Solution (Problem 8)
This question should take about 15 minutes to complete. It illustrates one use of fracture mechanics in materials processing, and requires some thought about the critical type of defect and its position in the component of interest.
Grinding wheels are fabricated, typically, from alumina powder, which is compacted and sintered at high temperature and pressure. The powder is sieved before compacting to remove impurities which may later act as defects in the
grinding wheel. Hence residual impurities are related in size to the sieve mesh
dimension.
One particular type of alumina wheel has a density of 3800 kg/m3,
a bore diameter of 140 mm and an outer diameter of 1.0 m. It spins at 3000 rpm.
The maximum stress in the wheel is given by:
Calculate the allowable size of the sieve mesh if the wheel is to have a factor of safety of two on critical defect size when operating at 3000 rpm.
Note that, for alumina, the fracture toughness R = 0.10 kJ/m2 and E = 371 GPa.You may assume plane strain conditions.
Answer: 1.1 mm
All we need to do is find the peak stress in the wheel and determine what type of crack will be critical. The importance of crack type arises because the Griffith equation will yield a value for ac, while one has
to consider that embedded cracks are defined in terms of 2a and surface cracks in terms of a. Thus if the mesh was sized for embedded cracks, and such a defect occurred at the surface, it would immediately cause fracture. Hence the sieve is sized on the basis of surface defects being critical.
Using the Griffith equation for plane strain:
Applying the factor of safety of 2 to crack size means that the sieve mesh must be no bigger than 1.1 mm in dimension.
Close Window