3. Analytical approach

the elastic analysis, the stress analysis and energy computations were organised in line with the linear relationship between the stress and the strain in coal and overburden properties. The peak and post-peak behaviour of coal and surrounding rock masses will be ignored. Therefore, in the current literature, the computed stress, strain and kinetic energy have been noticeably overestimated. At the second stage, a combination of the 2D and 3D discrete element models using UDEC and 3DEC was developed. Figure 2 illustrates the pillar model incorporating half of coal, roof and floor along the symmetrical centre-line of the pillar. The height of the roof and floor was 20 m and the mining height was fixed at 3 m, while the pillar widths varied in order to

A Mohr-Coulomb (MC) material that presents a constant strength after failure and a Mohr-Coulomb strain-softening material that can reach the peak strength and then decrease to a residual strength have been considered. A quasi-static loading condition as a velocity was applied on the top and bottom of the model. The applied velocity was started with a very small, constant velocity to represent a relative loading system to promote a model of a coal failure that progresses slowly. Simulating a proper loading/displacement condition is significantly crucial, specifically, gaining a sound understanding of the structural reaction of a single

simulate the pillars with width to height (w/h) ratios from 1 to 5.

210 Finite Element Method - Simulation, Numerical Analysis and Solution Techniques

Figure 2. Geometry and zoning of coal pillar model using UDEC.

An analytical method is developed to evaluate shear stress and strain distributions between the engaged surfaces throughout different joint layers by considering the beam theory method in different directions with respect to the different planes, where it can independently calculate shear forces between the different layers and shear strain as well as the curvature distribution along the different layers that have been extracted. The main concept to derive the following equations was extracted from [12, 13]. The cross-sectional analysis is based on the assumption of the Euler-Bernoulli beam model. The strain distribution across the section can be calculated by ε ¼ ε<sup>r</sup> � y � κ, where ε<sup>r</sup> is the strain at the reference point (which can be determined at any point), y is the distance between the selected point and location of the neutral axis of the cross-section and κ is the curvature across the section in different strata layers. A vector can be introduced by K Dð Þ which will be included in the internal action N (axial forces) and M (internal moment). External loads, which might be due to the effect of the self-weight of the strata layers as well as the possible applied forces due to the vertical or horizontal displacement in the different layers, can induce the external axial force Ne and external moment Me. The relationship between the internal and external actions can be presented by:

$$
\varepsilon = \begin{bmatrix} \varepsilon\_r \\\\ \kappa \end{bmatrix} \tag{1}
$$

All the equations can be re-presented in matrix format:

<sup>Δ</sup>εð Þ<sup>i</sup> <sup>¼</sup> <sup>Δ</sup><sup>ε</sup>

practical form, recalling the definitions of internal actions as:

rt <sup>ε</sup>ð Þ<sup>i</sup> � � <sup>¼</sup>

ð Þi r Δκð Þ<sup>i</sup> " # ∂N ε ð Þi <sup>r</sup> ; κð Þ<sup>i</sup> � �

∂M ε ð Þi <sup>r</sup> ; κð Þ<sup>i</sup> � �

> r ð Þi <sup>R</sup> <sup>¼</sup> <sup>N</sup>ð Þ<sup>i</sup> R Mð Þ<sup>i</sup> R

∂N ε ð Þi <sup>r</sup> ; κð Þ<sup>i</sup> � �

∂N ε ð Þi <sup>r</sup> ; κð Þ<sup>i</sup> � �

∂ε<sup>r</sup>

∂M ε ð Þi <sup>r</sup> ; κð Þ<sup>i</sup> � �

∂M ε ð Þi <sup>r</sup> ; κð Þ<sup>i</sup> � �

> ð ∂σ ∂ε � ∂ε ∂ε<sup>r</sup> dA ¼

> > ð ∂σ ∂ε �

> > > y � ∂σ ∂ε �

y � ∂σ ∂ε �

dA ¼ � <sup>ð</sup>

dA ¼ � <sup>ð</sup>

ð ∂σ ∂κ dA ¼

y � ∂σ ∂ε<sup>r</sup>

y � ∂σ ∂ε<sup>r</sup>

on the magnitude of the strain

∂ε<sup>r</sup>

∂N ε ð Þi <sup>r</sup> ; κð Þ<sup>i</sup> � �

¼ ð ∂σ ∂ε<sup>r</sup> dA ¼

∂κ ¼

¼ � <sup>ð</sup>

<sup>∂</sup><sup>κ</sup> ¼ � <sup>ð</sup>

∂N ε ð Þi <sup>r</sup> ; κð Þ<sup>i</sup> � �

∂M ε ð Þi <sup>r</sup> ; κð Þ<sup>i</sup> � �

∂M ε ð Þi <sup>r</sup> ; κð Þ<sup>i</sup> � �

∂ε<sup>r</sup>

∂ε<sup>r</sup>

∂ε<sup>r</sup>

∂ε<sup>r</sup>

" #

The partial derivatives of N and M with respect to ε<sup>r</sup> and κ can be re-arranged in a more

∂κ ¼

<sup>∂</sup><sup>κ</sup> ¼ � <sup>ð</sup>

where the values of the stress depend on the constitutive models adopted for the materials and

¼ ð ∂σ ∂ε<sup>r</sup>

¼ � <sup>ð</sup> y ∂σ ∂ε<sup>r</sup>

ð ∂σ ∂κ

> y ∂σ ∂κ

ð ∂σ ∂ε �

∂ð Þ ε<sup>r</sup> � y � κ ∂κ

∂ð Þ ε<sup>r</sup> � y � κ ∂ε<sup>r</sup>

dA ¼ � <sup>ð</sup>

∂ð Þ ε<sup>r</sup> � y � κ ∂ε<sup>r</sup>

∂ð Þ ε<sup>r</sup> � y � κ ∂κ

∂N ε ð Þi <sup>r</sup> ; κð Þ<sup>i</sup> � �

Numerically and Analytically Forecasting the Coal Burst Using Energy Based Approach Methods

∂M ε ð Þi <sup>r</sup> ; κð Þ<sup>i</sup> � �

∂κ

http://dx.doi.org/10.5772/intechopen.71879

ð Þ Changing strain and curvature (12)

dA (14)

dA (15)

dA (16)

dA (17)

dA ¼

y � ∂σ ∂ε

dA ¼ � <sup>ð</sup>

dA ¼ ð <sup>y</sup><sup>2</sup> � ∂σ ∂ε

ð ∂σ ∂ε

y � ∂σ ∂ε

dA (18)

dA (20)

dA (21)

dA (19)

(11)

213

(13)

∂κ

$$r(\varepsilon) = \begin{bmatrix} N \\ \mathbf{M} \end{bmatrix} \tag{2}$$

$$r\_{\epsilon} = \begin{bmatrix} N\_{\epsilon} \\ M\_{\epsilon} \end{bmatrix} \tag{3}$$

$$r(\varepsilon) = r\_{\varepsilon} \quad \text{(This is the vector for strain)}\tag{4}$$

By considering the nonlinear interactions, the presented equations can be re-written by:

$$r\left(\boldsymbol{\varepsilon}^{(i+1)}\right) = r\left(\boldsymbol{\varepsilon}^{(i)}\right) + r\_t\left(\boldsymbol{\varepsilon}^{(i)}\right) \times \Delta \boldsymbol{\varepsilon}^{(i)} = r\_t \tag{5}$$

$$r\_t(\boldsymbol{\varepsilon}^{(i)}) \times \Delta \boldsymbol{\varepsilon}^{(i)} = r\_\mathcal{R}^{(i)} \tag{6}$$

$$r\_R^{(i)} = r\_\varepsilon - r\left(\varepsilon^{(i)}\right) \tag{7}$$

$$\frac{\partial \text{N}\left(\varepsilon\_r^{(i)}, \kappa^{(i)}\right)}{\partial \varepsilon\_r} \times \Delta \varepsilon\_r^{(i)} + \frac{\partial \text{N}\left(\varepsilon\_r^{(i)}, \kappa^{(i)}\right)}{\partial \kappa} \times \Delta \kappa = \text{N}\_R^{(i)} \tag{8}$$

$$N\_{\mathcal{R}}^{(i)} = N\_{\varepsilon} - \mathcal{N}\left(\varepsilon\_{r}^{(i)}, \kappa^{(i)}\right) \tag{9}$$

$$M\_{\mathbb{R}}^{(i)} = M\_{\varepsilon} - M\left(\varepsilon\_{r}^{(i)}, \kappa^{(i)}\right) \tag{10}$$

Numerically and Analytically Forecasting the Coal Burst Using Energy Based Approach Methods http://dx.doi.org/10.5772/intechopen.71879 213

All the equations can be re-presented in matrix format:

method in different directions with respect to the different planes, where it can independently calculate shear forces between the different layers and shear strain as well as the curvature distribution along the different layers that have been extracted. The main concept to derive the following equations was extracted from [12, 13]. The cross-sectional analysis is based on the assumption of the Euler-Bernoulli beam model. The strain distribution across the section can be calculated by ε ¼ ε<sup>r</sup> � y � κ, where ε<sup>r</sup> is the strain at the reference point (which can be determined at any point), y is the distance between the selected point and location of the neutral axis of the cross-section and κ is the curvature across the section in different strata layers. A vector can be introduced by K Dð Þ which will be included in the internal action N (axial forces) and M (internal moment). External loads, which might be due to the effect of the self-weight of the strata layers as well as the possible applied forces due to the vertical or horizontal displacement in the different layers, can induce the external axial force Ne and external moment Me. The relationship between the internal and external actions

212 Finite Element Method - Simulation, Numerical Analysis and Solution Techniques

<sup>ε</sup> <sup>¼</sup> <sup>ε</sup><sup>r</sup> κ

rð Þ¼ ε

By considering the nonlinear interactions, the presented equations can be re-written by:

<sup>¼</sup> <sup>r</sup> <sup>ε</sup>ð Þ<sup>i</sup> � �

rt <sup>ε</sup>ð Þ<sup>i</sup> � �

r ð Þi

� <sup>Δ</sup>εð Þ<sup>i</sup> <sup>r</sup> þ

Nð Þ<sup>i</sup>

Mð Þ<sup>i</sup>

r εð Þ <sup>i</sup>þ<sup>1</sup> � �

∂N ε ð Þi <sup>r</sup> ; κð Þ<sup>i</sup> � �

∂ε<sup>r</sup>

re <sup>¼</sup> Ne Me

" #

N M

" #

<sup>þ</sup> rt <sup>ε</sup>ð Þ<sup>i</sup> � �

� <sup>Δ</sup>εð Þ<sup>i</sup> <sup>¼</sup> <sup>r</sup>

<sup>R</sup> <sup>¼</sup> re � <sup>r</sup> <sup>ε</sup>ð Þ<sup>i</sup> � �

∂N ε ð Þi <sup>r</sup> ; κð Þ<sup>i</sup> � �

<sup>R</sup> <sup>¼</sup> Ne � <sup>N</sup> <sup>ε</sup>ð Þ<sup>i</sup>

<sup>R</sup> <sup>¼</sup> Me � <sup>M</sup> <sup>ε</sup>ð Þ<sup>i</sup>

∂κ

<sup>r</sup> ; <sup>κ</sup>ð Þ<sup>i</sup> � �

<sup>r</sup> ; <sup>κ</sup>ð Þ<sup>i</sup> � �

ð Þi

rð Þ¼ ε re ð Þ This is the vector for strain (4)

� Δκ <sup>¼</sup> <sup>N</sup>ð Þ<sup>i</sup>

� <sup>Δ</sup>εð Þ<sup>i</sup> <sup>¼</sup> re (5)

<sup>R</sup> (6)

<sup>R</sup> (8)

" #

(1)

(2)

(3)

(7)

(9)

(10)

can be presented by:

$$r\_{l}\left(\boldsymbol{\varepsilon}^{(i)}\right) = \begin{bmatrix} \frac{\partial \mathcal{M}\left(\boldsymbol{\varepsilon}\_{r}^{(i)}, \boldsymbol{\kappa}^{(i)}\right)}{\partial \boldsymbol{\varepsilon}\_{r}} & \frac{\partial \mathcal{M}\left(\boldsymbol{\varepsilon}\_{r}^{(i)}, \boldsymbol{\kappa}^{(i)}\right)}{\partial \boldsymbol{\kappa}}\\ \frac{\partial \mathcal{M}\left(\boldsymbol{\varepsilon}\_{r}^{(i)}, \boldsymbol{\kappa}^{(i)}\right)}{\partial \boldsymbol{\varepsilon}\_{r}} & \frac{\partial \mathcal{M}\left(\boldsymbol{\varepsilon}\_{r}^{(i)}, \boldsymbol{\kappa}^{(i)}\right)}{\partial \boldsymbol{\kappa}} \end{bmatrix} \tag{11}$$

$$
\Delta \varepsilon^{(i)} = \begin{bmatrix} \Delta \varepsilon\_r^{(i)} \\ \Delta \kappa^{(i)} \end{bmatrix} \quad \text{(Changing strain and curvature)} \tag{12}
$$

$$r\_R^{(i)} = \begin{bmatrix} N\_R^{(i)} \\ M\_R^{(i)} \end{bmatrix} \tag{13}$$

The partial derivatives of N and M with respect to ε<sup>r</sup> and κ can be re-arranged in a more practical form, recalling the definitions of internal actions as:

$$\frac{\partial \mathcal{N}\left(\varepsilon\_r^{(i)}, \kappa^{(i)}\right)}{\partial \varepsilon\_r} = \int \frac{\partial \sigma}{\partial \varepsilon\_r} dA \tag{14}$$

$$\frac{\partial \mathcal{N}\left(\varepsilon\_r^{(i)}, \kappa^{(i)}\right)}{\partial \kappa} = \int \frac{\partial \sigma}{\partial \kappa} dA \tag{15}$$

$$\frac{\partial \mathcal{M}\Big(\varepsilon\_r^{(i)}, \kappa^{(i)}\Big)}{\partial \varepsilon\_r} = -\int y \frac{\partial \sigma}{\partial \varepsilon\_r} dA \tag{16}$$

$$\frac{\partial M\left(\varepsilon\_r^{(i)}, \kappa^{(i)}\right)}{\partial \kappa} = -\int y \frac{\partial \sigma}{\partial \kappa} dA \tag{17}$$

where the values of the stress depend on the constitutive models adopted for the materials and on the magnitude of the strain

$$\frac{\partial N\left(\varepsilon\_{r}^{(i)},\kappa^{(i)}\right)}{\partial \varepsilon\_{r}} = \int \frac{\partial \sigma}{\partial \varepsilon\_{r}} dA = \int \frac{\partial \sigma}{\partial \varepsilon} \times \frac{\partial \varepsilon}{\partial \varepsilon\_{r}} dA = \int \frac{\partial \sigma}{\partial \varepsilon} \times \frac{\partial (\varepsilon\_{r} - y \times \kappa)}{\partial \varepsilon\_{r}} dA = \int \frac{\partial \sigma}{\partial \varepsilon} dA \tag{18}$$

$$\frac{\partial \mathcal{N}\left(\varepsilon\_{r}^{(i)}, \kappa^{(i)}\right)}{\partial \kappa} = \int \frac{\partial \sigma}{\partial \kappa} dA = \int \frac{\partial \sigma}{\partial \varepsilon} \times \frac{\partial (\varepsilon\_{r} - y \times \kappa)}{\partial \kappa} dA = -\int y \times \frac{\partial \sigma}{\partial \varepsilon} dA \tag{19}$$

$$\frac{\partial M\left(\varepsilon\_r^{(i)}, \mathbf{x}^{(i)}\right)}{\partial \varepsilon\_r} = -\int y \times \frac{\partial \sigma}{\partial \varepsilon\_r} dA = -\int y \times \frac{\partial \sigma}{\partial \varepsilon} \times \frac{\partial (\varepsilon\_r - y \times \kappa)}{\partial \varepsilon\_r} dA = -\int y \times \frac{\partial \sigma}{\partial \varepsilon} dA \tag{20}$$

$$\frac{\partial M\left(\varepsilon\_r^{(i)}, \kappa^{(i)}\right)}{\partial \kappa} = -\int y \times \frac{\partial \sigma}{\partial \varepsilon\_r} dA = -\int y \times \frac{\partial \sigma}{\partial \varepsilon} \times \frac{\partial (\varepsilon\_r - y \times \kappa)}{\partial \kappa} dA = \int y^2 \times \frac{\partial \sigma}{\partial \varepsilon} dA \tag{21}$$

$$
\sigma = E \times \varepsilon \quad \text{for} \quad |\varepsilon| \le \varepsilon\_p \text{ (elastic strain)}\tag{22}
$$

$$
\sigma = f\_p \quad \text{for} \quad |\varepsilon| > \varepsilon\_p \quad \text{(plastic strain)}\tag{23}
$$

qe ¼ ð

u xð Þ

v xð Þ � � <sup>¼</sup> Nu1ð Þ<sup>x</sup>

2 4

0

B xð Þ¼

N0 <sup>u</sup>1ð Þx 0

2 4

0 N<sup>00</sup> <sup>v</sup>1ð Þx

L NT

0 Nv1ð Þx

<sup>e</sup> ð Þ<sup>x</sup> p xð Þdx <sup>¼</sup> <sup>L</sup>

0

Nv2ð Þx

2 XnG k¼1

Nu2ð Þx

0

Nu1ð Þ¼ <sup>x</sup> <sup>1</sup> � <sup>3</sup><sup>x</sup>

Nu2ð Þ¼ x

Nu3ð Þ¼� x

Nv1ð Þ¼ <sup>x</sup> <sup>1</sup> � <sup>3</sup>x<sup>2</sup>

Nv2ð Þ¼ <sup>x</sup> <sup>x</sup> � <sup>2</sup>x<sup>2</sup>

Nv3ð Þ¼ x

Nv4ð Þ¼� x

v xð Þ¼ Nv1ð Þx vL þ Nv2ð Þx θ<sup>L</sup> þ Nv3ð Þx vR þ Nv4ð Þx θ<sup>R</sup>

N0 <sup>u</sup>2ð Þx 0

> 3 L þ 4x

4 <sup>L</sup> � <sup>8</sup><sup>x</sup> N0 <sup>u</sup>3ð Þx 0

u xð Þ¼ Nu1ð Þx uL þ Nu2ð Þx uM þ Nu3ð Þx uR

0

N<sup>00</sup> <sup>v</sup>2ð Þx

N0

N0 <sup>u</sup>2ð Þ¼ x

<sup>u</sup>1ð Þ¼� x

wkNT

Nu3ð Þx

0

L þ

4x<sup>2</sup>

4x L þ

> x L þ 2x<sup>2</sup>

<sup>L</sup><sup>2</sup> <sup>þ</sup>

L þ x3

3x<sup>2</sup> <sup>L</sup><sup>2</sup> � <sup>2</sup>x<sup>3</sup>

> x2 L þ x3

2x<sup>2</sup>

2x<sup>2</sup>

<sup>e</sup> ð Þ xk p xð Þ<sup>k</sup> xk <sup>¼</sup> <sup>L</sup>

Numerically and Analytically Forecasting the Coal Burst Using Energy Based Approach Methods

0

0

3 5 �

Nv4ð Þx

Nv3ð Þx

<sup>2</sup> ð Þ xk <sup>þ</sup> <sup>1</sup> (35)

uL

¼ Neð Þx de

(36)

215

(44)

http://dx.doi.org/10.5772/intechopen.71879

vL θL uM uR vR

θR

<sup>L</sup><sup>2</sup> (37)

<sup>L</sup><sup>2</sup> (38)

<sup>L</sup><sup>2</sup> (39)

<sup>L</sup><sup>3</sup> (40)

<sup>L</sup><sup>2</sup> (41)

<sup>L</sup><sup>3</sup> (42)

<sup>L</sup><sup>2</sup> (43)

0

3

5 (45)

N<sup>00</sup> <sup>v</sup>4ð Þx

<sup>L</sup><sup>2</sup> (46)

<sup>L</sup><sup>2</sup> (47)

0

N<sup>00</sup> <sup>v</sup>3ð Þx

$$\frac{\partial \sigma}{\partial \varepsilon} = \frac{\partial (E \times \varepsilon)}{\partial \varepsilon} = E \quad \text{for} \quad |\varepsilon| \le \varepsilon\_p \text{ (elastic strain)}\tag{24}$$

$$\frac{\partial \sigma}{\partial \varepsilon} = \frac{\partial \begin{pmatrix} f\_p \\ \end{pmatrix}}{\partial \varepsilon} = 0 \quad \text{for} \quad |\varepsilon| > \varepsilon\_p \text{ (plastic strain)}\tag{25}$$

$$N\left(\boldsymbol{\varepsilon}\_{r}^{(i)},\boldsymbol{\kappa}^{(i)}\right) = \int \boldsymbol{\sigma} dA = \sum\_{j=1}^{n\_{\boldsymbol{\gamma}}} \sigma\left(\boldsymbol{y}\_{j},\boldsymbol{\varepsilon}\_{r}^{(i)},\boldsymbol{\kappa}^{(i)}\right) \times A\_{j} \tag{26}$$

$$M\left(\boldsymbol{\varepsilon}\_{r}^{(i)},\boldsymbol{\kappa}^{(i)}\right) = -\int \boldsymbol{y} \boldsymbol{\sigma} dA = -\sum\_{j=1}^{n\_{\parallel}} \boldsymbol{y}\_{j} \times \boldsymbol{\sigma}\left(\boldsymbol{y}\_{j},\boldsymbol{\varepsilon}\_{r}^{(i)},\boldsymbol{\kappa}^{(i)}\right) \times A\_{\dagger} \tag{27}$$

$$\frac{\partial \mathcal{N}\left(\varepsilon\_r^{(i)}, \kappa^{(i)}\right)}{\partial \varepsilon\_r} = \int \frac{\partial \sigma}{\partial \varepsilon} dA = \sum\_{j=1}^{n\_j} \frac{\partial \sigma \left(y\_j, \varepsilon\_r^{(i)}, \kappa^{(i)}\right)}{\partial \varepsilon} \times A\_j \tag{28}$$

$$\frac{\partial \mathcal{N}\left(\varepsilon\_r^{(i)}, \kappa^{(i)}\right)}{\partial \kappa} = -\int y \times \frac{\partial \sigma}{\partial \varepsilon\_r} dA = -\sum\_{j=1}^{n\_j} y\_j \times \frac{\partial \sigma\left(y\_j, \varepsilon\_r^{(i)}, \kappa^{(i)}\right)}{\partial \varepsilon} \times A\_j \tag{29}$$

$$\frac{\partial M\left(\varepsilon\_r^{(i)}, \kappa^{(i)}\right)}{\partial \varepsilon\_r} = -\int y \times \frac{\partial \sigma}{\partial \varepsilon\_r} dA = -\sum\_{j=1}^{n\_j} y\_j \times \frac{\partial \sigma\left(y\_j, \varepsilon\_r^{(i)}, \kappa^{(i)}\right)}{\partial \varepsilon} \times A\_j \tag{30}$$

$$\frac{\partial M\left(\varepsilon\_r^{(i)}, \kappa^{(i)}\right)}{\partial \kappa} = \int y^2 \times \frac{\partial \sigma}{\partial \varepsilon} dA = \sum\_{j=1}^{n\_l} y\_j^2 \times \frac{\partial \sigma \left(y\_j, \varepsilon\_r^{(i)}, \kappa^{(i)}\right)}{\partial \varepsilon} \times A\_j \tag{31}$$

$$\mathbf{r}\left(\mathbf{x},\mathbf{d}\_{\mathbf{e}}\right) = \begin{bmatrix} \mathbf{N}(\mathbf{x},\mathbf{d}\_{\mathbf{e}})\\ \mathbf{M}(\mathbf{x},\mathbf{d}\_{\mathbf{e}}) \end{bmatrix} = \begin{bmatrix} \int\_{\mathcal{A}} \sigma(\mathbf{x},\mathbf{y},\mathbf{d}\_{\mathbf{e}}) \mathbf{dA} \\\\ \cdot \int\_{\mathcal{A}} \mathbf{y}\sigma(\mathbf{x},\mathbf{y},\mathbf{d}\_{\mathbf{e}}) \mathbf{dA} \end{bmatrix} = \begin{bmatrix} \sum\_{j=1}^{n\_{\mathbf{j}}} \sigma(\mathbf{x},\mathbf{y},\mathbf{d}\_{\mathbf{e}}) \mathbf{A}\_{\mathbf{j}} \\\\ -\sum\_{j=1}^{n\_{\mathbf{j}}} \mathbf{y}\_{\mathbf{j}}\sigma(\mathbf{x},\mathbf{y},\mathbf{d}\_{\mathbf{e}}) \mathbf{A}\_{\mathbf{j}} \end{bmatrix} \tag{32}$$

$$I = \int\_{a}^{b} f(\mathbf{x})d\mathbf{x} = \left(\frac{b-a}{2}\right) \times \int\_{-1}^{1} f\left(\frac{a+b}{2} + \frac{b-a}{2} \times \overline{\mathbf{x}}\right) = \left(\frac{b-a}{2}\right) \times \sum\_{k=1}^{n\_G} w\_k \times f\left(\frac{a+b}{2} + \frac{b-a}{2} \times \overline{\mathbf{x}}\_k\right) \tag{33}$$

$$\mathbf{k}\_{\mathbf{e}}(\mathbf{x}, \mathbf{d}\_{\mathbf{e}}) = \int\_{L} \mathbf{B}^{\mathrm{T}}(\mathbf{x}) \mathbf{r}(\mathbf{x}, \mathbf{d}\_{\mathbf{e}}) \, d\mathbf{x} = \frac{\mathrm{L}}{2} \sum\_{k=1}^{n\_{\mathrm{G}}} w\_{k} \mathbf{B}^{\mathrm{T}}(\mathbf{x}\_{k}) \mathbf{r}(\mathbf{x}\_{k}, \mathbf{d}\_{\mathbf{e}}) \tag{34}$$

Numerically and Analytically Forecasting the Coal Burst Using Energy Based Approach Methods http://dx.doi.org/10.5772/intechopen.71879 215

$$\mathbf{q}\_{\mathbf{e}} = \int\_{L} \mathbf{N}\_{\mathbf{e}}^{\mathrm{T}}(\mathbf{x}) \mathbf{p}(\mathbf{x}) d\mathbf{x} = \frac{\mathrm{L}}{2} \sum\_{\mathbf{k}=1}^{n\_{\mathrm{G}}} w\_{\mathbf{k}} \mathbf{N}\_{\mathbf{e}}^{\mathrm{T}}(\mathbf{x}\_{\mathrm{k}}) p(\mathbf{x}\_{\mathrm{k}}) \quad \mathbf{x}\_{\mathrm{k}} = \frac{\mathrm{L}}{2} (\overline{\mathbf{x}\_{\mathrm{k}}} + 1) \tag{35}$$

$$
\begin{bmatrix} u(\mathbf{x}) \\ \mathbf{p}(\mathbf{x}) \end{bmatrix} = \begin{bmatrix} N\_{u1}(\mathbf{x}) & 0 & 0 & N\_{u2}(\mathbf{x}) & N\_{u3}(\mathbf{x}) & 0 & 0 \\ 0 & N\_{v1}(\mathbf{x}) & N\_{v2}(\mathbf{x}) & 0 & 0 & N\_{v3}(\mathbf{x}) \end{bmatrix} \times \begin{bmatrix} u\_L \\ v\_L \\ \theta\_L \\ u\_M \\ u\_R \\ v\_R \\ \theta\_R \end{bmatrix} = \mathbf{N}\_e(\mathbf{x}) \mathbf{d}\_e \tag{20}
$$

σ ¼ E � ε for j j ε ≤ ε<sup>p</sup> ð Þ elastic strain (22)

<sup>∂</sup><sup>ε</sup> <sup>¼</sup> <sup>E</sup> for j j <sup>ε</sup> <sup>≤</sup> <sup>ε</sup><sup>p</sup> ð Þ elastic strain (24)

; εð Þ<sup>i</sup> <sup>r</sup> ; <sup>κ</sup>ð Þ<sup>i</sup> � �

� � (23)

� � (25)

� Aj (26)

� Aj (27)

� Aj (28)

� Aj (29)

� Aj (30)

� Aj (31)

(32)

σ ¼ f <sup>p</sup> for j j ε > ε<sup>p</sup> plastic strain

<sup>σ</sup>dA <sup>¼</sup> <sup>X</sup>nj

<sup>y</sup>σdA ¼ �Xnj

<sup>∂</sup><sup>ε</sup> <sup>¼</sup> 0 for j j <sup>ε</sup> <sup>&</sup>gt; <sup>ε</sup><sup>p</sup> plastic strain

j¼1

j¼1

j¼1

j¼1 yj �

j¼1 yj �

dA ¼ �Xnj

dA ¼ �Xnj

dA <sup>¼</sup> <sup>X</sup>nj

σ x; y; de � �dA

yσ x; y; de � �dA

j¼1 y2 <sup>j</sup> �

dA <sup>¼</sup> <sup>X</sup>nj

σ yj ; εð Þ<sup>i</sup> <sup>r</sup> ; <sup>κ</sup>ð Þ<sup>i</sup> � �

yj � σ yj

∂σ yj ; ε ð Þi <sup>r</sup> ; κð Þ<sup>i</sup> � �

∂ε

∂σ yj ; ε ð Þi <sup>r</sup> ; κð Þ<sup>i</sup> � �

∂σ yj ; ε ð Þi <sup>r</sup> ; κð Þ<sup>i</sup> � �

∂σ yj ; ε ð Þi <sup>r</sup> ; κð Þ<sup>i</sup> � �

<sup>¼</sup> <sup>b</sup> � <sup>a</sup> 2 � �

> XnG k¼1

2

∂ε

∂ε

∂ε

Xnj j¼1

� Xnj j¼1 yj

�XnG k¼1

σ x; y; de � �Aj

> σ x; y; de � �Aj

wk � <sup>f</sup> <sup>a</sup> <sup>þ</sup> <sup>b</sup>

wkB<sup>T</sup>ð Þ xk <sup>r</sup> xk ð Þ ; de (34)

2 þ

b � a 2

� �

� xk

(33)

∂σ

∂σ ∂ε ¼

M εð Þ<sup>i</sup> <sup>r</sup> ; <sup>κ</sup>ð Þ<sup>i</sup> � �

∂N ε ð Þi <sup>r</sup> ; κð Þ<sup>i</sup> � �

∂M ε ð Þi <sup>r</sup> ; κð Þ<sup>i</sup> � �

∂ε<sup>r</sup>

∂M ε ð Þi <sup>r</sup> ; κð Þ<sup>i</sup> � �

r xð Þ¼ ; de

f xð Þdx <sup>¼</sup> <sup>b</sup> � <sup>a</sup>

2 � � � ð 1

keð Þ¼ x; de

�1

I ¼ ð b

a

∂N ε ð Þi <sup>r</sup> ; κð Þ<sup>i</sup> � �

∂ε<sup>r</sup>

∂κ ¼ �

∂κ ¼

N xð Þ ; de M xð Þ ; de � �

N εð Þ<sup>i</sup> <sup>r</sup> ; <sup>κ</sup>ð Þ<sup>i</sup> � �

<sup>∂</sup><sup>ε</sup> <sup>¼</sup> <sup>∂</sup>ð Þ <sup>E</sup> � <sup>ε</sup>

214 Finite Element Method - Simulation, Numerical Analysis and Solution Techniques

∂ f <sup>p</sup> � �

> ¼ ð

¼ ð ∂σ ∂ε

ð y � ∂σ ∂ε<sup>r</sup>

¼ � ð y � ∂σ ∂ε<sup>r</sup>

> ð <sup>y</sup><sup>2</sup> � ∂σ ∂ε

¼

<sup>f</sup> <sup>a</sup> <sup>þ</sup> <sup>b</sup> 2 þ

ð

L

ð

A

A

� �

b � a 2 � x

BTð Þ<sup>x</sup> r xð Þ ; de dx <sup>¼</sup> <sup>L</sup>

 ð

¼ � ð (36)

$$N\_{u1}(\mathbf{x}) = 1 - \frac{3\mathbf{x}}{L} + \frac{2\mathbf{x}^2}{L^2} \tag{37}$$

$$N\_{u2}(\mathbf{x}) = \frac{4\mathbf{x}}{L} + \frac{4\mathbf{x}^2}{L^2} \tag{38}$$

$$N\_{u3}(\mathbf{x}) = -\frac{\mathbf{x}}{L} + \frac{2\mathbf{x}^2}{L^2} \tag{39}$$

$$N\_{v1}(\mathbf{x}) = 1 - \frac{3\mathbf{x}^2}{L^2} + \frac{2\mathbf{x}^2}{L^3} \tag{40}$$

$$N\_{v2}(\mathbf{x}) = \mathbf{x} - \frac{2\mathbf{x}^2}{L} + \frac{\mathbf{x}^3}{L^2} \tag{41}$$

$$N\_{v3}(\mathbf{x}) = \frac{3\mathbf{x}^2}{L^2} - \frac{2\mathbf{x}^3}{L^3} \tag{42}$$

$$N\_{\mathbb{P}^4}(\mathbf{x}) = -\frac{\mathbf{x}^2}{L} + \frac{\mathbf{x}^3}{L^2} \tag{43}$$

$$\begin{aligned} u(\mathbf{x}) &= \mathbf{N}\_{u1}(\mathbf{x})u\_L + \mathbf{N}\_{u2}(\mathbf{x})u\_M + \mathbf{N}\_{u3}(\mathbf{x})u\_R \\ v(\mathbf{x}) &= \mathbf{N}\_{v1}(\mathbf{x})v\_L + \mathbf{N}\_{v2}(\mathbf{x})\theta\_L + \mathbf{N}\_{v3}(\mathbf{x})v\_R + \mathbf{N}\_{v4}(\mathbf{x})\theta\_R \end{aligned} \tag{44}$$

$$B(\mathbf{x}) = \begin{bmatrix} N\_{\imath 1}'(\mathbf{x}) & \mathbf{0} & \mathbf{0} & N\_{\imath 2}'(\mathbf{x}) & N\_{\imath 3}'(\mathbf{x}) & \mathbf{0} & \mathbf{0} \\\\ \mathbf{0} & N\_{\imath 1}''(\mathbf{x}) & N\_{\imath 2}''(\mathbf{x}) & \mathbf{0} & \mathbf{0} & N\_{\imath 3}''(\mathbf{x}) & N\_{\imath 4}''(\mathbf{x}) \end{bmatrix} \tag{45}$$

$$N\_{\rm ul}'(\mathbf{x}) = -\frac{\mathfrak{Z}}{L} + \frac{4\mathbf{x}}{L^2} \tag{46}$$

$$N\_{u2}'(\mathbf{x}) = \frac{4}{L} - \frac{8\mathbf{x}}{L^2} \tag{47}$$

$$N\_{\mathfrak{u}3}'(\mathbf{x}) = -\frac{1}{L} + \frac{4\mathbf{x}}{L^2} \tag{48}$$

<sup>A</sup> <sup>¼</sup> <sup>1</sup> 2

stage.

Figure 4. A typical stress-strain curve.

directly extracted from the simulated model.

the load and unload line) is the elastic energy index

4. Energy calculation based on the numerical approach

� <sup>∭</sup> <sup>σ</sup>xx � <sup>ε</sup>xx <sup>þ</sup> <sup>σ</sup>yy � <sup>ε</sup>yy <sup>þ</sup> <sup>σ</sup>zz � <sup>ε</sup>zz <sup>þ</sup> <sup>σ</sup>xy � <sup>ε</sup>xy <sup>þ</sup> <sup>σ</sup>xz � <sup>ε</sup>xz <sup>þ</sup> <sup>σ</sup>yz � <sup>ε</sup>yz dxdydz (57)

Numerically and Analytically Forecasting the Coal Burst Using Energy Based Approach Methods

where σxx � εxx, ……, σyz � εyz can be calculated, according to the principal of the virtual work and virtual deformation δA ¼ δR<sup>1</sup> þ δR2, when the induced stresses and strains cannot be

According to Xie et al. [14], the coal burst proneness of a coal can be determined by the coal burst proneness assessments. Special attentions were devoted by the number of researchers to develop coal burst proneness indexes, which are broadly utilised, such as elastic energy, impact energy, dynamic failure time as well as elastic deformation and stiffness ratio indexes. The elastic energy index WET is defined as the ratio of the elastic strain energy and the strain energy dissipation at point E (75–85% of the peak strength). As shown in Figure 4, the ratio of the area SEAC (between the unloaded line EA and the strain axis) and the area SEOA (between

WET <sup>¼</sup> SEAC

The impact energy index KE is defined as the ratio of the pre-peak area and the post-peak area, KE namely, the ratio of energy in the pre-peak stage and the energy released in the post-peak

SEOA (58)

http://dx.doi.org/10.5772/intechopen.71879

217

$$N\_{v1}''(x) = \frac{12x}{L^3} - \frac{6}{L^2} \tag{49}$$

$$N\_{v2}''(x) = \frac{6x}{L^2} - \frac{4}{L} \tag{50}$$

$$N\_{v3}''(\mathbf{x}) = \frac{6}{L^2} - \frac{12\mathbf{x}}{L^3} \tag{51}$$

$$N\_{v4}''(\mathbf{x}) = \frac{6\mathbf{x}}{L^2} - \frac{2}{L} \tag{52}$$

$$\mathbf{k}\_{\mathbf{e}}\left(\mathbf{x},\mathbf{d}\_{\mathbf{e}}^{(i)}\right) = \frac{\mathbf{L}}{2} \sum\_{\mathbf{k}=1}^{\mathrm{nc}} \mathbf{w}\_{\mathbf{k}} \mathbf{B}^{\mathrm{T}}(\mathbf{x}\_{\mathbf{k}}) \mathbf{r}^{(i)}\left(\mathbf{x}\_{\mathbf{k}},\mathbf{d}\_{\mathbf{e}}^{(i)}\right) = \frac{\mathbf{L}}{2} \sum\_{\mathbf{k}=1}^{\mathrm{nc}} \mathbf{w}\_{\mathbf{k}} \times \begin{bmatrix} N\_{\mathrm{u}1}'(\mathbf{x}\_{\mathbf{k}}) & \mathbf{0} \\ \mathbf{0} & N\_{\mathrm{v1}}'(\mathbf{x}\_{\mathbf{k}}) \\ \mathbf{0} & N\_{\mathrm{v2}}'(\mathbf{x}\_{\mathbf{k}}) \\ N\_{\mathrm{u2}}'(\mathbf{x}\_{\mathbf{k}}) & \mathbf{0} \\ N\_{\mathrm{u3}}'(\mathbf{x}\_{\mathbf{k}}) & \mathbf{0} \\ \mathbf{0} & N\_{\mathrm{v3}}'(\mathbf{x}\_{\mathbf{k}}) \\ \mathbf{0} & N\_{\mathrm{v4}}'(\mathbf{x}\_{\mathbf{k}}) \end{bmatrix} \times \begin{bmatrix} \mathcal{N}\left(\mathbf{x}\_{\mathbf{k}},\mathbf{d}\_{\mathbf{e}}^{(i)}\right) \\ \mathbf{0} \\ \mathbf{M}\left(\mathbf{x}\_{\mathbf{k}},\mathbf{d}\_{\mathbf{e}}^{(i)}\right) \end{bmatrix} \tag{53}$$

$$\mathbf{q}\_{\mathbf{e}} = \frac{\mathbf{L}}{2} \sum\_{\mathbf{k}=1}^{n\_{\mathbf{C}}} w\_{\mathbf{k}} \times \begin{bmatrix} N\_{\mathrm{u1}}(\mathbf{x}\_{\mathbf{k}}) & \mathbf{0} \\ \mathbf{0} & N\_{\mathrm{v1}}(\mathbf{x}\_{\mathbf{k}}) \\ \mathbf{0} & N\_{\mathrm{v2}}(\mathbf{x}\_{\mathbf{k}}) \\ N\_{\mathrm{u2}}(\mathbf{x}\_{\mathbf{k}}) & \mathbf{0} \\ N\_{\mathrm{u3}}(\mathbf{x}\_{\mathbf{k}}) & \mathbf{0} \\ \mathbf{0} & N\_{\mathrm{v3}}(\mathbf{x}\_{\mathbf{k}}) \\ \mathbf{0} & N\_{\mathrm{v4}}(\mathbf{x}\_{\mathbf{k}}) \end{bmatrix} \times \begin{bmatrix} n(\mathbf{x}\_{\mathbf{k}}) \\ w(\mathbf{x}\_{\mathbf{k}}) \\ w(\mathbf{x}\_{\mathbf{k}}) \\ \mathbf{0} \\ N\_{\mathrm{v4}}(\mathbf{x}\_{\mathbf{k}}) \end{bmatrix} \tag{54}$$

$$\mathbf{N}\left(\mathbf{x}\_{\mathbf{k}}, \mathbf{d}\_{\mathbf{e}}^{(\mathrm{i})}\right) = \sum\_{j=1}^{n\_{\mathrm{j}}} \sigma\left(\mathbf{x}\_{\mathbf{k}}, \mathbf{y}\_{\mathbf{i}}, \mathbf{d}\_{\mathbf{e}}^{(\mathrm{i})}\right) \mathbf{A}\_{\mathbf{j}} \tag{55}$$

$$\mathbf{M}\Big(\mathbf{x}\_{\mathbb{k}}, \mathbf{d}\_{\mathbf{e}}^{(\mathrm{i})}\Big) = -\sum\_{j=1}^{n\_{\mathrm{j}}} \mathbf{y}\_{\textnormal{j}} \sigma\Big(\mathbf{x}\_{\mathbb{k}}, \mathbf{y}\_{\textnormal{i}}, \mathbf{d}\_{\mathrm{e}}^{(\mathrm{i})}\Big) \mathbf{A}\_{\textnormal{j}} \tag{56}$$

Thus, by calculating stress and strain at the different points in the different layers of the overburden, the internal axial forces as well as internal moments can be calculated. It was assumed that the strain energy can be calculated by:

$$A = \frac{1}{2} \times \iiint (\sigma\_{xx} \times \varepsilon\_{xx} + \sigma\_{yy} \times \varepsilon\_{yy} + \sigma\_{zz} \times \varepsilon\_{zz} + \sigma\_{xy} \times \varepsilon\_{xy} + \sigma\_{xz} \times \varepsilon\_{xz} + \sigma\_{yz} \times \varepsilon\_{yz}) d\mathbf{x} dydz \tag{57}$$

where σxx � εxx, ……, σyz � εyz can be calculated, according to the principal of the virtual work and virtual deformation δA ¼ δR<sup>1</sup> þ δR2, when the induced stresses and strains cannot be directly extracted from the simulated model.

## 4. Energy calculation based on the numerical approach

N0

216 Finite Element Method - Simulation, Numerical Analysis and Solution Techniques

N<sup>00</sup> <sup>v</sup>1ð Þ¼ x

> N<sup>00</sup> <sup>v</sup>2ð Þ¼ x

N<sup>00</sup> <sup>v</sup>3ð Þ¼ x

> N<sup>00</sup> <sup>v</sup>4ð Þ¼ x

> > ¼ L 2 XnG k¼1

ke x;dð Þ<sup>i</sup> e � � ¼ L 2 XnG k¼1

wkBTð Þ xk <sup>r</sup>

qe <sup>¼</sup> <sup>L</sup> 2 XnG k¼1

assumed that the strain energy can be calculated by:

ð Þ<sup>i</sup> xk;dð Þ<sup>i</sup> e � �

wk �

N xk;dð Þ<sup>i</sup> e � �

M xk;dð Þ<sup>i</sup> e � �

<sup>u</sup>3ð Þ¼� x

1 L þ 4x

12x <sup>L</sup><sup>3</sup> � <sup>6</sup>

> 6x <sup>L</sup><sup>2</sup> � <sup>4</sup>

6 <sup>L</sup><sup>2</sup> � <sup>12</sup><sup>x</sup>

> 6x <sup>L</sup><sup>2</sup> � <sup>2</sup>

> > wk �

Nu1ð Þ xk 0

Nu2ð Þ xk 0 Nu3ð Þ xk 0

¼Xnj j¼1

¼ �Xnj j¼1 yj σ xk;yi

Thus, by calculating stress and strain at the different points in the different layers of the overburden, the internal axial forces as well as internal moments can be calculated. It was

0 Nv1ð Þ xk 0 Nv2ð Þ xk

0 Nv3ð Þ xk 0 Nv4ð Þ xk

σ xk;yi

;dð Þ<sup>i</sup> e � �

> ;dð Þ<sup>i</sup> e � �

N0

N0

N0

�

n xð Þ<sup>k</sup> w xð Þ<sup>k</sup> � �

<sup>u</sup>1ð Þ xk 0 0 N<sup>00</sup>

0 N<sup>00</sup>

<sup>u</sup>2ð Þ xk 0

<sup>u</sup>3ð Þ xk 0 0 N<sup>00</sup>

0 N<sup>00</sup>

<sup>L</sup><sup>2</sup> (48)

<sup>L</sup><sup>2</sup> (49)

<sup>L</sup> (50)

<sup>L</sup><sup>3</sup> (51)

<sup>L</sup> (52)

�

Aj (55)

Aj (56)

2 6 4 N xk;dð Þ<sup>i</sup> e � � 3 7 5

(53)

(54)

M xk;dð Þ<sup>i</sup> e � �

<sup>v</sup>1ð Þ xk

<sup>v</sup>2ð Þ xk

<sup>v</sup>3ð Þ xk

<sup>v</sup>4ð Þ xk

According to Xie et al. [14], the coal burst proneness of a coal can be determined by the coal burst proneness assessments. Special attentions were devoted by the number of researchers to develop coal burst proneness indexes, which are broadly utilised, such as elastic energy, impact energy, dynamic failure time as well as elastic deformation and stiffness ratio indexes. The elastic energy index WET is defined as the ratio of the elastic strain energy and the strain energy dissipation at point E (75–85% of the peak strength). As shown in Figure 4, the ratio of the area SEAC (between the unloaded line EA and the strain axis) and the area SEOA (between the load and unload line) is the elastic energy index

$$W\_{ET} = \left(\frac{S\_{EAC}}{S\_{EOA}}\right) \tag{58}$$

The impact energy index KE is defined as the ratio of the pre-peak area and the post-peak area, KE namely, the ratio of energy in the pre-peak stage and the energy released in the post-peak stage.

Figure 4. A typical stress-strain curve.

ued ¼

output results.

εð3

<sup>σ</sup>pd<sup>ε</sup> � � <sup>¼</sup> <sup>X</sup><sup>ε</sup><sup>3</sup>

ε2

εð Þ<sup>l</sup> � εð Þ <sup>l</sup>�<sup>1</sup> � � � σp lð Þ þ σp lð Þ �<sup>1</sup> 4

Tables 1 and 2 presents a comparison between the different elastic and post-failure energy components using UDEC and 3DEC output results as well as semi-close form solutions. As it can be found, there is a good agreement between the suggested semi-analytical methods as well as the calculated key energy components which were extracted from the UDEC and 3DEC

� � � � Residual elastic energy (62)

http://dx.doi.org/10.5772/intechopen.71879

219

Numerically and Analytically Forecasting the Coal Burst Using Energy Based Approach Methods

ε2

Elastic strain energy (kJ/m<sup>3</sup>

)

Dissipated elastic strain energy (kJ/m<sup>3</sup>

Figure 5. Analytically calculation of dissipated energy and released energy (Xie et al. [14]).

)

Table 1. A comparison between the different energy components (UDEC and the analytical solution).

Ratio (w/h) UDEC Analytic UDEC Analytic UDEC Analytic UDEC Analytic 1.56 1.63 0.78 0.77 3.61 3.73 12.47 10.59 1.5 1.92 1.35 0.8 0.78 7.89 7.44 12.48 11.03 2.0621 2.004 0.991 0.92 10.22 9.83 18.31 17.09 2.5 4.70 4.82 1.16 1.11 14.47 13.43 21.17 23.14 11.13 10.58 2.51 2.41 35.825 32.87 11.73 10.6 14.72 13.27 4.07 4.60 60.26 55.00 56.34 70.02 16.63 16.24 5.334 5.37 75.83 73.67 91.19 84.04

The amount of the energy in the pre-peak stage (kJ/m<sup>3</sup>

)

The energy released in the post-peak stage (kJ/m<sup>3</sup>

)

### 5. Energy calculation based on the analytical approach

According to Xie et al. [14], dissipated and released energy can play a significant role which may result in coal deformation and failure. Based on the failure mechanism, the fracture procedure of a coal mass might be started from a partial fracture which would be followed by local damage. This procedure will be finally resulting in collapsing the mining structures. The failure process is thermodynamically permanent, which includes released and dissipated energy. The dissipated energy can cause damage as well as a permanent deformation of the coal mass, which is followed by weakening of strength. A sudden release of the strain energy may lead to a catastrophic failure, which clearly indicates a certain condition where the coal mass collapses. The released and dissipated energy from the coal mass, individually, plays an essential role in the relevant sudden failure, which would be one of the major requirements to investigate the procedure of the deformation and failure of a coal mass. Figure 5 is a typical compression curve of coal under a constant displacement.

Figure 5 explicitly demonstrates the calculation of the dissipated, released and residual energies. With respect to the constant development of the inner micro-defects, energy dissipation is an indispensable characteristic of the deformation and failure of the coal mass. The evolution declines the strength of the coal, which may result in total failure. In this content, the dissipated energy is directly concerned with the damage as well as mitigating strength of the coal.

$$\mu\_{d1} = \int\_0^{\varepsilon\_1} (\sigma\_d d\varepsilon) = \sum\_{0}^{\varepsilon\_1} \left[ \left( \varepsilon\_{1(i)} - \varepsilon\_{1(i-1)} \right) \times \left( \frac{\sigma\_{d(i)} + \sigma\_{d(i-1)}}{4} \right) \right] \quad \text{Dissipated energy before peak (59)}$$

$$\mu\_{d2} = \int\_{\varepsilon\_{\beta}}^{\varepsilon\_{4}} (\sigma\_{\varepsilon} d\varepsilon) = \sum\_{\varepsilon\_{\beta}}^{\varepsilon\_{4}} \left[ \left( \varepsilon\_{(j)} - \varepsilon\_{(j-1)} \right) \times \left( \frac{\sigma\_{\varepsilon(j)} + \sigma\_{\varepsilon(j-1)}}{4} \right) \right] \quad \text{Dissipated energy after peak} \tag{60}$$

$$\mu\_r = \int\_{\varepsilon\_1}^{\varepsilon\_2} (\sigma\_m d\varepsilon) = \sum\_{\varepsilon\_1}^{\varepsilon\_2} \left[ \left( \varepsilon\_{(k)} - \varepsilon\_{(k-1)} \right) \times \left( \frac{\sigma\_{m(k)} + \sigma\_{m(k-1)}}{4} \right) \right] \quad \text{Released elastic energy} \tag{61}$$

Numerically and Analytically Forecasting the Coal Burst Using Energy Based Approach Methods http://dx.doi.org/10.5772/intechopen.71879 219

$$\mu\_{\rm{cl}} = \bigcap\_{\varepsilon\_2}^{\varepsilon\_3} \left( \sigma\_p d\varepsilon \right) = \sum\_{\varepsilon\_2}^{\varepsilon\_3} \left[ \left( \varepsilon\_{(l)} - \varepsilon\_{(l-1)} \right) \times \left( \frac{\sigma\_{p(l)} + \sigma\_{p(l-1)}}{4} \right) \right] \quad \text{Residual elastic energy} \tag{62}$$

Tables 1 and 2 presents a comparison between the different elastic and post-failure energy components using UDEC and 3DEC output results as well as semi-close form solutions. As it can be found, there is a good agreement between the suggested semi-analytical methods as well as the calculated key energy components which were extracted from the UDEC and 3DEC output results.

*<sup>K</sup>* <sup>=</sup> *<sup>S</sup>*

*OEFD* ( )

*S FDHG*

5. Energy calculation based on the analytical approach

218 Finite Element Method - Simulation, Numerical Analysis and Solution Techniques

compression curve of coal under a constant displacement.

ε<sup>1</sup>ð Þ<sup>i</sup> � ε<sup>1</sup>ð Þ <sup>i</sup>�<sup>1</sup> � � �

> εð Þ<sup>j</sup> � εð Þ <sup>j</sup>�<sup>1</sup> � � �

> > εð Þ<sup>k</sup> � εð Þ <sup>k</sup>�<sup>1</sup> � � �

*E*

ud<sup>1</sup> ¼

εð1

ð Þ¼ <sup>σ</sup>ad<sup>ε</sup> <sup>X</sup><sup>ε</sup><sup>1</sup>

ð Þ¼ <sup>σ</sup>cd<sup>ε</sup> <sup>X</sup><sup>ε</sup><sup>4</sup>

ð Þ¼ <sup>σ</sup>md<sup>ε</sup> <sup>X</sup><sup>ε</sup><sup>2</sup>

0

ε3

ε1

0

εð4

ε3

ε1

ud<sup>2</sup> ¼

ur ¼ εð2

*SOEFD*

**The amount of the energy in the Pre-peak Stage** 

**The energy released in** 

**the post-peak stage** 

**The impact energy index** 

According to Xie et al. [14], dissipated and released energy can play a significant role which may result in coal deformation and failure. Based on the failure mechanism, the fracture procedure of a coal mass might be started from a partial fracture which would be followed by local damage. This procedure will be finally resulting in collapsing the mining structures. The failure process is thermodynamically permanent, which includes released and dissipated energy. The dissipated energy can cause damage as well as a permanent deformation of the coal mass, which is followed by weakening of strength. A sudden release of the strain energy may lead to a catastrophic failure, which clearly indicates a certain condition where the coal mass collapses. The released and dissipated energy from the coal mass, individually, plays an essential role in the relevant sudden failure, which would be one of the major requirements to investigate the procedure of the deformation and failure of a coal mass. Figure 5 is a typical

Figure 5 explicitly demonstrates the calculation of the dissipated, released and residual energies. With respect to the constant development of the inner micro-defects, energy dissipation is an indispensable characteristic of the deformation and failure of the coal mass. The evolution declines the strength of the coal, which may result in total failure. In this content, the dissipated energy is directly concerned with the damage as well as mitigating strength of the coal.

> σa ið Þ þ σa ið Þ �<sup>1</sup> 4

σc jð Þ þ σc jð Þ �<sup>1</sup> 4

> σm kð Þ þ σm kð Þ �<sup>1</sup> 4

Dissipated energy before peak (59)

Dissipated energy after peak (60)

Released elastic energy (61)

� � � �

� � � �

� � � �

*S FDHG*

Figure 5. Analytically calculation of dissipated energy and released energy (Xie et al. [14]).


Table 1. A comparison between the different energy components (UDEC and the analytical solution).


coal pillar model based on the symmetrical boundary conditions with respect to the different width by height (w/h) ratios of 1–10 were developed. It was observed that when the w/h ratios are less than 4, the failure mode of pillar can be either a double or a single diagonal shear failure in which the trajectory cracking starts from the edges and progresses towards the centre of the pillar. While the w/h ratios are greater than 4, the obtained possible failure mode would be a combination of the shear and compression failure modes. Thus, the trajectory of the cracking due to the pure compression failure would be propagated from the centre to the

Numerically and Analytically Forecasting the Coal Burst Using Energy Based Approach Methods

http://dx.doi.org/10.5772/intechopen.71879

221

Analytical method is an important part of coal burst evaluation and forecasting. Analytical forecasting methods, either alone or combined with numerical simulations, can be used to estimate both in situ stress and induced stress, which leads to the prediction of failure-prone areas and calculation of critical values of the energies. The behaviour of a single pillar under different applied loads ranging from the quasi-static towards the dynamic loading conditions was simulated using commercial finite element package ABAQUS/Explicit. A strain-based failure condition was evaluated to determine the failure modes in a single pillar by respecting to the different w/h ratios. The observed numerical failure modes can be classified by shear and compression failures as well as a combination of both shear and compression were comprehensively illustrated. The released energy or residual energy is either transferred into kinetic energy or dissipated energy in non-elastic behaviour such as joint shear and fracturing. The unstable release of potential energy of the coal around the excavations, mainly in the form

Faham Tahmasebinia\*, Chengguo Zhang, Ismet Canbulat, Onur Vardar and Serkan Saydam

[1] Linkov AM. Rockbursts and the instability of rock masses. International Journal of Rock Mechanics and Mining Sciences & Geomechanics Abstracts. 1996;33(7):727-732

[2] Gong QM, Yin LJ, SY W, Zhao J, Ting Y. Rock burst and slabbing failure and its influence on TBM excavation at headrace tunnels in Jinping II hydropower station. Engineering

School of Mining Engineering, University of New South Wales, Sydney, Australia

\*Address all correspondence to: faham.tahmasebinia@sydney.edu.au

corners where a pillar gradually starts towards fully squashed (see Figure 6).

7. Remarkable conclusions

of kinetic energy, causes coal burst.

Geology. 2012;124(1):98-108

Author details

References

Table 2. A comparison between the different energy components (3DEC and the analytical solution).
