To prove this we have to compare the shaft bending stress developed for the same applied bending moment to the hollow shaft and solid shaft.

We will use the beam bending equation here for the calculation of bending moment stress:

The bending stress (**fs**) developed in a solid shaft is given by:

*fs=(32*M)/(∏*D^3)…..Eq.1*

And the bending stress (**fh**) developed by the same amount of bending moment (M) in a hollow shaft is given by:

*fh=(32*M*D0)/(∏*(D0^4-di^4))……..Eq.2*

*Where,*

*M – The applied bending moment for both the shafts*

*D – The diameter of the solid shaft*

*D0 – Outer diameter of the hollow shaft*

*di – The inner diameter of the hollow shaft*

Now, the ratio of **fh/fs** can be arrived by dividing the Eq.2 to the Eq.1 as:

*fh/fs=(D0*D^3)/(D0^4-di^4) =(D/D0)^3/(1-(di/D0)^4)…….Eq.3*

You can observe from the **Eq.3** that:

*as,** D0 >D and D0>di*** **

so the numerator of the** Eq.3** is much smaller than the denominator.

So, it can be said that the bending stress developed in a hollow shaft is lesser than that at a solid shaft of same weight or in other word, hollow shaft is stronger in bending than a same weight solid shaft.

### Shibashis Ghosh

Disclaimer: I work for Altair. mechGuru.com is my personal blog. Although i have tried to put my neutral opinion while writing about different competitor's technologies, still i would like you to read the articles by keeping my background in mind.

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Hey, rewrite the bending stress formula of hollow shafts to “shaft diameter =” and be my hero

Got it. Do= 3th squared root of( (32*M)/(PI(1-k^4)*fh) ) where k=Di/Do and k=assumed