Question

An aluminum part will be subjected to cyclic loading where the maximum stress will be 300 MPa and the minimum stress will be-100 MPa (compressive).
a) Mean stress, om =___________ MPa
b) Stress amplitude, a =_________ MPa
c) Stress ratio, R =________

Answer 1

Answer:

a) The mean stress experimented by the aluminium part is 100 megapascals, b) The stress amplitude of the aluminium part is 400 megapascals, c) The stress ratio of the aluminium part is 4.

Explanation:

a) The mean stress is determined by this expression:

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

Where:

$\sigma_{m}$ - Mean stress, measured in megapascals.

$\sigma_{min}$ - Minimum stress, measured in megapascals.

$\sigma_{max}$ - Maximum stress, measured in megapascals.

If we know that $\sigma_{min} = -100\,MPa$ and $\sigma_{max} = 300\,MPa$, the mean stress is:

$\sigma_{m} = \frac{-100\,MPa+300\,MPa}{2}$

$\sigma_{m} = 100\,MPa$

The mean stress experimented by the aluminium part is 100 megapascals.

b) The stress amplitude is given by the following difference:

$\sigma_{a} = |\sigma_{max}-\sigma_{min}|$

Where $\sigma_{a}$ is the stress amplitude, measured in megapascals.

If we know that $\sigma_{min} = -100\,MPa$ and $\sigma_{max} = 300\,MPa$, the stress amplitude is:

$\sigma_{a} = |300\,MPa-(-100\,MPa)|$

$\sigma_{a} = 400\,MPa$

The stress amplitude of the aluminium part is 400 megapascals.

c) The stress ratio ($R$) is the ratio of the stress amplitude to mean stress. That is:

$R = \frac{\sigma_{a}}{\sigma_{m}}$

If we know that $\sigma_{m} = 100\,MPa$ and $\sigma_{a} = 400\,MPa$, the stress ratio is:

$R = \frac{400\,MPa}{100\,MPa}$

$R = 4$

The stress ratio of the aluminium part is 4.