# How much pressure do wheels exert on the ground?

I would like to know how much pressure 4 "rigid" wheels (as opposed to tyres) would exert on the ground; I've been on the case for hours now and the best I've come across to calculate this was the Hertz pressure.

However when applying it to an empty server cabinet as an example (say, 100kg, 4 50x20mm wheels made out of aluminium, on concrete), the ground would crack (92MPa when 30 are allowed). That can't be right.

What is the correct way?

Hetrzian calculations assume infinite width for the parts and in real life tires have a finite width. What that means is the if the contact is line contact (like a cylinder on a plane) as opposed to a point contact (like a football on a plane) the pressure distribution is going to be abruptly interrupted at the ends, compared to an infinitely long line contact as Hertz predicted. Always read the assumptions stated clearly in wikipedia here first.

Since you probably do not know or don't care for the crowning of the wheels you can assume a finite cylinder on a plane calculation. This is very crude because a true cylinder on a plane produces infinite edge stresses thus requiring what is called end relief for them to roll efficiently and function properly.

This calculation for a width $ell$ on a wheel of diameter $d_1$ goes as follows:

Assume the road diameter being infinite $d_2=\infty$ and apply a normal load of $F$.

1. The contact patch is rectangular with length is $\ell$, and width $$b = K_b \sqrt{F}$$ where the material/geometry constant $$K_b = \sqrt{ \frac{2}{\pi \ell} \frac{(1-\nu_1^2)/E_1+(1-\nu_2^2)/E_2}{1/d_1+1/d_2}}$$ where $E_1$, $\nu_1$ is the Young's modulus and Poisson's ratio of the wheel, and $E_2$, $\nu_2$ of the road.
2. The Peak contact pressure is $$P=\frac{2 F}{\pi \ell b}$$

NOTE: That contact pressure is not equal to stress, but is only the normal stress component at the surface. There are additional calculations needed to get to the peak equivalent stress which occurs in the subsurface.

1. At each depth $z$ below the contact calculate the ratio $\zeta = \frac{z}{b}$
2. Stress components are \begin{aligned} \sigma_x &= -2 \nu_2 P \left( \sqrt{1+\zeta^2}-\zeta\right) \\ \sigma_y &= - P \left( \frac{1+z\zeta^2}{ \sqrt{1+\zeta^2}}-2 \zeta\right) \\ \sigma_z &= -P \frac{1}{\sqrt{1+\zeta^2}} \\ \tau_{max} & = \frac{\sigma_z-\sigma_x}{2} & \zeta &<0.436\\ \tau_{max} & = \frac{\sigma_z-\sigma_y}{2} & \zeta &\gt 0.436\\ \end{aligned}

Typically peak shear stress is $\tau_{peak} \approx 0.3 P$ located at $z=0.786 b$

• Thanks for the answer. Although very well presented, this is exactly what I have done, so I was wondering if hertz contacts were representative of the reality - for casters in this example (without crowns). Also, you mentioned that Hertz assumed infinite contact - is that what they mean by "half space"? Then why do the formulas include the width? Side note, for the time being I'm focusing on the maximum pressure at the surface and not in function of the depth, though it's still interesting. Nov 21 '14 at 22:32
• The reason a length value is needed is because the pressure distribution is assumed to be constant along the contact, then taking a small slice $\Delta x$ and applying the linear pressure $F/\Delta x$ yielding the width $b$. So the calculation assumes and infinitely long cylinder with linear pressure which is then trimmed to $\ell$. Nov 22 '14 at 0:57
• The actual distribution provides for infinite pressure at the ends (it follows a ${\rm asinh}(1/x)$ when $x \rightarrow 0$. So to get real-real you need to consider local plasticity and non-linear materials. Yikes! Nov 22 '14 at 1:00
• The solution people employ is is discritizing the domain into little rectangles and using the Love/Businesq formula for deflection and pressure $$\delta = \frac{1-\nu^2}{\pi E} \int \frac{P}{|r|}\,{\rm d}A$$ Nov 22 '14 at 1:03