Why a Hollow Shaft Stronger than a Same Weight Solid Shaft in Bending

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.

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Shibashis Ghosh

Hi, I am Shibashis, a blogger by passion and engineer by profession. I have written most of the articles for mechGuru.com. For more than a decades i am closely associated with the engineering design/manufacturing simulation technologies.
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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