# Shaft subject to Torque and Bending Moment ( ppi PE sample MD)



## Lily (Oct 3, 2010)

I did the Lindeburg PE sample exam (which I found very hard). and on problem 92, afternoon session asks for the minimum diameter of a shaft subject to both a bending moment and a Torque. I arrived to equivalent normal stress=r/J*(M^2+T^2)^0.5. and the solution arrived to stress=r/J*(4M^2+3T^2)^0.5. I have hard time understanding where the "3" came from. because for me bending stress=Mr/I and Shear stress =Tr/J and the equivalent normal stress = ( (bending stress/2)^2+(shear stress)^2)^0.5. Any ideas?

Thanks!


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## Shaggy (Oct 4, 2010)

Have a look in Shigley. In mine, it is section 18-4 (Shaft Design, Static Loading - Bending and Torsion).

Basically, the two formulas are using different failure theories. The first is maximum shear stress theory and the second is distortion energy (Von Mises). As for where the 4 and the 3 come from... I will leave that to someone else to derive.


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## Lily (Oct 5, 2010)

Shaggy said:


> Have a look in Shigley. In mine, it is section 18-4 (Shaft Design, Static Loading - Bending and Torsion).
> Basically, the two formulas are using different failure theories. The first is maximum shear stress theory and the second is distortion energy (Von Mises). As for where the 4 and the 3 come from... I will leave that to someone else to derive.


Thanks Shaggy, the 3 must then come from the (3^0.5)^2 in the distorsion energy theory.

So if no indication is given we must use the distortion energy failure theory? even though the maximum shear theory is more conservative?


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