math – vulcanhammer.net

Analysing a Gravity Wall Using Vector Analysis

Posted on 6 January 2025 by Don Warrington

One of the reasons I was interested in teaching Statics at Lee University was because I was continually disappointed at my students’ memory of their statics. Statics is crucial in the design and analysis of geotechnical structures, and most of the problems–at the undergraduate level at least–aren’t that involved, or at least I thought they weren’t. A great deal of the problem is that geotechnical statics usually involves converting distributed loads into resultants, which Statics–and Mechanics of Materials for that matter–generally associate this with beam problems, not always the case with geotechnical problems.

Another culprit is that Statics, in the U.S. at least, is a vector proposition from the start. At the University of Tennessee at Chattanooga where I taught, it was called “Vector Statics,” which gives the game away early. (At Lee we use the same book and teach the same material, but simply title it “Statics.”)

But what if we applied a vector approach to a simple geotechnical problem? That’s what we’re going to do here with a concrete gravity wall. I will use the method outlined in my post A Simplified Method to Design Cantilever Gravity Walls. You can refer to the theory there, I will try to keep it to a minimum. The wall is pictured at the top of the post, I will reproduce it below.

Analysing Overturning

We have three forces acting on the wall:

The weight of the gravity wall itself, W_conc
The weight of the soil trapped by the heel of the wall, W_soil
The lateral force of the soil on the wall, F_h

Forces 1 and 2 are determined by computing the cross-sectional area of the concrete and soil and multiplying each by the unit weight as shown above, and then converting the result to a vector force and placing it at the centroid of the area (another Statics topic.) Instead of the “manual” approach in A Simplified Method to Design Cantilever Gravity Walls, the was was drawn in CAD and both the areas and centroids were determined automatically. You can see the magnitudes and locations of those resultants above.

The lateral force of the soil is computed using Rankine’s theory. The first thing is to determine the working internal friction angle of the soil by applying the Shear Mobilisation Factor SMF. Assuming an SMF of 2/3, that friction angle changes from the 30 degree one shown above to a 21.05 degree one, which is applied to the formula for Rankine active pressures for level backfill,

$k_h = \frac{1-sin\phi}{1+sin\phi}$ (1)

Doing that results in a k_h = .471. The force on the wall is then determined by the formula

$F_h = \frac{k_h\gamma_{soil}H^2}{2}$ (2)

The division by two reflects the fact that soil effective stress (and thus lateral earth pressure) increases linearly with depth (like a fluid,) creating a triangular distribution (yet another concept from Statics.)

At this point there the resisting forces R and T are not defined. The forces themselves are easily computed by summing forces in the x and y directions. Doing this, we have

$R = -W_{conc}-W_{soil}$ (3a)
$T = -F_{h}$ (3b)

The location of F–along the surface of the footing–is evident. The location of R is not; it is some distance x from the toe (Point “A”) of the footing. We can obtain x by summing moments around Point “A,” and with a vector method that means taking cross products of the moment arms with the forces.

Converting both the moment arms r and the forces to vector notation yields the following:

Concrete Weight: r = 2.358 i + 4.049 j, W_con = −4.35 j
Soil Weight: r = 5.576 i + 8.264 j, W_soil = −6.30 j
Lateral Earth Pressure: r = 8 i + 4 j, F_h = −4.073261616 i
Vertical Footing Force: r = x i, R = 10.65 j (Equation (3a))

The force T does not enter into this because its line of action runs through Point “A,” thus its moment is zero as its moment arm is zero.

The cross product moments around the toe (Point “A” in the drawing) are as follows:

Concrete Weight: $\left [\begin {array}{ccc} i&j&k\\{\medskip} 2.358& 4.049&0 \\{\medskip}0&- 4.35&0\end {array}\right ] = -10.257 k$
Soil Weight: $\left [\begin {array}{ccc} i&j&k\\{\medskip} 5.576& 8.264&0\\{\medskip}0&- 6.3&0\end {array}\right ] = -35.129k$
Lateral Earth Pressure: $\left [\begin {array}{ccc} i&j&k\\{\medskip}8&4&0\\{\medskip}- 4.073&0&0\end {array}\right ] = 16.293k$
Vertical Footing Force: $\left [\begin {array}{ccc} i&j&k\\{\medskip}x&0&0\\{\medskip}0& 10.65&0\end {array}\right ] = 10.65 x k$

Summing these moments,

$29.093\,k+ 10.65\,xk=0$ (4)

Solving yields x = 2.732′. At this point we need to determine whether this is an acceptable location or not for the force. The goal is for the pressure to be positive (downward) along the entire surface of the footing. There are two ways of determining this:

Computing the pressures on each end of the footing, as is done in A Simplified Method to Design Cantilever Gravity Walls.
Determining whether the force is in the “middle third” of the foundation, as is discussed in the Soils and Foundations Reference Manual.

We will do the latter. The middle third of this foundation falls between 2.67′ < x < 5.33′, so the vertical footing force is within the middle third (barely.) As I noted in A Simplified Method to Design Cantilever Gravity Walls, “In this case we make a common assumption that, as long as the resultant force of the wall is within the kern and there are no negative pressures on the base, overturning will not be experienced. It is certainly possible to do an explicit overturning analysis to check this result.”

Analysing Sliding

With the lack of keys or deep foundations, the only lateral resistance to sliding is the friction force T. We computed that force based on Equation (3b,) but in reality that force cannot exceed–and there should be a factor of safety in that inequality–the frictional force possible, which is defined by the equation

$T_{max} = \mu R$ (5)

in which case

$FS_{Sliding} = \frac{T_{max}}{T}$ (6)

Equation (5) is written in “mechanical engineers format.” Geotechnical engineers understand all too well the concept of a friction angle. In my post Explaining the Relationship Between the Coefficient and the Angle of Friction I relate the two from a non-geotech standpoint; we can turn Equation (5) into a more “geotech-friendly” form by noting that

$\mu = \tan\phi$ (7)

Let us assume that the value of $\phi$ is the same under the wall as next to the wall, and let us also assume that the friction angle between the base and the soil is the same as the friction angle of the soil overall, as was done in A Simplified Method to Design Cantilever Gravity Walls. That being the case, $\mu = \tan(30) = 0.577$ . Substituting into Equation (5,) $T_{max} = 0.577 \times 10.65 = 6.15\,kips$ . The factor of safety from Equation (6) is thus $FS_{Sliding} = \frac{6.15}{4.073} = 1.51$ , which is barely over the minimum criterion for usual loads given in A Simplified Method to Design Cantilever Gravity Walls.

Observations

The use of vectors for this problem is overkill from a computational standpoint. It also requires locating the centroid/CG of the two regions in both the x- and y-directions, although with using CAD this is trivial. On the other hand doing it using vectors is more “bullet proof” in that the student is not required to “think” but just “plug and chug” without having to identify lines of action and perpendicular moment arms.
The fact that the word “barely” appears in both analyses should inspire some additional conservatism in the design. The simplest way to improve the situation would be to move the heel to the right, which would shift the resisting forces away from the toe (and thus increase their resisting moment) and also put the footing force resultant deeper into the middle third.
Both bearing capacity and settlement of the wall’s foundation, the methodology for which are discussed in A Simplified Method to Design Cantilever Gravity Walls, are beyond the scope of this post. Also beyond the scope of this post is the structural design of the wall and of course the global stability of the wall as well.

Explaining the Relationship Between the Coefficient and the Angle of Friction

Posted on 28 October 2024 by Don Warrington

One of the things that gets covered (if not very thoroughly) in Soil Mechanics is how friction is developed in soils. An analogy is made with the classic “block on a surface” problem we see in Statics, but the tie-in isn’t as strong as one would like.

The fact is that, for purely cohesionless soils, the friction between the particles and the friction between the surface and the block is basically the same Coulombic friction. As is usually the case in soil mechanics, how that actually plays out in soil properties has many complexities, but then again surface friction isn’t a simple or straightforward property in and of itself.

Another part of the problem is that, in Statics, friction isn’t taught with geotechnical considerations in mind, especially these days. This is a pity, not only for those of us in the geotechnical community but for those who work with granular materials on a production or use basis.

This is a brief treatment of the subject, basing the development of the topic from that in Movnin and Izrayelit (1970), which comes closer to relating the two quantities we see to define friction: the friction angle and the friction coefficient.

The Basics of the Friction Coefficient and Angle

Surface friction comes from the rubbing of two surfaces together, as shown at the right. We see the three forces with which the two surfaces interact: the normal force N, the resulting friction force F and the resultant of the two R. We also see that the addition of lubricant is important in that it separates the two surfaces and reduces the effect of the asperities on each other, something that contractor and engineer alike frequently overlook in both the maintenance and performance evaluation of the equipment.

The normal and frictional forces resisting the relative motion of the two surfaces is related by the equation

F = fN (1)

With granular materials, the main difference is that the surfaces of the particles aren’t straight at all but they do rub up against each other, the asperities on the particle surface contributing to the mutual resistance of the particles. Although water acts to a limited extent as a lubricant, its largest effect is the buoyant effect on the intergranular (effective) stress, as shown below.

Returning to the first diagram, without any mutual pressure of the surfaces (the normal force N) there is no friction force F tangential to the surface. Again in soil mechanics purely cohesionless (granular) soils have no frictional strength unless weight or other pressure is applied to them.

Diagram of forces on a body on a plane surface with friction, from Movnin and Izraelyt (1970)

Now let us consider the diagram at the right. The normal force N exerted by the surface on the block (caused by the force exerted on the block Q) and the frictional force F (caused by the force P which attempts to move the block) add vectorially to a resultant R, which in turn has an angle with the normal force N. The geometry of the forces and Equation (1) relate the angle to the friction factor as

f = F/N = tan (φ) (2)

Cone of Friction

Although F and N are related through both Equations (1) and (2), in reality F cannot exist without some tangential force pushing the block. This is the force P which is attempting to push the block along the plane. As P increases F increases until we get to a point where we have impending motion, beyond which the block moves and begins to accelerate. The value of f or φ when impending motion turns into actual motion is when we reach the ultimate value of f or φ, which we will designate as f₀ or φ₀.

These form a “cone of friction.” This cone of friction can be seen in the diagram at the left. As long as F < f₀ N (or F < tan (φ₀) N) and the resultant Q of N and F is within the cone, the block is motionless. Beyond that point it moves, and the coefficient of friction in motion can be different (usually smaller) than the coefficient of friction at the point of impending motion.

It is here that we can relate the friction factor f and the angle of friction φ can be related to each other and to concepts familiar to geotechnical people. When we construct the Mohr-Coulomb diagram, we define a failure envelope of legal stress states (within the envelope) and illegal stress states (outside the envelope.) We can see all of these with the failure function below. When the failure function is negative (1), we are within the envelope and failure does not take place. When the failure function is zero (2), we have impending failure. When the failure function is positive (3), we have failure and an illegal stress state.

Three-dimensional envelopes are certainly common in geotechnics, especially in finite elements. An example of this is shown below.

Determining the Friction Factor or Angle

Determining the angle of friction, from Movnin and Izraelit (1970)

To determine the friction angle, one simple way is to start with a block and a level surface and then raise the angle of the surface until the block moves. Such an apparatus is shown at the left.

As the angle α increases the direction of the weight G relative to the surface changes in can be divided into two parts: the normal force G₂ and the tangential force G₁. The latter will move the block down but it is resisted by the friction force F, which will resist until G₁ > F₀, at which point the block will start to move down the slope at a constant acceleration. By noting the angle at which this takes place, both f₀ or φ₀ = α₀ can be determined. The math for this is similar to the level surface and block.

The geotechnical counterpart to this is the angle of repose. Suppose we allow a small stream of sand to drop on a surface. Over time the sand will build up into a conical pile with the surface at an angle to the flat surface the sand is streamed onto. This angle is referred to as the angle of repose. In theory the angle of repose is equal to the friction angle of the soil, although with the usual complexities of geotechnics this isn’t always the case. There are clean sands with which we can use the angle of repose to estimate the internal friction angle of the soil. When I was teaching at UTC, some of the students were working on the ASCE MSE Wall project and needed a friction value for the sand being used in the box. While they were looking at direct shear or triaxial testing, I suggested using the angle of repose to get a “ballpark” value. They did this and it was helpful.

Some Comments

The use of the angle of friction has fallen out of favour in engineering education, which is one reason why it is difficult to relate friction as taught in Statics to friction as used in geotechnical engineering. That wasn’t always the case; one example from the early twentieth century is Tapered Keys and Their Use In Vulcan Hammers.
Hopefully this treatment of the subject will be useful to students to help them relate the concept of friction in statics to that in geotechnical engineering.

Academic Issues, Geotechnical Engineering

The 2:1 Method for Estimating Stresses Under Foundations

Posted on 29 April 202416 June 2025 by Don Warrington

Readers of my post Analytical Boussinesq Solutions for Strip, Square and Rectangular Loads and those related to it know that the math related to these methods can get complicated, and in any case the idea of a “purely elastic” soil response to load is purely theoretical. So is there a simpler way? The most common simplification used is the 2:1 Method, shown at the right.

The method basically assumes the following:

Uniformly loaded foundation
Only stresses of interest are under the centroid of the foundation, as shown, and only vertical stresses are considered
Stresses decrease as if there is a truncated pyramid below the foundation with a slope of 1H:2V.

As shown, to determine the additional vertical stress $\Delta \sigma_v$ at a distance $z$ below the centre of the foundation, the equation for a rectangular foundation of width B and length L (can be interchanged, but conventionally B is the smaller of the two) for a load Q on the surface is given by the equation

$\Delta \sigma_v = \frac{Q}{(B+z)(L+Z)}$ (1)

If the equation is written in terms of the unit load q, it becomes

$\Delta \sigma_v = \frac{q}{(1+\frac{z}{B})(1+\frac{z}{L})}$ (2)

Obviously for square foundations $B=L$ but the solution is the same.

Formulating the problem as shown in Equation (2) also allows us to apply the 2:1 method to continuous foundations, for as $L \rightarrow \infty$ the equation reduces to

$\Delta \sigma_v = \frac{q}{1+\frac{z}{B}}$ (2a, Continuous Foundations)

As an example, let us consider a foundation where $Q=100 kN,\,B=5m,\,L=8m,\,z=3m$ . By substitution into Equation (1), $\sigma_v = 1.14\,kPa$ . If we compute the unit load to be $q = \frac{100}{(5)(8)}=2.5\,kPa$ , substitution into Equation (2) yields the same result.

Although it is possible to apply this method to circular foundations, it is just as easy to use Boussinesq theory under the centre of the foundation. The equation for a circle of radius $r$ and all other variables the same is

$\Delta \sigma_v = q(1 - \frac{z^3}{(z^2+r^2)^\frac{3}{2}})$ (3)

The 2:1 method is the only method taught in Soils in Construction. It is generally used with shallow foundations, but can also be applied to pile groups in clay, as shown at the right. It’s simplicity and reasonable accuracy has brought it acceptance, and further information on it can be found in the Soils and Foundations Reference Manual. The concept has also been applied to pile toes, and you can see this in STADYN Wave Equation Program 10: Effective Hyperbolic Strain-Softened Shear Modulus for Driven Piles in Clay.