Problem 1:

Explain why the J-integral approach is not a good idea for cyclic loading. Why are there issues with the assumptions for the J-integral approach if the crack grows?

Solution:

For cyclic loading the unloading portion of the load-displacement behavior will be linear elastic instead of non-linear elastic as assumed:

reality

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For crack growth, the plastic zone will partially unload even if the overall load is maintained on the specimen. This violation is not too bad for small amounts of growth however.

Problem 2:

At room temperature, tensile specimens of polycarbonate show 60% elongation and no stress whitening, while thick compact specimens used in fracture toughness testing show stress whitening at the crack tip. Explain these observations.

Note: Polycarbonate is an amorphous glassy polymer at room temperature.

Solution:

Different stress states in the two specimens lead to different yielding mechanisms. This material  exhibits shear yielding when stressed uniaxially. No triaxial stresses are developing in the uniaxial sample in this material.

Under the triaxial tension stress state at the crack tip in the thick C(T) sample, the material yields by crazing. Voids form due to the hydrostatic tensile stresses and visibly whiten the material.

Problem 3:

The low strength, high toughness steel in the previous problem set required a huge sample to run a valid KIC test. Using the same material properties (Gys  = 350 MPa,  GUTS  = 650 MPa,  KIC  =

200 MPa√m), answer the following:

a)    If the plate thickness were 1.0 cm, would the thickness be sufficient for a JIc test?  Assume E =  170 GPa, v = 0.25.

b)   Estimate the specimen size requirements for a JIC test if the elastic properties are instead E =  207 GPa, v = 0.3.

c)    Comment on your answer to b) in comparison to the KIC  sample size you previously calculated.

Part a:

First we must convert the KIc value into an equivalent JIC value, we can do this since the original specimen from problem 4 Q7 was valid for LEFM testing.

KIC(2)          (200 MPa√m)2 (1 − 0.252)               kJ

E`                         170 GPa                                m2

Important, remember that JIC testing uses the flow stress, GY, which is the average of the yield strength (called GYS  in this course) and ultimate tensile strength (GUTS). This is because it is a    more accurate representation of the average stress in the yielded material.

Checking the size requirement:

B ≥ 25  = 25  = 11 mm

No, you need a plate thickness of at least 11 mm for a valid JIC test.

Part b:

First convert KIC to JIC  using the new elastic properties:

KIC(2)          (200 MPa√m)2 (1 − 0.32)               kJ

E`                        207 GPa                               m2

Remembering to use the flow stress GY, check the size requirement:

B ≥ 25  = 25  = 8.8 mm

So a sample with these elastic properties would only need to be 8.8mm thick.

Part c:

The question from the previous problem set was a high toughness low strength steel that would have required an impossibly large specimen (B>816 mm, W>1.6 m) to run a valid KIC test. By       using the JIC test instead, the sample and loads can be much more reasonably sized. This is          because JIC tests can accommodate significant plastic deformation before failure, and accurately predict failure.

Problem 4:

What is the minimum flow stress oY  needed to have a valid JIC test using the specimen dimensions and elastic constants for Problem set 4 Q9?

Solution:

Remembering to use the flow stress oY , we can start with the specimen size requirement and apply the relationship between KIC  and JIC :

B ≥ 25  = 25

Solving for oY and inserting the specimen size and material properties:

KIC(2) (1 − v2)               (35 MPa√m)2 (1 − 0.32)

oY   ≥ 25                                 = 25                                                              = 5.2 MPa

So we need oY  > 5.2 MPa to have a valid test.  About any metal should meet this requirement. This is a reasonable sample size as well.

Problem 5:

Several fracture toughness specimens have been loaded to various points and then unloaded.        Values of J and crack growth were measured in each specimen and are tabulated below. Using the basic test procedure, generate the J-R curve for this material and determine JQ , and, if possible,     JIC .

oYS  = 350 MPa        oTS  = 450 MPa       B = 25 mm         b0  = 22 mm

Solution:

We need to construct a plot like 7.24 in the book, with the power law expression Eq. 7.13 in book fit to our data:

J = C1(Δa)C2

But first, we need to determine what data to fit. According to the standard and lecture slides, the acceptable data falls in a region defined by five criteria:

1.   Jmax

2.    0.15 mm exclusion line

3.    1.5 mm exclusion line

4.   Δamin  exclusion line

5.   Δamax  exclusion line

Criteria 1 is meant to exclude data with way too much plasticity. Criteria 2 is meant to exclude      data during the blunting phase of the test. Criteria 3 excludes data with too much crack growth.   Criteria 4 and 5 are included for the same reasons as 2 and 3 respectively, but are evaluated after the first curve fit is complete. Once there is a fit to work with, it’s possible to be more precise        about what data is excluded.

The first three can be expressed as:

1.   Jmax  =  b0 GY  =  (22 mm) () = 1173

2.   J = 2 GY  (a − 0. 15) = 2 × 400 MPa (a − 0. 15)

3.   J = 2 GY  (a − 1.5) = 2 × 400 MPa (a − 1.5)

Notes:

•   The units all work for this: MPa × mm = 106 Pa × 10−3 m = 103   =  .

•    Lecture slides and the ASTM standard use Jmax  =  b0 GY, but Anderson 3rd edition lists Jmax  =  b0 GY . We are sticking with the ASTM value here.

All our data fits in this region (see plot at end), so we can move on to fitting the curve and          checking the last two requirements. This will generally require using a fancy calculator, plotting software (e.g., KaleidaGraph, Origin, etc.), mathematical software (e.g., Matlab, Mathematica,

etc.), or some other scientific software (e.g., Labview) to find the exact least squared fit. Note that this is an exact process; there is only one least-squares fit. For this data, the fitted equation is:

Jfit  = 215.2 (Δa)0.469

Now that we have the fit, we can draw the last two exclusion lines: Δamin  and Δamax . These are vertical lines drawn straight down from the intersection between the curve fit and the 0.15 mm  and 1.5 mm exclusion lines. The first point is just above Δamin , so nothing needs to be excluded.

The last thing we need to do is to find the intersection between the fit and another offset line like line #2 and #3, but this time starting from 0.2 mm. This offset procedure is similar to the way we  define an offset yield strength in a tensile test. Doing this with whatever mathematics package      you’re using, you’ll find that:

JQ  = 134.7

Plotting these data points, exclusion lines and fits, this problem looks like:

The 0.2 mm intersection line is in red, and all the exclusion lines are in black.

Last, we have enough data to check the sample size requirement:

25  = 25  = 8.2 mm < b0, B