**1. Introduction**

An instantaneous plane strain rigid plastic solution is obtained for compression of an infinite wedge of orthotropic material confined between two rough plates, inclined at angle 2α, and which intersect in a line. This boundary value problem is ideal for studying qualitative mathematical properties of boundary value problems, including constitutive equations and boundary conditions. For, exact analytical or semi-analytical solutions can be found for many constitutive equations. In particular, such solutions have been presented in [1,2] for isotropic viscoplastic materials and in [3] for the double slip and rotation model. A description of this model can be found in [4].

The present paper provides an analytic solution for rigid plastic orthotropic material. It is assumed that the principal axes of anisotropy are straight lines through the apex of the wedge and orthogonal curves, which are of course circular arcs. This type of orthotropy is of practical interest [5–8] among many others. The paper focuses on qualitative features of the solution such as non-existence of the solution, singularity in the stress and velocity fields, appearance of a rigid region near the plates and transition between the regimes of sticking and sliding. The effect of plastic anisotropy on these features is discussed.

The stress and velocity fields are singular if the regime of sliding occurs in the case of the maximum friction law. A detailed asymptotic analysis of the solution is performed for this case. In particular, it is shown that the asymptotic behavior of the solution is in agreemen<sup>t</sup> with the general theory developed in [9].

An applied aspect of the solution found, is that it can be used in conjunction with the method for analysis and the design of flat-rolling proposed in [10]. It is known that solutions, found by means of this method, show a good comparison with experiment [11–13], and are used for verifying solutions found by means of other approximate methods [14,15]. The importance of developing fast approximate methods for the analysis and design of the process of rolling has been emphasized in [16].

### **2. Statement of the Problem**

Two semi-infinite rough plates rotate towards each other with angular velocity of magnitude *ω* about an axis *O* and compress a wedge of polar orthotropic material. The plates are inclined to each other at an angle 2 α (Figure 1). The boundary value problem consists of the instantaneous plane strain deformation of the wedge. The problem is solved in a system of plane polar coordinates (*r*, *θ*) with its origin at *O* and with *θ* = 0, taken as the perpendicular bisector of the angle 2 α. It is assumed that the principal axes of anisotropy coincide with coordinate curves of the coordinate system chosen. Then, *θ* = 0 is an axis of symmetry for the flow and it is sufficient to find the solution in the region *θ* ≥ 0. The components of the stress tensor referred to the polar coordinate system are denoted as *σrr*, *σθθ* and *σrθ*; and the components of the velocity vector as *ur* and *uθ*. There is no material flux through *O*.

**Figure 1.** Geometry of the boundary value problem.

Therefore, the radial velocity should satisfy the following condition:

$$
u\_{\mathcal{I}} = 0\tag{1}$$

at *r* = 0. By symmetry,

$$u\_{\theta} = 0\tag{2}$$

and

$$
\sigma\_{r\emptyset} = 0 \tag{3}
$$

at *θ* = 0. The circumferential velocity should also satisfy the condition:

$$
\mu\_{\theta} = -\omega r \tag{4}
$$

at *θ* = *α*. Finally, the friction law is taken in the form:

$$\begin{cases} \ u\_r = 0 \text{if} |\sigma\_{r\theta}| \le \tau\_f \\ \ \sigma\_{r\theta} = -\tau\_f \text{otherwise} \end{cases} \tag{5}$$

at *θ* = *α*. Here *τf* > 0 denotes the frictional stress at sliding. The magnitude of *τf* will be specified later. The sense of *σrθ* in (5) is dictated by the condition that *ur* ≥ 0 at the plate.

It is assumed that the material obeys Hill's quadratic yield criterion [17] and its associated flow rule. The elastic portion of strain is neglected. In the case of plane strain deformation of a polar orthotropic material, whose principal axes of anisotropy coincide with the coordinate curves of the polar coordinate system, the constitutive equations of the model are:

$$\frac{\left(\sigma\_{rr} - \sigma\_{\theta\theta}\right)^2}{4(1-c)} + \sigma\_{r\theta}^2 = T^2 \tag{6}$$

and

$$
\zeta\_{\tau\tau}^{\tau} = \lambda \frac{\left(\sigma\_{\tau\tau} - \sigma\_{\theta\theta}\right)}{2(1-c)}, \quad \zeta\_{\theta\theta}^{\tau} = -\lambda \frac{\left(\sigma\_{\tau\tau} - \sigma\_{\theta\theta}\right)}{2(1-c)}, \zeta\_{r\theta}^{\tau} = \lambda \sigma\_{r\theta} \tag{7}
$$

where (6) in the yield criterion and (7) is the associated flow rule. The quantity *T* is the shear yield stress in the coordinate system chosen, *c* is a constitutive parameter, *λ* is a non-negative multiplier, *ξrr*, *ξθθ*, *ξr<sup>θ</sup>* denote the components of the strain rate tensor. The parameter *c* can be expressed in terms of the yield stresses in the directions of the principal axes of anisotropy and can vary (theoretically) in the range − ∞ < *c* < 1 [17]. Eliminating *λ* between the equations in (7) yields:

$$
\xi\_{rr}^{\mathfrak{x}} + \xi\_{\theta\theta}^{\mathfrak{x}} = 0, \quad \frac{\mathfrak{z}\_{r\theta}}{\mathfrak{z}\_{rr}^{\mathfrak{x}} - \xi\_{\theta\theta}^{\mathfrak{x}}} = \frac{(1 - c)\sigma\_{r\theta}}{\sigma\_{rr} - \sigma\_{\theta\theta}}. \tag{8}
$$

It is evident that the first equation here is the equation of incompressibility. The strain rate components are expressed in terms of the velocity components as

$$\xi\_{r\tau}^{\tau} = \frac{\partial u\_r}{\partial r}, \quad \xi\_{\theta\theta} = \frac{1}{r}\frac{\partial u\_\theta}{\partial \theta} + \frac{u\_r}{r}, \quad \xi\_{r\theta} = \frac{1}{2}\left(\frac{1}{r}\frac{\partial u\_r}{\partial \theta} + \frac{\partial u\_\theta}{\partial r} - \frac{u\_\theta}{r}\right). \tag{9}$$

The system of Equations (6), (8) and (9) are supplemented by the stress equilibrium equations:

$$\frac{\partial \sigma\_{rr}}{\partial r} + \frac{1}{r} \frac{\partial \sigma\_{r\theta}}{\partial \theta} + \frac{\sigma\_{rr} - \sigma\_{\theta\theta}}{r} = 0, \quad \frac{\partial \sigma\_{r\theta}}{\partial r} + \frac{1}{r} \frac{\partial \sigma\_{\theta\theta}}{\partial \theta} + \frac{2\sigma\_{r\theta}}{r} = 0. \tag{10}$$

In total, there are five unknowns (three components of the stress tensor and two components of the velocity vector). The equations to solve are (6), (8) and (10). It is understood here that the components of the strain rate tensor in (8) should be eliminated by means of (9). The solution should satisfy the conditions (1) to (5).

### **3. General Stress Solution**

The yield criterion (6) is satisfied by the following substitution:

$$
\sigma\_{rr} = \sigma + T\sqrt{1-c}\cos 2\varphi, \quad \sigma\_{\theta\theta} = \sigma - T\sqrt{1-c}\cos 2\varphi, \quad \sigma\_{r\theta} = -T\sin 2\varphi \tag{11}
$$

where *σ* and *ϕ* are new unknown functions of *r* and *θ*. The direction of flow dictates that *σrθ* ≤ 0 and *σrr* − *σθθ* ≥ 0. Then, it is immediate from (11) that:

$$0 \le \varphi \le \frac{\pi}{4}.\tag{12}$$

Using (11) and (12) the boundary condition (3) transforms to:

$$
\varphi = 0 \tag{13}
$$

at *θ* = 0. Substituting (11) into (10) gives:

$$\begin{cases} \frac{\partial \sigma}{\partial r} - 2T\sqrt{1-\varepsilon}\sin 2\varrho \frac{\partial \varrho}{\partial r} - \frac{2T\cos 2\varrho}{r} \frac{\partial \varrho}{\partial \theta} + \frac{2T\sqrt{1-\varepsilon}\cos 2\varrho}{r} = 0, \\\ -2T\cos 2\varrho \frac{\partial \varrho}{\partial r} + \frac{\partial \sigma}{r\partial \theta} - \frac{2T\sqrt{1-\varepsilon}\sin 2\varrho}{r} \frac{\partial \varrho}{\partial \theta} - \frac{2T\sin 2\varrho}{r} = 0. \end{cases} \tag{14}$$

A standard assumption made in similar problems of the classical theory of plasticity is that *ϕ* is independent of *r* [17]. In this case, the equations in (14) become:

$$\frac{\partial \sigma}{\partial r} - \frac{2T \cos 2\varrho}{r} \frac{d\varrho}{d\theta} + \frac{2T \sqrt{1 - c} \cos 2\varrho}{r} = 0, \quad \frac{\partial \sigma}{2T \partial \theta} - \sqrt{1 - c} \sin 2\varrho \frac{d\varrho}{d\theta} - \sin 2\varrho = 0. \tag{15}$$

The first equation can be immediately integrated to give:

$$\frac{\sigma}{2T} = \left(\frac{d\varphi}{d\theta} - \sqrt{1 - c}\right) \cos 2\varphi \ln\left(\frac{r}{r\_0}\right) + \frac{\sigma\_0(\theta)}{2T}.\tag{16}$$

Here *r*0 is a constant introduced for convenience and *<sup>σ</sup>*0(*θ*) is an arbitrary function of *θ*. Substituting (16) into the second equation in (15) yields:

$$\frac{d}{d\theta}\left[\left(\frac{d\rho}{d\theta} - \sqrt{1 - c}\right)\cos 2\rho\right] \ln\left(\frac{r}{r\_0}\right) = -\sqrt{1 - c}\sin 2\rho \frac{d\rho}{d\theta} + \sin 2\rho - \frac{d\sigma\_0}{2Td\theta}.\tag{17}$$

Since the right-hand side of this equation is independent of *r*, the coefficient of ln(*r*/*<sup>r</sup>*0) on the left-hand side must vanish. Then, Equation (17) results in the following two equations:

$$
\left(\frac{d\varphi}{d\theta} - \sqrt{1 - c}\right)\cos 2\varphi = K\rho\sqrt{1 - c}, \quad \frac{d\sigma\_0}{2Td\theta} = \left(-\sqrt{1 - c}\frac{d\varphi}{d\theta} + 1\right)\sin 2\varphi. \tag{18}
$$

Here *K*0 is a constant of integration. Equation (16) becomes:

$$\frac{\sigma}{2T} = K\_0 \sqrt{1 - c} \ln\left(\frac{r}{r\_0}\right) + \frac{\sigma\_0(\theta)}{2T}.\tag{19}$$

The second equation in (18) can be rewritten as:

$$\frac{d\sigma\_0}{2Td\rho} = \left(-\sqrt{1-c} + \frac{d\theta}{d\rho}\right)\sin 2\rho.$$

Eliminating in this equation the derivative *dθ*/*dϕ* by means of the first equation in (18) leads to:

$$
\frac{d\sigma\_0}{2Td\rho} = \frac{1}{\sqrt{1-c}} \left[ 1 - c + \frac{\cos 2\varrho}{(K\_0 + \cos 2\varrho)} \right] \sin 2\varrho.
$$

$$
\frac{\sigma\_0}{2T} = \frac{1}{2\sqrt{1-c}} [-\cos 2\varrho + K\_0 \ln(K\_0 + \cos 2\varrho) + K\_1] \tag{20}
$$

 

gives:

where *K*1 is constant of integration.

It is seen from (12) and (13) that *dϕ*/*dθ* > 0 at *ϕ* = 0. Therefore, it follows from the first equation in (18) that

$$K\_0 > -1.\tag{21}$$

The first equation in (18) can be integrated to give:

$$\theta \sqrt{1-c} = \varrho - K\_0 \text{arctanh}\left[\sqrt{\frac{1-K\_0}{1+K\_0}} \tan \varrho t\right] \left(1 - K\_0^2\right)^{-1/2} \tag{22}$$

if |*<sup>K</sup>*0| < 1,

Integrating

$$\theta \sqrt{1-c} = \phi - K\_0 \arctan\left[\sqrt{\frac{K\_0 - 1}{K\_0 + 1}} \tan \phi\right] \left(K\_0^2 - 1\right)^{-1/2} \tag{23}$$

if *K*0 > 1, and

$$
\theta \sqrt{1-c} = \varphi - \frac{\tan \varphi}{2} \tag{24}
$$

if *K*0 = 1. The solution for an important special case of (22), *K*0 = 0, is represented as

$$
\theta \sqrt{1-c} = \varphi.\tag{25}
$$

The constant *K*0 cannot be determined without the solution for velocity.

### **4. General Velocity Solution**

The velocity components may be represented as:

$$u\_r = \frac{\omega r}{2} \frac{d\lg(\varphi)}{d\theta} \quad \text{and} \qquad u\_\theta = -\omega r \lg(\varphi) \,. \tag{26}$$

The condition (1) and the first equation in (8) are then automatically satisfied for any choice of the function *g*(*ϕ*). Equations (9) and (26) combine to give:

$$\xi\_{rr} = \frac{\omega}{2} \frac{d\mathfrak{g}(\mathfrak{q})}{d\theta}, \quad \tilde{\xi}\_{\theta\theta} = -\frac{\omega}{2} \frac{d\mathfrak{g}(\mathfrak{q})}{d\theta}, \quad \tilde{\xi}\_{r\theta} = \frac{\omega}{4} \frac{d^2\mathfrak{g}(\mathfrak{q})}{d\theta^2}. \tag{27}$$

Substituting (11) and (27) in the second equation in (8) yields:

$$\frac{d^2\mathcal{S}}{d\theta^2} + 2\sqrt{1-c}\tan 2\varphi \frac{d\mathcal{g}}{d\theta} = 0\tag{28}$$

or

$$\frac{dG}{d\theta} + 2\sqrt{1 - c}\tan 2\varphi G = 0\tag{29}$$

where *G* = *dg*/*dθ*. Replacing in (29) differentiation with respect to *θ* with differentiation with respect to *ϕ* by means of the first equation in (18) results in:

$$\frac{dG}{d\varphi} = -\frac{2\sin 2\varphi}{(K\_0 + \cos 2\varphi)} G. \tag{30}$$

Integrating gives

$$G = G\_0 (K\_0 + \cos 2\varphi). \tag{31}$$

Here *G*0 is constant of integration. The definition for G and (31) combine to give:

$$\frac{d\mathbf{g}}{d\theta} = \mathbf{G}\_0 (\mathbf{K}\_0 + \cos 2\varphi). \tag{32}$$

Replacing here differentiation with respect to *θ* with differentiation with respect to *ϕ* by means of the first equation in (18) results in:

$$\frac{d\mathcal{S}}{d\boldsymbol{\varrho}} = \frac{G\_0 \cos 2\boldsymbol{\varrho}}{\sqrt{1-\boldsymbol{c}}}.\tag{33}$$

It is seen from (2), (13) and (26) that *g* = 0 at *ϕ* = 0. The solution of Equation (32) satisfying this condition is:

$$g = \frac{G\_0 \sin 2\varphi}{2\sqrt{1-\varepsilon}}.\tag{34}$$

Substituting (33) into (26) and then the resulting expression for the circumferential velocity into (4) yields:

$$G\_0 \sin 2\varphi\_w = 2\sqrt{1-c} \tag{35}$$

where *ϕw* is the value of *ϕ* at *θ* = *α*. The dependence of *ϕw* on *α* follows from the solution of the first equation in (18).

To complete the solution of the boundary value problem, it is necessary to satisfy the boundary condition (5).

### **5. Solution of the Boundary Value Problem**

The boundary condition (5) comprises two friction regimes, sticking and sliding. These regimes should be treated separately.

### *5.1. Regime of Sticking*

In this regime, the boundary condition (5) becomes *ur* = 0 at *θ* = *α*. It is seen from the definition for G and (26) that this condition is equivalent to the condition *G* = 0 at *θ* = *α*. Then, it follows from (31) that:

$$\mathbb{K}\_0 = -\cos 2\varphi\_w. \tag{36}$$

In this case the dependence of *θ* on *ϕ* is given by (22). Eliminating *K*0 in (22) by means of (36), it is possible to find that the argumen<sup>t</sup> of the inverse hyperbolic tangent function is equal to 1 at *θ* = *α*. Therefore, the left-hand side of (22) approaches infinity (or negative infinity) unless *K*0 = 0. In the latter case, it is more convenient to use the solution (25). It follows from this solution, (35) and (36) that

$$\varphi\_{\rm tr} = \frac{\pi}{4}, \quad \mathcal{G}\_0 = 2\sqrt{1-c} \text{ and } a = \frac{\pi}{4\sqrt{1-c}} = a\_{\rm cr}. \tag{37}$$

The solution at sticking is possible only if *α* and c satisfy the third equation. Another restriction on the existence of the solution at sticking is that the shear stress at *θ* = *α* is less or equal to *τf* involved in (5). It is seen from the first equation in (37) and (11) that *σrθ* = − *T* at *θ* = *α* if the regime of sticking occurs. Since *T* is the maximum possible value of the shear stress in the polar coordinate system, a necessary condition for the existence of the regime of sticking is that *τf* = *T*. If *τf* < *T* then no solution at sticking exists.

Assume that *τf* = *T*. The relation between *α* and *c* in (37) has been derived assuming that plastic yielding occurs in the region 0 ≤ *θ* ≤ *α*. In the case of rigid/plastic solids, rigid regions may appear. In the case under consideration, the solution at sticking is possible if *α* > *αcr* and the material in the region *α* ≥ *θ* ≥ *αcr* is rigid. It worthy of note that the stress solution at *K*0 = 0 given in Section 3 is valid in the rigid region. Therefore, the yield criterion is not violated in the range *α* ≥ *θ* ≥ *αcr* and the solution is complete.

### *5.2. Regime of Sliding*

It is convenient to consider two cases, *τf* = *T* and *τf* < *T*, separately. Assume that *τf* = *T* and *α* < *αcr*. Then, no solution at sticking exists and it is necessary to find the solution at sliding. It follows from (5), (11) and (35) that the first and second equations in (37) are valid. The equation for determining *K*0 follows from (22) or (23). It is however convenient to start with the special case *K*0 = 1. In this case Equation (24) is valid. Therefore, this special case occurs only if *α* and c satisfy the following equation:

$$\alpha = \frac{(\pi - 2)}{4\sqrt{1 - c}} = \alpha\_5. \tag{38}$$

It is evident from (37) and (38) that *αs* < *αcr*. Equation (22) is valid in the range *αs* < *α* < *αcr*. In this case, the equation for *K*0 is

$$\ln\sqrt{1-c} = \frac{\pi}{4} - K\_0 \operatorname{arctanh}\left[\sqrt{\frac{1-K\_0}{1+K\_0}}\right] \left(1-K\_0^2\right)^{-1/2}.\tag{39}$$

Equation (23) is valid in the range 0 < *α* < *αs*. In this case, the equation for *K*0 is

$$
\alpha \sqrt{1-c} = \frac{\pi}{4} - K\_0 \arctan\left[\sqrt{\frac{K\_0 - 1}{K\_0 + 1}}\right] \left(K\_0^2 - 1\right)^{-1/2}.\tag{40}
$$

Equations (39) and (40) should be solved numerically.

Prandtl's friction law reads *τf* = *mT* where 0 ≤ *m* ≤ 1. The case *m* = 1 has been treated above. Therefore, assume that *m* < 1. In this case, no solution at sticking exists. The friction law (5) becomes *σrθ* = −*mT* at *θ* = *α* (or *ϕ* = *ϕw*). Then, it follows from (11) that:

$$
\varphi\_{\mathcal{O}^{\mathcal{O}}} = \frac{1}{2} \arcsin m. \tag{41}
$$

The value of *G*0 is found from (35) and (41) as:

$$G\_0 = \frac{2\sqrt{1-c}}{m}.\tag{42}$$

The equation for determining *K*0 follows from (22) or (23). As before, it is more convenient to consider special cases first. The values of *αcr* and *αs* are now determined from (24) and (25) as:

$$
\mu\_{\mathcal{U}} = \frac{\varrho\_w}{\sqrt{1-\mathcal{c}}} \quad \text{and} \quad \mu\_s = \left(\varrho\_w - \frac{\tan \varrho\_w}{2}\right) \frac{1}{\sqrt{1-\mathcal{c}}}.\tag{43}
$$

In these equations, *ϕw* should be eliminated by means of (41). Equation (22) is valid in the ranges *αs* < *α* < *αcr* and *α* > *αcr*. In this case, the equation for *K*0 is:

$$\alpha \sqrt{1-c} = \rho\_w - K\_0 \operatorname{arctanh} \left[ \sqrt{\frac{1-K\_0}{1+K\_0}} \tan \phi\_w \right] \left( 1 - k\_0^2 \right)^{-1/2}. \tag{44}$$

The value of *K*0 is positive in the range *αs* < *α* < *αcr* and negative in the range *α* > *αcr*. Equation (23) is valid in the ranges 0 < *α* < *αs*. In this case, the equation for *K*0 is:

$$a\sqrt{1-c} = q\_{\overline{w}} - K\_0 \arctan\left[\sqrt{\frac{K\_0 - 1}{K\_0 + 1}} \tan \varphi\_w\right] \left(K\_0^2 - 1\right)^{-1/2}.\tag{45}$$
