Failure Theories-NEW PAGE

BSM: Nice work on this new page. A table at the end summarizing the three main theories and their failure conditions would be a nice addition.

Failure of a material depends on (1) nature of loading and (2) type of material. There are theories that can predict material failure for complex states.

5.6.1 Types of Material Failure - move here?

Maximum Shear Stress (Tresca) Criterion

If a material is ductile, failure is defined by yield stress ($ \sigma_Y $) and occurs at max shear stress ($ \tau_{max} $).

Uniaxial Tension

Taken from TAM251 Lecture Notes - L10S3

Consider a material subjected to uniaxial tension $ \sigma = \frac{P}{A} $ and $ P $ is loaded to the yield point, then the following is true:

$$ \sigma_1 = \sigma_Y\ $$

$$ \sigma_2 = 0\ $$

$$ \tau_{max} = \frac{\sigma_Y}{2}\ $$

If shear stress is responsible for causing the ductile material to yield, then

$$ \tau_{max} \ge \frac{\sigma_Y}{2}\ $$

where $ \sigma_Y $ is the tensile yield strength.

General 2D Loading State

$$ \sigma_z = \sigma_3 = 0\ $$

Failure still occurs at

$$ \tau_{max} = \frac{\sigma_Y}{2}\ $$

Taken from TAM251 Lecture Notes - L10S4

The equations for $ \sigma_1 $ and $ \sigma_2 $ can be plotted to show the failure surface. Loading conditions that occur outside of the surface are when the material fails. If $ \sigma_1 $ and $ \sigma_2 $ have the same signs

$$ |\sigma_1| = \sigma_Y\ $$

$$ |\sigma_2| = \sigma_Y\ $$

Taken from TAM251 Lecture Notes - L10S5

If $ \sigma_1 $ and $ \sigma_2 $ have opposite signs

Taken from TAM251 Lecture Notes - L10S5

$$ |\sigma_1 - \sigma_2| = \sigma_Y\ $$

Maximum Distortion Energy (Von Misses) Criterion

Ductile materials likely do not fail due to stresses that only result in a volume change. It is hypothesized that failure is driven by distortion strain energy.

Taken from TAM251 Lecture Notes - L10S6

All elastic deformations can be broken down into volumetric and distortional deformations. The total strain energy $ W $ in a material is broken into these same parts, resulting in

$$ W = W_v + W_d\ $$

For plane stress

$$ W_d = \frac{1+\nu}{3E}(\sigma_1^2 - \sigma_1 \sigma_2 + \sigma_2^2)\ $$

At the moment of yield

$$ W_{d,yield} = \frac{1+\nu}{3E}\sigma_Y^2\ $$

Equating these conditions gives

$$ \sigma_1^2 - \sigma_1 \sigma_2 + \sigma_2^2 = \sigma_Y^2\ $$

At yield

$$ \sigma_{VM}^2 = \sigma_Y^2\ $$

Failure occurs at

$$ \sigma_{VM} = \sqrt{\sigma_1^2 - \sigma_1 \sigma_2 + \sigma_2^2} \ge \sigma_Y\ $$

Taken from TAM251 Lecture Notes - L10S8

The equating equation be plotted to show the failure surface. The elliptical surface is the Von Mises surface overlaid with the Teresca surface. Loading conditions that occur outside of the surface are when the material fails.

Maximum Normal Stress Criterion

For brittle materials, failure is caused by the maximum tensile stress and NOT compressive stress.

$$ |\sigma_1| = \sigma_{ult}\ $$

$$ |\sigma_2| = \sigma_{ult}\ $$