Development of a generalized model using a new plastic modulus based on bounding surface plasticity

Abstract

Numerous attempts have been made to modify the generalized constitutive model and to introduce new constitutive models in the framework of generalized plasticity. The modified models can predict the behavior of sand fairly well, however, such models require many parameters and are difficult to calibrate. Moreover, it is highly desirable for a model to be able to reproduce soil behavior using a single set of parameters. In this paper, the constitutive model by Pastor and Zienkiewicz is further developed based on critical state and bounding surface models. The model is used to simulate the behavior of three types of sands under monotonic and cyclic loadings.

Appendix

The tensors $\bar{\sigma }_{ij}$, $\sigma_{ij}$ and $n_{ij}$ can be rewritten as (decomposed into) their volumetric and deviatoric components:

$$ \bar{\sigma }_{ij} = \bar{S}_{ij} + \bar{p}\delta_{ij} $$

(51)

$$ \sigma_{ij} = S_{ij} + p\delta_{ij} $$

(52)

$$ n_{ij} = n^{\prime}_{ij} + n^{\prime\prime}\delta_{ij} $$

(53)

Using Eqs. (51), (52), (53) and Eq. (27), it can be written:

$$ \begin{aligned} & \bar{S}_{ij} + \bar{p}\delta_{ij} = S_{ij} + p\delta_{ij} + bn^{\prime}_{ij} + bn^{\prime\prime}\delta_{ij} \\ & \quad \Rightarrow \bar{S}_{ij} + \bar{p}\delta_{ij} = (S_{ij} + bn^{\prime}_{ij} ) + (p + bn^{\prime\prime})\delta_{ij} \\ \end{aligned} $$

(54)

Therefore:

$$ \bar{S}_{ij} = (S_{ij} + bn^{\prime}_{ij} ) $$

(55)

$$ \bar{p} = (p + bn^{\prime\prime}) $$

(56)

On the other hand, extending the bounding surface equation (Eq. 23):

$$ \bar{\sigma }_{ij} \bar{\sigma }_{ij} - 2\bar{p}\delta_{ij} \bar{\sigma }_{ij} + \bar{p}^{2} \delta_{ij} \delta_{ij} = \frac{2}{3}(M\bar{p})^{2} $$

(57)

Now, Eq. (27) is inserted into Eq. (57):

$$ \begin{aligned} & (\sigma_{ij} + bn_{ij} )(\sigma_{ij} + bn_{ij} ) - 2\bar{p}\delta_{ij} (\sigma_{ij} + bn_{ij} ) \\ & \quad + \bar{p}^{2} \delta_{ij} \delta_{ij} = \frac{2}{3}(M\bar{p})^{2} \\ \end{aligned} $$

(58)

So:

$$ \begin{aligned} & \sigma_{ij} \sigma_{ij} + 2bn_{ij} \sigma_{ij} + b^{2} n_{ij} n_{ij} - 2\bar{p}\delta_{ij} \sigma_{ij} - 2b\bar{p}\delta_{ij} n_{ij} \\ & \quad + (\delta_{ij} \delta_{ij} - \frac{2}{3}M^{2} )\bar{p}^{2} = 0 \\ \end{aligned} $$

(59)

Inserting Eq. (56) into Eq. (59), it can be obtained:

$$ \begin{aligned} & \sigma_{ij} \sigma_{ij} + 2bn_{ij} \sigma_{ij} + b^{2} n_{ij} n_{ij} - 2(p + bn^{\prime\prime})\delta_{ij} \sigma_{ij} \\ & \quad - 2b(p + bn^{\prime\prime})\delta_{ij} n_{ij} + (\delta_{ij} \delta_{ij} - \frac{2}{3}M^{2} )(p + bn^{\prime\prime})^{2} = 0 \\ \end{aligned} $$

(60)

Using Eq. (33), it can be written:

$$ \begin{aligned} & \sigma_{ij} \sigma_{ij} + 2bn_{ij} \sigma_{ij} + b^{2} n_{ij} n_{ij} - 2p\delta_{ij} \sigma_{ij} - 2bn^{\prime\prime}\delta_{ij} \sigma_{ij} \\ & \quad - 2bp\delta_{ij} n_{ij} - 2b^{2} n^{\prime \prime } \delta_{ij} n_{ij} + z(p^{2} + 2bn^{\prime \prime } p + b^{2} n^{\prime \prime 2} ) = 0 \\ \end{aligned} $$

(61)

Then:

$$ \begin{aligned} & b^{2} (n_{ij} n_{ij} - 2n^{\prime \prime } \delta_{ij} n_{ij} + zn^{\prime \prime 2} ) \\ & \quad + 2b(n_{ij} \sigma_{ij} - n^{\prime\prime}\delta_{ij} \sigma_{ij} - p\delta_{ij} n_{ij} + zn^{\prime\prime}p) \\ & \quad + (\sigma_{ij} \sigma_{ij} - 2p\delta_{ij} \sigma_{ij} + zp^{2} ) = 0 \\ \end{aligned} $$

(62)

Hence, according to Eqs. (29), (30) and (31), a quadratic equation is reached:

$$ Ab^{2} + 2Bb + C = 0 $$

(63)

Then, two values for b are obtained with solving Eq. (63):

$$ b = \frac{{ -\, B + \sqrt {B^{2} - AC} }}{A} $$

(64)

$$ b = \frac{{ - \,B - \sqrt {B^{2} - AC} }}{A} $$

(65)

Using Eq. (64) to calculate b, and put it into Eq. (27), the image stress $\bar{\sigma }_{ij}$ will be achieved. If Eq. (65) is used to calculate b, Eq. (27) will give the supplement image stress $\bar{\sigma }^{\prime}_{ij}$.