THEORY OF ELASTICITY

G. I. Bykovtsev

ON THE COMPRESSION OF AN ANISOTROPICALLY HARDENING PLASTIC LAYER BY ROUGH PLATES

(Presented by Academician A. Yu. Ishlinskii, February 5, 1964)

The compression of an ideally plastic layer by rough plates was considered in \((^{1-3})\). The compression of a strip made of a material obeying the linearized relation of the theory of anisotropic hardening was considered in \((^4)\). Below a solution of this problem is given for a hardening rigid-plastic material, with the relations of the theory of anisotropic hardening \((^{5-7})\) being adopted.

The loading surface for an anisotropically hardening material, according to \((^{5-7})\), in the case of plane deformation can be represented in the form

\[ (\sigma_x-\sigma_y-c\varepsilon_x+c\varepsilon_y)^2+4(\tau_{xy}-c\varepsilon_{xy})^2=4k^2 . \tag{1} \]

Equality (1) is identically satisfied by the substitution

\[ \sigma_x=\sigma-k\sin 2\theta+c\varepsilon_x,\qquad \sigma_y=\sigma+k\sin 2\theta+c\varepsilon_y, \]
\[ \tau_{xy}=k\cos 2\theta+c\varepsilon_{xy}. \tag{2} \]

Substituting these equalities into the equilibrium equations, we obtain

\[ \frac{\partial\sigma}{\partial x} -2k\left(\cos 2\theta\,\frac{\partial\theta}{\partial x} +\sin 2\theta\,\frac{\partial\theta}{\partial y}\right) +c\left(\frac{\partial\varepsilon_x}{\partial x} +\frac{\partial\varepsilon_{xy}}{\partial y}\right)=0, \]
\[ \frac{\partial\sigma}{\partial y} -2k\left(\sin 2\theta\,\frac{\partial\theta}{\partial x} -\cos 2\theta\,\frac{\partial\theta}{\partial y}\right) +c\left(\frac{\partial\varepsilon_y}{\partial y} +\frac{\partial\varepsilon_{xy}}{\partial x}\right)=0. \tag{3} \]

Applying to (1) the associated flow law, we have

\[ \frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0,\qquad 2\frac{\partial u}{\partial x}\operatorname{ctg}2\theta +\frac{\partial u}{\partial y} +\frac{\partial v}{\partial x}=0, \tag{4} \]

where \(u\) and \(v\) are the projections of the velocity vector on the axes \(x\) and \(y\).

In what follows, when considering the deformation process, the Almansi strain tensor \((^7)\) is used, for which we have

\[ \frac{\partial\varepsilon_x}{\partial t} +\frac{\partial\varepsilon_x}{\partial x}u +\frac{\partial\varepsilon_x}{\partial y}v = \frac{\partial u}{\partial x}(1-2\varepsilon_x) -2\varepsilon_{xy}\frac{\partial v}{\partial x}, \]
\[ \frac{\partial\varepsilon_y}{\partial t} +\frac{\partial\varepsilon_y}{\partial x}u +\frac{\partial\varepsilon_y}{\partial y}v = \frac{\partial v}{\partial y}(1-2\varepsilon_y) -2\varepsilon_{xy}\frac{\partial u}{\partial y}, \]
\[ \frac{\partial\varepsilon_{xy}}{\partial t} +\frac{\partial\varepsilon_{xy}}{\partial x}u +\frac{\partial\varepsilon_{xy}}{\partial y}v = \frac{1}{2}\left(\frac{\partial u}{\partial y} +\frac{\partial v}{\partial x}\right) -\varepsilon_x\frac{\partial u}{\partial y} -\varepsilon_y\frac{\partial v}{\partial x}. \tag{5} \]

Consider a plastic layer situated symmetrically along the \(x\)-axis between two rough plates approaching each other with constant unit velocity. Let the thickness of the layer at time \(t\) be denoted by \(2h(t)\); then \(h(t)=h(0)-t\). On the plates the magnitude of the tangential stress is assumed constant and equal to \(T\), with \(T\ll k\).

In what follows it is convenient to introduce the dimensionless coordinates

\[ \alpha=\frac{x}{h(t)},\qquad \beta=\frac{y}{h(t)},\qquad t_1=\frac{h(t)}{h(0)}. \]

When considering the flow of material sufficiently far from the edges of the plate and from the center, one may assume that \(v\) is a function only of \(\beta\),

and \(\theta\) does not depend on \(\alpha\). Then from (4) it follows that

\[ \vartheta=-\beta,\qquad u=\alpha+\varphi(\beta,t_1),\qquad 2\operatorname{ctg}2\theta+\frac{\partial\varphi}{\partial\beta}=0. \tag{6} \]

From equations (5) we have

\[ \varepsilon_x=\frac{1}{2}(1-t_1^2),\qquad \frac{\partial\varepsilon_{xy}}{\partial t_1}=-\frac{1}{2}\frac{\partial\varphi}{\partial\beta}t_1, \]

\[ t_1\frac{\partial\varepsilon_y}{\partial t_1} =1-2\varepsilon_y+2\varepsilon_{xy}\frac{\partial\varphi}{\partial\beta}. \tag{7} \]

Passing in equations (3) to dimensionless coordinates, we obtain

\[ \sigma-2k\sin2\theta\,\frac{\partial\theta}{\partial\beta}\alpha +c\frac{\partial\varepsilon_{xy}}{\partial\beta} =f_1(\beta,t_1), \]

\[ \sigma+2k\sin2\theta+c\varepsilon_y=f_2(\alpha,t_1). \]

Comparing these equalities, we obtain

\[ c\frac{\partial\varepsilon_{xy}}{\partial\beta} -2k\sin2\theta\,\frac{\partial\theta}{\partial\beta} =\mu(t_1). \tag{8} \]

For the hydrostatic pressure we have

\[ \sigma=-\mu_1(t_1)\alpha-k\sin2\theta-\nu_1(t_1). \tag{9} \]

Integrating equality (9), we have

\[ \tau_{xy}=k\cos2\theta+c\varepsilon_{xy} =\mu(t_1)\beta+\chi(t_1). \tag{10} \]

Taking into account that for \(\beta=\pm1\) the quantity \(\tau_{xy}=\pm T\), we obtain

\[ \mu(t_1)=T,\qquad \chi(t_1)=0,\qquad c\varepsilon_{xy}+k\cos2\theta=T\beta. \tag{11} \]

Eliminating the quantities \(\varepsilon_{xy}\) and \(\varphi(\beta,t_1)\) from equalities (6), (7), and (11), we obtain

\[ t_1\operatorname{ctg}2\theta=\frac{2k}{c}\sin2\theta\,\frac{\partial\theta}{\partial t_1}. \]

Integrating this equation, we obtain

\[ \frac{c}{2k}t_1^2 =\frac{1}{2}\ln\frac{1+\sin2\theta}{1-\sin2\theta} -\sin2\theta+f(\beta). \]

Taking into account that for \(t_1=1\), \(\cos2\theta=T\beta/k\), we obtain

\[ \frac{c}{2k}(t_1^2-1) =\frac{1}{2}\ln\frac{1+\sin2\theta}{1-\sin2\theta} -\sin2\theta \]
\[ -\frac{1}{2}\ln \frac{1-\sqrt{1-\beta_1^2}}{1+\sqrt{1-\beta_1^2}} -\sqrt{1-\beta_1^2} \qquad \left(\beta_1=\frac{T\beta}{k}\right). \tag{12} \]

Equation (12) determines the dependence of \(\theta\) on \(\beta\) and \(t\). After \(\theta\) has been determined, the quantity \(\varepsilon_{xy}\) is determined from equation (11).

From equations (7) we have

\[ t_1\frac{\partial\varepsilon_y}{\partial t_1}+2\varepsilon_y = 1-\frac{4\varepsilon_{xy}}{t_1}\frac{\partial\varepsilon_{xy}}{\partial t_1}. \]

Integrating this equation, we obtain

\[ \varepsilon_y=\frac{t_1^2-1}{2t_1^2} -2\frac{\varepsilon_{xy}^2}{t_1^2}. \tag{13} \]

where in equality (13) it has been taken into account that for \(t_1=1\) the material is in an undeformed state, i.e. \(\varepsilon_y=\varepsilon_{xy}=0\).

We obtain the pressure distribution over the plate from (2); we have

\[ \sigma_y=-Ta-v(t_1)-c\frac{1-t_1^2}{2t_1^2}-2c\frac{\varepsilon_{xy}^2}{t_1^2}. \tag{14} \]

To determine the quantity \(v(t_1)\), note that if the edge of the plate \(x=0\) is free, then the stresses \(\sigma_x\) in the vertical section must be balanced by the tangential stresses at the bases of the layer, i.e.

Fig. 1

Fig. 2

\[ \int_0^1 \sigma_x\,d\beta+\int_0^a \tau_{xy}\big|_{\beta=1}\,d\alpha=0. \tag{15} \]

For \(\sigma_x\), from relations (2) and (9) we have

\[ \sigma_x=-Ta-2k\sin 2\theta-v_1(t_1)+\frac{c}{2}(1-t_1^2). \]

Taking into account that \(\tau_{xy}\big|_{\beta=1}=T\), from equation (15) we obtain

\[ v(t_1)=-2k\int_0^1 \sin 2\theta\,d\beta+\frac{c}{2}(1-t_1^2). \tag{16} \]

The dependence of \(v(t_1)/k\) on \(t_1\) for various \(T/k\) is shown in Fig. 1. Figure 2 presents the dependence of \(\varepsilon_{xy}\) on time for \(\beta=1\).

Voronezh State University

Received
31 I 1964

REFERENCES

1. L. Prandtl, in: Collected Translations. Theory of Plasticity, IL, 1950.
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