×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Are you an
Engineering professional?
Join Eng-Tips Forums!
• Talk With Other Members
• Be Notified Of Responses
• Keyword Search
Favorite Forums
• Automated Signatures
• Best Of All, It's Free!

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

#### Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

# Metal and Metallurgy engineering FAQ

## Fracture Mechanics

 What is Fracture Mechanics? by Maui faq330-1723 Posted: 28 May 11 (Edited 24 Aug 11) 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/2piwhere 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 or a ceramic, 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 adjacent atomic planes. The total surface energy required to form a crack of length 2a is        Ws = 4aYswhere 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)/EThe condition for 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)^0.5As 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)^0.5  or  S(pi*a)^0.5 = (EGc)^0.5where 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 a material. 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)^0.5as the stress intensity factor. Fracture occurs when the stress intensity factor K equals or exceeds the critical stress intensity factor KIC where        KIC = (EGc)^0.5KIC 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 it's value was set 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 psiThe 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. Back to Metal and Metallurgy engineering FAQ Index Back to Metal and Metallurgy engineering Forum

Close Box

# Join Eng-Tips® Today!

Join your peers on the Internet's largest technical engineering professional community.
It's easy to join and it's free.

Here's Why Members Love Eng-Tips Forums:

• Talk To Other Members
• Notification Of Responses To Questions
• Favorite Forums One Click Access
• Keyword Search Of All Posts, And More...

Register now while it's still free!