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Shaft Design Problem

  1. May 5, 2012 #1
    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 [itex]\tau[/itex]max=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 [itex]\tau[/itex]max 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.
     
    Last edited: May 5, 2012
  2. jcsd
  3. May 5, 2012 #2
  4. May 5, 2012 #3
    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 [itex]\tau[/itex]max 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?
     
  5. May 5, 2012 #4

    PhanthomJay

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    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.
     
  6. May 6, 2012 #5
    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.
     
  7. May 6, 2012 #6

    PhanthomJay

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    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.
     
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