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.

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## Geotechnical Engineering on the Moon? 50th Anniversary of Apollo 11 Landing — GeoPrac.net

Posted in Geotechnical Engineering

## Engineering Geology in the Civil Engineering Curriculum

For this post I’m featuring an article by J. David Rogers, Professor and Karl F. Hasselmann Chair in Geological Engineering, Department of Geosciences and Geological and Petroleum Engineering, Missouri University of Science and Technology, which he entitled  Disappearing Practice Opportunities: Why Are Owners And Engineers Taking Increased Risks? What Can Be Done To Counter This Threat?  It’s been around a while but bears repeating, especially for one thing: why we need to restore engineering geology to the civil engineering curriculum.

Posted in Geotechnical Engineering, Soil Mechanics

## p-q Diagrams and Mohr-Coulomb Failure

Students and practicioners of soil mechanics alike are used to seeing triaxial test results that look like this (from DM 7.01):

Ideally, the Mohr-Coulomb failure line should be straight, but with real soils it doesn’t have to be that way.  With the advent of finite element analysis we also have the failure function to consider, thus (from Warrington (2016)):

All of these involve constructing (or using) a line which is tangent to a circle at failure.  This can be confusing to understand completely.  The biggest problem from a “newbie” standpoint is that the maximum shear defined by the circle of stress (its radius) and the failure shear stress defined by the intersection of the circle with the Mohr-Coulomb failure envelope are not the same.

Is there a better graphical way to represent the interaction of stresses with the Mohr-Coulomb failure criterion?  The answer is “yes” and it involves the use of p-q diagrams.  These have been around for a long time and are used in such things as critical state soil mechanics and stress paths.  A broad explanation of these is found in our new publication, Geotechnical Site Characterization.  The purpose of this article is to present these as a purely mathematical transformation of the classic Mohr-Coulomb diagram.  This is especially important since their explanation is frequently lacking in textbooks.

## The Basics

Consider the failure function, which is valid throughout the Mohr-Coulomb plot.  It can be stated as follows:

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

(The main difference between the two formulations is multiplication by 2; the failure function can either be diametral or radial relative to Mohr’s Circle.  With a purely elasto-plastic model, the results are the same.)

Now let us define the following terms:

$p=1/2\,\sigma_{{1}}+1/2\,\sigma_{{3}}$

$q=1/2\,\sigma_{{1}}-1/2\,\sigma_{{3}}$

We should also define the following:

$\sin(\phi)=\tan(\delta)$

The physical significance of the last one is discussed in this post.  In any case we can start with $\phi$ and solve for $\delta$ or vice versa.  Solving for $\phi$ and substituting this and the equations for p and q into the failure functions yields

$f=2\,q-2\,c\sqrt {1-\left (\tan(\delta)\right )^{2}}-2\,p\tan(\delta)$

For the failure line, $f = 0$.  Let us set the p axis as the abscissa (x-axis) and the q axis as the ordinate (y-axis.)  For the failure line, if we substitute for $f$ and solve for q, we have

$q = p\tan(\delta) + c\sqrt {1-\left (\tan(\delta)\right )^{2}}$

This is a classic “slope-intercept” form like $y = mx + b$, where in this case $q = mp + b$, $m = \tan(\delta)$ and $b = c\sqrt {1-\left (\tan(\delta)\right )^{2}}$.  A sample plot of this kind is shown below.

### Some Observations

1. For the case of a purely cohesive soil, where $\phi = \delta = 0$, the failure envelope is horizontal, just like with a conventional Mohr-Coulomb diagram.
2. For the case of a purely cohesionless soil, where $c = 0$, the y-intercept is in both cases through the origin.
3. The two diagrams are thus very similar visually, it’s just that the p-q diagram eliminates the circles and tangents, reducing each case to a single point.

## Examples of Use

### Drained Triaxial Test in Clay

Consider the example of a drained triaxial test in clay with the following two data points:

1. Confining Pressure = 70 kPa; Failure Pressure = 200 kPa.
2. Confining Pressure = 160 kPa; Failure Pressure = 383.5 kPa.

Determine the friction angle and cohesion using the p-q diagram.

We first start by computing p and q for each case as follows:

$p_1 = 200/2+70/2 = 135\,kPa$

$p_2 = 383.5/2 + 160/2 = 271.75\,kPa$

$q_1 = 200/2-70/2 = 65\,kPa$

$q_2 = 383.5/2 - 160/2 = 111.75\,kPa$

The slope is simply

$m = \frac {q_2 - q_1}{p_2 - p_1} = \frac {111.75 - 65}{271.5 - 135} = 0.342 = \tan(\delta)$

from which

$\delta = 18.9^o$

$\phi = sin^{-1}(tan(\delta)) = sin^{-1}(0.342) = 20.03^o$

$b = q - mp = 65 - 0.342 \times 135 = 18.83$ (using values from the first point, just as easy to use the second one.)

$b = c\sqrt {1-\left (\tan(\delta)\right )^{2}} = c \sqrt {1-0.342^{2}} = 0.94 c$

$b = 18.83 = 0.94 c$

$c = 20.03\,kPa$

Use of this method eliminates the need to solve two equations in two unknowns, and the repetition of the quantity $tan(\delta)$ makes the calculations a little simpler.  When $c = 0$, the calculations are even simpler, as $p_1 = q_1 = 0$.

### Stress Paths

As mentioned earlier, p-q diagrams are commonly used with stress paths.  An example of this from DM 7.01 is shown below.

We note that p and q are defined here exactly as we have them above.  (That isn’t always the case; examples of other formulations of the p-q diagram are here.  We should note, however, that for this diagram $\phi" = \delta$)  With this we can track the stress state of a sample from the start (where the deviator stress is zero, at the start of the triaxial test) around to its various points of stress.

As an example, consider the stress path example from Verruijt, A., and van Bars, S. (2007). Soil Mechanics. VSSD, Delft, the Netherlands.  The basic data from Test 1 are below:

 $\sigma_3$ Deviator Stress Pore Water Pressure 40 0 0 40 10 4 40 20 9 40 30 13 40 40 17 40 50 21 40 60 25

Using the p-q diagram and performing some calculations (which are shown in the spreadsheet Stress Paths Verruijt Example)  the stress paths can be plotted as follows:

It’s worth noting that the q axis is unaffected by the drainage condition because the pore water pressures cancel each other out.  Only the p-axis changes.

## Conclusion

The p-q diagram is a method of simplifying the analysis of triaxial and other stress data which are commonly used in soil mechanics.  It can be used in a variety of applications and solve a range of problems.

Posted in Geotechnical Engineering

## An Updated Version of Our Mohr’s Circle Routine Available, with Documentation

In 2016 we posted (and updated two years later) about two- and three-dimensional Mohr’s Circle problems and their solution using a strictly linear algebra solution.  We’ve done two things recently to update that: we’ve resolved some of the limitations of the original method (especially with the eigenvectors/direction cosines) and we’ve put the routine online for your use.  Both can be accessed here: