# basic problem

Hi Experts,
I am reading "BASIC fracture mechanics" which is a very simple book. You must know this expression
sigmaY = KI/sqrt(2*pi*r)*cos(thita/2)*(1+sin(thita/2)*sin(3*thita/2)) and I think this is symmetric about thita = 0. My questions: Is the expression dirived from a rectangular (infinite or finite) plate with a crack at the center and under a symmetric load system case? There are different expressions For other cases, say non-symmetric cases?
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victor wrote:

Yes, there are dozens of solutions to a wide host of crack and loading geometries. If someone were to claim that there are over 100 distinct solutions, this claim would not be disputed at all.
To get these, more advanced books or original literature needs to be consulted.
The elastic crack has been of technical interest for most of a century, and there is nearly a century of work available for background and extensions.
Many PhD students, for example, have made contributions and refinements to this situation, and many older researchers also have contributed solutions and refinements too.
You can spend a lot of time and money just collecting the major contrubutions of others.
Jim
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Thanks a lot. Is the expression sigmaY = KI/sqrt(2*pi*r)*cos(thita/2)*(1+sin(thita/2)*sin(3*thita/2)) apply only on a rectangular (infinite or finite) plate with a crack at the center and under a symmetric tenson load?
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victor wrote:

What does the book say?
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Thank you. The book starts with problems of a clamped plate, but then just shows crack tip co-ordinates and some expressions include the above one. I want to know if the distribution of the stresses near the crack tip is the same or not for different loading and other conditions.
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victor wrote:

That is bad form that I wouldn't expect from that book.
Here is how you can figure it out.
1) the most commonly used crack tip stress geometry in text books is the plane 2-D crack in an infinite medium
2) the simplest solution is for the plane 2-D crack in an infinite medium
3) the only physical dimensions in your answer are the crack length.... there are no other dimensions for size of the specimen present, so there might well be no other dimensions in the problem.
4) (yes, I am getting to the point that if the test specimen containing the crack was of a finite size, then that size or length would enter into the solution somehow).
5) you can do a Google search on "Crack Tip Stress Intensity" in just a few seconds and you will learn a lot. Expecially, if you think about the answer given the above information.
Jim
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Thaks a lot Jim!
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I suddenly realized that the expression written as:
sigmaY = KI/sqrt(2*pi*r)*cos(thita/2)*(1+sin(thita/2)*sin(3*thita/2))
is about the most confusing way I have seen to express the nature of the crack tip stress fields and their physical dependence on parameters of loading and crack geometry -- especially to a beginner.
KI involves the far field applied stress (Sigma) and the square root of the crack length.
Depending upon how this is introduced, it can be a pretty confusing situation for the beginning introduction.
However, I don't have that text, and can't comment more deeply on how braintwisting that introduction to fracture mechanics can be, even though the equation is correct as an approximation.
Thank goodness my introduction to the area was less "simple" and proceeded with much more rigor.
Jim
victor wrote:

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Hi Jim,
I hope, the equation is correct as an approximation to any crack cases and all the differences for different cases are included in KI :) Can you suggest some tutorials in website or good books for a beginner who knows some mechanics including Elastic Theory?
Cheers.