Question

The theoretical failure strength of high-purity and fully-dense aluminum oxide (Al2O3, alumina) can be up to 4 GPa !

In other words, a pristine alumina sample is expected to support this stress, even in tension, since no stress magnifying flaws would be present.

Think about that for a second ... most metals yield (bend permanently) at stress levels on the order of 0.1 - 0.5 GPa ... and this is considered their upper limit for structural use.

Ceramics like alumina have such high bond strengths that they can theoretically be stretched in tension, elastically, to much higher stresses ... 4 GPa in this case.

Unfortunately, most ceramic specimens rarely exhibit their theoretical strength value due to microscopic flaws that are inevitably present in specimens that have been handled, transported, etc.

For an alumina specimen, what is the minimum stress at which failure is expected to occur, in MPa, if it features an elliptical surface scratch whose length is 2.43 mm and whose tip radius is 190 nm ?

Answer Format: X.X

Unit: MPa

Answer 1

Calculate the minimum stress at which failure occurs as follows: 0. = 0 min O min = 7 Here, o, is the failure strength of alumina, a is the length of surface scratch, and p, is the tip radius. 1 m 103 MPa Substitute 4 GPa | 1 GPa for G, 2.43 mm for a, and 1 mm 190 mm 10% nm for p 4x103 O min = 2.43x10-3 ) 1/2 11+21 190x10-9] Omin = 17.6071 MPa