3.11 Critical State Models for Soils

 

Soils consist of a two phase mixture of particles and water.  They exhibit very complex behavior in response to stress, and a number of different constitutive theories are used to model them $–$ in fact, entire books are devoted to critical state soil models.  Here, we outline a basic soil model known as `Cam-clay,’ developed at Cambridge (UK).   It uses many of the concepts that are used to model plastic deformation of metals, so you will find it helpful to review Sections 3.7 and 3.8 before reading this one.

 

 

 

3.11.1 Features of the behavior of soils

 

1. Soils cannot withstand significant tensile stress: we therefore focus on their response to combined pressure and shear loading. 

 

2. The behavior of a soil is very sensitive to its water content.   Two types of experiment are conducted on soils: in a `drained’ test, water is allowed to escape from the specimen as it is compressed (so the water pressure is zero); in an `undrained’ test, the volume of the specimen (water + soil particles) is held fixed.   In the latter test, the water pressure can be measured by means of a manometer connected to the pressurized cell.

 

3. Under combined pressure and shear loading, soil behaves like a frictional material.  In a drained test, the solid can support shear stresses $\tau <Mp/\sqrt{3}$ without excessive deformation, where M is a material property (analogous to friction coefficient).  If the shear stress reaches $\tau =Mp/\sqrt{3}$, the material collapses $–$ usually by shearing along one or more discrete shear planes parallel to the maximum resolved shear stress.  If the same test is conducted on an undrained specimen, shear failure occurs earlier, because the water supports part of the hydrostatic pressure.   In this case shear failure occurs when $\tau <M(p-p_w)/\sqrt{3}$, where p is the applied hydrostatic pressure, and $p_w$ is the water pressure.

 

4. If subjected to loads below those required to cause catastrophic collapse, soils show a complicated behavior that resembles that of a strain hardening metal $–$ except that soils compact in response to combined shear and pressure, whereas metals do not. In addition, the strain hardening occurs only as a result of the compaction: shear strain does not increase the solid’s strength.  For example, the figure shows the response of a soil sample to a test where in which specimen is subjected to a constant pressure, together with a steadily increasing shear stress.   The soil accumulates a permanent shear strain, and also compacts.   The strength of the solid increases up to the limiting shear stress $\tau =M(p-p_w)/\sqrt{3}$, at which point the compaction reaches a steady state, and the specimen continues to deform at constant shear stress.   

 

 

 

 

3.11.2 Constitutive equations for Cam-Clay

 

The constitutive equations for Cam-clay are very similar to the rate independent plastic equations in Section 3.7.  The main concepts are,

 

1. Strain rate decomposition into elastic and plastic parts;

 

2. Pressure decomposition into contributions from the water pressure (or ‘pore pressure’) and from the pressure supported by the soil particles.  The pore pressure must be calculated by modeling fluid seepage through the soil.

 

3. Elastic stress-strain law, which specifies the elastic part of the strain in terms of stress;

 

4. A yield criterion, which determines the magnitude of the plastic strain rate, given the stresses and the resistance of the material to flow.  Unlike metals, the yield criterion for a soil is a function of the hydrostatic stress, or pressure, in addition to shear stress. The yield criterion is expressed in terms of a State Variable which characterizes the resistance of the material to flow (analogous to yield stress). 

 

5. A Plastic flow rule that specifies the ratios of the plastic strain components under multi-axial stress

 

6. A hardening law, which specifies how the state variable evolves with plastic straining

 

7. The yield surface, flow rule and hardening law also define a critical state criterion for the solid.  The critical state criterion specifies the combination of stresses that lead to unconstrained collapse of the solid.

 

These are discussed in more detail below.

 

 

 

Strain rate decomposition

 

We assume small strains, so shape changes are characterized by ${\epsilon }_{ij}=\left(\partial u_i/\partial x_j+\partial u_j/\partial x_i\right)/2$. The strain is decomposed into elastic and plastic parts as

${\epsilon }_{ij}^{}={\epsilon }_{ij}^e+{\epsilon }_{ij}^p$

 

 

Pressure Decomposition

 

Assume that the soil is subjected to a stress ${\sigma }_{ij}$.  The pressure is $p=-{\sigma }_{kk}/3$ (note the negative sign).  In general, part of this pressure is supported by the water in the soil; while the rest is supported by the soil particles themselves.  The pressure is decomposed into two parts

$p=p_w+p_s$

where $p_s$ is the contribution to the pressure from the soil particles and $p_w$ is the contribution from the water.

 

When using the constitutive equation in a boundary value problem, the water pressure must be calculated as a separate problem, in addition to solving the usual mechanical field equations.  Here, we outline briefly a simple approximate description of fluid seepage through a soil.  More general treatments are also available, which include nonlinear versions of the flow law, finite strain effects, as well as the effects of fluid absorption by the soil particles to form a gel, the tendency of soil to absorb fluid due to capillarity, and the effects of partial soil saturation.

 

1. Fluid seepage through the soil is driven by gravity and by fluid pressure variations.  The driving force is quantified by the piezometric head, defined as

$\varphi =z+{\displaystyle \frac{p_w}{{\rho }_{w}g}}$

where z is the height above some arbitrary datum, $p_w$ is the fluid pressure (compressive pressure is positive); ${\rho }_w$ is the fluid density; and g is the acceleration due to gravity.

 

2. The volume of material flowing through unit area of solid in the $x_i$ direction per unit time obeys Darcy’s law

$q_i=-k{\displaystyle \frac{\partial \varphi }{\partial x_i}}$

where k is a material parameter, known as the permeability of the medium.

 

3. The fluid itself may be compressible, with bulk modulus $K_w$

 

4. The fluid can be absorbed in cavities in the soil.   The volume fraction of cavities n is defined as

$n={\displaystyle \frac{d{V}_c}{dV}}$

where $d{V}_c$ is the cavity volume in a volume of soil (including both cavities and soil particles) dV.

 

5. At time t=0 the solid starts with some cavity volume fraction $n_0$; this volume fraction evolves as the solid is deformed.  Usually the dominant contribution to the cavity volume change occurs as a result of plastic compaction of the soil (more sophisticated treatments include an elastic contribution).  The cavity volume fraction after the solid is subjected to an infinitesimal plastic strain ${\epsilon }_{ij}^p$ is

$n=n_0+{\epsilon }_{kk}^p$

 

6. At time t=0, a (possibly zero) fluid pressure $p_{w0}$ acts on the solid; and for t>0 the values of either fluid pressure, or volumetric flow rate must be specified on the boundary of the solid.

 

7. Finally, the rate of change of fluid pressure follows from conservation of fluid volume as

${\displaystyle \frac{n}{K_w}{\displaystyle \frac{\partial p_w}{\partial t}={\displaystyle \frac{\partial q_i}{\partial x_i}+{\displaystyle \frac{dn}{dt}}}}}$

 

 

Elastic constitutive equations

 

The elastic strains are related to the stresses using the standard linear elastic stress-strain law.   The elastic strain is related to stress by

${\sigma }_{ij}=C_{ijkl}{\epsilon }_{kl}^e$

where $C_{ijkl}$ are the components of the elastic compliance tensor.  For the special case of an isotropic material with Young’s modulus $E$ and Poisson’s ratio $\nu$

${\sigma }_{ij}={\displaystyle \frac{E}{1+\nu }\left\{{\epsilon }_{ij}+{\displaystyle \frac{\nu }{1-2\nu }{\epsilon }_{kk}{\delta }_{ij}}\right\}}$

 

 

Yield criterion and critical state surface

 

The yield criterion specifies the stresses that are required to cause plastic flow in the soil.  The concept is identical to the yield criterion used in metal plasticity, except that, unlike metals, hydrostatic pressure can cause yield in a soil.  The yield criterion is

$f({\sigma }_{ij})={\left({\displaystyle \frac{p_s}{a}-1}\right)}^2+{\left({\displaystyle \frac{{\sigma }_e}{Ma}}\right)}^2-1=0$

where

 

1.  $p_s=-{\sigma }_{kk}/3-p_w$ is the pressure exerted by the stresses on the soil skeleton.

 

2.  ${\sigma }_e=\sqrt{3S_{ij}{S}_{ij}/2}$ where $S_{ij}={\sigma }_{ij}-{\displaystyle \frac{1}{3}{\sigma }_{kk}{\delta }_{ij}}$ is the Von-Mises effective stress;

 

3.  M is a material property, whose physical significance was described in Sect 3.11.1.  Usually M<1.

 

4.  a is a state variable that quantifies the current yield strength of the soil. At time t=0, the soil has some finite strength $a=a_0$, which subsequently evolves with plastic straining, as described below.

 

 

The yield criterion is sketched in principal stress space in the figure below $–$ it resembles a football (if you are American $–$ or a rugby ball to the rest of the world) with its axis parallel to the line ${\sigma }_1={\sigma }_2={\sigma }_3$.  The shape of the football depends on M $–$ for M=1 it is a sphere; for M<1 it is stretched parallel to its axis.  The size of the yield locus is determined by a.

 

 

The yield criterion, together with the hardening law, also define a critical state surface, which determines the stress where unrestricted shear deformation can occur at constant shear stress (i.e. with zero hardening).  If the stresses lie inside the critical state surface (this is known as the `wet’ side of critical state), the material shows stable strain hardening behavior.  If the stresses lie outside the critical state surface (known as the `dry’ side of the critical state), the material softens with plastic straining, and so violates the Drucker stability condition.  Under these conditions the material is unstable, and plastic strain tends to localize into shear bands.

 

The critical state surface for Camclay is

$g({\sigma }_{ij})={\sigma }_e-M{p}_s$

The material is stable for $g<0$ and unstable for $g>0$.  The critical state surface is sketched in the figure: it is a cone, which cuts through the fattest part of the yield surface.

 

 

Flow law

 

The flow law specifies the plastic strain components under a multiaxial state of stress.  Like metal plasticity, the Cam-clay model bases the flow law on the yield criterion, so that

$d{\epsilon }_{ij}^p=d\lambda a{\displaystyle \frac{\partial f}{\partial {\sigma }_{ij}}=d\lambda \left(-{\displaystyle \frac{2}{3}\left[{\displaystyle \frac{p_s}{a}-1}\right]{\delta }_{ij}+3{\displaystyle \frac{S_{ij}}{M^{2}a}}}\right)}$

where $d\lambda$ is a dimensionless constant that depends on the increment of stress applied to the solid, and is proportional to the plastic strain magnitude. The procedure to calculate $d\lambda$ is discussed in more detail below: if the stress state lies inside the critical state surface  ( $g({\sigma }_{ij})<0$ ), then $d\lambda$ can be expressed in terms of the stress increment $d{\sigma }_{ij}$ and the fluid pressure increment $d{p}_w$ applied to the solid as

$d\lambda =\left\{\begin{array}{c}0f({\sigma }_{ij})<0\\ {\displaystyle \frac{〈-{\displaystyle \frac{1}{3a}\left[{\displaystyle \frac{p_s}{a}-1}\right]\left(d{\sigma }_{kk}+3d{p}_w\right)+{\displaystyle \frac{3}{2}{\displaystyle \frac{S_{ij}d{S}_{ij}}{{\left(Ma\right)}^2}}}}〉}{2c{\displaystyle \frac{p_s}{a}\left[{\displaystyle \frac{p_s}{a}-1}\right]}}f({\sigma }_{ij})=0}\end{array}\right.$

Here $〈x〉=x$ for $x>0$ and $〈x〉=0$ for $x<0$, and c is a material property governing the hardening rate, as defined below.  These expressions are valid only if the stress lies inside the critical state surface.  If the stress lies on or outside the critical state surface $g({\sigma }_{ij})\ge 0$ then $d\lambda$ cannot be determined from the stress increment. 

 

 

Hardening law

 

A soil becomes stronger if it is compacted to crush the soil particles together.   This is described in the constitutive law by making the state variable a evolve with plastic straining in some appropriate way.  A simple hardening law that captures the main features of experiments is

$da=-cad{\epsilon }_{kk}^p$

where c is a dimensionless material property, which determines the hardening rate.  Notice that, in this law, hardening occurs only as a result of compaction, and not as a result of shear deformation.

 

 

Calculating the plastic stress-strain relation

 

When using the constitutive equation, the formulas outlined in the preceding sections must be combined to predict the plastic strain $d{\epsilon }_{ij}^p$ resulting from an increment in stress $d{\sigma }_{ij}$ and fluid pressure increment $d{p}_w$.  This is done as follows:

 

1. Check the yield criterion.  If $f<0$, the plastic strain is zero $d{\epsilon }_{ij}^p=0$.

 

2. Check to see whether the stresses lie inside the critical state surface.  If $g({\sigma }_{ij})<0$ the material behaves like a stable, strain hardening plastic solid, and the plastic strain increment can be calculated by following steps 3-5 below.

 

3. Check for elastic unloading.   The solid will unload elastically, with $d{\epsilon }_{ij}^p=0$, if the stress increment brings the stress below yield. This is the case whenever

${\displaystyle \frac{\partial f}{\partial {\sigma }_{ij}}d{\sigma }_{ij}<0\Rightarrow \left(-{\displaystyle \frac{2}{3a}\left[{\displaystyle \frac{p_s}{a}-1}\right]\left(d{\sigma }_{kk}+3d{p}_w\right)+3{\displaystyle \frac{S_{ij}d{S}_{ij}}{{\left(Ma\right)}^2}}}\right)<0}$

 

4. If plastic strain does occur, the yield criterion must be satisfied throughout plastic straining.  This requires that

$df={\displaystyle \frac{\partial f}{\partial {\sigma }_{ij}}d{\sigma }_{ij}+{\displaystyle \frac{\partial f}{\partial a}da=0}}$

It is straightforward to show that

${\displaystyle \frac{\partial f}{\partial a}={\displaystyle \frac{-2p_s}{a^2}\left({\displaystyle \frac{p_s}{a}-1}\right)-2{\displaystyle \frac{{\sigma }_e^2}{M^2{a}^3}=-{\displaystyle \frac{2p_s}{a^2}}}}}$

 

5. The hardening law and plastic flow rule give

$da=-cad{\epsilon }_{kk}^p=-cad\lambda {\displaystyle \frac{\partial f}{\partial {\sigma }_{kk}}=2cd\lambda \left[{\displaystyle \frac{p_s}{a}-1}\right]}$

 

6. Finally, combining 3-5 leads to

$d\lambda ={\displaystyle \frac{\left(-{\displaystyle \frac{1}{3a}\left[{\displaystyle \frac{p_s}{a}-1}\right]\left(d{\sigma }_{kk}+3d{p}_w\right)+{\displaystyle \frac{3}{2}{\displaystyle \frac{S_{ij}d{S}_{ij}}{{\left(Ma\right)}^2}}}}\right)}{2c{\displaystyle \frac{p_s}{a}\left[{\displaystyle \frac{p_s}{a}-1}\right]}}}$

 

 

If the stress state is at yield $f({\sigma }_{ij})=0$ and also lies on the critical state $g({\sigma }_{ij})=0$ the material behaves like a perfectly plastic solid (with constant flow stress).  In this case,

 

1. The solid unloads elastically, with $d{\epsilon }_{ij}^p=0$ if

${\displaystyle \frac{\partial f}{\partial {\sigma }_{ij}}d{\sigma }_{ij}<0\Rightarrow \left(-{\displaystyle \frac{2}{3a}\left[{\displaystyle \frac{p_s}{a}-1}\right]\left(d{\sigma }_{kk}+3d{p}_w\right)+3{\displaystyle \frac{S_{ij}d{S}_{ij}}{{\left(Ma\right)}^2}}}\right)<0}$

 

2. If the solid deforms plastically, the stress state must satisfy

${\displaystyle \frac{\partial f}{\partial {\sigma }_{ij}}d{\sigma }_{ij}=0\Rightarrow \left(-{\displaystyle \frac{2}{3a}\left[{\displaystyle \frac{p_s}{a}-1}\right]\left(d{\sigma }_{kk}+3d{p}_w\right)+3{\displaystyle \frac{S_{ij}d{S}_{ij}}{{\left(Ma\right)}^2}}}\right)=0}$

In this case the plastic strain cannot be determined from the stress increment: any $d\lambda >0$ is admissible.  In a situation where the total strain of the solid is prescribed, the plastic strain increment can be determined by first solving for the elastic strain increment, and subtracting it from the total strain.

 

If the stress lies outside the critical state $g({\sigma }_{ij})>0$ and is at yield $f({\sigma }_{ij})=0$, the material softens.  In this case it is impossible to distinguish unambiguously between elastic unloading and plastic flow accompanied by strain softening.  The deformation in this regime usually consists of intense plastic shearing along one or more discrete shear bands, while the rest of the material unloads elastically.   Boundary value problems with material behavior in the unstable regime are generally ill-posed and cannot be solved uniquely. However, attempting to load a soil past the critical state usually results in catastrophic collapse (such as a landslide), so detailed solutions to boundary value problems in this regime are not of much practical interest.  The critical state surface can be used as a failure criterion to avoid collapse.

 

 

 

3.11.3 Application of the critical state equations to simple 2D loading

 

The constitutive equations for soils are complicated, and a simple 2D example helps to interpret them. To this end, consider a solid subjected to a 2D stress state of the form ${\sigma }_{ij}=-p{\delta }_{ij}+q({\delta }_{i1}{\delta }_{j2}+{\delta }_{i2}{\delta }_{j1})$, as illustrated in the figure below. Assume that the specimen is drained, so that the water pressure $p_w=0$.  In addition, assume that at time t=0 the solid has strength $a_0$. 

 

 

For this loading, the yield surface can be plotted in 2D, as a graph of the critical combinations of p and q that cause yield, as shown in the figure.  The yield surface is an ellipse, with semi-axes a and Ma.  The critical state surface is a straight line with slope M.

 

We can now examine the behavior of the solid as it is loaded.  Consider first the response to a constant pressure on the `wet’ side of critical state $p>a_0$, together with a steadily increasing shear stress q.  In this case

 

1. The solid first reaches yield when ${\left(p/a_0-1\right)}^2+3{\left(q/M{a}_0\right)}^2=1$

 

2. If the shear stress is raised beyond yield, the solid will deform plastically.  Since the flow law is derived from the yield criterion, the plastic strain direction is normal to the yield surface.  That is to say, if the solid experiences a plastic shear strain $d\gamma$ and volumetric strain $dv$, the vector ( $dv,d\gamma$ ) is normal to the yield surface, as shown in the picture.

 

 

3. On the `wet’ side of critical state, the volumetric plastic strain component dv is always compressive.  This means the solid compacts, and its strength increases (recall that $da=-cadv$ and $dv<0$ during compaction). 

 

4. As the yield surface expands, the volumetric strain component associated with an increase in shear stress $dq$ decreases (remember that we assume a constant pressure).    The hardening rate therefore decreases with strain, until the stress reaches the critical state.  At this point $dv=0$, so there is no further hardening.

 

 

Next, consider behavior on the `dry’ side of critical state. In this case

 

1. The solid first reaches yield when ${\left(p/a_0-1\right)}^2+3{\left(q/M{a}_0\right)}^2=1$

 

2. As before, the direction of the plastic strain is normal to the yield surface.

 

3. Notice that on the `dry’ side of critical state, the volumetric plastic strain component dv is always dilatational.  This means its strength decreases with plastic straining, as shown in the figure below (recall that $da=-cadv$ and $dv>0$ during dilation). 

 

4. The yield surface contracts during plastic straining, and this process continues until the stress reaches the critical state.  At this point the solid continues to deform at a constant shear stress.

 

 

 

 

3.11.4 Typical values of material properties for soils

 

Soil properties are highly variable, and for accurate predictions you will need to measure directly the properties of the soil you are intending to model.  In addition, soil models that are used in practice are somewhat more sophisticated than the simplified version given here. As a rough guide, material properites estimated from data in D.M. Wood, “Soil Behavior and Critical State Soil Mechanics,” Cambridge University Press, Cambridge, 1990 are listed in the table below.