Library / Terramechanics and off-road mobility / TER-44
Terramechanics and off-road mobility
Wheel torque required to overcome soil shear
Torque needed to balance the soil shear moment on a rigid wheel at constant angular speed, from the sampled shear stress over five nodes.
- Ref
- TER-44
- Rev
- v1
- Field
- Terramechanics and off-road mobility
- Result
- T [N m]
The soil shear stress on the rim acts at the radius r, so the torque the wheel needs to balance it is T = b·r²·∫τ dθ. The normal stress on a circular rim acts along the radius, so it has no moment about the wheel centre. The page integrates the sampled shear stress over five nodes; bearing and other losses are not included.
Torque needed to balance the soil shear moment on a rigid wheel at constant angular speed, from the sampled shear stress over five nodes. Normal stress on a rigid circular rim passes through the wheel centre and has no moment there. Bearing losses are excluded.
Calculator
- Integral of τ dθ2,000 Pa rad
- Equivalent tangential force, T/r120 N
- Mean shear stress over the arc, ∫τ dθ/(θ₅ − θ₁)5,000 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.
- The soil shear on the rim needs 36 N m of torque, equivalent to a tangential force of 120 N at the rim.
- Because the lever arm is r and the area scales with r too, a larger wheel needs more torque for the same stress pattern, with the square of the radius.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (N m) | Local step | Change in result |
|---|---|---|---|---|---|---|
| r Wheel radius | 0.05 to 1.5 m | 1 | 900 | 899 | +10% | +21 % |
| τ2 Shear stress, node 2 | 0 to 100,000 Pa | 18 | 378 | 360 | +10% | +5 % |
| b Wheel width | 0.02 to 2 m | 3.6 | 360 | 356.4 | +10% | +10 % |
| θ5 Contact angle, node 5 | 0 to 1.6 rad | −36 | 252 | 288 | +10% | +20 % |
| τ1 Shear stress, node 1 | 0 to 100,000 Pa | 36 | 216 | 180 | +10% | 0 % |
| τ3 Shear stress, node 3 | 0 to 100,000 Pa | 18 | 198 | 180 | +10% | +5 % |
| θ2 Contact angle, node 2 | 0 to 1.6 rad | −90 | 54 | 144 | +10% | −5 % |
| θ1 Contact angle, node 1 | 0 to 1.6 rad | −36 | 36 | 72 | +10% | 0 % |
| θ3 Contact angle, node 3 | 0 to 1.6 rad | −18 | 54 | 72 | +10% | −5 % |
| θ4 Contact angle, node 4 | 0 to 1.6 rad | 36 | 36 | 2.842e-14 | +10% | 0 % |
| τ4 Shear stress, node 4 | 0 to 100,000 Pa | 36 | 36 | 0 | +10% | 0 % |
| τ5 Shear stress, node 5 | 0 to 100,000 Pa | 36 | 36 | 0 | +10% | 0 % |
Sweep of Shear stress, node 3
| Shear stress, node 3 (Pa) | Wheel torque (N m) |
|---|---|
| 0 | 18 |
| 12,500 | 40.5 |
| 25,000 | 63 |
| 37,500 | 85.5 |
| 50,000 | 108 |
| 62,500 | 130.5 |
| 75,000 | 153 |
| 87,500 | 175.5 |
| 100,000 | 198 |
Sweep one variable
Variables and how each one contributes
Wheel torque
What the equation gives you. Everything else feeds it.
Wheel width
The wheel width, in m. Raises the torque in direct proportion. Raising it by 10 % (to 0.22) moves the result by +10 %, from 36 to 39.6 N m.
Wheel radius
The wheel radius, in m. Raises the torque with its square, because both the area and the lever arm grow. Raising it by 10 % (to 0.33) moves the result by +21 %, from 36 to 43.56 N m.
Contact angle, node 1
The contact angle of node 1, in rad. Moves the first end of the integration range. Raising it by 10 % (to 0) moves the result by +0 %, from 36 to 36 N m.
Contact angle, node 2
The contact angle of node 2, in rad. Changes the width of the first two trapezoids. Raising it by 10 % (to 0.22) moves the result by −5 %, from 36 to 34.2 N m.
Contact angle, node 3
The contact angle of node 3, in rad. Changes the width of the middle two trapezoids. Raising it by 10 % (to 0.44) moves the result by −5 %, from 36 to 34.2 N m.
Contact angle, node 4
The contact angle of node 4, in rad. Changes the width of the last two trapezoids. Raising it by 10 % (to 0.44) moves the result by +0 %, from 36 to 36 N m.
Contact angle, node 5
The contact angle of node 5, in rad. Moves the last end of the integration range. Raising it by 10 % (to 0.44) moves the result by +20 %, from 36 to 43.2 N m.
Shear stress, node 1
The shear stress at node 1, in Pa. Raises the torque through the first trapezoid. Raising it by 10 % (to 0) moves the result by +0 %, from 36 to 36 N m.
Shear stress, node 2
The shear stress at node 2, in Pa. Raises the torque through the first two trapezoids. Raising it by 10 % (to 5,500) moves the result by +5 %, from 36 to 37.8 N m.
Shear stress, node 3
The shear stress at node 3, in Pa. Raises the torque through the middle two trapezoids. Raising it by 10 % (to 11,000) moves the result by +5 %, from 36 to 37.8 N m.
Shear stress, node 4
The shear stress at node 4, in Pa. Raises the torque through the last two trapezoids. Raising it by 10 % (to 11,000) moves the result by +0 %, from 36 to 36 N m.
Shear stress, node 5
The shear stress at node 5, in Pa. Raises the torque through the last trapezoid. Raising it by 10 % (to 11,000) moves the result by +0 %, from 36 to 36 N m.
Worked example with the values above
- Shear stress at the nodes (θ rad, Pa): (0, 0) (0.2, 5,000) (0.4, 10,000) (0.4, 10,000) (0.4, 10,000).
- Trapezoidal integral: ∫τ dθ = 2,000 Pa rad.
- T = b r² ∫ = 0.2 × 0.09 × 2,000 = 36 N m.
Assumptions and limits
- A rigid wheel of constant radius and width, with stresses uniform across the width and constant driving slip.
- Contact angle measured from the downward vertical, with angles increasing from exit to entry, and an arc without soil bulging.
Common mistakes
- Entering angles in degrees instead of radians.
- Ordering the nodes from entry to exit, which reverses the sign of the integral.
- Mixing the normal-stress and shear-stress sign conventions.
Statistics and uncertainty
Stress sensors on a wheel rim read to perhaps 5 % to 10 %, and the trapezoidal sum over five nodes is coarse, so a smooth stress peak between nodes can be missed. Add nodes where stress changes quickly.
Where it is used
- Checking measured wheel loads and thrusts against force-cell readings.
- Building wheel–soil models from measured stress data.
Maths you need
- Contact angle: angle at the wheel centre from the vertical to a contact point
- Trapezoidal rule: area under sampled data
- Janosi–Hanamoto shear: τ = τmax(1 − e^(−j/K))