13th International Conference on Fracture June 16–21, 2013, Beijing, China -4- According to what discussed above, (0) f F ε = , ( ) y F l ε = . (7) The substitution of Eq. (7) into Eq. (6) provides Eq. (8), f a ε = , 1 ln y f b l ε ε = . (8) So, the elongation of the bar can be expressed as Eq. (9), 0 ( ) ln ln l f y f y l F y dy l ε ε ε ε − Δ = = − ∫ . (9) Thus, the ratio of elongation of the bar can be expressed as Eq. (10), ln ln f y f y l P l ε ε ε ε − Δ = = − . (10) Because fε and yε are material constants, P is a constant, too. That is to say, the dimension of plastic field left on specimen is in proportion to the plastic deformation. At the same time, because lΔ can be acquired from FRASTA, the plastic field left on specimen can be determined. Return to the state as shown in Fig. 1, ijδ represents the elongation of the bar at ( ix , j y ), where i and j represent the number of position along x and y axis respectively. The sizes of these bars along x and y direction (the direction of crack extension and specimen thickness respectively) are determined by the step lengths of laser microscope, which are represented by p and q respectively in this paper. Then, the original length ij l of each bar should be ij ij l P δ = , (11) the length of each bar after being broken is ' (1 ) ij ij l P P δ = + . (12) In order to calculate the load applied to the specimen during the course of fatigue, based on the above simple bar hypothesis, the fracture surfaces are assumed to be composed of independent rectangular bars. Once the normal stress on the cross section of each bar was obtained, the force applied on the bar can be determined by multiplying the normal stress with the cross section area. By this means, adding the force applied on each bar together, the load applied on the specimen can be obtained.
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