*Note for new readers, this is part of a series of posts on the history of fracture mechanics. You can find the first post in this series here.*

In my last post I talked about the energy balance during crack growth. I got some feedback that that part went a bit fast for some readers, so before continuing with the history of fracture mechanics I’ll discuss the energy balance in a bit more depth.

First let’s imagine we’re testing a specimen without any cracks, by pulling on it with a force that we steadily increase until it reaches the value P. When we pull on the specimen it will become longer; the harder we pull, the longer it gets. In this case we pull until the specimen’s length has increased by an amount d. This means we’re applying a force which is moving over some distance. In turn, this means we are doing work. In other words, we are transferring energy to the specimen. This energy is stored in the specimen as strain energy.

If we slowly reduce the force we are applying to the specimen, it will become shorter again^{*}; as we reduce the force to zero, the elongation of the specimen will reduce to zero as well. In this case work is still being done, because there is still a force being moved over a distance. However, we do not have to supply the energy required for this work ourselves. Instead, we can use the energy stored in the specimen while it was being stretched. In fact, there is a perfect balance: the amount of strain energy that is stored in the specimen when it is stretched is exactly equal to how much energy is required to return it to its original shape again.

With some calculus it’s quite easy to work out how much work you have performed, and also how much strain energy is stored in a specimen if you pull it to a certain length with a certain force. In fact, even if you don’t know calculus, you can just draw it. Simply plot the relationship between force and displacement on a graph. The surface area below that graph will be equal to the amount of work you have done. If we’re looking at a so-called linear elastic material, this relationship will be a straight line, and thus the surface will just be a triangle. This makes it really easy to work out how much strain energy is in a specimen: simply multiply the maximum force times the maximum displacement times one half (remember how to calculate the surface area of a triangle from your high-school geometry lessons).

Now let’s look what happens when there is a crack. Let’s assume we increase the load on the specimen until we reach force *P*. The specimen will elongate by the matching amount *d*. Again the work is stored as strain energy in the specimen. Now we let the crack grow, all the while still pulling with force *P*. As the crack gets longer, the specimen will become more flexible (there’s less material holding everything together). Since we’re still pulling with force *P*, the increased flexibility means the specimen will stretch slightly further, by an amount we’ll call *ε*. We’re moving a force over a distance, so again we’re doing work, which as you might expect will be stored as strain energy.

Now something strange has happened however. If we look at the amount of strain energy in the specimen before the crack grows (1/2 times *P* times *d*) and the amount of strain energy in the specimen after the crack grows (1/2 times *P* times *d*+*ε*), we would expect the difference to be equal to the amount of work done by moving the force from *d* to *d*+*ε* (the orange square in the figure below). After all, the energy should always balance. However when we work out the actual difference between the two energy states, we will discover that this difference is less than the extra work that was performed. So what is going on?

It turns out that some of the extra energy that is added by the extra work is actually released. As discussed in the previous posts, growing a crack requires energy. Instead of all the extra work remaining stored as strain energy in the specimen, some of the strain energy will instead be used to grow the crack.

We can work out how much extra work we have done because of the crack growth (*P* times *ε*). We can also work out the difference in stored strain energy (1/2 times *P* times *ε*). If we subtract this difference from the extra work, we know how much energy has been released.

According to Irwin & Kies, it is this energy release that powers the crack growth. If you divide the amount of released energy by the corresponding amount of crack growth, you get the quantity they called the *strain energy release rate* (or *G *for short). *G* is the amount of energy that is released when the crack grows by a short amount. If this is larger than the amount of energy required to create that amount of crack growth, then the crack doesn’t require an external energy input to keep growing; in other words it becomes unstable and could potentially keep growing until the specimen fails.

To summarise all this:

When the crack grows extra work is performed. Some of this work is stored as strain energy. The rest is released. The amount of energy that is released per unit of crack growth is called the strain energy release rate. If the strain energy release rate is greater than the amount of energy needed to create a unit of crack growth (aka the critical strain energy release rate) the crack growth will be unstable.

I hope that clears things up, if not, feel free to let me know in the comments below.

*If you remember the discussion on elastic and plastic deformation, you’ll realise this only applies in the case of elastic deformation. Plastic deformation dissipates energy, so if you plastically deform a specimen, the energy you require for that will not be stored. For the sake understanding the basics we can safely ignore plastic deformation in this discussion.