Without more detail it is difficult to know what you are after however the 3D stress has 9 components.
It is described by the stress tensor [tex]T[/tex]
Just as a one dimensional stress may be decomposed into normal and tengential (shear) stress so may the the three dimensional version be decomposed into a uniform 'hydrostatic' normal stress, [tex]{T_m}[/tex] and a set of 3D shear stresses, [tex]{T_d}[/tex]
The normal stress is often called the mean stress tensor and the shear stress tensor the deviator stress as it gives the deviation from uniformitivity, which is what I think you seek.
[tex]\begin{array}{l}<br />
{T_d}\quad = \quad \left( {\begin{array}{*{20}{c}}<br />
{\frac{{2{\sigma _{xx}} - {\sigma _{yy}} - {\sigma _{zz}}}}{3}} & {{\sigma _{xy}}} & {{\sigma _{xz}}} \\<br />
{{\sigma _{xy}}} & {\frac{{2{\sigma _{yy}} - {\sigma _{xx}} - {\sigma _{zz}}}}{3}} & {{\sigma _{yz}}} \\<br />
{{\sigma _{xz}}} & {{\sigma _{yz}}} & {\frac{{2{\sigma _{zz}} - {\sigma _{yy}} - {\sigma _{xx}}}}{3}} \\<br />
\end{array}} \right) \\ <br />
{T_m}\quad = \quad \left( {\begin{array}{*{20}{c}}<br />
{{\sigma _m}} & 0 & 0 \\<br />
0 & {{\sigma _m}} & 0 \\<br />
0 & 0 & {{\sigma _m}} \\<br />
\end{array}} \right) \\ <br />
T\quad = \quad {T_m}\quad + \quad {T_d} \\ <br />
\end{array}[/tex]