*3.1. Deformation Zone*

The sheet is deformed in the deformation zone between the entrance and the exit, where the initial thickness *h*0 is reduced to the final thickness *h*1, as shown in Figure 16. During symmetric rolling, as shown in Figure 16a, the circumferential speed of the bottom work roll *V*1 equals a circumferential speed of the top work roll *V*2, and the speed of the bottom surface of the sheet *V*1*s* equals the speed of the top surface of the sheet *V*2*s*. The neutral points on the top and bottom contact surfaces are at the same locations in the direction of rolling. Neutral points divide the deformation zone into two zones: a backward slip zone and a forward slip zone. In the backward slip zone, the contact friction forces are active, and in the forward slip zone, they are reactive. If pressure *p* or the shear strength *k* is known at the contact surface, the frictional stress *τ* can be expressed as *μp* in accordance with the Coulomb friction law or *mk* in accordance with the shear friction law. Here, *μ* is the Coulomb friction coefficient and *m* is the shear friction factor. During differential speed rolling when *V*1 is greater than *V*2, the neutral point associated with the slow work roll is shifted toward the entrance of the roll gap, while the neutral point associated with the fast work roll is moved toward the exit of the roll gap (Figure 16b). As a result, along with the backward slip zone and the forward slip zone, a third zone appears—the cross-shear zone—in which the forces of contact friction are oppositely directed.

Depending on length of the backward and forward slip zones from the side of the top and bottom work rolls, the following cases of differential speed rolling may occur:


(3) Limiting case, when only the cross-shear zone occupies the entire deformation zone. The first neutral point is at the entrance of the deformation zone, and the second neutral point is at the exit of the deformation zone. This case corresponds to the rolling–drawing process which requires a high front tension, as mentioned above (Figure 10b).

**Figure 16.** Schematic illustration of deformation zone during symmetric (**a**) and differential speed rolling (**b**).

In symmetric rolling, the pressure distribution at the work roll/sheet interface has a typical "friction hill". The rolling pressure increases gradually from the entrance of the deformation zone, reaches a maximum value corresponding to the neutral point and then gradually falls towards the exit of the deformation zone. In asymmetric rolling, the "friction hill" is cut off due to the cross-shear zone. As a result, rolling force is decreased. This is one of the main advantages of asymmetric rolling. Figure 17 shows pressure distribution along the deformation zone during symmetric and asymmetric rolling. Zuo et al. [55] reported that in the case of hot asymmetric rolling of aluminum alloy 5182, the rolling force was 5–30% lower than that of symmetric hot rolling. It was also confirmed by Tian et al. [56] that the extent of the cross-shear zone increases with the increase in the circumferential speed ratio, whereas the rolling force decreases.

**Figure 17.** Schematic illustration of pressure distribution along the deformation zone during symmetric and asymmetric rolling.

#### *3.2. Dissipation of Power and Speed Profiles in Deformation Zone*

The power provided by the faster work roll is represented in Figure 18 by the area of a-e-e'-d-a minus the area of e-b"-c-e'-e multiplied by the frictional stress, while that by the slower work roll is the area of a"-f-f'-d-a" minus the area of f-b'-c-f'-f multiplied by the frictional stress [57]. The power used for deformation by the faster work roll is the area of a'-g-e-e'-d-a' minus the area of e-b-c-e'-e multiplied by the frictional stress, while that by the slower work roll is the area of a'-f-f'-d-a' minus the area of f-g'-b-c-f'-f multiplied by the frictional stress. The power wasted by friction is the power provided by work rolls minus the power used for deformation. In particular, the power used for shear deformation due to the differential speed is represented by the area of a'-g-b-g'-a' multiplied by the frictional stress [57].

**Figure 18.** Schematic illustration of speed profiles in deformation zone during differential speed rolling, reproduced from [57], with permission from Elsevier, 2015.

Park [57] suggests Relation (4) that restricts the speed ratio to be less than the thickness ratio. If this relation is not satisfied, a portion of the power provided by the rolls becomes unnecessarily wasted as frictional heat, rather than used for deformation of the sheet. Additionally, an excessive tensile stress could be developed in the sheet, resulting in fracture if the material of the sheet is as brittle as a magnesium alloy.

$$\frac{V\_1}{V\_0} < \frac{h\_0}{h\_1} \tag{4}$$

where *V*0, *V*1 are the circumferential speeds of the work rolls; *h*0, *h*1 are the thickness of the sheet before and after rolling.

#### *3.3. Shear Strain and Equivalent Strain*

The level of a shear strain and equivalent strain plays a key role in terms of the possibility of using the asymmetric rolling process as a method of SPD. Two types of shear strain are known: pure shear and simple shear. Pure shear corresponds to symmetric rolling. During pure shear, the square is converted to a rectangle and the circle is converted to an ellipse (Figure 19). The thickness "AB" is compressed, and the length "AD" is elongated. The axes of the strain ellipsoid do not rotate (Figure 20).

**Figure 19.** Schematic illustration of pure shear during symmetric rolling.

**Figure 20.** FEM simulation (this work) of pure shear during symmetric rolling (**a**) and transformation of strain ellipsoid (in the middle layer) when passing through the deformation zone (**b**).

Simple shear corresponds to equal channel angular pressing (ECAP) (Figure 21). During simple shear, the square is converted to a parallelogram and the circle is converted to an ellipse (Figure 22). The thickness "AB" and the length "AD" are not changed. Simple shear involves rotation of the axes of the strain ellipsoid (Figure 23). Exact rotation leads to grain refinement and the formation of high angle boundaries during ECAP.

**Figure 21.** Schematic illustration of ECAP.

**Figure 22.** Schematic illustration of simple shear during ECAP.

**Figure 23.** FEM simulation (this work) of simple shear during ECAP (**a**) and rotation of the axes of the strain ellipsoid when passing through the deformation zone (**b**).

Simultaneous pure and simple shear combine the advantages of simple shear (rotation of the material as with ECAP) and the advantages of pure shear (compression and elongation of the material as with symmetric rolling). Simultaneous pure and simple shear (Figure 24) corresponds to asymmetric rolling. The square is converted to a parallelogram

and the circle is converted to an ellipse. The thickness "AB" is compressed, and the length "AD" is elongated. The ellipse is rotated (Figure 25) [58].

**Figure 25.** Results of FEM simulation (this work) of simultaneous pure and simple shear during asymmetric rolling (**a**) and rotation of the axes of the strain ellipsoid when passing through the deformation zone (**b**) (simulation conditions: room temperature, material AA5083, initial sheet thickness 1.0 mm, thickness reduction 60%, diameters of work rolls 500 mm, Coulomb friction coefficient 0.4, speed of the bottom work roll 43 mm/s, speed of the top work roll 100 mm/s, FEM code DEFORM 2D™).

The equivalent strain *ε* for asymmetric rolling process can be defined as von Mises strain [23,59–62]:

$$=\sqrt{\frac{2}{3}\mathfrak{e}\_{ij}\mathfrak{e}\_{ij}}\tag{5}$$

$$\overline{\varepsilon} = \sqrt{\frac{2}{9} \left[ \left( \varepsilon\_{11} - \varepsilon\_{22} \right)^2 + \left( \varepsilon\_{22} - \varepsilon\_{33} \right)^2 + \left( \varepsilon\_{33} - \varepsilon\_{11} \right)^2 + 6 \left( \varepsilon\_{12}^2 + \varepsilon\_{23}^2 + \varepsilon\_{31}^2 \right) \right]} \tag{6}$$

As, *ε*11 = *εxx*, *ε*22 = *<sup>ε</sup>yy*, *ε*33 = *εzz*, *ε*12 = *<sup>ε</sup>xy*, *ε*23 = *<sup>ε</sup>yz*, *ε*31 = *εzx*, then

*ε*

$$\mathbb{E} = \sqrt{\frac{2}{9} \left[ \left( \varepsilon\_{xx} - \varepsilon\_{yy} \right)^2 + \left( \varepsilon\_{yy} - \varepsilon\_{zz} \right)^2 + \left( \varepsilon\_{zz} - \varepsilon\_{xx} \right)^2 + 6 \left( \varepsilon\_{xy}^2 + \varepsilon\_{yz}^2 + \varepsilon\_{zx}^2 \right) \right]} \tag{7}$$

The deformation process during asymmetric rolling can be approximated by a twodimensional strain state of compressive strain along the normal direction *z* (*<sup>ε</sup>zz* and *εzz* = <sup>−</sup>*εxx*) together with simple shear strain along the rolling direction *x* (*<sup>ε</sup>xz*):

$$
\begin{bmatrix}
\varepsilon\_{xx} & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & \varepsilon\_{zz} = -\varepsilon\_{xx}
\end{bmatrix} + \begin{bmatrix}
0 & 0 & \varepsilon\_{xz} \\
0 & 0 & 0 \\
0 & 0 & 0
\end{bmatrix} = \begin{bmatrix}
\varepsilon\_{xx} & 0 & \varepsilon\_{xz} \\
0 & 0 & 0 \\
0 & 0 & \varepsilon\_{zz} = -\varepsilon\_{xx}
\end{bmatrix} \tag{8}
$$

Then, Equation (7) can be refined as:

$$\overline{\varepsilon} = \frac{2}{\sqrt{3}} \sqrt{\varepsilon\_{xx}^2 + \varepsilon\_{xz}^2} \tag{9}$$

The compressive strain component and simple shear strain component can be calculated, respectively, as follows: *εxx* = *ln h*0 *h*1 , *εxz* = *γxz* 2 = *γ*2 , where *h*0 is the initial sheet thickness, *h*1 is the final sheet thickness. Then, Equation (9) can be refined as:

$$\overline{\varepsilon} = \frac{2}{\sqrt{3}} \sqrt{\left(\ln \frac{h\_0}{h\_1}\right)^2 + \left(\frac{\gamma}{2}\right)^2} \tag{10}$$

If shear strain *γ* = 0, then Equation (10) transforms into the well-known equivalent strain *ε* for the symmetric rolling process:

$$\overline{\varepsilon} = \frac{2}{\sqrt{3}} \ln \frac{h\_0}{h\_1} \tag{11}$$

If compressive strain *εxx* = 0, when *h*0 = *h*1, then Equation (10) transforms into the well-known equivalent strain *ε* for simple shear:

$$
\overline{x} = \frac{\gamma}{\sqrt{3}}\tag{12}
$$

$$
\gamma = t \mathfrak{g} \mathfrak{p} \tag{13}
$$

where *γ* is the simple shear strain, *ϕ* is the shear angle (see Figure 22). Fueled by advances in severe plastic deformation, there has been much discussion about computing the equivalent strain *ε* in simple shear [63]. Since the increment in equivalent strain can be defined via the increment in plastic work, the authors of [63,64] argue for using the von Mises strain (11). On the other hand, Onaka [65] argues that one must use the Hencky relation [66]. The justification given by Onaka is the need to exclude rotations from the deformation gradient tensor in large simple shear deformations. Beygelzimer [67] showed that the equivalent strain of simple shear must be linear in *γ*. This approach gives evidence in favor of using the von Mises strain (12) for computing the equivalent strain in simple shear.

Saito et al. (1986) [68], Sakai et al. (1988) [69], Cui and Ohori (2000) [15] and Angella et al. (2013) [70] used the following equations for calculation of the equivalent strain *ε* and shear strain *γ* in asymmetric rolling:

$$\overline{\varepsilon} = \frac{2}{\sqrt{3}} \phi \ln \frac{1}{1 - r} \tag{14}$$

$$r = 1 - \frac{t\_1}{t\_0}, \phi = \sqrt{1 + \left\{ \frac{(1 - r)^2}{r(2 - r)} \tan \theta \right\}^2} \tag{15}$$

$$\gamma = 2\sqrt{\phi^2 - 1}\ln\frac{1}{1 - r} \tag{16}$$

where *t*0, *t*1 are the thickness of the sheet before and after rolling; *γ* is the shear strain; *θ* is the shear angle.

In 2005, Kang et al. [71] reported that the shear strain component *γ* imposed by asymmetric rolling with different diameters of the top and bottom work rolls could be calculated by the following equation:

$$\gamma = \frac{1}{\langle D \rangle} \left[ R\_1 \cos^{-1} \left( \frac{R\_1 - \Delta D / 2}{R\_1} \right) - R\_2 \cos^{-1} \left( \frac{R\_2 - \Delta D / 2}{R\_2} \right) \right] \tag{17}$$

where *D* = 12 (*<sup>D</sup>*1 + *<sup>D</sup>*2), Δ*D* = *D*1 − *D*2; *D*1 is the initial sheet thickness; *D*2 is the final sheet thickness; *R*1 is the larger radius of the work roll; *R*2 is the smaller radius of the work roll.

In 2008, Sidor et al. [60], by using the assumed shear strain *γ* during asymmetric rolling in accordance with Equation (17), reported that the equivalent von Mises strain could be calculated by the following equation:

$$\overline{\varepsilon} = \frac{\sqrt{2}}{3} \left[ 6 \left[ \ln \left( 1 - \frac{\left( h\_i - h\_f \right)}{h\_i} \right) \right]^2 + \frac{6}{\left( h\_i + h\_f \right)^2} \left[ R\_1 \cos^{-1} \left( \frac{2R\_1 - \left( h\_i - h\_f \right)}{2R\_1} \right) - R\_2 \cos^{-1} \left( \frac{2R\_2 - \left( h\_i - h\_f \right)}{2R\_2} \right) \right]^2 \right]^{1/2} \tag{18}$$

where *hi* is the initial sheet thickness, *hf* is the final sheet thickness, *R*1 is the larger radius of the work roll, *R*2 is the smaller radius of the work roll.

In 2016, Ma et al. [72] reported that the shear strain component *γ* imposed by asymmetric rolling with the same diameters of work rolls but different angular speeds could be calculated by the following equation:

$$\gamma = \frac{1}{h\_i + h\_f} R \cos^{-1} \left( 1 - \frac{h\_i - h\_f}{2R} \right) \left( 1 - \frac{v\_2}{v\_1} \right) \tag{19}$$

where *hi* is the initial sheet thickness, *hf* is the final sheet thickness, *R* is the radius of the work roll, *v*1 is the speed of the slower work roll, *v*2 is the speed of the faster work roll.

In 2015, Park J.-J. [57] reported that the shear strain component *γ* imposed by asymmetric rolling with different circumferential speeds could be calculated by the following equation:

$$\gamma = \frac{\Delta \mathbf{s}}{t\_{\text{ave}}} = \frac{(V\_{\text{us}} - V\_{\text{ls}})\Delta t}{t\_{\text{ave}}} = \frac{\mathbb{C}(V\_{\text{ur}} - V\_{\text{lr}})\Delta t'}{t\_{\text{ave}}} = \frac{\mathbb{C}'L}{t\_{\text{ave}}} \tag{20}$$

where Δ*s* is the displacement (see Figure 24); *tave* = *h*0+*h*1 2 is the average thickness of the sheet; *h*0 is the initial sheet thickness; *h*1 is the final sheet thickness; Δ*t* is the the time required for a material point on the fast surface of the sheet to pass through the deformation zone; *Vus*, *Vls* are the tangential speeds of the upper (faster) and lower (slower) surfaces of the sheet, respectively; *Vur*, *Vls* are the tangential speeds of the upper (faster) and lower (slower) work rolls, respectively; Δ*t* is the time required for a material point on the surface of the fast work roll to pass through the deformation zone; *L* is the length of the deformation zone; *C* varies from 0 to 0.7 and *C'* varies from 0 and 0.5 [57].

Substituting Equation (13) in Equation (10), the equivalent strain *ε* depending on the shear angle ϕ can be found for the asymmetric rolling process. Severe plastic deformation usually corresponds to equivalent strain *ε* ≥ 3 ... 4. This requires at least three or four passes of ECAP with a simple shear. This level can be obtained by single-pass asymmetric rolling with simultaneous pure and simple shear, when shear angle ϕ is no less than 80◦ (Figure 26) [73].

Taking *C'* = 0.5 in Equation (20) and simplifying that *L* = -*<sup>R</sup>*(*h*0 − *h*1), and equating Equations (13) and (20), the limiting shear angle *ϕmax* can be found:

$$\varphi\_{\text{max}} = \operatorname{arcctg} \frac{\sqrt{R \left(h\_0 - h\_1\right)}}{h\_0 + h\_1} \tag{21}$$

The influence of initial thickness *h*0 of the sheet and thickness reduction per pass ε on limiting shear angle *ϕ* during asymmetric rolling can be found with Equation (21) (Figure 27). From the graph, it follows that to create a shear angle ϕ of at least 80 degrees, it is necessary to make the thickness reduction per pass *ε* ≥ 50% when initial sheet thickness *h*0 = 1 mm, or *ε* ≥ 70% when *h*0 = 2 mm (in all cases, the diameter of work rolls is 300 mm).

The diameters of work rolls (cold rolling mills) at laboratory and industrial scales vary from 50 mm to 600 mm (Figure 28). The influence of diameter D of work rolls and thickness reduction per pass ε on limiting shear angle ϕ during asymmetric rolling is shown in Figure 29. From the graph, it follows that to create a shear angle *ϕ* of at least 80 degrees, it is necessary to make the thickness reduction per pass *ε* ≥ 50% when the diameter of work rolls D = 600 mm, or *ε* ≥ 70% when D = 300 mm (in all cases, the initial thickness of the sheet *h*0 = 2 mm).

**Figure 26.** Influence of the shear angle ϕ on the equivalent strain *ε* during symmetric rolling, asymmetric rolling and ECAP.

**Figure 27.** Influence of initial thickness *h*0 of the sheet and thickness reduction per pass *ε* on limiting shear angle *ϕ* during asymmetric rolling.

**Figure 28.** Diameter D (mm) of work rolls (cold rolling mills) at laboratory and industrial scales.

**Figure 29.** Influence of diameter D of work rolls and thickness reduction per pass *ε* on limiting shear angle *ϕ* during asymmetric rolling.

Hence, a very large equivalent strain *ε* ≥ 3 ... 4 during asymmetric rolling can be introduced with a large roll diameter (D ≥ 300 mm), high thickness reduction per pass (*ε* ≥ 50%) and thinner initial thickness of the sheet (*h*0 ≤ 2 mm). In this case, asymmetric rolling should be carried out on dry work rolls without lubrication (with high contact friction) and a high speed ratio of the work rolls (*SR* = 2 . . . 4).

### *3.4. Temperature Rise*

Since asymmetric rolling is performed with high contact friction and high thickness reduction per pass, the temperature increase in the deformation zone can be significant [74–76]. The temperature increase may affect the additional microstructural changes which are related to thermal softening deteriorating the mechanical properties. The model to predict the temperature increase of the sheet during differential speed rolling was proposed in [74]. The temperature increase might be affected by several factors (Table 1). The first factor is the total stored energy which is related to external deformation work. The second factor is the internal energy determined by the mechanical capability of materials, which is defined as the total area beneath the stress–strain curve. The other factors are the energy input caused by frictional heat and the energy loss by non-isothermal conditions. Considering the fact that 90% of the deformation work is converted into heat and 50% of the frictional heat is transferred to the sheet, the energy-based state equation during differential speed rolling is expressed as follows [74]:

$$
\rho \nabla V \Delta T = 0.9 \sigma \varepsilon V + 0.5 m \tau v A \Delta t - A h \Delta T \Delta t \tag{22}
$$

where *ρ* is the material density; *C* is the heat capacity; *V* is the main deformation volume; Δ*T* is the temperature increase in the main deformation volume; σ is the stress; *ε* is the strain; *m* is the friction factor; *τ* is the shear stress; *v* is the speed of the faster work roll; *A* is the contact area between the sheet and the faster work roll; Δ*t* is the deformation time; *h* is the heat transfer coefficient.

**Table 1.** Factors and their equations influencing the temperature increase of the sheet during differential speed rolling (based upon the energy-based state equation), reproduced from [74], with permission from Elsevier, 2014.


The temperature increase in the sheet during differential speed rolling can be represented as [74]:

$$
\Delta T = \frac{0.9\sigma\varepsilon + 0.5m\tau\upsilon\frac{A}{V}\Delta t}{\rho\text{C} + \frac{A}{V}h\Delta t} \tag{23}
$$

where *V* = *Lw<sup>h</sup>*0+*h*1 2 ; *A* = *Lw*; Δ*t* = *Lv* ; *h*0 is the initial sheet thickness; *h*1 is the final sheet thickness; *L* is the length of the deformation zone; *w* is the width of the sheet.

Since severe plastic deformation should be performed at relatively low temperatures *T* ≤ (0.3 ... 0.4)*Tmelt*, the temperature increase in the sheet during differential speed rolling, especially for Mg and Al alloys, must be taken into account.
