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Fracture Failure of Engineering Materials
Copyright © 1998-2005 by Pichai Rusmee
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This laboratory will explore the concept of fracture mechanics. Since the student will, by this time, be exposed to only one or two lectures on fracture mechanics, the emphasis of this laboratory will be more on the application of fracture analysis as well as reinforcing certain ideas pertaining to fracture mechanics. The student will determine the critical stress intensity factor, KIc, for an engineering material in the laboratory. Once they determine the necessary value, they will use it to analyze machine parts. Thickness effect on the stress intensity factor will be demonstrated in the lab.

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Concept of grain and crack
The concept of crack can be thought of in many ways. A material scientist or A metallurgist may look at cracks on a molecular level. An engineer will probably look at the applied mechanics or macro aspect of cracks.

On an atomic level, metals consist of crystal structure of various types such as Face Center Cubic, Body Center Cubic, or Hexagonal Close Pack. In each of the crystal structures, there are identifiable planes in which the individual atoms can slide pass each other relatively easily under some applied forces. Such action would be observed as yielding phenomenon. Instead of sliding pass each other, the atoms can also cleave apart. The void formed by this action is crack. On molecular level, a crack is a collection of dislocations splitting the material apart.


Figure 4.1 Crack formation in a single crystal lattice.

In real material, there will be more than one clump of material that share the same crystal orientation. This form the concept of grains. A grain is a collection of atoms that have the same crystallographic structure and orientation. Each grain will have its own cleavage plane where the crack prefers to propagate. As the crack crosses the boundary between two grains it must turn or change direction to align itself with the cleavage plane of the new grain. It follows that, the more turn the crack has to make the more energy it will take to make the crack propagates. Therefore, the material with small grain or material with crack propagating transverse to the grain will be able to absorb more loading energy before the crack start to propagate.


Figure 4.2 Crack propagation across grain boundaries.


Figure 4.3 Correlation of fracture stress with the grain size.

Applied mechanics
Applied mechanics would be less concern with the metallographic structure but will concentrate on the stress state in front of the crack tip or looking at crack growth in term of macro mechanics. From theory of elasticity, the stress in front of the crack tip in any direction is a function of the stress intensity factor, the distance from the crack tip, and the direction, θ:
This theory predicts that the stresses in front of the crack tip will be higher than the applied stress (as r approaches zero). In fact, σx and σzwould tend to infinity (Figure 4.4a). But in reality, the material can only be stress up to a finite value. Therefore, the infinite stress state is impossible since the material will yield or plastically deformed well before then. A yield zone will be directly in front of the crack tip could be estimated if a yield criterion is assumed. For mode I crack using the maximum shear stress criterion, the radius of the plastic or the yield zone is:
One result that arises from this analysis is that the maximum stress,σo, equals to σy - σx > σYS (Figure 4.4b). The stress in the y-direction is the apparent yield stress that is three times higher than the normal yield stress of the material. In term of the study of fracture mechanics, this is a beneficial stress state. As with notch, the presence of crack induces a triaxial stress state in front of the crack tip. Since this stress state raises the apparent yield stress of the material to be higher than the actual yield stress, the material will not shear as easily thus reducing the affect of shear failure.


Figure 4.4 Stress state in front of a crack tip a) according to theory of elasticity, b) with plastic zone.

Mode of fracture
There are three basic mode of fractures called Mode I, II, and III. They are fracture by pulling, pushing, and tearing respectively. Figure 4.5 demonstrate the three mode of fractures. Symbol a or c is usually used for crack length in case of an edge crack. For inside crack, a or c is half the total crack length.

FIG 4.5
Figure 4.5 Three mode of fractures.


Plane strain/plane stress
As in the composites lab, the stress or strain state is always in three dimension. But in most cases, they can be simplified to either plane strain or plane stress by ignoring either the out of plane strain or plane stress.

The fracture toughness experiment must be set up so that the plane strain condition is satisfied. This is because the plane strain will result in a lower maximum possible shear stress in the case of positive stresses. To illustrate one of the reason why this is the case, draw two sets of 3-D Mohr's circles. The plane stress condition will results in a higher out of plane shear stress than the plane strain condition for a given maximum stress. Again, the shear stress would promote yielding not fracture.

FIG 4.6
Figure 4.6 3-D Mohr's circles for plane stress and plane strain conditions.

The decisions to do a plane stress or plane strain analysis is subjective. Usually, a "thin" part is analyze as plane stress while a "thick" part as plane strain. However, ASTM E399 give a more specific condition in which a fracture toughness specimen will have an acceptable plane strain state. In order to have a valid test, the thickness of the specimen must be at least:
Grain directionality

Stress intensity factor

Geometric correction factor

Test results
This may be one standard that valid results could be impossible to obtain, i.e., satisfy all the conditions set forth in the ASTM E399. The thinner the specimen (up to certain thickness) will produce a higher KI value that the thicker specimen because the specimen would be in plane stress state. If we plot the KI's versus specimen thickness, KIc is the asymptote of the curve as the specimen gets thicker. The specimen in plane stress may show a telltale sign of its stress condition by exhibiting shear lips. Shear lips are ledges on the side of the specimen that make a 45° to the plane of fracture. Figure 4.7 shows the curve of KI versus specimen thickness and the extend of shear lips on the specimens of various thicknesses.


Figure 4.7 Stress intensity as a function of specimen thickness and the relative extend of the shear lips.

Using the result
For a given geometry and load condition, we calculate the KI for that case. If KI exceeds the previously known value of KIc than it is said that the crack will propagate. In the case where the sample in question is clearly in the state of plane stress, the KIc can still be used as long as it is understood that the result will probably be conservative.

We can attempt to combined the K's for different mode by using the weight average or various weighted sum methods. Two methods are commonly suggested.
   (Doesn't work very well)
or if mode I fracture dominates the other mode:
   (Doesn't work at all)
Caution: Just because a crack is presence doesn't mean that a sample will fail in fracture. It also doesn't mean that the other mode of failure will occur in the vicinity of the crack.

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Handout

Elementary strength of material texts usually assume that all materials are in continuous bulk, i.e., homogeneous without discontinuities, flaws, or imperfections. In reality, the opposite is often true. Fracture mechanics is a study of bodies containing such discontinuities or "defects."

An applied stress can be thought of as energy input to a body. This energy approach to fracture mechanics was proposed by Griffith using the first law of thermodynamic. A body under an applied stress is capable of absorbing some energy through elastic deformation. The excess energy that can not be absorbed must be released by some other means. If the loading condition does not preclude the shear stresses some of the energy can be released through yielding. Another way to release the energy is by creating new surfaces, i.e., crack growth. In this case, the controlling failure mode may be either yielding or fracture depending on the initial crack presence, load condition, etc.

One of the approaches in fracture mechanics is to keep track of energy release rates, G. It is assumed that a crack will propagate only when a certain energy level is reached. An alternative to the energy release rate approach is the stress intensity approach. The stress intensity factor, K, defines the magnitude of stresses in the vicinity of the crack tip. When a critical stress intensity, Kc, is reached, the crack propagation occurs. For linearly elastic materials, the energy release rates and the stress intensity factor are simply related:
    (1)
The general form of a stress intensity factor is given by:
    (2)
where σ = applied stress
a = crack length
f(α) = correction factor which varies with specimen geometry and crack geometry

Many handbooks contain f(α) values for various types of specimens and crack geometries.

In this laboratory, you are a design engineer in charge of evaluating the design limit of certain structural components. It is your responsibility to determine the maximum loads the parts (samples) can withstand without causing unacceptable failure as well as identifying the failure mode. Using ASTM E399 as a guide, you will carry out a plane strain fracture toughness experiment in order to obtain the necessary information for your investigation.

There are two structural components in question, Part A and Part B. Figure 1 shows a schematic drawing of Part A. It has been modeled as a simply supported beam with a load in the middle. The beam has a through the thickness edge crack underneath the load on the tension side. The following information is known about Part A:
  • The depth, d, is 0.1100 meters, the thickness, b, is 0.0254 meters and the length, L, is 1.0 meters.
  • The crack length, a, is 0.0007 m. in length.
  • For the case of an edge-cracked plate loaded in tension,[1]
       (3)
    In our case, assume that a«W

Figure 1 A machine component modeledas a simply support beam with a through the thickness edge crack.

The schematic of Part B is shown in Figure 2. It has similar dimension as part A but has been modeled as a cantilever beam. The through thickness edge crack for this part is 0.00021 m. in length on the tension side.


Figure 2 A machine component modeled as a cantilever beam with a through the thickness edge crack.

The tasks listed below outline the necessary steps to carry out the plane strain fracture toughness test using the compact tension (CT) specimens similar to that shown in Figure 3. It also points out factors effecting your experimental result that must be addressed in your report.

Task
  1. Load a precracked compact tension (CT) specimen to failure according the ASTM E399. Generate a plot of load versus crack opening displacement (COD).
  2. Calculate the initial slope of the plot you just generated using a linear least-squares curve fit forcing the curve fitted line to go through zero (i.e., in the form y = bx rather than y = a + bx). Do not forget to account for the start up or toe region of your plot. See the Appendix at the end of this section for additional help and information on curvefitting.
  3. On the same plot, draw a line through the origin at 95% of the slope you calculated in Task 2. The load at which this line intersects the curve is designated PQ unless there is a peak on the load-COD curve preceding the point of intersection. In this case, PQ is taken to be the load at the preceding peak.
  4. Calculate the first conditional result, KQ. For the CT specimen using:
        (3)
    Where α=a/W
  5. Is the KQ you calculated a valid KIc value? Check to see if all the criteria listed in the Appendix are satisfied. Include the summary of your finding in the result section.
  6. Define plane strain and plane stress.
  7. A stress element has a principal stress state where σx > σz ≤ 0 and σy = 1/2 σx Taking σz to be zero, draw Mohr's circles using σx, σy, and σz. These 3-D Mohr's circles show the principal stresses in a plane stress condition when τ's equal to zero. Make note of the maximum possible shear stress if we deviate from the principal stresses directions, i.e., rotate the stress element 45°

    Figure 3 Definition of symbols used for dimensions of CT Specimen.

  8. A stress element has a principal stress state where σx is as before but σy = 1/3 σx and σz = 1/2 σx. Draw another set of Mohr's circles using σx, σy, and σz. These Mohr's circles show the principal stresses in a plane strain condition when τ's equal to zero. Make note of the maximum possible shear stress if we deviate from the principal stresses directions, i.e., rotate the stress element 45°.
  9. Judging from the maximum possible shear stresses from Task 7 and 8, which condition, plane stress or plane strain, would promote yielding and not fracture? Can you explain why we perform a plane strain fracture toughness test and not plane stress fracture toughness?
  10. Will you be able to apply the test result to your machine component if they are in plane stress condition? Justify why or why not?
Reference
[1] Gross, B., Srawley, J.E., Stress Intensity Factors for a Single Notch Tension Specimen by Boundary Collocation of a Stress Function, NASA TN D-2395, 1964.

Appendix

TASK 2 (Least-Squares Approximation Technique)
Many times in engineering, it is necessary to approximate experimental data with some form of a curve fit. A method that lends itself nicely to linear elastic experimental data is the linear least-squares method of approximation. The method is simple to apply and gives useful results. We will first derive the necessary equations. (You will have to do a similar derivation for the equation y = bx for this task rather than y =a +bx.)

The idea behind the linear least-squares approximation technique is to assume a straight line in the form
    (4)
Next we sum the squares of the differences between this line and the experimental data for all the data points (m data points)
    (5)
The best line is achieved when the sum of the squareddifferences is as small as possible. Therefore we need to minimize φ by taking derivatives with respect to a and b and set them equal to zero.
    (6)
    (7)
We can rewrite (6) and (7) in the form:
    (8)
    (9)
We now have two equations and two unknowns. One easy way to solve for a and b is using Kramer's rule. We can rewrite (8) and (9) in matrix form
    (10)
Making the substitutions
    (11)
Where A, B, C, and D are as shown in (10). Then, a and b are
    (12)
    (13)
and remembering that
    (4)
We now have our approximation.

There are several ways available to solve for a and b. You can solve them by using brute force, spread sheet, or computer program. Although beyond the scope of this laboratory, this method can also be applied to obtain approximations other than linear equations. For example, to find an approximation of the form:
    (14)
This time you would need to take three derivatives instead of two and solve three equations for a, b, and c. You can also easily apply this method to exponential equations, logarithmic equations, sine equations, cosine equations, etc.

TASK 4 (Summary of ASTM E-399 Validity Criteria)
{Metric Equivalents Shown in [ ] } A- Precracking
  1. a > 2.5 ( KQ / σYS)^2      9.1.3
  2. 0.45 Wa ≤ 0.55 W      7.3.2.1
  3. |a4 - anotch| > 0.025 W       7.3.2.2
    |a4 - anotch| > 0.050 in [1.3 mm]
    |a5 - anotch| > 0.025 W
    |a5 - anotch| > 0.050 in [1.3 mm]
  4. Plane of precrack to be parallel with X and Z axes within ± 10°       8.2.4
  5. -1 ≤ R ≤ 0.1       A2.1.2
    where, R = Lower Load / Upper Load from data sheet.
  6. During last 2.5% of crack length a:       A2.3.3
    where Kmax is based on "Upper Load" from data sheet.
B- Loading rate
  1.            8.3
C- Test data
  1. Pmax / PQ < 1.10       9.1.2
  2. B > 2.5 ( KQ / σYS)^2       9.1.3
D- Fatigue precrack geometry |a1- a2| < 0.1 a       8.2.2
|a2 - a3| < 0.1 a
|a3 - a1| < 0.1 a
|a4 - a5| < 0.1 a
|a4 - a| < 0.15 a
|a5 - a| < 0.15 a

where,

|a1 - anotch| > 0.025 a and |a1 - anotch| > 0.050 in. [1.3 mm]
|a2 - anotch| > 0.025 a and |a2 - anotch| > 0.050 in. [1.3 mm]
|a3 - anotch| > 0.025 a and |a3 - anotch| > 0.050 in. [1.3 mm]

[Note: use the value of a as defined here to calculate KQ]

Figure 4 Schematic of crack dimensions for CT specimen.
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Lab work
Data Reduction
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Miscellaneous


Watch out for units especially the one inside the square root.

Unlike other previous laboratory exercises in this course, the emphasis of this laboratory will be different. This lab will be structured as an analytical assignment where the students are given certain machine part to be analyzes instead of the students reporting on the experimental results. The experiment part of the lab will be presented as one of the necessary step that the students need to perform in order to carry out their assignment. The same emphasis will continue to the Fatigue Crack Growth and the Pressure Vessel labs.

 


Last Modified
Sep 2005




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