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=c0+Hεpˉ,c=c_0+H\bar{\varepsilon_p},

in which c0c_0 is the initial cohesion (similar to the initial yield stress), HH is the hardening modulus, and εpˉ\bar{\varepsilon_p} is the accumulated plastic strain.

More

If one sets ηy=ηf=0\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(σ,c)=σ3ξc.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 3ξ2H3\xi^2H. In terms of total strain and stress, the hardening ratio is 3ξ2H/(E+3ξ2H)3\xi^2H/(E+3\xi^2H).

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

3ξ2HE+3ξ2H=0.06204607451.\dfrac{3\xi^2H}{E+3\xi^2H}=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

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