Blatz-Ko Model - LS-DYNA MAT_007

Constitutive Equation

Blatz-Ko energy function implemented in lsdyna is:

$W(\mathbf{\text{C}})=\frac{G}{2}[I_1-3+\frac{1}{\alpha }(I_3^{-\alpha }-1)]$

where $\mathbf{C}$ is the right Cauchy-Green tensor, $G$ is the shear modulus, $I_1$ and $I_2$ are the first and third invariants of $\mathbf{C}$ and

$\alpha =\frac{\nu }{1-2\nu }$

where $\nu$ is Poisson's ratio, which is internally set to $\nu =0.463$.

Using the above energy function, the second Piola-Kirchhoff stress is computed as

$\begin{array}{rl}\mathbf{\text{S}}& =2\frac{\partial W}{\partial \mathbf{\text{C}}}=2\frac{\partial W}{\partial I_1}\mathbf{\text{I}}+2\frac{\partial W}{\partial I_2}(I_1\mathbf{\text{I}}-\mathbf{\text{C}})+2\frac{\partial W}{\partial I_3}{I}_3{\mathbf{\text{C}}}^{-1}\\ & =G(\mathbf{\text{I}}-I_3^{-\alpha }{\mathbf{\text{C}}}^{-1})\end{array}$

where $\mathbf{\text{I}}$ is unit matrix.

Cauchy stress can be obtained from the above second Piola-Kirchhoff stress.

$\begin{array}{rl}\mathit{\sigma }& =J^{-1}{\mathbf{\text{FSF}}}^T=\frac{G}{J}[{\mathbf{\text{FF}}}^T-I_3^{-\alpha }\mathbf{\text{F}}({\mathbf{\text{F}}}^T\mathbf{\text{F}}{)}^{-1}{\mathbf{\text{F}}}^T]\\ & =\frac{G}{J}(\mathbf{\text{B}}-I_3^{-\alpha }\mathbf{\text{I}})\end{array}$

$\mathbf{\text{F}}$ and $\mathbf{\text{B}}$ are the deformation gradient and the left Cauchy Green tensor respectively.

Parameter Identification

Blatz-Ko model in lsdyna has the only one material constant, G. So we can easily obtain it from mill sheet.

First, the principal Cauchy stress can be expressed as below.

${\sigma }_i={\lambda }_i\frac{\partial W}{\partial {\lambda }_i}$

${\lambda }_i$ represents stretch ratio in principal directions. This partial derivative can be solved by applying chain rule in each direction. For the first principal direction, it gives

$\frac{\partial W}{\partial {\lambda }_1}=\frac{\partial W}{\partial I_1}\frac{\partial I_1}{\partial {\lambda }_1}+\frac{\partial W}{\partial I_3}\frac{\partial I_3}{\partial {\lambda }_1}$

Assuming incompressible, $I_3={\lambda }_1^2{\lambda }_2^2{\lambda }_3^2=1$, and $I_1={\lambda }_1^2+{\lambda }_2^2+{\lambda }_3^3$. Therefore, the Cauchy stress is

${\sigma }_1=G{\lambda }_1^2$

Since ${\lambda }_1=\lambda$, and engineering stress, ${\sigma }_{\text{Eng}}$ is $\frac{{\sigma }_1}{\lambda }$ for uniaxial tension, this equation is written as:

${\sigma }_{\text{Eng}}=G\lambda$

Finally, we can obtain G if we know the engineering stress at the stretch ratio for any material.

$G=\frac{{\sigma }_{\text{Eng}}}{\lambda }$

For example, if any material's tensile strength and elongation at break are 3.9MPa and 360.0% respectively, stretch ratio, $\lambda =1+{ϵ}_{\text{Eng}}=4.6$ and shear modulus, $G$ is calculated from the above formula.

$G=\frac{3.9}{4.6}=0.85\text{MPa}$

We can also get this analytical curve from the constitutive equation of Blatz-Ko model.

${\sigma }_{\text{Eng}}=\frac{\sigma }{\lambda }=G(\lambda -\frac{1}{\lambda })$