When dealing with geotechnical problems often is required that the initial stress state in the soil has to be known. In order to define the initial state at rest, the coefficient K0, called coefficient of earth pressure at rest, has to be calculated.
The best way I could think about to explain this coefficient is to have an analogy with the water. Let’s consider a swimming pool: the pressure on the wall at a specific depth y is given by the specific weight of the fluid multiplied by the depth. In reality, in this way we have found the vertical pressure, but because in a fluid such as water the pressure is the same in all directions, we have found also the horizontal pressure.
Now, what happens in the case of a soil?
Unlike water, the soil possesses resistance to shearing. This resistance leads to a reduced horizontal pressure. The more we have resistance to shearing, the less we have horizontal pressure.
How to calculate the horizontal pressure?
In order to calculate the horizontal pressure, we need to take into account the discussed resistance. This can be done using the coefficient of earth pressure at rest, as coined by Terzaghi (1920). It can be shown to be some multiple of the vertical stress at any points. Basically, it is accepted as the horizontal-to-vertical stress ratio in loose deposits and normally consolidated clays.
To define the relation, let Es and ν be the modulus of elasticity and Poisson’s ratio of the soil respectively. The lateral strain is:
But we know that:
Therefore, we can define the horizontal pressure as:
We can now define the quantity in the brackets as Ko, namely coefficient of earth pressure at rest:
The relation between horizontal and vertical pressure become simply:
If E, ν and γ are constants – hence, the soil properties are constants – the distribution of earth pressure at rest with depth is linear (hydrostatic nature).
If a retaining wall won’t yield – hence, there are no active or passive conditions – the total thrust on the wall is given by:
This thrust acts at 1/3 H above the base of the wall.
The problem in the application of the above definition is given by ν: choosing an appropriate value is not easy. This is a limit of the Terzaghi’s relation.
To solve these difficulties, various researchers proposed empirical relationships for K0, such as:
Jaky, 1944 – theoretical solution
The above was published in J. Jaky’s paper “The coefficient of earth pressure at rest”.
Jaky, 1944 – simplified solution
The above is a simplified version of the previous theoretical solution. The differences are of around 9 percent at low friction angles up to 16 percent at high friction angles. However, according to Wroth (1972), “considering the difficulty of making an appropriate choice for Φ’ for a given soil, this approximation is sufficiently accurate for most engineering purposes“.
Brooker and Ireland, 1965
The above solution from Brooker and Ireland is similar in appearance to Jaky’s simplified equation, but closer in results to his theoretical one.
In the above equations Φ’ represents the effective angle of friction of the soil and Ip the plasticity index.
In 1965, Brooker and Ireland recommended Jaky’s equation for cohesionless soils and their own equation for the cohesive soils.
In 1967, Alpan recommended Jaky’s equation for cohesionless soils and Kenney equation for cohesive soils. Kenney suggested the same in 1959.
The literature suggests that Jaky’s simplified equation provides the best agreement with experimental results from tests conducted on normally consolidated soils.
In the following, some values for the coefficient K0:
– Dry loose sand (e = 0.8) => K0 = 0.64;
– Saturated loose sand (e = 0.8) => K0 = 0.46;
– Dry dense sand (e = 0.6) => K0 = 0.49;
– Saturated dense sand (e = 0.6) => K0 = 0.36;
– Sand (compacted in layers) => K0 = 0.80;
– Soft clay (Ip = 30) => K0 = 0.60;
– Hard clay (Ip = 9) => K0 = 0.42;
– Undisturbed silty clay (Ip = 45) => K0 = 0.57.
For any issues or questions, you can contact the author at:
Eng. Onorio Francesco Salvatore