Posted in Academic Issues, Geotechnical Engineering

# The “Quick and Dirty” Way to Derive Boussinesq’s Point Load Stress Equations

In another post we show the formal, nice way to derive the equations for stresses under point loads. Here we’re going to show a “quick and dirty” way to derive them (or at least some of them.) It’s based on Tsytovich (1976) but I have made some changes to tighten up the theory behind them and make them a little more comprehensible. They don’t derive all of the equations, but the method gives a better physical understanding of what’s going on when we apply such a load to the ground surface.

It can be shown that this state exists due to static equilibrium. The point load P forms a sphere around itself; the principal stresses $\sigma_R$ radiate from the load, forming a sphere around the point of radius R. The magnitude of the stress is given by the equation

$\sigma_R = A \frac{\cos(\beta)}{R^2}$

By same static equilibrium, the vertical force of the stresses and P are thus

$P-\int_{0}^{2\,\pi}\!\sigma_{{R}}\cos(\beta){dF}=0$

The infinitesimal surface area of the stress can be defined as

$dF = 2\pi(R \sin\beta)(R d\beta)$

Substituting this into the integral yields

$P-\int_{0}^{1/2\,\pi}\!2\,\sigma_{{R}}\cos(\beta)\pi\,{R}^{2}\sin(\beta){d\beta}=0$

Making appropriate substitutions, the integral evaluates to

$P-2/3\,A\pi =0$

and thus

$A=3/2\,{\frac {P}{\pi }}$

Substituting,

$\sigma_R = 3/2\,{\frac {P\cos(\beta)}{\pi \,{R}^{2}}}$

We want the vertical and shear stresses at this point. What we need is a conversion from the polar to cylindrical coordinates, which are given by the equations (Timoshenko and Goodier (1951))

$\sigma_{{z}}=\sigma_{{R}}\left(\cos(\beta)\right)^{2}+\sigma_{{T}}\left(\sin(\beta)\right)^{2}$
$\tau_{{\it rz}}=\left(\sigma_{{R}}+\sigma_{{T}}\right)\sin(\beta)\cos(\beta)$

These are coordinate transformations for plane stress equations and are discussed in detail in Boresi et.al. (1993) in terms of the direction cosines of the stress vectors. It may seem odd to see sine terms for these but $\cos\beta = \sin(\beta-\frac{3\pi}{2})$. If we also assume that $\sigma_T = 0$, again substituting we have

$\sigma_{{z}}=3/2\,{\frac {P\left (\cos(\beta)\right )^{3}}{\pi \,{R}^{2}}}$
$\tau_{{{\it rz}}}=3/2\,{\frac {P\left (\cos(\beta)\right )^{2}\sin(\beta)}{\pi \,{R}^{2}}}$

Since

$\beta = \arccos({\frac {z}{\sqrt {{z}^{2}+{r}^{2}}}})$

and

$R = \sqrt {{z}^{2}+{r}^{2}}$

substituting both of these yields our desired result

$\sigma_{{z}}=3/2\,{\frac {P{z}^{3}}{\pi \,\left ({z}^{2}+{r}^{2}\right )^{5/2}}}$
$\tau_{{{\it rz}}}=3/2\,{\frac {P{z}^{2}r}{\pi \,\left ({z}^{2}+{r}^{2}\right )^{5/2}}}$

Many textbooks (such as Verruijt) state these equations in terms of R, but in general problems such as this are defined most simply in terms of the depth of the point of interest z and the horizontal distance from that point r.

For the vertical stresses, if we define an influence coefficient

$K=3/2\,{\frac {1}{\pi \,\left (1+(\frac{r}{z})^{2}\right )^{5/2}}}$

then

$\sigma_z = K \frac{P}{z^2}$

and we can use the following table to determine the influence coefficients K.

The reason we’ve skipped the lateral stresses is because they’re dependent upon the elastic properties of the soil, and also because the vertical stresses are of greater interest.

The point load problem is an important one because many of the area load problems are based on its solution. It can also be used in other ways in spite of the fact that the solution is singular at the point where the load is applied.

## Other References

• Boresi, A.P., Schmidt, R.J., and Sidebottom, O.M. (1993) Advanced Mechanics of Materials. Fifth Edition. New York: John Wiley and Sons.
• Timoshenko, S., and Goodier, J.N. (1951) Theory of Elasticity. New York: McGraww-Hill Book Company, Inc.