How to calculate Stress Intensity Factor?

In summary: You can use the equations for simple circular plates with point loads at the center to approximate the stresses in your problem. If the stress intensity is less than the critical stress intensity factor, then the material will not fail.
  • #1
saurabh anand
6
0
Hello everyone,
I am currently doing my masters research project on fracture mechanics. My problem is such that I have a flat circular disc(brittle) which is simply supported around its circumference. The top surface of the disc is subjected to a point load at its centre. the bottom surface has a crack of a given length and a depth of less that half the thickness of the plate ( part through crack). This crack passes through the centre of the plate. What kind of forces will the crack experience? and how can I calculate the stress intensity factor at any point on this crack. There has been a lot of work previously carried out for plates with remote tensile and bending loads on the edges of the plate. I am not sure which category my problem falls into because I am applying a local force at the centre. thanks in advance. Any help will be deeply appreciated.
 
  • #3
This book for under 40 dollars may help... Even if not, it is an excellent reference and tutorial for glass or ceramic fracture mechanics theory , applicable to 3 point disc supports:
http://spie.org/Publications/Book/2047829
 
  • #4
Thank you for the information. I will look into it.
 
  • #5
I am not very sure how to derive the mathematical equation for the stress field. But the way I would follow is to first look at and understand how the equations for simple supported circular plate with point load at the center are derived from first principles. What this will do is to give you a better perspective of the problem to be solved. I liked the book "Stresses in Plates and shells" by Ugural. You can perhaps look into that. Also Fracture Mechanics notes by Alan Zehnder from Cornell can be looked into. Then in the same problem without crack introduce a crack which is represented by zero traction boundary conditions.Try to solve for the stress function and solve with the s.s. plate with point load BC's.
Then may be you can solve the original problem using Ansys/Abaqus and look at the results.
 
  • #6
If crack runs along major diameter then it will experience bending forces tending to hinge open the crack .

If crack is of trivial depth compared to plate dimensions then stresses can be approximated using analysis of a solid plate and then treating crack as a stress raiser .

If crack is of significant depth then an analytic solution is only possible if shape and size of crack can be described by a limited range of mathematical functions .

Apart from a few trivial cases solving the resulting equations is very difficult indeed .
 
  • #7
Thankyou Koolraj09 the approach you suggested is exactly how I am going about it. thanks for the reference book. I will look into it.
 
  • #8
Nidum thanks for the reply. the crack in my case is of trivial depth.
 
  • #9
saurabh anand said:
Hello everyone,
I am currently doing my masters research project on fracture mechanics. My problem is such that I have a flat circular disc(brittle) which is simply supported around its circumference. The top surface of the disc is subjected to a point load at its centre. the bottom surface has a crack of a given length and a depth of less that half the thickness of the plate ( part through crack). This crack passes through the centre of the plate. What kind of forces will the crack experience? and how can I calculate the stress intensity factor at any point on this crack. There has been a lot of work previously carried out for plates with remote tensile and bending loads on the edges of the plate. I am not sure which category my problem falls into because I am applying a local force at the centre. thanks in advance. Any help will be deeply appreciated.
Here is what you need to do. Calculate the stress under point load with no crack- use a book like ROARK for this. The stress intensity is given as
stress = Ki/(y) (SQRT(C) where c is flaw depth and Y = 1.26, for penny crack. Thus Ki = (max stress without crack)(y)( sqrt(C)). If stress intensity is less than critical stress intensity factor, Kic. then it will not fail. Kic = inherent property of material which you need to know. For britle materials like glass Kic = about 1MPa (m^1/2). The stress is in Mpa and the flaw depth is in meters. Note that stress must be in tension for failure to occur.
Edit: I corrected the equations as now noted above.
 
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FAQ: How to calculate Stress Intensity Factor?

What is the definition of stress intensity factor?

The stress intensity factor (K) is a measure of the stress concentration at the tip of a crack or notch in a material under loading. It is used to predict the crack propagation and failure of a material.

How is stress intensity factor calculated?

The stress intensity factor can be calculated using a variety of methods, depending on the type of loading and geometry of the crack. The most common method is the stress-strain method, which involves calculating the stress intensity factor from the stress distribution around the crack tip using fracture mechanics principles.

What factors affect the stress intensity factor?

The stress intensity factor is affected by several factors, including the geometry of the crack, the loading conditions, the material properties, and the crack growth rate. It is also influenced by the presence of stress concentrations and the size and location of the crack within the structure.

Why is stress intensity factor important?

The stress intensity factor is an important parameter in predicting the failure of a material due to cracks or notches. It helps engineers determine the critical crack size and the load at which a material will fail, allowing for the design of safer and more reliable structures.

Are there any limitations to using stress intensity factor?

Yes, there are some limitations to using stress intensity factor. It is most accurate for linear elastic materials and small cracks. It may not accurately predict the failure of materials with complex microstructures or under high temperatures and loading rates. Additionally, it does not take into account the effects of material imperfections or residual stresses.

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