Perfect plasticity

A perfectly plastic material has a yield surface that does not change with accumulated plastic deformation. This removes hardening while retaining the distinction between elastic and plastic loading.

In antiplane perfect plasticity, velocity $w=v_3$ and shear stresses $a={\sigma }_{13},b={\sigma }_{23}$ depend only on $x_1,x_2$. Force balance is $a_{,1}+b_{,2}=0$, the fixed yield surface is $f(a,b)=k$, and the associated flow rule is $\nabla w=2\dot{\lambda }(f_a,f_b)$ with $\dot{\lambda }\ge 0$. Differentiating the yield equation in each coordinate and combining it with equilibrium shows that $a,b,w$ are constant along characteristic curves tangent to $(-f_b,f_a)$. For example $-f_b{a}_{,1}+f_a{a}_{,2}=f_b{b}_{,2}+f_a{a}_{,2}=0$. Nonzero regular yield gradient is needed for a characteristic direction.

A centred fan consists of straight characteristic curves emanating from one point. Stress and antiplane velocity are constant on each ray and depend on the polar angle $\varphi$. The radial tangent condition is $(\cos \varphi ,\sin \varphi )\parallel (-f_b,f_a)$, so $f_a\cos \varphi +f_b\sin \varphi =0$. A boundary ray carrying a traction-free crack-face stress $b=0$ has slope $\tan {\varphi }_b=-f_a/f_b$, interpreted as vertical when $f_b=0$. For the elliptic yield condition $a^2/A^2+b^2/B^2=1$, a forward fan has $a=-A^2\sin \varphi /H$, $b=B^2\cos \varphi /H$, $H=(A^2{\sin }^2\varphi +B^2{\cos }^2\varphi {)}^{1/2}$. The boundary rays are vertical; the adjacent rear regions have constant stresses $(\mp A,0)$. Velocity is a function of angle whose magnitude requires further loading data.

In plane strain perfect plasticity with ${\sigma }_{11}=-p+k\sin 2\varphi$, ${\sigma }_{22}=-p-k\sin 2\varphi$ and ${\sigma }_{12}=-k\cos 2\varphi$, equilibrium givesThe slip line tangent directions $(\cos \varphi ,\sin \varphi )$ and $(-\sin \varphi ,\cos \varphi )$ are eigenvectors with eigenvalues $+1,-1$. Taking directional derivatives proves the two Hencky stress relations. These sign choices must be changed consistently if the shear-stress convention changes.

For a symmetric reentrant notch opening towards negative $x_1$, let its semi-angle be $0\le \gamma \le \pi /2$. On the upper free face, tensile tangential stress selects $p=-k$, $\varphi =\pi /4-\gamma$. The adjacent centred fan has radial $\beta$ lines and circular $\alpha$ lines: $\varphi =\vartheta -\pi /2$, $p=-k+2k[\vartheta -(3\pi /4-\gamma )]$, for $\pi /4\le \vartheta \le 3\pi /4-\gamma$. It joins a constant stress region containing the forward axis, where $\varphi =-\pi /4$ and $p=-k(1+\pi -2\gamma )$. Therefore ${\sigma }_{22}=k(2+\pi -2\gamma )$ in that fully plastic notch-tip region. This is a local stress field, not a claim that an arbitrary specimen is plastically yielded everywhere.

For an associated flow rule in the reduced plane-strain yield coordinates, the rate-of-strain tensor has zero extension along both slip line directions. Write velocity as $u{e}_{\alpha }+v{e}_{\beta }$, with the local orthonormal axes turning through angle $\varphi$. Differentiating along an $\alpha$ curve gives longitudinal extension $du-vd\varphi$; along a $\beta$ curve it gives $dv+ud\varphi$. Setting these extensions to zero proves the relations for the corresponding anisotropic yield surface parameterization.

Parameterize a fixed yield surface by stress-space arc length $l$ with ${\xi }_{,l}=-\cos 2\varphi$ and ${\tau }_{,l}=-\sin 2\varphi$. For mean normal stress $\sigma$, force balance gives $\nabla \sigma =M\nabla l$, where $M=\left(\begin{array}{cc}\cos 2\varphi & \sin 2\varphi \\ \sin 2\varphi & -\cos 2\varphi \end{array}\right)$. Its eigenvectors along angles $\varphi$ and $\varphi +\pi /2$ have eigenvalues $+1$ and $-1$, proving the displayed slip line invariants.

A slip line is a characteristic curve of a yielded plane-strain plastic equilibrium system. The two orthogonal families carry stress and velocity compatibility relations. Their directions need not agree with principal stress directions.