2. General formulation for infinitesimal deformation

### 2.1 Governing equations and constitutive laws

Assuming no body force, governing equations for quasi-static deformation in isotropic and homogeneous materials are [38]:

$$
\sigma\_{\vec{\eta},j} = \mathbf{0} \tag{1}
$$

Under small deformation regime, strain displacement relation is given by

$$\varepsilon\_{ij} = \frac{1}{2} \left( u\_{i,j} + u\_{j,i} \right) \tag{2}$$

where i, j <sup>¼</sup> <sup>1</sup>, <sup>2</sup>, 3 and ðÞ<sup>j</sup> means differentiation with respect to jth spatial coordinate. σij are components of Cauchy stress tensor, ui are the components of displacement, εij are components of strains. For MGs, strain is decomposed as [21],

$$
\boldsymbol{\varepsilon}\_{\vec{\boldsymbol{\eta}}} = \boldsymbol{\varepsilon}\_{\vec{\boldsymbol{\eta}}}^{\boldsymbol{\epsilon}} + \boldsymbol{\varepsilon}\_{\vec{\boldsymbol{\eta}}}^{\boldsymbol{p}} + \frac{\mathbf{1}}{2} (\boldsymbol{\xi} - \boldsymbol{\xi}\_{0}) \boldsymbol{\delta}\_{\vec{\boldsymbol{\eta}}} \tag{3}
$$

where, ε<sup>e</sup> ij is the elastic strain, ε p ij is the deviatoric plastic strain and, ξ � ξ<sup>0</sup> ð Þ is the inelastic dilatation strain (strain due to excess free volume), where ξ is local concentration of free volume and ξ<sup>0</sup> is the free volume concentration at the reference state with no strain. Dividing the stress tensor into mean stress and deviatoric stress tensor,

$$
\sigma\_m = \frac{1}{3} (\sigma\_{11} + \sigma\_{22} + \sigma\_{33}) \tag{4}
$$

$$\mathcal{S}\_{\vec{\imath}\,} = \sigma\_{\vec{\imath}\,\rangle} - \sigma\_m \delta\_{\vec{\imath}\,\,} \tag{5}$$

The Mises effective shear stress is

$$
\pi\_{\epsilon} = \sqrt{\frac{1}{2} \mathcal{S}\_{\vec{\imath}\vec{\jmath}} \mathcal{S}\_{\vec{\imath}\vec{\jmath}}} \tag{6}
$$

Elastic strain and stress are relates to each other by Hooke's law,

$$
\sigma\_{\vec{v}\vec{j}} = 2\mu \left( \varepsilon\_{\vec{ij}}^{\epsilon} + \frac{\nu}{1 - 2\nu} \varepsilon\_{kk}^{\epsilon} \delta\_{\vec{v}\vec{j}} \right) \tag{7}
$$

where μ is shear modulus and ν is Poisson's ratio.

Following Mises theory [39] flow of deviatoric plastic strain ε p ij is assumed to be in the same direction as deviatoric stress tensor Sij, with its rate depending upon concentration of free volume ξ<sup>f</sup> , effective shear stress τ<sup>e</sup> and mean stress σ<sup>m</sup> as:

$$\frac{\partial \epsilon^{p}\_{\vec{\boldsymbol{\eta}}}}{\partial t} = f\left(\xi\_{\boldsymbol{f}}, \tau\_{\boldsymbol{e}}, \sigma\_{m}\right) \frac{S\_{\vec{\boldsymbol{\eta}}}}{2\tau\_{\boldsymbol{e}}} \tag{8}$$

The inelastic dilatation strain is associated with change of concentration of free volume. This change of concentration of free volume depends upon local concentration free volume itself ξ<sup>f</sup> , effective shear stress τ<sup>e</sup> and mean stress σ<sup>m</sup> [21].

$$\frac{\partial \xi\_f}{\partial t} = D\xi\_{\text{, }ii} + \mathbf{g}\left(\xi\_f, \tau\_{\text{e}}, \sigma\_m\right) \tag{9}$$

(11)

ideally ordered structure. Stress-driven creation, annihilation and diffusion are the

At sufficiently high stress an atom can be squeezed into a neighboring hole with a smaller volume. Due to this neighboring atom of new positions get displaced by some amount creating new free volume. Opposite to that annihilation process tries to reduce the total free volume and restore the system to the initial state. On the other hand diffusion process neither creates nor destroys free volume, but try to just redistribute it until it is uniformly distributed everywhere. Therefore the net rate of increase of free volume is must be the difference between the rate of creation of free

where nD is number of atomic jumps required to annihilate free volume equal

<sup>μ</sup><sup>∗</sup> <sup>¼</sup> <sup>2</sup> 3 μ 1 þ ν 1 � ν

By comparing Eqs. (10) and (13) with Eqs. (8) and (9), respectively,

ξf !

exp � <sup>Δ</sup>G<sup>m</sup> kBT � � 2αkBT

It can be seen from the above equations that the functions f and g are indepen-

In MGs width of the shear band is in the order of 10–20 nm. Heat gets generated

from plastic work done during shear banding. Yang et al. [25] have shown that 0.4 mm wide shear band can cause a temperature rise of about 0.25°C. As width decreases temperature rise increases because there is less space for heat to be dissipated. Also, thermal conductivity is less for MGs [40, 41], so temperature rise in shear bands can cause the glass to reach its glass transition temperature [42]. Hence it is needed to account for the heat induced in the shear band while modeling

Following Yang [23, 25, 26] thermal transport equation is given as,

∂2 T <sup>∂</sup>x<sup>2</sup> <sup>þ</sup> <sup>α</sup>TQ <sup>σ</sup>ij

where ρ is material density, Cp is the specific heat, k<sup>0</sup> is thermal conductivity and αTQ is Taylor-Quinney coefficient which represents the fraction of plastic work converted to heat. Coefficient of thermal expansion for metallic glasses is very small; therefore thermal strain is not comparable with total strain hence it is

∂ε p ij

ρCp ∂T <sup>∂</sup><sup>t</sup> <sup>¼</sup> <sup>k</sup><sup>0</sup> <sup>ξ</sup><sup>f</sup> <sup>μ</sup><sup>∗</sup> cosh <sup>τ</sup><sup>Ω</sup>

exp � <sup>Δ</sup>G<sup>m</sup> kBT

2kBT � �

� � sinh <sup>τ</sup><sup>Ω</sup>

<sup>ξ</sup><sup>f</sup> <sup>μ</sup><sup>∗</sup> cosh <sup>τ</sup><sup>Ω</sup>

� �

" #

� 1

2kBT

2kBT � �

<sup>∂</sup><sup>t</sup> (17)

� �

" #

� � (15)

� 1

� 1 nD

(13)

(14)

� 1 nD

(16)

three processes by which local free volume concentration can be changed.

volume and rate of the annihilation of free volume and is given as [17],

exp � <sup>Δ</sup>G<sup>m</sup> kBT � � 2αkBT

\_

to ϑ<sup>∗</sup> and

g ξ<sup>f</sup> , τ<sup>e</sup>

them in MGs.

neglected in this case.

69

<sup>ξ</sup><sup>f</sup> <sup>¼</sup> <sup>ϑ</sup><sup>∗</sup> <sup>ϑ</sup> exp � αϑ<sup>∗</sup>

ξf !

Adiabatic Shear Band Formation in Metallic Glasses DOI: http://dx.doi.org/10.5772/intechopen.87437

f ξ<sup>f</sup> , τ<sup>e</sup>

� � <sup>¼</sup> <sup>ϑ</sup><sup>∗</sup> <sup>ϑ</sup> exp � αϑ<sup>∗</sup>

2.4 Temperature evolution equation

dent of mean stress σm.

� � <sup>¼</sup> <sup>2</sup><sup>ϑ</sup> exp � αϑ<sup>∗</sup>

ξf !

where D is diffusivity which is assumed to be only dependent on temperature.

#### 2.2 Flow equation

Irreversible part of strain rate that is plastic strain rate can be represented by flow rule as,

$$\frac{\partial \eta^p}{\partial t} = 2\theta \exp\left(-\frac{a\theta^\*}{\xi\_f}\right) \exp\left(-\frac{\Delta G^m}{k\_B T}\right) \sinh\left(\frac{\pi\Omega}{2k\_B T}\right) \tag{10}$$

The above flow rule is based on the microscopic model for the shear strain rate in amorphous metals proposed by Spaepen [17]. By assuming that shear strain rate depends on three quantities as,

γ\_ ¼ ðstrain produced at each jump siteÞ � ð Þ fraction of potential jump site �ð Þ net number of forward jumps at each site per second

A potential site is a region in which the free volume is greater than some critical volume. The shear strain at each potential jump site is assumed to be 1. Fraction of potential jump site is the probability that any atom has free volume greater than critical free volume. Since MGs have amorphous structure of atoms probability is calculated by statistical distribution. Therefore the fraction of potential jump sites is exp �αϑ<sup>∗</sup> <sup>=</sup>ξ<sup>f</sup> � �, where <sup>α</sup> is geometrical factor of order 1, <sup>ϑ</sup><sup>∗</sup> is critical volume and ξ<sup>f</sup> is average free volume per atom.

By rate theory when there is no shear stress applied net number of forward jumps and backward jumps should be equal to each other and number of successful jumps per second will be equal to <sup>ϑ</sup> exp �ΔGm ð Þ <sup>=</sup>kBT , where <sup>ϑ</sup> is the frequency of atomic vibration, ΔG<sup>m</sup> is activation energy, kB is Boltzmann's constant and T is absolute temperature.

When shear stress is applied, shear strain allows to lower potential energy in one direction and hence system becomes biased. Due to this one more factor to be added to above equation. So net rate of forward jumps is,

$$=2\theta \exp\left(-\frac{\Delta G^{m}}{k\_{B}T}\right)\sinh\left(\frac{\pi\Omega}{2k\_{B}T}\right) \tag{12}$$

where Ω is the atomic volume and τ is the shear stress.

#### 2.3 Free volume change rate

Free volume is the excess volume where atoms can freely move. This is defined as the difference between average atomic volume and average atomic volume in an Adiabatic Shear Band Formation in Metallic Glasses DOI: http://dx.doi.org/10.5772/intechopen.87437

The inelastic dilatation strain is associated with change of concentration of free volume. This change of concentration of free volume depends upon local concentration free volume itself ξ<sup>f</sup> , effective shear stress τ<sup>e</sup> and mean stress σ<sup>m</sup> [21].

<sup>∂</sup><sup>t</sup> <sup>¼</sup> <sup>D</sup>ξ,ii <sup>þ</sup> <sup>g</sup> <sup>ξ</sup><sup>f</sup> , <sup>τ</sup>e, <sup>σ</sup><sup>m</sup>

where D is diffusivity which is assumed to be only dependent on temperature.

Irreversible part of strain rate that is plastic strain rate can be represented by

exp � <sup>Δ</sup>G<sup>m</sup> kBT � �

The above flow rule is based on the microscopic model for the shear strain rate in amorphous metals proposed by Spaepen [17]. By assuming that shear strain rate

γ\_ ¼ ðstrain produced at each jump siteÞ � ð Þ fraction of potential jump site

A potential site is a region in which the free volume is greater than some critical volume. The shear strain at each potential jump site is assumed to be 1. Fraction of potential jump site is the probability that any atom has free volume greater than critical free volume. Since MGs have amorphous structure of atoms probability is calculated by statistical distribution. Therefore the fraction of potential jump sites is

By rate theory when there is no shear stress applied net number of forward jumps and backward jumps should be equal to each other and number of successful jumps per second will be equal to <sup>ϑ</sup> exp �ΔGm ð Þ <sup>=</sup>kBT , where <sup>ϑ</sup> is the frequency of atomic vibration, ΔG<sup>m</sup> is activation energy, kB is Boltzmann's constant and T is

When shear stress is applied, shear strain allows to lower potential energy in one direction and hence system becomes biased. Due to this one more factor to be added

> kBT � �

Free volume is the excess volume where atoms can freely move. This is defined as the difference between average atomic volume and average atomic volume in an

, where α is geometrical factor of order 1, ϑ<sup>∗</sup> is critical volume and

sinh <sup>τ</sup><sup>Ω</sup> 2kBT � �

�ð Þ net number of forward jumps at each site per second

sinh <sup>τ</sup><sup>Ω</sup> 2kBT � �

� �

(9)

(10)

(11)

(12)

∂ξf

<sup>∂</sup><sup>t</sup> <sup>¼</sup> <sup>2</sup><sup>ϑ</sup> exp � αϑ<sup>∗</sup>

ξf !

2.2 Flow equation

∂γ<sup>p</sup>

depends on three quantities as,

ξ<sup>f</sup> is average free volume per atom.

to above equation. So net rate of forward jumps is,

<sup>¼</sup> <sup>2</sup><sup>ϑ</sup> exp � <sup>Δ</sup>G<sup>m</sup>

where Ω is the atomic volume and τ is the shear stress.

flow rule as,

Metallic Glasses

exp �αϑ<sup>∗</sup> <sup>=</sup>ξ<sup>f</sup> � �

absolute temperature.

2.3 Free volume change rate

68

ideally ordered structure. Stress-driven creation, annihilation and diffusion are the three processes by which local free volume concentration can be changed.

At sufficiently high stress an atom can be squeezed into a neighboring hole with a smaller volume. Due to this neighboring atom of new positions get displaced by some amount creating new free volume. Opposite to that annihilation process tries to reduce the total free volume and restore the system to the initial state. On the other hand diffusion process neither creates nor destroys free volume, but try to just redistribute it until it is uniformly distributed everywhere. Therefore the net rate of increase of free volume is must be the difference between the rate of creation of free volume and rate of the annihilation of free volume and is given as [17],

$$\dot{\xi}\_f = \theta^\* \,\theta \exp\left(-\frac{a\theta^\*}{\xi\_f}\right) \exp\left(-\frac{\Delta G^m}{k\_B T}\right) \left[\frac{2ak\_B T}{\xi\_f \mu^\*} \left\{\cosh\left(\frac{\pi\Omega}{2k\_B T}\right) - 1\right\} - \frac{1}{n\_D}\right] \tag{13}$$

where nD is number of atomic jumps required to annihilate free volume equal to ϑ<sup>∗</sup> and

$$
\mu^\* = \frac{2}{3} \mu \frac{\mathbf{1} + \nu}{\mathbf{1} - \nu} \tag{14}
$$

By comparing Eqs. (10) and (13) with Eqs. (8) and (9), respectively,

$$f\left(\xi\_f, \tau\_\epsilon\right) = 2\theta \exp\left(-\frac{a\theta^\*}{\xi\_f}\right) \exp\left(-\frac{\Delta G^m}{k\_B T}\right) \sinh\left(\frac{\pi\Omega}{2k\_B T}\right) \tag{15}$$

$$g\left(\xi\_f, \tau\_\epsilon\right) = \theta^\* \cdot \theta \exp\left(-\frac{a\theta^\*}{\xi\_f}\right) \exp\left(-\frac{\Delta G^m}{k\_B T}\right) \left[\frac{2ak\_B T}{\xi\_f \mu^\*} \left\{\cosh\left(\frac{\tau \Omega}{2k\_B T}\right) - 1\right\} - \frac{1}{n\_D}\right] \tag{16}$$

It can be seen from the above equations that the functions f and g are independent of mean stress σm.

#### 2.4 Temperature evolution equation

In MGs width of the shear band is in the order of 10–20 nm. Heat gets generated from plastic work done during shear banding. Yang et al. [25] have shown that 0.4 mm wide shear band can cause a temperature rise of about 0.25°C. As width decreases temperature rise increases because there is less space for heat to be dissipated. Also, thermal conductivity is less for MGs [40, 41], so temperature rise in shear bands can cause the glass to reach its glass transition temperature [42]. Hence it is needed to account for the heat induced in the shear band while modeling them in MGs.

Following Yang [23, 25, 26] thermal transport equation is given as,

$$
\rho C\_p \frac{\partial T}{\partial t} = k\_0 \frac{\partial^2 T}{\partial \mathbf{x}^2} + \alpha\_{TQ} \sigma\_{\ddagger j} \frac{\partial \varepsilon\_{\ddagger j}^p}{\partial t} \tag{17}
$$

where ρ is material density, Cp is the specific heat, k<sup>0</sup> is thermal conductivity and αTQ is Taylor-Quinney coefficient which represents the fraction of plastic work converted to heat. Coefficient of thermal expansion for metallic glasses is very small; therefore thermal strain is not comparable with total strain hence it is neglected in this case.

#### 2.5 Viscosity

Combining Eq. (10) and definition of stress driven viscosity [19],

$$\eta\_v = \frac{\tau}{\dot{\mathcal{Y}}^p} = \frac{\tau}{2 \sinh\left(\frac{\tau \Omega}{2k\_B T}\right)} \theta^{-1} \exp\left(-\frac{a \theta^\*}{\xi\_f}\right) \exp\left(-\frac{\Delta G^m}{k\_B T}\right) \tag{18}$$

Using Eq. (8) for plastic shear strain rate and using Hook's law Eq. (20) becomes

dt <sup>þ</sup> <sup>f</sup> <sup>ξ</sup><sup>f</sup> , <sup>τ</sup><sup>e</sup>

In the absence of any normal force acting on strip, force balance equation in

� � τ

τe

σð Þ x; t dx ¼ 0, (22)

dx (23)

5 (24)

<sup>∂</sup><sup>t</sup> (25)

(21)

∂γ ∂t ¼ 1 μ dτ

> ð h

�h

<sup>r</sup> <sup>¼</sup> <sup>1</sup> 2h ð h

> 2h ð h

dσ

�h

dt <sup>þ</sup> <sup>f</sup> <sup>ξ</sup><sup>f</sup> , <sup>τ</sup><sup>e</sup>

Integrating above Eq. (25) and by eliminating term of integration normal stress

f ξ<sup>f</sup> ; τ<sup>e</sup> � � σ

> ∂2 ξf

Free volume concentration is assumed to be changing only in x direction and

Lastly, assuming that temperature also can vary only in x direction, Eq. (17)

∂2 T <sup>∂</sup>x<sup>2</sup> <sup>þ</sup> <sup>α</sup>TQ τγ\_

Normalization of equations is done so that to get clearer picture and to deal with dimensionless quantities. Normalization is done as; stresses by shear modulus μ, free volume concentration by <sup>ϑ</sup><sup>∗</sup> , time by 1ð Þ <sup>=</sup><sup>R</sup> , temperature by the room

�h

Therefore by integrating both sides of Eq. (21) with respect to x from �h to h

∂γ ∂t

f ξ<sup>f</sup> ; τ<sup>e</sup> � � τ

� � σ

6τ<sup>e</sup> þ 1 3 ∂ξf ∂t

<sup>∂</sup>x<sup>2</sup> <sup>þ</sup> <sup>g</sup> <sup>ξ</sup><sup>f</sup> , <sup>τ</sup><sup>e</sup>

τe dx

6τ<sup>e</sup> þ 1 3 ∂ξf

3

� �dx (26)

� � (27)

<sup>p</sup> (28)

and the average shear strain rate is

Adiabatic Shear Band Formation in Metallic Glasses DOI: http://dx.doi.org/10.5772/intechopen.87437

dτ

∂ε ∂t

> ∂ε ∂t ¼ 1 2h ð h

over width by using Eq. (22),

dt <sup>¼</sup> <sup>μ</sup> <sup>r</sup> � <sup>1</sup>

Applying Eq. (3) for normal strain ε in x direction gives,

<sup>¼</sup> ð Þ <sup>1</sup> � <sup>2</sup><sup>ν</sup> 2μ

�h

uniform along y and z direction. Then Eq. (9) becomes as,

∂ξf <sup>∂</sup><sup>t</sup> <sup>¼</sup> <sup>D</sup>

ρCp ∂T <sup>∂</sup><sup>t</sup> <sup>¼</sup> <sup>k</sup><sup>0</sup>

2 4

x-direction gives,

gives us

becomes,

71

Hence change in free volume changes viscosity in exponential manner. Therefore as free volume increases viscosity decreases and softening occurs.

#### 3. One-dimensional shear problem

Figure 1 shows the geometry of thin strip of width 2h subjected to constant shear strain.

The dimensions of the strip in the direction normal to width are very large compared to h and hence assumed infinity. Shear strain rate is assumed to be very low so that it is under quasi-static range. Shear stress will lead to the creation of more free volume and the strip will dilate. If this creation of free volume is not uniform across the width of the layer, then due to geometric constraints there will be equal normal stresses in y and z direction. There is no restriction on the material to dilate along x direction, as normal stress is zero in this direction. In this case, effective shear stress is given by

$$
\pi\_{\epsilon} = \sqrt{\pi^2 + \frac{1}{3}\sigma^2} \tag{19}
$$

Also in this case shear strain decomposition can be written as

Figure 1. Shear problem geometry.

Adiabatic Shear Band Formation in Metallic Glasses DOI: http://dx.doi.org/10.5772/intechopen.87437

2.5 Viscosity

Metallic Glasses

strain.

Figure 1.

70

Shear problem geometry.

<sup>η</sup><sup>v</sup> <sup>¼</sup> <sup>τ</sup> γ\_

3. One-dimensional shear problem

effective shear stress is given by

<sup>p</sup> <sup>¼</sup> <sup>τ</sup> 2 sinh <sup>τ</sup><sup>Ω</sup> 2kBT

Combining Eq. (10) and definition of stress driven viscosity [19],

� � <sup>ϑ</sup>�<sup>1</sup> exp � αϑ<sup>∗</sup>

Figure 1 shows the geometry of thin strip of width 2h subjected to constant shear

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi τ<sup>2</sup> þ 1 3 σ2

<sup>e</sup> <sup>þ</sup> <sup>γ</sup>\_

The dimensions of the strip in the direction normal to width are very large compared to h and hence assumed infinity. Shear strain rate is assumed to be very low so that it is under quasi-static range. Shear stress will lead to the creation of more free volume and the strip will dilate. If this creation of free volume is not uniform across the width of the layer, then due to geometric constraints there will be equal normal stresses in y and z direction. There is no restriction on the material to dilate along x direction, as normal stress is zero in this direction. In this case,

τ<sup>e</sup> ¼

Also in this case shear strain decomposition can be written as

r

γ\_ ¼ γ\_

Hence change in free volume changes viscosity in exponential manner. Therefore as free volume increases viscosity decreases and softening occurs.

ξf !

exp � <sup>Δ</sup>G<sup>m</sup> kBT � �

<sup>p</sup> (20)

(18)

(19)

Using Eq. (8) for plastic shear strain rate and using Hook's law Eq. (20) becomes

$$\frac{\partial \chi}{\partial t} = \frac{1}{\mu} \frac{d\tau}{dt} + f\left(\xi\_f, \tau\_\epsilon\right) \frac{\tau}{\tau\_\epsilon} \tag{21}$$

In the absence of any normal force acting on strip, force balance equation in x-direction gives,

$$\int\_{-h}^{h} \sigma(\mathbf{x}, t) d\mathbf{x} = \mathbf{0},\tag{22}$$

and the average shear strain rate is

$$r = \frac{1}{2h} \int\_{-h}^{h} \frac{\partial \chi}{\partial t} d\chi \tag{23}$$

Therefore by integrating both sides of Eq. (21) with respect to x from �h to h gives us

$$\frac{d\boldsymbol{\pi}}{dt} = \mu \left[ \boldsymbol{r} - \frac{\mathbf{1}}{2h} \int\_{-h}^{h} \boldsymbol{f} \left( \boldsymbol{\xi}\_{f}, \boldsymbol{\pi}\_{\epsilon} \right) \frac{\boldsymbol{\pi}}{\boldsymbol{\pi}\_{\epsilon}} d\boldsymbol{\pi} \right] \tag{24}$$

Applying Eq. (3) for normal strain ε in x direction gives,

$$\frac{\partial \sigma}{\partial t} = \frac{(1 - 2\nu)}{2\mu} \frac{d\sigma}{dt} + f\left(\xi\_f, \tau\_\varepsilon\right) \frac{\sigma}{6\tau\_\varepsilon} + \frac{1}{3} \frac{\partial \xi\_f}{\partial t} \tag{25}$$

Integrating above Eq. (25) and by eliminating term of integration normal stress over width by using Eq. (22),

$$\frac{\partial \varepsilon}{\partial t} = \frac{1}{2h} \int\_{-h}^{h} \left[ f\left(\xi\_f, \tau\_\varepsilon\right) \frac{\sigma}{6\tau\_\varepsilon} + \frac{1}{3} \frac{\partial \xi\_f}{\partial t} \right] d\mathbf{x} \tag{26}$$

Free volume concentration is assumed to be changing only in x direction and uniform along y and z direction. Then Eq. (9) becomes as,

$$\frac{\partial \xi\_f}{\partial t} = D \frac{\partial^2 \xi\_f}{\partial x^2} + \mathbf{g} \left( \xi\_f, \,\tau\_\epsilon \right) \tag{27}$$

Lastly, assuming that temperature also can vary only in x direction, Eq. (17) becomes,

$$
\rho \mathbf{C}\_p \frac{\partial T}{\partial t} = k\_0 \frac{\partial^2 T}{\partial \mathbf{x}^2} + a\_{TQ} \tau \dot{\mathbf{y}}^p \tag{28}
$$

Normalization of equations is done so that to get clearer picture and to deal with dimensionless quantities. Normalization is done as; stresses by shear modulus μ, free volume concentration by <sup>ϑ</sup><sup>∗</sup> , time by 1ð Þ <sup>=</sup><sup>R</sup> , temperature by the room

temperature T<sup>0</sup> and space x by l [21, 43, 44]. Normalized shear stress, free volume concentration, time, temperature and space are denoted by τ^, ξ, ^t, T, ^ and x^, respectively, where R is,

$$R = \theta \exp\left(-\frac{\Delta G^{\rm m}}{k\_B T}\right) \tag{29}$$

#### 3.1 Isothermal homogeneous deformation

Here Eqs. (24) and (27) are applied to the case of homogeneous deformation. Free volume concentration is uniform throughout the width of strip. Hence normal stress will be zero and so the effective shear stress is equal to the shear stress. ð Þ τ<sup>e</sup> ¼jτj .

So Eq. (24) becomes,

$$\frac{d\pi}{dt} = \mu \left[ r - f\left(\xi\_f, \tau\_\epsilon\right) \right] \tag{30}$$

and Eq. (27) becomes

$$\frac{\partial \xi}{\partial t} = \mathbf{g}\left(\xi\_f, \tau\_e\right) \tag{31}$$

From Figure 2, initially, shear stress increases linearly with shear strain solely due to elastic deformation in the strip. After some time increased shear stress creates more and more free volume so that free volume concentration increases rapidly which eventually leads to softening of the material. This softening corresponds to a sharp turn in the stress-strain diagram of Figure 2a and shear stress continues to decrease afterward. This decay of shear stress now retards the creation of free volume and finally, the system gets stabilized and steady state is achieved. At steady state both free volume concentration and shear stress are constant and MG acts like a liquid. The value of peak stress and final free volume concentration at

Results for 1D shear problem with isothermal homogeneous deformation. (a) Normalized shear stress versus

Properties and parameters Notation Value Frequency of atomic vibration ϑ 10<sup>13</sup> s�<sup>1</sup> Room temperature T<sup>0</sup> 300 K Length l 9.4 μm Geometrical factor α 0.15 Atomic jumps to annihilate free volume of ϑ<sup>∗</sup> nD 3 Average applied strain rate r 0.2 s�<sup>1</sup> Initial free volume concentration ξ<sup>i</sup> 0.0075

In the previous section, shear band formation is considered as an isothermal process but in reality, there is very little time for the entire heat produced by friction to flow out. Therefore to take care of that it is better to consider shear banding as an adiabatic process. Therefore thermal transport equation should also

For homogeneous case there is no temperature variation over the width of strip and plastic work is solely due to shear stress. Therefore Eq. (28) will reduce to,

dt <sup>¼</sup> <sup>α</sup>TQ τγ\_

<sup>p</sup> (34)

steady state depends upon material parameters and applied strain rate.

ρCp dT

3.2 Adiabatic homogeneous deformation

shear strain, and (b) free volume concentration versus shear strain.

Mechanical properties and parameters for Vitreloy 1 [25, 31, 32].

Adiabatic Shear Band Formation in Metallic Glasses DOI: http://dx.doi.org/10.5772/intechopen.87437

be considered.

73

Figure 2.

Table 1.

After normalization, Eqs. (30) and (31) are,

$$\frac{d\hat{\boldsymbol{\pi}}}{d\hat{\boldsymbol{t}}} = \frac{\boldsymbol{r}}{R} - 2\exp\left(-\frac{a}{\xi}\right)\sinh\left(i\overline{\boldsymbol{\mu}}\right) \tag{32}$$

$$\frac{\partial \xi}{\partial \hat{t}} = \exp\left(-\frac{a}{\xi}\right) \left[\frac{a}{\beta \overline{\mu} \xi} \{\cosh\left(\hat{\tau}\overline{\mu}\right)\} - \frac{1}{n\_D}\right] \tag{33}$$

where μ ¼ μΩ=ð Þ 2kBT .

Equations (32) and (33) are solved numerically using fourth order Runge-Kutta scheme for shear strain rate r = 0.2 s�<sup>1</sup> to see variation of shear stress and free volume concentration with respect to time [45]. Here Vitreloy 1 BMG is taken as a model material. Mechanical properties and parameters for Vitreloy 1 are given in Table 1. Figure 2 shows the evolution of free volume and shear stress with shear strain. Initial free volume concentration is 0.0075 and initial stress is zero.


Adiabatic Shear Band Formation in Metallic Glasses DOI: http://dx.doi.org/10.5772/intechopen.87437


Table 1.

temperature T<sup>0</sup> and space x by l [21, 43, 44]. Normalized shear stress, free volume

<sup>R</sup> <sup>¼</sup> <sup>ϑ</sup> exp � <sup>Δ</sup>Gm

Here Eqs. (24) and (27) are applied to the case of homogeneous deformation. Free volume concentration is uniform throughout the width of strip. Hence normal stress will be zero and so the effective shear stress is equal to the shear stress.

dt <sup>¼</sup> <sup>μ</sup> <sup>r</sup> � <sup>f</sup> <sup>ξ</sup><sup>f</sup> , <sup>τ</sup><sup>e</sup>

¼ g ξ<sup>f</sup> , τ<sup>e</sup> � �

> ξ � �

Equations (32) and (33) are solved numerically using fourth order Runge-Kutta

βμξ f g cosh ð Þ <sup>τ</sup>^<sup>μ</sup> � <sup>1</sup>

� �

<sup>R</sup> � 2 exp � <sup>α</sup>

scheme for shear strain rate r = 0.2 s�<sup>1</sup> to see variation of shear stress and free volume concentration with respect to time [45]. Here Vitreloy 1 BMG is taken as a model material. Mechanical properties and parameters for Vitreloy 1 are given in Table 1. Figure 2 shows the evolution of free volume and shear stress with shear strain. Initial free volume concentration is 0.0075 and initial stress is zero.

Properties and parameters Notation Value Shear modulus μ 49.68 GPa Poisson's ratio ν 0.3 Density ρ 6810 kg/m3 Specific heat Cp 330 J/kgK Activation energy ΔG<sup>m</sup> 0.2–0.5 eV Average atomic volume Ω 20 A<sup>3</sup> Thermal conductivity k 20 W/mK Taylor-Quinney coefficient αTQ 0.9 Free volume diffusivity <sup>D</sup> <sup>2</sup> � <sup>10</sup>�<sup>16</sup>

h i � �

kBT � � ^ and x^, respec-

(29)

(30)

(31)

(33)

sinh ð Þ τ^μ (32)

nD

concentration, time, temperature and space are denoted by τ^, ξ, ^t, T,

dτ

After normalization, Eqs. (30) and (31) are,

∂ξ ∂^t dτ^ d^t ¼ r

<sup>¼</sup> exp � <sup>α</sup>

∂ξ ∂t

ξ � � α

3.1 Isothermal homogeneous deformation

tively, where R is,

Metallic Glasses

ð Þ τ<sup>e</sup> ¼jτj .

72

So Eq. (24) becomes,

and Eq. (27) becomes

where μ ¼ μΩ=ð Þ 2kBT .

Mechanical properties and parameters for Vitreloy 1 [25, 31, 32].

Figure 2.

Results for 1D shear problem with isothermal homogeneous deformation. (a) Normalized shear stress versus shear strain, and (b) free volume concentration versus shear strain.

From Figure 2, initially, shear stress increases linearly with shear strain solely due to elastic deformation in the strip. After some time increased shear stress creates more and more free volume so that free volume concentration increases rapidly which eventually leads to softening of the material. This softening corresponds to a sharp turn in the stress-strain diagram of Figure 2a and shear stress continues to decrease afterward. This decay of shear stress now retards the creation of free volume and finally, the system gets stabilized and steady state is achieved. At steady state both free volume concentration and shear stress are constant and MG acts like a liquid. The value of peak stress and final free volume concentration at steady state depends upon material parameters and applied strain rate.

#### 3.2 Adiabatic homogeneous deformation

In the previous section, shear band formation is considered as an isothermal process but in reality, there is very little time for the entire heat produced by friction to flow out. Therefore to take care of that it is better to consider shear banding as an adiabatic process. Therefore thermal transport equation should also be considered.

For homogeneous case there is no temperature variation over the width of strip and plastic work is solely due to shear stress. Therefore Eq. (28) will reduce to,

$$
\rho \mathbf{C}\_p \frac{dT}{dt} = a\_{TQ} \tau \dot{\mathbf{y}}^p \tag{34}
$$

After normalizing Eq. (34), it will become as

$$\frac{d\hat{T}}{dt} = \frac{\mu}{T\_0} \frac{a\_{TQ}}{\rho C\_p} \left( \hat{\mathbf{r}} \frac{d\hat{\mathbf{p}}^p}{dt} \right) \tag{35}$$

creates more free volume and after some point, strain softening behavior occurs. In the meantime, temperature increases rapidly as plastic work is done. The normalized temperature increase is plotted against shear strain in Figure 3c. This instantaneous temperature rise lowers the energy barrier and hence helps in shear band nucleation. After a sharp turn in shear stress, it acquires a steady state. On the other hand, the temperature keeps rising steadily which promotes annihilation of free

The main difference in the homogeneous isothermal model and adiabatic model

In homogeneous case, the initial free volume is assumed to be uniform through the width, but in practical cases, this will never happen. There will be always some disturbance in free volume due to quenching or thermal fluctuations. Therefore in this section, it is assumed that initial free volume has some non-uniformity over the

For simplicity, finite amplitude disturbance in the form of Gaussian distribution is assumed. This disturbance is added to initially assumed free volume concentra-

2

(38)

Δ2 " #

βμξ f g cosh ð Þ� <sup>τ</sup>^e<sup>μ</sup> <sup>1</sup> � <sup>1</sup>

� �sinh ð Þ ^τe<sup>μ</sup> ^<sup>τ</sup>

� �sinh ð Þ ^τe<sup>μ</sup> <sup>σ</sup>^

� � sinh ð Þ <sup>τ</sup>^e<sup>μ</sup> <sup>σ</sup>^

ξ

nD

dx^ (40)

<sup>∂</sup>^<sup>t</sup> (42)

� � (39)

^τe

^τe þ ∂ξ ∂^t

� �dx^ (41)

3τ^<sup>e</sup> þ 1 3 ∂ξ

<sup>ξ</sup>ð Þ¼ x, <sup>0</sup> <sup>ξ</sup><sup>i</sup> <sup>þ</sup> <sup>δ</sup> exp � ð Þ <sup>x</sup> � <sup>x</sup><sup>0</sup>

where ξ<sup>i</sup> is initially assumed free volume which is constant, δ is amplitude of disturbance, x<sup>0</sup> is location where this disturbance is going to add and Δ is characteristic half width. Now for following case parameters are taken as, ξ<sup>i</sup> = 0.0075, δ = 0.001, x<sup>0</sup> = 0, Δ = 100l and h = 2000l [21]. As x<sup>0</sup> is zero means disturbance will get added at midpoint of strip. Also it is assumed that Δ ≪ h so that there will be very small effect of disturbance on the both boundaries and that can be neglected. All

In inhomogeneous deformation normal stress will not be zero so effective shear stress will come into picture. Therefore Eqs. (24)–(27) should be solved simultaneously for variation of shear stress, normal stress, normal strain and free volume

> exp � <sup>α</sup> ξ

is that in the isothermal model free volume remains constant after some point wherein the adiabatic model it keeps on decreasing. This is due to temperature change which helps in the annihilation process and hence free volume decreases.

volume causing a continuous decrease in free volume concentration.

Shear stress after steady state is nearly the same in both the models.

3.3 Isothermal inhomogeneous deformation

Adiabatic Shear Band Formation in Metallic Glasses DOI: http://dx.doi.org/10.5772/intechopen.87437

stresses and strains are assumed to be zero initially.

∂ξ ∂^t ¼ ∂2 ξ

> d^τ d^t ¼ r <sup>R</sup> � <sup>1</sup> ð Þ h=l

dε d^t

dε d^t

75

<sup>¼</sup> <sup>1</sup> ð Þ 6h=l

<sup>¼</sup> ð Þ <sup>1</sup> � <sup>2</sup><sup>ν</sup> 2μ

concentration, respectively. After normalization equations are,

ξ � � α

> ð h=l

�h=l

exp � <sup>α</sup> ξ

<sup>þ</sup> exp � <sup>α</sup>

<sup>∂</sup>x^<sup>2</sup> <sup>þ</sup> exp � <sup>α</sup>

ð h=l

�h=l

∂σ ∂^t

width of the strip.

tion [21]. So to get,

Also for adiabatic case Eqs. (24) and (27) will become as

$$\frac{d\hat{\boldsymbol{\tau}}}{d\hat{t}} = \frac{1}{R} \left[ r - 2\theta \exp\left( -\frac{a}{\xi} \right) \exp\left( -\frac{\Delta G'''}{k\_B T\_0 \hat{T}} \right) \sinh\left( -\frac{\hat{\boldsymbol{\tau}} \overline{\mu}}{\hat{T}} \right) \right] \tag{36}$$

$$\frac{\partial \xi}{\partial \hat{t}} = \frac{1}{R} \theta \exp\left(-\frac{a}{\xi}\right) \exp\left(-\frac{\Delta G^{m}}{k\_{B}T\_{0}\hat{T}}\right) \left[\frac{a\hat{T}}{\beta\overline{\mu}\xi} \left\{\cosh\left(-\frac{\hat{t}\,\overline{\mu}}{\hat{T}}\right) - 1\right\} - \frac{1}{n\_{D}}\right] \tag{37}$$

Using numerical integration, Eqs. (35)–(37) are solved for variation of shear stress, free volume concentration and temperature with respect to applied strain for same initial conditions and strain rate.

Figure 3 shows the result of numerical solution of Eqs. (35)–(37). In Figure 3a and b, results are compared with the case of isothermal homogeneous deformation, where dash lines indicate the isothermal model and solid lines are for adiabatic model. When applied shear strain is small glass strip is in an elastic state. In this state, both results are nearly the same because plastic work causes temperature rise and in an elastic state plastic work is negligible. More and more increase in stress

#### Figure 3.

Results for 1D shear problem for adiabatic homogeneous deformation. (a) Normalized shear stress versus shear strain, (b) free volume concentration versus shear strain and (c) normalized temperature versus shear strain.

Adiabatic Shear Band Formation in Metallic Glasses DOI: http://dx.doi.org/10.5772/intechopen.87437

After normalizing Eq. (34), it will become as

dτ^ d^t ¼ 1 R

<sup>R</sup> <sup>ϑ</sup> exp � <sup>α</sup>

ξ � �

same initial conditions and strain rate.

∂ξ ∂^t ¼ 1

Metallic Glasses

Figure 3.

74

dT^ <sup>d</sup>^<sup>t</sup> <sup>¼</sup> <sup>μ</sup> T<sup>0</sup>

Also for adiabatic case Eqs. (24) and (27) will become as

ξ � �

kBT0T^ � � αT^

<sup>r</sup> � <sup>2</sup><sup>ϑ</sup> exp � <sup>α</sup>

exp � <sup>Δ</sup>G<sup>m</sup>

αTQ ρCp

Using numerical integration, Eqs. (35)–(37) are solved for variation of shear stress, free volume concentration and temperature with respect to applied strain for

Figure 3 shows the result of numerical solution of Eqs. (35)–(37). In Figure 3a and b, results are compared with the case of isothermal homogeneous deformation, where dash lines indicate the isothermal model and solid lines are for adiabatic model. When applied shear strain is small glass strip is in an elastic state. In this state, both results are nearly the same because plastic work causes temperature rise and in an elastic state plastic work is negligible. More and more increase in stress

Results for 1D shear problem for adiabatic homogeneous deformation. (a) Normalized shear stress versus shear strain, (b) free volume concentration versus shear strain and (c) normalized temperature versus shear strain.

τ^ dγ\_ p d^t � �

exp � <sup>Δ</sup>G<sup>m</sup>

� � � �

kBT0T^ � �

βμξ cosh � <sup>τ</sup>^<sup>μ</sup>

sinh � <sup>τ</sup>^<sup>μ</sup>

� 1

T^ � �

� �

" #

T^

� 1 nD (35)

(36)

(37)

creates more free volume and after some point, strain softening behavior occurs. In the meantime, temperature increases rapidly as plastic work is done. The normalized temperature increase is plotted against shear strain in Figure 3c. This instantaneous temperature rise lowers the energy barrier and hence helps in shear band nucleation. After a sharp turn in shear stress, it acquires a steady state. On the other hand, the temperature keeps rising steadily which promotes annihilation of free volume causing a continuous decrease in free volume concentration.

The main difference in the homogeneous isothermal model and adiabatic model is that in the isothermal model free volume remains constant after some point wherein the adiabatic model it keeps on decreasing. This is due to temperature change which helps in the annihilation process and hence free volume decreases. Shear stress after steady state is nearly the same in both the models.

#### 3.3 Isothermal inhomogeneous deformation

In homogeneous case, the initial free volume is assumed to be uniform through the width, but in practical cases, this will never happen. There will be always some disturbance in free volume due to quenching or thermal fluctuations. Therefore in this section, it is assumed that initial free volume has some non-uniformity over the width of the strip.

For simplicity, finite amplitude disturbance in the form of Gaussian distribution is assumed. This disturbance is added to initially assumed free volume concentration [21]. So to get,

$$\xi(\mathbf{x}, \mathbf{0}) = \xi\_i + \delta \exp\left[ -\frac{(\mathbf{x} - \mathbf{x}\_0)^2}{\Delta^2} \right] \tag{38}$$

where ξ<sup>i</sup> is initially assumed free volume which is constant, δ is amplitude of disturbance, x<sup>0</sup> is location where this disturbance is going to add and Δ is characteristic half width. Now for following case parameters are taken as, ξ<sup>i</sup> = 0.0075, δ = 0.001, x<sup>0</sup> = 0, Δ = 100l and h = 2000l [21]. As x<sup>0</sup> is zero means disturbance will get added at midpoint of strip. Also it is assumed that Δ ≪ h so that there will be very small effect of disturbance on the both boundaries and that can be neglected. All stresses and strains are assumed to be zero initially.

In inhomogeneous deformation normal stress will not be zero so effective shear stress will come into picture. Therefore Eqs. (24)–(27) should be solved simultaneously for variation of shear stress, normal stress, normal strain and free volume concentration, respectively. After normalization equations are,

$$\frac{\partial \xi}{\partial \hat{t}} = \frac{\partial^2 \xi}{\partial \hat{\alpha}^2} + \exp\left(-\frac{a}{\xi}\right) \left[\frac{a}{\beta \overline{\mu} \xi} \{\cosh\left(\hat{\tau}\_\epsilon \overline{\mu}\right) - 1\} - \frac{1}{n\_D}\right] \tag{39}$$

$$\frac{d\hat{\boldsymbol{\pi}}}{d\hat{\boldsymbol{t}}} = \frac{\boldsymbol{r}}{R} - \frac{\mathbf{1}}{(h/l)} \int\_{-h/l}^{h/l} \exp\left(-\frac{a}{\xi}\right) \sinh\left(\hat{\boldsymbol{\pi}}\_{\boldsymbol{\epsilon}}\overline{\boldsymbol{\mu}}\right) \frac{\hat{\boldsymbol{\pi}}}{\hat{\boldsymbol{\pi}}\_{\boldsymbol{\epsilon}}} d\hat{\boldsymbol{\varepsilon}}\tag{40}$$

$$\frac{d\varepsilon}{d\hat{t}} = \frac{1}{(\mathfrak{G}h/l)} \int\_{-h/l}^{h/l} \left[ \exp\left(-\frac{a}{\xi}\right) \sinh\left(\hat{x}\_{\varepsilon}\overline{\mu}\right) \frac{\hat{\sigma}}{\hat{x}\_{\varepsilon}} + \frac{\partial \xi}{\partial \hat{t}} \right] d\hat{\sigma} \tag{41}$$

$$\frac{d\varepsilon}{d\hat{t}} = \frac{(1 - 2\nu)}{2\mu} \frac{\partial \sigma}{\partial \hat{t}} + \exp\left(-\frac{a}{\xi}\right) \sinh\left(\hat{\tau}\_{\epsilon}\overline{\mu}\right) \frac{\hat{\sigma}}{3\hat{\tau}\_{\epsilon}} + \frac{1}{3} \frac{\partial \xi}{\partial \hat{t}}\tag{42}$$

Equation (39) is firstly converted into set of ordinary differential equations by using finite elements. Then all four equations are converted into algebraic equations using explicit method for time marching [46]. For explicit time marching method it is necessary to take very small time step, due to which computational time increases drastically. Integration with respect to x is carried out using modified trapezoidal rule. As there is negligible effect of disturbance in free volume at both boundaries and free volume is uniformly distributed near boundaries so it is assumed that, <sup>∂</sup>ξ=∂<sup>x</sup> <sup>¼</sup> 0 at both boundaries. Shear strain is calculated from Eq. (21) at each time step.

Results shown in Figure 4 are for different time steps where time is indicated by an average shear strain rate. Figure 4a shows a variation of free volume concentration over normalized distance means the width of the plate, while Figure 4b and c shows a variation of shear strain and normalized normal stress, respectively. Figure 5 shows the variation of normalized shear stress with respect to shear strain. In the initial stage, shear stress is very small, therefore free volume change occurs due to only diffusion and annihilation processes. As these processes tend to decrease the free volume, initially amplitude of disturbance of free volume concentration decreases. Hence deformation is elastic and tends to be nearly homogeneous. But with time shear stress increases and free volume start to increase due to the stressdriven creation process. After some time the creation process of the free volume becomes more dominating than diffusion and annihilation. At that time amplitude of disturbance starts to grow. As it is seen in Figure 4a free volume concentration

grows rapidly at the center of the disturbance. Meanwhile, shear strain increases rapidly where free volume concentration is increased, i.e., at the center of a plate, but decreases at the other locations as shown in Figure 4b. This develops inhomogeneous deformation in a plate. From Figure 4c, it can be seen that there is a drop in normal stress at the center but then it spikes up due to this inhomogeneous growth of free volume concentration. Here normal stress is smaller than shear stress. When there is instantaneous growth in free volume concentration shear stress falls abruptly as seen from Figure 5. But as shear stress falls, again creation process gets retarded and diffusion and annihilation processes dominate. This tends to reduce the localization effect. Eventually, the steady state is achieved where there

The dashed line in Figure 5 represents the normalized shear stress of homogeneous deformation solution. Initially, both solutions are nearly the same, but there is a difference after the peak value of shear stress is reached. Shear stress falls abruptly in case of inhomogeneous deformation due to an instantaneous increase in free volume concentration which is not the case for homogeneous deformation. Finally, inhomogeneous deformation gets converted into homogeneous deformation when the distribution of free volume becomes uniform. All results are very sensitive to several parameters like α, or initial free volume concentration.

In this section inhomogeneous deformation is modeled as adiabatic process. As it is clear from Section 3.2 temperature also affects creation of shear bands and have influence on softening it is must to consider its effect in case of inhomogeneous deformation. Now again considering Eq. (24)–(28) and by normalizing them

<sup>∂</sup>x^<sup>2</sup> <sup>þ</sup> exp � <sup>α</sup>

<sup>2</sup>ϑ<sup>0</sup> exp � <sup>α</sup>

ξ � � αT^

ξ

� � exp � <sup>Δ</sup>Gm

" #

βμξ cosh <sup>τ</sup>^<sup>μ</sup>

kBT0T^ � � sinh <sup>τ</sup>^<sup>μ</sup>

" # ( )

T^ � �

� �

� 1

T^ � � τ^ τ^e dx^

� 1 nD

(43)

(44)

is no variation in shear stress with respect to time.

Variation of shear stress for isothermal inhomogeneous deformation.

Adiabatic Shear Band Formation in Metallic Glasses DOI: http://dx.doi.org/10.5772/intechopen.87437

3.4 Adiabatic inhomogeneous deformation

ΔG<sup>m</sup> kBT<sup>0</sup> � � ∂<sup>2</sup>

2ð Þ h=l

ð<sup>h</sup>=<sup>l</sup> �h=l ξ

accordingly,

Figure 5.

<sup>¼</sup> exp � <sup>Δ</sup>G<sup>m</sup>

kBT0T^ � � exp

> ΔGm kBT<sup>0</sup>

� � <sup>r</sup> � <sup>1</sup>

∂ξ ∂^t

dτ^ d^t ¼ 1 ϑ0 exp

77

#### Figure 4.

Results for 1D shear problem with isothermal inhomogeneous deformation. Figure shows variation of (a) free volume concentration, (b) shear strain and (c) normalized normal stress over the normalized distance.

Adiabatic Shear Band Formation in Metallic Glasses DOI: http://dx.doi.org/10.5772/intechopen.87437

Equation (39) is firstly converted into set of ordinary differential equations by using finite elements. Then all four equations are converted into algebraic equations using explicit method for time marching [46]. For explicit time marching method it is necessary to take very small time step, due to which computational time increases drastically. Integration with respect to x is carried out using modified trapezoidal rule. As there is negligible effect of disturbance in free volume at both boundaries and free volume is uniformly distributed near boundaries so it is assumed that, <sup>∂</sup>ξ=∂<sup>x</sup> <sup>¼</sup> 0 at both boundaries. Shear strain is calculated from Eq. (21)

Results shown in Figure 4 are for different time steps where time is indicated by an average shear strain rate. Figure 4a shows a variation of free volume concentration over normalized distance means the width of the plate, while Figure 4b and c shows a variation of shear strain and normalized normal stress, respectively. Figure 5 shows the variation of normalized shear stress with respect to shear strain. In the initial stage, shear stress is very small, therefore free volume change occurs due to only diffusion and annihilation processes. As these processes tend to decrease the free volume, initially amplitude of disturbance of free volume concentration decreases. Hence deformation is elastic and tends to be nearly homogeneous. But with time shear stress increases and free volume start to increase due to the stressdriven creation process. After some time the creation process of the free volume becomes more dominating than diffusion and annihilation. At that time amplitude of disturbance starts to grow. As it is seen in Figure 4a free volume concentration

Results for 1D shear problem with isothermal inhomogeneous deformation. Figure shows variation of (a) free volume concentration, (b) shear strain and (c) normalized normal stress over the normalized distance.

at each time step.

Metallic Glasses

Figure 4.

76

Figure 5. Variation of shear stress for isothermal inhomogeneous deformation.

grows rapidly at the center of the disturbance. Meanwhile, shear strain increases rapidly where free volume concentration is increased, i.e., at the center of a plate, but decreases at the other locations as shown in Figure 4b. This develops inhomogeneous deformation in a plate. From Figure 4c, it can be seen that there is a drop in normal stress at the center but then it spikes up due to this inhomogeneous growth of free volume concentration. Here normal stress is smaller than shear stress. When there is instantaneous growth in free volume concentration shear stress falls abruptly as seen from Figure 5. But as shear stress falls, again creation process gets retarded and diffusion and annihilation processes dominate. This tends to reduce the localization effect. Eventually, the steady state is achieved where there is no variation in shear stress with respect to time.

The dashed line in Figure 5 represents the normalized shear stress of homogeneous deformation solution. Initially, both solutions are nearly the same, but there is a difference after the peak value of shear stress is reached. Shear stress falls abruptly in case of inhomogeneous deformation due to an instantaneous increase in free volume concentration which is not the case for homogeneous deformation. Finally, inhomogeneous deformation gets converted into homogeneous deformation when the distribution of free volume becomes uniform. All results are very sensitive to several parameters like α, or initial free volume concentration.

#### 3.4 Adiabatic inhomogeneous deformation

In this section inhomogeneous deformation is modeled as adiabatic process. As it is clear from Section 3.2 temperature also affects creation of shear bands and have influence on softening it is must to consider its effect in case of inhomogeneous deformation. Now again considering Eq. (24)–(28) and by normalizing them accordingly,

$$\frac{d\xi}{d\hat{t}} = \exp\left(-\frac{\Delta G^{m}}{k\_{B}T\_{0}\hat{T}}\right)\exp\left(\frac{\Delta G^{m}}{k\_{B}T\_{0}}\right)\left[\frac{\partial^{2}\xi}{\partial\hat{x}^{2}} + \exp\left(-\frac{a}{\xi}\right)\left\{\frac{a\hat{T}}{\beta\overline{\mu}\xi}\left\{\cosh\left(\frac{\hat{\tau}\mu}{\hat{T}}\right) - 1\right\} - \frac{1}{n\_{D}}\right\}\right],\tag{43}$$

$$\frac{d\hat{t}}{d\hat{t}} = \frac{1}{\theta\_0} \exp\left(\frac{\Delta G^m}{k\_B T\_0}\right) \left[r - \frac{1}{2(h/l)}\right]\_{-h/l}^{h/l} 2\theta\_0 \exp\left(-\frac{a}{\xi}\right) \exp\left(-\frac{\Delta G^m}{k\_B T\_0 \hat{T}}\right) \sinh\left(\frac{\tilde{\epsilon}\overline{\mu}}{\hat{T}}\right) \frac{\hat{t}}{\hat{t}\_\varepsilon} d\hat{\vec{x}}\tag{44}$$

Metallic Glasses

$$\frac{d\varepsilon}{d\hat{t}} = \exp\left(\frac{\Delta G^{m}}{k\_{B}T\_{0}}\right) \frac{1}{(\delta h/l)} \int\_{-h/l}^{h/l} \left[\exp\left(-\frac{a}{\xi}\right) \exp\left(-\frac{\Delta G^{m}}{k\_{B}T\_{0}\hat{T}}\right) \sinh\left(\frac{\hat{\tau}\overline{\mu}}{\hat{T}}\right) \frac{\hat{\sigma}}{\hat{\varepsilon}\_{e}}\right]$$

$$+\exp\left(-\frac{\Delta G^{m}}{k\_{B}T\_{0}\hat{T}}\right) \frac{\partial\xi}{\partial\hat{t}}\bigg|\_{} \tag{45}$$

$$\frac{d\hat{\sigma}}{d\hat{t}} = \frac{2}{(1-2\nu)} \left[ \frac{\partial \varepsilon}{\partial \hat{t}} - \frac{1}{3} \frac{\partial \tilde{\xi}}{\partial \hat{t}} - \frac{1}{3} \exp\left(-\frac{\Delta G'''}{k\_B T\_0 \hat{T}}\right) \exp\left(\frac{\Delta G'''}{k\_B T\_0}\right) \exp\left(-\frac{a}{\xi}\right) \sinh\left(\frac{\tilde{\tau}\tilde{\mu}}{\hat{T}}\right) \frac{\hat{\sigma}}{\hat{\varepsilon}\_{\varepsilon}}\right] \tag{46}$$

$$\frac{d\hat{T}}{d\hat{t}} = \exp\left(\frac{\Delta G^{m}}{k\_{B}T\_{0}}\right) \left[\frac{k}{\rho C\_{p}\theta\_{0}l^{2}}\frac{\partial^{2}\hat{T}}{\partial\hat{x}^{2}} + \frac{2\mu}{T\_{0}}\frac{\alpha\_{TQ}\hat{x}}{\rho C\_{p}}\exp\left(-\frac{\alpha}{\xi}\right)\exp\left(-\frac{\Delta G^{m}}{k\_{B}T\_{0}\hat{T}}\right)\sinh\left(\frac{\hat{\tau}\overline{\mu}}{\hat{T}}\right)\right]$$
 
$$\gamma \quad \therefore \quad \gamma$$

$$+\frac{2\mu}{T\_0} \frac{a\_{TQ}\hat{\sigma}}{\rho C\_p} \frac{\partial \epsilon^p}{\partial \hat{t}}\tag{47}$$

Equations (43)–(47) are again solved by finite element method by using explicit time marching scheme. Boundary conditions are same for free volume concentration as mentioned in isothermal inhomogeneous case and for temperature similar boundary conditions are assumed as, <sup>∂</sup>T=∂<sup>x</sup> <sup>¼</sup> 0 at both boundaries.

Figure 6 shows solution for adiabatic inhomogeneous case solved by explicit method with same parameters as used in earlier section. Figure 6a shows the distribution of free volume at different time steps, which are indicated by average shear strain. Figure 6b–d shows the distribution of normalized temperature, shear strain and the normalized normal stress, respectively.

As it is clear from previous sections initially when stress is low material is in an elastic state so the temperature remains almost constant. Free volume concentration initially decreases due to annihilation but then shoots up when stress increases with time. At the same time temperature inside the shear band increases very rapidly and can reach glass transition temperature. As initially, the temperature was uniform throughout the width it can be said that local heating must be caused due to increases in free volume. The temperature outside the shear band almost remains the same. This increase in free volume also softens the material causing the shear strain to increase in shear band region. But, as temperature increases suddenly annihilation process dominates over the creation process and large drop in free volume concentration is observed inside the shear band. Although free volume concentration drops, shear strain still increases as it depends on temperature also. Value of normalized normal stress shows a large increase in value near shear band but still, a pattern is the same as observed in the isothermal inhomogeneous case.

As shown in Figure 7 due to softening shear stress value drops abruptly which leads to a decrease in free volume concentration even more as the creation process gets retarded. For elastic region results matches exactly with an isothermal inhomogeneous case as the temperature remains same until plastic work is done. As the temperature keeps on increasing which leads to more drop in free volume concentration. After some point, free volume concentration value inside the shear band goes below than the value of free volume concentration outside the shear band as shown in Figure 8. This inhomogeneity in free volume cannot be removed as the generation process is retarded and annihilation process is in favor of inhomogeneity. Which results in a continuous decrease in shear stress value and steady state is not achieved. Still, shear strain is accommodated in the shear band as temperature increases to very high value.

These results fairly match with results obtained by Jiang and Dai [29], where the shear band is treated as an initial narrow zone of increased free volume concentration compared to the matrix surrounding it. This allows reduction of the governing

Results for 1D shear problem with adiabatic inhomogeneous deformation. Figure shows variation of (a) free volume concentration, (b) normalized temperature, (c) shear strain and (d) normalized normal stress over the

Variation of shear stress for inhomogeneous adiabatic deformation.

Adiabatic Shear Band Formation in Metallic Glasses DOI: http://dx.doi.org/10.5772/intechopen.87437

Figure 7.

79

Figure 6.

normalized distance.

#### Figure 6.

dε d^t

Metallic Glasses

<sup>¼</sup> <sup>2</sup> ð Þ 1 � 2ν

> þ 2μ T<sup>0</sup>

∂σ^ ∂^t

dT^ <sup>d</sup>^<sup>t</sup> <sup>¼</sup> exp

¼ exp

ΔG<sup>m</sup> kBT<sup>0</sup> � � 1

<sup>þ</sup> exp � <sup>Δ</sup>G<sup>m</sup>

∂ε ∂^t � 1 3 ∂ξ ∂^t � 1

ΔG<sup>m</sup> kBT<sup>0</sup> � � k

αTQ σ^ ρCp

increases to very high value.

78

∂ε<sup>p</sup>

ð Þ 6h=l

kBT0T^ � � ∂ξ

> ρCpϑ0l 2 ∂2 T^ <sup>∂</sup>x^<sup>2</sup> <sup>þ</sup>

ðh=<sup>l</sup> �h=l

∂^t #

<sup>3</sup> exp � <sup>Δ</sup>G<sup>m</sup>

2μ T<sup>0</sup>

boundary conditions are assumed as, <sup>∂</sup>T=∂<sup>x</sup> <sup>¼</sup> 0 at both boundaries.

strain and the normalized normal stress, respectively.

kBT0T^ � �

> αTQ τ^ ρCp

Equations (43)–(47) are again solved by finite element method by using explicit time marching scheme. Boundary conditions are same for free volume concentration as mentioned in isothermal inhomogeneous case and for temperature similar

Figure 6 shows solution for adiabatic inhomogeneous case solved by explicit method with same parameters as used in earlier section. Figure 6a shows the distribution of free volume at different time steps, which are indicated by average shear strain. Figure 6b–d shows the distribution of normalized temperature, shear

As it is clear from previous sections initially when stress is low material is in an elastic state so the temperature remains almost constant. Free volume concentration initially decreases due to annihilation but then shoots up when stress increases with time. At the same time temperature inside the shear band increases very rapidly and can reach glass transition temperature. As initially, the temperature was uniform throughout the width it can be said that local heating must be caused due to increases in free volume. The temperature outside the shear band almost remains the same. This increase in free volume also softens the material causing the shear strain to increase in shear band region. But, as temperature increases suddenly annihilation process dominates over the creation process and large drop in free volume concentration is observed inside the shear band. Although free volume concentration drops, shear strain still increases as it depends on temperature also. Value of normalized normal stress shows a large increase in value near shear band but still, a pattern is the same as observed in the isothermal inhomogeneous case. As shown in Figure 7 due to softening shear stress value drops abruptly which leads to a decrease in free volume concentration even more as the creation process gets retarded. For elastic region results matches exactly with an isothermal inhomogeneous case as the temperature remains same until plastic work is done. As the temperature keeps on increasing which leads to more drop in free volume concentration. After some point, free volume concentration value inside the shear band goes below than the value of free volume concentration outside the shear band as shown in Figure 8. This inhomogeneity in free volume cannot be removed as the generation process is retarded and annihilation process is in favor of inhomogeneity. Which results in a continuous decrease in shear stress value and steady state is not achieved. Still, shear strain is accommodated in the shear band as temperature

exp

� �

exp � <sup>α</sup> ξ � �

<sup>∂</sup>^<sup>t</sup> (47)

" # � �

�

exp � <sup>α</sup> ξ � � exp � <sup>Δ</sup>G<sup>m</sup>

ΔG<sup>m</sup> kBT<sup>0</sup> � �

kBT0T^ � �

> exp � <sup>α</sup> ξ � �

exp � <sup>Δ</sup>G<sup>m</sup>

kBT0T^ � �

dx^ (45)

sinh <sup>τ</sup>^<sup>μ</sup> T^ � � σ^ τ^e

> sinh <sup>τ</sup>^<sup>μ</sup> T^ � � σ^ τ^e

> > sinh <sup>τ</sup>^<sup>μ</sup> T^

(46)

Results for 1D shear problem with adiabatic inhomogeneous deformation. Figure shows variation of (a) free volume concentration, (b) normalized temperature, (c) shear strain and (d) normalized normal stress over the normalized distance.

#### Figure 7.

Variation of shear stress for inhomogeneous adiabatic deformation.

These results fairly match with results obtained by Jiang and Dai [29], where the shear band is treated as an initial narrow zone of increased free volume concentration compared to the matrix surrounding it. This allows reduction of the governing

Figure 8. Increased inhomogeneity in free volume concentration.

partial differential equations as a set of coupled ordinary differential equations which can be numerically integrated relatively easily. But the major issue with the approach adopted by Jiang and Dai [29] is that it can only be applied to the 1D problem whereas the approach used in present work can be readily extended to non-homogeneous solution in two/three dimensions [23].

the annihilation of free volume hence free volume always keeps on decreasing and

shear stress versus shear strain, and (b) free volume concentration versus shear strain.

Solution for isothermal homogeneous deformation case with various applied shear strain rates. (a) Normalized

Figure 10 shows solution for isothermal homogeneous deformation case with initial free volume concentration as 0.008 for different values of surrounding temperature. Figure 10a shows a variation of normalized shear stress with respect to shear strain while Figure 10b shows a variation of free volume concentration. From Figure 10, it can be seen that initially when shear strain is very low all the solutions match with each other. As surrounding temperature increases it supports annihilation process and hence free volume concentration decreases. But this decrease in free volume delays the softening and therefore maximum value of shear stress increases. Again this increased shear stress produces more and more free volume by dominating over annihilation process and hence the maximum value of free volume concentration is more in case of maximum surrounding temperature. In this case, also after softening steady state is achieved but values are different for different surrounding temperature. A similar effect will be seen in the case of inhomogeneous deformation. In the case of adiabatic deformation effect of change of applied strain rate will be similar but the effect of change of surrounding temperature may vary

Solution for isothermal homogeneous deformation case with various surrounding temperatures. (a) Normalized

shear stress versus shear strain, and (b) free volume concentration versus shear strain.

softening does not occur at all.

Adiabatic Shear Band Formation in Metallic Glasses DOI: http://dx.doi.org/10.5772/intechopen.87437

Figure 9.

by some amount.

Figure 10.

81

## 4. Parametric study

All the parameters used to solve the 1D shear problem in previous sections were obtained by literature review for the sake of comparison with existing results. Parameters selected are influenced by work of Huang et al. [21], Gao et al. [28] and Jiang and Dai [29]. Results are very sensitive to several parameters. In this section, it is tried to study and understand how results will alter if there is a change in some parameters. It is found that there are many number of parameters to which results are very sensitive like geometrical factor α, an initial value of free volume concentration ξi, shear strain rate r, surrounding temperature (in most cases room temperature) T0, etc. To keep the parametric study within the scope of this chapter, the effect of only two parameters is studied, namely: shear strain rate r and surrounding temperature T0.

Figure 9 shows solution for isothermal homogeneous deformation case with initial free volume concentration as 0.008 for different values of applied shear strain rate. Figure 9a shows a variation of normalized shear stress with respect to shear strain while Figure 9b shows a variation of free volume concentration. From Figure 9 it is seen that as the value of applied shear strain rate increases maximum value attained by shear stress also increases. This is due to the fact that as shear strain rate is increased there is less time for annihilation process to decrease the free volume and as stress increases rapidly large amount of free volume gets generated. This increased free volume later causes the softening. Steady state value of shear stress is not much affected by this change of shear strain rate but a steady state value of free volume concentration decreases as strain rate decreases. If the strain rate is very low then the rate of generation of free volume never exceeds the rate of Adiabatic Shear Band Formation in Metallic Glasses DOI: http://dx.doi.org/10.5772/intechopen.87437

Figure 9.

partial differential equations as a set of coupled ordinary differential equations which can be numerically integrated relatively easily. But the major issue with the approach adopted by Jiang and Dai [29] is that it can only be applied to the 1D problem whereas the approach used in present work can be readily extended to

All the parameters used to solve the 1D shear problem in previous sections were

Figure 9 shows solution for isothermal homogeneous deformation case with initial free volume concentration as 0.008 for different values of applied shear strain rate. Figure 9a shows a variation of normalized shear stress with respect to shear strain while Figure 9b shows a variation of free volume concentration. From Figure 9 it is seen that as the value of applied shear strain rate increases maximum value attained by shear stress also increases. This is due to the fact that as shear strain rate is increased there is less time for annihilation process to decrease the free volume and as stress increases rapidly large amount of free volume gets generated. This increased free volume later causes the softening. Steady state value of shear stress is not much affected by this change of shear strain rate but a steady state value of free volume concentration decreases as strain rate decreases. If the strain rate is very low then the rate of generation of free volume never exceeds the rate of

obtained by literature review for the sake of comparison with existing results. Parameters selected are influenced by work of Huang et al. [21], Gao et al. [28] and Jiang and Dai [29]. Results are very sensitive to several parameters. In this section, it is tried to study and understand how results will alter if there is a change in some parameters. It is found that there are many number of parameters to which results are very sensitive like geometrical factor α, an initial value of free volume concentration ξi, shear strain rate r, surrounding temperature (in most cases room temperature) T0, etc. To keep the parametric study within the scope of this chapter, the effect of only two parameters is studied, namely: shear strain rate r and surrounding

non-homogeneous solution in two/three dimensions [23].

Increased inhomogeneity in free volume concentration.

4. Parametric study

Figure 8.

Metallic Glasses

temperature T0.

80

Solution for isothermal homogeneous deformation case with various applied shear strain rates. (a) Normalized shear stress versus shear strain, and (b) free volume concentration versus shear strain.

the annihilation of free volume hence free volume always keeps on decreasing and softening does not occur at all.

Figure 10 shows solution for isothermal homogeneous deformation case with initial free volume concentration as 0.008 for different values of surrounding temperature. Figure 10a shows a variation of normalized shear stress with respect to shear strain while Figure 10b shows a variation of free volume concentration. From Figure 10, it can be seen that initially when shear strain is very low all the solutions match with each other. As surrounding temperature increases it supports annihilation process and hence free volume concentration decreases. But this decrease in free volume delays the softening and therefore maximum value of shear stress increases. Again this increased shear stress produces more and more free volume by dominating over annihilation process and hence the maximum value of free volume concentration is more in case of maximum surrounding temperature. In this case, also after softening steady state is achieved but values are different for different surrounding temperature. A similar effect will be seen in the case of inhomogeneous deformation. In the case of adiabatic deformation effect of change of applied strain rate will be similar but the effect of change of surrounding temperature may vary by some amount.

#### Figure 10.

Solution for isothermal homogeneous deformation case with various surrounding temperatures. (a) Normalized shear stress versus shear strain, and (b) free volume concentration versus shear strain.

As seen in this section solution varies drastically with the change in parameters, therefore a selection of parameters should be done very carefully.
