# Numerical Evaluation of Residual Stress Influence on SIF in CT Specimen
## 4 Stress Intensity Factor Calculation
### Results of Residual Stress and Stress Intensity Factor Calculation  
The first analysis was the analysis of residual stresses. Obtained stress profile (see Figure 10) is shown on the crack front. The stress profile in the CT specimen has a similar shape as the stress profile in the plate (see Section 3) but the actual values are different. In this case, the lowest stress value is \(-73\) MPa, in the middle of the CT specimen the stress reaches 43 MPa, and at both surfaces, stress values are 193 MPa. The residual stress value difference in the CT specimen compared to the plate model can be explained as a  
Figure 9: FE mesh of CT specimen. stress relaxation in the steel after machining the CT specimen and cutting out high-stress outer layers of the plate. When the CT specimen was machined from the pre-strained steel plate, 2.5 mm of the material together with high stresses were cut from both sides. After that, the inner compressed volumes of the steel were constrained by lower surface tension stresses and the steel could relax reducing inner volume compression. The distribution of the crack opening stress \(\sigma_{y}\) (direction TRD) in the CT specimen is presented in Figure 11. To understand how SIF distributes along the crack front due to residual stress, the SIF calculation was conducted. Figure 11: Distribution of crack opening stress \(\sigma_{y}\) (direction TRD) at residual stress case (MPa). Figure 10: Stress profile in the CT specimen (direction TRD). According to ASTM E399-12e3 standard [20], using standard shape and size specimen, SIF can be calculated analytically by the following equation:  
\[K_{I}=\Big{(}P_{Q}/BW^{1/2}\Big{)}\cdot f(a/W) \tag{2}\]  
where:  
\[f(a/W)=\frac{\big{(}2+\frac{a}{W}\big{)}}{\big{(}1-\frac{a}{W}\big{)}^{3/2}}. \bigg{[}0.886+4.64\frac{a}{W}-13.32\Big{(}\frac{a}{W}\Big{)}^{2}+14.72\Big{(} \frac{a}{W}\Big{)}^{3}-5.6\Big{(}\frac{a}{W}\Big{)}^{4}\bigg{]} \tag{3}\]  
here: \(K_{I}\)--stress intensity factor, MPa\(\cdot\)m\({}^{0.5}\); \(P_{Q}\)--load, N; \(B\) --specimen thickness, m; \(W\)--specimen width, m; \(a\)--crack length, m (see Figure 9). However, as it was proven earlier [21], FE programs usually have fairly accurate built-in scripts for numerical SIF calculation. Therefore, in this study, numerical simulation of SIF was chosen. For the accuracy verification of the created FE model, the comparison of numerically and analytically calculated SIF was conducted. The results are shown in Figure 12. As can be seen, numerically determined SIF values are a little bit higher than those calculated analytically using Equations (2) and (3). However, the difference is fairly small, and at 6 kN, load is around 3%. The load around 5800 N is reached at the crack opening displacement of 40 \(\upmu\)m (or 80 \(\upmu\)m for full-scale model).. According to the results, it can be concluded that the created FE model is sufficiently accurate and can be used for SIF calculation. The obtained numerically determined SIF results are presented in Figure 13. The figure shows that the shape of the SIF profile matches the shape of the stress profile. The SIF values reach 3.6 MPa\(\cdot\)m\({}^{0.5}\) on both sides of the specimen. However, deeper in the middle layers of the specimen, the SIF becomes negative. The lowest SIF value is \(-10.6\) MPa\(\cdot\)m\({}^{0.5}\), but in the very middle of the specimen, it is slightly higher and reaches \(-5.4\) MPa\(\cdot\)m\({}^{0.5}\). SIF is usually associated with the crack opening due to tension load and tension stress. In our case, residual stresses are both tensile and compressive. Compressive stress acts in the opposite way, i.e., tries to close the crack. In reality, crack closure would stop at the point when two crack surfaces meet each other. However, in our model, for simplification, only half of the CT specimen is created and no contact of the crack surface to a symmetry plane is modeled. Therefore, due to compressive residual stress, the crack surface can deform much more and pass the symmetry plane which results in negative SIF values.