Posted in Geotechnical Engineering

## General Geology – Lange, Ivanova, Lebedeva — Mir Books

In this post, we will see the book General Geology by O. Lange; M. Ivanova; N. Lebedeva. About the book The book is a basic introduction to geology. The first two chapters talk about the origin of the Earth and its properties, and the outer geospheres of earth: the atmosphere, hydrosphere, biosphere and lithosphere. The […]

General Geology – Lange, Ivanova, Lebedeva — Mir Books
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.
Posted in Geotechnical Engineering

## Going Around in Circles for Rigid and Flexible Foundations

In an earlier post Analytical Boussinesq Solutions for Strip, Square and Rectangular Loads, we discussed elastic solutions for these types of foundations. Most of the results shown were for perfectly elastic foundations. In this post we will concentrate on a) circular foundations and b) the difference between rigid and elastic foundations, and those which find themselves in between.

## Circular Foundations

Engineers have been familiar with charts such as this, from NAVFAC DM 7.01:

The influence coefficients (see diagram above) for the vertical, horizontal and shear stresses respectively directly under the centre of the load (x = 0) are not difficult to compute, being

$I_z = 1 - \frac{z^3}{(z^2+r^2)^\frac{3}{2}}$
$I_r = (1+\nu)\frac{z}{\sqrt{z^2+b^2}} - \frac{1}{2}(1 - \frac{z^3}{(z^2+r^2)^\frac{3}{2}})$
$I_{zr} = 0$

For just about everywhere else (except for the edge) closed form solutions are hard to come by. Why is this? Because, for most every other point, the solution for the influence coefficients involves the use of elliptical integrals. An illustration of the values of these is shown below.

Although mathematical packages such as Maple and Matlab are certainly capable of evaluating these, they are still not the common companions of engineers. Fortunately most of the questions about the stresses under circular foundations centre (sorry!) around the point under x = 0, so the above formulae are useful.

## Foundations Rigid and Flexible

Up to this point, we’ve considered for the most part the response of the soil–both stress and deflection–to a purely flexible foundation. For these foundations at the soil-foundation interface the pressure exerted on the foundation and the pressure the foundation exerts on the soil is the same. If there is any rigidity in the foundation–and virtually any foundation has some–then both deflections and stresses in the foundation are redistributed.

It’s probably useful to note that, for deflections in general, the formula we use is

$s = \frac{\omega p b (1-\nu^2)}{E}$

where

• $s =$ settlement of the foundation at the point of interest
• $\omega = I =$ influence factor
• $p =$ uniform pressure on the foundation
• $b = B =$ smaller dimension of rectangle or dimension of square side
• $\nu =$ Poisson’s Ratio of the soil
• $E =$ Modulus of elasticity of the soil

The values for $\omega$ are given below.

Turning to the flexible circular foundation, the value for $\omega$ for all of the radii can be computed using the formula (Timoshenko and Goodier (1951)):

$\omega=\frac{2}{\pi}\,\int_{0}^{1/2\,\pi}\!\sqrt{1-{\frac{{x}^{2}\left(\sin(\psi)\right)^{2}}{{r}^{2}}}}{d\psi}$

This still has a complete elliptical integral of the second kind, but it more manageable. It can be solved by applying the trapezoidal rule and using small integration increments over the interval. Values for $\omega$ for ratios of various values of x (see diagram above) to the actual radius of the circle are shown below.

Although many elastic calculations assume the flexible foundation, as noted earlier in reality foundations have rigidity. For a perfectly rigid foundation, the deflection of the entire foundation under a concentric load is uniform. The effect of this on the stresses can be seen below.

For the foundation in (a), if it is a circular foundation, the vertical stresses at the base can be computed by the formula

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

At the corners of the foundation, the stresses are theoretically infinite. This means that the lower bound solution for such a foundation is zero stress. In reality it is reasonable to assume that a) no foundation is perfectly rigid and b) the soil will proceed into plastic deformation, which will redistribute the stresses.

It is interesting to note that, for the strip loads we discussed previously, the distribution for a rigid strip is similar to the circle, thus

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

where the notation is as it was in that discussion.

The figure (b) above shows different vertical stress distributions for different flexibility ratios, given by the variable $\Gamma$. This can be approximated by the formula (Tsytovich (1976))

$\Gamma \cong 10\frac{E_s l^3}{E_f h_1^3}$

where

• $E_s =$ modulus of elasticity of the soil
• $l =$ half-width of the foundation, shown above
• $E_f =$ modulus of elasticity of the foundation
• $h_1 =$ height of the foundation

To get a better approximation would require plate theory, where the mathematics are very involved; an example of this, using a pile toe where the circle is loaded around the edge, can be found here.

The rigidity of the foundation also influences the distribution of stresses under the foundation, as shown below.

There’s certainly life beyond elastic theory. In reality the type of soil affects the distribution of contact pressure on the foundation, even with a uniformly loaded foundation, as shown below.

The clay distribution most resembles that of an elastic response of a soil with a rigid footing, as discussed earlier. The purely flexible foundation will have the same uniform reaction as the load, as noted earlier.

## Other Sources

• Foster, C.R. and Ahlvin, P.G. (1954) “Stresses and Deflections Induced by Uniform Circular Load.” Highway Research Board Proceedings, Highway Research Board, Washington, DC.
• Jahnke, E. and Emde, F. (1945) Tables of Functions with Formulae and Curves. New York: Dover Publishers.
• Timoshenko, S., and Goodier, J.N. (1951) Theory of Elasticity. New York: McGraww-Hill Book Company, Inc.
Posted in Academic Issues, Geotechnical Engineering

## Lower and Upper Bound Solutions for Bearing Capacity

Although today we have finite element methods which can combine elastic and plastic components of soil response to loading, the use of lower and upper bound plasticity is important in enhancing our understanding of plasticity in soils and many of the methods we use in geotechnical design. This is an overview of both lower and upper bound solutions to the classic bearing capacity problem. Much of this presentation is drawn from Tsytovich (1976) but the equations have been re-derived and checked.

## Definitions (from Verruijt)

1. Lower bound theorem.The true failure load is larger than the load corresponding to an equilibrium system.
1. Upper bound theorem.The true failure load is smaller than the load corresponding to a mechanism, if that load is determined using the virtual work principle.

For our purposes, since we’re assuming an elastic/perfectly plastic type of soil model, the lower bound solution is where the stress at some point reaches the elastic limit, while the upper bound solution has the stress fully plastic to the boundaries of the system, at which point the capacity of the system to resist further stress has been exhausted (reached its upper limit.)

## Assumptions

• Foundation is very rigid relative to the soil
• No sliding occurs between foundation and soil (rough foundation)
• Applied load is compressive and applied vertically to the centroid of the foundation
• No applied moments present
• Foundation is a strip footing (infinite length)
• Soil beneath foundation is homogeneous semi-infinite mass. For the derivations here, we additionally assume that the properties of the soil above the base of the foundation are the same as those below it
• Mohr-Coulomb model for soil
• General shear failure mode is the governing mode
• No soil consolidation occurs
• Soil above bottom of foundation has no shear strength; is only a surcharge load against the overturning load
• The effective stress of the soil weight acts in a hydrostatic fashion, i.e., the horizontal stresses are the same as the vertical ones.

These are fairly standard assumptions for basic bearing capacity theory; the “additions” from these are workarounds that have been developed. That includes the analysis of finite foundations (squares, rectangles, circles, etc.)

## Theory of Elasticity of Infinite Strip Footings

Let us begin by considering the system below of a strip footing with a uniform load. The variables are defined in the figure.

It can be shown that the stresses at a point of interest can be defined as follows:

$\sigma_{{z}}={\frac {p\left (\alpha+\sin(\alpha)\cos(2\,\beta)\right )}{\pi }}$ (1)
$\sigma_{{y}}={\frac {p\left (\alpha-\sin(\alpha)\cos(2\,\beta)\right )}{\pi }}$ (2)
$\tau={\frac {p\sin(\alpha)\sin(2\,\beta)}{\pi }}$ (3)

It can also be shown that the principal axis of the stresses at the point are along a line in the middle of the angle $\alpha$. This is the dashed line in the diagram above. Along this line the angle $\beta = 0$ (and thus $\frac{\alpha}{2}=-\beta'$) and the principal stresses due to the load become

$\sigma_{{1}}={\frac {p\left (\alpha+\sin(\alpha)\right )}{\pi }}$ (4)
$\sigma_{{3}}=-{\frac {p\left (-\alpha+\sin(\alpha)\right )}{\pi }}$ (5)

## Lower Bound Solution

Shallow foundations are seldom built with the base of the foundation at the same elevation as the groundline. They are customarily built to a depth from the surface, as shown below.

At this point, for analysis purposes, we transform the effect of the depth into an overburden stress, which is the product of the the unit weight of the soil $\gamma$ and the depth of the foundation base from the surface D (or h,) as shown below:

The effective stress at any point below the surface is given by the equation

$\sigma_{{0}}={\it \gamma}\,\left (h+z\right )$ (6)

At the point the hydrostatic stress assumption becomes important. The transformation from Equations (1-3) to (4-5) involved an axis rotation. Assuming the soil acts hydrostatically means that, no matter how we rotate the axis, the addition of the effective stress to the principal stress is independent of direction.

Doing just that yields the following:

$\sigma_{{1}}={\frac {\left (p-{\it \gamma}\,h\right )\left (\alpha+\sin(\alpha)\right )}{\pi }}+{\it \gamma}\,\left (h+z\right )$ (7)
$\sigma_{{3}}={\frac {\left (p-{\it \gamma}\,h\right )\left (\alpha-\sin(\alpha)\right )}{\pi }}+{\it \gamma}\,\left (h+z\right )$ (8)

At this point we state the failure function for Mohr-Coulomb theory:

$\sigma_{{1}}-\sigma_{{3}}-2\,c\cos(\phi)-\left (\sigma_{{1}}+\sigma_{{3}}\right )\sin(\phi)=0$ (9)

Substituting Equations (7) and (8) into Equation (9) yields

$-2\,{\frac {-p\sin(\alpha)+{\it \gamma}\,h\sin(\alpha)+c\cos(\phi)\pi+\sin(\phi)p\alpha-\sin(\phi){\it \gamma}\,h\alpha+\sin(\phi){\it \gamma}\,\pi \,h+\sin(\phi){\it \gamma}\,\pi \,z}{\pi }}=0$ (10)

Solving for z, we have

$z={\frac {p\sin(\alpha)}{\sin(\phi){\it \gamma}\,\pi }}-{\frac {h\sin(\alpha)}{\sin(\phi)\pi }}-{\frac {c\cos(\phi)}{\sin(\phi){\it \gamma}}}-{\frac {p\alpha}{{\it \gamma}\,\pi }}+{\frac {h\alpha}{\pi }}-h$ (11)

At this point we want to find the maximum value of z at which point plasticity first sets in. We do this by taking the derivative of z relative to $\alpha$ and setting it to zero, or

${\frac {p\cos(\alpha)}{\sin(\phi){\it \gamma}\,\pi }}-{\frac {h\cos(\alpha)}{\sin(\phi)\pi }}-{\frac {p}{{\it \gamma}\,\pi }}+{\frac {h}{\pi }}=0$ (12)

It can be shown that this condition is fulfilled when $\alpha = \frac{\pi}{2}$. Substituting that value back into Equation (11) gives us the value of z at which point plasticity is first induced, or

$z_{{\max}}={\frac {p\cos(\phi)}{\sin(\phi){\it \gamma}\,\pi }}-{\frac{h\cos(\phi)}{\sin(\phi)\pi }}-{\frac {c\cos(\phi)}{\sin(\phi){\it \gamma}}}-1/2\,{\frac {p}{{\it \gamma}}}+{\frac {p\phi}{{\it \gamma}\,\pi }}-1/2\,h-{\frac {h\phi}{\pi }}$ (13)

If we solve for the pressure $p$, that pressure will be in reality the critical pressure at which plasticity is first induced. Solving for that pressure,

$p_{{{\it cr}}}={\frac {2\,z_{{\max}}\sin(\phi){\it \gamma}\,\pi +2\,h\cos(\phi){\it \gamma}+2\,c\cos(\phi)\pi +\sin(\phi){\it \gamma}\,\pi\,h+2\,h\phi\,\sin(\phi){\it \gamma}}{2\,\cos(\phi)-\sin(\phi)\pi +2\,\phi\,\sin(\phi)}}$ (14)

At this point we need to face reality and note that, if the point we’re looking for is the point at which plastic deformation begins, then it cannot be at any depth other than the base of the foundation, or $z_{max} = 0$. Making that final substitution yields at last

$p_{{{\it cr}}}={\frac {2\,h\cos(\phi){\it \gamma}+2\,c\cos(\phi)\pi +\sin(\phi){\it \gamma}\,\pi \,h+2\,h\phi\,\sin(\phi){\it \gamma}}{2\,\cos(\phi)-\sin(\phi)\pi +2\,\phi\,\sin(\phi)}}$ (15)

## Upper Bound

The upper bound solution is a well-worn path in geotechnical engineering and only the highlights will be shown here.

In 1920-1 Prandtl and Reissener solved the problem for a soil by neglecting its own weight, i.e., Equation (6) They determined that the failure pattern and surface can be represented by the following configuration.

They determined that the upper bound critical pressure was given by the equation

$p_{{{\it cr}}}={\frac {\left (q+c\cot(\phi)\right )\left (1+\sin(\phi)\right ){e^{\pi \,\tan(\phi)}}}{1-\sin(\phi)}}-c\cot(\phi)$ (16)

If we define

$N_{{q}}={\frac {\left (1+\sin(\phi)\right ){e^{\pi \,\tan(\phi)}}}{1-\sin(\phi)}}$ (17)

then

$p_{{{\it cr}}}=qN_{{q}}+c\cot(\phi)\left (N_{{q}}-1\right )$ (18)

If we further define

$N_{{c}}=\left (N_{{q}}-1\right )\cot(\phi)$ (19)

we have

$p_{{{\it cr}}}=qN_{{q}}+cN_{{c}}$ (20)

The only thing missing from this equation is the effect of the weight of the soil bearing on the failure surface at the bottom of the failure region shown in Figure 3, and thus the bearing capacity equation can be written thus:

$p_{{{\it cr}}}=qN_{{q}}+cN_{{c}}+1/2\,{\it \gamma}\,bN_{{{\it \gamma}}}$ (21)

where

$N_{{{\it \gamma}}}=2\,\left (N_{{q}}+1\right )\tan(\phi)$ (22)

This last bearing capacity factor has been the subject of variable solutions over the years; the one shown here is that of Vesić, which is enshrined in FHWA/AASHTO recommended practice. Verruijt discusses this issue in detail.

## Worked Example

We can take an example from the Soils and Foundations Manual, shown below

It would probably be useful to state the bearing capacity equations in nomenclature that’s more consistent with American practice (and the diagram above.) In both cases this is, for the lower bound solution,

$p_{{{\it cr}}}={\frac {2\,D\cos(\phi){\it \gamma}+\sin(\phi){\it \gamma}\,\pi \,D+2\,D\phi\,\sin(\phi){\it \gamma}+2\,c\cos(\phi)\pi }{2\,\cos(\phi)-\sin(\phi)\pi +2\,\phi\,\sin(\phi)}}$ (15a)

and for the upper bound solution,

$p_{{{\it cr}}}=qN_{{q}}+cN_{{c}}+1/2\,{\it \gamma}\,BN_{{{\it \gamma}}}$ (21a)

One important practical difference between the two is the way the overburden is handled. With the lower bound solution, it is equal to $\gamma D$, while with the upper bound solution it is simply the pressure $q$. For a uniform soil above the foundation base with no water table to complicate things, $q = \gamma D = (125)(5) = 625\,psf$.

Direct substitution into Equation (15a) of all of the variables with show that the lower bound critical pressure is 4740.5 psf.

The upper bound is a little more complicated. The three bearing capacity factors are $N_q = 6.4,\,N_c = 14.8,\,and\,N_{\gamma} = 5.39$. Substituting these, q and the other variables yield an upper bound critical pressure of 13,436.8 psf.

If the lower bound is a reduction from the upper bound using a factor of safety, then the FS = 2.83. The lower bound solution is conservative.

## Conclusion

Although the lower bound solution may be too conservative for general practice, it is at least an interesting exercise to show the variations in critical pressure from the onset of plastic yielding to its final failed state.

## Presentation of “Estimating Load-Deflection Characteristics for the Shaft Resistance of Piles Using Hyperbolic Strain Softening”

Last year we posted the paper Estimating Load-Deflection Characteristics for the Shaft Resistance of Piles Using Hyperbolic Strain Softening. Today it’s presented at the University of Tennessee at Chattanooga’s Research Dialogues, and a slide show of the presentation is below.