Library / Terramechanics and off-road mobility / TER-42
Terramechanics and off-road mobility
Vertical wheel support from measured contact stresses
Vertical load supported by a rigid wheel of width b and radius r from the width-averaged normal and shear stresses σ and τ measured at increasing contact angle θ (rad), from exit to entry, by trapezoidal integration over five nodes.
- Ref
- TER-42
- Rev
- v1
- Field
- Terramechanics and off-road mobility
- Result
- W [N]
The vertical load on a rigid wheel is the vertical component of the soil stress integrated over the contact arc: W = b·r·∫(σ cos θ + τ sin θ) dθ. Normal stress σ acts along the radius, so its vertical component is σ cos θ, and the shear stress τ acts along the tangent, giving τ sin θ. The page uses five sampled nodes and the trapezoidal rule; repeat the last node to pad unused ones.
Vertical load supported by a rigid wheel of width b and radius r from the width-averaged normal and shear stresses σ and τ measured at increasing contact angle θ (rad), from exit to entry, by trapezoidal integration over five nodes. Unused nodes repeat the last angle so that they add nothing.
Calculator
- Integral of (σ cos θ + τ sin θ) dθ8,461 Pa rad
- Wheel load in kilonewtons0.5077 kN
- Angular span of the nodes, θ₅ − θ₁60 °
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 sampled stresses give a vertical load of 507.7 N, which is 51.75 kg of support.
- Only the nodes’ angles and stresses matter, so you can check a measured stress profile against a force-cell reading, and a large mismatch usually means the stress sensors, not the wheel, are wrong.
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 | 235.6 | 8,398 | 8,162 | +10% | +5.36 % |
| b Wheel width | 0.02 to 2 m | 50.77 | 5,077 | 5,026 | +10% | +10 % |
| σ1 Normal stress, node 1 | 0 to 300,000 Pa | 350.6 | 5,063 | 4,712 | +10% | +3.09 % |
| r Wheel radius | 0.05 to 1.5 m | 84.61 | 2,538 | 2,454 | +10% | +10 % |
| σ3 Normal stress, node 3 | 0 to 300,000 Pa | 429.1 | 2,785 | 2,356 | +10% | +1.55 % |
| τ2 Shear stress, node 2 | 0 to 100,000 Pa | 507.7 | 2,078 | 1,571 | +10% | 0 % |
| τ3 Shear stress, node 3 | 0 to 100,000 Pa | 507.7 | 1,868 | 1,360 | +10% | 0 % |
| θ1 Contact angle, node 1 | 0 to 1.6 rad | −59.88 | 507.7 | 567.6 | +10% | 0 % |
| θ5 Contact angle, node 5 | 0 to 1.6 rad | 36.45 | 586.2 | 549.8 | +10% | +5.61 % |
| θ2 Contact angle, node 2 | 0 to 1.6 rad | 387.9 | 507.7 | 119.8 | +10% | −0.146 % |
| θ3 Contact angle, node 3 | 0 to 1.6 rad | 471.2 | 511.4 | 40.2 | +10% | −0.621 % |
| θ4 Contact angle, node 4 | 0 to 1.6 rad | 507.7 | 507.7 | 1.137e-13 | +10% | 0 % |
| σ4 Normal stress, node 4 | 0 to 300,000 Pa | 507.7 | 507.7 | 0 | +10% | 0 % |
| σ5 Normal stress, node 5 | 0 to 300,000 Pa | 507.7 | 507.7 | 0 | +10% | 0 % |
| τ1 Shear stress, node 1 | 0 to 100,000 Pa | 507.7 | 507.7 | 0 | +10% | 0 % |
| τ4 Shear stress, node 4 | 0 to 100,000 Pa | 507.7 | 507.7 | 0 | +10% | 0 % |
| τ5 Shear stress, node 5 | 0 to 100,000 Pa | 507.7 | 507.7 | 0 | +10% | 0 % |
Sweep of Normal stress, node 3
| Normal stress, node 3 (Pa) | Vertical wheel load (N) |
|---|---|
| 0 | 429.1 |
| 37,500 | 723.7 |
| 75,000 | 1,018 |
| 112,500 | 1,313 |
| 150,000 | 1,607 |
| 187,500 | 1,902 |
| 225,000 | 2,196 |
| 262,500 | 2,491 |
| 300,000 | 2,785 |
Sweep one variable
Variables and how each one contributes
Vertical wheel load
What the equation gives you. Everything else feeds it.
Wheel width
The wheel width, in m. Raises the load in direct proportion. Raising it by 10 % (to 0.22) moves the result by +10 %, from 507.7 to 558.5 N.
Wheel radius
The wheel radius, in m. Raises the load in direct proportion at fixed stresses and angles. Raising it by 10 % (to 0.33) moves the result by +10 %, from 507.7 to 558.5 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 507.7 to 507.7 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 −0.146 %, from 507.7 to 506.9 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 −0.621 %, from 507.7 to 504.5 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 +0 %, from 507.7 to 507.7 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 +5.61 %, from 507.7 to 536.2 N.
Normal stress, node 1
The normal stress at node 1, in Pa. Raises the load through the first trapezoid. Raising it by 10 % (to 11,000) moves the result by +3.09 %, from 507.7 to 523.4 N.
Normal stress, node 2
The normal stress at node 2, in Pa. Raises the load through the first two trapezoids. Raising it by 10 % (to 11,000) moves the result by +5.36 %, from 507.7 to 534.9 N.
Normal stress, node 3
The normal stress at node 3, in Pa. Raises the load through the middle trapezoids. Raising it by 10 % (to 11,000) moves the result by +1.55 %, from 507.7 to 515.5 N.
Normal stress, node 4
The normal stress at node 4, in Pa. Raises the load through the last two trapezoids. Raising it by 10 % (to 11,000) moves the result by +0 %, from 507.7 to 507.7 N.
Normal stress, node 5
The normal stress at node 5, in Pa. Raises the load through the last trapezoid. Raising it by 10 % (to 11,000) moves the result by +0 %, from 507.7 to 507.7 N.
Shear stress, node 1
The shear stress at node 1, in Pa. Adds a small vertical component through sin θ. Raising it by 10 % (to 0) moves the result by +0 %, from 507.7 to 507.7 N.
Shear stress, node 2
The shear stress at node 2, in Pa. Adds a vertical component through sin θ. Raising it by 10 % (to 0) moves the result by +0 %, from 507.7 to 507.7 N.
Shear stress, node 3
The shear stress at node 3, in Pa. Adds a vertical component through sin θ. Raising it by 10 % (to 0) moves the result by +0 %, from 507.7 to 507.7 N.
Shear stress, node 4
The shear stress at node 4, in Pa. Adds a vertical component through sin θ. Raising it by 10 % (to 0) moves the result by +0 %, from 507.7 to 507.7 N.
Shear stress, node 5
The shear stress at node 5, in Pa. Adds a vertical component through sin θ. Raising it by 10 % (to 0) moves the result by +0 %, from 507.7 to 507.7 N.
Worked example with the values above
- Integrand σ cos θ + τ sin θ at the nodes (θ rad, Pa): (0, 10,000) (0.5236, 8,660) (1.047, 5,000) (1.047, 5,000) (1.047, 5,000).
- Trapezoidal integral: ∫ = 8,461.5 Pa rad.
- W = b r ∫ = 0.2 × 0.3 × 8,461.5 = 507.69 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))