Evaluating the material strength from fracture angle under uniaxial loading

Abstract

The most common experimental methods of measuring material strength are the uniaxial compressive and tensile tests. Generally, shearing fracture model occurs in both the tests. Compressive strength is higher than tensile strength for a material. Shearing fracture angle is smaller than 45° under uniaxial compression and greater than 45° under uniaxial tension. In this work, a unified relation of material strength under uniaxial compression and tension is developed by correlating the shearing fracture angle in theory. This constitutive relation is quantitatively illustrated by a function for analyzing the material strength from shear fracture angle. A computational simulation is conducted to validate this theoretical function. It is full of interest to give a scientific illustration for designing the high-strength materials and engineering structures.

Keywords

strength / fracture / mechanics

Introduction

Strength is a material capability to withstand deformation and fracture drived by an applied stress, that is an important parameter for engineering application and failure criterion evaluation [¹–³]. The most common experimental method of measuring material strength is the uniaxial compressive or tensile test. However, material strength somehow displays a difference under tensile and compressive loadings [⁴–⁸], which is related to the tension-compression asymmetry of material properties. This phenomenon significantly affects the engineering application and failure analysis.

The phenomenon of tension-compression asymmetry of strength was found in various materials, which attracts an extensive research interest [³,⁶,⁷]. In a single crystal NiTi alloy, the tension-compression asymmetry of the stress-strain response was reported [⁴], which is due to the dependency of the critical resolved shear stress required to initiate the deformation on both the stress direction and crystallographic orientation. The analogous phenomenon was also found in the single crystals of Cu, Au, Ni and Ni₃Al [⁵]. In various polycrystalline materials, the tension-compression asymmetry was found to be correlated to the Swift effect, because the sign of the axial plastic straining depends on the twisting direction [⁶]. In ultrafine-grained and nanocrystalline metals, the tension-compression asymmetry arises from the pressure or normal stress dependence of the dislocation self-energy during bowing out, which is a function of the material itself and grain size [⁷–¹⁰]. In amorphous metals, the tension-compression asymmetry in mechanical behaviour and fracture mechanisms was studied, which shows the combined effect of the normal stress and shear stress on the fracture plane resulting in the deviation of fracture angle from 45° [¹¹,¹²]. In addition, the tension-compression asymmetry is also found in composites [¹³–¹⁵], as well as in various materials under dynamic loading [¹⁶–²⁰]. Therefore, the tension-compression asymmetry in mechanical behaiviour is a universal phenomenon in various materials and loading condtions because of the different mechanisms [²¹–²⁴]. It is always related to the resolved shear stress and normal stress on a fracture plane, which induces the deviation of shearing fracture from the maximum shear stress direction.

Computational models and methods also have been developed to address the tension-compression asymmetry in mechanical behaviour of various materials [²⁵–³⁴]. Areias et al. reported an efficient algorithm for a finite element method to model the fracture of plates and shells on the basis of edge rotation and load control [³²]. An alternative crack propagation algorithm is proposed, which composes of two stages of I, producing an appropriate mesh without a localized limit, and II, using a staggered algorithm with screened Poisson equations [³³,³⁴]. Dynamic crack and shear band propagation are evaluated based on the extended finite element method (XFEM) [³⁵]. The meshless method can also be used for fracture analysis of a cracking process [³⁶]. A cracked particles method was developed which can simplify the meshfree method for arbitrary cracks [³⁷]. So, finite element method is available for simulating a cracking process.

In this paper, we propose an efficient relation for illustrating the strength and fracture angle of materials under both tensile and compreissive loadings. The phenomenon of tension-compression asymmetry is explained by a formula dedueced in theory. A computational model via finite element method is developed to valiate the theoretical relation.

Theoretical analysis

Schuh et al. provided an atomic-level explanation for pressure- or normal stress-dependent yield for amorphous metals in addition to nanostructured materials and amorphtous alloys [¹]. Mohr-Coulomb criterion has been comfirmed as a reasonable description of the material deformation and fracture behaviour under uniaxial compressive or tensile loading. Herein, it depends not only on the applied shear stress, but also on the stress normal to the shearing plane, {\sigma }^n, as below:

(1){{\displaystyle \tau }}_y={{\displaystyle \tau }}_{\mathrm{0}}-{\mathrm{\mu }\sigma }^n

Herein, {{\displaystyle \tau }}_y is the effective shear yield stress, {{\displaystyle \tau }}_{\mathrm{0}} is a constant for a given material, and \mu is a system specific coefficient that controls the strength of the normal stress effect. Afterwards, this theory harvests an extensive application in material failure analysis [³⁸–⁴³].

The field of material strength deals with the stress and deformation that result from their acting on a material. Under a uniaxial compressive loading, shearing or local shearing deformation occurs in the materials, with a shearing fracture angle of less than or equal to 45°, as schematically shown in Figure 1(a). Along the shearing fracture plane exist two component stresses caused by the applied uniaxial compressive stress, {{\displaystyle \sigma }}_c, i.e. shear stress, {{\displaystyle \sigma }}_c^s and normal stress, {{\displaystyle \sigma }}_c^n. Shear stress, {{\displaystyle \sigma }}_c^s is parallel to the shearing fracture plane, which is a driving stress for the shear crack initiation and propagation under the applied compressive stress, as shown in Figure 1(b). Normal stress, {{\displaystyle \sigma }}_c^n is perpendicular to the shearing fracture plane in a compressive model.

Similarly, under a uniaxial tensile loading (see Figures 1(c), (d)), shearing fracture angle is greater than or equal to 45°. Along the shearing fracture plane also exist two component stresses resulted from the applied uniaxial tensile stress, {{\displaystyle \sigma }}_t, i.e. shear stress, {{\displaystyle \sigma }}_t^s and normal stress, {{\displaystyle \sigma }}_t^n. Shear stress, {{\displaystyle \sigma }}_t^s is parallel to the shearing fracture plane, which is a driving stress for the shear crack initiation and propagation under the applied tensile stress. Normal stress, {{\displaystyle \sigma }}_t^n is perpendicular to the shearing fracture plane in a tensile model.

According to the stress analysis in Figure 1, we can know that shear stress and normal stress always exert on a shearing crack when the material is subjected to a uniaxial compressive or tensile loading, and the shear stress is the driving stress for the shear crack initiation and propagation. The difference is that the normal stress is in a compressive model under the uniaxial compressive loading and is in a tensile model under the uniaxial tensile loading, which causes the difference in shearing fracture angle and the tension-compression asymmetry of material strength. Subsequently, we will discuss the shear cracking process and the stress origin for shearing fracture under a uniaxial compressive and tensile loading, respectively.

When a material is subjected to a uniaxial compressive stress, {{\displaystyle \tau }}_C, according to the Eq. (1), the critical shear fracture stress, {{\displaystyle \tau }}_C is:

(2){{\displaystyle \tau }}_C={{\displaystyle \tau }}_{\mathrm{0}}+{{\displaystyle \mathrm{\mu }}}_C{{\displaystyle \sigma }}_C^n

According to the shear fracture criterion, the critical shear fracture condition under uniaxial compression can be obtained as:

(3){{\displaystyle \sigma }}_C^s\ge {{\displaystyle \tau }}_C={{\displaystyle \tau }}_{\mathrm{0}}+{{\displaystyle \mathrm{\mu }}}_C{{\displaystyle \sigma }}_C^n

When the material is subjected to a uniaxial tensile stress, {{\displaystyle \tau }}_T, according to the Eq. (1), the critical shear fracture stress, {{\displaystyle \tau }}_T is:

(4){{\displaystyle \tau }}_T={{\displaystyle \tau }}_{\mathrm{0}}-{{\displaystyle \mathrm{\mu }}}_T{{\displaystyle \sigma }}_T^n

According to the shear fracture criterion, the critical shear fracture condition under uniaxial tension can be obtained as:

(5){{\displaystyle \sigma }}_T^s\ge {{\displaystyle \tau }}_T={{\displaystyle \tau }}_{\mathrm{0}}-{{\displaystyle \mathrm{\mu }}}_T{{\displaystyle \sigma }}_T^n

Herein, for a given material, {{\displaystyle \tau }}_{\mathrm{0}} is a constant; {{\displaystyle \mu }}_C and {{\displaystyle \mu }}_T are the system specific coefficients that affects the strength caused by the normal stress under uniaxial compressive and tensile loading, respectively.

Under the uniaxial compressive or tensile stress, {{\displaystyle \sigma }}_{C/T}, according to the geometric relation as shown in Figures 1(a), (c), the shear stress, {{\displaystyle \sigma }}_{C/T}^s along the shearing fracture plane and the normal stress, {{\displaystyle \sigma }}_{C/T}^n perpendicular to the shearing fracture plane can be expressed:

(6){{\displaystyle \sigma }}_{C/T}^s={{\displaystyle \sigma }}_{C/T}\sin \mathrm{\theta }\cos \mathrm{\theta }

(7){{\displaystyle \sigma }}_{C/T}^n={{\displaystyle \sigma }}_{C/T}{\sin }^{\mathrm{2}}\mathrm{\theta }

According to the Eqs. (3), (5), (6), (7), the following relations can be achieved:

(8){{\displaystyle \sigma }}_C\ge {\displaystyle \frac{{{\displaystyle \tau }}_{\mathrm{0}}}{\sin \mathrm{\theta }(\cos \mathrm{\theta }-{{\displaystyle \mathrm{\mu }}}_C\sin \mathrm{\theta })}}

(9){{\displaystyle \sigma }}_T\ge {\displaystyle \frac{{{\displaystyle \tau }}_{\mathrm{0}}}{\sin \mathrm{\theta }(\cos \mathrm{\theta }+{{\displaystyle \mathrm{\mu }}}_T\sin \mathrm{\theta })}}

Thus, the critical uniaxial stresses for shear cracking process can be deduced:

(10){{\displaystyle \sigma }}_C={\displaystyle \frac{{{\displaystyle \tau }}_{\mathrm{0}}}{\sin \mathrm{\theta }(\cos \mathrm{\theta }-{{\displaystyle \mu }}_C\sin \mathrm{\theta })}}

(11){{\displaystyle \sigma }}_T={\displaystyle \frac{{{\displaystyle \tau }}_{\mathrm{0}}}{\sin \mathrm{\theta }(\cos \mathrm{\theta }+{{\displaystyle \mu }}_T\sin \mathrm{\theta })}}

When a material is subjected to the uniaxial compressive or tensile stress, crack will initiate and propagate at a favorite shearing fracture angle for contributing to the minimum applied stress. Thus, the shearing fracture angle, \theta can be considered as a variable parameter from 0° to 90°, and:

(12){\displaystyle \frac{\partial ({{\displaystyle \sigma }}_{C/T})}{\partial \mathrm{\theta }}}=\mathrm{0}

According to the Eqs. (10)-(12), we can obtain:

(13){\displaystyle \frac{\partial (\mathrm{1}/{{\displaystyle \sigma }}_C)}{\partial \mathrm{\theta }}}={\displaystyle \frac{\mathrm{1}}{{{\displaystyle \mathrm{2}\tau }}_{\mathrm{0}}}}(\cos \mathrm{2}\mathrm{\theta }-{{\displaystyle \mathrm{\mu }}}_C\sin \mathrm{2}\mathrm{\theta })=\mathrm{0}

(14){\displaystyle \frac{\partial (\mathrm{1}/{{\displaystyle \sigma }}_T)}{\partial \mathrm{\theta }}}={\displaystyle \frac{\mathrm{1}}{\mathrm{2}{{\displaystyle \tau }}_{\mathrm{0}}}}(\cos \mathrm{2}\mathrm{\theta }+{{\displaystyle \mathrm{\mu }}}_T\sin \mathrm{2}\mathrm{\theta })=\mathrm{0}

Then, the system specific coefficients that affects the strength caused by the normal stress, {{\displaystyle \mathrm{\mu }}}_C and {{\displaystyle \mathrm{\mu }}}_T are arrived as following relations:

(15){{\displaystyle \mathrm{\mu }}}_C={\displaystyle \frac{\cos \mathrm{2}\mathrm{\theta }}{\sin \mathrm{2}\mathrm{\theta }}}

(16){{\displaystyle \mathrm{\mu }}}_T=-{\displaystyle \frac{\cos \mathrm{2}\mathrm{\theta }}{\sin \mathrm{2}\mathrm{\theta }}}

From the fomula, it can be seen that the system specific coefficients are just related to the fracture angle \mathrm{\theta }. Thus, according to the Eqs. (10), (11), (15), (16), we can derive:

(17){{\displaystyle \sigma }}_{C/T}={\displaystyle \frac{{{\displaystyle \tau }}_{\mathrm{0}}}{\sin \mathrm{\theta }\left(\cos \mathrm{\theta }-{\displaystyle \frac{\cos \mathrm{2}\mathrm{\theta }}{\sin \mathrm{2}\mathrm{\theta }}}\sin \mathrm{\theta }\right)}}

Therefore, we will define:

(18){{\displaystyle \sigma }}_{C/T}=f(\mathrm{\theta }){{\displaystyle \tau }}_{\mathrm{0}}

(19){\displaystyle \frac{{{\displaystyle \sigma }}_{C/T}}{{{\displaystyle \tau }}_{\mathrm{0}}}}=F(\mathrm{\theta }){\displaystyle \frac{\mathrm{1}}{\sin \mathrm{\theta }\left(\cos \mathrm{\theta }-{\displaystyle \frac{\cos \mathrm{2}\mathrm{\theta }}{\sin \mathrm{2}\mathrm{\theta }}}\sin \mathrm{\theta }\right)}}

For a given material, Eq. (19) indicates that the material strength is related to {{\displaystyle \tau }}_{\mathrm{0}} and shearing fracture angle. So, the function of shearing fracture angle, f(\mathrm{\theta }) is key for evaluating the material strength. According to the Eq. (19), we can draw the curve of the function, f(\mathrm{\theta }), as shown in Figure 2. We can know that with the increase of shearing fracture angle, \mathrm{\theta }, the value of the function, f(\mathrm{\theta }) decreases, which leads to the decrease of material strength, {{\displaystyle \sigma }}_{C/T}. So, the compressive strength of a material is higher than the tensile strength, because the shearing fracture angle under uniaxial compressive loading is smaller than that under uniaxial tensile loading. It is the reason for the tension-compression asymmetry of material strength under an applied uniaxial loading.

Moreover, when the shearing fracture angle approaches to 0°, the value of function, f(\mathrm{\theta }) approaches to positive infinite, which means that the material fracture is not in a shearing model rather than in a cleavage model under an applied uniaxial compressive stress. Under this condition, material strength becomes the limited high for a given material. On the other hand, when the shearing fracture angle approaches to 90°, the value of function, f(\mathrm{\theta }) approaches to 0, which means that the material fracture is in a normal model rather than in a shearing model under an applied uniaxial tensile stress. Under this condition, material strength becomes the limited low for a given material. Therefore, for designing an advanced material with an improved strength under a uniaxial loading, decreasing the shearing fracture angle, \mathrm{\theta } (external phenomenon) and/or improving the {{\displaystyle \tau }}_{\mathrm{0}} (internal characteristic) are suggested to be an effective method. Herein, {{\displaystyle \tau }}_{\mathrm{0}} can be regarded as the critical shear fracture strength, which is the characteristic parameter of a material.

In particular, when shearing fracture angle is equal to 45°, the value of function, f(\theta ) is equal to 2. Thus, {{\displaystyle \sigma }}_{C/T}={{\displaystyle \mathrm{2}\tau }}_{\mathrm{0}}. That means under a uniaxial loading the shearing crack propagates along the direction of the maximum shear stress. In this case, the material has a high strength-toughness combination.

Computational simulation

This computational simulation is aim to validate the relation of strength and fracture angle, as shown in Figure 2. Five cases are constructed, which include the fracture angle of 30°, 37° and 45° under uniaxial compression and of 53° and 60° under uniaxial tension. By preinstalling the crack, shearing fracture direction is determined. Material parameters in these five cases are the same, which indicates the same value of {{\displaystyle \tau }}_{\mathrm{0}} in Eq. (19). By computational simulation using a cohesive element method, the displacement-force curves of these five cases are acquired, as shown in Figure 3. We can see that the max force at each case decreases with the increase of fracture angle.

Then, the stress-fracture angle relation can be achieved, as shown (red curve B) in Figure 4. It has a close trendency with the relation calculated by Eq. (19). Afterwards, the material parameter of {{\displaystyle \tau }}_{\mathrm{0}} is varied and the stress-fracture angle relation is corrected as shown (black curve A) in Figure 4. This corrected curve has a sound uniformity with the theoretic results. Therefore, by a step-by-step simulation, the computational results can arrive at the theoretic results, which validates the theoretic analysis.

Conclusions

To conclude, the tension-compression asymmetry of material strength under a uniaxial loading is analyzed in terms of shearing fracture angle. A unified relation of material strength under tension and compression loadings is illustrated by a constitutive function. Based on this function, a computational modeling is constructed for validating this unified relation. The theoretic strategies, i.e. decreasing the shearing fracture angle and/or increasing the critical shear fracture strength, are proposed for strengthening materials. Also, it is full of interest to develop a high-strength material structure by designing the shearing fracture direction.

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