Statistical nature and size dependence of strength

10.6.2 Statistical nature and size dependence of strength

The fracture strength of brittle materials generally exhibits two unusual characteristics, namely statis- tical variations and size dependence. Since the fracture stress is inversely proportional to the square root of the crack size (see Chapter 7), a sample containing larger cracks will have a lower strength, and vice versa. Thus, macroscopically similar samples do not necessarily exhibit the same fracture strength, since their flaw distributions may not be the same. The fracture strengths of ceramics and

542 Physical Metallurgy and Advanced Materials Table 10.8 Strengths of ceramics in fiber and polycrystalline forms ( from Richerson, 1992).

E (Gpa)

Ideal fracture strength

Measured strengths Measured strength

E/10 (GPa)

of fibers (GPa)

of polycrystalline specimen (GPa)

glasses are therefore not material constants, and on repeated measurements, they usually exhibit rather significant statistical scatter.

The fracture strength of a ceramic or glass also depends on the sample size. A larger sample has

a higher chance of containing larger cracks, and so in general is weaker in tension than a smaller sample. Ceramic and glass fibers with sub-millimeter diameters have very low chances of containing big flaws, and so they can have fracture strengths approaching the ideal value of E/10 (Chapter 7; see also Table 10.8). Fibers are therefore very strong materials, able to withstand tensile stresses up to two to three orders of magnitude higher than the yield strength of steel, for example. This is the reason why ceramic fibers are often used as the strengtheners in making load-bearing composite materials (see Chapter 11). Sintered, polycrystalline ceramics in bulk forms, however, have much lower fracture strengths due to the presence of porosity (Table 10.8).

The statistical scatter and size dependence of strength pose a problem in the mechanical design of ceramic components. In the laboratory, one can only test small pieces of samples, but since these would have strengths higher than the real component, which is usually bigger, the strength data from the small samples cannot be used directly for the design. The strength data would also depend on the test method employed. In a three-point bend test (Figure 10.22), the stress is compressive (and hence will not contribute to fracture) within the upper half-thickness of the test bar, and is tensile within the lower half-thickness and increasingly so towards the bottom of the sample. In a longitudinal direction, the stress increases linearly from zero at the two lower supports to a peak value at the mid-span of the sample. Thus, if the largest crack in the test bar happens to occur near the mid-span at the bottom of the specimen, then the MoR value from the bend test would be close to the fracture strength recorded if the same bar is subjected to a tensile test. However, the chance for the largest crack to situate exactly at the mid-span and at the bottom is low and, hence, for the same material, the MoR values obtained from bend tests in general are higher, and will scatter more, than the fracture strengths obtained from tensile tests.

Weibull (1951) proposed a statistical approach to deal with the above problems concerning the design of ceramic components. Instead of treating the fracture strength as a deterministic quantity, Weibull proposed the use of a survival probability to describe the chance a given piece of ceramic

will survive a given applied load. The survival probability was suggested to be

" m P # s (V ) = exp −V (σ/σ o ) ,

(10.1) where V is the volume of the ceramic, σ is the applied tensile stress, and σ o and m are material

constants. The choice of the exponential function in equation (10.1) is to ensure the obvious condi- tion P s (V 1 +V 2 ) =P s (V 1 ) ·P s (V 2 ) to be satisfied. To determine σ o and m for a given material, small ceramic pieces of identical size are tested and their strength data, which should scatter for the reason explained above, are ranked in an ascending order. The survival probability of each tested sample is then assigned according to P s = 1 − i/(N + 1), where i is the rank and N the total number of samples. A table between P s and the fracture strength σ is thus obtained for the N samples, and from equation (10.1), a plot of ln(1/P s ) versus ln σ should yield a straight line with slope and y-intercept

Non-metallics I – Ceramics, glass, glass-ceramics 543 equal to m and (ln V − m ln σ o ) respectively. The constants σ o and m can therefore be determined, since

the sample volume V is known. With σ o and m determined this way, the fracture strength σ of any sam- ple size V can be determined from equation (10.1) to correspond to a prescribed survival probability, say 0.95.

Equation (10.1) is for a uniform stress state σ, and in a varying stress situation, such as a three-point bend test, the Weibull formula can be generalized into

P s (V ) = exp ⎣− [σ(r)/σ o ] m d 3 r ⎦,

where σ(r) is the tensile stress at point r in the material volume V . Substituting in the known stress function for a three-point bend test, for example, the ratio of the MoR from the bend test to the fracture stress σ tensile from a tensile test at the same survival threshold can be shown to be (MoR) : σ tensile

2 = [2(m + 1) 1/m ] : 1. At a typical value of 10 for the Weibull modulus m, (MoR) : σ tensile = 1.73 : 1, i.e. the bend test will overestimate the fracture strength by 73%, as a result

of the small chance that the largest crack will occur near the mid-span and at the bottom of the test bar.

Worked example

Three-point bend tests in sintered Al 2 O 3 give MoR values in the range of 350–580 MPa. Tensile tests on similar samples give tensile strengths from 200 to 310 MPa. Estimate the Weibull modulus of this material.

Solution

The mean MoR value is (350

2 1/m 255 MPa. Thus, [2(m + 580)/2 = 465 MPa. The mean tensile strength is (200 + 310)/2 =

= 465/255 = 1.824. Numerically or graphically solving this gives m = 8.72.