dU/da = KI^2/E in Fracture Mechanics

Hi Experts,
It is me again, a beginner of Fracture Mechanics.
You know the expression of the released energe in the opening of a
U = sigma^2/(2E)*pi*a^2
derived from a case that a rectangular plate stretched by a line of
equal and opposite forces at each ends, and
KI = sigma*sqrt(pi*a)
is derived in the same (similar?) case. Then, from the above two
expressions, the book got
dU/da = KI^2/E
OK, fine. However, the book uses dU/da = KI^2/E in another case that a
system of equal and opposite forces applied to the crack surfaces. I
think dU/da = KI^2/E can only be used in the case that the forces
applied far away from the crack!
Can you explain why.....
Thank you in advance.
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The principle of superposition for linear equations is at work here.... and what is being applied is that you can represent the crack problem as two different boundary value problems.
It is fairly easy to draw it, but that isn't possible here. The drawing process is pretty persuasive to most readers, but there are some who won't accept the conclusions. Often, those readers aren't good at solving mathematical boundary value problems anyway.
The crack problem has two parts. 1) a uniform loading applied at a remote distance (infinity). 2) a discontinuity in materials (The crack) which introduces some stress -free boundaries (actually only the components of the stress tensor normal to the free surface are zero)
By convention, the crack is of finite width (x) and zero height (y) and this is a two dimensional solid (x and y).
Part of the solution is the far field solution sigmaY = constant (equivalent to loadY/area).
This part of the solution satisfies all of the equations of equilibrium everywhere and the boundary conditions on loading at infinity.
HOWEVER, this solution fails to satisfy the boundary conditions at the crack surface which requires that sigmaY = 0 on the crack face. instead, the far field simple solution gives sigmaY = constant.
Being clever, we say that what we will do is to just calculate the solution to the problem of a) no far field loading applied b) a crack is present c) we apply the loading sigmaY = -(constant) on the crack face... the pressurized crack....
Then we superimpose the stresses from the two problems.
We have the correct far field stress of sigmaY=constant (load /area).
On the crack face, the incorrect far field solutions stresses sigmaY=constant add to the pressurized crack stresses sigmaY=-constant to produce the correct boundary value conditions of sigmaY= 0 .
In other words, except for the uniform stress sigmaY=constant from the far field solution (for no crack), the pressurized crack produces exactly the same stresses as does the crack loaded at infinity.
Superposition of solutions gives correct boundary conditions everywhere on the boundaries..... even though neither of these solutions satisfied the boundary conditions everywhere on the coundaries.
The same principle is used in elasticity, heat flow, electrostatics, fluid mechanics and anywhere else that one has to solve boundary value problems.
It is a common advanced mathematical technique and did not originate in the elastic crack problem.
Fracture of Brittle Solids Brian Lawn, D. R. Clarke, S. Suresh, I. M. Ward, 03 June, 1993 ISBN: 0521409721
Analytical Fracture Mechanics David J. Unger, 10 July, 2001 ISBN: 0486417379
Failure of Materials in Mechanical Design: Analysis, Prediction, Prevention, 2nd Edition Jack A. Collins, 07 September, 1993 ISBN: 0471558915
Fracture Mechanics M. Janssen, J. Zuidema, R.J.H. Wanhill 2002 /
xii + 365 p. / ISBN 90-407-2221-8 / hardback /
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a few sample chapters.... and table of contents of what looks to be an excellent source from the solid mechaincs point of view.
My personal texts are all older than the above, but these will present the basics, the mechanics and some of the metallurgy of fracture mechanics.
Good luck. It happens that some who teach fracture mechanics are "light" on analytical mechanics skills. Also, most practical users of fracture mechanics are those who want to do, and may be "light" on analytical skills.
Reply to
Thanks a lot for your explanation which is quite clear on the principle of superposition. However, that didn't explain my question:( I am just wondering if the equation dU/da = KI^2/E can be used only in the case that a crack is oppend in a rectangular plate under two uniform distributed forces at both ends, or it can be used in other cases such that the loads is at the surfaces of a oppend crack?
Reply to
You missed the basic point. The pressurized crack and the crack in a plate are almost exactly equal in terms of anything related to the singularity of the crack tip stress singularity.
It also turns out that the strain energy release rates are identical for these two cases because they are so fundamentally close.
If you impose an arbitrary pressure (or surface traction) distribution on the crack face, then it has very little to do with the crack loaded by uniform stresses at infinity.
So ....... then what do you think? And why?
It seems as if this is a good time for you to get a better book that handles the mechanics with better rigour and better explanation.
If your descriptions are corrrect, the source you are citing is very poor in description......
But then, few text books are excellent.
Yours may have emphasis on things other than clear exposition of the basic analytical mechanics.
I have worked with a number of fracture mechanics experts whose analytical skills were poor, but then they didn't write books that purported to explain the analytical mechanics.
Reply to
I will get a good book....but, in the same time, I need a very basic book: I will upset whenever I find I cannot understand some points in the book. Do you know any online tutorial for Fracture Mechanics?
Thank you for all your helps.
Reply to
I will get a good book....but, in the same time, I need a very basic book: I will upset whenever I find I cannot understand some points in the book. Do you know any online tutorial for Fracture Mechanics?
Thank you for all your helps.
Reply to

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