Hencky Relations

Unless $$\tau_{x y}=\pm k_{i}$$, where $$k$$ is the shear yield stress, the $$x$$- and $$y$$-directions (or axes) will not correspond to the directions of the $$\alpha -$$ and $$\beta -$$ slip lines, which are themselves at ± 45° to the directions of principal stresses acting on the element. In general, we need to examine the stresses on a small curvilinear element in the $$x$$ - $$y$$ plane upon which a shear stress and a hydrostatic stress are acting, and where the principal stresses are $$– p – k$$ and $$– p + k$$ for a situation where there is plane strain compression, such as in forging or indentation in which $$k$$ is a constant but $$p$$ can vary from point to point: We can then identify on this diagram the directions of principal stress $$\sigma_1$$ and $$\sigma_2$$ (remembering that $$\sigma_1 > \sigma_2$$, and which of the lines are $$\alpha$$-lines and which are $$\beta$$-lines. We can also specify the angle $$\varphi$$ of the $$\alpha$$-lines with respect to the $$x$$-axis: Suppose that the $$\alpha$$-slip line passing through the element makes an angle $$\varphi$$ with respect to reference $$x$$- and $$y$$-axes, as in the above diagram. The $$\beta$$-slip line must then make an angle of 90° $$+ \varphi$$ with respect to the x-axis, so that an anticlockwise rotation from the $$\alpha$$-slip line to the $$\beta$$-slip line crosses the direction of maximum principal stress, $$\sigma_1$$ .

The direction parallel to the principal stress $$\sigma_1$$ makes an angle of 45° $$+\varphi$$ with respect to the $$x$$-axis and the direction parallel to the principal stress $$\sigma_2$$ makes an angle of $$135^{\circ}+\varphi^{\circ} 45^{\circ}-\varphi$$ with respect to the $$x$$-axis.

On a Mohr’s circle, this all looks like: where A and B represent the stress states along the $$\alpha$$- and $$\beta$$-lines respectively.

Hence, from the above,

$\sigma_{x x}=-p-k \sin 2 \varphi$

$\sigma_{y y}=-p+k \sin 2 \varphi$

$\tau_{x y}=k \cos 2 \varphi$

while the tensile stress in the $$z$$-direction in plain strain plastic yielding is simply:

$\frac{1}{2}\left(\sigma_{x x}+\sigma_{y y}\right)=-p$

Substituting these expressions into the equilibrium equations

$\frac{\partial \sigma_{x x}}{\partial x}+\frac{\partial \tau_{x y}}{\partial y}=0$

$\frac{\partial \tau_{x y}}{\partial x}+\frac{\partial \sigma_{y y}}{\partial y}=0$

and recognizing that $$k$$ is a constant independent of $$x$$ and $$y$$, we obtain two equations for $$p$$ and $$\varphi$$ as a function of $$x$$ and $$y$$:

$-\frac{\partial p}{\partial x}-2 k \cos 2 \varphi \frac{\partial \varphi}{\partial x}-2 k \sin 2 \varphi \frac{\partial \varphi}{\partial y}=0$

$-2 k \sin 2 \varphi \frac{\partial \varphi}{\partial x}-\frac{\partial p}{\partial y}+2 k \cos 2 \varphi \frac{\partial \varphi}{\partial y}=0$

For $$\varphi=0^{\circ}$$, in which case the $$\alpha$$ and $$\beta$$ lines coincide with the external $$x$$- and $$y$$-axes respectively at a particular position, these equations become

$\frac{\partial}{\partial x}(p+2 k \varphi)=0$

$\frac{\partial}{\partial y}(p-2 k \varphi)=0$

Integrating these equations we find

$p+2 k \varphi=f_{1}(y)+C_{1} \label{eq:1}$

$p-2 k \varphi=f_{2}(x)+C_{2} \label{eq:2}$

as the most general form of the solutions of these two partial differential equations. However, we know that when $$\varphi$$ is exactly zero, $$p$$ must have the same value in both Equations $$\ref{eq:1}$$ and Eq. $$\ref{eq:2}$$. Hence it follows that $$f_{1}(x)=f_{2}(y)=0$$.

In general for points in a slip-line field we have therefore proved that the Hencky relations have to be satisfied:

Hencky relations

The hydrostatic pressure $$p$$ varies linearly with the angle $$\varphi$$ turned by a slip line.

• $$p+2 k \varphi$$ is constant along an $$\alpha$$ line
• $$p-2 k \varphi$$ is constant along a $$\beta$$ line

where the angle φ is in radians.