Calibration of Stress-triaxiality Dependent Crack Formation Criteria: A New Hybrid Experimental–Numerical Method

Abstract

A new experimental technique has been developed to investigate the onset of fracture in metals at low and intermediate stress triaxialities. The gage section of a flat specimen has been designed such that cracks are most likely to initiate within the specimen center, remote from the specimen boundaries. Along with the specimen, a biaxial testing device has been built to apply a well-defined displacement field to the specimen shoulders. The stress state within the specimen is adjusted by changing the biaxial loading angle. Using this new experimental technique, the crack initiation in metals can be studied experimentally for stress triaxialities ranging from 0.0 to 0.6. The stress and strain fields within the specimen gage section are determined from finite element analysis. The reliability of the computational model of the test set-up has been verified by comparing the simulation results with laser speckle-interferometric displacement measurements during testing. Sample experiments have been performed on the Al-7Si-Mg gravity die casting alloy. A three-step hybrid experimental–numerical calibration procedure has been proposed and applied to determine a phenomenological crack formation criterion for the Al-7Si-Mg alloy.

Appendix

We briefly derive an approximate relationship between the biaxial loading angle and the stress triaxiality in the specimen for a Levy–von Mises plastic material. At a material point in the center of the specimen, the displacement loading may be interpreted as a combination of normal and shear strains. More specifically, we have the plastic normal strain along the y-axis, $\varepsilon ^{p}_{{yy}}$, and the (engineering) shear strain $\gamma ^{p}_{{xy}}$ in the x–y-plane, while the ratio of the two strain components is predetermined by the tangent of the biaxial loading angle α (Fig. 1):

$$\frac{{d\varepsilon ^{p}_{{yy}} }}{{d\gamma ^{p}_{{xy}} }} = \tan \alpha $$

(6)

Furthermore, we assume that the plastic flow along the x-axis be prohibited:

$$\varepsilon ^{p}_{{xx}} = 0$$

(7)

Summarizing of equations (6) and (7) yields in vector notation

$${\mathbf{d}}\varepsilon ^{p} = {\left( {\begin{array}{*{20}c} {{d\varepsilon ^{p}_{{xx}} }} \\ {{d\varepsilon ^{p}_{{yy}} }} \\ {{d\gamma ^{p}_{{xy}} }} \\ \end{array} } \right)} = {\mathbf{m}}d\gamma ^{p}_{{xy}} ,$$

(8)

where

$${\mathbf{m}} = {\left( {\begin{array}{*{20}c} {0} \\ {{\tan \alpha }} \\ {1} \\ \end{array} } \right)}.$$

(9)

The thickness of the specimen is small compared to the specimen width (x-axis) and height (y-axis). Therefore, we consider stress in thickness direction as negligibly small (σ _zz ≅ 0) and assume plane stress conditions throughout our analysis. For this special stress state, the flow rule relates the increments in plastic strains to the three non-zero components of the stress tensor (e.g. [12]):

$${\mathbf{d}}\varepsilon ^{p} = \delta {\mathbf{P\sigma }}$$

(10)

where δ ≥ 0 is the plastic consistency parameter,

$${\mathbf{P}} = \frac{1}{3}{\left[ {\begin{array}{*{20}c} {0} & {{ - 1}} & {0} \\ {{ - 1}} & {2} & {0} \\ {0} & {0} & {6} \\ \end{array} } \right]}\quad {\text{and}}\quad \sigma = {\left( {\begin{array}{*{20}c} {{\sigma _{{xx}} }} \\ {{\sigma _{{yy}} }} \\ {{\tau _{{yy}} }} \\ \end{array} } \right)}$$

(11)

Combining equations (8) and (10), we have

$${\mathbf{\sigma }} = \gamma {\mathbf{P}}^{{ - 1}} {\mathbf{d\varepsilon }}^{p} = \frac{{d\gamma ^{p}_{{xy}} }}{\delta }{\mathbf{P}}^{{ - 1}} {\mathbf{m}}$$

(12)

Next, we apply the definition of the von Mises equivalent stress, σ _v, which yields

$$\sigma _{v} = {\sqrt {\frac{3}{2}\sigma \cdot {\mathbf{P\sigma }}} } = {\left| {\frac{{d\gamma ^{p}_{{xy}} }}{\delta }} \right|}f{\left( \alpha \right)}$$

(13)

where

$$f{\left( \alpha \right)} = {\sqrt {\frac{3}{2}{\mathbf{P}}^{{{\text{ - 1}}}} {\mathbf{m}} \cdot {\mathbf{m}}} } = \frac{1}{2}{\sqrt {12\tan ^{2} \alpha + 3} }$$

(14)

Analogously, we find for the mean stress σ _m

$$\sigma _{m} = \frac{{\sigma _{{11}} + \sigma _{{22}} }}{3} = \frac{{d\gamma ^{p}_{{xy}} }}{\delta }\tan {\left( \alpha \right)}$$

(15)

And finally, using equations (13), (14) and (15) we can calculate the stress triaxiality as a function of the biaxial loading angle:

$$\frac{{\sigma _{m} }}{{\sigma _{v} }} = \frac{{{\text{sign}}{\left( u \right)}}}{{{\sqrt 3 }}}{\sqrt {\frac{1}{{1 + \frac{1}{{4\tan ^{2} \alpha }}}}} }\quad for\quad 0 \leqslant \alpha \leqslant 90^\circ $$

(16)