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## Analytical Boussinesq Solutions for Strip, Square and Rectangular Loads

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

1. They assume that the load is applied to a linear elastic, homogeneous semi-infinite space.
2. They assume that the foundation is completely flexible; rigid (or intermediate foundations) are not considered.
3. 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.
4. Only the vertical stresses will be considered.
5. 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 $p$ 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:

$\sigma_{z}=\frac{p}{\pi}\left[\left(\theta_{1}-\theta_{2}\right)+sin\theta_{1}cos\theta_{1}-sin\theta_{2}cos\theta_{2}\right]$

Verruijt notes that the area of greatest interest is directly under the centre of the foundation, where for any point $\theta_2 = - \theta_1$.  In this case the equation reduces to

$\sigma_{z}=\frac{2p}{\pi}\left[\theta_{1}+sin\theta_{1}cos\theta_{1}\right]$

At the base of the foundation, $\theta_1 = \frac{\pi}{2}$ for all points and this equation reduces to equating the soil stress to the foundation pressure $p$, 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 $2a$ is the total width of the foundation and $z$ is the depth from the base of the foundation, then the vertical stress under the foundation centre is

$\sigma_{z}=\frac{2p}{\pi}\left[arctan\left(\frac{a}{z}\right)+\frac{az}{a^{2}+z^{2}}\right]$

## 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 $m^{2}n^{2}>m^{2}+n^{2}+1$, then

$I = 1/4\,\left(2\,{\frac{mn\left({m}^{2}+{n}^{2}+2\right)}{\sqrt{{m}^{2}+{n}^{2}+1}\left({m}^{2}+{n}^{2}+1+{m}^{2}{n}^{2}\right)}}+\arctan(2\,{\frac{mn\sqrt{{m}^{2}+{n}^{2}+1}}{{m}^{2}+{n}^{2}+1-{m}^{2}{n}^{2}}})+\pi\right){\pi}^{-1}$

Otherwise, it is

$I = 1/4\,\left(2\,{\frac{mn\left({m}^{2}+{n}^{2}+2\right)}{\sqrt{{m}^{2}+{n}^{2}+1}\left({m}^{2}+{n}^{2}+1+{m}^{2}{n}^{2}\right)}}+\arctan(2\,{\frac{mn\sqrt{{m}^{2}+{n}^{2}+1}}{{m}^{2}+{n}^{2}+1-{m}^{2}{n}^{2}}})\right){\pi}^{-1}$

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 $\delta_v$ 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, $E_u$ is the modulus of elasticity of the soil (subject to the limitations discussed earlier) and $\nu$ 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,

$I = 1/2\,\left(2\,L\ln (\sqrt {{\frac {{L}^{2}}{{B}^{2}}}}\left (\sqrt {4+ {\frac {{L}^{2}}{{B}^{2}}}}-2\right )^{-1}){B}^{-1}+2\,\ln (\left( \sqrt {4+{\frac {{L}^{2}}{{B}^{2}}}}+{\frac {L}{B}}\right)\left ( \sqrt {4+{\frac {{L}^{2}}{{B}^{2}}}}-{\frac {L}{B}}\right )^{-1}) \right){\pi }^{-1}$

And at the corners,

$I = 1/2\,\left(2\,L\ln (\sqrt {4}\sqrt {{\frac {{L}^{2}}{{B}^{2}}}}\left ( \sqrt {4+4\,{\frac {{L}^{2}}{{B}^{2}}}}-2\right )^{-1}){B}^{-1}+2\, \ln (\sqrt {4}\left (\sqrt {4+4\,{\frac {{L}^{2}}{{B}^{2}}}}-2\,{ \frac {L}{B}}\right )^{-1})\right){\pi }^{-1}$

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

$I = 1/2\,\left(2\,L\ln (\left(\sqrt {1+{\frac {{L}^{2}}{{B}^{2}}}}+1 \right)\left (\sqrt {1+{\frac {{L}^{2}}{{B}^{2}}}}-1\right )^{-1}){B}^ {-1}+2\,\ln (\left(\sqrt {1+{\frac {{L}^{2}}{{B}^{2}}}}+{\frac {L}{B}} \right)\left (\sqrt {1+{\frac {{L}^{2}}{{B}^{2}}}}-{\frac {L}{B}} \right )^{-1})\right){\pi }^{-1}$

## Some Observations

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.

## References

• 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.

Posted in Soil Mechanics

## Breaking Down Sieve Data for both Lab and Textbook Problems

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:

1. 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.
2. 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.
3. 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:

1. It lines up the grain sizes and sieve designations for you.
2. 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.

Posted in Geotechnical Engineering, Soil Mechanics

## A Simple Example of Braced Cut Analysis

Most retaining walls are designed with active or passive earth pressures derived from Rankine, Coulomb or Log-Spiral theories.  One notable exception to that are braced cuts.  The development of the earth pressure distributions is attributable to Karl Terzaghi and Ralph Peck.  In the process of developing those, the way the wall is modelled was simplified to avoid statically indeterminate structures.  Although this is not the problem that it was in their day, the method is still dependent upon those statically determinate structures.

The example below is a simple example in that the supports are symmetrically placed and there is no sheeting toe penetrating the bottom of the excavation.  It’s primarily intended to illustrate the concepts, both geotechnical and structural, of the design of these structures.

# Overview of the Example

Let us consider a braced cut excavation which is 45′ deep and which has supports at a depth of 5′, 17′, 28′ and 40′.  The soil behind the wall is uniform with c = 1100 psf and γ = 110 pcf.  The water table is at the bottom of the excavation and does not enter into our calculations.  To show how this lays out we’ll use Pile Buck’s SPW 911 sheet pile software.  We’ll assume PZ-27 sheeting is being used, and that there is no surcharge on the wall.

The options for earth pressure distribution behind braced cuts are shown below, from NAVFAC DM 7.2. or Sheet Pile Design by Pile Buck.

We obviously have a clay soil, thus our selection will be either (b) or (c).  Whether the soil is soft to medium or stiff depends upon the stability number $N_o$, which is computed as follows:

$N_o = \frac{\gamma H}{c} = \frac{110 \times 45}{1100} = 4.5$

This is between (b) and (c), we are thus supposed to use the “larger” of the two diagrams.  The earth pressure coefficient for (b) is

$K_a= 1 - m \frac {4c}{\gamma H}$

Assuming m = 1,

$K_a = 1 - m \frac {4c}{\gamma H} = 1 - \frac{4 \times 1100}{110 \times 45} = 0.11$

and thus

$\sigma_h = K_a \gamma H = 0.11 \times 110 \times 45 = 550\,psf$

If we turn to Case (c) and assume that

$\sigma_h = 0.3 \gamma H = 0.3 \times 110 \times 45 = 1485\,psf$

this is obviously “larger” than Case (b), so we will use Case (c), even when using a “medium” case between the two extreme pressure profiles.

We thus have a pressure distribution that can be described as follows:

1. Beginning at the top, it linearly rises from zero to the maximum value of 1485 psf at a point a quarter down the wall, or 45/4 = 11.25′.
2. From that point until a quarter from the bottom of the wall, or 0.75 * 45 = 33.75′, it is a constant pressure of 1485 psf.
3. From that point until the bottom of the wall, it linearly decreases to a value of zero at the bottom of the wall.

# Guidelines for Structural Analysis of Wall

Turning to the structural aspects of the wall, the guidelines for dividing the wall up are as follows:

1. If the wall is cantilevered at either end, then the endmost support and the one next to it form a simply supported beam with a cantilever at one end and a distributed load.
2. Segments in the middle are analysed as simply supported beams with a distributed load.
3. If there’s a support at the top or the bottom of the wall, the beam at that location is analyzed as a simply supported beam.
4. Reactions are computed for each beam.  For supports where two segments meet, you simply add the two reactions from each beam for a total reaction for the support.
5. Maximum moments are computed for each beam; the largest of these maximum moments is the maximum moment of the system and the one used to size the sheeting.

This was Terzaghi and Peck’s attempt to make the calculations simple.  If the distributions are simple, then “handbook” type formulas can be used.  The trout in the milk takes place (as it does here) when the break points in the distribution don’t coincide with the supports, in which case you end up with a more complicated distribution.  There are two ways of dealing with this problem.

The first is to reduce the distributed loads to point load resultants.  This is a favourite tactic among geotechnical engineers and is used extensively with shallow foundations.  For purely hand calculations, it makes sense.  The moments will be higher (which is conservative) but the reactions will be identical, assuming the concentration of the moments went off according to plan.

The second is to employ beam software to analyse each segment.  Although there’s a lot of beam software out there, being the old coots we are, we’ll use CFRAME, a DOS program for two-dimensional structures.  It gets the job done and is fairly easy to use.  (Note: because of some bad interaction between CFRAME and DOSBox, we ran it on a Windows XP installation.  The manual for CFRAME: Computer Program with Interactive Graphics of Plane Frame Structures is here.)

# Implementation in CFRAME

The first thing we need to do is to specify the distributed loads.  CFRAME, like most finite element programs, considers the beam between each support (and the beams from the outermost supports to the cantilever element) as one element.  So there are six elements.  CFRAME asks us to specify the distributed load (constant or linearly varying) for each element, and requires us to specify the constant loads and the varying loads separately.

But here we run into something that trips up students.  Sheet piles are analysed as beams, but they’re “infinite” beams; we analyse them in terms of moment of inertia per length of wall, section modulus per length of wall, load per unit length of wall, etc.   The good news is that, for distributed loads, the pressure at any point is the load per unit length!  Pressure is expressed, in this case, as lb/ft^2 of wall, when in reality it’s lb/ft/ft of wall.  That makes things simpler; as long as we enter the moment of inertia and cross sectional area in terms of “per foot of wall” (which any US unit section should furnish us) then we’re good.  In this case for PZ-27 the moment of inertia is 184.2 in^4/ft of wall and the cross-sectional area is 7.94 in^2/ft of wall, and these are entered directly into CFRAME.

With that technicality out of the way, for are areas of constant earth pressure (the middle) we’re also good; it’s just 1485 psf, and we enter this directly into CFRAME.  With the ramped portions, they increase from the top and bottom of the wall at a rate of 1485/11.25 = 132 psf/ft from the end.  Looking at the topmost element, which we enter into CFRAME as (surprise!) element 1, the pressure at the topmost support is 132 * 5 = 660 psf, which we enter as the maximum pressure for the “triangle load” on the top element.

For element 2, we have two loads.  The first is a continuation of the ramped load from 660 psf at the top end of the beam to 1485 psf at a point 11.25′ from the top of the wall or 11.25′ – 5′ = 6.25′ from the top end of the beam.  The second load is simply a constant load to the bottom end of the beam.

The middle element 3 has a constant distribution across its entire length.  The bottom two elements are mirror images of the top two elements.

# Results from CFRAME

We entered the data into CFRAME via a small text file.  First we present the model itself.

Now we show the results.

The individual element results are shown below.  The tabular results of the program are here.

# Analysing the Results

First let’s look at the reactions at the supports, which come from the element results.  They are as follows:

1. Support 1 (Node 2):  The reaction/shear at that point from element 1 is 1650 lb/ft of wall and from element 2 7009 lb/ft of wall, summing it comes to 8659 lb/ft of wall.
2. Support 2 (Node 3): The reaction/shear at that point from element 2 is 8233 lb/ft and from element 3 8168 lb/ft, summing it comes to 16401 lb/ft.
3. Support 3 (Node 4) is the same as Node 3 by symmetry.
4. Support 4 (Node 5) is the same as Node 2 by symmetry.

Thus the maximum brace load is on Supports 2 and 3, 16401 lb/ft.  We have for convenience ignored the sign conventions and simply added the reactions, since they’re all in the same direction.

The maximum moment is actually in Element 2 (or 4,) and is 273,900 in-lb/ft of wall.  Since the elastic section modulus for PZ-27 is 30.2 in^3/ft of wall, the maximum bending stress is 273,900/30.2 =  9070 psi, which is well within most allowable specifications.  A lighter section can probably be employed, depending upon the allowable deflection and other requirements.

As a quick check, for a uniformly distributed load on a simply supported beam, the maximum moment is given by the equation

$M_{max} = \frac{wl^2}{8}$

Substituting the values for Element 3, we have

$M_{max} = \frac{wl^2}{8} = \frac {1485 \times 11^2}{8} = 22,461\,\frac{ft-lb}{ft}$

Now we compare these with SPW 911, whose output is as follows:

The differences are minor (SPW 911 and the hand calculation report the maximum moment in ft-lb/ft of wall, not in-lb/in of wall.)  Some discussion of eliminating the additional pins in the simply supported spans is given in Sheet Pile Design by Pile Buck.

Posted in Soil Mechanics

## Terzaghi “Low Walls” Curve Correlations

Practitioners who design gravity retaining walls are familiar with the existence of Terzaghi’s “low walls” curves to estimate the equivalent fluid pressure on the wall, horizontal and vertical.  The basic chart comes in several versions but the one (for straight backfill) from NAVFAC DM 7.02 is above.  The explanation for it (important when one is using it) is below.

In the “slide rule” days, reading charts like this was routine.  A better way now, with spreadsheets abounding, is to have a formula available.  Some least squares curve fitting correlations are shown below.  The variable $\beta$ is the slope angle as shown in the figure.  As is the case with the chart, the formulas return $K_v$ and $K_h$ in units of $\frac {psf}{ft}$ of wall length.

• $K_v$ (top chart)
• Soil 1 $K_v = 0.0127 \beta^{2.21}$
• Soil 2 $K_v = 0.043 \beta^{1.92}$
• Soil 3 $K_v = 0.109 \beta^{1.71}$
• $K_h$ (bottom chart)
• Soil 1 $K_h = 30 + \frac{exp^{0.139 \beta}}{4}$
• Soil 2 $K_h = 37 + \frac{exp^{0.138 \beta}}{4}$
• Soil 3 $K_h = 48 + \frac{exp^{0.148 \beta}}{5}$