Designing a Shaft for Specific Torsional Loads

[firebird90]

I need to design a shaft that have a specific torsional load acting on it. I have the safety factor of 2 and have a table of material properties (yield stress, ultimate stress, shear modulus, modulus of elasticity etc.).

I need to find the diameter of the shaft that resist the specifies torque. I have the formulas of \taumax=Tc/J and angle of twist formula. As the angle of twist formula is related to the length its useless because I don't have any specified. I need to get a \taumax by using material properties but I couldn't find anything to relate them up to now. I have searched many shaft design documents but no results and also I am confused right now.

 

[LawrenceC]

Maximum shear stress is related to yield stress.

http://www.eng-tips.com/viewthread.cfm?qid=155277

 

[firebird90]

Now I can make a proof for shear stress with max distortion energy theorem. So I am going to get a yield shear stress from the theorem and using it with factor of safety I will get my desired \taumax to use it in torsion formula, with using the yield stress to find diameter, I will get a diameter free of plastic deformation. Am I right with that?

 

[PhanthomJay]

I don't know why you need a proof for shear stress when it's max value can be looked up in a table of material properties. For steel, it's about 0.6 Fy, where Fy is the tensile yield stress. So with a SF of 2, allowable shear stress would be about 0.3Fy and you needn't worry about plastic deformation.

 

[firebird90]

The proof is my biggest problem with my work, Because it must be a project style showing how can I derived this 0.6Fy. I need a source to show it.

 

[PhanthomJay]

The Steel Code I use calculates the ultimate shear stress as the tensile yield stress divided by square root of 3, which rounds to 0.6 Fy. I do not know if that value comes from the distortion energy theorem to which you refer.