Question 7.8: Weibull Modulus Parameter Determination Seven silicon carbid......
Weibull Modulus Parameter Determination
Seven silicon carbide specimens were tested and the following fracture strengths were obtained: 23, 49, 34, 30, 55, 43, and 40 MPa. Estimate the Weibull modulus for the data by fitting the data to Equation 7-11.
\ln\left\{ \ln[ \frac{1}{1 \ – \ F(V)}] \right\}=m (\ln \ \sigma – \ln \ \sigma_{0} ) (7-11)
Discuss the reliability of the ceramic.
First, we point out that for any type of statistical analysis, we need a large number of samples. Seven samples are not enough. The purpose of this example is to illustrate the calculation.
One simple though not completely accurate method for determining the
behavior of the ceramic is to assign a numerical rank (1 to 7) to the specimens, with the specimen having the lowest fracture strength assigned the value 1. The total number of specimens is n (in our case, 7). The probability of failure F is then the numerical rank divided by n + 1 (in our case, 8). We can then plot ln{ln 1/[1 – F (V_{0})]} versus ln s. The following table and Figure 7-14 show the results of these calculations. Note that s is plotted on a log scale.
The slope of the fitted line, or the Weibull modulus m, is (using the two points indicated on the curve)
m = \frac{0.5 \ – \ (-2.0)}{ \ln(52) \ – \ \ln(23.5)}=\frac{2.5}{3.951 \ – \ 3.157}=3.15This low Weibull modulus of 3.15 suggests that the ceramic has a highly variable fracture strength, making it difficult to use reliably in load-bearing applications.
| ith Specimen | σ (MPa) | F (V_{0}) | ln{ln 1/[1 – (FV_{0})]} |
| 1 | 23 | 1/8 = 0.125 | -2.013 |
| 2 | 30 | 2/8 = 0.250 | -1.246 |
| 3 | 34 | 3/8 = 0.375 | -0.755 |
| 4 | 40 | 4/8 = 0.500 | -0.367 |
| 5 | 43 | 5/8 = 0.625 | -0.019 |
| 6 | 49 | 6/8 = 0.750 | +0.327 |
| 7 | 55 | 7/8 = 0.875 | +0.732 |
