Part a:

The growth rate relations are all in terms of K, so calculate for each point. Based on the information in the problem, use K = 1.12 o √几a

Crack growth plots are almost always plotted in log-log space:

This looks like a classic EAC crack growth curve from the lecture, with two regions, I and II. In this case, the fourth data point appears to be right at the transition between the regions, so we include it in Region I.

Now we can fit = C1 em K given in lecture to the Region I data (first four points). It is important to fit all the available data, not just between two points.

m = 0.966

C1 = 2.094 × 10− 11

Where is in and K is in MPa√m.

Plotting this fit vs. the data:

For region II, the growth rate is constant so dt  = 6.4 × 10      s .

Part b:

Yes, the relation changes from region I to region II. In region I, the chemical reaction rate controls the cracking rate, and the chemical reaction is driven by the stress intensity, so there is an exponential dependence on stress intensity. In region II, the cracking is limited by diffusion of reactants to or ahead of the crack tip. This means that the rate is not controlled by the applied     stress intensity, and so depends only on time.

Part c:

See answer to b), those data are for region II and diffusion of the environment to, or ahead of, the crack tip limits the cracking rate.

Note on exponential and power law curve fitting:

Fitting exponential and power laws to data is not as trivial as it seems. Depending on the exact details of the fit algorithm, a least-squares fit directly to an exponential ends up weighting the points at the upper end of the growth rates more heavily than the points at the lower end. This is accentuated for exponential and power law problems in crack growth because the growth rates can span several orders of magnitude.

Using an algorithm implemented like this gives you a fit that looks like this:

This is a notably worse fit at the low end of the range, generated from identical data. (To figure  out what your calculator/software is doing, compare to the constants for this fit method m=0.867 and C=4.33E-11, vs the ones given above)

There’s a better method (that Excel defaults to with the plot trend lines). Taking the log of both sides of the exponential used for EAC:

log = log(C1em K) = log C1  + m K

Since this is a simple line equation (m x + b), it can be fit much more accurately by any software. You just need to take the natural logarithm of your growth rates, do a linear fit vs K (not log K),   and turn the constant term (b in your linear fit) back into C1 for the power law by taking C1  = e b . Which logarithm base doesn’t matter so long as you’re consistent.

Which should you use? For our purposes, it’s more accurate to use the log-linear fit method described here. Again, this is what Excel defaults to with its trendlines, so you may have always   been using it without knowing. In general, it depends on the distribution of error for your measurements. If the expected error is relative, (e.g. ±10%), or if you have few data points, the log-linear method is more statistically sound. If the expected error is absolute (e.g. ±10), the      direct least-squares method can be more statistically sound.

Some of the problems later in the term will use a power law growth rate rather than an exponential. When using this method to fit with a power law, you need to take the log of both the growth rate and the stress intensity then convert both constants back.

Problem 3:

An investigation was made of the rate of crack growth in a 7079-T651 aluminium plate exposed to an aggressive environment under a static stress σ. A large test sample was used with a single-edge notch. As indicated in the accompanying table, the rate of crack growth under sustained loading was found to vary with the magnitude of the applied stress and the existing crack length. The material exhibits Regions I and II environment-assisted cracking but not region III.

If the KIC for the materials is 20 MPa√m, how long would it take to break a sample containing an edge crack 5 mm long under a load of 50 MPa?

Hint: First establish the crack growth rate relations.

Solution:

The first step in these problems is to check if the applied stress intensity is less than the threshold for crack growth. If it is below the threshold, there’s no crack growth and the life is infinite. We     aren’t given information about the threshold in this problem, so we assume there isn’t one.

Edge crack in a large sample: use K = 1.12 a √几a.

To find = f(K) relations, start by making a (log-log) plot:

Again, this looks like a standard EAC curve with both region I and II behaviour. Fitting =

C1  em K  using the first three points (Region I) and the log-linear method described above gives:

m = 1.3921                    C1  = 1.3529 × 10 − 12

Where is in and K is in MPa√m .

Since we are told there is no region III in this material, we can assume that crack propagates at

= 10 − 6 m/s

Plotting both regions with the data gives:

To find the lifetime, we must integrate:

tf  = jdt

First we need limits of integration:

a0 = 0.005 m

ac =几(1) ()2 = 0.0406 m

To decide if the crack is growing in either or both Regions I and II, we also need to know the

aI−II = 几(1) ( 1 1(KI)Ia(I))2 = 0.00983 m

It looks like our crack starts in Region I at 5 mm, transitions to Region II growth at 9.83 mm, and grows to failure at 40.6 mm. Our integral then becomes:

tf  = j−II dtRegion I + jaI(a)II dtRegion II

Converting our growth rate laws:

= C1 em K →       dt = = 几a

= C2        →       dt =

Inserting into the integrals:

tf  = tregion I + tregion II

tregion I  = j−几a                                     tregion II  = jaI(a)II

The integral on the left is not trivial. Assuming that Y is not a function of a (it is not for this

geometry), the general form of the solution is:

tregion I = 46,348 s

Note: Because the growth law expects MPa√m, you’ll need to keep the stress in MPa.