Fracture mechanics, stress intensity, fracture toughness

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SUMMARY

The discussion focuses on fracture mechanics, specifically the application of fracture toughness and stress intensity in assessing crack propagation in materials. The key parameters include a toughness of 170 MN m-3/2 and yield stress of 550 MPa for the crack arrest strips, compared to 55 MN m-3/2 and 320 MPa for the deck plate. The main inquiry revolves around determining the critical crack size that would lead to fracture and the optimal spacing for the crack arrest strips, utilizing the fracture intensity formula with the provided toughness values.

PREREQUISITES
  • Understanding of fracture mechanics principles
  • Knowledge of fracture toughness and stress intensity factors
  • Familiarity with material properties such as yield stress
  • Basic skills in applying engineering formulas for crack analysis
NEXT STEPS
  • Study the application of the fracture intensity formula in real-world scenarios
  • Research methods for calculating critical crack sizes in structural materials
  • Explore the design and placement of crack arrest strips in engineering applications
  • Learn about the impact of dynamic effects on fracture mechanics
USEFUL FOR

Engineers, material scientists, and students studying fracture mechanics who are involved in structural integrity assessments and crack prevention strategies.

DJ2019
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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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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.
 
haruspex said:
You might have better luck posting this in the engineering homework forum.
Moved. :smile:
 

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