Question

A material with a yield strength of 40 kpsi and an endurance strength of 20 kpsi is intended for an application with a von Mises stress that alternates between 17 kpsi and 3 kpsi. What is the factor of safety, N, for the design?

Answer 1

To solve this problem we will apply the concepts related to Soderberg's relation, which will allow us to find the safety factor based on yield stress, endurance strength, the mean and average stress. The mean and average stress values can be found through the alternating Stress previously given. We will proceed by defining the known values, that is

Yield stress $\sigma_y = 40ksi$

Endurance strength $\sigma_e = 20ksi$

The two values for alternating stress are

$\sigma_1 = 17ksi$

$\sigma_2 = 3ksi$

We know that mean stress is

$\sigma_m = \frac{\sigma_1+\sigma_2}{2}$

$\sigma_m = \frac{17+3}{2} = 10ksi$

And the average stress is

$\sigma_v = \frac{\sigma_1-\sigma_2}{2}$

$\sigma_v = \frac{17-3}{2} = 7ksi$

According to Soderberg relation

$\frac{1}{F.s} = \frac{\sigma_m}{\sigma_y}+\frac{\sigma_v}{\sigma_e}$

$\frac{1}{F.s} = \frac{10}{40}+\frac{7}{20}$

$\frac{1}{F.s} = 0.6$

$F.s = 1.66 \approx 2$

Therefore the factor of safety is 2