RDE: 4701 Concrete Validation with Kupfer Tests
Concrete validation with Kupfer tests.
Options and Keywords Used
 Concrete material law (/MAT/LAW24 (CONC), /MAT/LAW81)
 Brick elements
 Solid property (/PROP/TYPE14 (SOLID))
 Boundary condition (/BCS)
 Imposed displacement (/IMPDISP)
 Imposed velocity (/IMPVEL)
 Pressure load (/PLOAD)
Input Files
Model Description
For stability reasons, 1 element models must use a time step scale factor of 0.1.
 q_{a} = 1.1 and q_{b}= 0.05 (default values)
 I_{solid}= 24
 I_{frame}= 2 (corotational formulation)
 I_{strain}= 1 (to posttreat stains)
In this example, two material laws, /MAT/LAW24 and /MAT/LAW81, will be compared to the experiment data.
The following system is used: mm, ms, g, MPa
 Concrete Material Law (/MAT/LAW24)
 Initial density
 0.0022 $\left[\frac{g}{m{m}^{3}}\right]$
 Concrete elasticity Young's modulus
 ${E}_{c}=31700[\mathrm{MPa}]$
 Poisson's ratio
 $\nu =0.22$
 Concrete plasticity initial value of hardening parameter
 ${k}_{y}=0.35$
 Concrete plasticity dilatancy factor at yield
 ${\alpha}_{y}=0.6$
 Concrete plasticity dilatancy factor at failure
 ${\alpha}_{f}=0.2$
 Data Read Kupfer Experimental Data
 Concrete uniaxial compression strength
 ${f}_{c}=32.22[\mathrm{MPa}]$
 Concrete uniaxial tension strength
 0.01 ${f}_{c}$ , then set $\raisebox{1ex}{${f}_{t}$}\!\left/ \!\raisebox{1ex}{${f}_{c}$}\right.=0.1$ (Default=0.1 in LAW24)
 Concrete biaxial strength
 1.15 ${f}_{c}$ , then set $\raisebox{1ex}{${f}_{b}$}\!\left/ \!\raisebox{1ex}{${f}_{c}$}\right.=1.15$
 Concrete Material Law (/MAT/LAW81)
 Initial density
 0.0022 $\left[\frac{g}{m{m}^{3}}\right]$
 Bulk modulus
 $K=\frac{{E}_{c}}{3\left(12\nu \right)}=\text{18869}\text{.048[MPa]}$
 Young's modulus
 ${E}_{c}=31700[\mathrm{MPa}]$
 Poisson's ratio
 $\nu =0.22$
 Shear modulus
 $G=\frac{{E}_{c}}{2\left(1+\nu \right)}=\text{12991}\text{.8[MPa]}$
 Friction angle
 $\varphi =\text{68}{\text{.35}}^{\circ}$
 Ratio
 $\alpha =\text{0}\text{.4186898}$
 Cap limit pressure set constant
 ${P}_{b}=0.838\text{,}\text{\hspace{0.17em}}{f}_{c}\text{=27[MPa]}$
 Cap beginning pressure
 ${P}_{a}=\alpha \cdot {P}_{b}\text{=0}\text{.351,}\text{\hspace{0.17em}}{f}_{c}=11.305\text{[MPa]}$
 Material cohesion set constant
 $c=0.169175\text{,}\text{\hspace{0.17em}}{f}_{c}=5.4508\text{[MPa]}$
Simulation Iterations
Test  Principle Stress  Triaxiality  Failure Stress 

T000 Uniaxial tension 
${\sigma}_{1}=0$ ; ${\sigma}_{2}=0$ ; ${\sigma}_{3}=1$  1/3  0.1 ${f}_{c}$ 
C000 Uniaxial compression 
${\sigma}_{1}=1$ ; ${\sigma}_{2}=0$ ; ${\sigma}_{3}=0$  1/3  ${f}_{c}$ 
CC00 Biaxial compression 
${\sigma}_{1}=1$ ; ${\sigma}_{2}=1$ ; ${\sigma}_{3}=0$  2/3  1.15 ${f}_{c}$ 
CC01 Compression/Compression 
${\sigma}_{1}=0.052$ ; ${\sigma}_{2}=0$ ; ${\sigma}_{3}=1$  0.5849  1.22 ${f}_{c}$ 
TC01 Compression/Tension 
${\sigma}_{1}=0.052$ ; ${\sigma}_{2}=0$ ; ${\sigma}_{3}=1$  0.3077  0.8 ${f}_{c}$ 
TC02 Compression/Tension 
${\sigma}_{1}=0.102$ ; ${\sigma}_{2}=0$ ; ${\sigma}_{3}=1$  0.2838  0.6 ${f}_{c}$ 
TC03 Compression/Tension 
${\sigma}_{1}=0.204$ ; ${\sigma}_{2}=0$ ; ${\sigma}_{3}=1$  0.2377  0.35 ${f}_{c}$ 
The triaxiality can be computed using the principal stresses:
with ${\text{\sigma}}_{m}=p=\frac{1}{3}\left({\text{\sigma}}_{1}+{\text{\sigma}}_{2}+{\text{\sigma}}_{3}\right)$ and ${\text{\sigma}}_{VM}=\sqrt{\frac{1}{2}\left[{\left({\text{\sigma}}_{1}{\text{\sigma}}_{2}\right)}^{2}+{\left({\text{\sigma}}_{2}{\text{\sigma}}_{3}\right)}^{2}+{\left({\text{\sigma}}_{3}{\text{\sigma}}_{1}\right)}^{2}\right]}$ .
Results
Failure Results with LAW24 and LAW81
The failure curve in LAW24 is:
With $b=\frac{1}{2}({b}_{c}+{b}_{t})$

$p\le {P}_{a}$
It is linear with $p\mathrm{tan}\varphi +c$
The failure is ${\sigma}_{m}\mathrm{tan}({68.35}^{\circ})+5.4508$

${P}_{a}<p\le {P}_{b}$
(cap)
The cap curve is:
$$\sqrt{1{\left(\frac{p{p}_{a}}{{p}_{b}{p}_{a}}\right)}^{2}}\cdot \left(p\mathrm{tan}\varphi +c\right)$$The failure is:
$$\sqrt{1{\left(\frac{{\sigma}_{m}27}{2711.305}\right)}^{2}}\cdot {\sigma}_{m}\cdot \mathrm{tan}({68.35}^{\circ})+5.4508$$
Results for Concrete Tension Tests
Concrete does not support very much load in tension. In LAW24 the uniaxial tensile failure (modeled by stress) and elastic modulus softening behavior is defined by ${H}_{t},\text{\hspace{0.05em}}\text{\hspace{0.17em}}{D}_{\mathrm{sup}},{\epsilon}_{\mathrm{max}}$ . The softening modulus ${H}_{t}={E}_{c}$ (default) for tension is set. The peak for the above curve is at 0.1 where it is defined by $\raisebox{1ex}{${f}_{t}$}\!\left/ \!\raisebox{1ex}{${f}_{c}$}\right.=0.1$ (default) in input.
Conclusion
Under complex loading, the concrete mechanic failure behavior is shown using two Radioss material models LAW24 and LAW81 and results compared to experiments. For LAW24 the default values are a good choice, if no experimental data is available. For LAW81, the material parameters $\varphi ,c,\alpha ,{P}_{b}$ need to be calculated with curve fitting using at least four experimental tests.