Solution
This question should take about 10 minutes to complete. It is an introductory level question.
A sheet of glass measuring 2 m by 200 mm by 2 mm contains a central slit parallel to the 200 mm side. The sheet is restrained at one end and loaded in tension with a mass of 500 kg.
What is the maximum allowable length of slit before fracture occurs?
Assume the following material property values: E = 60 GPa, surface energy is 0.5 J/m2, Poisson's ratio = 0.25 and the fracture stress of sound glass is 170 MPa.
Answer: 0.254 mm
This is a simple substitution into the Griffiths equation, i.e.:
where, to get the fracture stress in MPa (the standard SI engineering unit), the critical strain energy release rate is in N/m, E is in N/m2, and a is in m. This provides an answer in N/m2 (Pa), which needs to be
divided by 106. Thus:
In this calculation ,we have converted from the applied mass to a load using the gravitational constant g and the area of the plate (calculating the area as
though the crack is not there, because stress intensity factors are defined on the basis of nominal stress in the absence of a crack). The last thing to be done is to realise that this is a central crack in a plate and the full slit length is then, by definition 2a.
Thus the maximum length of slit which can be supported is 0.254 mm.
As the applied stress is only 12.26 MPa, this indicates how
critical even relatively small defects are in brittle materials, as this value of applied
stress can be compared with the fracture stress of sound (uncracked) glass of 170 MPa.
This illustration is the purpose of the apparently superfluous information!
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