**1 Answer**

written 5.6 years ago by |

There are two special cases of the fluctuating stresses:-

Completely reversed stress

Repeated stress

*1. Completely Reversed stress* :-

It has zero mean stress.

In this case half portion of the cycle consists of tensile stress and remaining half of compressive stress. therefore, mean stress is zero.

For a completely reversed stress, $\sigma_{min}$ = -$\sigma_{max}$

$\sigma_{m}$ = $\frac{\sigma_{max} + \sigma_{min}}{2}$ = $\frac{\sigma_{max} + - (\sigma_{max})}{2}$ = 0

$\sigma_a$ = $\frac{\sigma_{max} - (\sigma_{min})}{2}$ = $\frac{\sigma_{max} - - (\sigma_{max})}{2}$ = $\sigma_{max}$

$\therefore \sigma_m = 0\ \text{and} \ \sigma_a = \sigma_{max}$

**2. Repeated stress:**

- The stresses which vary from zero to a certain maximum value are called repeated stresses.
The maximum stress is zero in this case and therefore, amplitude stress and mean stress are equal.

$\sigma_{m}$ = $\frac{\sigma_{max} + \sigma_{min}}{2}$ = $\frac{\sigma_{max} + 0}{2}$ = $\frac{\sigma_{max}}{2}$

$\sigma_a$ = $\frac{\sigma_{max} - \sigma_{min}}{2}$ = $\frac{\sigma_{max} - 0}{2}$ = $\frac{\sigma_{max}}{2}$

$\therefore \sigma_m = \frac{\sigma_{max}}{2}\ \text{and} \ \sigma_a = \frac{\sigma_{max}}{2}$