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.

stress-element-2-dv-a

Summing forces in the x’ direction:

\begin{equation}
\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
\end{equation}

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

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

Summing forces in the y’ direction:

\begin{equation}
\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
\end{equation}

Again setting the equation equal to 0 and dividing each term by A and rearranging:
\begin{equation}
\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)
\end{equation}

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

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

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.

\begin{equation}
\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)
\end{equation}

Then, using the relations:

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

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

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

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

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:

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

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

\begin{equation}
\sin2\alpha=2\sin\alpha\cos\alpha
\end{equation}

\begin{equation}
\cos2\alpha=\cos^2\alpha-\sin^2\alpha
\end{equation}

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

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

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

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

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

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

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

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

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 \(\sigma_{x’}\):

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

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

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

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

tan_relation_for_normal_stress_eq_1-dv-a

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

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

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

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

\begin{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}}
\end{equation}

\begin{equation}
\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}}
\end{equation}

\begin{equation}
\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}}
\end{equation}

\begin{equation}
\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}
\end{equation}

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

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

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

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

tan_relation_for_shear_stress_eq_1-dv-a

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

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

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

\begin{equation}
\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}}
\end{equation}

\begin{equation}
\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}}
\end{equation}

\begin{equation}
\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}}
\end{equation}

\begin{equation}
\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}}
\end{equation}

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

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

mohr_circle_2d-7-dv-a

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.