Issue 30

L. Guerra Rosa et alii, Frattura ed Integrità Strutturale, 30 (2014) 438-445; DOI: 10.3221/IGF-ESIS.30.53 441 (9) where  1 and  2 are the mean strengths of specimens of type 1 and 2 (which may have different sizes and stress distributions), V E1 and V E2 are the effective volumes, and m is the Weibull modulus. Similarly, the following relationships can be derived from Eq. (4), (6) and (8): (10) (11) The above two Eq. (10) and (11) confirmed that the bending strength is higher than the tension strength. If the Weibull modulus m is equal to 10, the 3-point bending strength is 1.45 times the tension strength, and the 4-point bending strength is 1.73 times the tension strength. In view of the fact that the Weibull modulus m is usually assumed to be a constant for a given material, only the characteristic strength  0 is needed to be extrapolated from laboratory specimen test data to components. For a component with a varying stress field  , an effective surface area, A eff , may be computed using the following relationship: (12) Then the characteristic strength σ 0 for the component can be calculated from the data of specimen as: (13) In service, the components are generally subjected to multiaxial loading conditions, hence, we need to analyze the effect of multi-axial tensile stresses on flaws and determining one equivalent stress based on the selected multiaxial criterion. Then, the equivalent stress can be assumed to be the applied stress σ in the above equations. Glasses can demonstrate a loss of strength over time. This phenomenon is a kind of stress corrosion and it is known as static fatigue of glass. Chemical attack by water vapour (or other media) permits a pre-existing flaw to grow to critical dimensions and cause spontaneous crack propagation as shown in the following Fig. 1. Figure 1 : Regions of a typical log V versus log K plot. m E E V V /1 1 2 2 1              m t p m /12 3 1 2       m t p m m /1 2 4 2 1 4            dA A m eff         max   m component specimen specimen component A A /1 0 0         