Problem: To determine the lateral (horizontal) force exerted on a retaining wall by the backfill behind it, under specified conditions. You will be going through a simplified version of the original work by C. A. Coulomb in the 18th Century. That was before he started studying electricity.*

Figure 1 shows a concrete retaining wall of height H. It is called a gravity wall because it depends only on its weight for stability. The zone behind it is backfilled with clean sand. Your common sense will tell you that the sand will exert some lateral force on the wall. You can also see that the only cause of this lateral force has to be the weight of the retained sand, or the force of gravity.

If the sand were a fluid, it would be a simple problem of hydrostatics. The lateral pressure would be the same as the vertical pressure. It would increase directly with depth, so that a diagram of pressure would be a triangle. As will be developed further on, the sand will also cause a triangular horizontal pressure. However, it has an internal strength which makes the lateral pressure less than the vertical pressure. You can conceptualize this from common sense.

Special conditions, to make the problem easier to solve, are:

  1. The wall rests on a smooth rock surface, on which it can slide. Assume that no friction is mobilized along the bottom of the wall. In other words, the wall can move ahead of the sand and will not resist it, although obviously the wall wonít move any more than it has to. Under these conditions, the only force the sand can put on the wall is that due to its weight and in excess of what it can restrain by its own internal strength. This is the minimum possible force the sand can put on the wall, called by engineers the active pressure. The wall has to move only a very small amount, smaller than you can see, to reduce the lateral pressure to this minimum. Retaining walls are often designed to resist only this force, because that makes them economical. Obviously, this would not be acceptable if even small movements of the wall could not be tolerated. You can also see intuitively that if the wall could not move, called the at-rest condition, the pressure would be greater than for the active condition. It would greatly increase if the wall were pushed into the soil, called the passive condition. But for this problem, we will stick to the minimum, or active, pressure.
  2. The sand is dry, so you wonít have to consider any water pressure.
  3. The back of the wall is vertical and very slippery, so the sand canít put any upward or downward frictional force on the wall. It has to push horizontally.
  4. The backfill surface is level for a long distance back from the top of the wall.
  5. The dimension of the wall perpendicular to the paper is one foot. In effect, the forces you determine will be the forces on each lineal foot of a very long wall.
  6. The sliding surface (discussed below) is a plane. This has been found not to be true, but it makes the geometry easier to deal with and causes very little error in the results.

 


Figure 2 illustrates the mechanism by which, for this analysis, the sand crowds against the wall. A plane of sliding forms at some angle theta with the horizontal. The sand between this plane and the wall forms a sliding wedge, and slides down and toward the wall, causing the wall to displace to the left. The amount of movement is greatly exaggerated for illustration.

Figure 3 shows the same thing as figure 2, except that it called a free-body diagram. The total, or resultant, force from the sand is represented by P. The rollers between P and the wall represent the slippery condition on the back of the wall, assuring that P can only be applied perpendicular to the wall. The rollers under the wall represent the concept that it can easily move ahead of the sand.

 

 

Figure 4 shows a free-body diagram of the sliding wedge, not the wall, and the forces acting on it. You may not have had this in physics yet, but you can get enough from these sheets to solve the problem. You will study this in more depth in E. Mech. 011, Statics, next year. The forces are:

  1. P, which is the same as the P of Figure 3, except that it is the force the wall exerts on the sand. For equilibrium, the wall and the sand must exert the same force on each other.
  2. W, which is the force due to the weight of the sand. This simply the area of the wedge times the sand density, gamma, pounds per cubic foot. W must of course act vertically.
  3. R, which is the sandís internal frictional resistance to sliding. R acts at an angle phi with the perpendicular, or normal, to the plane of sliding. More about R and phi in Figure 5.

Figure 5 is inserted to show to show the significance of the angle phi. It is called the angle of internal friction and is an important parameter for geotechnical engineers. It is more or less unique to each type of soil, kind of dependent on the roughness of the individual soil grains. It is determined through laboratory tests. Values for sand are in the range of 25 to 35 degrees.

 

Figure 5 is drawn with the slope angle theta equal to phi, and that the wedge is sitting there with no force P holding it. At this angle, R = W which means the friction just balances the sliding force. R can be broken down into two components: N perpendicular to the sliding surface and F parallel to it. The trig relations between W, N, and F are shown. If you have had the sliding block friction part of physics then you will recognize a similar situation here. Obviously, if theta is less than or equal to phi, no sliding will occur and there will be no P. As theta increases beyond phi, W will overcome R, the block will slide, and P will develop. R. can only resist so much because it is limited by phi. Plus, as theta increases, N and F decrease. However, as discussed further on, the highest theta may not give the greatest P because then the sliding wedge, and W, get smaller.

 

Figure 6 is like Figure 4, except that it shows more detail. You will learn next year in Statics that, for equilibrium, the sum of the forces in any direction must equal zero. For convenient analysis, we will consider forces in the horizontal (x) and vertical (y) directions. R has to be broken down into its vertical and horizontal components. P and W donít have to be broken down because they act only horizontally and vertically, respectively. The following equations express this and enable solution for P in terms of W, phi, and theta.

For equilibrium:

Sum of the vertical forces = 0,

so W = R cos( q Ė f ) can be solved for R in terms of W, q, f.

Sum of the horizontal forces = 0,

so P = R sin( q Ė f ) and R is known and can be solved for P in terms of W, q, f.

W = A g, where A = Area of the sliding wedge, and g = Density of sand.

In Statics you will also learn that the sum of the moments about any point must also be zero. If it were true here, it would give another equation to work with. Itís not exactly true here, but we donít need another equation and not much error is introduced if we ignore it. This is an example of approximations geotechnical engineers have to make if they are to arrive at practical answers, making this kind of engineering as much an art as a science.


Figure 7, introduces an additional variable, X. X is obviously dependent on theta and H, but it will be a convenient variable to use in the trial procedure discussed below. You can see that there are many possible Pís, one for each possible theta. Also, there wonít be any P unless theta is greater than phi. As mentioned above, you can see intuitively that, as theta rises above phi, P will increase because the failure plane slopes more steeply. However, as theta approaches 90 degrees the sliding wedge of sand gets very small, reducing W, and P decreases. P reaches a maximum somewhere in between and thatís what you have to find.

First, work with the equations to get one for P in terms of phi, gamma, H, theta, and X. Then write a routine to make a series of trial solutions for P, with plugged values for phi, gamma, and H, and letting X increase. Essentially, the routine should calculate theta, and then P, for each trial X. Some good values to work with are: phi between 25 and 35 degrees, gamma between 90 and 130 pounds per cu. ft., and H between 8 and 16 feet. Trial X should start at 0.5 ft., increase by 0.5 ft. increments, and stop as the increasing X causes theta to reduce to phi. Itís important to use a small X increment so you donít miss the peak.

As implied above, X was introduced as the key variable when it would appear simpler to just vary theta, but it is easier that way. It will especially be easier to use X as the variable in the more complicated (optional) case at the end where the backfill slopes up behind the wall.

The routine should stop each time when it finds the maximum P. To do this have the computer compare P with the P from the previous trial, and select the previous P when the new P gets to be smaller.

 

 

Once you get your routine working, you can in a very short time find maximum Pís for a wide range of phiís gammaís, and Hís. Print out a table with the following columns: phi, gamma, H, X, theta, and maximum P. The X which goes with the maximum P will give you an idea of the size of the sliding wedge.

As noted earlier, like fluids, soils under these conditions produce triangular lateral pressure diagrams. You could demonstrate this using the theory you are developing, but it would make the problem too long, so just accept it for now. Since P is the resultant of the pressure, it equals the area of the pressure diagram. It acts at the centroid of pressure, which for a triangle is one-third up from the base. You are probably working on this in Physics now, or soon will be.

All this allows dimensioning of Figure 8 as shown. The parameter k is the slope of the pressure diagram, and is called the coefficient of earth pressure. It is an important concept, because, if you knew k, you could quickly develop the pressure diagram and find maximum P for any H and gamma. The coefficient k can be thought of as the ratio between the horizontal and vertical pressure at any point down the back of the wall. Typical kís for the active condition are between 0.2 and 0.4.

Modify your program to add a calculation for k each time after you find the maximum P, and add a column for k in the table. This shows that, with a given set of conditions, k depends only on phi. See how it works out.

This last part is optional. If this has been a very easy concept to understand and no trouble to program, try it with a sloping backfill behind the wall, as shown in Figure 9. Finding the wedge area is no more complicated than for the horizontal backfill case. Finding theta in terms of X is a little more complicated. This is done partly to illustrate why X is easier to use as a variable than theta. You should find that k increases with beta, which makes sense.

* This problem was designed by Ralph E. Curtiss, P.E., Allegheny Power, Greensburg, Pa


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