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Resistance Curve

Zhigang Suo's picture

These notes were initially written when I taught fracture mechanics in spring 2010, and were updated when I taught the course again in 2014.

You can access all notes for the course on fracture mechanics

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Why the resistance curve does not take a steady state under wedge loading in materials not pocessing small-scale yielding (like cohesive zones with long bridging length and under small loading rate)? If the bonding length is long enough, will it approach a steady state when crack propagates?

Zhigang Suo's picture

Here is a discussion on the large-scale bridging condition:

Z. Suo, G. Bao and B. Fan, Delamination R-curve phenomena due to damage, J. Mech. Phys. Solids. 40, 1-16 (1992).

As illustrated in Fig. 2, a staedy-state is still possible even under the large-scale bridging condition.  To attain the steady state, you need to use a long and thin beam. 

You may also take a look at the comments on experiemnts by Bent Sorensen.

In the page 13 of the same article, you mentioned it does not take a steady state under Wedge forces and my question was regarding that.

Zhigang Suo's picture

I'm sorry to miss your question.  Under the large-scale bridging condition, the length of the bridging zone is comparable or even exceeds the size of the specimen (e.g., the thickness of the beam).  Consequently, it is difficult to convert the experiemntally determined relation between the load and the crack extension into a R-curve.

In the case of wedge force, the configuration does not have a translational symmetry.  Steady state is impossible, regardless of the size of the bridging zone.

By contrast, when a moment is applied, a steady state can be attained. 

 

I1: Under the large-scale bridging condition, it is difficult to convert the experiemntally determined relation between the load and the crack extension into a R-curve.
Q1
: You mean crack length is hard to be experimentaly measured in the presense of large bridging?

I2: In the case of wedge force, the configuration does not have a
translational symmetry. Steady state is impossible, regardless of the
size of the bridging zone. By contrast, when a moment is applied, a steady state can be attained.
Q2: I guess you mean material response is translational symmetric under the case of pure moment and/or shear (in the direction of the interface) forces, i.e. it develops along the interface (in weak adhesives) while crack is extending and different pre-bonded points are assumed to provide same material response. But I am looking for the reason of the case under wedge forces. Is that because when the crack slightly extends, the moment at the tip of the crack increases while the wedge force at the load-point is constant, so that the rate of energy release gradually increases?

Zhigang Suo's picture

I1:  The dificulty lies in the determination of G or K in terms of the applied load.  Under the large-scale bridging condition, the bridging zone is so large that one may not neglect it in solving the boundary-value problem.  However, the bridging law is typically unknown prior to the experiment, so one does not not know what to use to set up the boundary condition. 

One may choose to neglect this inconvenience, neglect the bridging zone, and use the linear elastic elastic solution to relate G to the applied load.  But this practice has no reason to be valid.  Fig. 10 in the 1992 paper illustrates this point.

I2:  By translational symmetry I mean that when the bridging zone is translated along the beam, the field remains unchanged. 

I1: While this is true in a strict sense, in practice a lot
depends on data reduction methods. We have performed similar
simulations in one paper
and compared two data reduction methods with FEM calculations. For one
method the same results like in your paper were obtained, but other was
very close to FE results (the error was within few percent). Applying
this method to the experimental data, we got steady state value of G
for two specimen's thicknesses (for third thickness the specimens were
too short).

I2:  If the specimen is sufficiently long, we will get translational symmetry for wedge force.

Zhigang Suo's picture

Dear Sergej:  Thank you very much for your comments.  I've just read your paper.  I really like it.  In particular, I am intrigued by your Figure 7, which shows that Eq. (9) can be used to evaluate J.

As you pointed out, the result is significant in practice.  It enables one to use DCB to determine the bridging law from experimental data.

Do you know why Eq. (9) works, while Eq. (7), (8) and (10) do not?

Dear Zhigang,

I think that strictly speaking they all do not work, Eq. (9) just has much better accuracy than others. As it was shown in our next paper
(Fig. 6), Eq. 9 in general is very accurate even for short cracks. The
reason for this - it does not contain crack length. Williams in 1989
have shown that for short cracks one needs to correct crack length to
get accurate results in DCB test. There is no crack length in Eq. 9, no
corrections are needed.

But it is also possible to use angle
method (Williams 1987, Paris & Paris 1988, Nilsson 2006), which
should incorporate all bridging effects, since it is based on
J-integral computation along exterior boundary of the specimen.  

Bent F. Sørensen's picture

As pointed out by Sergej, the J integral evaluated along the external boundaries of the DCB loaded with wedge forces is (per beam) force per unit width multiplied by the rotation angle at the pint where the force is applied (Paris and Paris, 1988). It is then easy to understand why the LEFM equation for  G  overestimates J (see Fig. 10 in Suo, Bao, Fan, 1992): The bridging tractions reduce the deflection and the rotation of the beam, so that the rotation at the point where the wedge force applies is less than what is predicted using beam theory - the beam theory equation for the energy release rate  G  can be considered as being the J integral solution with the rotation angle of an unbridged beam.

This may also be the reason why equation (9) in Sergej paper works better that equation (7). Bridging tractions will reduce both the defection and rotation of the beams. Thus, if the deflection is measured (but not the rotation) the reduce deflection used in (9) will reflect the reduced rotation.

The J integral solution described above also explains why the so-called root rotation is important for the wedge loaded DCB; it gives an additional rotation also at the point where the forces apply and thus influences the J value.

Ulf Stighs group in Sweden has successfully used a shaft encoder to measure the beam-end rotation of DCB-specimen loaded with wedge forces for the measurement of J, see e.g.

Andersson T. and Stigh U., 2004, "The stress-elongation relation for and adhesive layer loaded in peel using equilibrium of energetic forces", International Journal of Solids and Structures 41, 413-434.

I prefer testing DCB-specimens using pure bending moments. As discussed in Suo, Bao and Fan (1992), the J integral is determined by the applied moments. There are no effects of root rotation. No additional measurements (crack length, rotations or deflections) are needed for the J integral determination. However, a special loading device is needed for the creation of pure bending moments. But I think it is better to perform experiments that are difficult to perform (requiring special loading devices), but are easy to model and interpret than visa versa.
 
The J integral approach can be generalized to mixed mode cracking. Assuming that the fracture resistance is due to a mixed mode bridging law that can be derived from a potential function, the bridging law can be determined by partial differentiation. We have used that approach for the determination of mixed mode bridging laws from DCB specimens loaded with uneven bending moments.

Sørensen, B. F., and Jacobsen, T. K., 2009, "Delamination of fibre composites: determination of mixed mode cohesive laws", Composite Science and Technology 69, 445-56.

Zhigang Suo's picture

Dear Sergej and Bent:  Thank you so much for your comments.  I have been reading the papers that you suggested, and found them very helpful.  My notes posted here was for one lecture in the course on fracture mechanics.  In this lecture, I elected to introduce R-curve, without specifying any miscrospcopic mechanism. 

As listed on the website of the course, I will later give lectures on crack bridging.  Your input is helping me to prepare these lectures.     

P. F. C.'s picture

Dear Zhigang,

Thanks for this helpful lecture

I have a question regarding the stability of an advancing crack. When
G<sigma0, there is no propagation, the crack is stable, I can
understand it. It means a crack of length a<a0 (where G(a0)=sigma0)
won't grow.

But what I don't understand is how a crack could stop when it has
started to propagate. Indeed, when the crack starts to propagate from a
length a1, it reaches a new length a2. From this new length, the R-curve
will show us there is a propagation again, to a new length a3 and so
on.

And G(a) being an increasing function of a,as long as G(a) is greater
than sigma0, I don't understand why an advancing crack should stop when
under a constant load. To my mind, it will propagate inevitably until
the critical length and then propagate brutally.

Would you mind to explain to me what is actually happening ? I think my reasoning is biased but I cannot fix it.

 

Thanks in advance, best regards.

Paul

Zhigang Suo's picture

Dear Paul:  Please take a look at the drawing in the section "stable extension of a crack".  At load lambda_2, the crack has extended to a length a.  If the load is fixed, the crack will not extend further, because the loading curve is below the resistance curve.

Also, not all loading curves are increasing functions of the length of the crack.  Consider a double cantilever beam under the displacement-controlled conditon. 

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