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# Determination of the principal stresses

The following stress state is given:

$\newcommand{\myvec}[1]{{\begin{bmatrix}#1\end{bmatrix}}}$

$S = \myvec{2 & 1 & 0\\1 & 1 & 0\\0 & 0 & 0} \cdot 100 \, Nmm^{-2}$

Is it an uniaxial, plane or spatial stress state? Determine the principal stresses and their directions!

## Solution

An uniaxial stress state exists if only σx or σy or σz is given (an exercise for the uniaxial stress state is available here). A plane stress state can have two principal stresses and one shear stress. This is the case here, so it is a plane stress state.

The principal stresses σ1 and σ2 and the angle φ0 can be determined with Mohr’s circle of stress, which leads to the following representation:

If the stresses σx, σy and τxy are plotted to scale, the principal stresses and the angle can be measured from Mohr’s circle of stresses.

Alternatively, the values ​​can be calculated as follows.

$\tag{1} \sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left( \frac{\sigma_x -\sigma_y}{2}\right)^2 + \tau_{xy}^2}$

$\tag{2} \sigma_{1} = 262\,Nmm^{-2}$

$\tag{3} \sigma_{2} = 38\,Nmm^{-2}$

For the angle φ 0 applies

$\tag{4} tan2\phi_0 = \frac{2 \tau_{xy}}{\sigma_x -- \sigma_y}$

$\tag{5} \phi_0 = \frac{1}{2} \cdot arctan \left( \frac{2 \tau_{xy}}{\sigma_x -- \sigma_y} \right)$

$\tag{6} \phi_0 = 31.7°$

φ0 is the direction of the principal stress σ1. The direction for σ2 is φ0 + π / 2. The principal stresses and their directions are thus determined for the plane stress state present here.