In our last post on the subject we discussed the concept of resultants as replacements for simplicity of calculations. We’ll expand on the topic here by discussing two examples: surcharge retaining loads and loads on shallow foundations.
Retaining Wall Surcharge Loads
I discuss the topic of surcharge loads behind a retaining wall in my post The Difference Between Flamant’s and Terzaghi’s Solution for Line Loads Behind Retaining Walls. I show that the loads can either be applied to the wall as a non-uniform, distributed load or a single resultant, as is shown below.
Obviously condensing that kind of distributed load into a resultant requires more serious integration that you have for a uniform or triangular load. In the past using a resultant was common in hand solutions of sheet pile problems; with software, if it has the option for something other than uniform surcharge (SPW2006 for example does not) it generally reproduces the distributed load.
One other use of a resultant concerns strip vs. line loads. You’ll notice that the information for line loads is more extensive than strip loads. One reason is that engineers commonly convert a strip (distributed) load into a line (resultant) load by multiplying the width of the load by its pressure and placing the load in the middle of the strip (assuming, of course, the distributed load is uniform.)
Shallow foundations offer an interesting situation because in some ways we reverse the process, starting with a resultant and end up with a distributed load. We’ll restrict the discussion to the simplest case, a continous foundation, in this case under a gravity retaining wall. Other cases are discussed in more detail here.
You may recall that, for a uniform pressure distribution, the load is in the middle of the distribution. If a shallow foundation is loaded concentrically, the load is through the centroid, and this is the case. That is designated by the centreline. If the load is concentric, the eccentricity e = 0 and the two pressures q’min and q’max are both the same, as we see in the equations above, namely the load divided by the width of the foundation.
If the load is eccentric, e is nonzero. As e increases, q’max increases and q’min decreases. When we reach the point where e = B/6, q’min = 0. Beyond that point we get liftoff from one end of the foundation, which is generally not an acceptable result, especially since foundations cannot transmit tension between the soil and the foundation. As long as -B/6 < e < B/6, both q’min and q’max are positive and at least this criterion is met. This region is referred to as the “middle third” and is crucual to the success of shallow foundations. (Food for thought: how does that related to the “triangle rule” that we use with retaining walls?)
Eccentricity in loads is not strictly the result of off-centroid loads. It can also be due to a moment on the foundation, in which case e = M/N’ (without any other factors to make the load eccentric.)
As mentioned earlier, this is the simplest case. Things get complicated when we actually implement this concept, and these are discussed in Foundation Design and Analysis: Shallow Foundations, Bearing Capacity II.