Library / Terramechanics and off-road mobility / TER-45
Terramechanics and off-road mobility
End pressures beneath an eccentrically loaded flat track
Rear-end contact pressure under a rectangular track patch of width b and length L carrying load W at an offset e forward of mid-length, assuming a linear pressure profile.
- Ref
- TER-45
- Rev
- v1
- Field
- Terramechanics and off-road mobility
- Result
- prear [Pa]
A linear pressure distribution under a rectangular patch of width b and length L, with load W at offset e from mid-length, has end pressures p = (W/(bL))(1 ± 6e/L). It is exact for a rigid rectangular plate on a Winkler-type or elastic bed, and the rear value becomes zero when e = L/6, the edge of the middle third. Beyond it contact is partial and the formula gives a meaningless negative pressure.
Rear-end contact pressure under a rectangular track patch of width b and length L carrying load W at an offset e forward of mid-length, assuming a linear pressure profile. The front value is in the derived rows. Full contact needs |e| ≤ L/6; beyond that a reduced-contact model is needed.
Calculator
- Front-end pressure, (W/(bL))(1 + 6e/L)30,000 Pa
- Mean pressure, W/(bL)20,000 Pa
- Largest offset before the rear edge lifts, L/60.5 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.
- The rear end carries 10,000 Pa against 30,000 Pa at the front, around a mean of 20,000 Pa.
- Contact stays complete up to an offset of 0.5 m; this load is at 50 % of that limit, so the formula is valid.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (Pa) | Local step | Change in result |
|---|---|---|---|---|---|---|
| W Normal load | 1,000 to 1,000,000 N | 416.7 | 416,700 | 416,200 | +10% | +10 % |
| e Load offset forward of mid-length | −0.5 to 0.5 m | 0 | 40,000 | 40,000 | +0.02 m | −8 % |
| b Track width | 0.1 to 2 m | 2,000 | 40,000 | 38,000 | +10% | −9.09 % |
| L Contact length | 3 to 10 m | 5,100 | 10,000 | 4,900 | +10% | −0.826 % |
Sweep of Load offset forward of mid-length
| Load offset forward of mid-length (m) | Rear-end pressure (Pa) |
|---|---|
| −0.5 | 40,000 |
| −0.375 | 35,000 |
| −0.25 | 30,000 |
| −0.125 | 25,000 |
| 0 | 20,000 |
| 0.125 | 15,000 |
| 0.25 | 10,000 |
| 0.375 | 5,000 |
| 0.5 | 0 |
Sweep one variable
Variables and how each one contributes
Rear-end pressure
What the equation gives you. Everything else feeds it.
Normal load
The normal load on the track patch, in N. Raises both end pressures in direct proportion. Raising it by 10 % (to 26,400) moves the result by +10 %, from 10,000 to 11,000 Pa.
Track width
The track width, in m. A wider track lowers the pressures in inverse proportion. Raising it by 10 % (to 0.44) moves the result by −9.09 %, from 10,000 to 9,091 Pa.
Contact length
The contact length, in m. A longer patch lowers the mean pressure and the offset effect, since 6e/L falls. Raising it by 10 % (to 3.3) moves the result by −0.826 %, from 10,000 to 9,917 Pa.
Load offset forward of mid-length
The load offset forward of mid-length, in m. A forward offset raises the front pressure and lowers the rear pressure; at e = L/6 the rear pressure reaches zero. Raising it by 10 % (to 0.275) moves the result by −10 %, from 10,000 to 9,000 Pa.
Worked example with the values above
- Mean pressure: W/(bL) = 24,000 / (0.4 × 3) = 20,000 Pa.
- Offset factor: 1 − 6e/L = 1 − 6 × 0.25 / 3 = 0.5.
- p_rear = 20,000 × 0.5 = 10,000 Pa.
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 formula when |e| exceeds L/6, where the rear edge has already lifted.
- Measuring e from the wrong end of the patch.
- Using the width of one track for a load carried by two.
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