Question 7.7: Strength of Ceramics and Probability of Failure An advanced ......

We assume all samples tested had the same volume; thus the size of the sample will not be a factor in this case. We can use the symbol V for sample volume instead of V_{0} . We are dealing with a brittle material so we begin with Equation 7-9:

F(V_{0})= 1 – P(V_{0}) = 1- \exp [-(\frac{\sigma}{\sigma_{0}})^m]   (7-9)
F(V)= 1 – P(V) = 1- \exp [-(\frac{\sigma}{\sigma_{0}})^m]
or
1- F(V)=  \exp [-(\frac{\sigma}{\sigma_{0}})^m]

Taking the natural logarithm of both sides,

Taking natural logarithms again,
ln\left\{ – \ln[1 – F(V)] \right\} = m (\ln \ \sigma  -m \ln \ \sigma_{0})     (7-10)

We can eliminate the minus sign on the left-hand side of Equation 7-10 by rewriting it as

\ln\left\{ \ln[ \frac{1}{1 \ – \ F(V)}] \right\}=m (\ln \ \sigma  – \ln \ \sigma_{0} )    (7-11)

For F = 0.4, σ = 250 MPa, and m = 9 in Equation 7-11, we have

\ln [ \ln( \frac{1}{1 \ – \ 0.4}) ]=9 (\ln \ 250  – \ln \ \sigma_{0} )     (7-12)

Therefore,

ln[ln (1/0.6)] = -0.67173
= 9(5.52146 – ln \sigma_{0} )

Therefore, ln \sigma_{0} = 5.52146 + 0.07464 = 5.5961. This gives us a value of \sigma_{0} = 269.4 MPa. This is the characteristic strength of the ceramic. For a stress level of 269.4 MPa, the probability of survival is 0.37 (or the probability of failure is 0.63).
Now, we want to determine the value of s for F = 0.1. As the required probability of failure (F) decreases, the stress level to which the ceramic can be subjected (σ) also decreases. We know that m = 9 and \sigma_{0} = 269.4 MPa. We substitute these values into Equation 7-11:

ln[ln(\frac{1}{1 \ – \ 0.1})] = 9(ln σ – ln 269.4)
-2.25037 = 9(ln σ – 5.5961)
∴ -0.25004 = ln σ – 5.5961
ln σ = 5.3461

or σ = 209.8 MPa. As expected, as we lowered the probability of failure to 0.1, we also decreased the level of stress that can be supported.