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Students and practitioners alike of geotechnical engineering have learned and used Boussinesq elastic solutions for stresses and deflections induced in a semi-infinite space by structures at the surface. Although these solutions are very idealized and have many limitations, they’re still useful.
For the most part engineers have implemented these solutions–especially for loads other than point or line loads–using charts. This chart, from Naval Facilities Engineering Command (1986)–DM 7.01, Soil Mechanics, shows the isobars for stresses induced by strip and square foundations.
In addition to being hard to read (a fault which has been fixed in many of the books that have cribbed this chart) it requires a great deal of interpolation to use it and many others. In the past, the computational demands of using analytical solutions put them out of reach for practical use and educational purposes. That’s no longer the case; however, some of those solutions are difficult to find. This piece attempts to bridge that cap and set forth analytical solutions that can be used, along with a spreadsheet to implement at least some of them.
Assumptions of the Solutions Presented
- They assume that the load is applied to a linear elastic, homogeneous semi-infinite space.
- They assume that the foundation is completely flexible; rigid (or intermediate foundations) are not considered.
- They do not consider strain-softening hyperbolic effects. These are extensively discussed in this monograph. It is more than likely that, for the cases presented here, a homogenized value for the modulus of elasticity can be arrived at, perhaps by using the methods used in the linked monograph for the toe.
- Only the vertical stresses will be considered.
- All of the loads on the foundation are uniform.
Stresses Under Strip Loads
The simplest case for this set of foundation geometries is the strip load, which reduces a three-dimensional problem to a two-dimensional one. The problem is illustrated (and a point under consideration located) in the figure below, taken from Verruijt, A., and van Bars, S. (2007). Soil Mechanics. VSSD, Delft, the Netherlands..
where is the uniform pressure on the foundation in load per unit area, and the two angles from the two corners define the location of the point in question. Probably the mathematically simplest representation is also found in Verruijt, as follows:
Verruijt notes that the area of greatest interest is directly under the centre of the foundation, where for any point . In this case the equation reduces to
At the base of the foundation, for all points and this equation reduces to equating the soil stress to the foundation pressure , as we would expect.
The tricky part of this is in determining the two angles from the geometry of the system and the desired location of the point in question. For students probably the simplest way of doing this is to use CAD software. However, the formula for the stress under the foundation centre can also be written in a more convenient form. If is the total width of the foundation and is the depth from the base of the foundation, then the vertical stress under the foundation centre is
Stresses under Square and Rectangular Loads
These are well known, and most engineers and engineering students have used the “Fadum charts” as shown below to obtain the solution.
Both the strength and the weakness of the charts is that the stresses computed are under the corner of the rectangle/square. It is a very specific position, but by using superposition (permissible with elastic, path-independent solutions) we can add and subtract rectangles to obtain the stress at just about any point under or near the structure in question.
There are many expressions of the equations that generated these charts; the following is derived from Bowles (1996). We will use the notation shown above. If , then
Otherwise, it is
We used the Rectangular Elastic Solutions Spreadsheet to generate our own version of the Fadum chart, shown below. The spreadsheet also includes the numerical values which were used to plot the results, which could save some uncertainty in interpolating the results.
The Rectangular Elastic Solutions Spreadsheet also has an example of how superposition works with two rectangular foundations with two different loads and a point away from both. It is taken from Verruijt, A., and van Bars, S. (2007). Soil Mechanics. VSSD, Delft, the Netherlands., who solve the problem using Newmark’s Method. The diagram for the problem is shown at the right. The superposition method is illustrated below. Using the formulae the superposition method is more precise than using the charts, which makes it a more interesting alternative to the Newmark charts in cases such as this.
Deflections of Squares and Rectangles
Elastic solutions can be used to predict both stresses and deflections. Most engineers are familiar with tables such as this, and these are still used for initial deflections and deflections in media such as intermediate geomaterials (IGMs.)
So how does one compute these values? Let’s start by using the same coordinate system as is used with the Fadum chart above, only putting the origin at the centre of the foundation. In this way the long edges have the coordinates x = +-B/2 and the short edges y = +- L/2, and the corners the four combinations of the two. The center is obviously x =y = 0. First, the deflection at any point on the surface of the foundation can be computed by the equation
where is the deflection at a given point, q is the uniform load on the foundation (the same as p above,) B is the short length of the foundation, is the modulus of elasticity of the soil (subject to the limitations discussed earlier) and is Poisson’s Ratio for the soil. The last variable I is the influence coefficient. For square or rectangular foundations, the influence coefficient (again using the limitations discussed earlier) is computed by the following equation (Perloff and Baron, 1976):
As an example of how this looks over an entire foundation, consider the case of B=L=1. For the case of -B/2 < x < B/2 and -L/2 < y < L/2, the influence coefficient of the foundation can be plotted as follows:
It is easy to see what is meant by “flexible” foundation.
The problem with this formula is that, if blindly followed mathematically (just inserting the variables) singularities quickly arise both along the edges or at the corners. Symbolically solving (and taking a few limits) get around this. For the mid-point of the edges,
And at the corners,
A plot of the same functions mentioned in the table above from the formulae is below.
The center deflection can be found by either substituting x=0, y=0 into the first equation or by using the equation
Use of theory of elasticity in this way has been employed in foundation design for a long time, even with the inherent limitations of the method. It gives reasonable approximations for either initial deflections or for deflections of IGM’s. For implementation on a recurring basis, use of the formulas allows a more precise implementation of these methods is not necessarily a more accurate one, and eliminates the errors inherent in reading charts.
In our opinion it is possible to improve the accuracy of the method by improving our understanding of the elastic modulus of the soil, and in particular strain-softening near the foundation itself.
The deflections are probably the less satisfactory products of this theory than the stresses. No foundation is either purely flexible or rigid, and using a purely flexible foundation produces larger variations in deflections than one would expect in reality. Also, the typical rule of thumb that foundations with an aspect ratio larger than 10 can be treated as continuous/infinite foundations is reasonable for stresses but not for deflections, and in fact the DM 7 chart shown above was truncated from its source. Whether this is reflected in reality is another question, and this too doubtless relates to the flexibility of the foundation.
- Bowles, J.E. (1996) Foundation Analysis and Design. Fifth Edition. New York: McGraw-Hill.
- Perloff, W.H., and Baron, W. (1976) Soil Mechanics: Principles and Applications. New York: Ronald Press.
One of the core things students learn in a basic Soil Mechanics course is how to analyze and chart the results of sieve tests on soils. Texts such as Soils and Foundations Reference Manual and Verruijt, A., and van Bars, S. (2007). Soil Mechanics. VSSD, Delft, the Netherlands. usually present the concept but don’t always give a detailed explanation of how these things are actually analyzed. The example below is from Materials Testing and is based on the use of their forms DD-1206 and DD-1207, which of course we furnish as well.
Let’s consider this example, presented below:
This is a sample sieve analysis result. For the lab (especially when sample washing is involved) it can get this complex, and the procedure is described in Materials Testing. To look at this more schematically (and this is common for most “textbook” problems) the following simplifications are done:
- There are no losses in the process of sieve analysis. This is obviously unrealistic but these should be kept to a minimum. For textbook type problems this means that Blocks 8 and 23 are the same.
- The tare (sieve or pan) weight is removed from consideration. This means that typically Column 15 is the first column in the problem, given the soil weights retained on each sieve. The weight that “hits the pan” is in Block 22.
- Washing losses and errors are not considered, which means that Blocks 21 and 24 are set to zero.
With these out of the way, we can proceed to the analysis. The top sieve in the stack (the 2″ sieve) retained no soil (it doesn’t always happen this way) so it has zero soil retained and all passing. The next sieve (the 1 1/2″ sieve) retained 83.7 g (Column 15.) For the cumulative weight retained (Column 16) we add the value of Column 15 just above it to this one, thus 0 + 83.7 = 83.7 g. The next sieve (the 3/4″ sieve) retained 161 g, so for the cumulative weight retained for this sieve we add in the same way, thus 83.7 + 161 = 244.7 g. We keep going in the same way “down the stack” until we have all of the cumulative weights retained for each sieve computed.
This is terrific, but what we really need is the percentages of the total sample each cumulative retained value represents. This is what goes in Column 17. These values are obtained by dividing each result in Column 16 by the total sample weight in Block 23 and multiplying them by 100 to obtain a percentage. In this way, the cumulative percent retained for the 2″ sieve is obviously zero, for the 1 1/2″ sieve 83.7/4381.4 x 100 = 1.9%, for the 3/4″ sieve 244.7/4381.4 x 100 = 3.7%, and so on.
We can also compute the percent passing. An easy way to do this is to start by noting that 100% passes the whole sieve stack. We can than successively subtract the cumulative retained percentage as we go down. Thus the percentage of the 2″ sieve is again obviously 100 – 0 = 100%, for the 1 1/2″ sieve 100-1.9 = 98.1%, for the 3/4″ sieve 98.1-3.7 = 94.4% (note that we use the percentage passing from the previous sieve each time) and so on.
We then plot the results on the sieve curve as follows:
It’s a temptation these days to use a spreadsheet and its semi-logarithmic plotting features. I would avoid this: using a dedicated form like this has two advantages:
- It lines up the grain sizes and sieve designations for you.
- It can be plotted either from the percent passing size (left) or percent retained side (right.)
It’s also possible to use a tool such as the Spears Lab Spreadsheet, but this takes a lot of practice and is a little tricky to use, especially for “textbook” type problems where the tare is not considered. It’s easy to get a really stupid looking result; I have seen quite a few over the years.
While looking through some files, I found these from the original STADYN project, from the comparison case with GRLWEAP. I’m passing these along to give you an idea of the graphical output of this program. My thanks to Jonathan Tremmier of Pile Hammer Equipment for allowing me to use this copy of GRLWEAP.