1 − B2 = 100,000.
- kc = 100,000 / 5 = 20,000.
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
- Mixing plate widths, or using width where the radius is needed.
- Using test points from different soil states or different loading branches.
- Using a shear point at or beyond the plateau, where the deformation length cannot be found.
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