An Analytic Solution for an Expanding/Contracting Strain-Hardening Viscoplastic Hollow Cylinder at Large Strains and Its Application to Tube Hydroforming Design

Abstract

This paper presents a semi-analytic rigid/plastic solution for the expansion/contraction of a hollow cylinder at large strains. The constitutive equations comprise the yield criterion and its associated flow rule. The yield criterion is pressure-independent. The yield stress depends on the equivalent strain rate and the equivalent strain. No restriction is imposed on this dependence. The solution is facilitated using the equivalent strain rate as an independent variable instead of the polar radius. As a result, it reduces to ordinary integrals. In the course of deriving the solution above, the transformation between Eulerian and Lagrangian coordinates is used. A numerical example illustrates the solution for a material model available in the literature. A practical aspect of the solution is that it readily applies to the preliminary design of tube hydroforming processes.

1. Introduction

It is rare in plasticity theory to find an analytic solution to boundary value problems, especially if strains are large. Even in the case of a thick-walled sphere loaded by internal and external pressures uniformly distributed over its inner and outer surfaces, it is sometimes concluded that a complete solution should be obtained by an iterative or successive approximation method [1]. For strain-hardening materials, a method to find semi-analytic solutions at large strains for a fully plastic sphere was proposed in [2]. A different method that allows for determining the elastic/plastic boundary was developed in [3]. Both methods take advantage of the dependence of the equivalent stress on a single quantity, namely, the equivalent strain. In particular, the natural space coordinate is replaced with the equivalent stress or the equivalent strain. This is the key point for deriving semi-analytic solutions. However, the accurate description of material behavior under certain conditions requires the assumption that the equivalent stress depends on the equivalent strain and the equivalent strain rate ([4,5,6] among many others). It is worth noting that these general relationships involve temperature. However, if the process is isothermal, then the equivalent stress is only a function of the equivalent strain and the equivalent strain rate. In this case, the methods proposed in [2,3] cannot be used directly. A modification of these methods for finding a semi-analytic solution for an infinite rigid/plastic hollow cylinder loaded by inner pressure is developed in the present paper. A numerical technique is only required for evaluating an ordinary integral. No restriction is imposed on the dependence of the equivalent stress on the equivalent strain and the equivalent strain rate.

Several solutions to the above problem are available in the literature. The authors of [7] adopt a model of steady-state creep in both the linear and power-law ranges. The equivalent stress is independent of the equivalent strain. A viscoelastic creep model for a cylinder made of polymeric materials was used in [8]. Sequential limit analysis was applied in [9,10,11] to study hollow cylinders’ finite strain expansion/contraction. The constitutive equations considered in [9] account for viscoplasticity and nonlinear isotropic strain hardening. A model of orthotropic strain-hardening material was adopted in [10]. Hill’s quadratic yield criterion describes plastic anisotropy [12]. The solution in [11] is a model that combines viscoplasticity and kinematic hardening. The authors of [13] present an analytic solution for a model of strain-hardening viscoplastic material. This solution is most relevant to the solution in the present paper. However, an essential difference is that the former is for a specific material model, whereas the latter is for an arbitrary dependence of the equivalent stress on the equivalent strain and the equivalent strain rate. Sequential limit analysis was used in [14] to solve the boundary value problem formulated in [13].

The solutions for hollow cylinders loaded by internal and external pressures allow us to find the failure pressure of cylindrical vessels, which is an important design parameter [15,16,17]. Another area of application is the preliminary design of tube hydroforming processes. Recent reviews on this technology are provided in [18,19]. Despite the wide use of numerical methods for hydroforming process designs [20,21], simple semi-analytic solutions are essential for estimating the optimal values of the process parameters [22,23]. In particular, it has been shown in [24] that a simple plane-strain solution for the expansion of a two-layer tube supplies an accurate prediction for the maximum internal pressure value. Apart from the intrinsic utility above, the present solution is also useful for testing the accuracy of numerical solutions. The latter is a necessary step before using numerical codes [25,26].

The novelty of the present paper is that the solution is valid for an arbitrary dependence of the yield stress on the equivalent strain and the equivalent strain rate. The originality of the solution method is that the equivalent strain rate is used as an independent variable instead of the polar radius, even though the yield stress depends on the two kinematic quantities as stated above.

2. Statement of the Problem

Consider a hollow cylinder whose initial inner and outer radii are $a_0$ and $b_0$, respectively. The state of strain is plane. The cylinder’s inner and outer radii increase/decrease in the course of deformation. The current radii are a and b. The cylindrical coordinate system $\left(r,\hspace{0.17em}\theta ,\hspace{0.17em}z\right)$ with its z-axis coinciding with the cylinder’s axis of symmetry is adopted. The flow everywhere is orthogonal to the z-axis. Let ${\sigma }_r$ and ${\sigma }_{\theta }$ be the radial and circumferential stresses, respectively. These stresses are the principal stresses. The pressure-independent plane-strain yield criterion is:

$$\left|{\sigma }_{\theta }-{\sigma }_r\right|=\frac{2}{\sqrt{3}}{\sigma }_Y$$

(1)

Here, ${\sigma }_Y$ is the yield stress in tension/compression. The plastic flow rule associated with the yield criterion of Equation (1) results in the equation of incompressibility and the condition that the plastic work rate is positive. Let ${\xi }_r$ and ${\xi }_{\theta }$ be the radial and circumferential strain rates, respectively. The equation of incompressibility is

$${\xi }_r+{\xi }_{\theta }=0$$

(2)

The yield stress depends on both the equivalent strain rate, ${\xi }_{eq}$, and the equivalent strain, ${\epsilon }_{eq}$. In the case under consideration, these quantities are defined as

$${\xi }_{eq}=\sqrt{\frac{2}{3}}\sqrt{{\xi }_r^2+{\xi }_{\theta }^2}$$

(3)

and

$$\frac{d{\epsilon }_{eq}}{dt}={\xi }_{eq}$$

(4)

Here, t is the time and $d/dt$ denotes the material derivative. The yield criterion of Equation (1) can be represented in the form

$$\frac{{\sigma }_{\theta }-{\sigma }_r}{{\sigma }_0}=\frac{2m}{\sqrt{3}}\mathsf{\Phi }\left({\epsilon }_{eq},\hspace{0.17em}{\xi }_{eq}\right)$$

(5)

where ${\sigma }_0$ is a reference stress and $\mathsf{\Phi }\left({\epsilon }_{eq},\hspace{0.17em}{\xi }_{eq}\right)$ is an arbitrary function of the equivalent strain and the equivalent strain rate. Moreover, here and in what follows, $m=1$ in the case of expansion and $m=-1$ in the case of contraction. Equations (2) and (3) combine to give

$${\xi }_{eq}=\frac{2m}{\sqrt{3}}{\xi }_{\theta }$$

(6)

The only stress equilibrium equation which is not identically satisfied is

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

(7)

It is assumed that the circumferential velocity everywhere vanishes. Then,

$${\xi }_r=\frac{\partial u}{\partial r},\hspace{1em}{\xi }_{\theta }=\frac{u}{r}.$$

(8)

Here, u is the radial velocity.

The velocity boundary condition is

$$u=mU$$

(9)

for $r=a$. Here $U>0$. The stress boundary condition is

$${\sigma }_r=0$$

(10)

for $r=b$.

3. General Solution

Equations (2) and (8) combine to give $\partial u/\partial r+u/r=0$. The solution of this equation satisfying the boundary condition Equation (9) is

$$u=mU\frac{a}{r}$$

(11)

By definition,

$$\begin{gathered} \partial r/\partial t=u \\ \partial a/\partial t=mU \end{gathered}$$

(12)

Using these equations and Equation (11) one gets $\partial r/\partial a=a/r$. This equation can be immediately integrated to yield

$$r^2=a^2+R^2-a_0^3$$

(13)

Here, R is the Lagrangian coordinate such that $r=R$ at the initial instant. One can solve Equation (13) for R. As a result,

$$R^2=r^2-a^2+a_0^2$$

(14)

The equivalent strain rate is determined from Equations (6), (8), (11) and (13) as

$${\xi }_{eq}=\frac{2Ua}{\sqrt{3}{r}^2}=\frac{2Ua}{\sqrt{3}\left(a^2+R^2-a_0^2\right)}$$

(15)

Substituting Equation (15) into Equation (5) and taking into account Equation (12) yields

$$\frac{\partial {\epsilon }_{eq}}{\partial a}=\frac{2m}{\sqrt{3}}\frac{a}{\left(a^2+R^2-a_0^2\right)}$$

(16)

The initial condition to this equation is ${\epsilon }_{eq}=0$ at $a=a_0$. The solution of Equation (16) satisfying this condition is

$${\epsilon }_{eq}=\frac{m}{\sqrt{3}}\ln \left(\frac{a^2-a_0^2+R^2}{R^2}\right)$$

(17)

Eliminating R here by means of Equation (14) leads to

$${\epsilon }_{eq}=\frac{m}{\sqrt{3}}\ln \left(\frac{r^2}{r^2-a^2+a_0^2}\right)$$

(18)

The solution is facilitated by replacing the independent variable r with the equivalent strain rate. In this case,

$$\frac{\partial {\sigma }_r}{\partial r}=\frac{\partial {\sigma }_r}{\partial {\xi }_{eq}}\frac{\partial {\xi }_{eq}}{\partial r}.$$

(19)

The derivative $\partial {\xi }_{eq}/\partial r$ is readily determined from Equation (15) as $\frac{\partial {\xi }_{eq}}{\partial r}=-\frac{4}{\sqrt{3}}\frac{Ua}{r^3}$.

Substituting this equation into Equation (19), then, the resulting equation into Equation (7) and using Equation (5) yields

$$\frac{\partial {\sigma }_r}{{\sigma }_0\partial {\xi }_{eq}}=-\frac{m}{\sqrt{3}}\frac{\mathsf{\Phi }\left({\epsilon }_{eq},\hspace{0.17em}{\xi }_{eq}\right)}{{\xi }_{eq}}$$

(20)

Here, Equation (15) has been used to eliminate r.

It is convenient to introduce the following dimensionless quantities:

$$\begin{gathered} c_0=\frac{a_0}{b_0}, \\ \alpha =\frac{a}{b_0}, \\ \rho =\frac{r}{b_0}, \\ \gamma =\frac{{\xi }_{eq}}{{\xi }_0}, \\ \nu =\frac{\sqrt{3}{\xi }_0{b}_0}{2U} \end{gathered}$$

(21)

Here, ${\xi }_0$ is a characteristic strain rate. Equation (20) becomes

$$\frac{\partial {\sigma }_r}{{\sigma }_0\partial \gamma }=-\frac{m}{\sqrt{3}}\frac{\mathsf{\Phi }\left({\epsilon }_{eq},\hspace{0.17em}{\xi }_0\gamma \right)}{\gamma }$$

(22)

Let ${\gamma }_b$ be the value of $\gamma$ at $r=b$. Since $R=b_0$ at $r=b$, it follows from Equations (15) and (21) that

$${\gamma }_b=\frac{\alpha }{\nu \left({\alpha }^2+1-c_0^2\right)}$$

(23)

The boundary condition Equation (10) becomes ${\sigma }_r=0$ for $\gamma ={\gamma }_b$. The solution of Equation (22) satisfying this boundary condition is

$$\frac{{\sigma }_r}{{\sigma }_0}=-\frac{m}{\sqrt{3}}{\displaystyle \underset{{\gamma }_b}{\overset{\gamma }{\int }}\frac{\mathsf{\Phi }\left({\epsilon }_{eq},\hspace{0.17em}{\xi }_0\chi \right)}{\chi }d\chi }$$

(24)

Here, $\chi$ is a dummy variable of integration. Let ${\gamma }_a$ be the value of $\gamma$ at $r=a$. It follows from Equations (15) and (21) that

$${\gamma }_a=\frac{1}{\nu \alpha }$$

(25)

The pressure over the inner radius of the cylinder, P, is determined from Equation (24) at $\gamma ={\gamma }_a$. Then,

$$p=\frac{P}{{\sigma }_0}=\frac{m}{\sqrt{3}}{\displaystyle \underset{{\gamma }_b}{\overset{{\gamma }_a}{\int }}\frac{\mathsf{\Phi }\left({\epsilon }_{eq},\hspace{0.17em}{\xi }_0\chi \right)}{\chi }d\chi }$$

(26)

Here, p is the dimensionless pressure.

Eliminating r between Equations (15) and (18) and using Equation (21) yields

$${\epsilon }_{eq}=-\frac{m}{\sqrt{3}}ln\left[1+\nu \gamma \left(\frac{c_0^2}{\alpha }-\alpha \right)\right]$$

(27)

Then, the function $\mathsf{\Phi }\left({\epsilon }_{eq},\hspace{0.17em}{\xi }_0\gamma \right)$ can be represented as

$$\mathsf{\Phi }\left({\epsilon }_{eq},\hspace{0.17em}{\xi }_0\gamma \right)=\mathsf{\Lambda }\left(\alpha ,\hspace{0.17em}{\xi }_0\gamma \right)$$

(28)

It is understood here that ${\epsilon }_{eq}$ on the left-hand side is to be eliminated using Equation (27). Then Equations (24) and (26) become

$$\frac{{\sigma }_r}{{\sigma }_0}=-\frac{m}{\sqrt{3}}{\displaystyle \underset{{\gamma }_b}{\overset{\gamma }{\int }}\frac{\mathsf{\Lambda }\left(\alpha ,\hspace{0.17em}{\xi }_0\chi \right)}{\chi }d\chi }$$

(29)

and

$$p=\frac{m}{\sqrt{3}}{\displaystyle \underset{{\gamma }_b}{\overset{{\gamma }_a}{\int }}\frac{\mathsf{\Lambda }\left(\alpha ,\hspace{0.17em}{\xi }_0\chi \right)}{\chi }d\chi }$$

(30)

If the function $\mathsf{\Phi }\left({\epsilon }_{eq},\hspace{0.17em}{\xi }_{eq}\right)$ is given, the integrals in Equations (29) and (30) can be evaluated with no difficulty. Equation (30) provides the variation of p with a. Equations (15) and (21) combine to yield

$$\rho =\sqrt{\frac{\alpha }{\nu \gamma }}$$

(31)

This equation and Equation (29) give the radial distribution of the radial stress at a fixed value of a in parametric form, with $\gamma$ being the parameter. The radial distribution of the circumferential stress is readily determined from Equations (5), (28) and (29).

4. Numerical Example

Comprehensive overviews of material models adopted in the present paper have been provided in [6,27]. At a constant temperature, one of the widely used functions $\mathsf{\Phi }\left({\epsilon }_{eq},\hspace{0.17em}{\xi }_{eq}\right)$ is as follows:

$$\mathsf{\Phi }\left({\epsilon }_{eq},\hspace{0.17em}{\xi }_{eq}\right)=\left(1+\mu {\epsilon }_{eq}^n\right)\left[1+\lambda \ln \left(\frac{{\xi }_{eq}}{{\xi }_0}\right)\right]$$

(32)

Here, $\mu$, $\lambda$, ${\xi }_0$, and n are material constants. For mild steels $\mu =8.38$, $\lambda =0.0362$, ${\xi }_0=0.001$, and $n=0.316$. This set of parameters is used below. Also, $c_0=1/2$ and $m=1$ in all calculations.

Equations (27), (28) and (32) combine to give

$$\mathsf{\Lambda }\left(\alpha ,\hspace{0.17em}{\xi }_0\gamma \right)=\langle 1+\mu {\left\{-\frac{m}{\sqrt{3}}ln\left[1+\nu \gamma \left(\frac{c_0^2}{\alpha }-\alpha \right)\right]\right\}}^n\rangle \left(1+\lambda \ln \gamma \right)$$

(33)

Substituting Equation (33) into Equations (29) and (30) allows for the radial stress and pressure to be calculated. The material model adopted is rate-dependent. Therefore, the pressure required to deform the cylinder depends on the speed of expansion/contraction. The dimensionless parameter $\nu$ controls this dependence. Figure 1 illustrates the variation in the dimensionless pressure with $\nu$ at several stages of the process. It is seen from this figure that the effect of $\nu$ on the pressure is most significant if $\nu$ is small enough. It is worth noting that $\nu$ decreases as U increases, as follows from Equation (21). Therefore, the behavior of the curves in Figure 1 is in agreement with physical expectations.

Figure 2 shows the variation in the dimensionless pressure with the dimensionless inner radius at several values of $\nu$. In all cases, the pressure sharply increases at the beginning of the process, attains a maximum, and gradually decreases. The existence of the maximum is associated with thinning of the cylinder’s wall as the deformation proceeds. The radial distributions of the radial and circumferential stresses at $\alpha =1$ (i.e., when the inner radius of the cylinder has increased two times) and several values of $\nu$ are depicted in Figure 3 and Figure 4, respectively. At the same polar radius, both $\left|{\sigma }_r\right|$ and $\left|{\sigma }_{\theta }\right|$ increase as $\nu$ decreases. The through-thickness variation in the circumferential stress is not significant. The radial distributions of the radial and circumferential stresses at $\nu =10$ and several values of $\alpha$ are depicted in Figure 5 and Figure 6, respectively. The behavior of the curves in Figure 5 is in agreement with Figure 2. Figure 6 shows that the circumferential stress is a practically linear function of $\rho$ if $\alpha \le 1$ (approximately).

5. Conclusions

The present paper analyses the expansion/contraction of a hollow cylinder under plane-strain conditions. The constitutive equations include the yield stress that depends on the equivalent strain and the equivalent strain rate. No restriction is imposed on this dependence. Nevertheless, the solution is semi-analytic. A numerical procedure is only necessary to evaluate ordinary integrals. The solution was facilitated using Lagrangian coordinates and the equivalent strain rate as an independent variable instead of the polar radius.

A significant difference in this approach from similar available approaches is that the former allows one to consider the dependence of the yield stress on two quantities (the equivalent strain and the equivalent strain rate). The available approaches apply to constitutive equations that include a yield stress that depends on one quantity. The general structure of the solution found here suggests that the approach developed should apply to a class of boundary value problems (i.e., plane-strain bending and plane-strain bending under tension).

The solution for the cylinder’s expansion readily applies to tube hydroforming designs. In particular, Figure 2 demonstrates that the internal pressure attains a maximum at a certain stage of the process. The maximum value of this pressure depends on the dimensionless parameter $\nu$ introduced in Equation (21). The speed of expansion controls the value of this parameter. On the other hand, the maximum value of the internal pressure is an essential design parameter [22,23].