# Two Dimensional Stress Analysis

## Uniaxial Stress

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

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:

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

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:

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

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 $$( x',y')$$:

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