Here we will derive the principal stresses in a two-dimensional system. Starting with a unit cube subjected to combined tensile and shear stresses, take a cut at an arbitrary angle to make available the rotated normal and shear forces.

Summing forces in the x’ direction:

\sum F_{x’}=A\sigma_{x’}-\sigma_x \cos\alpha A \cos\alpha – \tau_{xy} \sin\alpha A \cos\alpha – \tau_{xy}\cos\alpha A\sin\alpha – \sigma_y \sin\alpha A\sin\alpha

Setting the equation equal to 0 and dividing each term by A, then solving for $$\sigma_{x’}$$:

\sigma_{x’} = \sigma_x cos^2(\alpha) + 2\tau_{xy} \sin\alpha \cos\alpha+\sigma_y sin^2(\alpha)

Summing forces in the y’ direction:

\sum F_{y’} = A\tau_{x’y’}+\sigma_x \sin\alpha A\cos\alpha – \sigma_y \cos\alpha A\sin\alpha – \tau_{xy}\cos\alpha A\cos\alpha +\tau_{xy} \sin\alpha A\sin\alpha

Again setting the equation equal to 0 and dividing each term by A and rearranging:

\tau_{x’y’} = -\sigma_x \sin\alpha \cos\alpha +\sigma_y \cos\alpha \sin\alpha + \tau_{xy}cos^2(\alpha) – \tau_{xy}sin^2(\alpha)

\tau_{x’y’} = \sigma_y \cos\alpha \sin\alpha – \sigma_x \sin\alpha \cos\alpha + \tau_{xy}(cos^2(\alpha) – sin^2(\alpha))

\tau_{x’y’} = (\sigma_y -\sigma_x)\cos\alpha \sin\alpha – \tau_{xy}(\cos^2\alpha – \sin^2\alpha)

To obtain similar relations for $$\sigma_{y’}$$, we need only to evaluate $$\sigma_{x’}$$at $$\alpha+\pi/2$$, since $$\sigma_{x’}$$ and $$\sigma_{y’}$$ are orthogonal to each other and thus 90 degrees apart.

\sigma_{y’}=\sigma_x\cos^2\left(\alpha+\frac{\pi}{2}\right)+2\tau_{xy}\sin\left(\alpha+\frac{\pi}{2}\right)\cos\left(\alpha+\frac{\pi}{2}\right)+\sigma_y\sin^2\left(\alpha+\frac{\pi}{2}\right)

Then, using the relations:

\sin\left(\alpha+\frac{\pi}{2}\right)=\cos\alpha

\cos\left(\alpha+\frac{\pi}{2}\right)=-\sin\alpha

The equation for $$\sigma_{y’}$$ reduces to:

\sigma_{y’}=\sigma_x\sin^2\alpha-2\tau_{xy}\cos\alpha\sin\alpha+\sigma_y\cos^2\alpha

Next, we put these equations into forms that are functions of $$2\alpha$$ instead of $$\alpha$$. To do this we will apply the following trig identities:

\sin^2\alpha = \frac{1-\cos2\alpha}{2}

\cos^2\alpha = \frac{1+\cos2\alpha}{2}

\sin2\alpha=2\sin\alpha\cos\alpha

\cos2\alpha=\cos^2\alpha-\sin^2\alpha

Incorporating these into the equations for $$\sigma_{x’}$$, $$\sigma_{y’}$$, and $$\tau_{y’}$$:

\sigma_{x’}=\sigma_x\frac{1+\cos2\alpha}{2}+\tau_{xy}\sin2\alpha+\sigma_y\frac{1-\cos2\alpha}{2}

\sigma_{x’}=\frac{\sigma_x+\sigma_y}{2} + \frac{\sigma_x-\sigma_y}{2}\cos2\alpha+\tau_{xy}\sin2\alpha

\sigma_{y’}=\sigma_x\frac{1-\cos2\alpha}{2} – \tau_{xy}\sin2\alpha + \sigma_y\frac{1+\cos2\alpha}{2}

\sigma_{y’}=\frac{\sigma_x+\sigma_y}{2} + \frac{\sigma_y-\sigma_x}{2}\cos2\alpha-\tau_{xy}\sin2\alpha

\tau_{x’y’}=\frac{\sigma_y\sin2\alpha}{2}-\frac{\sigma_x\sin2\alpha}{2}+\tau_{xy}\cos2\alpha

\tau_{x’y’}=\frac{\sigma_y-\sigma_x}{2}\sin2\alpha+\tau_{xy}\cos2\alpha

\tau_{x’y’}=-\frac{\sigma_x-\sigma_y}{2}\sin2\alpha+\tau_{xy}\cos2\alpha

To determine the principal stresses, we differentiate each equation with respect to $$\alpha$$ and set equal to 0 to find the max and min values. These equations allow the stress components to be determined in any desired rotated coordinate frame of angle $$\alpha$$.

Starting with $$\sigma_{x’}$$:

\frac{d}{d\alpha}\sigma_{x’} = -(\sigma_x-\sigma_y)\sin2\alpha+2\tau_{xy}\cos2\alpha=0

(\sigma_x-\sigma_y)\sin2\alpha=2\tau_{xy}\cos2\alpha=0

\tan2\alpha=\frac{2\tau_{xy}}{\sigma_x-\sigma_y}=\frac{\tau_{xy}}{\frac{\sigma_x-\sigma_y}{2}}

Using the above we can construct three equations based on the tangent definition and the following figure.

\alpha=\frac{1}{2}\text{atan}\left(\frac{2\tau_{xy}}{\sigma_x-\sigma_y}\right)

\sin2\alpha=\pm\frac{\tau_{xy}}{\sqrt{\tau_{xy}^2+\left(\frac{\sigma_x-\sigma_y}{2}\right)^2}}

\cos2\alpha=\pm\frac{\sigma_x-\sigma_y}{\sqrt{\tau_{xy}^2+\left(\frac{\sigma_x-\sigma_y}{2}\right)^2}}

Then we insert these into our equation for $$\sigma_{x’}$$ to arrive at the principal stress equation:

\sigma_{x’}=\frac{\sigma_x+\sigma_y}{2}\pm\frac{\sigma_x-\sigma_y}{2}\frac{\frac{\sigma_x-\sigma_y}{2}}{\sqrt{\tau_{xy}^2+\left(\frac{\sigma_x-\sigma_y}{2}\right)^2}} \pm \tau_{xy}\frac{\tau_{xy}}{\sqrt{\tau_{xy}^2+\left(\frac{\sigma_x-\sigma_y}{2}\right)^2}}

\sigma_{x’}=\frac{\sigma_x+\sigma_y}{2}\pm\frac{\tau_{xy}^2+\left(\frac{\sigma_x-\sigma_y}{2}\right)^2}{\sqrt{\tau_{xy}^2+\left(\frac{\sigma_x-\sigma_y}{2}\right)^2}}

\sigma_{x’}=\frac{\sigma_x+\sigma_y}{2}\pm\frac{ \left( \tau_{xy}^2+\left(\frac{\sigma_x-\sigma_y}{2}\right)^2 \right)^1}{\left(\tau_{xy}^2+\left(\frac{\sigma_x-\sigma_y}{2}\right)^2 \right)^{1/2}}

\sigma_1\text{,}\sigma_2=\frac{\sigma_x+\sigma_y}{2}\pm\sqrt{\tau_{xy}^2+\left(\frac{\sigma_x-\sigma_y}{2}\right)^2}

We repeat the above procedure to obtain the principal shear. Differentiating $$\tau_{x’y’}$$ with respect to $$\alpha$$,

\frac{d}{d\alpha}\tau_{x’y’}=-(\sigma_x-\sigma_y)\cos2\alpha-2\tau_{xy}\sin2\alpha=0

2\tau_{xy}\sin2\alpha=-(\sigma_x-\sigma_y)\cos2\alpha

\tan2\alpha=-\frac{\sigma_x-\sigma_y}{2\tau_{xy}}=-\frac{\dfrac{\sigma_x-\sigma_y}{2}}{\tau_{xy}}

\alpha=\frac{1}{2}\arctan\left(-\frac{\sigma_x-\sigma_y}{2\tau_{xy}}\right)

\sin2\alpha = \pm\frac{\dfrac{\sigma_x-\sigma_y}{2}}{\sqrt{\tau_{xy}^2+\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2}}

\cos2\alpha = \pm\frac{\tau_{xy}}{\sqrt{\tau_{xy}^2+\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2}}

\tau_{x’y’} = \pm \frac{\sigma_x-\sigma_y}{2}\frac{\dfrac{\sigma_x-\sigma_y}{2}}{\sqrt{\tau_{xy}^2+\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2}} \pm \tau_{xy}\frac{\tau_{xy}}{\sqrt{\tau_{xy}^2+\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2}}

\tau_{x’y’} = \pm \frac{\sigma_x-\sigma_y}{2} \frac{\dfrac{\sigma_x-\sigma_y}{2}}{\sqrt{\tau_{xy}^2+\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2}} \pm \frac{\tau_{xy}^2}{\sqrt{\tau_{xy}^2+\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2}}

\tau_{x’y’} = \pm\frac{\tau_{xy}^2+\left(\frac{\sigma_x-\sigma_y}{2}\right)^2}{\sqrt{\tau_{xy}^2+\left(\frac{\sigma_x-\sigma_y}{2}\right)^2}}

\tau_{x’y’} = \pm\frac{\left(\tau_{xy}^2+\left(\frac{\sigma_x-\sigma_y}{2}\right)^2\right)^1}{\left(\tau_{xy}^2+\left(\frac{\sigma_x-\sigma_y}{2}\right)^2\right)^{1/2}}

\tau_{max}=\sqrt{\tau_{xy}^2+\left(\frac{\sigma_x-\sigma_y}{2}\right)^2}

Mohr’s circle can be constructed from the stress components to graphically represent the stress state at the point.

### References

1. Dowling, Norman E. Mechanical Behavior of Materials. s.l. : Prentice Hall, 1993.
2. Beer, Ferdinand P. and Johnston, Russell E. Mechanics of Materials. s.l. : McGraw-Hill, 1981.