BilinearDP

Bilinear Drucker-Prager Model

Syntax

material BilinearDP (1) (2) (3) (4) (5) (6) (7) (8) [9]
# (1) int, unique material tag
# (2) double, elastic modulus
# (3) double, poissons ratio
# (4) double, \eta_y
# (5) double, \eta_f
# (6) double, \xi
# (7) double, initial cohesion, c_0
# (8) double, hardening ratio/modulus, H
# [9] double, density, default: 0.0

Theory

See more details on the formulation in the parent page.

Hardening Function

The cohesion develops linearly with the accumulated plastic strain,

$$c=c_0+H\overline{{\epsilon }_p},$$

in which $c_0$ is the initial cohesion (similar to the initial yield stress), $H$ is the hardening modulus, and $\overline{{\epsilon }_p}$ is the accumulated plastic strain.

More

If one sets ${\eta }_y={\eta }_f=0$, the model effectively becomes the von Mises model with the associative plasticity. In a uniaxial loading case, the yield function is then

$$F(\sigma ,c)=\sigma -\sqrt{3}\xi c.$$

This leads to a yield stress ${\sigma }_y=\sqrt{3}\xi c_0$. The plastic hardening modulus is $3{\xi }^{2}H$. In terms of total strain and stress, the hardening ratio is $3{\xi }^{2}H/(E+3{\xi }^{2}H)$.

For the following model, one can compute the hardening ratio to be

$${\displaystyle \frac{3{\xi }^{2}H}{E+3{\xi }^{2}H}}=0.06204607451.$$

One can validate this value by plotting the response.

node 1 5 -5 -5
node 2 5 5 -5
node 3 -5 5 -5
node 4 -5 -5 -5
node 5 5 -5 5
node 6 5 5 5
node 7 -5 5 5
node 8 -5 -5 5

material BilinearDP 1 1E4 .3 0 0 .58461851886189 5 645.1584849161

element C3D8 1 1 2 3 4 5 6 7 8 1

fix 1 1 1 2 5 6
fix 2 2 1 4 5 8
fix 3 3 1 2 3 4

displacement 1 0 -.2 3 5 6 7 8

hdf5recorder 1 Element E 1
hdf5recorder 2 Element S 1

step static 1
set fixed_step_size 1
set ini_step_size 1E-2
set symm_mat 0

analyze

save recorder 1 2

exit