Hyperplasticity Geomechanics – critical state 1

13/1

Critical state models are some of the most important in geomechanics, and a key advantage ofhyperplasticity is that it can be used to formulate these models.

Hyperplasticity Geomechanics – critical state 2

13/2

In this lecture we first introduce the so-called Cambridge “triaxial” variables which are usuallyused to express critical state models.

We then formulate the Modified Cam Clay (MCC) model within hyperplasticity, in two differentways.

This leads to the result that different forms of energy/dissipation functions can lead to thesame constitutive response ... this observation has very far-reaching consequences because itmeans that it is a proof (an existence proof) that the energy functions and dissipation are not“observable”.

Finally we pursue briefly a simple physical analogy of consolidation.

Hyperplasticity Geomechanics – critical state 3

13/3

As critical state models were originally based on the results of triaxial tests they are usually firstexpressed in terms of the Cambridge triaxial variables.

p and q essentially measure the mean stress and the deviator stress. The conjugate strainvariables are epsilon_v and epsilon_s.

p and q are related to the invariants of the stress tensor....

... and the equivalent strain variables to the invariants of the strain tensor.

Just for clarity:

(a) in this section we shall follow soil mechanics convention and take compressive stresses andstrains as positive

(b) all stresses here are effective stresses – but we drop the prime notation for simplicity

Hyperplasticity Geomechanics – critical state 4

13/4

As well as p and q etc we shall need the plastic strain equivalents of epsilon_v and epsilon_s, andwe shall use alpha_v and alpha_s for these. The corresponding generalised stresses we shall callchi_p and chi_q.

So the example here gives the Gibbs free energy function and the dissipation. The standarddifferentials take their usual form

Hyperplasticity Geomechanics – critical state 5

13/5

Here in diagrammatic form is the Modified Cam Clay model (MCC)

There is an elliptical yield surface in p,q space, passing through the origin and intercepting the paxis also at p_c. The critical state line, at slope M, passes through the peak of the ellipse.

The consolidation behaviour is experessed in log(v) - log (p) space. (v is the specific volume =volume of voids/ volume of solids = 1 + e). This is a very small difference from the conventionalform of the model in which the consolidation behaviour is expressed in e – log(p), or V – log(p)which gives the same slopes of the lines. Th lambda and kappa in our model therefore have aslightly different meaning from in the usual model – they differ by a factor V = 1+e.

“Consolidation” lines are at slope lambda and “swelling” lines at slope kappa. As consolidationoccurs p_c increases.

The critical state line is parallel to the normal consolidation line.

Hyperplasticity Geomechanics – critical state 6

13/6

rather than developing the model step by step, which is a little tedious, I shall skip straight to theresult, and explain the various terms that appear in the energy and dissipation equations. I havetaken the Gibbs free energy version of the model as the example.

In g, the first two terms define elasticity. They take a slightly unusual form – the first termensures the straight swelling lines in log(V) – log(p) space, which amounts to the same thing asthe bulk modulus proportional to pressure. Note the p on the bottom line of the second term –this makes the shear modulus G also proportional to pressure (g_s is a dimensionless constant).

The third term is a kinematic hardening term – present because the centre of the ellipse movesat the same time as the ellipse expands.

The dissipation includes the (familiar) root-sum-of-squares of the plastic strain rates that isrequired for Modified Cam Clay, but note the difference from the conventional formulation thatthe multiplier on this is not the pressure but a term that turns out to be proportional to thepreconsolidation pressure. This is important – this form results in “associated flow”. If we haddissipation proportional to pressure we would have friction and non-associated flow.

I include here all the standard differentials of the original energy and dissipation expressions. Wecan then manipulate these to demonstrate that the model we derive is indeed Modified CamClay.

Hyperplasticity Geomechanics – critical state 7

13/7

... here is some of the development. We can obtain the expressions for the strain rate – elasticplus plastic strain rates. Note the presence of the off-diagonal term in the elastic compliancematrix (we shall return to this issue in a later unit).

The preconsolidation pressure p_c is just related to the plastic volumetric strain through anexponential expression – this gives the straight consolidation lines in log(V) – log(p) space.

We can derive the yield surface expression (in true stress space) – the familiar ellipse.

Hyperplasticity Geomechanics – critical state 8

13/8

However, note that there is an alternative form of expressions from which we can deriveidentically the same model!

I have highlighted the differences here. First of all the kinematic hardening term is missing (i.e.we only have isotropic hardening) ...

... and secondly there is an extra term in the dissipation expression. This term is a little odd, as atfirst sight of course it might be negative. However, the square root expression is always biggerthan alpha_v^dot, so overall the dissipation is always positive.

Conceptually one can think of the models as in the two diagrams. our first model involvedisotropic expansion of the yield surface about its centre, coupled with a translation of the centre(kinematic hardening). the second model does not involve any translation, but the focus for theisotropic expansion is not the centre of the ellipse but the origin.

It is a matter of taste which of the two models you prefer!

The really important point though is that two different sets of energy and dissipation expressionsresult in the same model behaviour. This is an existence proof that we cannot observe stressstrain behaviour and (with certainty) deduce what the energy and dissipation expressions mustbe. We can find some possible expressions, but we cannot prove they are “right” because theymay not be unique. So the energy and dissipation expressions are “unobservable”.

Hyperplasticity Geomechanics – critical state 9

13/9

Her we emphasise that important general conclusion from the last slide.

Hyperplasticity Geomechanics – critical state 10

13/10

But we do notice that whilst the two MCC models are different, they have a lot in common. Socan we determine what they must have in common?

Consider two models defined by g_1 and g2 and corresponding d_1 and d_2. We assume thatg_2 differs from g_1 in the way shown, and d_2 and d_1 also.

We carry out the derivation for model 2, and when we obtain the expressions for the strain andfor the chi_bar – chi = 0 expression, we discover that these expressions are identical to what wewould have obtained from model 1, so the models produce the same result.

So we can add any q(alpha) to the g expression, as long as we subtract dq/dalpha timesalpha^dot from the d expression.

BUT (and this is important), both the original and new dissipations must always be positive.

The quantity that is invariant across the two models is (dg/dalpha)alpha^dot + d

Hyperplasticity Geomechanics – critical state 11

13/11

In carrying out model development, it is often a distraction to be following through various non-linear terms when wants just to examine the overall structure of the model. Just for the record Itherefore give hear the “linearised” version of the first of the two MCC models that weconsidered. This retains the same basic structure bat has easier maths.

Hyperplasticity Geomechanics – critical state 12

13/12

Finally, it is sometimes useful to bear in mind a simple physical model that gives rise to a certainalgebraic model. Consider the system shown in the diagram. It is straightforward to write downthe terms that appear in the Helmholtz free energy and the dissipation, and these can be seen tohave the same structure as those dealing with consolidation in the MCC model.

If we follow through the derivation we can see that if we set mu = s we obtain a model thatreproduces exactly the volumetric aspects of MCC.

This then serves as a conceptual model for the processes going on in consolidation. We canimagine that the harder the material is pushed together (positive alpha_v) the greater the forcein the spring holding down the block and the more the dissipation. The dissipation isproportional to the equivalent of p_c.

Hyperplasticity Geomechanics – critical state 13

13/13

... and if we imagine also a movement of the block epsilon_s into the page, and a spring ofstiffness 3G attached to the block in that direction, then the expressions become as show. Thismodel would reproduce exactly the (linearised) MCC model, with just the small detail that itimplies M = 1.

That final detail could be resolved by imagining that there was an anisotropy to the coefficientsof friction of the block on the plane in the directions parallel to alpha_v and alpha_s, but as thisis a conceptual model only we do not pursue that detail.

Hyperplasticity Geomechanics – critical state 14

13/14

In summary, we have formulated Modified Cam Clay within the hyperplasticity framework.

This can be done in one of two different ways, and importantly this proves the non-uniquenessof the energy and dissipation expressions. This proves to be an interesting observation,although in the end it does not prove to be too much of a disadvantage.

However, the knowledge that these functions cannot be measured at least stops us fromwasting our time trying to devise experiments that measure them!