## Chapter 7

## Q. 7.2

Properties of SiAlON Ceramics

Assume that an advanced ceramic sialon (acronym for SiAlON or silicon aluminum oxynitride) has a tensile strength of 60,000 psi. Let us assume that this value is for a flaw-free ceramic. (In practice, it is almost impossible to produce flaw-free ceramics.) A thin crack 0.01 in. deep is observed before a sialon part is tested. The part unexpectedly fails at a stress of 500 psi by propagation of the crack. Estimate the radius of the crack tip.

## Step-by-Step

## Verified Solution

The failure occurred because the 500 psi applied stress, magnified by the stress concentration at the tip of the crack, produced an actual stress equal to the ultimate tensile strength. From Equation 7-3,

\sigma_{actual }\cong 2 \sigma \sqrt{a/r} (7-3)

\sigma_{actual }=2 \sigma \sqrt{a/r}

60,000 \ psi=(2) (500 \ psi) \sqrt{0.01 \ in./r}

\sqrt{0.01 /r}=60 or 0.01/r = 3600

r = 2.8 × 10^{-6} in. = 7.1 × 10^{-6} cm = 710 \mathring{A}

The likelihood of being able to measure a radius of curvature of this size by any method of nondestructive testing is virtually zero. Therefore, although Equation 7-3 may help illustrate the factors that influence how a crack propagates in a brittle material, it does not help predict the strength of actual ceramic parts.