Question

A 50 mm diameter shaft is subjected to a static axial load of 160 kN. If the yield stress of the material is 350 MPa, the ultimate tensile stress is 400 MPa and fatigue limit (endurance limit) for R = -1.0 is 320 MPa, what is the amplitude of the largest fully-reversed bending moment that can be applied to this shaft so that fatigue dose not occur?

Answer 1

In order to develop this problem it is necessary to take into account the concepts related to fatigue and compression effort and Goodman equation, i.e, an equation that can be used to quantify the interaction of mean and alternating stresses on the fatigue life of a materia.

With the given data we can proceed to calculate the compression stress:

`\sigma_c = (P)/(A)`

`\sigma_c = (160*10^3)/(\pi/4*0.05^2)`

`\sigma_c = 81.5MPa`

Through Goodman's equations the combined effort by fatigue and compression is expressed as:

`(\sigma_a)/(S_e)+(\sigma_c)/(\sigma_u)=(1)/(Fs)`

Where,

`\sigma_a=`Fatigue limit for comined alternating and mean stress

`S_e =`Fatigue Limit

`\sigma_c=`Mean stress (due to static load)

`\sigma_u =` Ultimate tensile stress

`Fs =`Security Factor

We can replace the values and assume a security factor of 1, then

`(\sigma_a)/(320)+(81.5)/(400)=(1)/(1)`

Re-arrenge for
`\sigma_a`

`\sigma_a = 254.8Mpa`

We know that the stress is representing as,

`\sigma_a = (M_c)/(I)`

Then,

Where
`M_c`=Max Moment

I= Intertia

The inertia for this object is

`I=(\pi d^4)/(64)`

Then replacing and re-arrenge for
`M_c`

`M_c = (\sigma_a*\pi*d^3)/(32)`

`M_c = (260.9*10^6*\pi*0.05^3)/(32)`

`M_c = 3201.7N.m`

Thereforethe moment that can be applied to this shaft so that fatigue does not occur is 3.2kNm