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Linear Elastic Fracture Mechanics
4

Linear Elastic Fracture Mechanics

Linear Elastic Fracture Mechanics

(OP)
It baffles me how accounting for a little crack on a beam using linear elastic fracture mechanics can give you a bending moment resistance that is almost half of that of the code. Someone want to explain that to me?

Clansman

If a builder has built a house for a man and has not made his work sound, and the house which he has built has fallen down and so caused the death of the householder, that builder shall be put to death." Code of Hammurabi, c.2040 B.C.

RE: Linear Elastic Fracture Mechanics

Materials do not behave the same in the presence of defects. Especially those that are inherently prone to brittle fracture. That's why so much study has been devoted to fracture mechanics, damage tolerance, etc.

RE: Linear Elastic Fracture Mechanics

The results from LEFM are in ksi.square root inch; not ksi. That is ,it contains a geometric factor for the flaw/crack size. (for mode 1, modes 2 & 3 are way too complicated for me).

RE: Linear Elastic Fracture Mechanics

4
It's probably best for you to educate yourself on this topic. A good place to start is an engineering text or link that explains the theoretical aspects of fracture mechanics, such as

http://school.mech.uwa.edu.au/~dwright/DANotes/fracture/LEFM/LEFM.html

I'll attempt a quick summary for you here. Plastic deformation occurs by a process of shear. For most engineering materials, the actual shear strengths are much lower than the theoretically predicted values. This occurs because defects in the crystalline structure called dislocations begin to move when the resolved shear stress exceeds a certain value. If a similar calculation is performed to determine the critical tensile stress Sc required to separate adjacent atomic planes by breaking the atomic bonds we find

        Sc = E/2pi

where pi=3.1415... For a steel alloy where the elastic modulus is usually 30X10^6 psi, this amounts to a critical tensile stress of 4.8x10^6 psi. The actual measured fracture strengths of steel alloy specimens are several orders of magnitude lower than this. Defects in the material cause fracture to occur at applied stress levels that are far below the theoretical values. Examples of such defects are non-metallic inclusions, voids, sharp cracks and notches. One of the theories used to model the impact that these defects have on the load carrying capacity of materials is called Linear Elastic Fracture Mechanics (LEFM).

Consider a perfectly elastic material, such as glass, which contains a sharp crack. The crack grows when the atomic bonds at the crack tip are broken. Work must be done to break these bonds and separate the atomic planes. The total energy required to form a crack of length 2a is

        Ws = 4aYs

where 4a is the total surface area of the crack per unit thickness and Ys is the surface energy of the material. As the crack grows, more surface area is created and so more work is done by the applied forces. The total energy per unit thickness required to produce a crack of length 2a under an applied tensile stress S is

        Wtotal = 4aYs - pi(S^2)(a^2)/E

The condition for unstable crack growth is obtained by taking the derivative of Wtotal with respect to crack length and setting the resulting expression equal to zero. We find
    
                dWtotal/da = 0

                4Ys - 2pi(S^2)a/E = 0
             
                S = [2EYs/pi*a]^0.5  or  S(pi*a)^0.5 = (2EYs)^0.5

As the length of a pre-existing crack increases, the stress required for fracture decreases. This equation is known as the Griffith criterion for fracture. It often appears in the form

        S = (EGc/pi*a)^0.52  or  S(pi*a)^0.5 = (EGc)^0.5

where Gc is called the critical strain energy release rate, or the total work of fracture. This equation can be used to predict the critical values of stress and crack length that are required for a crack to grow in an unstable manner. When the term S(pi*a)^0.5 reaches the critical value (EGc)^0.5, the crack will begin to grow. In this context, it is convenient to treat S(pi*a)^0.5 as a measure of the driving force for crack propagation. It is common practice to define

        K = S(pi*a)^0.5

as the stress intensity factor. Fracture occurs when the stress intensity factor K equals or exceeds the critical stress intensity factor KIC where

        KIC = (EGc)^0.5

KIC is usually referred to as the fracture toughness, and values for this material property are usually only well defined under plane strain conditions. There is also a rather complicated geometry factor that comes into play in performing this type of analysis, but for convenience I set it equal to 1 during the derivation. More advanced derivations will always include it.

As an example, suppose an inspection technique for finding cracks in a high strength steel landing gear has a resolution of 0.1 inches. What tensile stress level will the landing gear be able to support without breaking? For this type of steel we can expect Gc = 300 lbs/in., and if E = 30x10^6 psi, then

    S = (EGc/pi*a)^0.5

    S = [(30x10^6 psi)(300 lbs/in.)/pi*(0.1 in.)]^0.5

    S = 1.7x10^5 psi

The landing gear should be able to support a stress of 170,000 psi. What happens if a crack is detected in a used landing gear component that is 1.0 inches in length? Performing the same calculation, we find that S = 54,000 psi. As this example demonstrates, a crack of significant length can dramatically reduce the load carrying capacity of a critical component.

Does this answer your question?

Maui
  

RE: Linear Elastic Fracture Mechanics

The OP hasn't responded yet, but thanks, Maui, for this valuable post (*).

RE: Linear Elastic Fracture Mechanics

Thanks Bestwrench.
 

RE: Linear Elastic Fracture Mechanics

I recommend 2 books by the same author Prof. J E Gordon both excellently written with the layman in mind

New Science of Strong Materials (or why you dont fall through the floor)  2nd ed  1976

Structures (or why things dont fall down) 1978

The 1st is published by Penguin, the 2nd by Pelican

They are very enjoyable reads as he illustrates concepts with practical examples, contemporary and historical and frommany different disciplines - the art of bow-making to dress design.

RE: Linear Elastic Fracture Mechanics

(OP)
Maui,

Thank you so much for that response!

Clansman

If a builder has built a house for a man and has not made his work sound, and the house which he has built has fallen down and so caused the death of the householder, that builder shall be put to death." Code of Hammurabi, c.2040 B.C.

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