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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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About the author

Shibashis

Shibashis is a production engineer by profession and writer/blogger by passion. Apart from maintaining this blog, Shibashis also get some freelance work like: Mechanical design, CAD, CAE, Assignment report writing, Website design. Read More>>

  • Andre

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

  • http://www.facebook.com/people/Rob-de-H/100002923106839 Rob de H

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

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