Introduction to Static Failure Theories

When designing mechanical components, engineers must ensure that the applied stresses do not cause the material to fail. For ductile materials (like most structural steel and aluminum alloys) under static loading, yielding is the primary mode of failure.

To evaluate complex loading conditions, we first extract the principal stresses ($\sigma_1, \sigma_2, \sigma_3$), which are the maximum and minimum normal stresses acting on an element where the shear stress is zero. Once these are known, we use a static failure criterion to determine if the material will yield.

Maximum Shear Stress (Tresca) Criterion

The Maximum Shear Stress (MSS) theory, or Tresca criterion, states that yielding begins when the maximum shear stress in a complex state of stress reaches the maximum shear stress at yielding in a simple tension test.

For a principal stress state where $\sigma_1 \ge \sigma_2 \ge \sigma_3$, the criterion for safe design is:

$\sigma_1 - \sigma_3 < \frac{S_y}{n}$

where:

• $S_y$ = yield strength of the material [MPa or psi]
• $n$ = design factor of safety

Distortion Energy (von Mises) Criterion

The Distortion Energy theory, commonly known as the von Mises criterion, provides a more accurate prediction for ductile yielding. It states that yielding occurs when the distortion energy per unit volume reaches the distortion energy at yield in simple tension.

The equivalent von Mises stress ($\sigma'$), also called effective stress, is calculated as:

$\sigma' = \sqrt{\frac{(\sigma_1 - \sigma_2)^2 + (\sigma_2 - \sigma_3)^2 + (\sigma_3 - \sigma_1)^2}{2}}$

For a safe design, the equivalent stress must be less than the allowable stress:

$\sigma' < \frac{S_y}{n}$

Worked Example

Problem: A ductile steel bracket ($S_y$ = 250 MPa) is subjected to a state of plane stress. The principal stresses at the critical point are calculated as $\sigma_1 = 150$ MPa, $\sigma_2 = 50$ MPa, and $\sigma_3 = 0$ MPa. Calculate the factor of safety using the von Mises criterion.

Solution:

1. Identify the principal stresses: $\sigma_1 = 150$, $\sigma_2 = 50$, $\sigma_3 = 0$.
2. Calculate the von Mises equivalent stress $\sigma'$: $\sigma' = \sqrt{\frac{(150 - 50)^2 + (50 - 0)^2 + (0 - 150)^2}{2}}$ $\sigma' = \sqrt{\frac{(100)^2 + (50)^2 + (-150)^2}{2}}$ $\sigma' = \sqrt{\frac{10000 + 2500 + 22500}{2}} = \sqrt{17500} \approx 132.28 \text{ MPa}$
3. Calculate the factor of safety $n$: $n = \frac{S_y}{\sigma'} = \frac{250}{132.28} \approx 1.89$

Engineering Check

The von Mises criterion generally predicts a slightly larger safe region than the Tresca criterion. When evaluating brittle materials, or cyclic loads (fatigue), entirely different theories (such as Mohr's theory or Goodman diagrams) must be used.\n