Elastic solutions to stresses induced by loads at the surface have their limitations, but they allow the use of the principle of superposition. The principle of superposition states that you can add the effects of different loads on a single point. Superposition requires that the stress state of the point be path independent, which is the case with elastic conditions. No matter how you load and unload a point in a system, if elasticity is maintained the result will be the same for a given set of loads.
The stresses that result from each load affect the total stress at the point of interest. They can be computed and added together. So, since “point” loads are physically impossible, is their computation be useful? The answer to this question is “yes” but it takes some judgement, like so many things in geotechnical engineering.
Let us consider the following case, from NAVFAC DM 7.01.
To solve this problem, DM 7 converted the square footings to circular ones and then used the chart shown in the post Going Around in Circles for Rigid and Flexible Foundations. This chart, like others, is hard to read. Is it possible to use point loads as a substitute?
There are three things we need to note here. The first is that the load on each column is 27 tons, and is the same for each column.
The second is that the “r” shown in the table above is not the same as it is for the point loads. The variable “r” for the point loads is the horizontal distance from the load to the point of interest.
The third is that there are three column positions shown in the diagram, with three corresponding values of r:
- The column on top of the load, Column B2
- The columns in the mid-point of the edges, Columns A2, B1, B3, and C2. These have an value of r of 15′ from the point of interest.
- The columns in the corners of the square, Columns A1, A3, C1, and C3. These have a value of r of 21.2′ from the point of interest.
Now, instead of the chart, we apply the formula derived earlier for the influence coefficient, which is
from which the stress is computed by the equation
Using this formula, we can construct the table below for this problem.
|(1) Z, ft||(2) r/Z for B2||(3) r/Z for A2, B1, B3, C2||(4) r/z for A1, A3, C1, C3||(5) K for B2||(6) K for A2, B1, B3, C2||(7) K for A1, A3, C1, C3||(8) Stress for B2, tsf||(9) Stress for each of A2, B1, B3, C2, tsf||(10) Stress for each of A1, A3, C1, C3, tst||(11) Total Stress at Z, tsf|
We could have opted to add the influence coefficients and then compute the stresses since, for each elevation Z, both the elevation and the column load were the same. We did not for clarity; it is certainly possible to have columns of different loads.
The results are conservative, they tend to be higher, especially at the lower value elevations. It’s worth noting that the total stress at Z = 2′ is higher than the distributed loads on the footings. One way to make this better is to use the center formulae for stress under circles for Column B2 and point loads for the rest. Given, however, the limitations of the method in general, and the considerably lower effort in obtaining the influence coefficients, the method is a reasonable one to use.
Superposition doesn’t only apply to point loads; the example given in the post Analytical Boussinesq Solutions for Strip, Square and Rectangular Loads uses it for rectangular and square loads. Nevertheless, with appropriate engineering judgement, using point loads in place of distributed loads can be a viable option.