Library / Terramechanics and off-road mobility / TER-43
Terramechanics and off-road mobility
Net horizontal wheel force from contact stresses
Net horizontal force on a rigid wheel from the same sampled stress profile: the shear term gives thrust and the normal-stress projection opposes forward motion over the front arc.
- Ref
- TER-43
- Rev
- v1
- Field
- Terramechanics and off-road mobility
- Result
- Fx [N]
The net horizontal force on a rigid wheel is the horizontal component of the stresses integrated over the contact arc: Fx = b·r·∫(τ cos θ − σ sin θ) dθ. The shear stress pushes the wheel forward, and the normal stress pushes it back over the front arc, which is the soil compaction resistance. The sign is positive forwards, so a negative result means the wheel is being retarded.
Net horizontal force on a rigid wheel from the same sampled stress profile: the shear term gives thrust and the normal-stress projection opposes forward motion over the front arc. Subtract other vehicle losses separately. Same nodes, angles and units as the vertical-load equation.
Calculator
- Integral of (τ cos θ − σ sin θ) dθ−654.5 Pa rad
- Gross thrust, b r ∫τ cos θ dθ253.8 N
- Compaction resistance, b r ∫σ sin θ dθ293.1 N
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 net horizontal soil force is −39.27 N, so the wheel is retarded by the soil, which is a net resistance.
- Net thrust is a small difference between large shear and compaction terms, so it is sensitive to stress errors: a few per cent on either term can flip its sign.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (N) | Local step | Change in result |
|---|---|---|---|---|---|---|
| σ2 Normal stress, node 2 | 0 to 300,000 Pa | −4,595 | 117.8 | 4,712 | +10% | +40 % |
| σ3 Normal stress, node 3 | 0 to 300,000 Pa | −3,984 | 96.77 | 4,081 | +10% | +34.6 % |
| τ2 Shear stress, node 2 | 0 to 100,000 Pa | −175.3 | 2,545 | 2,721 | +10% | −34.6 % |
| τ1 Shear stress, node 1 | 0 to 100,000 Pa | −117.8 | 1,453 | 1,571 | +10% | −20 % |
| τ3 Shear stress, node 3 | 0 to 100,000 Pa | −78.54 | 706.9 | 785.4 | +10% | −10 % |
| b Wheel width | 0.02 to 2 m | −392.7 | −3.927 | 388.8 | +10% | +10 % |
| θ5 Contact angle, node 5 | 0 to 1.6 rad | −309.6 | 68.66 | 378.3 | +10% | +106 % |
| θ1 Contact angle, node 1 | 0 to 1.6 rad | −108.4 | 241.8 | 350.2 | +10% | 0 % |
| r Wheel radius | 0.05 to 1.5 m | −196.3 | −6.545 | 189.8 | +10% | +10 % |
| θ2 Contact angle, node 2 | 0 to 1.6 rad | −41.63 | 23.55 | 65.18 | +10% | +2.01 % |
| θ3 Contact angle, node 3 | 0 to 1.6 rad | −41.18 | −10.76 | 30.43 | +10% | −6.26 % |
| θ4 Contact angle, node 4 | 0 to 1.6 rad | −39.27 | −39.27 | 2.842e-14 | +10% | −2.22e-14 % |
| σ1 Normal stress, node 1 | 0 to 300,000 Pa | −39.27 | −39.27 | 0 | +10% | 0 % |
| σ4 Normal stress, node 4 | 0 to 300,000 Pa | −39.27 | −39.27 | 0 | +10% | 0 % |
| σ5 Normal stress, node 5 | 0 to 300,000 Pa | −39.27 | −39.27 | 0 | +10% | 0 % |
| τ4 Shear stress, node 4 | 0 to 100,000 Pa | −39.27 | −39.27 | 0 | +10% | 0 % |
| τ5 Shear stress, node 5 | 0 to 100,000 Pa | −39.27 | −39.27 | 0 | +10% | 0 % |
Sweep of Shear stress, node 3
| Shear stress, node 3 (Pa) | Net horizontal force (N) |
|---|---|
| 0 | −78.54 |
| 12,500 | 19.63 |
| 25,000 | 117.8 |
| 37,500 | 216 |
| 50,000 | 314.2 |
| 62,500 | 412.3 |
| 75,000 | 510.5 |
| 87,500 | 608.7 |
| 100,000 | 706.9 |
Sweep one variable
Variables and how each one contributes
Net horizontal force
What the equation gives you. Everything else feeds it.
Wheel width
The wheel width, in m. Raises the force in direct proportion. Raising it by 10 % (to 0.22) moves the result by +10 %, from −39.27 to −43.2 N.
Wheel radius
The wheel radius, in m. Raises the force in direct proportion at fixed stresses and angles. Raising it by 10 % (to 0.33) moves the result by +10 %, from −39.27 to −43.2 N.
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 −39.27 to −39.27 N.
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 0.576) moves the result by +2.01 %, from −39.27 to −40.06 N.
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 1.152) moves the result by −6.26 %, from −39.27 to −36.81 N.
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 1.152) moves the result by −2.22×10−14 %, from −39.27 to −39.27 N.
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 1.152) moves the result by +106 %, from −39.27 to −80.93 N.
Normal stress, node 1
The normal stress at node 1, in Pa. Adds a backward push through sin θ. Raising it by 10 % (to 11,000) moves the result by +0 %, from −39.27 to −39.27 N.
Normal stress, node 2
The normal stress at node 2, in Pa. Adds a backward push through sin θ. Raising it by 10 % (to 11,000) moves the result by +40 %, from −39.27 to −54.98 N.
Normal stress, node 3
The normal stress at node 3, in Pa. Adds a backward push through sin θ. Raising it by 10 % (to 11,000) moves the result by +34.6 %, from −39.27 to −52.87 N.
Normal stress, node 4
The normal stress at node 4, in Pa. Adds a backward push through sin θ. Raising it by 10 % (to 11,000) moves the result by +0 %, from −39.27 to −39.27 N.
Normal stress, node 5
The normal stress at node 5, in Pa. Adds a backward push through sin θ. Raising it by 10 % (to 11,000) moves the result by +0 %, from −39.27 to −39.27 N.
Shear stress, node 1
The shear stress at node 1, in Pa. Adds forward thrust through cos θ. Raising it by 10 % (to 5,500) moves the result by −20 %, from −39.27 to −31.42 N.
Shear stress, node 2
The shear stress at node 2, in Pa. Adds forward thrust through cos θ. Raising it by 10 % (to 5,500) moves the result by −34.6 %, from −39.27 to −25.67 N.
Shear stress, node 3
The shear stress at node 3, in Pa. Adds forward thrust through cos θ. Raising it by 10 % (to 5,500) moves the result by −10 %, from −39.27 to −35.34 N.
Shear stress, node 4
The shear stress at node 4, in Pa. Adds forward thrust through cos θ. Raising it by 10 % (to 5,500) moves the result by +0 %, from −39.27 to −39.27 N.
Shear stress, node 5
The shear stress at node 5, in Pa. Adds forward thrust through cos θ. Raising it by 10 % (to 5,500) moves the result by +0 %, from −39.27 to −39.27 N.
Worked example with the values above
- Integrand τ cos θ − σ sin θ at the nodes (θ rad, Pa): (0, 5,000) (0.5236, −669.9) (1.047, −6,160) (1.047, −6,160) (1.047, −6,160).
- Trapezoidal integral: ∫ = −654.5 Pa rad.
- Fx = b r ∫ = 0.2 × 0.3 × −654.5 = −39.27 N.
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))