J controlled fracture
============================

### HRR
```{note}
The HRR fields were published in 1968 by (J.R. Rice and G.F> Rosengren)[https://doi.org/10.1016/0022-5096(68)90013-6] and by (J.W. Hutchinson)[https://doi.org/10.1016/0022-5096(68)90014-8]
```

The **HRR** field, named after Hutchinson Rice and Rosengren is the solution for the mechanical fields at the tip of a stationary crack in a power law material. 

The load is assumed to increase monotonically, thus allowing the use of the *deformation theory* of plasticity.  As we mentioned previously, the *deformation theory* theory can be thought of as non-linear elasticity with the relation.

$$
\frac{\epsilon}{\epsilon_y} = \alpha \left ( \frac{\sigma}{\sigma_y}\right)^n
$$

Writing down the $J$ integral for $r \to 0$ :

$$
J=\int_{-\pi}^{\pi} W n_1 - T_i \frac{\partial u_i} { \partial x_1} rd \theta
$$ 

We can estimate that the integrand have the form of $\frac{f(\theta)}{r}$

This implies that 

$$
\sigma \cdot \epsilon \propto \frac{g(\theta)}{r}
$$

if we expand $\sigma_{ij}$ into a series around $r \to 0$ such that 

$$
\sigma_{ij}(r,\theta) = \sum_s r^s\sigma_{ij}^*(\theta,s)
$$

we can obtain for $r \to 0$ where the plastic strains are much greater than the elastic ones:

$$
\epsilon_{ij} =\epsilon^p_{ij} =r^{sn}\gamma^*_{ij}(\theta,s)
$$

Since we required earlier that $\sigma \cdot \epsilon \propto r^{-1}$ it implies that $s+sn=-1$ and thus $s$ which will lead to the strongest singularity is 

$$
s=\frac{-1}{1+n}
$$

Now, the **HRR** stress and strain fields (for $r \to 0$) are given by:

$$
\sigma_{ij} = \sigma_y k_n r^{-1/(n+1)}\sigma_{ij}^*(\theta,n) \\
\epsilon_{ij} = \epsilon_y k_n^n r^{-n/(n+1)}\epsilon_{ij}^*(\theta,n) 
$$

```{note}
For $n \to 1$ we obtain a $r^{-1/2}$ singularity
```

Using the **HRR** fields in the $J$ integral will lead to 

$$
J=\sigma_y \epsilon_y \alpha k_n^{n+1} \int_{-\pi}^{\pi} f_{ij}^*(\theta,n)d\theta = \sigma_y \epsilon_y \alpha k_n^{n+1} I_n
$$

With $k_n$ being:

$$
k_n=\left( \frac{J}{\alpha \sigma_y \epsilon_y I_N}\right)^{1/(n+1)}
$$

and finally 

$$
\sigma_{ij} = \sigma_y \left( \frac{J}{r\alpha \sigma_y \epsilon_y I_N}\right)^{1/(n+1)}\sigma_{ij}^*(\theta,n) \\
\epsilon_{ij} = \epsilon_y\left( \frac{J}{r\alpha \sigma_y \epsilon_y I_N}\right)^{n/(n+1)}\epsilon_{ij}^*(\theta,n) 
$$

```{image} ../EPFM/regs.png
:alt: Kirsch's plate
:width: 600px
:align: center
```



**It is time to find the angular dependency of the stress and strain (but we wont)** 

To do so, one need to  introduce a stress function $\phi$ whose derivatives define $\sigma$. 

### Crack growth resistance curves


```{image} ../EPFM/ex.png
:alt: Jda procedure
:width: 600px
:align: center
```
The value of the tearing modulus is calculated by applying:

$$
T_R = \frac{E}{\sigma_0^2}\frac{dJ}{da}
$$

on the polynomial fit to the data points between the exclusion lines. 

Similar to the way we handeled $G$ we can examine the stability of a growing crack by comparing the driving force for crack growth as :

$$
T_{app}=\frac{E}{\sigma_0^2}\left( \frac{dJ}{da}\right )_{\Delta_T} \\
\\ \ \ \\
\Delta T = \color{red}{\Delta} + \color{blue}{C_mP}
$$

with $\color{red}{\Delta}$ being the applied displacement and $\color{blue}{C_m,P}$ being the machine compliance and applied load. 

We can easily obtain 

$$
\left( \frac{dJ}{da}\right )_{\Delta_T} = \left( \frac{dJ}{da}\right )_{P}-\left( \frac{dJ}{dP}\right )_{a}\left( \frac{d\Delta}{da}\right )_{P} \left [ C_m+\left(\frac{\partial \Delta}{\partial P }\right)_a \right]^-1
$$

When the machine is infinitely stiff ($\color{blue}{C_m} \to \infty$ - load control) we are left with 

$$
\color{blue}{\left( \frac{dJ}{da}\right )_{\Delta_T} = \left( \frac{dJ}{da}\right )_{P}}
$$

A crack will grow in a stable manner when $J=J_R$ and $T_{app} \leq T_R$

A crack will grow in an unstable manner when $J=J_R$ and $T_{app} \gt T_R$