Endurance Limit Modifying Factors

Se’ vs Se

The actual endurance limit for a part may not match its nominal or empirical value

Generally if
$S_{ut}<200kpsi$ $S_e^{\prime }=0.5S_{ut}$

Modifying Factors

• Material composition
• Manufacturing method, heat treatment, stress concentration
• Its environment
• Its design

Marin’s Equation

This provides a correction factor for the modifying factors mentioned above
$S_e=k_a{k}_b{k}_c{k}_d{k}_e{S}_e^{\prime }$
Where

• a = surface factor
• b = size factor
• c = load factor
• d = temperature factor
• e = reliability factor
• Se’ = test endurance limit
• Se = actual endurance limit

Surface Factor

• Stresses are highest at the surface and surface roughness can caused localized stress concentrations $k_a=a{S}_{ut}^b$

Basically you get a and b from a table depending on the surface finish of your material

Size Factor

Fatigue is controlled by the weakest link in the material

As size increases, the endurance limit slightly decreases

For round rotating bars subject to bending and torsion

$$k_b=\{\begin{array}{ll}{\left(\frac{d}{0.3}\right)}^{-0.107}=0.879d^{-0.107}& 0.3\le d\le 2\text{in}\\ 0.91d^{-0.157}& 2<d\le 10\text{in}\\ {\left(\frac{d}{7.62}\right)}^{-0.107}=1.24d^{-0.107}& 7.62\le d\le 51\text{mm}\\ 1.51d^{-0.157}& 51<d\le 254\text{mm}\end{array}$$

For axial loading
$k_b=1.0$
For non-round parts, you need to find an equivalent diameter
To do so, you equate the volume of a material stressed at and above 95% of the max stress to the same volume

As a reference For a rotating round section
$A_{0.95\sigma }=0.0766d^2$
For a non-rotating round section
$A_{0.95\sigma }=0.01046d^2$
and the effective diameter is therefore
$d_e=0.370d$
For non-round parts, there are tables for different shapes that yield A_0.95

Load Factor

$$k_c=\{\begin{array}{ll}1& \text{bending}\\ 0.85& \text{axial}\\ 0.59& \text{torsion}\end{array}$$

Temperature Factor

You need to look at the ratio of the tensile strengths at room temperatures (ST) and operating temperatures (SRT)

$k_d=\frac{S_T}{S_{RT}}$
For steels:

$$\frac{S_T}{S_{RT}}=\{\begin{array}{l}0.98+3.5(10^{-4})T_F-6.3(10^{-7})T_F^2\\ 0.99+5.9(10^{-4})T_C-2.1(10^{-6})T_C^2\end{array}$$

In most cases the required information to solve this stuff isn’t given so IN GENERAL $k_d=1$

Reliability Factor

Accounts for scatter in endurance limit factor
$k_e=1-0.08z_a$
where z_a comes from a table and is dependant on how reliable you want your design to be

k_e also comes from a table but if you assume 50% confidence then $k_e=1$

Other Factors

Marin’s equation doesn’t account for everything!!!

Solving Strategy

• We need to find all the aspects of Marin’s Equation
• First find Se’ using The Stress-Life Method and S-N graph which is based off of ultimate strength
• Then find the k’s according to above
• Use the k’s and Se’ to find Se
• Later on you can use Se in questions to solve for things