13th International Conference on Fracture June 16–21, 2013, Beijing, China -4- 0 eff Ieff a a P K f B W W ⎛ +Δ ⎞ = ⋅ ⎜ ⎟ ⋅ ⎝ ⎠ , (3) Eq. (3) differs from the classically used nominal stress intensity factor, which is calculated using only the initial crack size (a0). The nominal stress intensity factor under-predicts the true effective crack driving force and should not be used in the assessment of quasi-brittle materials. Combining Eqs. (2) and (3) leads to a relation between load and effective crack growth, Eq. (4). 1 0 m mm eff eff K a B W P a a f W ⋅Δ ⋅ ⋅ = ⎛ +Δ ⎞ ⎜ ⎟ ⎝ ⎠ , (4) When the derivative of Eq. (4) (dP/da) is 0, the load and effective crack length correspond to maximum load. This leads to the relation between power m and effective crack extension in the form of Eq. (5) [8]. 0 max max 0 max ' effP effP effP a a f a W m a a W f W ⎛ +Δ ⎞ ⎜ ⎟ Δ ⎝ ⎠ = ⋅ ⎛ +Δ ⎞ ⎜ ⎟ ⎝ ⎠ , (5) Eq. (5) is general and can be applied to any geometry for which the shape function f(a/W) is known. A numerical inversion of Eq. (5) is quite simple. One can tabulate a set of a0, m and ΔaeffPmax values for a desired geometry and then a specific equation may be fitted to the data, or the information can be interpolated from the table. As an example, the solutions for the standard SE(B) and C(T) specimens are shown graphically in Figs 4a and b. An approximate solution for the standard SE(B) specimen with span width S/W = 4 is given by Eq. (6) and for the standard C(T) specimen with H/W = 0.6 is given in Eq. (7). 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 m = 0.05 m = 0.1 m = 0.15 m = 0.2 Δa effPmax/W a0/W m = 0.25 SE(B) specimen S/W = 4 a) 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 b) m = 0.05 m = 0.1 m = 0.15 m = 0.2 Δa effPmax/W a0/W m = 0.25 C(T) specimen Figure 4. Effective crack growth corresponding to maximum load for standard SE(B) and C(T) specimens
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