# On the application of the goodman equation to a multiaxial stress state

1. Dec 10, 2008

I don't quite understand how the modified goodman equation can be applied to a multiaxial stress state. The explanation given in my stress analysis class has been quite confusing and verbose so I've come here to see if I can't get a better understanding.

First I'll lay out what I think to be true:

Utilizing the Von-Mises failure criterion in place of the uni-axial stresses in the goodman equation should be able to account for a multiaxial stress state AND fatigue. Here's how I think it should work

Uni-axial stress amplitude is replaced with von-mises stress amplitude (same equation, different stresses)

The fatigue limit at whatever number of cycles the designer is concerned with is replaced with the Von-Mises stress at that stress amplitude. Or: (Uni-Axial Fatigue limit at X cycles)*(1/3)=fully reversed stress amplitude (or SIGMAar in the good man equation).

Mean stress is replaced with mean von-mises stress (same equation, different stresses)

Ultimate stress is replaced with (sqrt(2)/3)*SIGMA(u) or the von mises stress at failure.

Equations:

That last equation should be sqrt(2)/3. Made a mistake when writing the equations. Thanks.
Is this correct?

2. Jan 2, 2009

### binoyjjohn

Actually you got the equations for stress amplitude (tau_a) and mean stress (tau_m) swapped, here.

3. Apr 4, 2009

### nvn

Whitebread wrote: "Utilizing the von Mises failure criterion in place of the uniaxial stresses in the [modified] Goodman equation should be able to account for a multiaxial stress state and fatigue. Here's how I think it should work. Uniaxial stress amplitude is replaced with von Mises stress amplitude."

Agreed.

Whitebread wrote: "The fatigue limit at whatever number of cycles the designer is concerned with is replaced with the von Mises stress at that stress amplitude. Or, (uniaxial fatigue limit at N cycles)*(1/3) = fully reversed stress amplitude (or sigma_ar in the [modified] Goodman equation)."

Disagree. Fatigue strength is a material property, not a von Mises stress. The fatigue strength should not be adjusted.

Whitebread wrote: "Mean stress is replaced with mean von Mises stress."

Agreed.

Whitebread wrote: "Ultimate stress is replaced with (sqrt(2)/3)*sigma_u, or the von Mises stress at failure."

Disagree. Tensile ultimate (mean) strength, Stu, is a material property, not a von Mises stress, and should not be adjusted.

A similar question is posted at thread 304749.

4. Jun 10, 2009