section epub:type=”chapter”> The fracture toughness KC of ceramics and rigid polymers is very low compared to that of metals and composites. Cement and ice have the lowest KC at ≈0.2 MN m −3/2. Even engineering ceramics such as silicon nitride, alumina, and silicon carbide only reach ≈ 4 MNm−3/2. Rigid polymers (thermoplastics below the glass transition temperature or heavily cross-linked thermosets such as epoxy) have Kc ≈0.5–4 MN m−3/2. This low fracture toughness makes ceramics and rigid polymers very vulnerable to the presence of crack-like defects. They are defect-sensitive materials liable to fail by fast fracture from defects well before they can yield. Unfortunately, many manufactured ceramics contain cracks and flaws left by the production process. The defects are worse in cement because of the rather crude nature of the mixing and setting process. Ice usually contains small bubbles of trapped air. Molded polymer components often contain small voids. And all but the hardest brittle materials accumulate additional defects when they are handled or exposed to an abrasive environment. The chapter discusses the probabilistic fracture of brittle materials, the statistics of strength, and the Weibull distribution. The chapter concludes with a discussion of the modulus of rupture. We saw in Chapter 14 that the fracture toughness Kc of ceramics and rigid polymers was very low compared to that of metals and composites. Cement and ice have the lowest Kc at ≈0.2 MN m−3/2. Traditional manufactured ceramics (brick, pottery, china, porcelain) and natural stone or rock are better, at ≈ 0.5–2 MN m−3/2. But even engineering ceramics such as silicon nitride, alumina, and silicon carbide only reach ≈ 4 MN m−3/2. Rigid polymers (thermoplastics below the glass transition temperature, or heavily cross-linked thermosets such as epoxy) have Kc ≈ 0.5–4 MN m−3/2. This low fracture toughness makes ceramics and rigid polymers very vulnerable to the presence of crack-like defects. They are defect-sensitive materials, liable to fail by fast fracture from defects well before they can yield. Unfortunately, many manufactured ceramics contain cracks and flaws left by the production process (e.g., the voids left between the particles from which the ceramic was fabricated). The defects are worse in cement, because of the rather crude nature of the mixing and setting process. Ice usually contains small bubbles of trapped air (and, in the case of sea ice, concentrated brine). Molded polymer components often contain small voids. And all but the hardest brittle materials accumulate additional defects when they are handled or exposed to an abrasive environment. The design strength of a brittle material in tension is therefore determined by its low fracture toughness in combination with the lengths of the crack-like defects it contains. If the longest microcrack in a given sample has length 2amax, then the tensile strength is simply given by from the fast fracture equation. Some engineering ceramics have tensile strengths about half that of steel—around 200 MN m−2. Taking a typical fracture toughness of 2 MN m−3/2, the largest microcrack has a size of 60 μm, which is of the same order as the original particle size. Pottery, brick, and stone generally have tensile strengths that are much lower than this—around 20 MN m−2. This indicates defects of the order of 2 mm for a typical fracture toughness of 1 MN m−3/2. The tensile strength of cement and concrete is even lower—2 MN m−2 in large sections—implying the presence of at least one crack 6 mm or more in length for a fracture toughness of 0.2 MN m−3/2. The chalk with which I write on the board when I lecture is a brittle solid. Some sticks of chalk are weaker than others. On average, I find (to my slight irritation), that about 3 out of 10 sticks break as soon as I start to write with them; the other 7 survive. The failure probability, Pf, for this chalk, loaded in bending under my (standard) writing load, is 3/10, that is, When you write on a board with chalk, you are not unduly inconvenienced if 3 pieces in 10 break while you are using them; but if 1 in 2 broke, you might seek an alternative supplier. So the failure probability, Pf, of 0.3 is acceptable (just barely). If the component were a ceramic cutting tool, a failure probability of 1 in 100 (Pf = 10−2) might be acceptable, because a tool is easily replaced. But if it were the glass container of a cafetiere, the failure of which could cause injury, one might aim for a Pf of 10−4. When using a brittle solid under load, it is not possible to be certain that a component will not fail. But if an acceptable risk (the failure probability) can be assigned to the function filled by the component, then it is possible to design so this acceptable risk is met. This chapter explains why ceramics have this dispersion of strength; and shows how to design components so they have a given probability of survival. The method is an interesting one, with application beyond ceramics to the malfunctioning of any complex system in which the breakdown of one component will cause the entire system to fail. Chalk is a porous ceramic. It has a fracture toughness of 0.9 MN m−3/2 and, being poorly consolidated, is full of cracks and angular holes. The average tensile strength of a piece of chalk is 15 MN m−2, implying an average length for the longest crack of about 1 mm (calculated from Equation (16.1)). But the chalk itself contains a distribution of crack lengths. Two nominally identical pieces of chalk can have tensile strengths that differ greatly—by a factor of 2 or more. This is because one was cut so that, by chance, all the cracks in it are small, whereas the other was cut so that it includes one of the longer flaws of the distribution. Figure 16.1 illustrates this: if the block of chalk is cut into pieces, piece A will be weaker than piece B because it contains a larger flaw. It is inherent in the strength of brittle materials that there will be a statistical variation in strength. There is no single “tensile strength” but there is a certain, definable, probability that a given sample will have a given strength. The distribution of crack lengths has other consequences. A large sample will fail at a lower stress than a small one, on average, because it is more likely that it will contain one of the larger flaws (Figure 16.1). So there is a volume dependence of the strength. For the same reason, a brittle rod is stronger in bending than in simple tension: in tension the entire sample carries the tensile stress, while in bending only a thin layer close to one surface (and thus a relatively smaller volume) carries the peak tensile stress (Figure 16.2). The Swedish engineer, Weibull, invented the following way of handling the statistics of strength. He defined the survival probability Ps(V0) as the fraction of identical samples, each of volume V0, which survive loading to a tensile stress σ. He then proposed that where σ0 and m are constants. This equation is plotted in Figure 16.3(a). When σ = 0 all the samples survive, of course, and Ps(V0) = 1. As σ increases, more and more samples fail, and Ps(V0) decreases. Large stresses cause virtually all the samples to break, so Ps(V0) → 0 and σ → ∞. If we set σ = σ0 in Equation (16.2) we find that Ps(V0) = 1/e (= 0.37). So σ0 is simply the tensile stress that allows 37 % of the samples to survive. The constant m tells us how rapidly the strength falls as we approach σ0 (see Figure 16.3(b)). It is called the Weibull modulus. The lower m, the greater the variability of strength. For ordinary chalk, m is about 5, and the variability is great. Brick, pottery, and cement are like this too. The engineering ceramics (e.g., SiC, Al2O3, and Si3N4) have values of m of about 10; for these, the strength varies much less. Even steel shows some variation in strength, but it is small: it can be described by a Weibull modulus of about 100. Figure 16.3(b) shows that, for m greater than about 20, a material can be treated as having a single, well-defined failure stress. So much for the stress dependence of Ps. But what of its volume dependence? We have already seen that the probability of one sample surviving a stress σ is Ps(V0). The probability that a batch of n such samples all survive the stress is just {Ps(V0)}n. If these n samples were stuck together to give a single sample of volume V = nV0 then its survival probability would still be {Ps(V0)}n. So This is equivalent to or The Weibull distribution (Equation (16.2)) can be rewritten as If we insert this result into equation (16.3) we get or
Fracture Probability of Brittle Materials
Publisher Summary
16.1 Introduction
16.2 Statistics of Strength
16.3 Weibull Distribution
