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MECH4900
Problem 1:
To develop an expression for the crack tip opening displacement (CTOD) , the book starts with
an equation for the displacement at the crack tip ! based on an effective crack length of + ! :
! = 4′"+!2
The book then substitutes in the plane stress equation for the Irwin plastic zone size, ! to derive
an equation for = 2!.
Develop an expression for using the plane strain equation for ! found in your textbook.
Solution:
This problem follows the Wells derivation that the book uses, and so looks different than what was
presented in lecture.
CTOD is derived by looking at the displacement (in the y direction) ahead of the crack tip by an
amount !. This is easily estimated using LEFM, and is given in Equation 3.1 in Anderson 3rd ed. and
in the problem:
! = 4′"+!2 ′ is the effective modulus, and for the plane strain case, # = $%&'!.
The last part of the puzzle is to find ! for the plane strain case, which is given in Equation 2.68 in
Anderson 3rd ed.:
! )*+,- ./0+1, = 16 0 "2324
Now some algebra:
= 2! = 84 1 − 47"8
16 4 "23742 = 4(1 − 4)√3 "4!
Which is your answer. If you want it in terms of :
= 4√3 23
It is worth noting that:
)*+,- 5 = )*+,- 6 1 − 4√3 1 − 4√3 < 1 → )*+,- 5 < )*+,- 6
Which is what we expect to see: the plane strain case has less deformation than the plane stress
case. For an average metal, by about a factor of 1/2.
Problem 2:
Use the iterative method in the book to calculate -77, known as the Irwin correction, for a
through crack in a plate of width 2W. Assume plane stress conditions and the following stress
intensity relationship and values:
-77 = A-77 + 4-772 7 = 250 !. = 350 2 = 203 2 = 50.8
Solution:
This calculation requires an iterative approach because a and K are interrelated. The Irwin approach
accounts for the plastic zone by adding extra crack length equal to ry:
The problem is that adding the extra crack length changes the effective stress intensity, which
changes the ry, which changes the effective crack length, which changes the stress intensity………
Following the iterative procedure outlined in the book:
1. -77 is estimated from the nominal crack size. -77 ./-) % = √ + 4 27 = 73.5 √
2. An estimate of -77is made by adding the new plastic zone size estimate ! (based on -77 ./-) %) to the original crack length: -77 ./-)4 = + ! = + 12πP-77 ./-) %! Q4 = 32.4
3. -77 is calculated from the -77from step 2. -77 ./-) 8 = A-77 ./-) 4 + 4-77 ./-) 42 7 = 85.2 √
4. The estimate of -77is updated by using the new -77 from step 3.
-77 ./-) 9 = + ! = + 12πP-77 ./-) 8! Q4 = 34.8
5. Repeat this cycle until -77 stops changing…
This is a bit tedious, but it converges fairly quickly:
Iteration New !"" (MPa√) New !""(mm)
1 73.5 32.4
2 85.2 34.8
3 89.3 35.8
4 90.8 36.1
5 91.4 36.3
6 91.7 36.3
Plotted, this looks like:
Since -77 has stabilized, six iterations is good enough. Your final answer is 91.7 MPa√.
Problem 3:
For an infinite plate with a through crack 50.8 mm long, complete the table below of -77values
calculated using the LEFM, Irwin correction, and strip yield (Dugdale and Barenblatt) models.
Describe when you would use each of these methods.
Note: with this geometry, you can use the closed form Irwin correction formula rather than the
iterative method from Problem 1. Assume !. = 250 .
Applied stress
(MPa)
-77 √
LEFM Irwin correction Strip yield model
25
50
100
150
200
225
249
250
Solution:
This is a fair amount of calculation, so you should really use Excel or a mathematics package
(Mathematica, MatLab, Maple…) for this.
The LEFM column is easy. LEFM assumes no plastic zone so -77 = ,:0;+* <$=> = √
The Irwin correction column is a LOT easier than in Problem 4 because there is a closed-form
solution available for this geometry. This is Equation 2.70 in Anderson 3rd ed:
-77 = √S1 − 12 0 !.24
The strip yield model equation to use is Equation 2.81 in Anderson 3rd ed:
-77 = !.√ T 84 ln sec P 2!.QZ%/4
Throwing all that at your preferred software yields the following table:
Applied stress
(MPa)
!"" √
LEFM Irwin correction Strip yield model
25 7.06 7.08 7.08
50 14.1 14.3 14.2
100 28.2 29.5 29.3
150 42.4 46.8 46.3
200 56.5 68.5 68.9
225 63.6 82.4 86.6
249 70.3 99.1 143
250 70.6 99.9 ∞
If you plot these three, you get:
The three solutions agree up to ~ 6"#4 , so at or below this, we really don’t need a plasticity
correction. Above this, the Irwin and Strip Yield models are worthwhile; they predict an increase
in Keff due to plasticity, whereas LEFM does not. The two models agree to about 0.85 !., but the
accuracy of either correction should be questioned at high fractions of the yield stress, particularly
since as → !., the strip yield model goes to ∞.
Problem 4:
A material exhibits the following crack growth resistance behaviour: = 6.95( − @)@.B
where @ is the initial crack size. has units of /4 and crack size is in mm. Consider a wide
plate made from this material with a through thickness crack ( << ). = 207 , and you
can assume plane stress conditions.
a) If this plate fractures at 138 MPa, compute the following:
a. The half crack size at failure C
b. The amount of stable crack growth at each crack tip that precedes failure C − @
b) If this plate has an initial crack length 2@ of 50.8 mm and the plate is loaded to failure,
compute the following:
a. The stress at failure
b. The half crack size at failure C
c. The stable crack growth at each crack tip
Part a:
First, a few notes:
• The units used in this problem are unusual. is in DE;! but is in . This means that the
coefficient on the right side has units of DE;!√;;. If you want to change any units, you must
very carefully adjust the coefficient. However, all of the values given are in the correct
units for this to ‘just work’. Why do this and not use a consistent unit system? R-curve
data is often plotted as a function of or because of the size scales involved, so it
is easier to keep the mixed units.
• It is only a coincidence that the exponent in the R equation is 0.5. This exponent can be
any real number between 0 and 1 that fits the measured R curve. This solution keeps the
decimal value instead of using radicals in recognition of this.
• No initial crack length @ is given for the first part of this problem. That’s OK.
The problem says to consider this an infinite plate ( ≪ ) in plane stress, so:
= √ = 4 = 4
We know that unstable fracture occurs when: = = 4 = 4 = 6.95( − @)@.B = 6.952 ( − @)&@.B
We can equate these and solve for G − @. (Note: → G in these equations because this is the
crack length at final fracture.)
6.952 (G − @)&@.B = 4 → G − @ = P2 4 6.95 Q& %@.B = P 2 (138 )4 6.95 × 207,000 Q&4 G − @ = 144.6
This is the amount of subcritical crack extension before failure and the answer to the second part
of a).
To get the crack length at failure G , we need to take advantage of the other thing we know about
R curves at final fracture: = when → G .
6.95(G − @)@.B = 4 G
G = 6.95(G − @)@.B4 = 207,000 × 6.95(144.6 )@.B /4(138 )4 = 289.1
Which is the answer to the first part of a).
Graphically, this solution looks like:
The two relations we used to solve this problem come from the point of tangency between the
and curves. At that point, the value of each is equal ( = ), and the slope of the curve is equal
as well 4HH+ = HJH+7.
() ()
@ G
Part b:
In this part, @ is given, but the stress at fracture () and the final crack length (G) are unknown.
The same two equations will give the answer, but since both unknowns appear in both equations,
they need to be solved simultaneously: = = 4 = 6.952 (G − @)&@.B 4 G = 6.95(G − @)@.B
Solving the first of these for 4:
4 = 6.95 2 (G − @)&@.B
Substituting into the second equation: 6.95 2 (G − @)&@.B × G = 6.95(G − @)@.B
Rearranging and cancelling: G2 (G − @)&@.B = (G − @)@.B → G2 = G − @ G = 2@ = 2 × 25.4 = 50.8
This is the final half-crack size at failure that the problem asks for.
To find the stress, we can use the equation for 4 from earlier:
= S6.95 2 (G − @)&@.B = S6.95 × 207000 2 (50.8 − 25.4 )&@.B = 213
The problem also asks for the stable crack growth at each crack tip, which is just G − @: G − @ = 50.8 − 25.4 = 25.4
Problem 5:
Suppose that a double cantilever beam (DCB) specimen is fabricated from a material that exhibits
R curve behaviour. Using the given information, and assuming plane strain, calculate:
a) load at failure
b) the amount of stable crack growth
= 25.4 ℎ = 12.7 0 = 152
= 7.15( − @)@.9K 4 ℎ = 207 = 0.27
Solution:
This problem uses the same basic strategy as the previous: use both criteria to find the point of
tangency. The first thing needed is an expression for G. From the SIF sheet:
= 2√3 ℎ8 4L → = 4′ = 12 4 4 ′ 4 ℎ8
Keep in mind # = $%&'! for plane strain. It is convenient to leave it as ′ for now.
Putting this into the two criteria at final fracture:
= → 12 4 G4 ′ 4 ℎ8 = = 7.15(G − @)@.9K () = → 24 4 G ′ 4 ℎ8 = 2G = 0.49 ∗ 7.15(G − @)&@.B% ()
These equations need to be solved simultaneously. Noticing that HH+ = 4+ allows us to use an
algebraic trick for this. Dividing equation A by equation B yields: G2 = 10.49 (G − @) → G = 11 − 0.492 @ = 201.3
From here, it’s simple to get the answer to part b: ℎ = G − @ = 201.3 − 152 = 49.3
Either equation A or B can be used to get the load. Using A and solving for P, and substituting in
the definition of # for plane strain:
= S 4 ℎ8(1 − 4) 12 G4 7.15(G − @)@.9K = 5.41