This article has been translated and posted by Desdeelmurete, the original Spanish version can be found in the following link “Empujes del terreno según Rankine-Desdeelmurete.com”.
There were many authors who got involved into the mess of calculating the Earth Pressure, but in this post, we are going to explain only the Rankine's theory. Hope it's useful.
Basically, and even at the risk of oversimplifying and making a mistake, this man, Rankine, considers that when the ground stresses are in the "Mohr's envelope", the ground is at the failure point and therefore completely plasticized.
Earth pressures - Active earth pressure
Let's first look at active earth pressure. As we can see in the Mohr-Coulomb Model, the failure surface will form an angle of π/2 + φ with respect to the vertical stress (the greatest) and π/4 + φ/2 with respect to the horizontal stress, being the family of characteristic lines such that:
Figure 1. Failure surface – Angle respect the horizontal
The value of the earth pressure per linear meter of the wall will be the force exerted by the "wedge" of horizontal tensions at the height of the wall. Attention! The earth pressure is a force. It’s true that we always push on a surface (in this case on a line), but what we exert is a force.
Where: γ is the bulk density and H is the height of the wall. Ka is the active earth pressure coefficient. The following picture provides the order of magnitude. It is also presented the kp (passive earth pressure coefficient) that will appear later. The formula for both can be found in the link “Active and passive earth coefficients - Coulomb and Rankine”.
Figure 2. Active and passive earth coefficient depending on x/H extracted from the CTE (Código Técnico de la Edificación)
So far, we have assumed that the ground has no cohesion. What would happen if it has cohesion? Common sense seems to tell us that the active (destabilizing) ground pressure will be reduced, and the passive (compensating) ones will be increased. Let's see why.
What has been considered is that the active ground pressure produces tractions that generate cracks in the upper part of the ground, so we consider that the ground pressure at the top would be null.
Figure 3. Hypothesis: Active ground pressure produces tractions, appearing cracks and top pressure would be null
Finally, we can define the active ground pressure as:
A second member show up (it is deduced from the analysis of Mohr's circles) that reduces the active ground pressure.
Earth pressures - Passive earth pressure
Similarly, we could define the passive ground pressure, specifying that:
- The greatest stress is the horizontal, not the vertical, having its implications in Mohr's circles.
Figure 4. Mohr’s circle presenting the horizontal stress as the greatest one
- Because of the above: the failure surface will form an angle of π/2 - φ with respect to the vertical stress and π/4 - φ/2 with respect to the horizontal stress. The family of characteristic lines will be such that:
Figure 5. Passive earth pressure characteristic lines
- There are no cracks.
We can, therefore, define the passive earth pressure as:
In this article we have learned two very interesting concepts that came in handy, per example, for Terzaghi:
1. The geometry of the characteristic lines of the active and passive ground pressure.
2. The quantitative assessment of the active and passive ground pressure.
Thank you very much for your time!