**2. Statement of the Problem**

What is considered is the process of bending under tension due to large strains under plane strain conditions in the classical formulation proposed in [28], where a model of rigid perfectly plastic material was adopted. A schematic diagram of the process is shown in Figure 1. The cross section of a sheet is rectangular at the initial instant. The initial thickness of the sheet is *H*. As the deformation proceeds, the initially straight lines *A*1*B*<sup>1</sup> and *C*1*D*<sup>1</sup> become circular arcs *AB* and *CD*, respectively. The initially straight lines *B*1*C*<sup>1</sup> and *A*1*D*<sup>1</sup> remain straight (lines *BC* and *AD*, respectively). The radii of circular arcs *AB* and *CD* are denoted as *rAB* and *rCD*, respectively. A Cartesian coordinate system (*x*, *y*) is introduced. Its *x*-axis coincides with the axis of symmetry of the process. Its origin coincides with the intersection of the outer radius of the sheet and the axis of symmetry. Due to symmetry, it is sufficient to consider the domain *y* ≥ 0. The orientation of line *BC* relative to the *x*-axis is *θ*0. The actual distribution of stresses over *BC* is replaced with the bending moment *M* and the tensile force *F*. Both are per unit length. The surface *AB* is traction-free. Equilibrium requires that the tensile force is compensated with pressure over the surface *CD*. It is assumed that this pressure is uniformly distributed. ‐ i ‐ ‐ ‐ ‐ ‐ ‐

**Figure 1.** Schematic diagram of the bending process: (**a**) initial configuration and (**b**) intermediate and final configurations.

Then,

$$P = \frac{F}{r\_{\rm CD}}.\tag{1}$$

The constitutive equations include an isotropic pressure-independent yield criterion (i.e., the linear invariant of the stress tensor does not affect plastic yielding) and its associated flow rule. The elastic portion of the strain tensor is neglected. The present paper assumes that the tensile yield stress is a function of the equivalent strain, *εeq*, and the equivalent strain rate, *ξeq*, obtained by fitting any experimental data. Then, the plane-strain yield criterion can be represented as

$$|\sigma\_1 - \sigma\_2| = \frac{2\sigma\_0}{\sqrt{3}} \Phi(\varepsilon\_{\epsilon\eta}, \mathfrak{z}\_{\epsilon\eta}). \tag{2}$$

Here *σ*<sup>1</sup> and *σ*<sup>2</sup> are the principal stresses, *σ*<sup>0</sup> is a reference stress, and Φ *εeq*, *ξeq* is an arbitrary function of its arguments. The associated flow rule is equivalent to the equation of incompressibility and the condition that the stress and strain rate tensors are coaxial. In the case under consideration, the equivalent strain rate is determined as

$$
\xi\_{eq} = \sqrt{\frac{2}{3}} \sqrt{\xi\_1^2 + \xi\_2^2} \tag{3}
$$

where *ξ*<sup>1</sup> and *ξ*<sup>2</sup> are the principal strain rates. The equivalent plastic strain is given by

$$\frac{d\varepsilon\_{\epsilon\eta}}{dt} = \mathfrak{f}\_{\epsilon\eta}.\tag{4}$$

here *d*/*dt* denotes the convected derivative.

## **3. Kinematics**

The approach for solving plane-strain bending problems proposed in [13] consists of two major steps. The first step describes the kinematics of the process. This step is the same for all isotropic incompressible materials. The second step concerns the determination of the stresses, bending moment, and tensile force. This step depends on the constitutive equations chosen. The present section outlines the basic relations from the first step derived in [13]. These relations are necessary for the second step.

Let (*ζ*, *η*) be the Lagrangian coordinate system satisfying the conditions *x* = *Hζ* and *y* = *Hη* at the initial instant. Then, *ζ* = 0 on *AB* and *ζ* = −1 on *CD* throughout the process of deformation (Figure 1). The transformation equations between the (*ζ*, *η*)− and (*x*, *y*)− coordinate systems are

$$\frac{x}{H} = \sqrt{\frac{\zeta}{a} + \frac{s}{a^2}} \cos(2a\eta) - \frac{\sqrt{s}}{a} \quad \text{and} \quad \frac{y}{H} = \sqrt{\frac{\zeta}{a} + \frac{s}{a^2}} \sin(2a\eta). \tag{5}$$

Here *a* is a time-like variable and *s* is a function of *a*. At the initial instant, *a* = 0. The function *s*(*a*) should be found from the solution. This function must satisfy the condition

$$s = \frac{1}{4} \tag{6}$$

at *a* = 0. The coordinate lines of the Lagrangian coordinate system coincide with principal strain rate trajectories [13]. Then, the coordinate lines of the Lagrangian coordinate system coincide with principal stress trajectories for the coaxial models. Therefore, the contour of the cross-section is free of shear stresses throughout the process of deformation.

It is convenient to introduce a cylindrical coordinate system (*r*, *θ*) by the transformation equations:

$$\frac{x}{H} + \frac{\sqrt{s}}{a} = \frac{r\cos\theta}{H} \quad \text{and} \quad \frac{y}{H} = \frac{r\sin\theta}{H}.\tag{7}$$

Equations (5) and (7) combine to give

$$
\frac{r}{H} = \sqrt{\frac{\zeta}{a} + \frac{s}{a^2}} \quad \text{and} \quad \theta = 2a\eta. \tag{8}
$$

Therefore, the radial, *σr* , and circumferential, *σ<sup>θ</sup>* , stresses are the principal stresses. In what follows, it is assumed that *σ<sup>r</sup>* ≡ *σ*<sup>1</sup> and *σ<sup>θ</sup>* ≡ *σ*2. Accordingly, *ξ<sup>r</sup>* ≡ *ξ*<sup>1</sup> and *ξ<sup>θ</sup>* ≡ *ξ*2. Here *ξr* is the radial strain rate and *ξ<sup>θ</sup>* is the circumferential strain rate. It follows from (8) that

$$\frac{r\_{AB}}{H} = \frac{\sqrt{s}}{a} \quad \text{and} \quad \frac{r\_{CD}}{H} = \sqrt{\frac{s}{a^2} - \frac{1}{a}}.\tag{9}$$

It is straightforward to find the principal strain rates using (5). In particular,

$$\mathfrak{f}\_r = -\frac{(\zeta + ds/da)}{2(\zeta a + s)} \frac{da}{dt} \quad \text{and} \quad \mathfrak{f}\_\theta = \frac{(\zeta + ds/da)}{2(\zeta a + s)} \frac{da}{dt} . \tag{10}$$

Equations (3) and (10) combine to give

$$
\tilde{\zeta}\_{\varepsilon q} = \frac{|\zeta + d\mathbf{s}/da|}{\sqrt{3}(\zeta a + \mathbf{s})} \frac{da}{dt}. \tag{11}
$$

The equivalent strain rate vanishes at the neutral line. Therefore, it follows from (11) that the *ζ*− coordinate of the neutral line is

$$
\zeta = \zeta\_n = -\frac{ds}{da}.\tag{12}
$$

The neutral line moves through the material as the deformation proceeds. It is known that it moves towards *CD* in the case of strain-hardening materials. Assume that it is true in the case of the constitutive equations under consideration. Specific calculations below will verify this assumption. The initial condition for Equation (4) is

$$
\varepsilon\_{\epsilon\eta} = 0 \tag{13}
$$

at *a* = 0. Let *ζn*<sup>0</sup> be the value of *ζ<sup>n</sup>* at *a* = 0. It is necessary to consider three regions for determining the equivalent strain. These regions are: 0 ≥ *ζ* ≥ *ζn*<sup>0</sup> (Region 1), *ζn*<sup>0</sup> ≥ *ζ* ≥ *ζ<sup>n</sup>* (Region 2), and *ζ<sup>n</sup>* ≥ *ζ* ≥ 0 (Region 3). In Region 1, *ζ* − *ζ<sup>n</sup>* = *ζ* + *ds*/*da* ≥ 0 throughout the process of deformation. Therefore, one can substitute (11) into (4) and immediately integrate the resulting equation using (13) and (6) to get

$$
\varepsilon\_{\epsilon\eta} = \frac{1}{\sqrt{3}} \ln[4(\zeta a + s)]. \tag{14}
$$

In Region 3, *ζ* − *ζ<sup>n</sup>* = *ζ* + *ds*/*da* ≤ 0 throughout the process of deformation. Therefore, one can substitute (11) into (4) and immediately integrate the resulting equation using (13) and (6) to get

$$
\varepsilon\_{\epsilon\eta} = -\frac{1}{\sqrt{3}} \ln[4(\zeta a + s)].\tag{15}
$$

In Region 2, *ζ* − *ζ<sup>n</sup>* = *ζ* + *ds*/*da* ≥ 0 instantly. However, strain reversal occurs in this region. Consider any *ζ*− curve of Region 2. Let *a<sup>c</sup>* be the value of *a* at which this curve coincides with the neutral line and *s<sup>c</sup>* is the corresponding value of *s*. Evidently, both *a<sup>c</sup>* and *sc* are functions of *ζ*. Then, the equivalent strain in Region 2 is given by

$$\varepsilon\_{eq} = \frac{1}{\sqrt{3}} \ln \left\{ \frac{\zeta a + s}{4 \left[ \zeta a\_c(\zeta) + s\_c(\zeta) \right]^2} \right\}. \tag{16}$$

#### **4. Stress Solution**

The only stress equilibrium equation that is not satisfied automatically in the cylindrical coordinate system is

$$\frac{\partial \sigma\_r}{\partial r} + \frac{\sigma\_r - \sigma\_\theta}{r} = 0.\tag{17}$$

The yield criterion (2) can be represented as

$$
\sigma\_{\theta} - \sigma\_{r} = \frac{2\beta\sigma\_{0}}{\sqrt{3}} \Phi(\varepsilon\_{e\eta}, \xi\_{e\eta}).\tag{18}
$$

Here and in what follows, *β* = +1 in regions 1 and 2, and *β* = −1 in region 3. Equations (17) and (18) combine to give

$$\frac{\partial \sigma\_r}{\partial \sigma\_0 \partial r} = \frac{2\beta}{\sqrt{3}r} \Phi(\varepsilon\_{eq}, \mathfrak{f}\_{eq}). \tag{19}$$

Using (8), one can replace differentiation with respect to *r* with differentiation with respect to *ζ*. As a result,

$$\frac{\partial \sigma\_r}{\partial \sigma\_0 \partial \zeta} = \frac{\beta a}{\sqrt{3}(\zeta a + s)} \Phi(\varepsilon\_{\epsilon \eta \prime}, \mathfrak{z}\_{\epsilon \eta}) . \tag{20}$$

One can rewrite Equation (11) as

$$\chi = \frac{\oint \gamma(\zeta + ds/da)}{\sqrt{\Im}(\zeta a + s)}. \tag{21}$$

Here *χ* = *ξeq*/*ξ*0, *ξ*<sup>0</sup> is a characteristic strain rate, and *γ* = *ξ* −1 0 *da*/*dt*. Equations (14) and (15) can be rewritten as

$$
\varepsilon\_{\epsilon\eta} = \frac{\beta}{\sqrt{3}} \ln[4(\zeta a + s)]. \tag{22}
$$

One can eliminate *ζ* in (21) using (22). As a result,

$$\chi = \frac{\beta \gamma \left[ \exp \left( \sqrt{3} \beta \varepsilon\_{eq} \right) - 4s + 4ads / da \right]}{\sqrt{3} a \exp \left( \sqrt{3} \beta \varepsilon\_{eq} \right)}. \tag{23}$$

Using (22), one can replace differentiation with respect to *ζ* with differentiation with respect to *εeq*. Then, Equation (20) becomes

$$\frac{\partial \sigma\_r}{\sigma\_0 \partial \varepsilon\_{\varepsilon q}} = \Phi(\varepsilon\_{\varepsilon q}, \,\, \xi\_{\varepsilon q}). \tag{24}$$

Using (23), one can represent the function Φ *εeq*, *ξeq* as

$$\Phi(\varepsilon\_{eq}, \mathfrak{z}\_{eq}) = \Phi\left[\varepsilon\_{eq}, \mathfrak{z}\_0 \chi(\varepsilon\_{\mathfrak{e}q})\right] = \Lambda(\varepsilon\_{\mathfrak{e}q}).\tag{25}$$

Equations (24) and (25) combine to give

$$\frac{\partial \sigma\_r}{\partial \sigma\_0 \partial \varepsilon\_{\epsilon q}} = \Lambda \left( \varepsilon\_{\epsilon q} \right). \tag{26}$$

Let *ε*<sup>1</sup> be the value of *εeq* on surface *AB* (Figure 1). It follows from (14) that

$$
\varepsilon\_1 = \frac{1}{\sqrt{3}} \ln(4s). \tag{27}
$$

Since *σ<sup>r</sup>* = 0 on surface *AB*, the distribution of the radial stress in region 1 is determined from (26) and (27) as

$$\frac{\sigma\_r}{\sigma\_0} = \int\_{\varepsilon\_1}^{\varepsilon\_{c\eta}} \Lambda\_1(\mu) d\mu. \tag{28}$$

Here *µ* is a dummy variable of integration and Λ<sup>1</sup> *εeq* denotes the function Λ *εeq* when *β* = +1 in (23). The value of *εeq* at *ζ* = *ζn*<sup>0</sup> is determined from (14) as

$$
\varepsilon\_0 = \frac{1}{\sqrt{3}} \ln[4(\zeta\_{n0}a + s)].\tag{29}
$$

Then, the radial stress at *ζ* = *ζn*<sup>0</sup> is found from (28) in the form

$$\frac{\sigma\_r^{(0)}}{\sigma\_0} = \int\_{\varepsilon\_1}^{\varepsilon\_0} \Lambda\_1(\varepsilon\_{eq}) d\varepsilon\_{eq}. \tag{30}$$

Let *ε*<sup>3</sup> be the value of *εeq* on surface *CD* (Figure 1). It follows from (15) that

$$
\varepsilon\_3 = -\frac{1}{\sqrt{3}} \ln[4(s-a)].\tag{31}
$$

Using (9), one can represent (1) as

$$p = \frac{P}{\sigma\_0} = \frac{fa}{\sqrt{s - a}}\tag{32}$$

where *f* = *F*/(*σ*0*H*). It follows from (26), (31), and (32) that the distribution of the radial stress in region 3 is

$$\frac{\sigma\_r}{\sigma\_0} = \int\_{\varepsilon\_3}^{\varepsilon\_{tq}} \Lambda\_3(\mu) d\mu - \frac{fa}{\sqrt{s-a}}.\tag{33}$$

Here Λ<sup>3</sup> *εeq* denotes the function Λ *εeq* when *β* = −1 in (23). The value of *εeq* at *ζ* = *ζ<sup>n</sup>* is determined from (12) and (15) as

$$\varepsilon\_{\rm ll} = -\frac{1}{\sqrt{3}} \ln \left[ 4 \left( s - a \frac{ds}{da} \right) \right]. \tag{34}$$

Then, the radial stress at *ζ* = *ζ<sup>n</sup>* is found from (33) in the form

$$\frac{\sigma\_r^{(n)}}{\sigma\_0} = \int\_{\varepsilon\_3}^{\varepsilon\_n} \Lambda\_3(\varepsilon\_{eq}) d\varepsilon\_{eq} - \frac{fa}{\sqrt{s-a}}.\tag{35}$$

In region 2, Equation (20) becomes

$$\frac{\partial \sigma\_r}{\partial \sigma\_0 \partial \zeta} = \frac{a \Lambda\_2(\zeta)}{\sqrt{3}(\zeta a + s)}. \tag{36}$$

Here the function Λ2(*ζ*) is determined by replacing *εeq* and *χ* in Φ *εeq*, *ξ*0*χ* with the right-hand sides of Equations (16) and (21), respectively. The radial stress must be continuous at *ζ* = *ζ<sup>n</sup>* and *ζ* = *ζn*0. Therefore, using (35), one can represent the solution of Equation (36) as

$$\frac{\sigma\_r}{\sigma\_0} = \frac{a}{\sqrt{3}} \int\_{\frac{\pi}{6}a}^{\frac{\pi}{3}} \frac{\Lambda\_2(\mu)}{(\mu a + s)} d\mu + \int\_{\varepsilon\_3}^{\varepsilon\_n} \Lambda\_3(\varepsilon\_{eq}) d\varepsilon\_{eq} - \frac{fa}{\sqrt{s - a}}.\tag{37}$$

Equations (30) and (37) combine to give

$$\int\_{\varepsilon\_1}^{\varepsilon\_0} \Lambda\_1(\varepsilon\_{eq}) d\varepsilon\_{eq} - \frac{a}{\sqrt{3}} \int\_{\frac{\varepsilon\_0}{\varepsilon\_0}}^{\zeta\_{m0}} \frac{\Lambda\_2(\zeta)}{(\zeta a + s)} d\zeta - \int\_{\varepsilon\_3}^{\varepsilon\_0} \Lambda\_3(\varepsilon\_{eq}) d\varepsilon\_{eq} + \frac{fa}{\sqrt{s - a}} = 0. \tag{38}$$

This equation connects *a*, *s*, and *ds*/*da*. Therefore, it can be considered an ordinary differential equation to find *s* as a function of *a*. Its solution must satisfy the condition (6).

Once Equation (38) has been solved, the distribution of the radial stress in all three regions is readily determined from (28), (33), and (37). Equations (28) and (33) should be used in conjunction with Equations (14) and (15), respectively. The distribution of the circumferential stress is found from (18). The bending moment is given by

$$M = \int\_{r\_{CD}}^{r\_{AB}} \left(\sigma\_{\theta} - \frac{F}{h}\right) r dr. \tag{39}$$

Here *h* is the current thickness of the sheet, *h* = *rAB* − *rCD* (Figure 1). Using (8), one can rewrite (39) as

$$m = \frac{2\sqrt{3}M}{\sigma\_0 H^2} = \frac{\sqrt{3}}{a} \int\_{-1}^{0} \left(\frac{\sigma\_\theta}{\sigma\_0} - f\frac{H}{h}\right) d\zeta. \tag{40}$$

Note that *m* = 1 in the case of rigid perfectly plastic material if *f* = 0 [28].

#### **5. Numerical Method**

It has been noted that Equation (38) is an ordinary differential equation. Several well-established methods are available for solving such equations numerically (see, for example, [29]). However, the form of the differential equation in Equation (38) is nonstandard. Therefore, a special numerical method should be developed to solve this equation. The method below is a modification of the method developed in [30] for solving a system of partial differential equations.

Equation (38), together with Equation (6), constitutes the initial value problem. However, a difficulty is that the equation provides no information at the initial instant. In particular, each term of this equation vanishes at *a* = 0 since *εeq* = 0 everywhere. It is a consequence of rigid plasticity. Such difficulty does not arise in elastic/plastic solutions (see, for example, [31]). Below, the difficulty noted is resolved by finding the state of stress at the initial instant without using Equation (38) and developing a general iterative procedure.

Assume that *s* = *s<sup>i</sup>* and *ds*/*da* = *q<sup>i</sup>* at *a* = *a<sup>i</sup>* , and *s<sup>i</sup>* and *q<sup>i</sup>* are known. Let ∆*a* be a small increment of *a*. Then, *ai*+<sup>1</sup> = *a<sup>i</sup>* + ∆*a*, and the value of *s* at *a* = *ai*+<sup>1</sup> can be approximated as

$$s\_{i+1} = s\_i + \frac{1}{2}(q\_i + q\_{i+1})\Delta a.\tag{41}$$

Here *qi*+<sup>1</sup> is the value of the derivative *ds*/*da* at *a* = *ai*+<sup>1</sup> , and its value is unknown. Using (27), (29), (31), (34), and (41), one can express the limits of integration involved in (38) in terms of *qi*+<sup>1</sup> and known quantities. Thus Equation (38) determines *qi*+<sup>1</sup> . The procedure above can be repeated for *a* = *ai*+<sup>2</sup> = *ai*+<sup>1</sup> + ∆*a* and subsequent values of *a*.

It remains to find *s* and *ds*/*da* at *a* = 0 to start the iterative procedure above. The required value of *s* is given by (6). The initial cross-section of the sheet is shown in Figure 1a. The through-thickness stress vanishes at this instant. The equivalent strain vanishes everywhere. Equation (21) becomes

$$\chi = \begin{cases} \frac{4\gamma(\zeta + q\_0)}{\sqrt{3}} & \text{in the range } -q\_0 \le \zeta \le 0\\ -\frac{4\gamma(\zeta + q\_0)}{\sqrt{3}} & \text{in the range } -1 \le \zeta \le -q\_0 \end{cases} \tag{42}$$

Here *q*<sup>0</sup> is the value of the derivative *ds*/*da* at *a* = 0. Equations (2) and (42) combine to give

$$\frac{\sigma\_{\eta}}{\sigma\_{0}} = \begin{cases} \frac{2}{\sqrt{3}} \Phi \Big[ 0, \frac{4\xi\_{0}\gamma(\zeta+q\_{0})}{\sqrt{3}} \Big] & \text{in the range } -q\_{0} \le \zeta \le 0\\ -\frac{2}{\sqrt{3}} \Phi \Big[ 0, -\frac{4\gamma(\zeta+q\_{0})}{\sqrt{3}} \Big] & \text{in the range } -1 \le \zeta \le -q\_{0} \end{cases} \tag{43}$$

The tensile force at the initial instant is determined as

$$F = H \int\_0^1 \sigma\_\eta d\zeta. \tag{44}$$

Substituting (43) into (44) gives

$$f = \frac{2}{\sqrt{3}} \int\_{-q\_0}^{0} \Phi\left[0, \frac{4\xi\_0 \gamma(\zeta + q\_0)}{\sqrt{3}}\right] d\zeta - \frac{2}{\sqrt{3}} \int\_{-1}^{-q\_0} \Phi\left[0, -\frac{4\xi\_0 \gamma(\zeta + q\_0)}{\sqrt{3}}\right] d\zeta. \tag{45}$$

This equation determines *q*0. In particular, *q*<sup>0</sup> = 1/2 if *f* = 0 (pure bending). The dimensionless bending moment can be found using (40), where *σ<sup>θ</sup>* should be replaced with *ση*.

## **6. Numerical Example**

The numerical results below are for pure bending. Several functions Φ *εeq*, *ξeq* are available in the literature [18–24]. Since bending of strain-hardening materials, including quantitative results, has been investigated in many papers, the numerical example below emphasizes the effect of loading speed on process parameters.

In most cases, the function Φ *εeq*, *ξeq* is represented as a product of two functions. One of these functions depends on the equivalent strain and the other on the equivalent strain rate. The illustrative example in this section assumes that Φ *εeq*, *ξeq* is the product of Swift's strain-hardening law and Herschel–Bulkley's viscoplastic law. Then,

$$\Phi(\varepsilon\_{\epsilon q}, \mathfrak{f}\_{\epsilon q}) = \left(1 + \lambda \varepsilon\_{\epsilon q}\right)^{\vee} \left[1 + \left(\frac{\mathfrak{f}\_{\epsilon q}}{\mathfrak{f}\_0}\right)^{\wedge}\right].\tag{46}$$

In all calculations, *λ* = 0.73, *ν* = 0.17, and *n* = 0.2. The characteristic strain rate is involved in the solution only through *γ*. In particular, one can rewrite (46) as

$$
\Phi(\varepsilon\_{\epsilon\eta\prime}\chi\xi\_0) = \left(1 + \lambda\varepsilon\_{\epsilon\eta}\right)^{\nu} (1 + \chi^{\eta}).\tag{47}
$$

The numerical solution has been found using the procedure described in the previous section. It demonstrates the effect of *γ* on the through-thickness distribution of stresses and the bending moment. Figures 2 and 3 depict the through-thickness distributions of the radial and circumferential stresses, respectively, at several process stages identified by the inner radius of the sheet. These distributions correspond to *γ* = 0.1. In these figures,

$$Z = \frac{r - r\_{\rm CD}}{H}.\tag{48}$$

‐

‐ *γ* **Figure 2.** Through-thickness distribution of the radial stress at *γ* = 0.1 and several process stages.

‐ *γ*

‐ *γ*

‐

‐

‐ *<sup>γ</sup>* **Figure 3.** Through-thickness distribution of the circumferential stress at *γ* = 0.1 and several process stages.

Thus *Z* = 0 corresponds to the inner radius of the sheet and *Z* = *h*/*H* to its outer radius. Using (8) and (9), one transforms Equation (48) to

$$Z = \sqrt{\frac{\zeta}{a} + \frac{s}{a^2}} - \sqrt{\frac{s}{a^2} - \frac{1}{a}}.\tag{49}$$

At the final stage of the process, the inner radius of the sheet is rather small (smaller than the initial thickness of the sheet). Figures 4 and 5 show the same distributions at *γ* = 1. The circumferential stress is discontinuous at the neutral line. Figures 3 and 5 show that the neutral line moves to the inner surface of the sheet as the deformation proceeds. Thus the assumption made in Section 3 has been confirmed. The magnitude of the radial stress is quite high near the neutral line, except at the very beginning of the process (see Table 1). Therefore, the assumption used in simplified solutions [32,33] that the radial stress is negligible is not valid. The rightmost points of the curves in Figures 2–5 correspond to *Z* = *h*/*H*. Therefore, it is seen from these figures that the change in the thickness is negligible.

‐ *γ* **Figure 4.** Through-thickness distribution of the radial stress at *γ* = 1 and several process stages.

‐ *γ*

γ ‐

‐ *γ*

‐ *γ* **Figure 5.** Through-thickness distribution of the circumferential stress at *γ* = 1 and several process stages. ‐ *γ*

**Table 1.** Radial stress at the neutral line.


Figure 6 demonstrates the effect of *γ* on the dimensionless bending moment. The broken line corresponds to the rate-independent material model (i.e., the material obeys Swift's strain-hardening law). The bending moment increases as *γ* increases, which is in agreement with physical expectations. The increase in the bending moment is sharper for smaller values of *γ*. It is worthy of note that the bending moment is nearly constant for a given value of *γ*. ‐ *γ*

 γ ‐ **Figure 6.** Variation of the dimensionless bending moment as the deformation proceeds at several values of γ. The broken line corresponds to the rate-independent material.
