The Energy Release Rate 
============================
Let us consider a more generalized case. 


```{image} ../LEFM/Load_Disp.png
:alt: general loading
:width: 700px
:align: center
```

We will now try to generalize Griffith's energy balance by considering a body subjected to <span style="color:red">load </span> or <span style="color:green"> displacement </span> boundary conditions (or a combination of both). 

Similar to Griffith's energy balance, we will focus on the **driving force** for crack extension resulting from the re-distribution of energy in the system. 

Let us define $G$ as the amount of energy being released from the system due to an increment in the crack length of magnitude $\Delta a$ :

$$
G = -\frac{d\Pi}{da}
$$

Here, $\Pi$ is the potential energy of the loaded (elastic) body  and can be written as :

```{math}
\begin{align*}
&\Pi = \int_VWdV - \int_{A}t\cdot\delta u dA - \int_{V}b \cdot \delta u dV \\
&\delta U = \int_{A}t\cdot\delta u dA + \int_{V}b \cdot \delta u dV  \text{  :Work done on the body } \\
&W = \int_0^{\epsilon}\sigma_{ij}d\epsilon_{ij} \\

&\Omega = \int_V \Omega dV \ \  \text{:strain energy} \\
\end{align*}
```

Let us assume a generalized measure of load (<span style="color:red"> $P$ </span>) and displacement (<span style="color:green"> $q$ </span>). 

For the case of <span style="color:red">load control </span>, $\delta U$ can be written as:

$$
\color{red}{\delta U = P\delta q}
$$

And for the case of <span style="color:green"> displacement </span> controlled loading we shall obtain the same result: 

$$
\color{green}{\delta U = P\delta q}
$$

For a solid elastic body under equilibrium conditions, it is easy to show that 

$$
 \delta U = \delta \Omega 
 $$

Consider now the bodies in the figure above.  
When subjected to<span style="color:red"> predetermined load $P$ </span> which will lead to a crack extension of $\delta a$  the change in strain energy can be written as :

$$ 
\color{red}{ \delta \Omega = \frac{\partial \Omega}{\partial q}\frac{\partial q}{\partial P}\delta P +\left (\frac{\partial \Omega}{\partial q}\frac{\partial q}{\partial a}+\frac{\partial \Omega}{\partial a}|_q \right )\delta a }
$$

Remembering that the applied load is fixed, we can use $\delta P =0$ leading to:

$$
 \color{red}{ \delta \Omega = P\delta q + \frac{\partial \Omega}{\partial a}|_q  \delta a} 
$$

The expression we obtain ties the change in stored energy to the work done on the system and the work invested in growing the crack by $\delta a$. rewriting the second term we obtain:

$$
 \color{red}{ \delta \Omega = P\delta q - G  \delta a} 
$$
Using 

$$
 \delta (Pq) = P\delta q + q\delta P
$$ 
we arrive at :

$$
 \color{red}{ G = -\frac{\partial}{\partial a}(\Omega -Pq)}
$$

Now looking at the case where the <span style="color:green"> displacement is prescribed </span> and the load is only measured, we can write:

$$
\color{green}{\delta \Omega = \frac{\partial\Omega}{\partial a}\delta a + \frac{\partial\Omega}{\partial q}\delta q} 
$$

After the displacement is prescribed and the crack is allowed to grow we have 

$$
\color{green}{\delta \Omega =  \frac{\partial\Omega}{\partial a}\delta a } 
$$ 

which in turns, taking use of our definition for $G$ result in 

$$
 \color{green}{\delta \Omega = -G\delta a}
$$

```{admonition} Note:
:class: tip
The two expressions we obtained for $G$ are quite different.
What are the implications?
```

Looking at the two expressions in a graphical manner may make it easier to understand:



```{image} ../LEFM/visual.png
:alt: energies visualized
:width: 600px
:align: center
```
The two extreme cases can be generalized by considering a mixed loading, where the load is applied through a spring. 

The overall displacement $\Delta X$ can be written as :

$$
 \Delta X = \Delta + KP
 $$

The derivative of the displacement is $0$ and $\Delta X$ can be taken to be a function of $(P,a)$ as can $G$ resulting in 

$$
\left(\frac{dG}{da} \right)|_{\Delta x} = \left(\frac{dG}{da} \right)|_{P}+\left(\frac{dG}{dP} \right)|_{a}  \left (\frac{dP}{da} \right)|_{\Delta X}
$$
and using the derivative of $\Delta X$ we obtain 

$$

\left(\frac{dG}{da} \right)|_{\Delta x} = \left(\frac{dG}{da} \right)|_{P}-\left(\frac{dG}{dP} \right)|_{a}  \left (\frac{d\Delta}{da} \right)|_{P}\left[ K + \frac{\partial \Delta}{\partial P} |_{a} \right]^{-1}  

$$ (label)

## Crack extension and G

From the simple analysis above, we saw that for cracks to grow, the system must have sufficient energy to support the crack extension process. In other words, $G$ needs to obtain a critical value $G_c$ for cracks to grow in a loaded solid. 

Next, we will discuss how to measure $G_c$. Let us start with a simple example. 

Consider a DCB specimen:



```{image} ../LEFM/DCB.png
:alt: DCB specimen
:width: 400px
:align: center
```
The moment of Inertia for this geometry is known and given by:

$$
I=\frac{B\color{blue}{h}^3}{12}
$$
With, $B$ being the DCB thickness. 
Similarly, using beam theory we can tie the <span style="color:green"> displacement $\Delta$ </span> to the applied <span style="color:red"> load $P$ </span> using :

$$
{\color{green}\Delta} = 2\frac{{\color{red}{P}}a^3}{3EI}
$$

The compliance is thus:

$$
C=\frac{\color{green}{\Delta}}{\color{red}{P}} = \frac{2a^3}{3EI}
$$

And now we can calculate the driving force for crack extension using (7.1) we obtain:

$$
G = \frac{{\color{red}{P}}^2}{2B}\frac{dC}{da}
$$ (label)

```{admonition} We can now answer the previous question:
:class: tip
The two expressions we obtained for $G$ are quite different.
What are the implications?
```
When deriving (7.2) we did not pay any special attention to the way load is applied (i.e. load controlled or displacement controlled).  
Starting with the <span style="color:red"> load controlled scenario </span> we can differentiate $G$ with respect to the crack length. This procedure will help us identify the variation in crack extension driving force as a function of the crack growth. 

$$
\left( \frac{dG}{da}\right)_{\color{red}{P}} = \frac{2{\color{red}{P}}^2a}{BEI} = \frac{2G}{a}
$$ (label)

Next, we will replace <span style="color:red"> P</span> in (7.2) with 

$$
{\color{red}{P}}=\frac{2{\color{green}{\Delta}} E I}{2a^3}
$$

So that we can take the derivative of (7.2) with respect to $a$ under a <span style="color:green"> constant displacement $\Delta$</span> leading to :


$$
\left( \frac{dG}{da} \right) _{{\color{green}{\Delta}}} = - \frac{2{\color{green}{\Delta}}^2EI}{Ba^5} = - \frac{4G}{a}
$$ (label)

Looking at <span style="color:red"> (7.3) </span> and <span style="color:green"> (7.4) </span> it is evident that under <span style="color:red"> load control the driving force for crack extension increases with crack growth  </span>. On the contrary,    <span style="color:green"> for displacement controlled scenarios, the driving force for crack growth has a negative slope and thus decreases as the crack grows </span> . On

## The crack growth resistance curve


```{image} ../LEFM/stab.png
:alt: R and G
:width: 600px
:align: center
```

We can write a simple condition for crack growth stability:

$$
\begin{align*}
&\color{red}{G}=\color{green}{R} \\
&\frac{d\color{red}{G}}{da} \le \frac{d\color{blue}{R}}{da}
\end{align*}
$$

As long as as

$$
\frac{d\color{red}{G}}{da} \gt \frac{d\color{blue}{R}}{da}
$$

Crack growth is unstable since the driving force for crack extension increase with the crack growth process itself. 

```{admonition} Points for discussion:
:class: tip
What is the meaning of the R curve being flat?

What can cause the R curve to have a positive slope?

```

```{image} ../LEFM/stab2.png
:alt: R and G
:width: 900px
:align: center
```