Consider a point 
in a continuum under a state of plane stress, or plane strain, with stress components 
and all other stress components equal to zero (Figure 7.1, Figure 8.1). From static equilibrium of an infinitesimal material element at (Figure 8.2), the normal stress 
and the shear stress 
on any plane perpendicular to the 
-
plane passing through with a unit vector 
making an angle of 
with the horizontal, i.e. 
is the direction cosine in the direction, is given by:
These equations indicate that in a plane stress or plane strain condition, one can determine the stress components at a point on all directions, i.e. as a function of , if one knows the stress components on any two perpendicular directions at that point. It is important to remember that we are considering a unit area of the infinitesimal element in the direction parallel to the -
plane.
Figure 8.1 – Stress transformation at a point in a continuum under plane stress conditions.
Figure 8.2 – Stress components at a plane passing through a point in a continuum under plane stress conditions.
The principal directions (Figure 8.3), i.e. orientation of the planes where the shear stress components are zero, can be obtained by making the previous equation for the shear stress equal to zero. Thus we have:
and we obtain
This equation defines two values 
which are 
apart (Figure 8.3). The same result can be obtained by finding the angle which makes the normal stress a maximum, i.e. 
The principal stresses 
and 
, or minimum and maximum normal stresses 
and 
, respectively, can then be obtained by replacing both values of into the previous equation for . This can be achieved by rearranging the equations for and , first transposing the first term in the first equation and squaring both sides of each of the equations then adding them. Thus we have
![\begin{align}
\left[ \sigma_\mathrm{n} - \tfrac{1}{2} ( \sigma_x + \sigma_y )\right]^2 + \tau_\mathrm{n}^2 &= \left[\tfrac{1}{2}(\sigma_x - \sigma_y)\right]^2 + \tau_{xy}^2 \\
(\sigma_\mathrm{n} - \sigma_\mathrm{avg})^2 + \tau_\mathrm{n}^2 &= R^2 \end{align}\,\!](http://upload.wikimedia.org/math/1/2/9/129d8cd805c51a932abec44d9278cf67.png)
where
![R = \sqrt{\left[\tfrac{1}{2}(\sigma_x - \sigma_y)\right]^2 + \tau_{xy}^2} \quad \text{and} \quad \sigma_\mathrm{avg} = \tfrac{1}{2} ( \sigma_x + \sigma_y )\,\!](http://upload.wikimedia.org/math/f/5/0/f50295adc0dd48728357f04de663c5dc.png)
which is the equation of a circle of radius 
centered at a point with coordinates ![[\sigma_\mathrm{avg}, 0]\,\!](http://upload.wikimedia.org/math/b/6/4/b6468372d603f706ec51b7825dc6e679.png)
, called Mohr’s circle. But knowing that for the principal stresses the shear stress 
, then we obtain from this equation:
![\sigma_1 =\sigma_\mathrm{max} = \tfrac{1}{2}(\sigma_x + \sigma_y) + \sqrt{\left[\tfrac{1}{2}(\sigma_x - \sigma_y)\right]^2 + \tau_{xy}^2}\,\!](http://upload.wikimedia.org/math/d/a/f/dafae3184798e2ade5e5334d39e943d0.png)
![\sigma_2 =\sigma_\mathrm{min} = \tfrac{1}{2}(\sigma_x + \sigma_y) - \sqrt{\left[\tfrac{1}{2}(\sigma_x - \sigma_y)\right]^2 + \tau_{xy}^2}\,\!](http://upload.wikimedia.org/math/9/b/b/9bb8c47e56d3a4cf0c697dac33084684.png)
Figure 8.3 – Transformation of stresses in two dimensions, showing the planes of action of principal stresses, and maximum and minimum shear stresses.
When 
the infinitesimal element is oriented in the direction of the principal planes, thus the stresses acting on the rectangular element are principal stresses: 
and 
. Then the normal stress and shear stress as a function of the principal stresses can be determined by making . Thus we have
Then the maximum shear stress 
occurs when 
, i.e. 
(Figure 8.3):
Then the minimum shear stress 
occurs when 
, i.e. 
(Figure 8.3):
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