Fracture mechanics, stress intensity, fracture toughness

In summary, the conversation discusses the use of stiffeners to increase the resistance of material and prevent further fracture. The formula for fracture intensity using the fracture toughness value as K is suggested as a solution to the problem. The conversation also mentions the use of longitudinal strips of high toughness material for crack arrest purposes on a ship's deck and poses two questions related to this scenario. The first question asks for the size of a crack that would cause fracture in the deck plate, while the second question asks for the safe spacing between the crack arrest strips. The assumption of ignoring dynamic effects and assuming a wide deck is also mentioned.
  • #1
DJ2019
2
0
Homework Statement
7. A ship’s deck is provided with longitudinal strips of high toughness material for crack arrest purposes. The strips are welded in the deck and of equal thickness to it. The strip material has toughness 170 MN m-3/2 and yield stress 550 MPa, while the deck plate has toughness 55 MN m-3/2 and yield stress 320 MPa. If the highest “yearly wave” encountered produces a deck stress of 95 MPa:
(i) What size crack in the deck plate will cause fracture thereof?
(ii) How far apart should the crack arrest strips be placed to be safe? Ignore dynamic effects and assume the deck is wide.
Relevant Equations
Fracture toughness
I've been giving this some thought. It's clear that the stiffners will increase the resistance of the material so that the energy release is no longer high enough to cause further fracture. I'm just not sure what formula I can use to take into account the new resistance. I suspect part 1 of the question is just looking for use of the fracture intensity formula using the fracture toughness value as K. Could it be that simple?
 
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  • #2
DJ2019 said:
Homework Statement:: 7. A ship’s deck is provided with longitudinal strips of high toughness material for crack arrest purposes. The strips are welded in the deck and of equal thickness to it. The strip material has toughness 170 MN m-3/2 and yield stress 550 MPa, while the deck plate has toughness 55 MN m-3/2 and yield stress 320 MPa. If the highest “yearly wave” encountered produces a deck stress of 95 MPa:
(i) What size crack in the deck plate will cause fracture thereof?
(ii) How far apart should the crack arrest strips be placed to be safe? Ignore dynamic effects and assume the deck is wide.
Homework Equations:: Fracture toughness

I've been giving this some thought. It's clear that the stiffners will increase the resistance of the material so that the energy release is no longer high enough to cause further fracture. I'm just not sure what formula I can use to take into account the new resistance. I suspect part 1 of the question is just looking for use of the fracture intensity formula using the fracture toughness value as K. Could it be that simple?
You might have better luck posting this in the engineering homework forum.
 
  • #3
haruspex said:
You might have better luck posting this in the engineering homework forum.
Moved. :smile:
 

1. What is fracture mechanics and why is it important?

Fracture mechanics is a field of study that deals with the behavior and properties of materials under stress, particularly in relation to the formation and propagation of cracks. It is important because understanding how materials fracture can help engineers design safer and more durable structures.

2. What is stress intensity and how is it calculated?

Stress intensity is a measure of the severity of stress at the tip of a crack in a material. It is calculated using the stress distribution around the crack, the crack length, and the material's fracture toughness. The equation for stress intensity is K = σ√πa, where σ is the applied stress, and a is the crack length.

3. What is fracture toughness and why is it important?

Fracture toughness is a material's resistance to brittle fracture when a crack is present. It is an important property to consider in engineering design as it determines the maximum stress a material can withstand before it fractures. It is also a critical factor in determining the safety and reliability of structures.

4. How does temperature affect fracture mechanics?

Temperature can greatly influence the behavior of materials under stress and can affect fracture mechanics in several ways. For example, low temperatures can make materials more brittle, while high temperatures can cause them to deform and fail in a ductile manner. Additionally, temperature changes can affect the rate of crack propagation and can alter the material's fracture toughness.

5. How is fracture toughness measured?

Fracture toughness is typically measured through specialized tests, such as the Charpy or Izod impact tests, where a notched specimen is subjected to a sudden impact and the energy absorbed by the material is measured. Other methods, such as the three-point bend test, can also be used to determine fracture toughness. The results of these tests are then used to calculate the material's fracture toughness value.

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