A coupled numerical–experimental study of the breakage process of brittle substances

Abstract

Pre-existing cracks in brittle substances seem to be the main cause of their breakage under various loading conditions. In the present paper, a coupled numerical–experimental analysis of crack propagation, cracks coalescence, and breakage process of brittle substances such as rocks and rock-like samples have been studied. The numerical analyses are accomplished using a numerical code based on the Higher order Displacement Discontinuity Method for Crack (HDDMCR^2D) analysis. A quadratic displacement discontinuity variation along each boundary element is assumed to evaluate the Mode I and Mode II stress intensity factors. Based on the linear elastic fracture mechanics theory, the maximum tangential stress criterion (i.e., a mixed mode fracture criterion) is implemented in the HDDMCR^2D code for predicting the crack initiation and its direction of propagation (cracks propagation path). Some numerical and analytical problems in finite and infinite planes are solved numerically by the proposed numerical method, and the results are compared in different tables illustrating the accuracy and validity of the numerical results. Experimental tests are also being done to evaluate the final breakage path and cracks initiation and cracks coalescence stresses in rock-like specimens containing two random cracks. The numerical and experimental results obtained from the tested specimens show a good agreement between the corresponding values and demonstrate the accuracy and effectiveness of the approach.

Appendices

Appendix 1

The integrals and their derivatives used for quadratic displacement discontinuity elements (with equal sub-elements) for finite and infinite plane fracture mechanics problems

Starting from the common potential function F(x,y) expressed by Marji et al. (2006) for the solution of stress and displacement fields at the discretized boundaries using the displacement discontinuity function, DD  _j (δ) given in Eq. (1):

$$ F\left(x,y\right)=\frac{-1}{4\pi \kern0.24em \left(1-\nu\;\right)}{\displaystyle \underset{-\mathcal{l}}{\overset{\mathcal{l}}{\int }}{\operatorname{DD}}_j\left(\delta \right) \ln {\left[{\left(x-\delta \right)}^2+{y}^2\right]}^{\frac{1}{2}}\operatorname{d}\delta, \kern0.5em j=x,y\kern0.24em } $$

(9)

Inserting the common displacement discontinuity function DD  _j (δ) (Eq. 1) in Eq. (9) gives:

$$ \begin{array}{l}F\left(x,y\right)=\frac{-1}{4\pi \kern0.24em \left(1-\nu\;\right)}\left\{\left[{\displaystyle \underset{-\mathcal{l}}{\overset{\mathcal{l}}{\int }}{\varGamma}_1\left(\delta \right) \ln {\left[{\left(x-\delta \right)}^2+{y}^2\right]}^{\frac{1}{2}}\operatorname{d}\delta}\right]\right.{\operatorname{DD}}_j^1+\\ {}\left[{\displaystyle \underset{-\mathcal{l}}{\overset{\mathcal{l}}{\int }}{\varGamma}_2\left(\delta \right) \ln }{\left[{\left(x-\delta \right)}^2+{y}^2\right]}^{\frac{1}{2}}\operatorname{d}\delta \right]{\operatorname{DD}}_j^2\kern0.36em +\left[{\displaystyle \underset{-\mathcal{l}}{\overset{\mathcal{l}}{\int }}{\varGamma}_3\left(\delta \right) \ln }{\left[{\left(x-\delta \right)}^2+{y}^2\right]}^{\frac{1}{2}}\operatorname{d}\delta \right]{\operatorname{DD}}_j^3,j=x,y\kern0.6em \\ {}\kern0.96em \end{array} $$

(10)

Inserting the shape functions Γ ₁(δ), Γ ₂(δ), and Γ ₃(δ) in Eq. (10) after some manipulations and rearrangements the following three special integrals are deduced:

$$ {I}_1\left(x,y\right)={{\displaystyle \underset{-\mathcal{l}}{\overset{\mathcal{l}}{\int }} \ln \left[{\left(x-\delta \kern0.24em \right)}^2+{y}^2\right]}}^{\frac{1}{2}}\operatorname{d}\delta =y\left({\phi}_1-{}_2\right)-\left(x-\mathcal{l}\right) \ln \left({\eta}_1\right)+\left(x+\mathcal{l}\right) \ln \left({\eta}_2\right)-2\mathcal{l} $$

(11)

$$ {I}_2\left(x,y\right)={{\displaystyle \underset{-\mathcal{l}}{\overset{\mathcal{l}}{\int }}\delta \kern0.36em \ln \left[{\left(x-\delta \kern0.24em \right)}^2+{y}^2\right]}}^{\frac{1}{2}}\operatorname{d}\delta = xy\left({\phi}_1-{\phi}_2\right)+0.5\left({y}^2-{x}^2+{\mathcal{l}}^2\right) \ln \frac{\eta_1}{\eta_2}-\mathcal{l}x $$

(12)

$$ \begin{array}{c}\hfill {I}_3\left(x,y\right)={{\displaystyle \underset{-\mathcal{l}}{\overset{\mathcal{l}}{\int }}{\delta}^{\;2} \ln \left[{\left(x-\delta\;\right)}^2+{y}^2\right]}}^{\frac{1}{2}}\operatorname{d}\delta =\frac{y}{3}\left(3{x}^2-{y}^2\right)\left({\phi}_1-{\phi}_2\right)+\frac{1}{3}\left(3x{y}^2-{x}^3+{a}^3\right) \ln \left({\eta}_1\right)\hfill \\ {}\hfill -\frac{1}{3}\left(3x{y}^2-{x}^3-{\mathcal{l}}^3\right) \ln \left({\eta}_2\right)-\frac{2\mathcal{l}}{3}\left({x}^2-{y}^2+\frac{{\mathcal{l}}^2}{3}\right)\hfill \end{array} $$

(13)

Where ϕ ₁, ϕ ₂, η ₁, and η ₂ can be defined as:

$$ {\phi}_1= \arctan \left(\frac{y}{x-\mathcal{l}}\right),\kern0.72em {\phi}_2= \arctan \left(\frac{y}{x+\mathcal{l}}\right),{\eta}_1={\left[{\left(x-\mathcal{l}\right)}^2+{y}^2\right]}^{\frac{1}{2}}\ \mathrm{and}\kern0.84em {\eta}_2={\left[{\left(x+\mathcal{l}\right)}^2+{y}^2\right]}^{\frac{1}{2}} $$

(14)

Appendix 2

The integrals and their derivatives used for three special crack tip elements of equal length for finite and infinite plane fracture mechanics problems

Starting from the common special potential function F _C (x,y) expressed by Marji et al. (2006) for the solution of stress and displacement fields at the crack tip using the displacement discontinuity function, DD  _j (δ) given in Eq. (4):

$$ {F}_C\left(x,y\right)=\frac{-1}{4\pi \kern0.24em \left(1-\nu \right)\;}{\displaystyle \underset{-\mathcal{l}}{\overset{\mathcal{l}}{\int }}{\operatorname{DD}}_j\left(\delta \right) \ln {\left[{\left(x-\delta \right)}^2+{y}^2\right]}^{\frac{1}{2}}\operatorname{d}\delta, \kern0.5em j=x,y\kern0.24em } $$

(15)

Inserting the common displacement discontinuity function, DD  _j (δ) (Eq. 3) in Eq. (15) gives:

$$ \begin{array}{l}\begin{array}{c}\hfill {F}_C\left(x,y\right)=\frac{-1}{4\pi \kern0.24em \left(1-\nu \right)\;}\left\{\left[{\displaystyle \underset{-\mathcal{l}}{\overset{\mathcal{l}}{\int }}{\varGamma}_{C1}\left(\delta \right) \ln {\left[{\left(x-\delta \right)}^2+{y}^2\right]}^{\frac{1}{2}}\operatorname{d}\delta}\right]{\operatorname{DD}}_j^1+\right.\hfill \\ {}\hfill \left[{\displaystyle \underset{-\mathcal{l}}{\overset{\mathcal{l}}{\int }}{\varGamma}_{C2}\left(\delta \right) \ln }{\left[{\left(x-\delta \right)}^2+{y}^2\right]}^{\frac{1}{2}}\operatorname{d}\delta \right]{\operatorname{DD}}_j^2\kern0.36em +\left[{\displaystyle \underset{-\mathcal{l}}{\overset{\mathcal{l}}{\int }}{\varGamma}_{C3}\left(\delta \right) \ln }{\left[{\left(x-\delta \right)}^2+{y}^2\right]}^{\frac{1}{2}}\operatorname{d}\delta \right]{\operatorname{DD}}_j^3,j=x,y\hfill \end{array}\\ {}\;\end{array} $$

(16)

Inserting the shape functions Γ _C1(δ), Γ _C2(δ), and Γ _C3(δ) in Eq. (16) after some manipulations and rearrangements the following three special integrals are deduced:

$$ \begin{array}{c}\hfill {I}_{C1}\left(x,y\right)={\displaystyle \underset{-\mathcal{l}}{\overset{\mathcal{l}}{\int }}{\delta}^{\frac{1}{2}} \ln {\left[{\left(x-\delta \right)}^2+{y}^2\right]}^{\frac{1}{2}}\operatorname{d}\delta },\kern0.5em \hfill \\ {}\hfill {I}_{C2}\left(x,y\right)={\displaystyle \underset{-\mathcal{l}}{\overset{\mathcal{l}}{\int }}{\delta}^{\frac{3}{2}} \ln {\left[{\left(x-\delta \right)}^2+{y}^2\right]}^{\frac{1}{2}}\operatorname{d}\delta}\hfill \\ {}\hfill {I}_{C3}\left(x,y\right)={\displaystyle \underset{-\mathcal{l}}{\overset{\mathcal{l}}{\int }}{\delta}^{\frac{5}{2}} \ln {\left[{\left(x-\delta \right)}^2+{y}^2\right]}^{\frac{1}{2}}\operatorname{d}\delta}\hfill \end{array} $$

(17)

The derivatives of the integrals, I _C1, I _C2, are given by Marji et al. (2006) and the first two derivatives of I _C3 (for three special crack tip element case) can be expressed as:

$$ \begin{array}{c}\hfill {I}_{C,x}^3=x{\displaystyle \underset{0}{\overset{2\mathcal{l}}{\int }}\frac{\delta^{\frac{5}{2}}}{\left[{\left(x-\delta \right)}^2+{y}^2\right]}}\operatorname{d}\delta -{\displaystyle \underset{0}{\overset{2\mathcal{l}}{\int }}\frac{\delta^{\frac{7}{2}}}{\left[{\left(x-\delta \right)}^2+{y}^2\right]}}\operatorname{d}\delta =x{\varOmega}_3-{\varOmega}_4\hfill \\ {}\hfill {I}_{C,y}^3=y{\displaystyle \underset{0}{\overset{2\mathcal{l}}{\int }}\frac{\delta^{\frac{5}{2}}}{\left[{\left(x-\delta \right)}^2+{y}^2\right]}}\operatorname{d}\delta =y{\varOmega}_3\hfill \end{array} $$

(18)

Where

$$ {\varOmega}_3={\displaystyle \underset{0}{\overset{2\mathcal{l}}{\int }}\frac{\delta^{\frac{5}{2}}}{\left[{\left(x-\delta \right)}^2+{y}^2\right]}}\operatorname{d}\delta =\frac{2}{3}{(2a)}^{\frac{3}{2}}+2x{\varOmega}_2-\left({x}^2+{y}^2\right)\varOmega {}_1 $$

(19)

where Ω ₁, Ω ₂, and the derivatives of Ω ₁, are defined by Marji et al. (2006) as:

$$ \begin{array}{l}\begin{array}{c}\hfill {\varOmega}_1={\displaystyle \underset{-\mathcal{l}}{\overset{\mathcal{l}}{\int }}\frac{\delta^{\frac{1}{2}}}{\left[{\left(x-\delta \right)}^2+{y}^2\right]}\operatorname{d}\delta }=\hfill \\ {}\hfill {\lambda}^{-1}\left[\begin{array}{l}0.5\left( \cos \varphi -\left(\frac{x}{y}\right) \sin \beta \right) \ln \frac{2\mathcal{l}-2\sqrt{2\mathcal{l}}\;\rho \cos \beta +{\lambda}^2}{2\mathcal{l}+2\sqrt{2\mathcal{l}}\;\rho \cos \beta +{\lambda}^2}+\left( \sin \beta +\left(\frac{x}{y}\right) \cos \beta \right)\times \\ {} \arctan \left(\frac{2\sqrt{2\mathcal{l}}\;\rho \sin \beta }{\lambda^2-2\mathcal{l}}\right)\end{array}\right]\hfill \end{array}\\ {}\kern5.12em \end{array} $$

(20)

$$ \begin{array}{c}\hfill {\varOmega}_2={\displaystyle \underset{-\mathcal{l}}{\overset{\mathcal{l}}{\int }}\frac{\delta^{\frac{3}{2}}}{\left[{\left(x-\delta \right)}^2+{y}^2\right]}\operatorname{d}\delta }=\hfill \\ {}\hfill\ \lambda\;\left[\begin{array}{l}0.5\left( \cos \beta +\left(\frac{x}{y}\right) \sin \beta \right) \ln \frac{2\mathcal{l}-2\sqrt{2\mathcal{l}}\;\lambda \cos \beta +{\lambda}^2}{2\mathcal{l}+2\sqrt{2\mathcal{l}}\;\lambda \cos \beta +{\lambda}^2}+\left( \sin \beta +\left(\frac{x}{y}\right) \cos \beta \right)\times \\ {} \arctan \left(\frac{2\sqrt{2\mathcal{l}}\;\rho \sin \beta }{\lambda^2-2\mathcal{l}}\right)\end{array}\right]\hfill \end{array} $$

(21)

where $\lambda ={\left({x}^2+{y}^2\right)}^{\frac{1}{4}},\mathrm{and}\ \beta =0.5 \arctan \left(y/x\right)$,

and finally

$$ {\varOmega}_4={\displaystyle \underset{0}{\overset{2\mathcal{l}}{\int }}\frac{\delta^{\frac{7}{2}}}{\left[{\left(x-\delta \right)}^2+{y}^2\right]}}\operatorname{d}\delta =\frac{2}{5}{\left(2\mathcal{l}\right)}^{\frac{5}{2}}+2x{\varOmega}_3-\left({x}^2+{y}^2\right)\varOmega {}_2 $$

(22)