Two Dimensional Stress Analysis

Uniaxial Stress

A solid beam under tensile stress will eventually fracture due to a shearing effect in an internal plane.

shear

Here, \( \sigma _x \) is the tensile stress acting in the same axis as the beam, \( \tau _ \theta \) is the resultant shear stress acting at an angle \( \theta \) to the vertical axis.

To apply force equilibrium, the stresses are converted to forces:

sforces

Resolving the \( \theta \) forces into vertical and horizontal components and equating the horizontal forces acting on the beam yields

$$ \sigma _ \theta = \sigma _x + \tau _ \theta tan \theta $$

Equation the vertical forces yields:

$$ \sigma _ \theta tan \theta = - \tau _ \theta $$

From these two equilibriums, expressions to determine the \( \theta \) stresses can be derived:

$$ \sigma _ \theta = \sigma _x cos ^2 \theta $$
$$ \tau _ \theta = - \frac{\sigma _x}{2} sin 2 \theta $$


Biaxial Stress

biaxial

For normal stress acting in two dimensions, the same equilibrium relations can be resolved to find the internal \( \theta \) forces:

$$ \sigma _ \theta = \frac{\sigma _x + \sigma _y}{2} + \frac{\sigma _x - \sigma _y}{2} cos 2 \theta $$
$$ \tau _ \theta = - \frac{\sigma _x - \sigma _y}{2} sin 2 \theta $$

For a body under pure shearing stress as shown:

pshear

$$ \sigma _ \theta = \tau _{xy} sin 2 \theta $$
$$ \tau _ \theta = \tau _{xy} cos 2 \theta $$

fshear

For a body under both shear and normal stress, the above two equilibriums can be combined additively:

$$ \sigma _ \theta = \frac{\sigma _x + \sigma _y}{2} + \frac{\sigma _x - \sigma _y}{2} cos 2 \theta + \tau _{xy} sin 2 \theta $$
$$ \tau _ \theta = - \frac{\sigma _x - \sigma _y}{2} sin 2 \theta + \tau _{xy} cos 2 \theta $$

Stress Transformations

To transform the orthogonal stresses to a different orientation, shown \( (x,y) \) to \( ( x',y') \):

transform

$$ \sigma _{x'} = \frac{\sigma _x + \sigma _y}{2} + \frac{\sigma _x - \sigma _y}{2} cos 2 \theta + \tau _{xy} sin 2 \theta $$
$$ \sigma _{y'} = \frac{\sigma _x + \sigma _y}{2} - \frac{\sigma _x - \sigma _y}{2} cos 2 \theta + \tau _{xy} sin 2 \theta $$
$$ \tau _{x'y'} = - \frac{\sigma _x - \sigma _y}{2} sin 2 \theta + \tau _{xy} cos 2 \theta $$