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The Sorry State of Compression Coefficients

Posted on 27 May 2023 by Don Warrington

I’ve dealt with the issue of consolidation extensively since my first post on the subject, From Elasticity to Consolidation Settlement: Resolving the Issue of Jean-Louis Briaud’s “Pet Peeve”. His problem was the lack of relationship between the way we handle consolidation settlement vs. elastic settlement. In this post I plan to look at a different problem, i.e. the way we express the relationship between soil pressure and consolidation settlement, or settlement by rearrangement of the particles.

Up to now…

Let’s start with the diagram at the right, from Broms (as will be the case with the graphics we use.) Soil is made up of a combination of soil particles and voids between them. The voids can be filled with air, water or (God forbid) something else. For saturated soils water, for practical purposes, fills all of the voids.

In any case, for illustrative purposes we can “melt” the solids into a continuous solid and leave the rest as a void. We assume that the solids do not compress during the application of pressure and thus their volume is constant. From the first state (on the left) to the second state (on the right) additional pressure is applied. All the change of the volume must take place in the void; the equation at the bottom is based purely on the geometry, where $H_o$ is the height of the layer being compressed, $e_o$ is the initial void ratio of the soil, $\Delta e$ is the change in void ratio during compression, and $\delta_p$ is the primary settlement of the soil.

Unfortunately, as discussed elsewhere on this site, the relationship between the increase in pressure and the settlement/change in the volume of the voids isn’t linear but (empirically) logarithmic. That is shown in the graphic on the left; once the pressures get past the ambient effective stress, the settlement takes places according to the relationship shown at the bottom of the graphic. Here $C_c$ is the compression coefficient, $p'_o$ is the effective stress, and $\Delta p$ is the change in pressure on the soil at a given point.

Combining the two equations in the two graphics yields the “accepted” form of the consolidation settlement equation for normally consolidated soils, thus

$\delta_p = \frac{C_c H_o}{1+e_o} \log_{10} (\frac{p'_o + \Delta p}{p'_o})$ (1)

To this deceptively absolute state of affairs Verruijt has the following objections:

We should be using natural logarithms instead of common ones. Common logarithms date from the days when engineers used logarithmic and semi-logarithmic paper, determining $C_c$ graphically. The compression coefficients would be changed by multiplying or dividing (depending on the form, more on that shortly) by a factor of 2.3. In an era of spreadsheets and MATLAB, natural logarithms would make more sense (and reduce student mistakes,) but I don’t see that changing.
We should use the strain rather than the void ratio. Actually, as Verruijt points out, this is done in Continental Europe. In the U.S. and Scandinavia, void ratio is used as the parameter of deflection. To change this would require a change in the compression coefficient, and that leads to…
…his preferred form of the compression equation, which would look like

$\epsilon = \frac{1}{C_{10}} \log_{10} (\frac{p'_o + \Delta p}{p'_o})$ (2)

where $C_{10}$ is another form of the compression coefficient. (Well, actually, he’d prefer natural logarithms, but as I said let’s put that aside.)

Multiplying both sides of Equation (2) by $H_o$ gives us

$\delta_p = \frac{H_o}{C_{10}} \log_{10} (\frac{p'_o + \Delta p}{p'_o})$ (3)

The two compression coefficients are related as follows:

$\frac{1}{C_{10}} = \frac{C_c}{1+e_o}$ (4)

Actually a variant of Equation (4) finds its way into American practice in Hough’s Method for sands, which is described in the Soils and Foundations Reference Manual.

Enter NAVFAC DM 7.1

The “New” NAVFAC DM 7.1 (Soil Mechanics) is an excellent compendium of the current state of geotechnical practice relating to soil mechanics. In the process of discussing consolidation settlement, it highlights some recent changes that promise to add to the confusion described above.

For normally consolidated soils, Equations (1) and (3) are written as follows

$\delta_p = C_{\epsilon c} H_o \log_{10} (\frac{p'_o + \Delta p}{p'_o})$ (5)

where $C_{\epsilon c}$ is the modified compression index. This means that Equation (4) can be expanded as follows:

$\frac{1}{C_{10}} = \frac{C_c}{1+e_o} = C_{\epsilon c}$ (6)

Whether we can dispense with the initial void ratio is a separate topic. Assuming that we can, what we have is a situation with three different compression coefficients, all designated with some form of $C_x$ , and none of them the same. (If we threw in natural logarithms, we’d have six.) The potential for confusion is evident, no where than when two of the three coefficients end up in the same table:

And Secondary Compression…

Secondary compression has had the problem for much longer. If we look at Graphic 3 on the right, we see that we have a secondary compression coefficient $C_{\alpha}$ . The presentation is a little hard to follow but the secondary compression equation is

$\delta_s = \frac{C_{\alpha}H_o}{1+e_o} \log_{10}\frac{t_{life}}{t_{100}}$ (7)

where $\delta_s$ is the amount of secondary compression, $C_{\alpha}$ is the coefficient of secondary compression, $t_{life}$ is the life of the structure and $t_{100}$ is the time at which 100% of primary compression has taken place. (Of course that’s a source of confusion in itself because, in theory, 100% primary compression is never achieved, something that buffaloed many of my students on a test last semester.)

However, as NAVFAC DM 7.1 points out, we can also write this as

$\delta_s = C_{e \alpha}H_o \log_{10}\frac{t_{life}}{t_{100}}$ (8)

where the modified secondary compression coefficient is

$C_{e \alpha} = \frac{C_{\alpha}H_o}{1+e_o}$ (9)

So what is to be done?

My advise to students and practitioners alike is to be vigilant and careful. Make sure you understand which coefficient is being called for. For software, make sure you completely understand which coefficient is being used by the software; otherwise, you will have the classic “garbage in/garbage out” result. Verruijt hoped that we would come to uniform practice but we can’t wait for this; we have to get our work done, and we need to do it carefully.

Can Any Alpha Method be Converted to a Beta Method?

Posted on 26 May 202326 May 2023 by Don Warrington

It’s been a favourite topic of this site to consider the issue of alpha vs. beta methods for deep foundations (both driven and bored piles.) In our post Shaft Friction for Driven Piles in Clay: Alpha or Beta Methods? we show that the Kolk and van der Velde method for driven piles in clay can be converted from an alpha method to a beta one by some simple math. The key to this success is that the ratio of undrained shear strength to effective stress is at the core of the method.

If we want to simplify things further, we can consider this, from the “new” NAVFAC DM 7.1, originally from Skemption:

$\frac{c}{\sigma'_o} = 0.11+0.0037PI$ (1)

where

$c =$ undrained shear strength of the soil
$\sigma'_o$ = vertical effective stress of the soil
$PI =$ plasticity index of the soil

The relationship between undrained shear strength and vertical effective stress in a qualitative sense is illustrated by the diagram at the right, from Broms.

Substituting this into our derived value for $\beta$ in the Kolk and van der Velde method yields

$\beta = 0.9 (\frac{L-z}{d})^{-0.2}(0.11+0.0037PI)^{0.7}$ (2)

where

$\beta =$ ratio of the vertical stress to the horizontal friction on the pile shaft
$L =$ length of the pile
$z =$ distance from the soil surface
$d =$ diameter of the pile

This makes the $\beta$ factor simply a function of the pile geometry and the plasticity index at a depth $z$ .

But can this be done for methods where the relationship between undrained shear strength and the effective stress? The answer is “sort of,” and this post will explore that possibility.

Let us consider an example from the Dennis and Olson method for driven piles. It is a classic “alpha-beta” type of formulation; we will only consider the alpha method portion of the method. For a beta method to be equivalent to an alpha method, the following must hold:

$f_s = \alpha c = \beta \sigma'_o$ (3)

We should note that, for the beta side of the method,

$\beta = F_{SD} K \tan \delta$ (4)

where

$F_{SD} =$ geometry factor based on the aspect ratio of the pile
$K =$ lateral earth pressure coefficient
$\delta =$ friction angle of the pile-soil interface

We will not consider this computation further, but only assume that

$f_s = \beta \sigma'_o$ (5)

For the shaft resistance in clay

$f_s = \alpha \overline{c} F_c F_L$ (6)

The two F constants are defined in the original monograph. The relationship between $\alpha$ and $c F_c$ is shown below.

Figure 1 Relationship of c Fc with alpha for Dennis and Olson Method

This is more complicated than, say the O’Neill and Reese method for drilled shafts. But the idea is the same. Our goal is basically to convert the values of alpha (where c is an independent variable) to use as a beta method.

We start by modifying Equation (3) for the Dennis and Olson method thus:

$f_s = \alpha \overline{c} F_c F_L = \beta \sigma'_o$ (7)

Solving for $\beta$ ,

$\beta = \frac {\alpha \overline{c} F_c F_L}{\sigma'_o}$ (8)

Substituting Equation (1) into Equation (8) yields

$\beta=\alpha\,\left ( .11+ .0037\,{\it PI}\right ){\it F_c}\,{\it F_L}$ (9)

The remaining difficulty is that $\alpha$ is a function of $c$ . This can be dealt with by manipulating Equation (1) to read

$\overline c = (0.11+0.0037PI)\sigma'_o$ (10)

in which case

$\overline c F_c = (0.11+0.0037PI)\sigma'_o F_c$ (11)

The left hand side is the independent variable of the graph above; the right hand side can be computed to substitute for that same independent variable.

Let us consider an example, namely the one used in the Dennis and Olson example:

The problem here is that we are given an undrained shear strength value for the clay layer but not a plasticity index. We are given a unit weight for the clay layer (not automatic for problems like this.) So we can compute the ratio of the undrained shear strength to the effective stress. For the top layer, the midpoint effective stress is 900 psf, and the undrained shear strength 2000 psf. The ratio is thus 2000/900 = 2.22. From Equation (1), the plasticity index is about 571. This, of course, is highly unlikely, and illustrates an important point about academically formulated problems: they’re not always realistic in their parameters. For the effective stress levels we have, it is likely that the undrained shear strength needs to be considerably lower than is given in the problem.

In any case substituting $F_c$ and $\sigma'_o$ from the original data and $PI$ from the current data yields $c F_c = 1400\,psf$ , which is the same as the original. From here we can compute $\alpha = 0.49$ and, substituting into Equation (9), we obtain $\beta = 0.76$ . Multiplying this by the effective stress of 900 psi yields the same result of $f_s = 685\,psf$ .

Conclusion

Getting rid of $\alpha$ altogether is hampered by the fact that there is not an analytic function for $\alpha$ in the first place. The Dennis and Olson method is not unique in this regard.
For methods such as Kolk and van der Velde where the ratio of undrained shear stress and vertical effective stress are important parts of the method, applying correlations such as Equation (1) is fairly simple. When this is not the case then things are more complicated.
Using Equation (1) is doubtless a good check on values of $c$ , which when applied in an alpha method implicitly contains the effects of effective stress.
Going forward, probably the best way to “close the loop” and make all methods beta methods is to formulate the method for clays in terms of $\frac {c}{\sigma'_o}$ as is the case with Kolk and van der Velde. Doing this would be an important step in moving static methods forward.

Academic Issues, Civil Engineering

Floating or Compensated Foundations

Posted on 5 April 20235 April 2023 by Don Warrington

One type of foundation that needs some explanation are floating or compensated foundations. Since they are sometimes referred to as “floating,” some fluid mechanics background is in order.

Fluid Mechanics

For ships to float, they obey Archimedes’ Law, where the weight of the ship is equal to the weight of water displaced by the hull of the ship. This is more thoroughly explained in my handout Buoyancy and Stability: An Introduction. I also go through all this in this video:

If the hull of the ship is rectangular, it’s also possible to compute the upward force of the water–which equals the downward force of the weight–by multiplying the hydrostatic pressure by the plan area of the ship, as is shown below. As the ship settles further and further into the water, the hydrostatic pressure increases until equilibrium is reached.

Illustrating Water Pressure Increasing in Proportion to the Draught

This last will be useful when we consider soils because, although box shaped ships are not so common, box shaped buildings and foundations are.

Turning to buildings, soils are an intermediate material between pure fluids and solids. Some are obviously more intermediate than others, but in softer soils they are more “fluid-like.” Let us consider the multi-storey building at the right.

If we consider that the soil acts as a fluid, then for the building to “float” in the soil the weight of the soil displaced must be greater than or equal to the weight of the building. The difference between ships and buildings is twofold. One, it is possible for a building to weigh less than the weight of the soil displaced and not get shoved upward until equilibrium is reached. The second is that, frequently, we use a “per unit area” approach to balance the equation and come up with the “draught” D of the building.

In this case we have a three-storey building where each storey has a unit weight of 10 kPa, or 10 kN per square metre of area. Multiplying the number of storeys n by the unit weight Δq yields 30 kPa. The soil weight is 18 kN/m³, or otherwise put the displaced soil exerts an “upward force” of 18 kPa/m of depth. Dividing the downward pressure by the unit weight yields a foundation depth/draught D = 1.67 m.

At this point it is worth noting that, depending upon the properties of the soil, it is not always necessary for the soil displaced to equal in weight to the building, but can be less. This is because soils, unlike fluids, have shear strength when not moving, an issue I discuss in my monograph Variations in Viscosity. An illustration of this is at the left.

Here we have a building with eight storeys and 10 kPa/floor, for a total pressure of 80 kPa. On the soil side we have a unit weight of 19 kN/m³ (after eliminating those pesky kilogram force units) and a foundation depth of 4 m, which results in an upward pressure of 76 kPa. The difference between the two is 4 kPa, not much but still enough to reduce the depth of the basement if the soil were a true fluid.

It’s first worth noting that an alternative way to look at the problem is that we are computing the total stress of the foundation at the base and then comparing it with the downward pressure of the building. That works for box like structures such as we are dealing with. If we have a more complex structure such as is shown at the right, we will have to adjust our strategy.

Beyond that, soils are routinely called upon to handle normal and shear stresses induced by the pressure exerted on the foundation. How well they do this is at the core of geotechnical foundation design. We must consider whether foundations will fail in bearing capacity, settlement or both. Bearing capacity is not as great of a problem with “large” structures such as mat foundations as it is with spread footings. Settlement, whether elastic or consolidation, is a major issue, and is something else that separates soils from fluids: rearrangement of the particles during volume change of the soil.

How much net pressure that is permissible is something that needs to be considered once it is established. Nevertheless, it is possible to use the soil’s own weight to help balance and support the structure during its useful life.

Note: graphics are from Bengt Broms, and more of these can be found in the post Bengt Broms Geotechnical Slides.

Academic Issues, Civil Engineering

When Semi-Infinite Spaces Aren’t, and When Foundations are Neither Rigid Nor Flexible

Posted on 4 April 202329 April 2023 by Don Warrington

In our posts Analytical Boussinesq Solutions for Strip, Square and Rectangular Loads and Going Around in Circles for Rigid and Flexible Foundations we discussed foundations on an elastic, homogeneous half-space, mostly purely flexible. In the latter post we ventured into rigid foundations but stuck with the semi-infinite spaces. In this post we’re going to explore cases where either one of the other or both of these aren’t true.

Some Review: Purely Rigid Foundations

In Going Around in Circles for Rigid and Flexible Foundations we looked at stress distributions under circles (and strip loads for that matter.) The results can be summarised graphically below, from Tsytovich (1976).

Figure 1. Isobars in Soil under Foundations a)absolutely rigid foundation, b)flexible foundation (from Tsytovich (1976))

View (a) shows a point load on a rigid foundation; it could be a distributed one too, as long as the load is concentric. In any case at the corners the vertical stress is infinite. In the real world one would expect the soil to go plastic long before that and the stresses to redistribute themselves, but we’ll stick with pure elasticity for the moment.

View (b) shows a distributed load on a flexible foundation. At the interface between the foundation and the soil the vertical stress is the load p, and it decreases the further you get away from the foundation. The strip load version of this is used to find the lower bound solution for bearing capacity in Lower and Upper Bound Solutions for Bearing Capacity.

This is the state of affairs for foundations which are either perfectly flexible or perfectly rigid. The truth is that neither one of these extreme approximations is really true. This is illustrated in the figure below.

Figures 2. Diagrams of contact pressures a) under an absolutely rigid foundation, b) under foundations of various flexibilities. (From Tsytovich (1976))

View (a) shows the rigid foundation with the stresses at the base of the foundation as they would be in elastic theory (solid line) and those with some “real world” plasticity thrown in (dashed line.)

If the rigid foundation is circular, for a semi-infinite, elastic homogeneous space, the stress distribution is as follows:

$\sigma_{z}=\frac{p}{2\sqrt{1-\left(\frac{x}{r}\right)^{2}}}$ (1)

where

$\sigma_z =$ vertical stress in soil
$p =$ uniform pressure on foundation. With rigid foundations we can have a point load P and obtain the same result as long as the load is at the centroid of the foundation
$x =$ distance from centroid
$r =$ radius of foundation

The relationship between a distributed load and a point load at the centroid is

$P = p \pi r^2$ (2)

For a rigid strip foundation,

$\sigma_{z}=\frac{2p}{\pi\sqrt{1-\left(\frac{2y}{b}\right)^{2}}}$ (3)

where

$y =$ distance from centreline of strip load
$b =$ width of foundations

If we define, as is done in Figure 1, the half width of the foundation as

$b_1 = \frac{b}{2}$ (4)

then Equation (3) becomes

$\sigma_{z}=\frac{2p}{\pi\sqrt{1-\left(\frac{y}{b_1}\right)^{2}}}$ (5)

The line load can be computed as follows:

$P = p b$ (6)

View (b) shows a foundation with varying flexibility and the effect that has on stress distribution at the base. The flexibility of the foundation is described by the variable $\Gamma$ . We’ll discuss how that’s calculated later but $\Gamma$ is a measure of the flexibility of the foundation.

$\Gamma = 0$ is a totally inflexible (rigid) foundation
$\Gamma = \infty$ is a totally flexible foundation.

Before we get to that, let’s take a look at the other problem, that of a non-semi-infinite half space.

Non-Infinite Spaces and Flexible Foundations

We’ll start with this problem, which is a progression from what we saw in Analytical Boussinesq Solutions for Strip, Square and Rectangular Loads and Going Around in Circles for Rigid and Flexible Foundations. Unfortunately the state of some of our usual sources forced us to make the treatment of this topic longer than we thought, so you can see it in our post Strange Results: The Case of Settlements on Non-Infinite Elastic Half Spaces and Flexible Foundations.

Non-Infinite Spaces and Rigid Foundations

We now turn to the case of non-infinite spaces and rigid foundations. To deal with this problem we first present this table, from Tsytovich (1976):

Table 1. Contact Pressures under a Rigid Foundation on a Soil Layer of Limited Thickness (fractions of p, from Tsytovich (1976))

The simplest way to show how this is used is through an example. Consider the case of a rigid strip load 3m wide and having a uniform pressure p = 50 kPa. Determine the stress distribution under the soil if the layer under it (until is encounters a hard layer) is 3 m.

We start by considering Figure 1 and computing b₁ = b/2 = 1.5m. We can thus say that h/b₁ = 3/1.5 = 2. The ratio y/b₁ varies from zero (at the centre of the foundation) to 0.95 (almost to the edge of the foundation, where the stress is infinite.) We compute the results for both the limited layer depth case (using Table 1) to the semi-infinite elastic space (using Equation 5) and tabulate the results below.

y/b₁	y, m	Pressure Ratio (from chart)	Pressure, Limited Layer Depth, kPa	Pressure Ratio, Semi-Infinite Half-Space	Pressure, Semi-Infinite Half Space, kPa
0	0	0.705	35.25	0.637	31.831
0.1	0.15	0.707	35.35	0.640	31.991
0.2	0.3	0.714	35.7	0.650	32.487
0.3	0.45	0.725	36.25	0.667	33.368
0.4	0.6	0.744	37.2	0.695	34.730
0.5	0.75	0.773	38.65	0.735	36.755
0.6	0.9	0.818	40.9	0.796	39.789
0.7	1.05	0.891	44.55	0.891	44.572
0.8	1.2	1.029	51.45	1.061	53.052
0.9	1.35	1.366	68.3	1.461	73.025
0.95	1.425	1.869	93.45	2.039	101.941

Table 2. Results of Rigid Strip Load Example

The effect of the limited layer depth is primarily to flatten the pressure distribution across the base of the foundation. The pressures are greater for the limited layer depth case in the centre and less towards the edges. Inspection of Table 1 will show that this effect will become more pronounced as the layer below the foundation becomes thinner.

It is interesting to note that, while the right column is very close to Equation (5), it is not identical. The solution is shown in detail in Elastic Solutions Spreadsheet.

Foundation Flexibility

We have discussed the foundation flexibility coefficient $\Gamma$ . A general formulation of this is

$\Gamma =1/4\,{\frac {\pi \,E_{{s}}{l}^{3}b\left (1-\nu_{{f}}\right )^{2}}{\left (1-\nu_{{s}}\right )^{2}E_{{f}}I_{{f}}}}$ (7)

where

$E_s, \,\nu_s=$ Young’s Modulus and Poisson’s Ratio of the soil
$E_f, \,\nu_f =$ Young’s Modulus and Poisson’s Ratio of the foundation
$l =$ Half length of foundation
$b =$ Width of foundation
$I_f =$ moment of inertia of foundation

If we substitute

$I_f = \frac{b h^3}{12}$ (8)

then

$\Gamma=3\,{\frac {\pi \,E_{{s}}{l}^{3}\left (1-\nu_{{f}}\right )^{2}}{\left (1-\nu_{{s}}\right )^{2}E_{{f}}{h}^{3}}}$ (9)

Making common substitutions of $\nu_f = \nu_s = \frac{1}{3}$ yields

$\Gamma = 9.425\,{\frac {E_{{s}}{l}^{3}}{E_{{f}}{h}^{3}}} \approx 10 {\frac {E_{{s}}{l}^{3}}{E_{{f}}{h}^{3}}}$ (10)

which we will use in our subsequent calculations.

At this point it’s probably worth noting that relative flexibility between foundation and soil is most important in mat foundations. These days most of these will be designed using finite element analysis or some other numerical method, and rightly so. If the flexibility is more than rigid ( $\Gamma > 0$ ) the distribution of the load will come into play, and it is seldom that a foundation is uniformly loaded. In the case of eccentrically loaded foundations, even with rigid foundations the load is redistributed.

Nevertheless some kind of “back of the envelope” exercise is useful, not only for educational purposes but also for purposes of preliminary calculations. This is what we will do for stress distribution under a foundation with flexibility of $0 < \Gamma < \infty$ . To begin we will present the following table, and as before we will illustrate its use with an example.

Table 2. Relative Pressures for Half-Span l of Flexible Uniformly Loaded Beams on a Soil Layer of Limited Thickness H, with a Distribution Diagram (from Tsytovich (1976))

Let’s first dispense with the columns labelled $\Gamma = \infty$ . These are purely flexible foundations, the pressure on the soil is the same as the pressure on the foundation. The rest of these are for foundations with varying degrees of rigidity, from purely rigid foundations ( $\Gamma = 0$ ) to those where, as $\Gamma$ increases, the flexibility of the foundation does also.

Since we are dealing with rectangular foundations, with a uniform pressure p the stress distribution is symmetrical about the centroidal axes of the foundation. The ratio $\xi = \frac{x}{l}$ is the fraction of the distance between the centroidal axis and the long end of the foundation, and in this case is divided into eight equal segments.

For an example, let us consider the foundation we looked at in Strange Results: The Case of Settlements on Non-Infinite Elastic Half Spaces and Flexible Foundations. It is shown below; the deflection result shown was discussed in that post.

Figure 3. Example of Shape and Rigidity Factors I for Calculating Settlements of Points on Loaded Area at the Surface of an Elastic Half-Space (from NAVFAC DM 7.01)

The parameters necessary are as follows:

The Young’s Modulus for concrete is approximately 720,000 ksf.
We will assume that the foundation is 0.45′ (5.4″) thick for reasons that will become apparent.
The half length $l$ of the foundation is 10′, which means that, for Table 2, $\frac{l}{h} = \frac{10}{10} = 1$ .
We will use the approximate value of $\Gamma$ in Equation 10. Substituting, $\Gamma = \frac{20 \times 10^3}{720000 \times 0.45^3} \approx 3$ , which avoids a great deal of interpolation.

Using Table 2 and making the appropriate substitutions yields the following results.

\xi — Table 3 Results of flexibility study

\Gamma = 3 — Table 3 Results of flexibility study

From this result we note the following:

As a practical matter, the results of the rigid foundation and that for $\Gamma = 3$ are not that different, but they are different from the flexible foundation ( $\Gamma = \infty$ .)
The foundation is fairly thin to be considered “rigid.” One possibility is that the Young’s Modulus for the soil is very low. If we were to increase this by a factor of 10 to 200 ksf, we would achieve the same value for $\Gamma$ with a foundation 1′ thick, which is still rigid relative to the soil.
By the time $\Gamma = 10$ the foundation is approaching being purely flexible.

The solution is shown in detail in Elastic Solutions Spreadsheet.

Other Representations of Relative Rigidity

Although it would be nice to be able to determine the soil stress distribution under a foundation, for preliminary purposes it is probably not necessary since other methods of analysis must be done. Nevertheless the rigidity coefficient $\Gamma$ is potentially useful as a starting point to determine whether a foundation can be considered rigid or flexible.

One such use is shown in NAVFAC DM 7.1:

Table 3. Relative Mat Stiffness and Behavior (after Brown (1969,) Frazer and Wardle (1976))

A different notation for the stiffness factor is noted, but the similarity between the equations (especially that of the rectangular foundation) and Equation 10 is unmistakable. This is because there are common sources to both. For rectangular foundations the relationship between the two can be found by the equation

$K_m = \frac{\pi}{2\Gamma}$ (11a)

or conversely

$\Gamma = \frac{\pi}{2 K_m}$ (11b)

Thus, considering rectangular foundations in Table 3, a foundation is flexible if $\Gamma = 31$ , rigid if $\Gamma = 0.157$ , and intermediate between these values. For practical purposes, an intermediate foundation has $0.16 < \Gamma < 31$ , it is rigid below this and flexible above.

One interesting difference is that Table 3 uses the short dimension B while Equation 10 uses the long half dimension l. For the square foundation in the example this isn’t a problem. However, it makes sense that the longer dimension drives the flexibility–and the bending moments–of the foundation.

In any case the behavior of the foundation can be affected by the relative rigidity of the mat and the soil under it. As NAVFAC DM 7.1 notes:

As indicated in Table 5-8, mats with low stiffness ratios can be considered completely flexible. Flexible mats will apply a relatively uniform pressure distribution, and the center, edges, and corners will settle differentially. Mats with high values of $K_m$ * will act in a rigid manner and will tend to settle uniformly.
* Or low values of $\Gamma$

Two other factors need to be considered: the bending stresses in the mat (which is also affected by the reinforcement scheme) and the maximum stresses in the soil. The bending stresses in the mat needs to be considered on a case-by-case basis. Conventional wisdom may indicate that rigid mats would have larger bending stresses, but flexible mats are probably relatively thin and bending stresses may increase in these cases. The maximum stress in the soil immediately around the mat are higher with rigid mats than with flexible ones, especially along the edges. However the soil stresses that most influence the behaviour of the mat may be those which induce the largest settlements, such as those in, say, soft clay layers.

Academic Issues, Geotechnical Engineering

Strange Results: The Case of Settlements on Non-Infinite Elastic Half Spaces and Flexible Foundations

Posted on 30 March 20233 April 2023 by Don Warrington

In our earlier post Analytical Boussinesq Solutions for Strip, Square and Rectangular Loads we discussed the stress under and settlement of foundations (mostly flexible) on a semi-infinite half space. Usually, though, a hard/competent layer intervenes to mess things up. Some of the books offered on this site–in print and download–have solutions for this problem. Unfortunately unexpected things happen when we consider these things carefully.

To illustrate this, let’s start with the table and diagram below, from NAVFAC DM 7.01.

Table 1. Shape and Rigidity Factors I for Calculating Settlements of Points on Loaded Area at the Surface of a Limited Elastic Half-Space (from NAVFAC DM 7.01)

The settlement at the centre of the full foundation (sum of the corners of the divided foundation, see below) is given by the equation

$\delta_v = NqB' \frac {1-\nu_s^2}{E_s} I$ (1)

where

$N =$ number of partial foundations used to compute total settlement (with centre settlements, usually $N = 4$
$q =$ uniform load on foundation
$B' =$ small dimension of partial foundation
$\nu_s =$ Poisson’s Ratio of soil
$E_s =$ Young’s Modulus of soil
$I =$ influence coefficient

Note that note that we’re now dealing with settlements. The soil being compressed is limited to a height H from the surface to the “rigid base.” In this diagram we do this corner by corner, as we did for the stresses in Analytical Boussinesq Solutions for Strip, Square and Rectangular Loads. In both this and Going Around in Circles for Rigid and Flexible Foundations we used another chart, which is shown below.

Table 2. Values for the influence coefficients omega (from Tsytovich (1976))

The two charts differ as follows:

Both use Equation (1), albeit differently, as described immediately below.
The basic formula is the same but the influence coefficient notation is different; the DM 7.01 chart uses I while the Tsytovich chart uses $\omega$ .
The settlement is presented differently; the DM 7.01 chart shows it at the centre of the circular foundation and the corner of the rectangles while the Tsytovich charts shows an average settlement. For the Tsytovich chart, the dimension B’ should be replaced by b, the full small dimension of the foundation, and L’ by l, the full large dimension of the foundation. In this case $\alpha = \frac{l}{b}$ .

Let us look at an example, from NAVFAC DM 7.01.

Table 3. Example of Shape and Rigidity Factors I for Calculating Settlements of Points on Loaded Area at the Surface of an Elastic Half-Space (from NAVFAC DM 7.01)

As was the case with the stresses in Analytical Boussinesq Solutions for Strip, Square and Rectangular Loads, we use the corners and divide up the entire foundation into four (4) identical foundations. The influence coefficients shown in the DM 7.01 chart above are used. The maximum deflection (at the centre) is thus four times the partial corner deflections.

Here’s where we run into the first problem: the example is wrong because it only considers the corner deflection of one partial foundation. The problem statement implies that, if we add all four partial foundations touching the corners of the partial foundations (which are all at the centre of the full one) we would get the complete settlement at the centre. Doing this yields $\delta_{tot} = 4 \times 0.225 = 0.9'$ .

If we use the Tsytovich chart we can compute the average deflection of the foundation, and we can use the entire foundation at one time. We first note that $\alpha = \frac{l}{b} = \frac{20}{20} = 1$ . We then note that $\frac{h}{b} = \frac{10}{20} = 0.5$ . From the Tsytovich chart $\omega_{mh} = 0.39$ and the settlement as follows:

$\delta_{mh} = 4 \times 20 \times \frac{1-0.5^2}{20} \times 0.39 = 1.17'$ (2)

Although “average” results can be different based on the method of averaging (something students frequently overlook,) it makes sense that a average result should be somewhat below the deflection at the centre. That’s not the case here.

So let’s turn to the newer NAVFAC DM 7.1. They replaced the above chart with the following for settlements:

Figure 1. Elastic Influence Factors for Poisson’s Ratio = 1/2. (a) for mu1, (b) for mu0.

To start with, Figure 5-6 (and the accompanying text) really don’t say whether it’s settlement at the corners, centre or an average settlement. Giroud (1972), the source of Figure 5-6a, does say that it is a corner settlement similar in concept to the old DM 7.01, but the new document does not make this clear. From this, H/B = 10/10 = 1 and L/B = 1. Looking at the chart $\mu_1 = 0.35$ and summing the corner displacements of the partial foundations,

$\rho = 4 \times 1 \times 0.35 \times \frac {4 \times 10}{20} = 2.8'$ (3)

This is significantly different than the old DM 7.01. It is larger than the average settlement shown in the Tsytovich table. But can it be checked against another method?

The answer is “yes,” and to do so we turn to Das (2007). Let us begin by defining the reduced foundation dimensions as B’ and L’, which are obviously half each of B and L. The displacement at the centre of the foundation (the corners of the reduced foundations added together) is thus

$\delta_v = NqB' \frac {1-nu_s^2}{E_s} I_s I_f$ (4)

In this case there are two influence factors, and they correspond with those given in NAVFAC DM 7.1 Figure 5-6: $I_s$ with $\mu_1$ and $I_f$ with $\mu_0$ . We can dispense with the latter by noting that, for the case with no embedment ( $D_f = 0 ,\,I_f = 1$ in a similar way to $\mu_0$ above. (Das gives charts, which we do not reproduce here.) Let us then define

$m = \frac{L'}{B'}$ (5)

and

$n = \frac{H}{B'}$ (6)

Using these ratios, we can define two quantities

$F_1 = \left(m\ln ({\frac {\left (1+\sqrt {{m}^{2}+1}\right )\sqrt {{m}^{2}+{n}^{2}}}{m\left (1+\sqrt {{m}^{2}+{n}^{2}+1}\right )}})+\ln ({\frac {\left (m+\sqrt {{m}^{2}+1}\right )\sqrt {1+{n}^{2}}}{m+\sqrt {{m}^{2}+{n}^{2}+1}}})\right){\pi }^{-1}$ (7)

$F_2 = 1/2\,n\arctan({\frac {m}{n\sqrt {{m}^{2}+{n}^{2}+1}}}){\pi }^{-1}$ (8)

From these quantities,

$I_s = F_1 + \frac{2-\nu}{1-\nu}F_2$ (9)

For our example m = 10/10 = 1 and n = 10/10 = 1. For $\nu = 0.5$ (problem statement,) substituting and solving into Equations (4-8) yields the following:

$F_1 = 0.142$
$F_2 = 0.083$
$I_s = 0.392$

Substituting this yields 2.352′, which is reasonably close to the NAVFAC DM 7.1 solution, and still greater than the solution from Tsytovich.

Notes

The NAVFAC DM 7.1 solution is restricted to values of $\nu = 0.5 $. This is unreasonable; this assumes that the soil is a fluid. It also created a singularity in The Equivalent Thickness Method for Estimating Elastic Settlements.
This shows that even with as venerable a document as NAVFAC DM 7.01 errors can arise, and this should be considered with any book or paper. It also shows that, even with a “cut and dried” topic like theory of elasticity, variations can arise.
All of these solutions are shown in our Elastic Solutions Spreadsheet.
The charts in Das for $I_f = \mu_0$ are similar to the second chart in Figure 1, except that they vary for Poisson’s Ratio and the aspect ratio for the foundation.

References

Das, B.M. (2007) Principles of Foundation Engineering. Sixth Edition. Toronto, Ontario: Thompson.
Giroud, J.-P. 1972. “Settlement of Rectangular Foundation on Soil Layer.” Journal of the Soil Mechanics and Foundations Division, 98(SM1), 149-154.

	Fraction of Pressure			Soil Vertical Reaction, ksf
$\xi$	Rigid Foundation	$\Gamma = 3$	Flexible Foundation	Rigid Foundation	$\Gamma = 3$	Flexible Foundation
0.0625	0.828	0.876	1	3.312	3.504	4
0.1875	0.829	0.874	1	3.316	3.496	4
0.3125	0.836	0.874	1	3.344	3.496	4
0.4375	0.837	0.866	1	3.348	3.464	4
0.5625	0.857	0.87	1	3.428	3.48	4
0.6875	0.899	0.893	1	3.596	3.572	4
0.8125	0.987	0.958	1	3.948	3.832	4
0.9375	1.927	1.791	1	7.708	7.164	4

Up to now…

Enter NAVFAC DM 7.1

And Secondary Compression…

So what is to be done?

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Fluid Mechanics

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Some Review: Purely Rigid Foundations

Non-Infinite Spaces and Flexible Foundations

Non-Infinite Spaces and Rigid Foundations

Foundation Flexibility

Other Representations of Relative Rigidity

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Notes

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