Problem 1:

To develop an expression for the crack tip opening displacement (CTOD) 6, the book starts with    an equation for the displacement at the crack tip uy based on an effective crack length of a + Ty :

uy = KIl2几(Ty)

The book then substitutes in the plane stress equation for the Irwin plastic zone size, Ty to derive an equation for 6 = 2uy .

Develop an expression for 6 using the plane strain equation for Ty 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 Ty . This is easily estimated using LEFM, and is given in Equation 3.1 in Anderson 3rd  ed. and in the problem:

uy = KIl2几(Ty)

E′ is the effective modulus, and for the plane strain case, E ′ = .

The last part of the puzzle is to find Ty for the plane strain case, which is given in Equation 2.68 in Anderson 3rd  ed.:

Ty plane strain =6几(1) ()2

Now some algebra:

6 = 2uy = KI|6几(1) (KIGYS2几)2 = 4(几(1)2)

Which is your answer. If you want it in terms of G:

4    G

6 =

It is worth noting that:

6plane e = 6plane G

< 1 → 6plane e < 6plane G

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 Keff, 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:

Keff = o(几aeff lsecant (几2(a)w(eff))

o = 250 MPa

oys = 350 MPa 2w = 203 mm 2a = 50.8 mm

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.   Keff is estimated from the nominal crack size.

Keff step 1 = G√几a lsecant ( 2(几)w(a)) = 73.5 MPa√m

2.   An  estimate  of  aeff is  made  by  adding  the  new  plastic  zone  size  estimate  ry  (based  on Keff step 1) to the original crack length:

aeff step2 = a + ry = a + ()  = 32.4 mm

3.   Keff is calculated from the aefffrom step 2.

Keff step 3 = G(几aeff step 2 lsecant (几a) = 85.2 MPa√m

4.   The estimate of aeff is updated by using the new Keff from step 3.

aeff step 4 = a + ry = a + ()  = 34.8 mm

5.    Repeat this cycle until Keff stops changing…

This is a bit tedious, but it converges fairly quickly:

Plotted, this looks like:

Since aeff has stabilized, six iterations is good enough. Your final answer is 91.7 MPa√m.

Problem 3:

For an infinite plate with a through crack 50.8 mm long, complete the table below of Keffvalues 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 oys = 250 MPa.

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

Keff = Knormal LEFM = o√冗a

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:

Keff = l1 冗(s )2

The strip yield model equation to use is Equation 2.81 in Anderson 3rd  ed:

1/2

Throwing all that at your preferred software yields the following table:

If you plot these three, you get:

The three solutions agree up to ~ , 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. Above about 0.85 ays, the accuracy of either     correction should be questioned, particularly since as a → ays, the strip yield model goes to ∞ .

Problem 4:

A material exhibits the following crack growth resistance behaviour:

6.95

2

(aC − a0)− 0.5 =

→ aC − a0 = ()− = ()−2

aC − a0 = 144.6 mm

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 aC , we need to take advantage of the other thing we know about R curves at final fracture: R = G when a → aC .

6.95(aC − a0)0.5 =

E 6.95(aC − a0)0.5 207,000 MPa × 6.95(144.6 mm)0.5 kJ/m2

The two relations we used to solve this problem come from the point of tangency between the G and R curves. At that point, the value of each is equal (G = R), and the slope of the curve is equal

as well ( = ).

Part b:

In this part, a0 is given, but the stress at fracture (o) and the final crack length (aC) are unknown. The same two equations will give the answer, but since both unknowns appear in both equations, they need to be solved simultaneously:

6.95 =   2

=

− 0.5

and

and

R = G

o2 E(叮) aC = 6.95(aC − a0)0.5

Solving the first of these for o2 :

o 叮(5)E

Substituting into the second equation:

(aC − a0)      ×         = 6.95(aC − a0)0.5

Rearranging and cancelling:

= (aC − a0)0.5

→ = aC − a0

aC = 2a0 = 2 × 25.4 mm = 50.8 mm

This is the final half-crack size at failure that the problem asks for.

To find the stress, we can use the equation for o2 from earlier:

o = l6.2(9)叮(5)E (aC − a0)− 0.5 = l6.95 × 2 2(0)叮(70)00 MPa (50.8 mm − 25.4 mm)−0.5 = 213 MPa The problem also asks for the stable crack growth at each crack tip, which is just aC − a0:

aC − a0 = 50.8 mm − 25.4 mm = 25.4 mm

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

B = 25.4 mm ℎ = 12.7 mm     a0 = 152 mm

R = 7.15(a − a0)0.49 wℎen a is in mm

E = 207 GPa    v = 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 :

K = 2√3 → G = = 12

Keep in mind E ′ = 1 2 for plane strain. It is convenient to leave it as E′ for now.

Putting this into the two criteria at final fracture:

G = R

=