INVESTIGATION OF PLASTIC DEFORMATION BY HARDNESS MEASUREMENT

G. D. DEL

(Presented by the scientific seminar of the Department of Strength of Materials).

It is known [1, 2, 3] that in the process of plastic deformation the hardness of the deformed body increases. It has been established [2] that the curves “intensity of shear stresses \(\kappa\)—hardness \(H\),” constructed from the results of testing a material in tension, compression, and torsion, coincide. Hence the assumption was made that \(\kappa\) can be determined from hardness also in the case of a complex state of stress.

In support of this assumption one may also cite Bridgman’s experimental results [1], which showed that hydrostatic pressure has practically no effect on hardness.

The existence of a relation between stresses and hardness is already used quite widely in the investigation of plastic deformation. For this purpose, by means of a tension, compression, or torsion test and a parallel measurement of hardness, a calibration curve “hardness—stress” is constructed. Then, at various stages of deformation, hardness is measured at different points of the body or model under study. From the calibration curve the corresponding intensity of shear stresses is obtained, and from the flow curve—the corresponding intensity of strains. In the case of a linear state of stress, complete information is obtained about the stress–strain state. In the case of a complex state of stress, the information is very limited.

1. A decoding method is proposed which, from the intensity of shear stresses \(\kappa\) known at various points of the body under study, as established by hardness measurement, makes it possible to determine the stressed and kinematic state of the deformed body.

The stress components satisfy the equation

\[ (\sigma_x-\sigma_y)^2+(\sigma_y-\sigma_z)^2+(\sigma_z-\sigma_x)^2 +6(\tau_{xy}^{\,2}+\tau_{yz}^{\,2}+\tau_{xz}^{\,2}) =6\kappa^2(x,y,z,q). \tag{1.1} \]

Here the parameter \(q\) characterizes the stage of deformation of the body under study. For example, in axial compression, the reduction, variable height, etc., may serve as \(q\). In what follows we shall consider the function \(\kappa(x,y,z,q)\) to be known.

Together with (1.1), it is proposed to consider the system of differential equations of equilibrium:

\[ \left. \begin{aligned} \frac{\partial \sigma_x}{\partial x}+\frac{\partial \tau_{xy}}{\partial y}+\frac{\partial \tau_{xz}}{\partial z}&=0,\\ \frac{\partial \tau_{xy}}{\partial x}+\frac{\partial \sigma_y}{\partial y}+\frac{\partial \tau_{yz}}{\partial z}&=0,\\ \frac{\partial \tau_{xz}}{\partial x}+\frac{\partial \tau_{yz}}{\partial y}+\frac{\partial \sigma_z}{\partial z}&=0 \end{aligned} \right\} \tag{1.2} \]

and the relations of the theory of flow

\[ \left. \begin{aligned} \frac{\partial v_x}{\partial x}&=\lambda(\sigma_x-\sigma),\ldots\\ \frac{1}{2}\left(\frac{\partial v_x}{\partial y}+\frac{\partial v_y}{\partial x}\right)&=\lambda \tau_{xy},\ldots \end{aligned} \right\} \tag{1.3} \]

Here \(v_x, v_y, v_z\) are the components of the particle displacement-velocity vector, \(\sigma\) is the hydrostatic pressure, and \(\lambda\) is an unknown parameter.

To determine the 10 unknowns \(\sigma_x, \sigma_y, \sigma_z, \tau_{xy}, \tau_{xz}, \tau_{yz}, v_x, v_y, v_z, \lambda\), we have 10 equations (1.1)—(1.3).

Let us point out a certain analogy with the optical method, in which the magnitude of the difference and the direction of the principal stresses are established experimentally. To determine the state of stress, the differential equations of equilibrium are solved for these data. The strains are found from Hooke’s law.

Since the systems (1.1)—(1.3) practically have to be solved for a fixed parameter \(q\), the problem under consideration formally coincides with the problem of an inhomogeneous ideally plastic body under the Mises plasticity condition. The homogeneous non-hardening body corresponds to the special case when \(k=\textit{const}\).

Let us note that instead of relations (1.3), when this proves justified, the corresponding relations of deformation theories may be used.

Let us proceed to the consideration of special cases.

2. Torsion. In the case of pure torsion \(\sigma_x=\sigma_y=\sigma_z=\tau_{xy}=0\), and therefore equations (1.1), (1.2) take the form

\[ \left. \begin{aligned} \tau_{xz}^{2}+\tau_{yz}^{2}&=k^{2}(x,y,q)\\ \frac{\partial \tau_{xz}}{\partial x}+\frac{\partial \tau_{yz}}{\partial y}&=0 \end{aligned} \right\} \tag{2.1} \]

(the \(z\)-axis is aligned with the axis of the rod being twisted).

Introduce the stress function \(F\), putting

\[ \tau_{xz}=\frac{\partial F}{\partial y},\qquad \tau_{yz}=-\frac{\partial F}{\partial x}. \tag{2.2} \]

Then one of equations (2.1) is satisfied automatically, and the other takes the form

\[ \left(\frac{\partial F}{\partial x}\right)^2+\left(\frac{\partial F}{\partial y}\right)^2=k^2. \tag{2.3} \]

As in the case of an ideally plastic body, equation (2.3) is an equation of hyperbolic type.

Lines \(F=\mathrm{const}\) are called stress lines. A stress line, in particular, is the contour of a twisted rod. Lines normal to the stress lines are called normals. The normals are characteristics of equation (2.3).

Having at hand a system of stress lines, it is easy to obtain the shear stresses at each point of the transverse section of the rod:

\[ \tau_{xz}=\pm k\sin\varphi,\qquad \tau_{yz}=\mp k\cos\varphi, \tag{2.4} \]

where \(\varphi\) is the angle between the normal and the \(x\)-axis. The choice of sign is determined by the direction of twisting. Therefore, for interpretation in the present case it is sufficient to construct, on the transverse section of the rod, a system of stress lines. The normals are the projections onto the transverse section of the lines of steepest descent of the surface \(z=F(x,y)\). The tangents to the lines of steepest descent form with the plane of the transverse section an angle \(\alpha\), whose tangent, according to (2.3), is \(\operatorname{tg}\alpha=k\). This circumstance can be used to construct a family of stress lines.

Let us construct the contour of the transverse section of the twisted specimen. Let \(\Delta F\) be the difference in the values of the function \(F\) between neighboring stress lines. At various points of the contour we draw normals to it and lay off along them, in the direction toward the center of the section, segments of length

\[ h=\frac{\Delta F}{\operatorname{tg}\alpha}=\frac{\Delta F}{k}; \tag{2.5} \]

Connecting the ends of these segments with a smooth curve, we obtain a stress line. By subsequently repeating this operation, one can construct a family of stress lines.

Fig. 1. Torsion of a triangular prism.

Having constructed a family of stress lines for various values of the relative angle of twist, which in this case may serve as the parameter \(q\), one can obtain an idea of the stress state in the torsion of strain-hardening rods.

To determine the deformed state in the present case, it proves possible to use the relations of deformation theories [4]. Figure 1 shows the distribution of \(k(x,y)\) under torsion of a triangular prism made of steel Kh13 up to a relative angle of twist of \(15\ \mathrm{m}^{-1}\), and the corresponding system of stress lines. The stress surface differs substantially from the pyramid that occurs in the absence of hardening. The stress lines are curved, and the distance between them increases as one moves toward the center. The calculation was checked by the twisting moment. The moment found by approximate integration,

\[ M=2\iint F\,dx\,dy, \tag{2.6} \]

differed from the actual value by \(3\%\).

3. Plane deformation.

In plane deformation, displacements of the particles of the body occur in one plane. Let us direct the \(z\)-axis along the normal to this plane. Then the velocity \(v_z\) and the stress components \(\tau_{xz}, \tau_{yz}\) will be equal to zero, while the nonzero components of the velocity vector and of the stress tensor do not depend on \(z\).

From the condition of incompressibility it follows [4]

\[ \sigma_z=\frac{1}{2}(\sigma_x+\sigma_y)=\sigma . \tag{3.1} \]

In view of this, equation (1.1) is simplified to

\[ (\sigma_x-\sigma_y)^2+4\tau_{xy}^2=4k^2(x,y,q). \tag{3.2} \]

The equilibrium equations in the case of plane deformation are

\[ \frac{\partial \sigma_x}{\partial x}+\frac{\partial \tau_{xy}}{\partial y}=0,\qquad \frac{\partial \tau_{xy}}{\partial x}+\frac{\partial \sigma_y}{\partial y}=0. \tag{3.3} \]

Let us consider a method for solving the system (3.2)—(3.3) by the slip-line method.

Slip lines are lines tangent at every point to the planes of maximum shear stress. There are two orthogonal families of slip lines. We shall call one of them the family \(\alpha\), and the other the family \(\beta\). We shall fix the direction of the lines \(\alpha, \beta\) so that they form a right-handed coordinate system in which the shear stress has a positive direction. The angle of inclination of the tangent to the line \(\alpha\), measured in the positive direction from the \(x\)-axis, will be denoted by \(\Theta\).

Using the known relations

\[ \sigma_x=\sigma-k\sin 2\Theta,\qquad \sigma_y=\sigma+k\sin 2\Theta,\qquad \tau_{xy}=k\cos 2\Theta \tag{3.4} \]

we pass to the new variables \(\sigma\) and \(\Theta\). Equation (3.2) is then satisfied automatically, while (3.3) take the form:

\[ \left. \begin{aligned} \frac{\partial \sigma}{\partial x} -2k\cos 2\Theta\,\frac{\partial \Theta}{\partial x} -2k\sin 2\Theta\,\frac{\partial \Theta}{\partial y} &= \sin 2\Theta\,\frac{\partial k}{\partial x} -\cos 2\Theta\,\frac{\partial k}{\partial y}, \\[4pt] \frac{\partial \sigma}{\partial y} -2k\sin 2\Theta\,\frac{\partial \Theta}{\partial x} +2k\cos 2\Theta\,\frac{\partial \Theta}{\partial y} &= -\cos 2\Theta\,\frac{\partial k}{\partial x} -\sin 2\Theta\,\frac{\partial k}{\partial y}. \end{aligned} \right\} \tag{3.5} \]

(3.5) is a system of hyperbolic type; its characteristics coincide with the slip lines [5].

Let us pass to the local coordinate system formed by the tangents to the slip lines. In this case \(\Theta=0\), and therefore we have

\[ \frac{\partial}{\partial \alpha}(\sigma-2k\Theta) = -\frac{\partial k}{\partial \beta}, \qquad \frac{\partial}{\partial \beta}(\sigma+2k\Theta) = -\frac{\partial k}{\partial \alpha}, \tag{3.6} \]

where \(\dfrac{\partial}{\partial \alpha},\ \dfrac{\partial}{\partial \beta}\) are derivatives along the lines \(\alpha, \beta\). Consequently, along the slip lines \(\alpha, \beta\) we have, respectively:

\[ \frac{\sigma}{2k}-\Theta=-A_\alpha, \qquad \frac{\sigma}{2k}+\Theta=-A_\beta. \tag{3.7} \]

Here it is assumed that

\[ A_\alpha=\frac{1}{2k}\int_{\alpha}\frac{\partial k}{\partial \beta}\,d\alpha, \qquad A_\beta=\frac{1}{2k}\int_{\beta}\frac{\partial k}{\partial \alpha}\,d\beta. \tag{3.8} \]

Let us consider methods for the numerical solution of the basic boundary-value problems, based on relations (3.7) and on replacing the integrals (3.8) by a finite sum. In doing so, we shall assume the derivatives \(\partial \kappa/\partial \alpha\) and \(\partial \kappa/\partial \beta\) to be continuous.

1. The Cauchy problem. The functions \(\sigma\) and \(\Theta\) are prescribed on some arc of a smooth curve \(AB\), which has no characteristic directions at any point and lies in the plane of deformation. It is required to find a solution of system (3.5) satisfying, on the arc \(AB\), the prescribed values of \(\sigma\) and \(\Theta\). The required solution exists and is unique in the triangular region bounded by the arc \(AB\) (Fig. 2) and by the slip lines \(\alpha,\beta\) issuing from its ends.

Fig. 2.                Fig. 3.

Divide the arc \(AB\) into a number of sufficiently small parts by the points \(1,2,3,\ldots,6\). From (3.7) we find \(A_\alpha\) and \(A_\beta\) at these points. Starting from the known \(\Theta\), graphically or from the equations

\[ y_7-y_2=-\operatorname{ctg}\Theta_2\,(x_7-x_2), \qquad y_7-y_1=\operatorname{tg}\Theta_1\,(x_7-x_1) \tag{3.9} \]

we find the positions of points \(7,8,\ldots,11\). Next we find (see Fig. 3)

\[ \left. \begin{aligned} (A_\alpha)_7 &= (A_\alpha)_1+\frac{\kappa_A-\kappa_C}{2\kappa_7\Delta l}\,l_{1-7},\\[4pt] (A_\beta)_7 &= (A_\beta)_2+\frac{\kappa_F-\kappa_D}{2\kappa_7\Delta l}\,l_{2-7}. \end{aligned} \right\} \tag{3.10} \]

Here \(l_{1-7},\,l_{2-7}\) are the distances between points 1 and 7, and 2 and 7. The segments \(AC\) and \(DF\) are perpendicular to \(1\!-\!7\) and \(2\!-\!7\) and pass through their midpoints, with

\[ AB=BC=DE=EF=\frac{\Delta l}{2}. \]

By the formulas

\[ \sigma=\kappa(-A_\alpha-A_\beta), \qquad \Theta=\frac{1}{2}(A_\alpha-A_\beta), \tag{3.11} \]

which follow from (5.7), we find \(\sigma_7\) and \(\Theta_7\). In the same way we find \(\sigma\) and \(\Theta\) at the points \(8,9,10,11\). Now the row of points \(7,8,9,10,11\) is regarded as the initial one, and we continue the solution, and so on, over the whole region.

The obtained solution can be refined by introducing in (3.9), instead of the angles \(\Theta_1\) and \(\Theta_2\), the mean angles \(\frac{1}{2}(\Theta_1+\Theta_7)\) and \(\frac{1}{2}(\Theta_2+\Theta_7)\).

1. Initial characteristic problem. It is required to find the solution of the system (3.5) if the unknown functions are given along two orthogonal segments of the slip lines \(OA\) and \(OB\) (Fig. 4). The solution is determined in the region \(OACB\).

Fig. 4.

Divide the segments \(OA\) and \(OB\) into a number of sufficiently small parts. In Fig. 3 this is done by the points \(1, 2, 3,\ldots,10\). From (3.7) we find \(A_\alpha\) and \(A_\beta\) at these points. The position of point 11 and the values of \(\varepsilon\), \(\Theta\), \(A_\alpha\), \(A_\beta\) at it are found from the known \(\Theta\), \(A_\alpha\), \(A_\beta\) at points 4 and 6, as in the case of the Cauchy problem.

Similarly, the points \(12, 13,\ldots,16, 17\), etc., are determined throughout the whole region.

Consider the degenerate case of this problem, when the segment of the slip line \(OB\) contracts to the point \(O\) (Fig. 5). The angle of opening at the point \(O\) (i.e., the angle \(AOC\)) and, along the segment of the slip line \(OA\), \(\sigma\) and \(\Theta\) are given. It is required to find the solution of the system (3.5) satisfying these data. The solution is determined in the region \(OAE\).

Fig. 5.

Let \(OA\) be a segment of the slip line \(\alpha\). Then at the point \(O\) the radius of curvature of the line \(\beta\), \(R_\beta = 0\). Since

\[ \frac{1}{R_\beta}=-\frac{\partial \Theta}{\partial \beta}, \tag{3.12} \]

it follows from (3.6) that at the point \(O\) \(\varepsilon\) is discontinuous.

At the point \(N\), according to (3.8),

\[ (A_\beta)_N=(A_\beta)_K+\frac{1}{2\kappa_N}\int_K^N \frac{\partial \kappa}{\partial \alpha}\,d\beta . \tag{3.13} \]

Let the point \(K\) tend to the node \(O\). Then, by virtue of the continuity of \(d\kappa \cdot d\alpha\), the second term in (3.13) tends to zero. Consequently, inte-

\(A_z\) is equal at point \(O\) for all slip lines \(z\) passing through this point. If a segment of a slip line \(z\) is contracted into a point, the integral \(A_\alpha\) is common.

Let us divide the segment \(OA\) into a number of sufficiently small points \(0,1,5,\ldots\). We find \(A_z\) at these points. We divide the angle \(OAC\) into a number of small rays \(01, 02,\ldots\). We determine the angles \(\Theta\) corresponding to these rays at the point \(O\). From them and from the known \(A_z\), we find \(A_\alpha\) at this point for each ray. Then, from the \(\Theta\) known at point 1, we find the position of point 2. By formulas analogous to (5.10), we find at this point \(A_\alpha, A\), and from (3.11), \(\Theta\) and \(z\). The points 3 and 4 are determined analogously. Thereafter the problem is solved as in the general case of the initial characteristic problem.

Fig. 6.

1. Mixed problem. The problem consists in finding the solution of system (3.5), if along the segment of the slip line \(OA\) (Fig. 6) \(\sigma\) and \(\Theta\) are given, and along the curve \(OB\), which has no characteristic directions, the angle \(\Theta\) is prescribed. The solution of the problem exists and is unique in the region \(OA_1CB\).

From the value of \(\Theta\) at the point \(O\) of the segment \(OB\), we determine the angle \(AOA_1\) (it is assumed that it is acute). In the region of the triangle \(AOA_1\), the solution is constructed as in the degenerate case of the initial characteristic problem. Then we divide the segment \(OA_1\) into a number of small parts by the points \(0, 1, 2,\ldots\). We begin the construction from point 1. Drawing from it a ray at an angle \(\frac{1}{2}(\Theta_1+\Theta_6)\) to the \(y\)-axis (in the general case, when \(\Theta\) varies along \(OB\), this can be achieved by a number of successive approximations), we find the position of point 6. We determine \(A_z\) at it from the known \(A_z\) at point 1. From the known \(\Theta\) and \(A_z\) we find \(A_\alpha\) and \(z\). Point 7 is determined as in the case of the initial characteristic problem. Point 11 is determined as point 6, and so on.

The velocity field is determined from the known stress field in the same way as in the case of an ideally plastic body. If one starts from the hypothesis of a single flow curve “the intensity of tangential stresses \(k\)—the intensity of strains \(e_i\),” then the magnitude \(e_i\) can be determined from the hardness. Having the known \(e_i\) and proceeding from the condition of incompressibility, one can find at any point the magnitude of the principal logarithmic strains \(e_1\) and \(e_2\). Indeed, from the system

\[ \frac{\sqrt{2}}{3}\sqrt{(e_1-e_2)^2+e_1^2+e_2^2}=e_i,\qquad e_1+e_2=0 \tag{3.14} \]

we find

\[ e_1=-e_2=\frac{\sqrt{3}}{2}\,e_i . \tag{3.15} \]

To determine the directions of these strains, additional conditions are required. The directions of the principal strains are known, na—

for example, along the axes of symmetry—one of the directions coincides with the axis, the other is perpendicular to it.

The developed method has been successfully tested in the investigation of the stress–strain state of a cylinder compressed in the radial direction.

4. Axisymmetric deformation. Axisymmetric deformation arises when a body having an axis of symmetry is loaded by an axisymmetric load.

In a cylindrical coordinate system (the \(z\)-axis coincides with the axis of symmetry) the components of stresses and velocities do not depend on the polar angle \(\varphi\), and \(\tau_{r\varphi}=\tau_{\varphi z}=v_{\varphi}=0\). The problem is statically indeterminate, and its solution is very difficult. We shall restrict ourselves to considering a method for investigating the stressed state under the Haar–Karman assumption that the circumferential stress is equal to one of the principal stresses in the meridional planes \((\sigma_{\varphi}=\sigma_2)\), which is widely used in the engineering theory of plasticity. Under this assumption, A. Yu. Ishlinskii [6] and later Shield [7] investigated the problem of axisymmetric deformation of an ideally plastic material.

Equation (1.1), written in terms of the principal stresses, takes the form

\[ (\sigma_1-\sigma_2)^2=3\kappa^2 . \tag{4.1} \]

After substituting the known relations for the principal stresses, we have

\[ (\sigma_z-\sigma_r)^2+4\tau_{rz}^2=3\kappa^2(r,z,q) \tag{4.2} \]

The equilibrium equations for axisymmetric deformation are

\[ \frac{\partial\sigma_r}{\partial r}+\frac{\partial\tau_{rz}}{\partial z}+\frac{\sigma_r-\sigma_\varphi}{r}=0,\qquad \frac{\partial\tau_{rz}}{\partial r}+\frac{\partial\sigma_z}{\partial z}+\frac{\tau_{rz}}{r}=0. \tag{4.3} \]

According to the relations [7],

\[ \sigma_r=p-\frac{\sqrt{3}}{2}\kappa\sin 2\Theta,\qquad \sigma_z=p+\frac{\sqrt{3}}{2}\kappa\sin 2\Theta, \]

\[ \tau_{rz}=\frac{\sqrt{3}}{2}\kappa\cos 2\Theta,\qquad \sigma_\varphi=p\pm\frac{\sqrt{3}}{2}\kappa \tag{4.4} \]

we pass to new variables—the normal pressure

\[ p=\frac{1}{2}(\sigma_r+\sigma_z) \tag{4.5} \]

and the angle \(\Theta\) between the tangent to the slip line \(\alpha\) and the \(x\)-axis. Equation (4.2) is then satisfied identically, and (4.3) take the form

\[ \left. \begin{aligned} \frac{\partial p}{\partial r} &-\sqrt{3}\kappa\cos 2\Theta\,\frac{\partial\Theta}{\partial r} -\sqrt{3}\kappa\sin 2\Theta\,\frac{\partial\Theta}{\partial z} \\ &=\frac{\sqrt{3}}{2}\sin 2\Theta\,\frac{\partial\kappa}{\partial r} -\frac{\sqrt{3}}{2}\cos 2\Theta\,\frac{\partial\kappa}{\partial z} +\frac{\sqrt{3}\kappa}{2r}(\pm 1+\sin 2\Theta), \\[1.2em] \frac{\partial p}{\partial z} &-\sqrt{3}\kappa\sin 2\Theta\,\frac{\partial\Theta}{\partial r} +\sqrt{3}\kappa\cos 2\Theta\,\frac{\partial\Theta}{\partial z} \\ &=-\frac{\sqrt{3}}{2}\cos 2\Theta\,\frac{\partial\kappa}{\partial r} -\frac{\sqrt{3}}{2}\sin 2\Theta\,\frac{\partial\kappa}{\partial z} -\frac{\sqrt{3}\kappa}{2r}\cos 2\Theta . \end{aligned} \right\} \tag{4.6} \]

Equations (4.6) differ from the analogous equations written for an ideally plastic material only in the right-hand side. Since the right-hand side of differential equations affects neither the type of the equations nor the direction of the characteristics, system (4.6), as in the case of an ideally plastic material, is a system of hyperbolic type. Its characteristics coincide with the slip lines.

In a local coordinate system formed by tangents to the slip lines, we have

\[ \left. \begin{aligned} \frac{\partial}{\partial \alpha}\left(p-\sqrt{3}\varkappa\Theta\right) &= \frac{\sqrt{3}}{2}\frac{\varkappa}{r}(\sin\Theta \pm \cos\Theta) - \frac{\sqrt{3}}{2}\frac{\partial \varkappa}{\partial \beta}, \\ \frac{\partial}{\partial \beta}\left(p+\sqrt{3}\varkappa\Theta\right) &= -\frac{\sqrt{3}}{2}\frac{\varkappa}{r}(\cos\Theta \pm \sin\Theta) - \frac{\sqrt{3}}{2}\frac{\partial \varkappa}{\partial \alpha}. \end{aligned} \right\} \tag{4.7} \]

Integrating equations (4.7), we obtain the relations on the slip lines:

\[ \frac{p}{\sqrt{3}\varkappa}-\Theta=-A_{\alpha}, \qquad \frac{p}{\sqrt{3}\varkappa}+\Theta=-A_{\beta}, \tag{4.8} \]

where

\[ \left. \begin{aligned} A_{\alpha} &= \frac{1}{2\varkappa} \int_{\alpha} \left[ \frac{\partial \varkappa}{\partial \beta} - \frac{\varkappa}{r}(\sin\Theta \pm \cos\Theta) \right]\,d\alpha, \\ A_{\beta} &= \frac{1}{2\varkappa} \int_{\beta} \left[ \frac{\partial \varkappa}{\partial \alpha} + \frac{\varkappa}{r}(\cos\Theta \pm \sin\Theta) \right]\,d\beta . \end{aligned} \right\} \tag{4.9} \]

The solution of boundary-value problems in the case under consideration can be constructed in the same way as in the case of plane deformation. Instead of relations (3.10), the following may be recommended:

\[ \left. \begin{aligned} (A_{\alpha})_{7} &= (A_{\alpha})_{1} + \frac{l_{1-7}}{2\varkappa_{7}} \left[ \frac{\varkappa_{A}-\varkappa_{C}}{\Delta l} - \frac{\varkappa_{B}}{r_{B}}(\sin\Theta_{1}\pm\cos\Theta_{1}) \right], \\ (A_{\beta})_{7} &= (A_{\beta})_{2} + \frac{l_{2-7}}{2\varkappa_{7}} \left[ \frac{\varkappa_{F}-\varkappa_{D}}{\Delta l} + \frac{\varkappa_{E}}{r_{E}}(\cos\Theta_{2}\pm\sin\Theta_{2}) \right]. \end{aligned} \right\} \tag{4.10} \]

The normal pressure and the angle \(\Theta\) are determined from the known \(A_{\alpha}\), \(A_{\beta}\):

\[ p=\sqrt{3}\varkappa(-A_{\alpha}-A_{\beta}), \qquad \Theta=\frac{1}{2}(A_{\alpha}-A_{\beta}). \tag{4.11} \]

By this method, stress diagrams were obtained in the neck of a stretched specimen.

Thus, having the necessary set of boundary conditions and having determined at various points the yield limit increased due to work hardening, one can, in the cases considered, determine the stress-strain state under plastic deformation of a hardening body.

REFERENCES

1. P. Bridgman. Investigation of Large Plastic Deformations and Rupture. Moscow, 1955.
2. A. M. Rozenberg, L. A. Khvorostukhin. ZhTF, vol. XXV, no. 2, 1955.
3. G. A. Smirnov-Alyaev. Strength of Materials under Plastic Deformation. Moscow, 1962.
4. L. M. Kachanov. Fundamentals of the Theory of Plasticity. Moscow, 1956.
5. A. I. Kuznetsov. Vestnik LGU, no. 13, 1958.
6. A. Yu. Ishlinskii. Applied Mathematics and Mechanics, vol. VII, issue 3, 1944.
7. Shield. Mechanics, 1, 41, 1957.