Journal of the European Ceramic Sociery 28(2008)1551-1556 Table 1 Parameters of the Weibull distributions discussed in this paper Characteristic strength [MPa Weibull modulus Threshold stress [MPaI Evaluation of individual batches A and L, two-parameter Weibull curve fitting Alumina(batch A) 492±1 104士26 Laminate(batch L) 650±15 Evaluation of a combined batch A+L, three-parameter Weibull curve fittin 92±1 11.5±2.2 Laminate(batch L) 492±1 115±2.2 The upper and lower limits refer to estimates of the 90% confidence intervals. from small flaws in the surface layer, the Weibull distribution of Eq (1)can be modified to account for the action of the residual stress by setting a=σr The results of the fracture experiments can be explained on the basis of this idea. The characteristic strength(in terms of applied stress)of the laminate(650+ 15 MPa) is significantly higher than that of the A-specimens(492+ 17 MPa). Following the above argument the difference(-158+32 MPa)is caused by the compressive residual stress in the first layer of the laminate O A+. This fits well to the values for the residual compressive stress △ It is assumed that in terms of total stress the strength dis- tributions of both materials are identical, since they are based applied stress [MPa on the same flaw population. In other words the data sets A Fig1.Probability of failure versus applied rupture stress for alumina specimens and L are samples out of the same parent distribution. Since (O)and specimens of the alumina-alumina/zirconia laminate(4)in a Weibull evaluation uncertainties arising from the sampling procedure plot. The lines represent best fits of a two-parameter Weibull distribution to the are smaller for a large than for a small number of tests sample a common data evaluation will result in more precise Weibull parameters. Fig 3 shows the Weibull statistics of a set and A-specimens at the position of the fracture origin was the of tests which consists of the data of batch A and L.For the occurrence of compressive residual stresses in the surface region residual stress of the data in batch L the value a- -_158MPa of the laminate specimens. In the following, the significant dif- ference in the mechanical behaviour between the monoliths(see Fig. 1)is put down to these stresses. The total stress(od)is the sum of the applied stress(oa)and the residual stress(ores): ot=a+res In the monolithic ceramic the residual stress is zero. In the laminate it is compressive and almost constant in the surface region, where fracture initiates. It is assumed in the following that for the laminates. which fail tensile side ▲ L shifted 600700800 total stress [MPa ongin alumina specimen. Fig 3. Same data as in Fig. I plotted versus the sum of applied and residual The same type of defects was responsible for failure in the laminates. stress:ot=Oa+Ores. The line represents a common evaluation of both batches.J. Pascual et al. / Journal of the European Ceramic Society 28 (2008) 1551–1556 1553 Table 1 Parameters of the Weibull distributions discussed in this paper Characteristic strength [MPa] Weibull modulus Threshold stress [MPa] Evaluation of individual batches A and L, two-parameter Weibull curve fitting Alumina (batch A) 492 ± 17 10.4 ± 2.6 0 Laminate (batch L) 650 ± 15 18.1 ± 5.1 0 Evaluation of a combined batch A + L, three-parameter Weibull curve fitting Alumina (batches A + L) 492 ± 11 11.5 ± 2.2 0 Laminate (batch L) 492 ± 11 11.5 ± 2.2 158 ± 32 The upper and lower limits refer to estimates of the 90% confidence intervals. Fig. 1. Probability of failure versus applied rupture stress for alumina specimens () and specimens of the alumina–alumina/zirconia laminate () in a Weibull plot. The lines represent best fits of a two-parameter Weibull distribution to the data. and A-specimens at the position of the fracture origin was the occurrence of compressive residual stresses in the surface region of the laminate specimens. In the following, the significant dif￾ference in the mechanical behaviour between the laminates and the monoliths (see Fig. 1) is put down to these stresses. The total stress (σt) is the sum of the applied stress (σa) and the residual stress (σres): σt =σa + σres. In the monolithic ceramic the residual stress is zero. In the laminate it is compressive and almost constant in the surface region, where fracture initiates.30 It is assumed in the following that for the laminates, which fail Fig. 2. A large alumina grain acting as fracture origin in an alumina specimen. The same type of defects was responsible for failure in the laminates. from small flaws in the surface layer, the Weibull distribution of Eq. (1) can be modified to account for the action of the residual stress by setting σ = σt: F(σ) = 1 − exp − σa + σres σ0 m . (2) The results of the fracture experiments can be explained on the basis of this idea. The characteristic strength (in terms of applied stress) of the laminate (650 ± 15 MPa) is significantly higher than that of the A-specimens (492 ± 17 MPa). Following the above argument the difference (−158 ± 32 MPa) is caused by the compressive residual stress in the first layer of the laminate. This fits well to the values for the residual compressive stress mentioned above. It is assumed that in terms of total stress, the strength dis￾tributions of both materials are identical, since they are based on the same flaw population. In other words the data sets A and L are samples out of the same parent distribution. Since evaluation uncertainties arising from the sampling procedure are smaller for a large than for a small number of tests in a sample a common data evaluation will result in more precise Weibull parameters. Fig. 3 shows the Weibull statistics of a set of tests which consists of the data of batch A and L. For the residual stress of the data in batch L the value σres = −158 MPa Fig. 3. Same data as in Fig. 1 plotted versus the sum of applied and residual stress: σt = σa + σres. The line represents a common evaluation of both batches.